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A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Blood as a Buffer
Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly.
• Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution.
• How Does A Buffer Maintain pH?
A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.
• Introduction to Buffers
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Preparing Buffer Solutions
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Thumbnail: Simulated titration of an acidified solution of a weak acid (pKa = 4.7) with alkali. (Public Domain; Lasse Havelund).
1.27: Lewis Acids and Bases
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. According to Lewis,
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Proton-Transfer Reactions Involve Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note carefully that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Acid-base Reactions without Transferring Protons
The major utility of the Lewis definition is that it extends the concept of acids and bases beyond the realm of proton transfer reactions. The classic example is the reaction of boron trifluoride with ammonia to form an adduct:
$\ce{BF_3 + NH_3 \rightarrow F_3B-NH_3}$
One of the most commonly-encountered kinds of Lewis acid-base reactions occurs when electron-donating ligands form coordination complexes with transition-metal ions.
Exercise $1$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $\ce{Al(OH)_3 + OH^{–} \rightarrow Al(OH)_4^–}$
2. $\ce{SnS_2 + S^{2–} \rightarrow SnS_3^{2–}}$
3. $\ce{Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}}$
4. $\ce{AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–}$
5. $\ce{Fe^{2+} + NO \rightarrow Fe(NO)^{2+}}$
6. $\ce{Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}}$
Applications to organic reaction mechanisms
Although organic chemistry is beyond the scope of these lessons, it is instructive to see how electron donors and acceptors play a role in chemical reactions. The following two diagrams show the mechanisms of two common types of reactions initiated by simple inorganic Lewis acids:
In each case, the species labeled "Complex" is an intermediate that decomposes into the products, which are conjugates of the original acid and base pairs. The electric charges indicated in the complexes are formal charges, but those in the products are "real".
In reaction 1, the incomplete octet of the aluminum atom in $\ce{AlCl3}$ serves as a better electron acceptor to the chlorine atom than does the isobutyl part of the base. In reaction 2, the pair of non-bonding electrons on the dimethyl ether coordinates with the electron-deficient boron atom, leading to a complex that breaks down by releasing a bromide ion.
Non-aqueous Protonic Acid-Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. However, many other familiar substances can also serve as the basis of protonic solvent systems as Table $1$ indicates:
Table $1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKap
water 2 H2O → H3O+ + OH 14
ammonia 2 NH3 → NH4+ + NH2 33
acetic acid 2 CH3COOH → CH3COOH2+ + CH3COO 13
ethanol 2 C2H5OH → C2H5OH2+ + C2H5O 19
hydrogen peroxide 2 HO-OH → HO-OH2+ + HO-O 13
hydrofluoric acid 2 HF → H2F+ + F 10
sulfuric acid 2 H2SO4 → H3SO4+ + HSO4 3.5
One use of nonaqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/01%3A_Electronic_Structure_and_Bonding_(Acids_and_Bases)/1.26%3A_Buffer_Solutions.txt |
Ethers are compounds having two alkyl or aryl groups bonded to an oxygen atom, as in the formula R1–O–R2. The ether functional group does not have a characteristic IUPAC nomenclature suffix, so it is necessary to designate it as a substituent. To do so the common alkoxy substituents are given names derived from their alkyl component (below):
Alkyl Group Name Alkoxy Group Name
CH3 Methyl CH3O– Methoxy
CH3CH2 Ethyl CH3CH2O– Ethoxy
(CH3)2CH– Isopropyl (CH3)2CHO– Isopropoxy
(CH3)3C– tert-Butyl (CH3)3CO– tert-Butoxy
C6H5 Phenyl C6H5O– Phenoxy
Ethers can be named by naming each of the two carbon groups as a separate word followed by a space and the word ether. The -OR group can also be named as a substituent using the group name, alkox
Example \(1\)
CH3-CH2-O-CH3 is called ethyl methyl ether or methoxyethane.
The smaller, shorter alkyl group becomes the alkoxy substituent. The larger, longer alkyl group side becomes the alkane base name. Each alkyl group on each side of the oxygen is numbered separately. The numbering priority is given to the carbon closest to the oxgen. The alkoxy side (shorter side) has an "-oxy" ending with its corresponding alkyl group. For example, CH3CH2CH2CH2CH2-O-CH2CH2CH3 is 1-propoxypentane. If there is cis or trans stereochemistry, the same rule still applies.
Examples \(2\)
• \(CH_3CH_2OCH_2CH_3\), diethyl ether (sometimes referred to as just ether)
• \(CH_3OCH_2CH_2OCH_3\), ethylene glycol dimethyl ether (glyme).
Exercises \(2\)
Try to name the following compounds using these conventions:
Try to draw structures for the following compounds:
• 2-pentyl 1-propyl ether J
• 1-(2-propoxy)cyclopentene J
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers".
• anisole (try naming anisole by the other two conventions. J )
• oxirane
1,2-epoxyethane, ethylene oxide, dimethylene oxide, oxacyclopropane,
• furan (this compound is aromatic)
tetrahydrofuran
oxacyclopentane, 1,4-epoxybutane, tetramethylene oxide,
• dioxane
1,4-dioxacyclohexane
Exercise \(2\)
Try to draw structures for the following compounds-
• 3-bromoanisole J
• 2-methyloxirane J
• 3-ethylfuran J
Heterocycles
In cyclic ethers (heterocycles), one or more carbons are replaced with oxygen. Often, it's called heteroatoms, when carbon is replaced by an oxygen or any atom other than carbon or hydrogen. In this case, the stem is called the oxacycloalkane, where the prefix "oxa-" is an indicator of the replacement of the carbon by an oxygen in the ring. These compounds are numbered starting at the oxygen and continues around the ring. For example,
If a substituent is an alcohol, the alcohol has higher priority. However, if a substituent is a halide, ether has higher priority. If there is both an alcohol group and a halide, alcohol has higher priority. The numbering begins with the end that is closest to the higher priority substituent. There are ethers that are contain multiple ether groups that are called cyclic polyethers or crown ethers. These are also named using the IUPAC system.
Sulfides
Sulfur analogs of ethers (R–S–R') are called sulfides, e.g., (CH3)3C–S–CH3 is tert-butyl methyl sulfide. Sulfides are chemically more reactive than ethers, reflecting the greater nucleophilicity of sulfur relative to oxygen.
Problems
Name the following ethers:
(Answers to problems above: 1. diethyl ether; 2. 2-ethoxy-2-methyl-1-propane; 3. cis-1-ethoxy-2-methoxycyclopentane; 4. 1-ethoxy-1-methylcyclohexane; 5. oxacyclopropane; 6. 2,2-Dimethyloxacyclopropane)
Exercise \(3\)
Answer
A one-eyed one-horned flying propyl people ether | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.05%3A_The_Nomenclature_of_Ethers.txt |
In the IUPAC system of nomenclature, functional groups are normally designated in one of two ways. The presence of the function may be indicated by a characteristic suffix and a location number. This is common for the carbon-carbon double and triple bonds which have the respective suffixes -ene and -yne. Halogens, on the other hand, do not have a suffix and are named as substituents, for example: (CH3)2C=CHCHClCH3 is 4-chloro-2-methyl-2-pentene.
Alcohols are usually named by the first procedure and are designated by an -ol suffix, as in ethanol, CH3CH2OH (note that a locator number is unnecessary on a two-carbon chain). On longer chains the location of the hydroxyl group determines chain numbering. For example: (CH3)2C=CHCH(OH)CH3 is 4-methyl-3-penten-2-ol. Other examples of IUPAC nomenclature are shown below, together with the common names often used for some of the simpler compounds. For the mono-functional alcohols, this common system consists of naming the alkyl group followed by the word alcohol. Alcohols may also be classified as primary, , secondary, , and tertiary, , in the same manner as alkyl halides. This terminology refers to alkyl substitution of the carbon atom bearing the hydroxyl group (colored blue in the illustration).
Many functional groups have a characteristic suffix designator, and only one such suffix (other than "-ene" and "-yne") may be used in a name. When the hydroxyl functional group is present together with a function of higher nomenclature priority, it must be cited and located by the prefix hydroxy and an appropriate number. For example, lactic acid has the IUPAC name 2-hydroxypropanoic acid.
2.07: The Nomenclature of Amines
In the IUPAC system of nomenclature, functional groups are normally designated in one of two ways. The presence of the function may be indicated by a characteristic suffix and a location number. This is common for the carbon-carbon double and triple bonds which have the respective suffixes ene and yne. Halogens, on the other hand, do not have a suffix and are named as substituents, for example: (CH3)2C=CHCHClCH3 is 4-chloro-2-methyl-2-pentene. If you are uncertain about the IUPAC rules for nomenclature you should review them now.
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. The nomenclature of amines is complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. Furthermore, the terms primary (1º), secondary (2º) & tertiary (3º) are used to classify amines in a completely different manner than they were used for alcohols or alkyl halides. When applied to amines these terms refer to the number of alkyl (or aryl) substituents bonded to the nitrogen atom, whereas in other cases they refer to the nature of an alkyl group. The four compounds shown in the top row of the following diagram are all C4H11N isomers. The first two are classified as 1º-amines, since only one alkyl group is bonded to the nitrogen; however, the alkyl group is primary in the first example and tertiary in the second. The third and fourth compounds in the row are 2º and 3º-amines respectively. A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a 4º-ammonium cation. For example, (CH3)4N(+) Br(–) is tetramethylammonium bromide.
• The IUPAC names are listed first and colored blue. This system names amine functions as substituents on the largest alkyl group. The simple -NH substituent found in 1º-amines is called an amino group. For 2º and 3º-amines a compound prefix (e.g. dimethylamino in the fourth example) includes the names of all but the root alkyl group.
• The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram.
• Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine. These are the names given in the last row (colored black).
Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently. Since these names are not based on a rational system, it is necessary to memorize them. There is a systematic nomenclature of heterocyclic compounds, but it will not be discussed here.
2.08: The Structures of Alkyl Halides Alcohols Ethers and A
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). The halogen atoms significantly alters the physical properties of the molecules including electronegativity, bond length, bond strength, and molecular size. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.06%3A_The_Nomenclature_of_Alcohols.txt |
Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). The halogen atoms significantly alters the physical properties of the molecules including electronegativity, bond length, bond strength, and molecular size.
Boiling Point and Water Solubility
It is instructive to compare the boiling points and water solubility of amines with those of corresponding alcohols and ethers. The dominant factor here is hydrogen bonding, and the first table below documents the powerful intermolecular attraction that results from -O-H---O- hydrogen bonding in alcohols (light blue columns). Corresponding -N-H---N- hydrogen bonding is weaker, as the lower boiling points of similarly sized amines (light green columns) demonstrate. Alkanes provide reference compounds in which hydrogen bonding is not possible, and the increase in boiling point for equivalent 1º-amines is roughly half the increase observed for equivalent alcohols.
Compound CH3CH3 CH3OH CH3NH2 CH3CH2CH3 CH3CH2OH CH3CH2NH2
Mol.Wt. 30 32 31 44 46 45
Boiling
Point ºC
-88.6º 65º -6.0º -42º 78.5º 16.6º
The second table illustrates differences associated with isomeric 1º, 2º & 3º-amines, as well as the influence of chain branching. Since 1º-amines have two hydrogens available for hydrogen bonding, we expect them to have higher boiling points than isomeric 2º-amines, which in turn should boil higher than isomeric 3º-amines (no hydrogen bonding). Indeed, 3º-amines have boiling points similar to equivalent sized ethers; and in all but the smallest compounds, corresponding ethers, 3º-amines and alkanes have similar boiling points. In the examples shown here, it is further demonstrated that chain branching reduces boiling points by 10 to 15 ºC.
Compound CH3(CH2)2CH3 CH3(CH2)2OH CH3(CH2)2NH2 CH3CH2NHCH3 (CH3)3CH (CH3)2CHOH (CH3)2CHNH2 (CH3)3N
Mol.Wt. 58 60 59 59 58 60 59 59
Boiling
Point ºC
-0.5º 97º 48º 37º -12º 82º 34º
The water solubility of 1º and 2º-amines is similar to that of comparable alcohols. As expected, the water solubility of 3º-amines and ethers is also similar. These comparisons, however, are valid only for pure compounds in neutral water. The basicity of amines (next section) allows them to be dissolved in dilute mineral acid solutions, and this property facilitates their separation from neutral compounds such as alcohols and hydrocarbons by partitioning between the phases of non-miscible solvents.
Basicity of Amines
A review of basic acid-base concepts should be helpful to the following discussion. Like ammonia, most amines are Brønsted and Lewis bases, but their base strength can be changed enormously by substituents. It is common to compare basicity's quantitatively by using the pKa's of their conjugate acids rather than their pKb's. Since pKa + pKb = 14, the higher the pKa the stronger the base, in contrast to the usual inverse relationship of pKa with acidity. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their water solutions are basic (have a pH of 11 to 12, depending on concentration). The first four compounds in the following table, including ammonia, fall into that category.
The last five compounds (colored cells) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In pyridine the nitrogen is sp2 hybridized, and in nitriles (last entry) an sp hybrid nitrogen is part of the triple bond. In each of these compounds (shaded red) the non-bonding electron pair is localized on the nitrogen atom, but increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton.
Compound
NH3 CH3C≡N
pKa 11.0 10.7 10.7 9.3 5.2 4.6 1.0 0.0 -1.0 -10.0
Secondly, aniline and p-nitroaniline (first two green shaded structures) are weaker bases due to delocalization of the nitrogen non-bonding electron pair into the aromatic ring (and the nitro substituent). This is the same delocalization that results in activation of a benzene ring toward electrophilic substitution. The following resonance equations, which are similar to those used to explain the enhanced acidity of ortho and para-nitrophenols illustrate electron pair delocalization in p-nitroaniline. Indeed, aniline is a weaker base than cyclohexyl amine by roughly a million fold, the same factor by which phenol is a stronger acid than cyclohexanol. This electron pair delocalization is accompanied by a degree of rehybridization of the amino nitrogen atom, but the electron pair delocalization is probably the major factor in the reduced basicity of these compounds. A similar electron pair delocalization is responsible for the very low basicity (and nucleophilic reactivity) of amide nitrogen atoms (last green shaded structure). This feature was instrumental in moderating the influence of amine substituents on aromatic ring substitution, and will be discussed further in the section devoted to carboxylic acid derivatives.
Conjugated amine groups influence the basicity of an existing amine. Although 4-dimethylaminopyridine (DMAP) might appear to be a base similar in strength to pyridine or N,N-dimethylaniline, it is actually more than ten thousand times stronger, thanks to charge delocalization in its conjugate acid. The structure in the gray box shows the locations over which positive charge (colored red) is delocalized in the conjugate acid. This compound is often used as a catalyst for acyl transfer reactions.
Finally, the very low basicity of pyrrole (shaded blue) reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Indole (pKa = -2) and imidazole (pKa = 7.0), see above, also have similar heterocyclic aromatic rings. Imidazole is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity.
Although resonance delocalization generally reduces the basicity of amines, a dramatic example of the reverse effect is found in the compound guanidine (pKa = 13.6). Here, as shown below, resonance stabilization of the base is small, due to charge separation, while the conjugate acid is stabilized strongly by charge delocalization. Consequently, aqueous solutions of guanidine are nearly as basic as are solutions of sodium hydroxide.
The relationship of amine basicity to the acidity of the corresponding conjugate acids may be summarized in a fashion analogous to that noted earlier for acids:
Strong bases have weak conjugate acids, and weak bases have strong conjugate acids.
Acidity of Amines
We normally think of amines as bases, but it must be remembered that 1º and 2º-amines are also very weak acids (ammonia has a pKa = 34). In this respect it should be noted that pKa is being used as a measure of the acidity of the amine itself rather than its conjugate acid, as in the previous section. For ammonia this is expressed by the following hypothetical equation:
NH3 + H2O ____> NH2(–) + H2O-H(+)
The same factors that decreased the basicity of amines increase their acidity. This is illustrated by the following examples, which are shown in order of increasing acidity. It should be noted that the first four examples have the same order and degree of increased acidity as they exhibited decreased basicity in the previous table. The first compound is a typical 2º-amine, and the three next to it are characterized by varying degrees of nitrogen electron pair delocalization. The last two compounds (shaded blue) show the influence of adjacent sulfonyl and carbonyl groups on N-H acidity. From previous discussion it should be clear that the basicity of these nitrogens is correspondingly reduced.
Compound C6H5SO2NH2
pKa 33 27 19 15 10 9.6
The acids shown here may be converted to their conjugate bases by reaction with bases derived from weaker acids (stronger bases). Three examples of such reactions are shown below, with the acidic hydrogen colored red in each case. For complete conversion to the conjugate base, as shown, a reagent base roughly a million times stronger is required.
C6H5SO2NH2 + KOH C6H5SO2NH(–) K(+) + H2O a sulfonamide base
(CH3)3COH + NaH (CH3)3CO(–) Na(+) + H2 an alkoxide base
(C2H5)2NH + C4H9Li (C2H5)2N(–) Li(+) + C4H10 an amide base
Important Reagent Bases
The significance of all these acid-base relationships to practical organic chemistry lies in the need for organic bases of varying strength, as reagents tailored to the requirements of specific reactions. The common base sodium hydroxide is not soluble in many organic solvents, and is therefore not widely used as a reagent in organic reactions. Most base reagents are alkoxide salts, amines or amide salts. Since alcohols are much stronger acids than amines, their conjugate bases are weaker than amide bases, and fill the gap in base strength between amines and amide salts. In the following table, pKa again refers to the conjugate acid of the base drawn above it.
Base Name Pyridine Triethyl
Amine
Hünig's Base Barton's
Base
Potassium
t-Butoxide
Sodium HMDS LDA
Formula (C2H5)3N (CH3)3CO(–) K(+) [(CH3)3Si]2N(–) Na(+) [(CH3)2CH]2N(–) Li(+)
pKa 5.3 10.7 11.4 14 19 26 35.7
Pyridine is commonly used as an acid scavenger in reactions that produce mineral acid co-products. Its basicity and nucleophilicity may be modified by steric hindrance, as in the case of 2,6-dimethylpyridine (pKa=6.7), or resonance stabilization, as in the case of 4-dimethylaminopyridine (pKa=9.7). Hünig's base is relatively non-nucleophilic (due to steric hindrance), and like DBU is often used as the base in E2 elimination reactions conducted in non-polar solvents. Barton's base is a strong, poorly-nucleophilic, neutral base that serves in cases where electrophilic substitution of DBU or other amine bases is a problem. The alkoxides are stronger bases that are often used in the corresponding alcohol as solvent, or for greater reactivity in DMSO. Finally, the two amide bases see widespread use in generating enolate bases from carbonyl compounds and other weak carbon acids.
Nonionic Superbases
An interesting group of neutral, highly basic compounds of nitrogen and phosphorus have been prepared, and are referred to as superbases. To see examples of these compounds Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.09%3A_The_Physical_Properties_of_Alkanes_Alkyl_Halides_Alco.txt |
Before we begin our exploration of stereochemistry and chirality, we first need to consider the subject of conformational isomerism, which has to do with rotation about single bonds.
We learned in section 2.1 that single bonds in organic molecules are free to rotate, due to the 'end-to-end' (sigma) nature of their orbital overlap. Consider the carbon-oxygen bond in ethanol, for example: with a 180o rotation about this bond, the shape of the molecule would look quite different:
Or ethane: rotation about the carbon-carbon sigma bond results in many different possible three-dimensional arrangements of the atoms.
These different arrangements, resulting from sigma bond rotation, are referred to in organic chemistry as conformations. Any one specific conformation is called a conformational isomer, or conformer.
In order to better visualize different conformations of a molecule, it is convenient to use a drawing convention called the Newman projection. In a Newman projection, we look lengthwise down a specific bond of interest – in this case, the carbon-carbon bond in ethane. We depict the ‘front’ atom as a dot, and the ‘back’ atom as a larger circle.
The six carbon-hydrogen bonds are shown as solid lines protruding from the two carbons. Note that we do not draw bonds as solid or dashed wedges in a Newman projection.
Looking down the C-C bond in this way, the angle formed between a C-H bond on the front carbon and a C-H bond on the back carbon is referred to as a dihedral angle. (The dihedral angle between the hour hand and the minute hand on a clock is 0o at noon, 90o at 3:00, and so forth).
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation: all of the dihedral angles are 60o, and the distance between the front and back C-H bonds is maximized.
If we now rotate the front CH3 group 60° clockwise, the molecule is in the highest energy ‘eclipsed' conformation, where the dihedral angles are all 0o (we stagger the bonds slightly in our Newman projection drawing so that we can see them all).
The energy of the eclipsed conformation, where the electrons in the front and back C-H bonds are closer together, is approximately 12 kJ/mol higher than that of the staggered conformation.
Another 60° rotation returns the molecule to a second staggered conformation. This process can be continued all around the 360° circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of conformations in between these two extremes.
Now let's consider butane, with its four-carbon chain. There are now three rotating carbon-carbon bonds to consider, but we will focus on the middle bond between C2 and C3. Below are two representations of butane in a conformation which puts the two CH3 groups (C1 and C4) in the eclipsed position, with the two C-C bonds at a 0o dihedral angle.
If we rotate the front, (blue) carbon by 60° clockwise, the butane molecule is now in a staggered conformation.
This is more specifically referred to as the gauche conformation of butane. Notice that although they are staggered, the two methyl groups are not as far apart as they could possibly be.
A further rotation of 60° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
One more 60 rotation produces another staggered conformation called the anti conformation, where the two methyl groups are positioned opposite each other (a dihedral angle of 180o).
As with ethane, the staggered conformations of butane are energy 'valleys', and the eclipsed conformations are energy 'peaks'. However, in the case of butane there are two different valleys, and two different peaks. The gauche conformation is a higher energy valley than the anti conformation due to steric strain, which is the repulsive interaction caused by the two bulky methyl groups being forced too close together. Clearly, steric strain is lower in the anti conformation. In the same way, steric strain causes the eclipsed A conformation - where the two methyl groups are as close together as they can possibly be - to be higher in energy than the two eclipsed B conformations.
The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
Because the anti conformation is lowest in energy (and also simply for ease of drawing), it is conventional to draw open-chain alkanes in a 'zigzag' form, which implies anti conformation at all carbon-carbon bonds. The figure below shows, as an example, a Newman projection looking down the C2-C3 bond of octane.
Exercise 3.1: Draw Newman projections of the lowest and highest energy conformations of propane.
Exercise 3.2: Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below (C2 should be your front carbon).
Solutions to exercises
Online lectures from Kahn Academy
Newman projections part I
Newman projections part II
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.10%3A_Rotation_Occurs_About_Carbon-Carbon_Single_Bonds.txt |
Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp3 tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as angle strain. Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called torsional strain. These destabilizing effects, angle strain and torsional strain are known together as ring strain. The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while cyclohexane is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry.
Ring Strain and the Structures of Cycloalkanes
There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon.
Chair Conformation of Cyclohexane - Equitorial and Axial
There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation.
• The chair conformation drawing is more favored than the boat because of the energy, the steric hindrance, and a new strain called the transannular strain.
• The boat conformation is not the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular strain, which means that the strain results from steric crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain.
• When we twist the C-C bond in a boat, it becomes a twisted boat.
Some Conformations of Cyclohexane Rings. (William Reusch, MSU)
Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram.
Conformational Energy Profile of Cyclohexane. (William Reusch, MSU).
The transition state structure is called a half chair. This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation.
This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help.
These are hydrogens in the axial form.
These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form.
Ring Strain
Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy.
With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain.
There are different types of ring strain:
• Transannular strain isdefined as the crowding of the two groups in a ring.
• Eclipsing strain, also known as torsional strain, is intramolecular strain due to the bonding interaction between two eclipsed atoms or groups.
• Bond angle strain is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic/hybrid orbitals.
Cyclohexane
Most of the time, cyclohexane adopts the fully staggered, ideal angle chair conformation. In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°.
Cyclohexane in the chair conformation. (William Reusch, MSU).
In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called axial. Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called equatorial.
Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a ring flip.
Methylcyclohexane
Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a 1,3-diaxial interaction (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation.
Methylcyclohexane in the chair conformation. (William Reusch, MSU).
Problems
1. Trans- 1,2-dimethylcyclopropane is more stable than cis-1,2-dimethylcyclopropane. Why? Drawing a picture of the two will help your explanation.
2. Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain.
3. Which of these chair conformations are the most stable and why?
3.
1. What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains?
2. Why is that the bigger rings have lesser strains compared to smaller rings?
Answers
1. The cis isomer suffers from steric hindrance and has a larger heat of combustion.
2. Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free.
3. Top one is more stable because it is in an equitorial conformation. When assembling it with the OChem Molecular Structure Tool Kit, equitorial formation is more spread out.
4. When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable.
5. Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.11%3A_Some_Cycloalkanes_Have_Angle_Strain.txt |
Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp3 tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as angle strain. Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called torsional strain. These destabilizing effects, angle strain and torsional strain are known together as ring strain. The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while cyclohexane is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry.
Ring Strain and the Structures of Cycloalkanes
There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon.
Chair Conformation of Cyclohexane - Equitorial and Axial
There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation.
• The chair conformation drawing is more favored than the boat because of the energy, the steric hindrance, and a new strain called the transannular strain.
• The boat conformation is not the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular strain, which means that the strain results from steric crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain.
• When we twist the C-C bond in a boat, it becomes a twisted boat.
Some Conformations of Cyclohexane Rings. (William Reusch, MSU)
Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram.
Conformational Energy Profile of Cyclohexane. (William Reusch, MSU).
The transition state structure is called a half chair. This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation.
This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help.
These are hydrogens in the axial form.
These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form.
Ring Strain
Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy.
With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain.
There are different types of ring strain:
• Transannular strain isdefined as the crowding of the two groups in a ring.
• Eclipsing strain, also known as torsional strain, is intramolecular strain due to the bonding interaction between two eclipsed atoms or groups.
• Bond angle strain is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic/hybrid orbitals.
Cyclohexane
Most of the time, cyclohexane adopts the fully staggered, ideal angle chair conformation. In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°.
Cyclohexane in the chair conformation. (William Reusch, MSU).
In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called axial. Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called equatorial.
Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a ring flip.
Methylcyclohexane
Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a 1,3-diaxial interaction (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation.
Methylcyclohexane in the chair conformation. (William Reusch, MSU).
Problems
1. Trans- 1,2-dimethylcyclopropane is more stable than cis-1,2-dimethylcyclopropane. Why? Drawing a picture of the two will help your explanation.
2. Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain.
3. Which of these chair conformations are the most stable and why?
3.
1. What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains?
2. Why is that the bigger rings have lesser strains compared to smaller rings?
Answers
1. The cis isomer suffers from steric hindrance and has a larger heat of combustion.
2. Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free.
3. Top one is more stable because it is in an equitorial conformation. When assembling it with the OChem Molecular Structure Tool Kit, equitorial formation is more spread out.
4. When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable.
5. Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.12%3A_Conformers_of_Cyclohexane.txt |
Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures.
Compounds between Metals and Nonmetals (Cation and Anion)
Compounds made of a metal and nonmetal are commonly known as Ionic Compounds, where the compound name has an ending of –ide. Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na+ is paired with one Cl-; one Ca2+ is paired with two Br-. There are two rules that must be followed through:
• The cation (metal) is always named first with its name unchanged
• The anion (nonmetal) is written after the cation, modified to end in –ide
Table 1: Cations and Anions:
+1 Charge +2 Charge -1 Charge -2 Charge -3 Charge -4 Charge
Group 1A elements Group 2A elements Group 7A elements Group 6A elements Group 5A elements Group 4A elements
Hydrogen: H+ Beryllium: Be2+ Hydride: H- Oxide: O2- Nitride: N3- Carbide: C4-
Lithium: Li+ Magnesium: Mg2+ Fluoride: F- Sulfide: S2- Phosphide: P3-
Soduim: Na+ Calcium: Ca2+ Chloride: Cl-
Potassium: K+ Strontium: Sr2+ Bromide: Br-
Rubidium: Rb+ Barium: Ba2+ Iodide: I-
Cesium: Cs+
Example 1
Na+ + Cl- = NaCl; Ca2+ + 2Br- = CaBr2
Sodium + Chlorine = Sodium Chloride; Calcium + Bromine = Calcium Bromide
The transition metals may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe2+ and Fe3+. To distinguish the difference, Fe2+ would be named iron (II) and Fe3+ would be named iron (III).
Table of Transition Metal and Metal Cations:
+1 Charge +2 Charge +3 Charge +4 Charge
Copper(I): Cu+ Copper(II): Cu2+ Aluminum: Al3+ Lead(IV): Pb4+
Silver: Ag+ Iron(II): Fe2+ Iron(III): Fe3+ Tin(IV): Sn4+
Cobalt(II): Co2+ Cobalt(III): Co3+
Tin(II): Sn2+
Lead(II): Pb2+
Nickel: Ni2+
Zinc: Zn2+
Example 2
Ions: Fe2++ 2Cl- Fe3++ 3Cl-
Compound: FeCl2 FeCl3
Nomenclature Iron (II) Chloride Iron (III) Chloride
However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with -ous and the one with the the higher charge has a Latin name ending with -ic. The most common ones are shown in the table below:
Transition Metal Ion with Roman Numeral Latin name
Copper (I): Cu+ Cuprous
Copper (II): Cu2+ Cupric
Iron (II): Fe2+ Ferrous
Iron (III): Fe3+ Ferric
Lead (II): Pb2+ Plumbous
Lead (IV): Pb4+ Plumbic
Mercury (I): Hg22+ Mercurous
Mercury (II): Hg2+ Mercuric
Tin (II): Sn2+ Stannous
Tin (IV): Sn4+ Stannic
Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in –ous or –ic.
• -ous ending is used for the lower oxidation state
• -ic ending is used for the higher oxidation state
Example 3
Compound Cu2O CuO FeCl2 FeCl3
Charge Charge of copper is +1 Charge of copper is +2 Charge of iron is +2 Charge of iron is +3
Nomenclature Cuprous Oxide Cupric Oxide Ferrous Chloride Ferric Chloride
However, this -ous/-ic system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H2SO4 is commonly known as Sulfuric Acid, and H2SO3 is known as Sulfurous Acid.
Compounds between Nonmetals and Nonmetals
Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as Molecular Compounds, where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so prefixes are used to distinguish them.
# of Atoms 1 2 3 4 5 6 7 8 9 10
Prefixes Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Example 4
CO = carbon monoxide BCl3 = borontrichloride
CO2 = carbon dioxide N2O5 =dinitrogen pentoxide
The prefix mono- is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element.
Binary Acids
Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An acid is a substance that dissociates into hydrogen ions (H+) and anions in water. A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix hydro- is placed in front of the nonmetal modified to end with –ic. The state of acids is aqueous (aq) because acids are found in water.
Some common binary acids include:
HF (g) = hydrogen fluoride -> HF (aq) = hydrofluoric acid
HBr (g) = hydrogen bromide -> HBr (aq) = hydrobromic acid
HCl (g) = hydrogen chloride -> HCl (aq) = hydrochloric acid
H2S (g) = hydrogen sulfide -> H2S (aq) = hydrosulfuricacid
It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with –ide.
Example 5
hypo____ite ____ite ____ate per____ate
ClO- ClO2- ClO3- ClO4-
hypochlorite chlorite chlorate perchlorate
---------------->
As indicated by the arrow, moving to the right, the following trends occur:
Increasing number of oxygen atoms
Increasing oxidation state of the nonmetal
(Usage of this example can be seen from the set of compounds containing Cl and O)
This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used.
Polyatomic Ions
In polyatomic ions, polyatomic (meaning two or more atoms) are joined together by covalent bonds. Although there may be a element with positive charge like H+, it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below:
Table: Common Polyatomic ions
Name: Cation/Anion Formula
Ammonium ion NH4+
Hydronium ion
H3O+
Acetate ion
C2H3O2-
Arsenate ion
AsO43-
Carbonate ion
CO32-
Hypochlorite ion
ClO-
Chlorite ion
ClO2-
Chlorate ion
ClO3-
Perchlorate ion
ClO4-
Chromate ion
CrO42-
Dichromate ion
Cr2O72-
Cyanide ion
CN-
Hydroxide ion
OH-
Nitrite ion
NO2-
Nitrate ion
NO3-
Oxalate ion
C2O42-
Permanganate ion
MnO4-
Phosphate ion
PO43-
Sulfite ion
SO32-
Sulfate ion
SO42-
Thiocyanate ion
SCN-
Thiosulfate ion
S2O32-
To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H2SO4) or carbonic acid (H2CO3). To name them, follow these quick, simple rules:
1. If the ion ends in -ate and is added with an acid, the acid name will have an -ic ending. Examples: nitrate ion (NO3-) + H+ (denoting formation of acid) = nitric acid (HNO3)
2. If the ion ends in -ite and is added with an acid, then the acid name will have an -ous ending. Example: nitite ion (NO2-) + H+ (denoting formation of acid) = nitrous acid (HNO2)
Problems
1. What is the correct formula for Calcium Carbonate?
a. Ca+ + CO2-
b. CaCO2-
c. CaCO3
d. 2CaCO3
2. What is the correct name for FeO?
a. Iron oxide
b. Iron dioxide
c. Iron(III) oxide
d. Iron(II) oxide
3. What is the correct name for Al(NO3)3?
a. Aluminum nitrate
b. Aluminum(III) nitrate
c. Aluminum nitrite
d. Aluminum nitrogen trioxide
4. What is the correct formula of phosphorus trichloride?
a. P2Cl2
b. PCl3
c. PCl4
d. P4Cl2
5. What is the correct formula of lithium perchlorate?
a. Li2ClO4
b. LiClO2
c. LiClO
d. None of these
6. Write the correct name for these compounds.
a. BeC2O4:
b. NH4MnO4:
c. CoS2O3:
7. What is W(HSO4)5?
8. How do you write diphosphorus trioxide?
9. What is H3P?
10. By adding oxygens to the molecule in number 9, we now have H3PO4? What is the name of this molecule?
Answer
1.C; Calcium + Carbonate --> Ca2+ + CO32- --> CaCO3
2.D; FeO --> Fe + O2- --> Iron must have a charge of +2 to make a neutral compound --> Fe2+ + O2- --> Iron(II) Oxide
3.A; Al(NO3)3 --> Al3+ + (NO3-)3 --> Aluminum nitrate
4.B; Phosphorus trichloride --> P + 3Cl --> PCl3
5.D, LiClO4; Lithium perchlorate --> Li+ + ClO4- --> LiClO4
6. a. Beryllium Oxalate; BeC2O4 --> Be2+ + C2O42- --> Beryllium Oxalate
b. Ammonium Permanganate; NH4MnO4 --> NH4+ + MnO4- --> Ammonium Permanganate
c. Cobalt (II) Thiosulfate; CoS2O3 --> Co + S2O32- --> Cobalt must have +2 charge to make a neutral compund --> Co2+ + S2O32- --> Cobalt(II) Thiosulfate
7. Tungsten (V) hydrogen sulfate
8. P2O3
9. Hydrophosphoric Acid
10. Phosphoric Acid
Contributors and Attributions
• Pui Yan Ho (UCD), Alex Moskaluk (UCD), Emily Nguyen (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/02%3A_An_Introduction_to_Organic_Compounds-_Nomenclature_Physical_Properties_and_Representation_of_Structure/2.15%3A_Fused_Cyclohexane_Rings.txt |
There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $\pi$ bonds and/or cyclic rings.
Saturated vs. Unsaturated Molecules
In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.
CH3CH2CH3 1-methyoxypentane
Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).
CH3CH=CHCH3 3-chloro-5-octyne
Calculating Degrees of Unsaturation (DoU)
Degree of Unsaturation (DoU) is also known as Double Bond Equivalent. If the molecular formula is given, plug in the numbers into this formula:
$DoU= \dfrac{2C+2+N-X-H}{2}$
• $C$ is the number of carbons
• $N$ is the number of nitrogens
• $X$ is the number of halogens (F, Cl, Br, I)
• $H$ is the number of hydrogens
As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.
For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.
The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.
• One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 $\pi$ bond).
• Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 $\pi$ bonds).
DoU
Possible combinations of rings/ bonds
# of rings
# of double bonds
# of triple bonds
1
1
0
0
0
1
0
2
2
0
0
0
2
0
0
0
1
1
1
0
3 3 0 0
2 1 0
1 2 0
0 1 1
0 3 0
1 0 1
Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.
Example: Benzene
What is the Degree of Unsaturation for Benzene?
Solution
The molecular formula for benzene is C6H6. Thus,
DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.
Problems
1. Are the following molecules saturated or unsaturated:
1. (b.) (c.) (d.) C10H6N4
2. Using the molecules from 1., give the degrees of unsaturation for each.
3. Calculate the degrees of unsaturation for the following molecular formulas:
1. (a.) C9H20 (b.) C7H8 (c.) C5H7Cl (d.) C9H9NO4
4. Using the molecular formulas from 3, are the molecules unsaturated or saturated.
5. Using the molecular formulas from 3, if the molecules are unsaturated, how many rings/double bonds/triple bonds are predicted?
Answers
1.
(a.) unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)
(b.) unsaturated
(c.) saturated
(d.) unsaturated
2. If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.
(a.) 2
(b.) 2 (one double bond and the double bond from the carbonyl)
(c.) 0
(d.) 10
3. Use the formula to solve
(a.) 0
(b.) 4
(c.) 2
(d.) 6
4.
(a.) saturated
(b.) unsaturated
(c.) unsaturated
(d.) unsaturated
5.
(a.) 0 (Remember-a saturated molecule only contains single bonds)
(b.) The molecule can contain any of these combinations (i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds
(c.) (i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds
(d.) (i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... (As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)
Contributors
• Kim Quach (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/03%3A_Alkenes-_Structure_Nomenclature_and_an_Introduction_to_Reactivity__Thermodynamics_and_Kinetics/3.01%3A_Molecular_Formulas_and_the_Degree_of_Unsaturation.txt |
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, i.e. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the SN1 vs. SN2 character of a nucleophilic substitution reaction.
8.4A: Steric effects on electrophilicity
Consider two hypothetical SN2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon.
Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (SN2) displacement.
The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a concerted SN2 reaction. Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (SN1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an SN2 displacement, when the electrophile being attacked is a sp3-hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4o. Remember that in an SN1 mechanism, the nucleophile attacks an sp2-hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles.
With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.
8.4B: Stability of carbocation intermediates
We know that the rate-limiting step of an SN1 reaction is the first step - formation of the this carbocation intermediate. The rate of this step – and therefore, the rate of the overall substitution reaction – depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms. According to Hammond’s postulate (section 6.2B), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (SN1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.
The critical question now becomes, what stabilizes a carbocation?
Think back to Chapter 7, when we were learning how to evaluate the strength of an acid. The critical question was “how stable is the conjugate base that results when this acid donates its proton"? In many cases, this conjugate base was an anion – a center of excess electron density. Anything that can draw some of this electron density away– in other words, any electron withdrawing group – will stabilize the anion.
So if it takes an electron withdrawing group to stabilize a negative charge, what will stabilize a positive charge? An electron donating group!
A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups – methyl, ethyl, and the like – are weak electron donating groups, and thus stabilize nearby carbocations. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is SN1 or SN2.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Example 8.10
Exercise 8.10: In which of the structures below is the carbocation expected to be more stable? Explain.
For the most part, carbocations are very high-energy, transient intermediate species in organic reactions. However, there are some unusual examples of very stable carbocations that take the form of organic salts. Crystal violet is the common name for the chloride salt of the carbocation whose structure is shown below. Notice the structural possibilities for extensive resonance delocalization of the positive charge, and the presence of three electron-donating amine groups.
Example 8.11
Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
Solution
When considering the possibility that a nucleophilic substitution reaction proceeds via an SN1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an SN1 mechanism must be considered unlikely, and the reaction probably proceeds by an SN2 mechanism. In the next chapter we will see several examples of biologically important SN1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure.
Example 8.12
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
4.13: The Relative Stabilities of Alkenes
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Tutorial | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/04%3A_The_Reactions_of_Alkenes/4.02%3A_Carbocation_Stability_Depends_on_the_Number_of_Alkyl_Groups_Attached_to_the_Positively_Charged_Carbon.txt |
Oxidation-Reduction Reactions, or redox reactions, are reactions in which one reactant is oxidized and one reactant is reduced simultaneously. This module demonstrates how to balance various redox equations.
Identifying Redox Reactions
The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction. This requires that one and typically more species changing oxidation states during the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components. These are often separated into independent two hypothetical half-reactions to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:
$\ce{ Cu (s) + 2 Ag^+ (aq) \rightarrow Cu^{2+} (aq) + 2 Ag (s)} \nonumber$
The first step in determining whether the reaction is a redox reaction is to split the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:
$\ce{ Cu (s) \rightarrow Cu^{2+}(aq)} \nonumber$
The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in $\ce{Cu}$ to +2 in $\ce{Cu^{2+}}$. Now consider the silver atoms
$\ce{ 2 Ag^+ (aq) \rightarrow 2 Ag (s)} \nonumber$
In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an pure element. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.
Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.
Balancing Redox Reactions
Balancing redox reactions is slightly more complex than balancing standard reactions, but still follows a relatively simple set of rules. One major difference is the necessity to know the half-reactions of the involved reactants; a half-reaction table is very useful for this. Half-reactions are often useful in that two half reactions can be added to get a total net equation. Although the half-reactions must be known to complete a redox reaction, it is often possible to figure them out without having to use a half-reaction table. This is demonstrated in the acidic and basic solution examples. Besides the general rules for neutral conditions, additional rules must be applied for aqueous reactions in acidic or basic conditions.
One method used to balance redox reactions is called the Half-Equation Method. In this method, the equation is separated into two half-equations; one for oxidation and one for reduction.
Half-Equation Method to Balance redox Reactions in Acidic Aqueous Solutions
Each reaction is balanced by adjusting coefficients and adding $\ce{H2O}$, $\ce{H^{+}}$, and $\ce{e^{-}}$ in this order:
1. Balance elements in the equation other than $\ce{O}$ and $\ce{H}$.
2. Balance the oxygen atoms by adding the appropriate number of water ($\ce{H2O}$) molecules to the opposite side of the equation.
3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding $\ce{H^{+}}$ ions to the opposite side of the equation.
4. Add up the charges on each side. Make them equal by adding enough electrons ($\ce{e^{-}}$) to the more positive side. (Rule of thumb: $\ce{e^{-}}$ and $\ce{H^{+}}$ are almost always on the same side.)
5. The $\ce{e^{-}}$ on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.
6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.
The equation can now be checked to make sure that it is balanced.
Half-Equation Method to Balance redox Reactions in Basic Aqueous Solutions
If the reaction is being balanced in a basic solution, the above steps are modified with the the addition of one step between #3 and #4:
3b Add the appropriate number of $\ce{OH^{-}}$ to neutralize all $\ce{H^{+}}$ and to convert into water molecules.
The equation can now be checked to make sure that it is balanced.
Neutral Conditions
The first step to balance any redox reaction is to separate the reaction into half-reactions. The substance being reduced will have electrons as reactants, and the oxidized substance will have electrons as products. (Usually all reactions are written as reduction reactions in half-reaction tables. To switch to oxidation, the whole equation is reversed and the voltage is multiplied by -1.) Sometimes it is necessary to determine which half-reaction will be oxidized and which will be reduced. In this case, whichever half-reaction has a higher reduction potential will by reduced and the other oxidized.
Example $1$: Balancing in a Neutral Solution
Balance the following reaction
$\ce{Cu^+(aq) + Fe(s) \rightarrow Fe^{3+} (aq) + Cu (s)} \nonumber$
Solution
Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:
$\ce{Cu^+ (aq) + e^{-} \rightarrow Cu(s)} \nonumber$
and
$\ce{Fe^{3+} (aq) + 3e^{-} \rightarrow Fe(s)} \nonumber$
The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:
$\ce{Cu^+ (aq) + e^{-} \rightarrow Cu(s)} \nonumber$
and
$\ce{Fe (s) \rightarrow Fe^{3+}(aq) + 3e^{-}} \nonumber$
Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire $Cu^+(aq) + e^{-} \rightarrow Cu(s)$ half-reaction by 3 and leaving the other half reaction as it is. This gives:
$\ce{3Cu^+(aq) + 3e^{-} \rightarrow 3Cu(s)} \nonumber$
and
$\ce{Fe(s) \rightarrow Fe^{3+}(aq) + 3e^{-}} \nonumber$
Step 3: Adding the equations give:
$\ce{3Cu^+(aq) + 3e^{-} + Fe(s) \rightarrow 3Cu(s) + Fe^{3+}(aq) + 3e^{-}} \nonumber$
The electrons cancel out and the balanced equation is left.
$\ce{3Cu^{+}(aq) + Fe(s) \rightarrow 3Cu(s) + Fe^{3+}(aq)} \nonumber$
Acidic Conditions
Acidic conditions usually implies a solution with an excess of $\ce{H^{+}}$ concentration, hence making the solution acidic. The balancing starts by separating the reaction into half-reactions. However, instead of immediately balancing the electrons, balance all the elements in the half-reactions that are not hydrogen and oxygen. Then, add $\ce{H2O}$ molecules to balance any oxygen atoms. Next, balance the hydrogen atoms by adding protons ($\ce{H^{+}}$). Now, balance the charge by adding electrons and scale the electrons (multiply by the lowest common multiple) so that they will cancel out when added together. Finally, add the two half-reactions and cancel out common terms.
Example $2$: Balancing in a Acid Solution
Balance the following redox reaction in acidic conditions.
$\ce{Cr_2O_7^{2-} (aq) + HNO_2 (aq) \rightarrow Cr^{3+}(aq) + NO_3^{-}(aq) } \nonumber$
Solution
Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.
$\ce{Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq) } \nonumber$
$\ce{HNO_2 (aq) \rightarrow NO_3^{-}(aq)} \nonumber$
Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:
$\ce{Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq)} \nonumber$
and
$\ce{HNO_2(aq) \rightarrow NO_3^{-}(aq)} \nonumber$
Step 3: Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7 $\ce{H2O}$ molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:
$\ce{Cr_2O_7^{2-} (aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} \nonumber$
and
$\ce{HNO_2(aq) + H_2O(l) \rightarrow NO_3^{-}(aq) } \nonumber$
Step 4: Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.
$\ce{14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l)} \nonumber$
and
$\ce{HNO_2 (aq) + H2O (l) \rightarrow 3H^+(aq) + NO_3^{-}(aq)} \nonumber$
Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:
$\ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)} \nonumber$
For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:
$\ce{HNO_2(aq) + H_2O(l) \rightarrow 3H^+(aq) + NO_3^{-}(aq) + 2e^{-}} \nonumber$
Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:
$\ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 2Cr^{3+} (aq) + 7H_2O(l).} \nonumber$
and
\begin{align*} 3 \times \big[ \ce{HNO2 (aq) + H2O(l)} &\rightarrow \ce{3H^{+}(aq) + NO3^{-} (aq) + 2e^{-}} \big] \[4pt] \ce{3HNO2 (aq) + 3H_2O (l)} &\rightarrow \ce{9H^{+}(aq) + 3NO_3^{-}(aq) + 6e^{-}} \end{align*} \nonumber
Step 7: Add the reactions and cancel out common terms.
\begin{align*} \ce{3HNO_2 (aq) + 3H_2O (l)} &\rightarrow \ce{9H^+(aq) + 3NO_3^{-}(aq) + 6e^{-} } \[4pt] \ce{6e^{-} + 14H^+(aq) + Cr_2O_7^{2-}(aq)} &\rightarrow \ce{2Cr^{3+}(aq) + 7H_2O(l)} \[4pt] \hline \ce{3HNO_2 (aq)} + \cancel{\ce{3H_2O (l)}} + \cancel{6e^{-}} + \ce{14H^+(aq) + Cr_2O_7^{2-} (aq)} &\rightarrow \ce{9H^+(aq) + 3NO_3^{-}(aq)} + \cancel{6e^{-}} + \ce{2Cr^{3+}(aq)} + \cancelto{4}{7}\ce{H_2O(l)} \end{align*}
The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:
$\ce{3HNO_2(aq) + 5H^+(aq) + Cr_2O_7^{2-} (aq) \rightarrow 3NO_3^{-}(aq) + 2Cr^{3+}(aq) + 4H_2O(l)} \nonumber$
Basic Conditions
Bases dissolve into $\ce{OH^{-}}$ ions in solution; hence, balancing redox reactions in basic conditions requires $\ce{OH^{-}}$. Follow the same steps as for acidic conditions. The only difference is adding hydroxide ions to each side of the net reaction to balance any $\ce{H^{+}}$. $\ce{OH^{-}}$ and $\ce{H^{+}}$ ions on the same side of a reaction should be added together to form water. Again, any common terms can be canceled out.
Example $1$: Balancing in Basic Solution
Balance the following redox reaction in basic conditions.
$\ce{ Ag(s) + Zn^{2+}(aq) \rightarrow Ag_2O(aq) + Zn(s)} \nonumber$
Solution
Go through all the same steps as if it was in acidic conditions.
Step 1: Separate the half-reactions.
$\ce{Ag (s) \rightarrow Ag_2O (aq)} \nonumber$
$\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber$
Step 2: Balance elements other than O and H.
$\ce{ 2Ag (s) \rightarrow Ag_2O (aq)} \nonumber$
$\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber$
Step 3: Add H2O to balance oxygen.
$\ce{H_2O(l) + 2Ag(s) \rightarrow Ag_2O(aq)} \nonumber$
$\ce{Zn^{2+}(aq) \rightarrow Zn(s)} \nonumber$
Step 4: Balance hydrogen with protons.
$\ce{H_2O (l) + 2Ag (s) \rightarrow Ag_2O (aq) + 2H^+ (aq)} \nonumber$
$\ce{Zn^{2+} (aq) \rightarrow Zn (s)} \nonumber$
Step 5: Balance the charge with e-.
$\ce{H_2O (l) + 2Ag (s) \rightarrow Ag_2O (aq) + 2H^+ (aq) + 2e^{-}} \nonumber$
$\ce{Zn^{2+} (aq) + 2e^{-} \rightarrow Zn (s)} \nonumber$
Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.
Step 7: Add the reactions and cancel the electrons.
$\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H^+(aq). } \nonumber$
Step 8: Add OH- to balance H+. There are 2 net protons in this equation, so add 2 OH- ions to each side.
$\ce{H_2O(l) + 2Ag(s) + Zn^{2+}(aq) + 2OH^{-}(aq) \rightarrow Zn(s) + Ag_2O(aq) + 2H^+(aq) + 2OH^{-}(aq).} \nonumber$
Step 9: Combine OH- ions and H+ ions that are present on the same side to form water.
$\ce{\cancel{H2O(l)} + 2Ag(s) + Zn^{2+}(aq) + 2OH^{-}(aq) -> Zn(s) + Ag_2O(aq) + \cancel{2}H_2O(l)} \nonumber$
Step 10: Cancel common terms.
$\ce{2Ag(s) + Zn^{2+}(aq) + 2OH^{-} (aq) \rightarrow Zn(s) + Ag_2O(aq) + H_2O(l)} \nonumber$ | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/04%3A_The_Reactions_of_Alkenes/4.14%3A_Reactions_and_Synthesis.txt |
Several organic compounds may have identical compositions but will have widely different physical and chemical properties because the arrangement of the atoms is different. Isomers and identical compounds both have the same number of each kind of element in a formula. A simple count will establish this fact.
Introduction
As a consequence of the double bond, some alkene compounds exhibit a unique type of isomerism. Rotation around a single bond occurs readily, while rotation around a double bond is restricted. The pi bond prevents rotation because of the electron overlap both above and below the plane of the atoms.
A single bond is analogous to two boards nailed together with one nail. A double bond is analogous to two boards nailed together with two nails. In the first case you can twist the boards, while in the second case you cannot twist them.
Geometric Isomers are compounds with different spatial arrangements of groups attached to the carbons of a double bond. In alkenes, the carbon-carbon double bond is rigidly fixed. Even though the attachment of atoms is the same, the geometry (the way the atoms "see" each other) is different.
When looking for geometric isomers, a guiding principle is that there MUST BE TWO DIFFERENT "GROUPS" ON EACH CARBON OF THE DOUBLE BOND. A "group" can be hydrogen, alkyls, halogens, etc.
Identical compounds may appear to have different arrangements as written, but closer examination by rotation or turning will result in the molecules being superimposed. If they are super impossible or if they have identical names, then the two compounds are in fact identical.
Isomers of compounds have a different arrangement of the atoms. Isomer compounds will differ from identical compounds by the arrangement of the atoms. See example below.
Both compounds have the same number of atoms, C5H12. They are isomers because in the left molecule the root is 4 carbons with one branch. In the right molecule, the root is 3 carbons with 2 branches. They are isomers because they have the same number of atoms but different arrangements of those atoms.
Completely different compounds: If the number of each element is different, the two compounds are merely completely different. A simple count of the atoms will reveal them as different.
1,2-dichlorethene
In the example on the left, the chlorine atoms can be opposite or across from each other in which case it is called the "trans" isomer. If the the chlorine atoms are next to or adjacent each other, the isomer is called " cis".
If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc. there cannot be any geometric isomers.
2-butene
Consider the longest chain containing the double bond: If two groups (attached to the carbons of the double bond) are on the same side of the double bond, the isomer is a cis alkene. If the two groups lie on opposite sides of the double bond, the isomer is a trans alkene. One or more of the "groups" may or may not be part of the longest chain. In the case on the left, the "group" is a methyl - but is actually part of the longest chain.
A common mistake is to name this compound as 1,2-dimethylethene. Look at all carbons for the longest continuous chain - the root is 4 carbons - butene.
Problems
For the structures below:
a. Draw both cis/trans isomers, if any, of the structure based upon the name.
b. Look at the graphic and state whether the compound is cis, trans, or not cis/trans isomers.
• 2-methyl-2-butene
• 3-methyl-2-pentene
• 1-pentene
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.01%3A_Cis-Trans_Isomers_Result_from_Restricted_Rotation.txt |
To better understand the molecular origins of the ideal gas law,
$PV=nRT$
the basics of the Kinetic Molecular Theory of Gases (KMT) should be understood. This model is used to describe the behavior of gases. More specifically, it is used to explain macroscopic properties of a gas, such as pressure and temperature, in terms of its microscopic components, such as atoms. Like the ideal gas law, this theory was developed in reference to ideal gases, although it can be applied reasonably well to real gases.
In order to apply the kinetic model of gases, five assumptions are made:
1. Gases are made up of particles with no defined volume but with a defined mass. In other words their volume is miniscule compared to the distance between themselves and other molecules.
2. Gas particles undergo no intermolecular attractions or repulsions. This assumption implies that the particles possess no potential energy and thus their total energy is simply equal to their kinetic energies.
3. Gas particles are in continuous, random motion.
4. Collisions between gas particles are completely elastic. In other words, there is no net loss or gain of kinetic energy when particles collide.
5. The average kinetic energy is the same for all gases at a given temperature, regardless of the identity of the gas. Furthermore, this kinetic energy is proportional to the absolute temperature of the gas.
Temperature and KMT
The last assumption can be written in equation form as:
$KE = \dfrac{1}{2}mv^2 = \dfrac{3}{2}k_BT$
where
• $k_B$ is Boltzmann's constant (kB = 1.381×10-23 m2 kg s-2 K-1) and
• $T$ is the absolution temperature (in Kelvin)
This equation says that the speed of gas particles is related to their absolute temperature. In other words, as their temperature increases, their speed increases, and finally their total energy increases as well. However, it is impossible to define the speed of any one gas particle. As such, the speeds of gases are defined in terms of their root-mean-square speed.
Pressure and KMT
The macroscopic phenomena of pressure can be explained in terms of the kinetic molecular theory of gases. Assume the case in which a gas molecule (represented by a sphere) is in a box, length L (Figure 1). Through using the assumptions laid out above, and considering the sphere is only moving in the x-direction, we can examine the instance of the sphere colliding elastically with one of the walls of the box.
The momentum of this collision is given by p=mv, in this case p=mvx, since we are only considering the x dimension. The total momentum change for this collision is then given by
$mv_x - m(-v_x) = 2mv_x$
Given that the amount of time it takes between collisions of the molecule with the wall is L/vx we can give the frequency of collisions of the molecule against a given wall of the box per unit time as vx/2L. One can now solve for the change in momentum per unit of time:
$(2mv_x)(v_x/2L) = mv_x^2/L$
Solving for momentum per unit of time gives the force exerted by an object (F=ma=p/time). With the expression that F=mvx2/L one can now solve for the pressure exerted by the molecular collision, where area is given as the area of one wall of the box, A=L2:
$P=\dfrac{F}{A}$
$P=\dfrac{mv_x^2}{[L(L^2)}$
The expression can now be written in terms of the pressure associated with collisions from N number of molecules:
$P=\dfrac{Nmv_x^2}{V}$
This expression can now be adjusted to account for movement in the x, y and z directions by using mean-square velocity for three dimensions and a large value of N. The expression now is written as:
$P={\dfrac{Nm\overline{v}^2}{3V}}$
This expression now gives pressure, a macroscopic quality, in terms of atomic motion. The significance of the above relationship is that pressure is proportional to the mean-square velocity of molecules in a given container. Therefore, as molecular velocity increases so does the pressure exerted on the container.
Contributors and Attributions
• Sevini Shahbaz, Andrew Cooley | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.05%3A_Asymmetric_Centers_and_Stereocenters.txt |
When examining the ideal gas laws in conjunction with the kinetic theory of gases, we gain insights into the behavior of ideal gas. We can then predict how gas particles behaviors such as gas molecular speed, effusion rates, distances traveled by gas molecules. Graham's Law, which was formulated by the Scottish physical chemist Thomas Graham, is an important law that connects gas properties to the kinetic theory of gases.
Introduction
The Kinetic Molecular Theory states that the average energy of molecules is proportional to absolute temperature as illustrated by the following equation:
$e_K=\dfrac{3}{2}\dfrac{R}{N_A}T$
where
• ek is the average translation kinetic energy,
• R is the gas constant,
• NA is Avogadro's number, and
• T is temperature in Kelvins.
Since R and NA are constants, this means that the Kelvin temperature (T) of a gas is directly proportional to the average kinetic energy of its molecules. This means that at a given temperature, different gases (for example He or O2) will the same average kinetic energy.
Graham's Law
Gas molecules move constantly and randomly throughout the volume of the container they occupy. When examining the gas molecules individually, we see that not all of the molecules of a particular gas at a given temperature move at exactly the same speed. This means that each molecule of a gas have slightly different kinetic energy. To calculate the average kinetic energy (eK) of a sample of a gas, we use an average speed of the gas, called the root mean square speed (urms).
$e_K=\dfrac{1}{2}m{u_{rms}^2}$
with
• eK is the kinetic energy measures in Joules
• m is mass of a molecule of gas (kg)
• urms is the root mean square speed (m/s)
The root mean square speed, urms, can be determined from the temperature and molar mass of a gas.
$u_{rms}=\sqrt{\dfrac{3RT}{M}}$
with
• R the ideal gas constant (8.314 kg*m2/s2*mol*K)
• T temperature (Kelvin)
• M molar mass (kg/mol)
When examining the root mean square speed equation, we can see that the changes in temperature (T) and molar mass (M) affect the speed of the gas molecules. The speed of the molecules in a gas is proportional to the temperature and is inversely proportional to molar mass of the gas. In other words, as the temperature of a sample of gas is increased, the molecules speed up and the root mean square molecular speed increases as a result.
Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation:
$\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{(u_rms)_A}{(u_rms)_B}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
In according with the Kinetic Molecular Theory, each gas molecule moves independently. However, the net rate at which gas molecules move depend on their average speed. By examining the equation above, we can conclude that the heavier the molar mass of the gas molecules slower the gas molecules move. And conversely, lighter the molar mass of the gas molecules the faster the gas molecules move.
Limitations of Graham's Law
Graham's Law can only be applied to gases at low pressures so that gas molecules escape through the tiny pinhole slowly. In addition, the pinhole must be tiny so that no collisions occur as the gas molecules pass through. Since Graham's Law is an extension of the Ideal Gas Law, gases that follows Graham's Law also follows the Ideal Gas Law.
Molecular Effusion
The random and rapid motion of tiny gas molecules results in effusion. Effusion is the escape of gas molecules through a tiny hole or pinhole.
The behavior of helium gas in balloons is an example of effusion. The balloons are made of latex which is porous material that the small helium atom can effuse through. The helium inside a newly inflated balloon will eventually effuse out. This is the reason why balloons will deflate after a period of time. Molecular speeds are also used to explain why small molecules (such as He) diffuse more rapidly than larger molecules (O2). That is the reason why a balloon filled with helium gas will deflate faster than a balloon filled with oxygen gas.
The effusion rate, r, is inversely proportional to the square root of its molar mass, M.
$r\propto\sqrt{\dfrac{1}{M}}$
When there are two different gases the equation for effusion becomes
$\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}$
$M_A$ is the molar mass of gas A, $M_B$ is the molar mass of gas B, $T$ Temperature in Kelvin, $R$ is the ideal gas constant.
From the equation above,
Rates of effusion of two gases
The relative rates of effusion of two gases at the same temperature is given as:
$\dfrac{\it{Rate\;of\;effusion\;of\;gas\;}\mathrm 1}{\it{Rate\;of \;effusion\;of\;gas\;}\mathrm 2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}$
Units used to express rate of effusion includes: moles/seconds, moles/minutes, grams/seconds, grams/minutes.
Relative distances of two gases
The relative distances traveled by the two gases is given as:
$\dfrac{Distance\;traveled\;by\;gas\;1}{Distance\;traveled\;by\;gas\;2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}$
By examine the Graham's law as stated above, we can conclude that a lighter gas will effuse or travel more rapidly than a heavier gas. Mathematically speaking, a gas with smaller molar mass will effuse faster than a gas with larger molar mass under the same condition.
Molecular Diffusion
Similar to effusion, the process of diffusion is the spread of gas molecules through space or through a second substance such as the atmosphere.
Diffusion has many useful applications. Here is an example of diffusion that is use in everyday households. Natural gas is odorless and used commercially daily. An undetected leakage can be very dangerous as it is highly flammable and can cause an explosion when it comes in contact with an ignition source. In addition, the long term breathing of natural gas can lead to asphyxiation. Fortunately, chemists have discovered a way to easily detect natural gas occur leak by adding a small quantity of a gaseous organic sulfur compound named methyl mercaptan, CH3SH, to the natural gas. When a leak happens, the diffusion of the odorous methyl mercaptan in the natural gas will serve as a sign of warning.
General guideline in solving effusion problems
The following flowchart outlines the steps in solving quantitative problems involving Graham's Law. It can be used as a general guideline.
Example $1$
Calculate the root mean square speed, $u_{rms}$, in m/s of helium at $30 ^o C$.
Solution
Start by converting the molar mass for helium from g/mol to kg/mol.
$M=(4.00\;g/mol)\times\dfrac{1\;kg}{1000\;g}$
$M=4.00\times10^{-3}\;kg/mol$
Now, using the equation for $u_{rms}$ substitute in the proper values for each variable and perform the calculation.
$u_{rms}=\sqrt{\dfrac{3RT}{M}}$
$u_{rms}=\sqrt{\dfrac{3 \times (8.314\; kg\;m^2/s^2*mol*K)(303K)}{4.00 \times 10^{-3}kg/mol}}$
$u_{rms}=1.37 \times 10^3 \;m/s$
Example $2$
What is the ratio of $u_{rms}$ values for helium vs. xenon at $30^oC$. Which is higher and why?
Solution
There are two approaches to solve this problem: the hard way and the easy way
Hard way:
The $u_{rms}$ speed of helium is calculated from the above example.
First convert the molar mass of xenon from g/mol to kg/mol as we did for helium in example 1
$M_{Xe}=(131.3\;g/mol)\times\dfrac{1\;kg}{1000\;g}$
$M=0.1313\;kg/mol$
Now, using the equation for the urms, insert the given and known values and solve for the variable of interest.
$u_{rms}=\sqrt{\dfrac{3(8.314\; kg\;m^2/s^2*mol*K)(303\;K)}{0.1313\;kg/mol}}$
$u_{rms}=2.40 \times 10^2\;m/s$
Compare the two values for xenon and helium and decide which is greater.
• $u_{Xe}=2.4 \times 10^2 \;m/s$
• $u_{He}=1.37 \times10^3 \;m/s$
So the ratio of RMS speeds is
$\dfrac{u_{Xe}}{u_{He}} \approx 0.18$
Helium has the higher $u_{rms}$ speed. This is in according with Graham's Law, because helium atoms are much lighter than xenon atoms.
Easy Way:
Since the temperature is the same for both gases, only the square root of the ratio of molar mass is needed to be calculated.
$\sqrt{\dfrac{M_{He}}{M_{Xe}}} = \sqrt{\dfrac{4.00 \; g/mol}{131.3\;g/mol }} \approx 0.18$
In either approach, helium has a faster RMS speed than xenon and this is due exclusively to its smaller mass.
Example $3$
If oxygen effuses from a container in 5.00 minutes, what is the molecular weight of a gas with the same given quantity of molecules effusing from the same container in 4.00 min?
Solution
Let oxygen be gas A and its rate is 1/5.00 minutes because it takes that much time for a certain quantity of oxygen to effuse and its molecular weight is 32 grams/mole (O2 (g)). Let the unknown gas be B and its rate is 1/4.00 minutes and Mb be the molecular weight of the unknown gas.
First, choose the appropriate equation,
$\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem.
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
$\sqrt{M_B}=\dfrac{(1/5.00\;minutes)(\sqrt{32\;grams/mole})}{1/4.00\;minutes}$
$\sqrt{M_B}=4.525$
${M_B}=(4.525)^2=20.5\;grams/mole$
Example $4$
Oxygen, O2 (g), effuses from a container at the rate of 3.64 mL/sec, what is the molecular weight of a gas effusing from the same container under identical conditions at 4.48 mL/sec?
Solution
Let label oxygen as gas A. Therefore, rate of effusion of gas A is equaled to 3.64 mL/sec and the molecular weight is 32 grams/mole. The unknown gas is B and its rate of effusion is 4.48 mL/sec. We are solving for the molecular weight of gas B which is labeled as MB.
First, choose the appropriate equation,
$\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem.
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A})(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
$\sqrt{M_B}=\dfrac{(3.64\;mL/sec)(\sqrt{32\;grams/mole})}{4.48\;mL/sec}$
$\sqrt{M_B}=4.50$
${M_B}=(4.50)^2=20.2\;grams/mole$
Example $5$
Which answer(s) are true when comparing 1.0 mol O2 (g) at STP (Standard Temperature and Pressure) and 0.50 mole of S2 (g) at STP?
The two gases have equal through the same orifice, a tiny opening:
1. average molecular kinetic energies
2. root-mean-square speeds
3. masses
4. volumes
5. densities
6. effusion rates
Solution
The average kinetic energy of gas molecules depends on only the Temperature as exemplifies in this equation:
$e_K=\dfrac{3}{2}\dfrac{R}{N_A}T$
In addition, the mass of 0.50 mole of S2 is the same as that of 1.0 mole of O2. The rest of the choices are false.
Therefore, the correct answers are (a) and (c).
• Tram Anh Dao | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.06%3A_How_to_Draw_Enantiomers.txt |
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature.
Introduction
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule, at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
Stereocenters are labeled R or S
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.
Consider the first picture: a curved arrow is drawn from the highest priority (1) substituent to the lowest priority (4) substituent. If the arrow points in a counterclockwise direction (left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,(Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. For example: (R)-2-Bromobutane and (S)-2,3- Dihydroxypropanal.
Sequence rules to assign priorities to substituents
Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:
Rule 1
First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number.
1. When dealing with isotopes, the atom with the higher atomic mass receives higher priority.
2. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away.
3. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.
When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated.
Remember that
• Wedges indicate coming towards the viewer.
• Dashes indicate pointing away from the viewer.
Rule 2
If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.
If the chains are similar, proceed down the chain, until a point of difference.
For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.
Rule 3
If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to.
• If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority
• If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority.
Example 2
A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:
However:
Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above.
Caution!!
Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.
When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.
After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S.
1. Put the lowest priority substituent in the back (dashed line).
2. Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3)
3. Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise.
i) If it is clockwise it is R.
ii) if it is counterclockwise it is S.
USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).
IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.
If you have a modeling kit use it to help you solve the following practice problems.
Exercise \(1\)
Are the following R or S?
Answer
1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH3 is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH2NH2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH2NH2, you are turning right, giving you a R. (5)
5. S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.07%3A_Naming_Enantiomers_by_the_RS_System.txt |
What do you notice about these three pictures? Count the number of left gloves and right gloves.
6 left and 6 right gloves, correct?
What about this one:
I count 8 right gloves, 4 left gloves. So there’s a slight excess of right gloves here.
Finally, this figure:
ONLY right hand gloves here. 12 right gloves, zero left gloves.
Application to organic chemistry?
Gloves are chiral objects. That is, they lack an internal plane of symmetry. Left gloves and right gloves are mirror images of each other, but they can’t be superimposed. In chemistry, there’s a word we have to describe a pair of non-superimposable mirror images – they’re called enantiomers.
Tying it back to the drawings, we can have three types of situations.
1. Racemic Mixture: In the first drawing, we have an equal number of left and right gloves (i.e. enantiomers). This is called a racemic mixture of enantiomers.
2. Enantiomeric excess: In the second drawing, we have an excess of right gloves compared to left gloves. In a situtation like this we can say we have an “enantiomeric excess” of gloves, or alternatively, the mixture is “enantioenriched” in the right-hand glove. [We can also calculate the "excess" here: the mixture is 66% right and 33% left - so we have a 33% "excess" of the right-hand enantiomer].
3. Enantiomeric pure: In the third drawing, we have only right-hand gloves. This is said to be an “enantiomerically pure” mixture of gloves, since we have only one enantiomer present.
To tie it back to chemistry, let’s say we have a solution of a chiral molecule, like 2-butanol, which can exist as either the (R)-enantiomer or the (S)-enantiomer.
• A solution containing equal amounts of (R)-2-butanol and (S)-2-butanol is a racemic mixture.
• A solution containing an excess of either the (R)-enantiomer or the (S)-enantiomer would be enantioenriched.
• A solution containing only the (R)-enantiomer or the (S)-enantiomer will be enantiomerically pure.
Contributors
James Ashenhurst (MasterOrganicChemistry.com)
• A big thanks to Agnieszka at IlluScientia for the glove drawings.
4.5.1 Enantiomeric Excess and Optical Purity
Enantiomeric Excess
For non-racemic mixtures of enantiomers, one enantiomer is more abundant than the other. The composition of these mixtures is described by the enantiomeric excess, which is the difference between the relative abundance of the two enantiomers. Therefore, if a mixture contains 75% of the R enantiomer and 25% S, the enantiomeric excess if 50%. Similarly, a mixture that is 95% of one enantiomer, the enantiomeric excess is 90%, etc.
Enantiomeric excess is useful because it reflects the optical activity of the mixture. The standard optical rotation by the mixture ($[\alpha]_{mix}$) is equal to the product of the standard optical rotation of the major isomer ($[\alpha]_{major}$) and the enantiomeric excess ($EE$):
$[\alpha]_{mix} = EE \times [\alpha]_{major}$
In the same way, the enantiomeric excess in a mixture can be measured if the optical rotation of the pure enantiomer is known.
Diastereomeric Excess
A similar approach can be used to describe mixtures of diastereomers, resulting in the diastereomeric excess.
4.5.2 Resolution of Enantiomers
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
To learn more about chemical procedures for achieving resolution Click Here.
Cliffs Notes
Web Pages
Further Reading on Enantiomeric Excess and Optical Purity
Carey 4th Edition On-Line Activity
5.13: How to Name Isomers with More than One Asymmetric Center
Diastereomers are stereoisomers that are not related as object and mirror image and are not enantiomers. Unlike enatiomers which are mirror images of each other and non-sumperimposable, diastereomers are not mirror images of each other and non-superimposable. Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities. They have two or more stereocenters.
Introduction
It is easy to mistake between diasteromers and enantiomers. For example, we have four steroisomers of 3-bromo-2-butanol. The four possible combination are SS, RR, SR and RS (Figure 1). One of the molecule is the enantiomer of its mirror image molecule and diasteromer of each of the other two molecule (SS is enantiomer of RR and diasteromer of RS and SR). SS's mirror image is RR and they are not superimposable, so they are enantiomers. RS and SR are not mirror image of SS and are not superimposable to each other, so they are diasteromers.
Figure 1
Diastereomers vs. Enantiomers vs. Meso Compounds
Tartaric acid, C4H6O6, is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (figure 2).
(R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius.
Figure 2
To identify meso, meso compound is superimposed on its mirror image, and has an internal plane that is symmetry (figure 3). Meso-tartaric acid is achiral and optically unactive.
Problems
Identify which of the following pair is enantiomers, diastereomers or meso compounds.
Answer
1. Diasteromers
2. Identical
3. Meso
4. Enantiomers | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.10%3A_Enantiomeric_Excess.txt |
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
Introduction
The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion.
Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone ($O_3$) to produce $O_2$ molecules; however, $NO$ molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for $NO$ to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with $O_3$ and destroy the molecule. This process is called the Rowland-Molina theory of $O_3$ depletion.
The Chapman Cycle
The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O2) to create ozone (O3) and atomic oxygen (O). This process is called the Chapman cycle.
Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu + O_2 \rightarrow 2O^. \nonumber$
Step 2: Oxygen radicals then react with molecular oxygen to produce ozone:
$O_2 + O^. \rightarrow O_3 \nonumber$
Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen:
$O_3 + O^. \rightarrow 2O_2 \nonumber$
Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon:
$O_3 + h\nu \rightarrow O_2 + O^. \nonumber$
It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O2 is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself.
Chemistry of Ozone Depletion
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:
$Cl + O_3 \rightarrow ClO + O_2 \tag{step 1}$
$ClO + O^. \rightarrow Cl + O_2 \tag{step 2}$
$O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction}$
Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface.
From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's.
The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction.
• Because Antarctica is surrounded by water, winds over the continent blow in a unique clockwise direction creating a so called "polar vortex" that effectively contains a single static air mass over the continent. As a result, air over Antarctica does not mix with air in the rest of the earth's atmosphere.
• Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a conglomeration of frozen H2O and HNO3. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds
As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O3. Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process.
Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink
Ozone Depleting Substances
The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act:
Table $1$: Ozone Depleting Substances And Their Ozone-Depletion Potential. Taken directly from the Clean Air Act, as of June 2010.
Substance Ozone- depletion potential
chlorofluorocarbon-11 (CFC–11) 1.0
chlorofluorocarbon-12 (CFC–12) 1.0
chlorofluorocarbon-13 (CFC–13) 1.0
chlorofluorocarbon-111 (CFC–111) 1.0
chlorofluorocarbon-112 (CFC–112) 1.0
chlorofluorocarbon-113 (CFC–113) 0.8
chlorofluorocarbon-114 (CFC–114) 1.0
chlorofluorocarbon-115 (CFC–115) 0.6
chlorofluorocarbon-211 (CFC–211) 1.0
chlorofluorocarbon-212 (CFC–212) 1.0
chlorofluorocarbon-213 (CFC–213) 1.0
chlorofluorocarbon-214 (CFC–214) 1.0
chlorofluorocarbon-215 (CFC–215) 1.0
chlorofluorocarbon-216 (CFC–216) 1.0
chlorofluorocarbon-217 (CFC–217) 1.0
halon-1211 3.0
halon-1301 10.0
halon-2402 6.0
carbon tetrachloride 1.1
methyl chloroform 0.1
hydrochlorofluorocarbon-22 (HCFC–22) 0.05
hydrochlorofluorocarbon-123 (HCFC–123) 0.02
hydrochlorofluorocarbon-124 (HCFC–124) 0.02
hydrochlorofluorocarbon-141(b) (HCFC–141(b)) 0.1
hydrochlorofluorocarbon-142(b) (HCFC–142(b)) 0.06
General Questions
• What are the causes of the depletion of our ozone layer?
• the release of free radicals, the use of CFC's, the excessive burning of fossil fuels
• What is the chemical reaction that displays how ozone is created?
• UV + O2 -> 2O + heat, O2 + O -> O3, O3 + O -> 2O2
• Which reactions demonstrate the destruction of the ozone layer?
• Cl + O3 ------> ClO + O2 and ClO + O ------> Cl + O
• How do CFC's destroy the ozone layer?
• the atomic chlorine freed from CFC reacts in a catalytic manner with ozone and atomic oxygen to make more oxygen molecules
• Why should regulations be enforced now in regards to pollution and harmful chemicals?
• without regulation, the production and use of chemicals will run out of hand and do irreversible damage to the stratosphere
• What type of atom in the CFC molecule is most destructive to the ozone?
• chlorine
• In which layer of the atmosphere does the ozone layer?
• the stratosphere, the second closest to the Earth's surface
• What cycle is responsible for ozone in the stratosphere?
• the Chapman cycle
• What factor is responsible for breaking up stable molecules?
• ultraviolet rays from the sun | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.14%3A__Reactions_of_Compounds_that_Contain_an_Asymmetric_Center.txt |
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
Introduction
The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion.
Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone ($O_3$) to produce $O_2$ molecules; however, $NO$ molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for $NO$ to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with $O_3$ and destroy the molecule. This process is called the Rowland-Molina theory of $O_3$ depletion.
The Chapman Cycle
The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O2) to create ozone (O3) and atomic oxygen (O). This process is called the Chapman cycle.
Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu + O_2 \rightarrow 2O^. \nonumber$
Step 2: Oxygen radicals then react with molecular oxygen to produce ozone:
$O_2 + O^. \rightarrow O_3 \nonumber$
Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen:
$O_3 + O^. \rightarrow 2O_2 \nonumber$
Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon:
$O_3 + h\nu \rightarrow O_2 + O^. \nonumber$
It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O2 is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself.
Chemistry of Ozone Depletion
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:
$Cl + O_3 \rightarrow ClO + O_2 \tag{step 1}$
$ClO + O^. \rightarrow Cl + O_2 \tag{step 2}$
$O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction}$
Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface.
From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's.
The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction.
• Because Antarctica is surrounded by water, winds over the continent blow in a unique clockwise direction creating a so called "polar vortex" that effectively contains a single static air mass over the continent. As a result, air over Antarctica does not mix with air in the rest of the earth's atmosphere.
• Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a conglomeration of frozen H2O and HNO3. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds
As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O3. Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process.
Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink
Ozone Depleting Substances
The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act:
Table $1$: Ozone Depleting Substances And Their Ozone-Depletion Potential. Taken directly from the Clean Air Act, as of June 2010.
Substance Ozone- depletion potential
chlorofluorocarbon-11 (CFC–11) 1.0
chlorofluorocarbon-12 (CFC–12) 1.0
chlorofluorocarbon-13 (CFC–13) 1.0
chlorofluorocarbon-111 (CFC–111) 1.0
chlorofluorocarbon-112 (CFC–112) 1.0
chlorofluorocarbon-113 (CFC–113) 0.8
chlorofluorocarbon-114 (CFC–114) 1.0
chlorofluorocarbon-115 (CFC–115) 0.6
chlorofluorocarbon-211 (CFC–211) 1.0
chlorofluorocarbon-212 (CFC–212) 1.0
chlorofluorocarbon-213 (CFC–213) 1.0
chlorofluorocarbon-214 (CFC–214) 1.0
chlorofluorocarbon-215 (CFC–215) 1.0
chlorofluorocarbon-216 (CFC–216) 1.0
chlorofluorocarbon-217 (CFC–217) 1.0
halon-1211 3.0
halon-1301 10.0
halon-2402 6.0
carbon tetrachloride 1.1
methyl chloroform 0.1
hydrochlorofluorocarbon-22 (HCFC–22) 0.05
hydrochlorofluorocarbon-123 (HCFC–123) 0.02
hydrochlorofluorocarbon-124 (HCFC–124) 0.02
hydrochlorofluorocarbon-141(b) (HCFC–141(b)) 0.1
hydrochlorofluorocarbon-142(b) (HCFC–142(b)) 0.06
General Questions
• What are the causes of the depletion of our ozone layer?
• the release of free radicals, the use of CFC's, the excessive burning of fossil fuels
• What is the chemical reaction that displays how ozone is created?
• UV + O2 -> 2O + heat, O2 + O -> O3, O3 + O -> 2O2
• Which reactions demonstrate the destruction of the ozone layer?
• Cl + O3 ------> ClO + O2 and ClO + O ------> Cl + O
• How do CFC's destroy the ozone layer?
• the atomic chlorine freed from CFC reacts in a catalytic manner with ozone and atomic oxygen to make more oxygen molecules
• Why should regulations be enforced now in regards to pollution and harmful chemicals?
• without regulation, the production and use of chemicals will run out of hand and do irreversible damage to the stratosphere
• What type of atom in the CFC molecule is most destructive to the ozone?
• chlorine
• In which layer of the atmosphere does the ozone layer?
• the stratosphere, the second closest to the Earth's surface
• What cycle is responsible for ozone in the stratosphere?
• the Chapman cycle
• What factor is responsible for breaking up stable molecules?
• ultraviolet rays from the sun | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.15%3A_Using_Reactions_that_Do_Not_Break_Bonds_to_an_Asymmetric_Center.txt |
We turn now to concept of chirality that formed the basis of the story about Louis Pasteur in the beginning of this chapter. Recall that the term chiral, from the Greek work for 'hand', refers to anything which cannot be superimposed on its own mirror image. Your hands, of course, are chiral - you cannot superimpose your left hand on your right, and you cannot fit your left hand into a right-handed glove (which is also a chiral object). Another way of saying this is that your hands do not have a mirror plane of symmetry: you cannot find any plane which bisects your hand in such a way that one side of the plane is a mirror image of the other side. Chiral objects do not have a plane of symmetry.
Your face, on the other hand is achiral - lacking chirality - because, some small deviations notwithstanding, you could superimpose your face onto its mirror image. If someone were to show you a mirror image photograph of your face, you could line the image up, point-for-point, with your actual face. Your face has a plane of symmetry, because the left side is the mirror image of the right side.
What Pasteur, Biot, and their contemporaries did not yet fully understand when Pasteur made his discovery of molecular chirality was the source of chirality at the molecular level. It stood to reason that a chiral molecule is one that does not contain a plane of symmetry, and thus cannot be superimposed on its mirror image. We now know that chiral molecules contain one or more chiral centers, which are almost always tetrahedral (sp3-hybridized) carbons with four different substituents. Consider the cartoon molecule A below: a tetrahedral carbon, with four different substituents denoted by balls of four different colors (for the time being, don't worry about exactly what these substituents could be - we will see real examples very soon).
The mirror image of A, which we will call B, is drawn on the right side of the figure, and an imaginary mirror is in the middle. Notice that every point on A lines up through the mirror with the same point on B: in other words, if A looked in the mirror, it would see B looking back.
Now, if we flip compound A over and try to superimpose it point for point on compound B, we find that we cannot do it: if we superimpose any two colored balls, then the other two are misaligned.
A is not superimposable on its mirror image (B), thus by definition A is a chiral molecule. It follows that B also is not superimposable on its mirror image (A), and thus it is also a chiral molecule. Also notice in the figure below (and convince yourself with models) that neither A nor B has an internal plane of symmetry.
A and B are stereoisomers: molecules with the same molecular formula and the same bonding arrangement, but a different arrangement of atoms in space. There are two types of stereoisomers: enantiomers and diastereomers. Enantiomers are pairs of stereoisomers which are mirror images of each other: thus, A and B are enantiomers. It should be self-evident that a chiral molecule will always have one (and only one) enantiomer: enantiomers come in pairs. Enantiomers have identical physical properties (melting point, boiling point, density, and so on). However, enantiomers do differ in how they interact with polarized light (we will learn more about this soon) and they may also interact in very different ways with other chiral molecules - proteins, for example. We will begin to explore this last idea in later in this chapter, and see many examples throughout the remainder of our study of biological organic chemistry.
Diastereomers are stereoisomers which are not mirror images of each other. For now, we will concentrate on understanding enantiomers, and come back to diastereomers later.
We defined a chiral center as a tetrahedral carbon with four different substituents. If, instead, a tetrahedral carbon has two identical substituents (two black atoms in the cartoon figure below), then of course it still has a mirror image (everything has a mirror image, unless we are talking about a vampire!) However, it is superimposable on its mirror image, and has a plane of symmetry.
This molecule is achiral (lacking chirality). Using the same reasoning, we can see that a trigonal planar (sp2-hybridized) carbon is also not a chiral center.
Notice that structure E can be superimposed on F, its mirror image - all you have to do is pick E up, flip it over, and it is the same as F. This molecule has a plane of symmetry, and is achiral.
Let's apply our general discussion to real molecules. For now, we will limit our discussion to molecules with a single chiral center. It turns out that tartaric acid, the subject of our chapter introduction, has two chiral centers, so we will come back to it later.
Consider 2-butanol, drawn in two dimensions below.
Carbon #2 is a chiral center: it is sp3-hybridized and tetrahedral (even though it is not drawn that way above), and the four things attached to is are different: a hydrogen, a methyl (-CH3) group, an ethyl (-CH2CH3) group, and a hydroxyl (OH) group. Let's draw the bonding at C2 in three dimensions, and call this structure A. We will also draw the mirror image of A, and call this structure B.
When we try to superimpose A onto B, we find that we cannot do it. A and B are both chiral molecules, and they are enantiomers of each other.
2-propanol, unlike 2-butanol, is not a chiral molecule. Carbon #2 is bonded to two identical substituents (methyl groups), and so it is not a chiral center.
Notice that 2-propanol is superimposable on its own mirror image.
When we look at very simple molecules like 2-butanol, it is not difficult to draw out the mirror image and recognize that it is not superimposable. However, with larger, more complex molecules, this can be a daunting challenge in terms of drawing and three-dimensional visualization. The easy way to determine if a molecule is chiral is simply to look for the presence of one or more chiral centers: molecules with chiral centers will (almost always) be chiral. We insert the 'almost always' caveat here because it is possible to come up with the exception to this rule - we will have more to say on this later, but don't worry about it for now.
Here's another trick to make your stereochemical life easier: if you want to draw the enantiomer of a chiral molecule, it is not necessary to go to the trouble of drawing the point-for-point mirror image, as we have done up to now for purposes of illustration. Instead, keep the carbon skeleton the same, and simply reverse the solid and dashed wedge bonds on the chiral carbon: that accomplishes the same thing. You should use models to convince yourself that this is true, and also to convince yourself that swapping any two substituents about the chiral carbon will result in the formation of the enantiomer.
Here are four more examples of chiral biomolecules, each one shown as a pair of enantiomers, with chiral centers marked by red dots.
Here are some examples of achiral biomolecules - convince yourself that none of them contain a chiral center:
When looking for chiral centers, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not chiral centers – look, for example, at the drawings of glycine and citrate in the figure above.
Can a chiral center be something other than a tetrahedral carbon with four different substituents? The answer to this question is 'yes' - however, these alternative chiral centers are very rare in the context of biological organic chemistry, and outside the scope of our discussion here.
You may also have wondered about amines: shouldn't we consider a secondary or tertiary amine to be a chiral center, as they are tetrahedral and attached to four different substituents, if the lone-pair electrons are counted as a 'substituent'? Put another way, isn't an amine non-superimposable on its mirror image?
The answer: yes it is, in the static picture, but in reality, the nitrogen of an amine is rapidly and reversibly inverting, or turning inside out, at room temperature.
If you have trouble picturing this, take an old tennis ball and cut it in half. Then, take one of the concave halves and flip it inside out, then back again: this is what the amine is doing. The end result is that the two 'enantiomers' if the amine are actually two rapidly interconverting forms of the same molecule, and thus the amine itself is not a chiral center. This inversion process does not take place on a tetrahedral carbon, which of course has no lone-pair electrons.
Exercise 3.8: Locate all of the chiral centers (there may be more than one in a molecule). Remember, hydrogen atoms bonded to carbon usually are not drawn in the line structure convention - but they are still there!
Exercise 3.9:
a) Draw two enantiomers of i) mevalonate and ii) serine.
b) Are the two 2-butanol structures below enantiomers?
Exercise 3.10: Label the molecules below as chiral or achiral, and locate all chiral centers.
Solutions to exercises
Kahn Academy video tutorials
Chirality
Enantiomers
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.17%3A_Nitrogen_and_Phosphorus_Atoms_Can_Be_Asymmetric_Centers.txt |
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5.20: The Stereochemistry of Enzyme-Catalyzed Reactions
While challenging to understand and visualize, the stereochemistry concepts we have explored in this chapter are integral to the study of living things. The vast majority of biological molecules contain chiral centers and/or stereogenic alkene groups. Most importantly, proteins are chiral, which of course includes all of the the enzymes which catalyze the chemical reactions of a cell, the receptors which transmit information within or between cells, and the antibodies which bind specifically to potentially harmful invaders. You know from your biology classes that proteins, because they fold up into a specific three dimensional shape, are able to very specifically recognize and bind to other organic molecules. The ligand or substrate bound by a particular protein could be a small organic molecule such as pyruvate all the way up to a large biopolymer such as a specific region of DNA, RNA, or another protein. Because they are chiral molecules, proteins are very sensitive to the stereochemistry of their ligands: a protein may bind specifically to (R)-glyceraldehyde, for example, but not bind to (S)-glyceraldehyde, just as your right hand will not fit into a left-handed baseball glove (see end of chapter for a link to an animation illustrating this concept).
The over-the-counter painkiller ibuprofen is currently sold as a racemic mixture, but only the S enantiomer is effective, due to the specific way it is able to bind to and inhibit the action of prostaglandin H2 synthase, an enzyme in the body's inflammation response process.
The R enantiomer of ibuprofen does not bind to prostaglandin H2 synthase in the same way as the S enantiomer, and as a consequence does not exert the same inhibitory effect on the enzyme's action (see Nature Chemical Biology 2011, 7, 803 for more details). Fortunately, (R)-ibuprofen apparently does not cause any harmful side effects, and is in fact isomerized gradually by an enzyme in the body to (S)-ibuprofen.
Earlier in this chapter we discussed the tragic case of thalidomide, and mentioned that it appears that it is specifically the S enantiomer which caused birth defects. Many different proposals have been made over the past decades to try to explain the teratogenic (birth defect-causing) effect of the drug, but a clear understanding still evades the scientific community. In 2010, however, a team in Japan reported evidence that thalidomide binds specifically to a protein called 'thereblon'. Furthermore, when production of thereblon is blocked in female zebra fish, developmental defects occur in her offspring which are very similar to the defects caused by the administration of thalidomide, pointing to the likelihood that thalidomide binding somehow inactivates the protein, thus initiating the teratogenic effect. (http://news.sciencemag.org/2010/03/t...-partner-crime)
You can, with a quick trip to the grocery store, directly experience the biological importance of stereoisomerism. Carvone is a chiral, plant-derived molecule that contributes to the smell of spearmint in the R form and caraway (a spice) in the S form.
Although details are not known, the two enantiomers presumably interact differently with one or more smell receptor proteins in your nose, generating the transmission of different chemical signals to the olfactory center of your brain.
Exercise 3.28: Ephedrine, found in the Chinese traditional medicine ma huang, is a stimulant and appetite suppressant. Both pseudoephedrine and levomethamphetamine are active ingredients in over-the-counter nasal decongestants. Methamphetamine is a highly addictive and illegal stimulant, and is usually prepared in illicit 'meth labs' using pseudoephedrine as a starting point.
What is the relationship between ephedrine and pseudoephedrine? Between methamphetamine and levomethamphetamine? Between pseudoephedrine and methamphetamine? Your choices are: not isomers, constitutional isomers, diastereomers, enantiomers, or same molecule.
Solutions to exercises
Enzymes are very specific with regard to the stereochemistry of the reactions they catalyze. When the product of a biochemical reaction contains a chiral center or a stereogenic alkene, with very few exceptions only one stereoisomer of the product is formed. In the glycolysis pathway, for example, the enzyme triose-phosphate isomerase catalyzes the reversible interconversion between dihydroxyacetone (which is achiral) and (R)-glyceraldehyde phosphate. The (S)-glyceraldehyde enantiomer is not formed by this enzyme in the left-to-right reaction, and is not used as a starting compound in the right-to-left reaction - it does not 'fit' in the active site of the enzyme.
In the isoprenoid biosynthesis pathway, two five-carbon building-block molecules combine to form a ten-carbon chain containing an E-alkene group. The enzyme does not catalyze formation of the Z diastereomer.
In chapters 9-17 of this book, and continuing on into your study of biological and organic chemistry, you will be learning about how enzymes are able to achieve these feats of stereochemical specificity. If you take a more advanced class in organic synthesis, you will also learn how laboratory chemists are figuring out ingenious ways to exert control over the stereochemical outcomes of nonenzymatic reactions, an area of chemistry that is particularly important in the pharmaceutical industry.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.19%3A_The_Stereochemistry_of_Electrophilic_addition_Reactions_of_Alke.txt |
Chiral molecules, as we learned in the introduction to this chapter, have an interesting optical property. You may know from studying physics that light waves are oscillating electric and magnetic fields. In ordinary light, the oscillation is randomly oriented in an infinite number of planes. When ordinary light is passed through a polarizer, all planes of oscillation are filtered out except one, resulting in plane-polarized light.
A beam of plane-polarized light, when passed through a sample of a chiral compound, interacts with the compound in such a way that the angle of oscillation will rotate. This property is called optical activity.
If a compound rotates plane polarized light in the clockwise (+) direction, it is said to be dextrorotatory, while if it rotates light in the counterclockwise (-) direction it is levorotatory. (We mentioned L- and D-amino acids in the previous section: the L-amino acids are levorotatory). The magnitude of the observed optical activity is dependent on temperature, the wavelength of light used, solvent, concentration of the chiral sample, and the path length of the sample tube (path length is the length that the plane-polarized light travels through the chiral sample). Typically, optical activity measurements are made in a 1 decimeter (10 cm) path-length sample tube at 25 °C, using as a light source the so-called “D-line” from a sodium lamp, which has a wavelength of 589 nm. The specific rotation of a pure chiral compound at 25° is expressed by the expression:
. . . where alpha(obs) is the observed rotation, l is path length in decimeters, and c is the concentration of the sample in grams per 100 mL. In other words, the specific rotation of a chiral compound is the optical rotation that is observed when 1 g of the compound is dissolved in enough of a given solvent to make 100 mL solution, and the rotation is measured in a 1 dm cuvette at 25 oC using light from a sodium lamp.
Every chiral molecule has a characteristic specific rotation, which is recorded in the chemical literature as a physical property just like melting point or density. Different enantiomers of a compound will always rotate plane-polarized light with an equal but opposite magnitude. (S)-ibuprofen, for example, has a specific rotation of +54.5o (dextrorotatory) in methanol, while (R)-ibuprofen has a specific rotation of -54.5o. There is no relationship between chiral compound's R/S designation and the direction of its specific rotation. For example, the S enantiomer of ibuprofen is dextrorotatory, but the S enantiomer of glyceraldehyde is levorotatory.
A 50:50 mixture of two enantiomers (a racemic mixture) will have no observable optical activity, because the two optical activities cancel each other out. In a structural drawing, a 'squigly' bond from a chiral center indicates a mixture of both R and S configurations.
Chiral molecules are often labeled according to whether they are dextrorotatory or levorotatory as well as by their R/S designation. For example, the pure enantiomers of ibuprofen are labeled (S)-(+)-ibuprofen and (R)-(-)-ibuprofen, while (±)-ibuprofen refers to the racemic mixture, which is the form in which the drug is sold to consumers.
Exercise 3.14: The specific rotation of (R)-limonene is +11.5o in ethanol. What is the expected observed rotation of a sample of 6.00 g (S)-limonene dissolved in ethanol to a total volume of 80.0 mL in a 1.00 dm (10.0 cm) pathlength cuvette?
Exercise 3.15: The specific rotation of (S)-carvone is +61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a mixture of R and S carvone is measured at -23°. Which enantiomer is in excess in the mixture?
Solutions to exercises
All of the twenty natural amino acids except glycine have a chiral center at their alpha-carbon (recall that basic amino acid structure and terminology was introduced in section 1.3). Virtually all of the amino acids found in nature, both in the form of free amino acids or incorporated into peptides and proteins, have what is referred to in the biochemical literature as the 'L' configuration:
The 'L' indicates that these amino acid stereoisomers are levorotatory. All but one of the 19 L-amino acids have S stereochemistry at the a-carbon, using the rules of the R/S naming system.
D-amino acids (the D stands for dextrorotatory) are very rare in nature, but we will learn about an interesting example of a peptide containing one D-amino acid residue later in chapter 12.
Exercise 3.16: Which of the 20 common L-amino acids found in nature has the R configuration? Refer to the amino acid table for structures.
Solutions to exercises
Khan Academy video tutorial on optical acitivity
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
6.12: Designing a Synthesis I- An Introduction to Multistep Synthesis
Wikipedia Link: Energy Development | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/05%3A_Stereochemistry-_The_Arrangement_of_Atoms_in_Space_The_Stereochemistry_of_Addition_Reactions/5.21%3A__Enantiomers_Can_Be_Distinguished_by_Biological_Molecules.txt |
Kekulé was the first to suggest a sensible structure for benzene. The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.
This diagram is often simplified by leaving out all the carbon and hydrogen atoms! In diagrams of this sort, there is a carbon atom at each corner. You have to count the bonds leaving each carbon to work out how many hydrogens there are attached to it. In this case, each carbon has three bonds leaving it. Because carbon atoms form four bonds, that means you are a bond missing - and that must be attached to a hydrogen atom.
Problems with the Kekulé structure
Although the Kekulé structure was a good attempt in its time, there are serious problems with it regarding chemistry, structure and stability.
The Kekulé structure has problems with the chemistry. Because of the three double bonds, you might expect benzene to have reactions like ethene - only more so! Ethene undergoes addition reactions in which one of the two bonds joining the carbon atoms breaks, and the electrons are used to bond with additional atoms. Benzene rarely does this. Instead, it usually undergoes substitution reactions in which one of the hydrogen atoms is replaced by something new.
The Kekulé structure has problems with the shape. Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths.
• C-C (0.154 nm)
• C=C (0.134 nm)
That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.
The Kekulé structure has problems with the stability of benzene. Real benzene is a lot more stable than the Kekulé structure would give it credit for. Every time you do a thermochemistry calculation based on the Kekulé structure, you get an answer which is wrong by about 150 kJ mol-1. This is most easily shown using enthalpy changes of hydrogenation. Hydrogenation is the addition of hydrogen to something. If, for example, you hydrogenate ethene you get ethane:
$\ce{CH_2=CH_2 + H_2 \rightarrow CH_3CH_3} \tag{1}$
In order to do a fair comparison with benzene (a ring structure) we're going to compare it with cyclohexene. Cyclohexene, C6H10, is a ring of six carbon atoms containing just one C=C.
When hydrogen is added to this, cyclohexane, C6H12, is formed. The "CH" groups become CH2 and the double bond is replaced by a single one. The structures of cyclohexene and cyclohexane are usually simplified in the same way that the Kekulé structure for benzene is simplified - by leaving out all the carbons and hydrogens.
In the cyclohexane case, for example, there is a carbon atom at each corner, and enough hydrogens to make the total bonds on each carbon atom up to four. In this case, then, each corner represents CH2. The hydrogenation equation could be written:
The enthalpy change during this reaction is -120 kJ mol-1. In other words, when 1 mole of cyclohexene reacts, 120 kJ of heat energy is evolved. Where does this heat energy come from? When the reaction happens, bonds are broken (C=C and H-H) and this costs energy. Other bonds have to be made, and this releases energy. Because the bonds made are stronger than those broken, more energy is released than was used to break the original bonds and so there is a net evolution of heat energy.
If the ring had two double bonds in it initially (cyclohexa-1,3-diene), exactly twice as many bonds would have to be broken and exactly twice as many made. In other words, you would expect the enthalpy change of hydrogenation of cyclohexa-1,3-diene to be exactly twice that of cyclohexene - that is, -240 kJ mol-1.
In fact, the enthalpy change is -232 kJ mol-1 - which isn't far off what we are predicting. Applying the same argument to the Kekulé structure for benzene (what might be called cyclohexa-1,3,5-triene), you would expect an enthalpy change of -360 kJ mol-1, because there are exactly three times as many bonds being broken and made as in the cyclohexene case.
In fact what you get is -208 kJ mol-1 - not even within distance of the predicted value! This is very much easier to see on an enthalpy diagram. Notice that in each case heat energy is released, and in each case the product is the same (cyclohexane). That means that all the reactions "fall down" to the same end point.
Heavy lines, solid arrows and bold numbers represent real changes. Predicted changes are shown by dotted lines and italics. The most important point to notice is that real benzene is much lower down the diagram than the Kekulé form predicts. The lower down a substance is, the more energetically stable it is. This means that real benzene is about 150 kJ mol-1 more stable than the Kekulé structure gives it credit for. This increase in stability of benzene is known as the delocalization energy or resonance energy of benzene.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.01%3A_Delocalized_Electrons_Explain_Benzenes_Structur.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.02%3A_The_Bonding_in_Benzene.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.03%3A_Resonance_Contributors_and_the_Resonance_Hybrid.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.04%3A_How_to_Draw_Resonance_Contributors.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.05%3A_The_Predicted_Stabilities_of_Resonance_Contribu.txt |
Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.06%3A_Delocalized_Energy_Is_the_Additional_Stability_.txt |
In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, i.e. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the SN1 vs. SN2 character of a nucleophilic substitution reaction.
8.4A: Steric effects on electrophilicity
Consider two hypothetical SN2 reactions: one in which the electrophile is a methyl carbon and another in which it is tertiary carbon.
Because the three substituents on the methyl carbon electrophile are tiny hydrogens, the nucleophile has a relatively clear path for backside attack. However, backside attack on the tertiary carbon is blocked by the bulkier methyl groups. Once again, steric hindrance - this time caused by bulky groups attached to the electrophile rather than to the nucleophile - hinders the progress of an associative nucleophilic (SN2) displacement.
The factors discussed in the above paragraph, however, do not prevent a sterically-hindered carbon from being a good electrophile - they only make it less likely to be attacked in a concerted SN2 reaction. Nucleophilic substitution reactions in which the electrophilic carbon is sterically hindered are more likely to occur by a two-step, dissociative (SN1) mechanism. This makes perfect sense from a geometric point of view: the limitations imposed by sterics are significant mainly in an SN2 displacement, when the electrophile being attacked is a sp3-hybridized tetrahedral carbon with its relatively ‘tight’ angles of 109.4o. Remember that in an SN1 mechanism, the nucleophile attacks an sp2-hybridized carbocation intermediate, which has trigonal planar geometry with ‘open’ 120 angles.
With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.
8.4B: Stability of carbocation intermediates
We know that the rate-limiting step of an SN1 reaction is the first step - formation of the this carbocation intermediate. The rate of this step – and therefore, the rate of the overall substitution reaction – depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms. According to Hammond’s postulate (section 6.2B), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (SN1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.
The critical question now becomes, what stabilizes a carbocation?
Think back to Chapter 7, when we were learning how to evaluate the strength of an acid. The critical question was “how stable is the conjugate base that results when this acid donates its proton"? In many cases, this conjugate base was an anion – a center of excess electron density. Anything that can draw some of this electron density away– in other words, any electron withdrawing group – will stabilize the anion.
So if it takes an electron withdrawing group to stabilize a negative charge, what will stabilize a positive charge? An electron donating group!
A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be destabilized by an electron withdrawing group.
Alkyl groups – methyl, ethyl, and the like – are weak electron donating groups, and thus stabilize nearby carbocations. What this means is that, in general, more substituted carbocations are more stable: a tert-butyl carbocation, for example, is more stable than an isopropyl carbocation. Primary carbocations are highly unstable and not often observed as reaction intermediates; methyl carbocations are even less stable.
Alkyl groups are electron donating and carbocation-stabilizing because the electrons around the neighboring carbons are drawn towards the nearby positive charge, thus slightly reducing the electron poverty of the positively-charged carbon.
It is not accurate to say, however, that carbocations with higher substitution are always more stable than those with less substitution. Just as electron-donating groups can stabilize a carbocation, electron-withdrawing groups act to destabilize carbocations. Carbonyl groups are electron-withdrawing by inductive effects, due to the polarity of the C=O double bond. It is possible to demonstrate in the laboratory (see section 16.1D) that carbocation A below is more stable than carbocation B, even though A is a primary carbocation and B is secondary.
The difference in stability can be explained by considering the electron-withdrawing inductive effect of the ester carbonyl. Recall that inductive effects - whether electron-withdrawing or donating - are relayed through covalent bonds and that the strength of the effect decreases rapidly as the number of intermediary bonds increases. In other words, the effect decreases with distance. In species B the positive charge is closer to the carbonyl group, thus the destabilizing electron-withdrawing effect is stronger than it is in species A.
In the next chapter we will see how the carbocation-destabilizing effect of electron-withdrawing fluorine substituents can be used in experiments designed to address the question of whether a biochemical nucleophilic substitution reaction is SN1 or SN2.
Stabilization of a carbocation can also occur through resonance effects, and as we have already discussed in the acid-base chapter, resonance effects as a rule are more powerful than inductive effects. Consider the simple case of a benzylic carbocation:
This carbocation is comparatively stable. In this case, electron donation is a resonance effect. Three additional resonance structures can be drawn for this carbocation in which the positive charge is located on one of three aromatic carbons. The positive charge is not isolated on the benzylic carbon, rather it is delocalized around the aromatic structure: this delocalization of charge results in significant stabilization. As a result, benzylic and allylic carbocations (where the positively charged carbon is conjugated to one or more non-aromatic double bonds) are significantly more stable than even tertiary alkyl carbocations.
Because heteroatoms such as oxygen and nitrogen are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations. In fact, the opposite is often true: if the oxygen or nitrogen atom is in the correct position, the overall effect is carbocation stabilization. This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful. (We previously encountered this same idea when considering the relative acidity and basicity of phenols and aromatic amines in section 7.4). Consider the two pairs of carbocation species below:
In the more stable carbocations, the heteroatom acts as an electron donating group by resonance: in effect, the lone pair on the heteroatom is available to delocalize the positive charge. In the less stable carbocations the positively-charged carbon is more than one bond away from the heteroatom, and thus no resonance effects are possible. In fact, in these carbocation species the heteroatoms actually destabilize the positive charge, because they are electron withdrawing by induction.
Finally, vinylic carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any reaction.
Example 8.10
Exercise 8.10: In which of the structures below is the carbocation expected to be more stable? Explain.
For the most part, carbocations are very high-energy, transient intermediate species in organic reactions. However, there are some unusual examples of very stable carbocations that take the form of organic salts. Crystal violet is the common name for the chloride salt of the carbocation whose structure is shown below. Notice the structural possibilities for extensive resonance delocalization of the positive charge, and the presence of three electron-donating amine groups.
Example 8.11
Draw a resonance structure of the crystal violet cation in which the positive charge is delocalized to one of the nitrogen atoms.
Solution
When considering the possibility that a nucleophilic substitution reaction proceeds via an SN1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate. If this intermediate is not sufficiently stable, an SN1 mechanism must be considered unlikely, and the reaction probably proceeds by an SN2 mechanism. In the next chapter we will see several examples of biologically important SN1 reactions in which the positively charged intermediate is stabilized by inductive and resonance effects inherent in its own molecular structure.
Example 8.12
State which carbocation in each pair below is more stable, or if they are expected to be approximately equal. Explain your reasoning.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
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Resonance effects involving aromatic structures can have a dramatic influence on acidity and basicity. Notice, for example, the difference in acidity between phenol and cyclohexanol.
Looking at the conjugate base of phenol, we see that the negative charge can be delocalized by resonance to three different carbons on the aromatic ring.
Although these are all minor resonance contributors (negative charge is placed on a carbon rather than the more electronegative oxygen), they nonetheless have a significant effect on the acidity of the phenolic proton. Essentially, the benzene ring is acting as an electron-withdrawing group by resonance.
As we begin to study in detail the mechanisms of biological organic reactions, we’ll see that the phenol side chain of the amino acid tyrosine (see table 5 at the back of the book), with its relatively acidic pKaof 9-10, often acts as a catalytic proton donor/acceptor in enzyme active sites.
Exercise 7.4.1
Draw the conjugate base of 2-napthol (the major resonance contributor), and on your drawing indicate with arrows all of the atoms to which the negative charge can be delocalized by resonance.
Solution
The base-stabilizing effect of an aromatic ring can be accentuated by the presence of an additional electron-withdrawing substituent, such as a carbonyl. For the conjugate base of the phenol derivative below, an additional resonance contributor can be drawn in which the negative formal charge is placed on the carbonyl oxygen.
Now the negative charge on the conjugate base can be spread out over two oxygens (in addition to three aromatic carbons). The phenol acid therefore has a pKa similar to that of a carboxylic acid, where the negative charge on the conjugate base is also delocalized to two oxygen atoms. The ketone group is acting as an electron withdrawing group - it is 'pulling' electron density towards itself, through both inductive and resonance effects.
Exercise 7.4.2
The position of the electron-withdrawing substituent relative to the phenol hydroxyl is very important in terms of its effect on acidity. Which of the two substituted phenols below is more acidic? Use resonance drawings to explain your answer.
Solution
Exercise 7.4.3
Rank the four compounds below from most acidic to least.
Solution
Exercise 7.4.4
Nitro groups are very powerful electron-withdrawing groups. The phenol derivative picric acid (2,4,6 -trinitrophenol) has a pKa of 0.25, lower than that of trifluoroacetic acid. Use a resonance argument to explain why picric acid has such a low pKa.
Solution
Consider the acidity of 4-methoxyphenol, compared to phenol:
Notice that the methoxy group increases the pKa of the phenol group - it makes it less acidic. Why is this? At first inspection, you might assume that the methoxy substituent, with its electronegative oxygen, would be an electron-withdrawing group by induction. That is correct, but only to a point. The oxygen atom does indeed exert an electron-withdrawing inductive effect, but the lone pairs on the oxygen cause the exact opposite effect – the methoxy group is an electron-donating group by resonance. A resonance contributor can be drawn in which a formal negative charge is placed on the carbon adjacent to the negatively-charged phenolate oxygen.
Because of like-charge repulsion, this destabilizes the negative charge on the phenolate oxygen, making it more basic. It may help to visualize the methoxy group ‘pushing’ electrons towards the lone pair electrons of the phenolate oxygen, causing them to be less 'comfortable' and more reactive. The example above is a somewhat confusing but quite common situation in organic chemistry - a functional group, in this case a methoxy group, is exerting both an inductive effect and a resonance effect, but in opposite directions (the inductive effect is electron-withdrawing, the resonance effect is electron-donating). As a general rule a resonance effect is more powerful than an inductive effect - so overall, the methoxy group is acting as an electron donating group. A good rule of thumb to remember:
When resonance and induction compete, resonance usually wins!
Exercise 7.4.5
Rank the three compounds below from lowest pKa to highest, and explain your reasoning. Hint - think about both resonance and inductive effects!
Solution
Next section⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.09%3A_How_Delocalized_Electrons_Affect_pKa_Values.txt |
We end this chapter with a discussion of a type of reaction that is different from anything we have seen before. In the Diels-Alder cycloaddition reaction, a conjugated diene reacts with an alkene to form a ring structure.
In a Diels-Alder reaction, the alkene reacting partner is referred to as the dienophile. Essentially, this process involves overlap of the 2p orbitals on carbons 1 and 4 of the diene with 2p orbitals on the two sp2-hybridized carbons of the dienophile. Both of these new overlaps end up forming new sigma bonds, and a new pi bond is formed between carbon 2 and 3 of the diene.
One of the most important things to understand about this process is that it is concerted – all of the electron rearrangement takes place at once, with no carbocation intermediates.
The Diels-Alder reaction is enormously useful for synthetic organic chemists, not only because ring-forming reactions are useful in general but also because in many cases two new stereocenters are formed, and the reaction is inherently stereospecific. A cis dienophile will generate a ring with cis substitution, while a trans dienophile will generate a ring with trans substitution:
In order for a Diels-Alder reaction to occur, the diene molecule must adopt what is called the s-cis conformation:
The s-cis conformation is higher in energy than the s-trans conformation, due to steric hindrance. For some dienes, extreme steric hindrance causes the s-cis conformation to be highly strained, and for this reason such dienes do not readily undergo Diels-Alder reactions.
Cyclic dienes, on the other hand, are ‘locked’ in the s-cis conformation, and are especially reactive. The result of a Diels-Alder reaction involving a cyclic diene is a bicyclic structure:
Here, we see another element of stereopecificity: Diels-Alder reactions with cyclic dienes favor the formation of bicyclic structures in which substituents are in the endo position.
The endo position on a bicyclic structure refers to the position that is inside the concave shape of the larger (six-membered) ring. As you might predict, the exo position refers to the outside position.
The rate at which a Diels-Alder reaction takes place depends on electronic as well as steric factors. A particularly rapid Diels-Alder reaction takes place between cyclopentadiene and maleic anhydride.
We already know that cyclopentadiene is a good diene because of its inherent s-cis conformation. Maleic anhydride is also a very good dienophile, because the electron-withdrawing effect of the carbonyl groups causes the two alkene carbons to be electron-poor, and thus a good target for attack by the pi electrons in the diene.
In general, Diels-Alder reactions proceed fastest with electron-donating groups on the diene (eg. alkyl groups) and electron-withdrawing groups on the dienophile.
Alkynes can also serve as dienophiles in Diels-Alder reactions:
Below are just three examples of Diels-Alder reactions that have been reported in recent years:
The Diels-Alder reaction is just one example of a pericyclic reaction: this is a general term that refers to concerted rearrangements that proceed though cyclic transition states. Two well-studied intramolecular pericyclic reactions are known as the Cope rearrangement . . .
. . .and the Claisen rearrangement (when an oxygen is involved):
Notice that the both of these reactions require compounds in which two double bonds are separated by three single bonds.
Pericyclic reactions are rare in biological chemistry, but here is one example: the Claisen rearrangement catalyzed by chorismate mutase in the aromatic amino acid biosynthetic pathway.
The study of pericyclic reactions is an area of physical organic chemistry that blossomed in the mid-1960s, due mainly to the work of R.B. Woodward, Roald Hoffman, and Kenichi Fukui. The Woodward-Hoffman rules for pericyclic reactions (and a simplified version introduced by Fukui) use molecular orbital theory to explain why some pericyclic processes take place and others do not. A full discussion is beyond the scope of this text, but if you go on to study organic chemistry at the advanced undergraduate or graduate level you are sure to be introduced to this fascinating area of inquiry.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/07%3A_Delocalized_Electrons_and_Their_Effect_on_Stability_Reactivity_and_pKa_(More_About_Molecular_Orbital_Theory)/7.12%3A_The_Diels-Adler_Reaction_Is_a_14-Addition_React.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
Influence of the solvent in an SN2 reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction. SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
The SN1 mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
Exercise
Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
Influence of the solvent in an SN1 reaction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2, but it does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.01%3A_The_Mechanism_For_an_%28S_N2%29_Reaction.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
Influence of the solvent in an SN2 reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction. SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
The SN1 mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
Exercise
Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
Influence of the solvent in an SN1 reaction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2, but it does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.02%3A_Factors_That_Affect_%28S_N2%29_Reactions.txt |
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor in the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Introduction
When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge.
Dipole Moment
When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, an electric dipole is established. The size of a dipole is measured by its dipole moment ($\mu$). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals $3.34 \times 10^{-30}\; C\, m$). The dipole moment of a molecule can be calculated by Equation $\ref{1}$:
$\vec{\mu} = \sum_i q_i \, \vec{r}_i \label{1}$
where
• $\vec{\mu}$ is the dipole moment vector
• $q_i$ is the magnitude of the $i^{th}$ charge, and
• $\vec{r}_i$ is the vector representing the position of $i^{th}$ charge.
The dipole moment acts in the direction of the vector quantity. An example of a polar molecule is $\ce{H_2O}$. Because of the lone pair on oxygen, the structure of $\ce{H_2O}$ is bent (via VSEPR theory), which means that the vectors representing the dipole moment of each bond do not cancel each other out. Hence, water is polar.
The vector points from positive to negative, on both the molecular (net) dipole moment and the individual bond dipoles. Table A2 shows the electronegativity of some of the common elements. The larger the difference in electronegativity between the two atoms, the more electronegative that bond is. To be considered a polar bond, the difference in electronegativity must be large. The dipole moment points in the direction of the vector quantity of each of the bond electronegativities added together.
It is relatively easy to measure dipole moments: just place a substance between charged plates (Figure $2$); polar molecules increase the charge stored on the plates, and the dipole moment can be obtained (i.e., via the capacitance of the system). Nonpolar $\ce{CCl_4}$ is not deflected; moderately polar acetone deflects slightly; highly polar water deflects strongly. In general, polar molecules will align themselves: (1) in an electric field, (2) with respect to one another, or (3) with respect to ions (Figure $2$).
Equation $\ref{1}$ can be simplified for a simple separated two-charge system like diatomic molecules or when considering a bond dipole within a molecule
$\mu_{diatomic} = Q \times r \label{1a}$
This bond dipole is interpreted as the dipole from a charge separation over a distance $r$ between the partial charges $Q^+$ and $Q^-$ (or the more commonly used terms $δ^+$ - $δ^-$); the orientation of the dipole is along the axis of the bond. Consider a simple system of a single electron and proton separated by a fixed distance. When the proton and electron are close together, the dipole moment (degree of polarity) decreases. However, as the proton and electron get farther apart, the dipole moment increases. In this case, the dipole moment is calculated as (via Equation $\ref{1a}$):
\begin{align*} \mu &= Qr \nonumber \[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align*}
The Debye characterizes the size of the dipole moment. When a proton and electron are 100 pm apart, the dipole moment is $4.80\; D$:
\begin{align*} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \[4pt] &= 4.80\; D \label{3} \end{align*}
$4.80\; D$ is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. If the charge separation is increased then the dipole moment increases (linearly):
• If the proton and electron are separated by 120 pm: $\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \label{4a}$
• If the proton and electron are separated by 150 pm: $\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \label{4b}$
• If the proton and electron are separated by 200 pm: $\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \label{4c}$
Example $1$: Water
The water molecule in Figure $1$ can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference in electronegativity is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).
The bond moment of the O-H bond =1.5 D, so the net dipole moment is
$\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber$
Polarity and Structure of Molecules
The shape of a molecule and the polarity of its bonds determine the OVERALL POLARITY of that molecule. A molecule that contains polar bonds might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar). If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide (Figure $\PageIndex{3a}$). This is a linear molecule and each C=O bond is, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygen atoms a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. the vector addition of the dipoles equals zero) and the overall molecule has a zero dipole moment ($\mu=0$).
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
For $AB_n$ molecules, where $A$ is the central atom and $B$ are all the same types of atoms, there are certain molecular geometries which are symmetric. Therefore, they will have no dipole even if the bonds are polar. These geometries include linear, trigonal planar, tetrahedral, octahedral and trigonal bipyramid.
Example $3$: $\ce{C_2Cl_4}$
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and $\ce{Cl_2C=CCl_2}$ does not have a net dipole moment.
Example $3$: $\ce{CH_3Cl}$
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
\begin{align*} \mu &= \dfrac{178}{100}(4.80\; D) \nonumber \[4pt] &= 8.54\; D \nonumber \end{align*}
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
Table $1$: Relationship between Bond length, Electronegativity and Dipole moments in simple Diatomics
Compound Bond Length (Å) Electronegativity Difference Dipole Moment (D)
HF 0.92 1.9 1.82
HCl 1.27 0.9 1.08
HBr 1.41 0.7 0.82
HI 1.61 0.4 0.44 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.03%3A_The_Reversibility_of_an_%28S_N2%29_Reaction_Depends_on_the_Basicities_of_the_Leaving_Groups_in_the_Forward_and_Rever.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
Influence of the solvent in an SN2 reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction. SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
The SN1 mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
Exercise
Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
Influence of the solvent in an SN1 reaction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2, but it does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.04%3A_The_Mechanism_for_an_%28S_N1%29_Reaction.txt |
The SN2 mechanism
There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent).
This is called an 'SN2' mechanism. In the term SN2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products.
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way.
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane.
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the SN2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile.
Exercise
Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
Influence of the solvent in an SN2 reaction
The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of protic solvents (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction. SN2 reactions are faster in polar, aprotic solvents: those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory:
These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. By using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
The SN1 mechanism
A second model for a nucleophilic substitution reaction is called the 'dissociative', or 'SN1' mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches:
This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp2 hybrid orbitals is an empty, unhybridized p orbital.
In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry.
We saw that SN2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of SN1 reactions? In the model SN1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp2-hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an SN1 reaction with water as the incoming nucleophile.
Exercise
Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above.
The SN1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the SN1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an SN1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states.
Exercise
Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions.
Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an SN1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
Exercise
Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is SN2, and reaction B is SN1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
Influence of the solvent in an SN1 reaction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2, but it does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step. Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction.The SN1 reaction of allyl bromide in methanol is an example of what we would call methanolysis, while if water is the solvent the reaction would be called hydrolysis:
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
Exercise
Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above.
One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon::
Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3-hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module).
Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.05%3A_Factors_That_Affect_%28S_N1%29_Reactions.txt |
Frontside vs. Backside Attacks
A biomolecular nucleophilic substitution (SN2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group. It is possible for the nucleophile to attack the electrophilic center in two ways.
• Frontside Attack: In a frontside attack, the nucleophile attacks the electrophilic center on the same side as the leaving group. When a frontside attack occurs, the stereochemistry of the product remains the same; that is, we have retention of configuration.
• Backside Attack: In a backside attack, the nucleophile attacks the electrophilic center on the side that is opposite to the leaving group. When a backside attack occurs, the stereochemistry of the product does not stay the same. There is inversion of configuration.
The following diagram illustrates these two types of nucleophilic attacks, where the frontside attack results in retention of configuration; that is, the product has the same configuration as the substrate. The backside attack results in inversion of configuration, where the product's configuration is opposite that of the substrate.
Experimental Observation: All SN2 Reactions Proceed With Nucleophilic Backside Attacks
Experimental observation shows that all SN2 reactions proceed with inversion of configuration; that is, the nucleophile will always attack from the backside in all SN2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic center, and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism.
SN2 Reactions Are Stereospecific
The SN2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an R enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the S enantiomer.
Conversely, if the substrate is an S enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the R enantiomer.
In conclusion, SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
Next section: SN2 Reactions-The Leaving Group
8.07: Benzylic Halides Allylic Halides Vinylic Halides and Aryl Halides
The functional group of alkyl halides is a carbon-halogen bond, the common halogens being fluorine, chlorine, bromine and iodine. With the exception of iodine, these halogens have electronegativities significantly greater than carbon. Consequently, this functional group is polarized so that the carbon is electrophilic and the halogen is nucleophilic, as shown in the drawing below.
Two characteristics other than electronegativity also have an important influence on the chemical behavior of these compounds. The first of these is covalent bond strength. The strongest of the carbon-halogen covalent bonds is that to fluorine. Remarkably, this is the strongest common single bond to carbon, being roughly 30 kcal/mole stronger than a carbon-carbon bond and about 15 kcal/mole stronger than a carbon-hydrogen bond. Because of this, alkyl fluorides and fluorocarbons in general are chemically and thermodynamically quite stable, and do not share any of the reactivity patterns shown by the other alkyl halides. The carbon-chlorine covalent bond is slightly weaker than a carbon-carbon bond, and the bonds to the other halogens are weaker still, the bond to iodine being about 33% weaker. The second factor to be considered is the relative stability of the corresponding halide anions, which is likely the form in which these electronegative atoms will be replaced. This stability may be estimated from the relative acidities of the H-X acids, assuming that the strongest acid releases the most stable conjugate base (halide anion). With the exception of HF (pKa = 3.2), all the hydrohalic acids are very strong, small differences being in the direction HCl < HBr < HI.
The characteristics noted above lead us to anticipate certain types of reactions that are likely to occur with alkyl halides. The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red.
Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Ionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH → CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.06%3A_More_About_the_Stereochemistry_of_%28S_N2%29_and_%28S_N1%29_Reactions.txt |
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
8.09: The Role of the Solvent in (S N2) and (S N1) Reactions
Nucleophilicity
Recall the definitions of electrophile and nucleophile:
Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in forming a covalent bond to an electrophile (or Lewis acid).
If we use a common alkyl halide, such as methyl bromide, and a common solvent, ethanol, we can examine the rate at which various nucleophiles substitute the methyl carbon. Nucleophilicity is thereby related to the relative rate of substitution reactions at the halogen-bearing carbon atom of the reference alkyl halide. The most reactive nucleophiles are said to be more nucleophilic than less reactive members of the group. The nucleophilicities of some common Nu:(–) reactants vary as shown in the following
Nucleophilicity: CH3CO2(–) < Cl(–) < Br(–) < N3(–) < CH3O(–) < CN(–) ≈ SCN(–) < I(–) < CH3S(–)
The reactivity range encompassed by these reagents is over 5,000 fold, thiolate being the most reactive. Note that by using methyl bromide as the reference substrate, the complication of competing elimination reactions is avoided. The nucleophiles used in this study were all anions, but this is not a necessary requirement for these substitution reactions. Indeed reactions 6 & 7, presented at the beginning of this section, are examples of neutral nucleophiles participating in substitution reactions. The cumulative results of studies of this kind has led to useful empirical rules pertaining to nucleophilicity:
1. (i) For a given element, negatively charged species are more nucleophilic (and basic) than are equivalent neutral species.
2. (ii) For a given period of the periodic table, nucleophilicity (and basicity) decreases on moving from left to right.
3. (iii) For a given group of the periodic table, nucleophilicity increases from top to bottom (i.e. with increasing size), although there is a solvent dependence due to hydrogen bonding. Basicity varies in the opposite manner.
Solvent Effects
Solvation of nucleophilic anions markedly influences their reactivity. The nucleophilicities cited above were obtained from reactions in methanol solution. Polar, protic solvents such as water and alcohols solvate anions by hydrogen bonding interactions, as shown in the diagram below.
These solvated species are more stable and less reactive than the unsolvated "naked" anions. Polar, aprotic solvents such as DMSO (dimethyl sulfoxide), DMF (dimethylformamide) and acetonitrile do not solvate anions nearly as well as methanol, but provide good solvation of the accompanying cations. Consequently, most of the nucleophiles discussed here react more rapidly in solutions prepared from these solvents. These solvent effects are more pronounced for small basic anions than for large weakly basic anions. Thus, for reaction in DMSO solution we observe the following reactivity order:
Nucleophilicity: I(–) < SCN(–) < Br(–) < Cl(–) ≈ N3(–) < CH3CO2 (–) < CN(–) ≈ CH3S(–) < CH3O(–)
Note that this order is roughly the order of increasing basicity. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.08%3A_Competition_Between_%28S_N2%29_and_%28S_N1%29_Reactions.txt |
Lewis Structures are visual representations of the bonds between atoms and illustrate the lone pairs of electrons in molecules. They can also be called Lewis dot diagrams and are used as a simple way to show the configuration of atoms within a molecule. Between 1916 and 1919, Gilbert Newton Lewis, Walther Kossel, and Irving Langmuir came up with a theory to explain chemical bonding. This theory would be later called Lewis Theory and it is based on the following principles:
1. Valence electrons, or the electrons in the outermost electron shell, have an essential role in chemical bonding.
2. Ionic bonds are formed between atoms when electrons are transferred from one atom to another. Ionic bond is a bond between nonmetals and metals .
3. Covalent bonds are formed between atoms when pairs of electrons are shared between atoms. A covalent bond is between two nonmetals.
4. Electrons are transferred/shared so that each atom may reach a more stable electron configuration i.e. the noble gas configuration which contains 8 valence electrons. This is called octet rule.
Lewis Symbols and Lewis Structures
A Lewis Symbol for an element is composed of a chemical symbol surrounded by dots that are used to represent valence electrons. An example of a Lewis symbol is shown below with the element Carbon, which has the electron configuration of 1s22s22p2:
This Lewis symbol shows that carbon has four valence electrons in its outer orbital and these four electrons play a major role in bonding of carbon molecules.
Lewis symbols differ slightly for ions. When forming a Lewis symbol for an ion, the chemical symbol is surrounded by dots that are used to represent valence electrons, and the whole structure is placed in square brackets with superscript representing the charge of the ion. An example of a Lewis symbol for the cation and anion of Carbon is shown below:
Cation of Carbon
Anion of Carbon
Constructing Lewis Structures
To construct Lewis Structures one can generally abide by the following steps:
1. Find how many valence electrons (N) are in the molecule that needs to be shown on the Lewis Structure by using the periodic table. Find the charge, add an electron for every negative charge and remove an electron for every positive charge.
2. Draw out the single bonds and initial framework, called the skeleton, of the molecule.
3. Complete the octets around the non-central atoms i.e. the terminal atoms by using the lone-pairs of electrons.
4. Compare the number of electrons currently depicted to the number needed (N) in the central atom and add electrons to it if less the number is less than N.
5. If there are extra lone-pair electrons and the octet rule is not filled for the central atom, use the extra electrons to form double or triple bonds around the central atom.
6. Check the formal charge of each atom (Formal Charge explained below).
When constructing the structures keep in mind the following:
• The dots surrounding the chemical symbol are the valence electrons, and each dash represents one covalent bond (consisting of two valence electrons)
• Hydrogen is always terminal in the structure
• The atom with the lowest ionization energy is typically the central atom in the structure
• The octet rule means there are 8 valence electrons around the atoms, but for hydrogen the maximum is 2 electrons
Lewis Structures can differ based on whether the electrons are shared through ionic or covalent bonds.
Example 1: Ionic Bonding in NaCl
An example of ionic bonding can be seen below in the instance of the reaction of Sodium and Chlorine:
$2Na_{(s)} + 2Cl_{2 (g)} \rightarrow 2NaCl_{s}$
Sodium has one valence electron and Chlorine has seven valence electrons; the two elements together form the noble gas configuration. The Chlorine atom takes the valence electron from the Sodium atom leaving the Chlorine atom with one extra electron and thus negatively charged and the Sodium atom without an electron and thus positively charged. The two atoms then become ions and because of their opposite charges the ions are held together in an ionic bond.
An example of covalent bonding can be seen below with the reaction of Hydrogen and Fluorine:
+=
Hydrogen has one valence electron and Fluorine has seven valence electrons; together the elements form the noble gas configuration. The Hydrogen atom shares its electron with Fluorine atom so that the Hydrogen atom has 2 electrons and the Fluorine atom has 8 electrons. Therefore both atoms have their outermost shells completely filled.
Example 2: The Chlorate Ion
Try the Chlorate ion: (ClO3-)
SOLUTION
First, lets find the how many valence electrons chlorate has:
ClO3- : 7 e-(from Cl) + 3(6) e-(from 3 O atoms) + 1 (from the total charge of -1) = 26
There are 26 valence electrons.
Next lets draw the basic framework of the molecule:
The molecule uses covalent bonds to hold together the atoms to the central Chlorine. The remaining electrons become non-bonding electrons. Since 6 electrons were used for the bonds, the 20 others become those un-bonding electrons to complete the octet:
The oxygen atom's shells fill up with 18 electrons, and the other 2 complete Chlorine's octet.
Example 3: Formaldehyde
Constructing the Lewis Structure of the formaldehyde (H2CO) molecule.
SOLUTION
First find the valence electrons:
H2CO: 2(1) e- (from the H atoms) + 4 e- (from the C atom) + 6 e- (from the O atom)
There are 12 valence electrons. Next draw out the framework of the molecule:
To satisfy the octet of Carbon, one of the pairs of electrons on Oxygen must be moved to create a double bond with Carbon. Therefore our Lewis Structure would look as it does below:
The Hydrogen atoms are each filled up with their two electrons and both the Carbon and the Oxygen atoms' octets are filled.
Formal Charge
The charge on each atom in a molecule is called the formal charge. The formal charge can be calculated if the electrons in the bonds of the molecule are equally shared between atoms. This is not the same thing as the net charge of the ion.
In calculating formal charge, the following steps can be extremely helpful:
1. Determine the number of valence electrons that should be present for each atom in the structure.
2. Count the electrons around each atom in the structure (each lone pair = 2 electrons, each single bond =1 electron, each double bond = 2 electrons, each triple bond = 3 electrons).
3. Subtract the number of valence electrons that should be present (from step 1) from the electrons counted in step 2 for each atom. This is the formal charge for each atom.
4. Check that the formal charges add up to equal the overall charge of the molecule.
Formal charge = (number of valence electrons) - (number of non-bonding electrons + 1/2 number of bonding electrons)
In Lewis structures, the most favorable structure has the smallest formal charge for the atoms, and negative formal charges tend to come from more electronegative atoms.
An example of determining formal charge can be seen below with the nitrate ion, NO3-:
• The double bonded O atom has 6 electrons: 4 non-bonding and 2 bonding (one electron for each bond). Since O should have 6 electrons, the formal charge is 0.
• The two singly bonded O atoms each have 7 electrons: 6 non-bonding and 1 bonding electron. Since O should have 6 electrons, and there is one extra electron, those O atoms each have formal charges of -1.
• The N atom has 4 electrons: 4 bonding and 0 non-bonding electrons. Since N should have 5 electrons and there are only 4 electrons for this N, the N atom has a formal charge of +1.
• The charges add up to the overall charge of the ion. 0 + (-1) + (-1) + 1 = -1. Thus, these charges are correct, as the overall charge of nitrate is -1.
Resonance
There are times when more than one acceptable Lewis structure can be drawn for a molecule and no single structure can represent the molecule entirely. When this occurs the molecule/ion is said to have resonance. The combination of the various plausible Lewis structures is called a resonance hybrid.
Some rules for drawing resonance structures are as follows:
1. The same number of electrons must be present for all resonance structures.
2. The octet rule must be obeyed.
3. Nuclei (or the chemical symbol in the structure's representation) cannot be rearranged; only the valence electrons differ for resonance structures.
Example 4: Nitrate Ion
Consider the nitrate ion, NO3-
• Each structure differs based on the movement of two electrons.
• From structure 1 to structure 2 one of the lone pairs on the blue O moves to form a double bond with N. One of the electrons shared in the double bond between the red O and N then moves to be a lone pair on the red O because the N cannot have 10 electrons surrounding it.
• This is the same process that occurs from structure 2 to structure 3 except the changes occur in the green O and blue O.
Problems
1. Draw the Lewis structure for water.
2. Find the formal charge for each of the atoms in water.
3. Draw the Lewis structure for ammonia.
4. Find the formal charge for each of the atoms in ammonia.
Answers:
1.
2. All atoms in water have a formal charge of 0.
3.
4. The formal charge for each H is 0 and for N is 0. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.10%3A_Intermolecular_Versus_Intramolecular_Reactions.txt |
9.1A: SAM methyltransferases
Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Exercise 9.1: SAM is formed by a nucleophilic substitution reaction between methionine and adenosine triphosphate (ATP). Propose a mechanism for this reaction.
Solution
9.1B: Synthetic parallel – the Williamson ether synthesis
Synthetic organic chemists often use reactions in the laboratory that are conceptually very similar to the SAM-dependent methylation reactions described above. In what is referred to as "O-methylation", an alcohol is first deprotonated by a strong base, typically sodium hydride (this is essentially an irreversible deprotonation, as the hydrogen gas produced can easily be removed from the reaction vessel).
The alkoxide ion, a powerful nucleophile, is then allowed to attack the electrophilic carbon in iodomethane, displacing iodide in an SN2 reaction. Recall (section 7.3A) that iodide ion is the least basic - and thus the best leaving group - among the halogens commonly used in the synthetic lab. CH3I is one of the most commonly used lab reagents for methyl transfer reactions, and is the lab equivalent of SAM. In the synthesis of a modified analog of the signaling molecule myo-inisitol triphosphate, an alcohol group was methylated using sodium hydride and iodomethane (Carbohydrate Research 2004, 339, 51):
Methylation of amines (N-methylation) by iodomethane is also possible. A recent study concerning the optimization of peptide synthesis methods involved the following reaction (J. Org. Chem. 2000 65, 2309):
Template:ExampleStart
Exercise 9.2: Notice that in the N-methylation reaction a weak base (potassium carbonate) is used, rather than the very strong base (sodium hydride) that was used in the o-methylation reaction. Why is this weak base sufficient for N-methylation but not for o-methylation? In your answer, draw a step-by step mechanism for both reactions, and think carefully about pKa values and where in each mechanism the deprotonation step occurs.
Solution
The methylation of an alcohol by iodomethane is just one example of what is generally referred to as the Williamson ether synthesis, in which alcohols are first deprotonated by a strong base and then allowed to attack an alkyl halide electrophile. Below we see the condensation between ethanol and benzyl bromide:
It is important to note that this reaction does not work well at all if secondary or tertiary alkyl halides are used: remember that the alkoxide ion is a strong base as well as a nucleophile, and elimination will compete with nucleophilic substitution (Section 8.5D).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.11%3A_Biological_Methylating_Reagents_Have_Good_Leaving_Groups.txt |
E2 reactions are typically seen with secondary and tertiary alkyl halides, but a hindered base is necessary with a primary halide. The mechanism by which it occurs is a single step concerted reaction with one transition state. The rate at which this mechanism occurs is second order kinetics, and depends on both the base and alkyl halide. A good leaving group is required because it is involved in the rate determining step. The leaving groups must be coplanar in order to form a pi bond; carbons go from sp3 to sp2 hybridization states.
General Reaction
In this reaction Ba represents the base and X represents a leaving group, typically a halogen. There is one transition state that shows the concerted reaction for the base attracting the hydrogen and the halogen taking the electrons from the bond. The product be both eclipse and staggered depending on the transition states. Eclipsed products have a synperiplanar transition states, while staggered products have antiperiplanar transition states. Staggered conformation is usually the major product because of its lower energy confirmation.
An E2 reaction has certain requirements to proceed:
• Secondary and tertiary alkyl halides will proceed with E2 in the presence of a base (OH-, RO-, R2N-)
• Both leaving groups should be on the same plane, this allows the double bond to form in the reaction. In the reaction above you can see both leaving groups are in the plane of the carbons.
• Follows Zaitsev's rule, the most substituted alkene is usually the major product.
• Hoffman Rule, if a sterically hindered base will result in the least substituted product.
Reaction Coordinate
Problems
1.
2. What is the major product and why?
3. What is the major prodcut when 2-bromo-2-methylbutane reacts with with sodium ethoxide?
4.
5.
9.02: The E Reaction is Regioselective
NS13. Regiochemistry in Elimination
Sometimes, an elimination reaction could lead to formation of a double bond in more than one place. If the halide is on one carbon and there are protons that could be removed on either side, then taking one proton or the other might lead to two different products. This reaction could have different regiochemical outcomes, meaning it could happen at two different places in the molecule.
What factors might influence which product forms? We might think about product stability, in case there are corresponding differences in barriers leading to those products. We already know about stereochemical effects in alkene stability, but what about other effects?
It is well-established that alkene stability is influenced by degree of substitution of the double bond. The greater the number of carbons attached to the double bond, the more stable it is.
That effect is related to hyperconjugation. Specifically, it's an interaction between bonding orbitals and antibonding orbitals on neighbouring carbons. The interaction allows the bonding electrons to drop a little lower in energy through delocalization. The interaction also pushes the antibonding orbitals a little higher in energy, but since they have no electrons, they don't contribute to the real energy of the molecule. Overall, the molecules goes down in energy.
• In most cases, the most-substituted alkene results from elimination reactions.
However, alkene stability isn't the only factor that plays a role in elimination. Steric hindrance can play a role, too.
In a case in which there are two different hydrogens from which to select, the one leading to the more-substituted double bond is sometimes a little bit crowded. That leaves the base with fewer viable pathways to approach the proton.
On the other hand, the removal of a proton leading to the less stable alkene is often less crowded, allowing the base to approach much more easily from a number of angles.
• In some cases, especially with very bulky bases, the least substituted alkene forms, even though it is less stable. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/09%3A_Elimination_Reactions_of_Alkyl_Halides_(Competition_between_Substitution_and_Elimination)/9.01%3A_The_E_Reaction.txt |
Unimolecular Elimination (E1) is a reaction in which the removal of an HX substituent results in the formation of a double bond. It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate. Also, the only rate determining (slow) step is the dissociation of the leaving group to form a carbocation, hence the name unimolecular. Thus, since these two reactions behave similarly, they compete against each other. Many times, both these reactions will occur simultaneously to form different products from a single reaction. However, one can be favored over another through thermodynamic control. Although Elimination entails two types of reactions, E1 and E2, we will focus mainly on E1 reactions with some reference to E2.
General Reaction
An E1 reaction involves the deprotonation of a hydrogen nearby (usually one carbon away, or the beta position) the carbocation resulting in the formation of an alkene product. In order to accomplish this, a Lewis base is required. For a simplified model, we’ll take B to be a Lewis base, and LG to be a halogen leaving group.
As can be seen above, the preliminary step is the leaving group (LG) leaving on its own. Because it takes the electrons in the bond along with it, the carbon that was attached to it loses its electron, making it a carbocation. Once it becomes a carbocation, a Lewis Base (\(B^-\)) deprotonates the intermediate carbocation at the beta position, which then donates its electrons to the neighboring C-C bond, forming a double bond. Unlike E2 reactions, which require the proton to be anti to the leaving group, E1 reactions only require a neighboring hydrogen. This is due to the fact that the leaving group has already left the molecule. The final product is an alkene along with the HB byproduct.
Reactivity
Due to the fact that E1 reactions create a carbocation intermediate, rules present in \(S_N1\) reactions still apply.
As expected, tertiary carbocations are favored over secondary, primary and methyl’s. This is due to the phenomena of hyperconjugation, which essentially allows a nearby C-C or C-H bond to interact with the p orbital of the carbon to bring the electrons down to a lower energy state. Thus, this has a stabilizing effect on the molecule as a whole. In general, primary and methyl carbocations do not proceed through the E1 pathway for this reason, unless there is a means of carbocation rearrangement to move the positive charge to a nearby carbon. Secondary and Tertiary carbons form more stable carbocations, thus this formation occurs quite rapidly.
Secondary carbocations can be subject to the E2 reaction pathway, but this generally occurs in the presence of a good / strong base. Adding a weak base to the reaction disfavors E2, essentially pushing towards the E1 pathway. In many instances, solvolysis occurs rather than using a base to deprotonate. This means heat is added to the solution, and the solvent itself deprotonates a hydrogen. The medium can effect the pathway of the reaction as well. Polar protic solvents may be used to hinder nucleophiles, thus disfavoring E2 / Sn2 from occurring.
Acid catalyzed dehydration of secondary / tertiary alcohols
We’ll take a look at a mechanism involving solvolysis during an E1 reaction of Propanol in Sulfuric Acid.
• Step 1: The OH group on the pentanol is hydrated by H2SO4. This allows the OH to become an H2O, which is a better leaving group.
• Step 2: Once the OH has been hydrated, the H2O molecule leaves, taking its electrons with it. This creates a carbocation intermediate on the attached carbon.
• Step 3: Another H2O molecule comes in to deprotonate the beta carbon, which then donates its electrons to the neighboring C-C bond. The carbons are rehybridized from sp3 to sp2, and thus a pi bond is formed between them.
Mechanism for Alkyl Halides
This mechanism is a common application of E1 reactions in the synthesis of an alkene.
Once again, we see the basic 2 steps of the E1 mechanism.
1. The leaving group leaves along with its electrons to form a carbocation intermediate.
2. A base deprotonates a beta carbon to form a pi bond.
In this case we see a mixture of products rather than one discrete one. This is the case because the carbocation has two nearby carbons that are capable of being deprotonated, but that only one forms a major product (more stable).
How are Regiochemistry & Stereochemistry involved?
In terms of regiochemistry, Zaitsev's rule states that although more than one product can be formed during alkene synthesis, the more substituted alkene is the major product. This infers that the hydrogen on the most substituted carbon is the most probable to be deprotonated, thus allowing for the most substituted alkene to be formed.
Unlike E2 reactions, E1 is not stereospecific. Thus, a hydrogen is not required to be anti-periplanar to the leaving group.
In this mechanism, we can see two possible pathways for the reaction. One in which the methyl on the right is deprotonated, and another in which the CH2 on the left is deprotonated. Either one leads to a plausible resultant product, however, only one forms a major product. As stated by Zaitsev's rule, deprotonation of the most substituted carbon results in the most substituted alkene. This then becomes the most stable product due to hyperconjugation, and is also more common than the minor product.
Outside Sources
1. Vollhardt, K. Peter C., and Neil E. Schore. Organic Chemistry Structure and Function. New York: W. H. Freeman, 2007.
2. McMurry, J., Simanek, E. Fundamentals of Organic Chemistry, 6th edition. Cengage Learning, 2007.
Problems
1) Which of these steps is the rate determining step (A or B)?
What is the major product formed (C or D)?
2) In order to produce the most stable alkene product, from which carbon should the base deprotonate (A, B, or C)?
If the carbocation were to rearrange, on which carbon would the positive charge go onto without sacrificing stability (A, B, or C)?
3) Predict the major product of the following reaction.
4) (True or False) – There is no way of controlling the product ratio of E1 / Sn1 reactions.
5) Explain why the presence of a weak base / nucleophile favors E1 reactions over E2.
Answers
1. A , C
2. B, B
3.
4. False - They can be thermodynamically controlled to favor a certain product over another.
5. By definition, an E1 reaction is a Unimolecular Elimination reaction. This means the only rate determining step is that of the dissociation of the leaving group to form a carbocation. Since E2 is bimolecular and the nucleophilic attack is part of the rate determining step, a weak base/nucleophile disfavors it and ultimately allows E1 to dominate. (Don't forget about Sn1 which still pertains to this reaction simultaneously).
Contributors
• Satish Balasubramanian | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/09%3A_Elimination_Reactions_of_Alkyl_Halides_(Competition_between_Substitution_and_Elimination)/9.03%3A_The_E1_Reaction.txt |
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/09%3A_Elimination_Reactions_of_Alkyl_Halides_(Competition_between_Substitution_and_Elimination)/9.04%3A_Competition_between_E_and_E_Reactions.txt |
NS14. Stereochemistry in Elimination
Sometimes, elimination reactions may lead to multiple stereoisomers; that is, they could lead to either the cis or the trans isomer, or in more complicated structures, either the Z or the E isomer.
Of course, if there were some inherent stability difference between these isomers, that could be a factor that plays a role in influencing the outcome. Elimination reactions aren't generally reversible, so products are not directly determined by alkene isomer stabilities. Nevertheless, sometimes the barrier leading to a more stable product is a little lower than the barrier leading to a less stable product.
We do know that in simple cis vs. trans cases, the trans isomer is generally lower in energy because of fewer steric interactions between the substituents on the double bond. In the absence of other information, we could take that as a starting point. Let's see whether elimination reactions generally lead to trans isomers.
It turns out that sometimes this is true: eliminations often lead to the more stable product. Sometimes it isn't true, though. The answer depends on the mechanism.
• E1 eliminations generally lead to the more stable stereochemistry.
• E2 eliminations may or may not lead to the more stable stereochemistry.
Instead, in an E2 reaction, stereochemistry of the double bond -- that is, whether the E or Z isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. In other words, if the carbon with the hydrogen and the carbon with the halogen are both chiral, then one diastereomer will lead to one product, and the other diastereomer will lead to the other product.
The following reactions of potassium ethoxide with dibromostilbene (1,2-dibromo-1,2-diphenylethane) both occurred via an E2 mechanism. Two different diastereomers were used. Two different stereoisomers (E vs. Z) resulted.
Problem NS14.1.
Provide the stereochemical configurations of the following compounds from the above reactions:
1. 1,2-dibromo-1,2-diphenylethane (upper example)
2. 1-bromo-1,2-diphenylethene (upper example)
3. 1,2-dibromo-1,2-diphenylethane (lower example)
4. 1-bromo-1,2-diphenylethene (lower example)
In E2 eliminations, the spatial relationship between the proton and leaving group determines the product stereochemistry. That's because pi bond formation happens at the same time that the halide leaves and at the same time that the base removes the proton. All of these events have to be coordinated together. The central, tricky event is the pi bond formation. The leaving group can leave in any direction, and the base can approach from many directions, but unless the pi bond is ready to form, nothing else happens.
Let's slow the reaction down and imagine it takes place in slightly different stages.
As the leaving group leaves, it takes its electrons with it. It begins to leave a positive charge behind. That positive charge will be centered on the carbon from which the halide is departing. That carbocation, if it fully formed, would have only three neighbours to bond with. It would be trigonal planar. It would have an unoccupied, non-bonding p orbital.
As the base takes the proton, the hydrogen leaves behind the electrons from the C-H bond that held it in place. These electrons stay behind on the carbon atom. They are left in a non-bonding carbon valence orbital, a p orbital or something quite like it.
Now we have a filled p orbital next to an empty p orbital. They overlap to form a pi bond.
Of course, in an E2 reaction, things don't happen in stages. Everything happens at once. That means that, as the base removes the proton, the pi bond must already start forming. Because a pi bond requires parallel alignment of two p orbitals, and the p orbitals are forming from the C-H and C-LGp bonds, then those bonds must line up in order for the elimination to occur.
So let's look again at that dibromostilbene example.
In the first case, we need to spin the molecule so that we can see how the H on one carbon and the Br on the other are aligned and ready to eliminate via an E2 reaction. The substituents coming towards us in the reactant will still be coming towards us in the product. The substituents pointing away from us in the reactant will still be pointing away from us in the product.
So the relationships between the substituents on the nascent double bond are determined by their relationship once the reactant is aligned for the E2 reaction.
In the second case, we can spin the molecule but quickly realize the C-H and C-Br bonds are not lined up in this conformer. We need a bond rotation. Once we have made a conformational change, the C-H and C-Br bonds line up. It doesn't matter if this conformer is not favoured; if there is going to be any E2 reaction at all, this is the conformer it will have to go through.
Again, the relationships between on the new double bond are determined by their relationship once the reactant is aligned for the E2 reaction.
Problem NS14.2.
Sometimes it is easier to see the relationships between substituents by using a Newman projection. Draw Newman projections showing how the two isomers above proceed to different products in an E2 reaction.
Problem NS14.3.
Predict the product of each of the following E2 reactions. Note that the compounds differ in the incorporation of a 2H isotope (deuterium, or D) in place of a regular 1H isotope (protium, or H).
Problem NS14.4.
Conformational analysis of cyclohexanes requires the use of diamond lattice projections ("chairs"). Show why elimination can't proceed through the E2 mechanism in the following compound.
Problem NS14.5.
Predict the E2 elimination products from the following compounds.
In contrast to E2 reactions, E1 reactions do not occur in one step. That means there is time for reorganization in the intermediate. Once the leaving group leaves, the cation can sort itself into the most stable conformer. When the proton is taken, generally the most stable stereoisomer results because it comes from the most stable conformer of the cation. Any steric interactions in the alkene would also have occurred in the cation, so this interaction would have been sorted out at that point.
Thus, in the case of the dibromostilbenes examined before, E1 elimination would result in the same product in either case.
• E1 reactions can, in principle, lead to either stereochemistry of alkene.
• Free rotation around bonds in the carbocation intermediate allows the cation to adopt either conformer prior to elimination.
• However, steric interactions will lead to a preponderance of one conformer.
• The more stable conformer will lead to the more stable alkene. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/09%3A_Elimination_Reactions_of_Alkyl_Halides_(Competition_between_Substitution_and_Elimination)/9.05%3A_E_and_E_Reactions_are_Stereoselective.txt |
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9.08: Competition between Substitution and Elimination
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant.In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination.
The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both SN2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine).
The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents.
Nucleophile
Anionic Nucleophiles
( Weak Bases: I, Br, SCN, N3,
CH3CO2 , RS, CN etc. )
pKa's from -9 to 10 (left to right)
Anionic Nucleophiles
( Strong Bases: HO, RO )
pKa's > 15
Neutral Nucleophiles
( H2O, ROH, RSH, R3N )
pKa's ranging from -2 to 11
Alkyl Group
Primary
RCH2
Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g.
ClCH2CH2Cl + KOH ——> CH2=CHCl
SN2 substitution. (N ≈ S >>O)
Secondary
R2CH–
SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O)
In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly.
Tertiary
R3C–
E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed.
Allyl
H2C=CHCH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Benzyl
C6H5CH2
Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/09%3A_Elimination_Reactions_of_Alkyl_Halides_(Competition_between_Substitution_and_Elimination)/9.06%3A_Elimination_from_Substituted_Cyclohexanes.txt |
When alcohols react with a hydrogen halide, a substitution occurs, producing an alkyl halide and water:
Scope of Reaction
• The order of reactivity of alcohols is 3° > 2° > 1° methyl.
• The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is generally unreactive).
The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or NaI. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid:
Mechanisms of the Reactions of Alcohols with HX
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation in an \(S_N1\) reaction with the protonated alcohol acting as the substrate.
The \(S_N1\) mechanism is illustrated by the reaction tert-butyl alcohol and aqueous hydrochloric acid (\(H_3O^+\), \(Cl^-\) ). The first two steps in this \(S_n1\) substitution mechanism are protonation of the alcohol to form an oxonium ion. Although the oxonium ion is formed by protonation of the alcohol, it can also be viewed as a Lewis acid-base complex between the cation (\(R^+\)) and \(H_2O\). Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group (\)H_2O\_), which makes the dissociation step of the \(S_N1\) mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
When we convert an alcohol to an alkyl halide, we perform the reaction in the presence of acid and in the presence of halide ions and not at elevated temperatures. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and because halide ions are present in a high concentration, most of the carbocations react with an electron pair of a halide ion to form a more stable species, the alkyl halide product. The overall result is an \(S_n1\) reaction.
Primary Alcohols
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.
In these reactions, the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
The role of acid catalysis
Acid protonates the alcohol hydroxyl group, making it a good leaving group. However, other strong Lewis acids can be used instead of hydrohalic acids. Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride does not react with primary or secondary alcohols unless zinc chloride or a similar Lewis acid is added to the reaction mixture as well. Zinc chloride, a good Lewis acid, forms a complex with the alcohol through association with an unshared pair of electrons on the oxygen atom. This enhances the hydroxyl’s leaving group potential sufficiently so that chloride can displace it.
Rearrangement
As we might expect, many reactions of alcohols with hydrogen halides, particularly those in which carbocations are formed, are accompanied by rearrangements. The general rule is that if rearrangement CAN OCCUR (to form more stable or equally stable cations), it will! In these reactions, mixtures of products can be formed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.01%3A_Nucleophilic_Substitution_Reactions_of_Alcohols-_Forming_Alkyl_Halides.txt |
This page looks at reactions in which the -OH group in an alcohol is replaced by a halogen such as chlorine or bromine. It includes a simple test for an -OH group using phosphorus(V) chloride. The general reaction looks like this:
$ROH + HX \rightarrow RX + H_2O$
Reaction with hydrogen chloride
Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid, but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance. A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature. A tertiary halogenoalkane (haloalkane or alkyl halide) is formed.
Replacing -OH by bromine
Rather than using hydrobromic acid,the alcohol is typically treated with a mixture of sodium or potassium bromide and concentrated sulfuric acid. This produces hydrogen bromide, which reacts with the alcohol. The mixture is warmed to distil off the bromoalkane.
$CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O \label{1.1.2}$
Replacing -OH by iodine
In this case, the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H3PO4, and the iodoalkane is distilled off. The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide, which reacts with the alcohol.
$CH_3CH_2OH + HI \rightarrow CH_3CH_2I + H_2O \label{1.1.3}$
Phosphoric(V) acid is used instead of concentrated sulfuric acid because sulfuric acid oxidizes iodide ions to iodine and produces hardly any hydrogen iodide. A similar phenomenon occurs to some extent with bromide ions in the preparation of bromoalkanes but not enough to interfere with the main reaction. There is no reason why you could not use phosphoric(V) acid in the bromide case instead of sulfuric acid if desired.
Reacting Alcohols with Phosphorus Halides
Alcohols react with liquid phosphorus(III) chloride (also called phosphorus trichloride) to yield chloroalkanes.
$3CH_3CH_2CH_2OH + PCl_3 \rightarrow 3CH_3CH_2CH_2Cl + H_3PO_3 \label{1.1.3a}$
Alcohols also violently react with solid phosphorus(V) chloride (phosphorus pentachloride) at room temperature, producing clouds of hydrogen chloride gas. While it is not a good approach to make chloroalkanes, it is a good test for the presence of -OH groups. To show that a substance was an alcohol, you would first have to eliminate all the other groups that also react with phosphorus(V) chloride. For example, carboxylic acids (containing the -COOH group) also react with it (because of the -OH in -COOH) as does water (H-OH).
If you have a neutral liquid not contaminated with water, and clouds of hydrogen chloride are produced when you add phosphorus(V) chloride, then you have an alcohol group present.
$CH_3CH_2CH_2OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl + POCl_3 + HCl \label{1.1.4}$
There are also side reactions involving the $POCl_3$ reacting with the alcohol.
Other reactions involving phosphorus halides
Instead of using phosphorus(III) bromide or iodide, the alcohol is usually heated under reflux with a mixture of red phosphorus and either bromine or iodine. The phosphorus first reacts with the bromine or iodine to give the phosphorus(III) halide.
$2P_{(s)} + 3Br_2 \rightarrow 2PBr_3\label{1.1.5}$
$2P_{(s)} + 3I_2 \rightarrow 2PI_3 \label{1.1.6}$
These then react with the alcohol to give the corresponding halogenoalkane, which can be distilled off.
$3CH_3CH_2CH_2OH + PBr_3 \rightarrow 3CH_3CH_2CH_2Br + H_3PO_3 \label{1.1.7}$
$3CH_3CH_2CH_2OH + PI_3 \rightarrow 3CH_3CH_2CH_2I + H_3PO_3 \label{1.1.8}$
Reacting alcohols with Thionyl Chloride
Sulfur dichloride oxide (thionyl chloride) has the formula SOCl2. Traditionally, the formula is written as shown, despite the fact that the modern name writes the chlorine before the oxygen (alphabetical order). The sulfur dichloride oxide reacts with alcohols at room temperature to produce a chloroalkane. Sulfur dioxide and hydrogen chloride are given off. Care would have to be taken because both of these are poisonous.
$CH_3CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2Cl + SO_2 + HCl \label{1.1.9}$
The advantage that this reaction has over the use of either of the phosphorus chlorides is that the two other products of the reaction (sulfur dioxide and HCl) are both gases. That means that they separate themselves from the reaction mixture.
Contributors
If There is one thing you learn how to do well in Org 1, it’s make alcohols. Let’s count the ways: hydroboration, acid-catalyzed hydration, oxymercuration for starters, and then substitution of alkyl halides with water or $HO^-$. If you want to extend it even further, There is dihydroxylation (to make diols) using $OsO_4$ or cold $KMnO_4$, and even opening of epoxides under acidic or basic conditions to give alcohols.
There is just one issue here and it comes up once you try to use alcohols in synthesis. Let’s say you want to use that alcohol in a subsequent substitution step, getting rid of the $HO^-)\ and replacing it with something else. See any problems with that? Remember that good leaving groups are weak bases – and the hydroxide ion, being a strong base, tends to be a pretty bad leaving group. So what can we do? What you want to do is convert the alcohol into a better leaving group. One way is to convert the alcohol into a sulfonate ester – we talked about that with \( TsCl$ and $MsCl$. Alternatively, alcohols can be converted into alkyl chlorides with thionyl chloride ($SOCl_2$). This is a useful reaction, because the resulting alkyl halides are versatile compounds that can be converted into many compounds that are not directly accessible from the alcohol itself.
If you take an alcohol and add thionyl chloride, it will be converted into an alkyl chloride. The byproducts here are hydrochloric acid ($HCl$) and sulfur dioxide ($SO_2$). Note: there are significant differences in how this reaction is taught at different schools. Consult your instructor to be 100% sure that this applies to your course). See post here
Example $1$: Conversion of Alcohols to Alkyl Chlorides
There is one important thing to note here: see the stereochemistry? It’s been inverted.*(white lie alert – see below) That’s an important difference between $SOCl_2$ and TsCl, which leaves the stereochemistry alone. We’ll get to the root cause of that in a moment, but in the meantime, can you think of a mechanism which results in inversion of configuration at carbon?
As an extra bonus, thionyl chloride will also convert carboxylic acids into acid chlorides (“acyl chlorides”). Like alcohols, carboxylic acids have their limitations as reactants: the hydroxyl group interferes with many of the reactions we learn for nucleophilic acyl substitution (among others). Conversion of the OH into Cl solves this problem.
Mechanism
As you might have guessed, conversion of alcohols to alkyl halides proceeds through a substitution reaction – specifically, an $S_N2$ mechanism. The first step is attack of the oxygen upon the sulfur of $SOCl_2$, which results in displacement of chloride ion. This has the side benefit of converting the alcohol into a good leaving group: in the next step, chloride ion attacks the carbon in $S_N2$ fashion, resulting in cleavage of the C–O bond with inversion of configuration. The $HOSCl$ breaks down into $HCl$ and sulfur dioxide gas, which bubbles away.
Formation of Alkyl Chlorides
Since the reaction proceeds through a backside attack ($S_N2$), there is inversion of configuration at the carbon
The mechanism for formation of acid chlorides from carboxylic acids is similar. The conversion of caboxylic acids to acid chlorides is similar, but proceeds through a [1,2]-addition of chloride ion to the carbonyl carbon followed by [1,2]-elimination to give the acid chloride, $SO_2$ and $HCl$
In the laboratory
Like many sulfur-containing compounds, thionyl chloride is noseworthy for its pungent smell. Thionyl chloride has a nauseating sickly-sweet odor to it that imprints itself forever upon your memory. One accident that occurred during my time as a TA involved a student dropping a flask with 5 mL of thionyl chloride into a rotovap bath outside the fume hood. The cloud of $SO_2$ and $HCl$ that formed cleared the teaching lab for half an hour, so you can imagine what thionyl chloride would do if exposed to the moisture in your lungs. Treat with caution, just as you would if you were working with phosgene.
*Here’s the white lie. Although it’s often taught that $SOCl_2$ leads to 100% inversion of configuration, in reality it’s not always that simple. Inversion of configuration with $SOCl_2$ is very solvent dependent. Depending on the choice of solvent, one can get either straight inversion, or a mixture of retention and inversion. For the purposes of beginning organic classes, most students can ignore this message. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.02%3A_Other_Methods_Used_to_Convert_Alcohols_into_Alkyl_Halide.txt |
Nucleophilic Substitution of the Hydroxyl Group
The chemical behavior of alkyl halides can be used as a reference in discovering analogous substitution and elimination reactions of alcohols. The chief difference, of course, is a change in the leaving anion from halide to hydroxide. Because oxygen is slightly more electronegative than chlorine (3.5 vs. 2.8 on the Pauling scale), the C-O bond is expected to be more polar than a C-Cl bond. Furthermore, an independent measure of the electrophilic characteristics of carbon atoms from their NMR chemical shifts (both 13C and alpha protons) indicates that oxygen and chlorine substituents exert a similar electron-withdrawing influence when bonded to sp3 hybridized carbon atoms. Despite this promising background evidence, alcohols do not undergo the same SN2 reactions commonly observed with alkyl halides. For example, the rapid SN2 reaction of 1-bromobutane with sodium cyanide, shown below, has no parallel when 1-butanol is treated with sodium cyanide. In fact, ethyl alcohol is often used as a solvent for alkyl halide substitution reactions such as this.
CH3CH2CH2CH2–Br + Na(+) CN(–) CH3CH2CH2CH2–CN + Na(+) Br(–)
CH3CH2CH2CH2–OH + Na(+) CN(–) No Reaction
The key factor here is the stability of the leaving anion (bromide vs. hydroxide). HBr is a much stronger acid than water (by more than 18 orders of magnitude), and this difference is reflected in reactions that generate their respective conjugate bases. The weaker base, bromide, is more stable, and its release in a substitution or elimination reaction is much more favorable than that of hydroxide ion, a stronger and less stable base.
A clear step toward improving the reactivity of alcohols in SN2 reactions would be to modify the –OH functional group in a way that improves its stability as a leaving anion. One such modification is to conduct the substitution reaction in a strong acid, converting –OH to –OH2(+). Because the hydronium ion (H3O(+)) is a much stronger acid than water, its conjugate base (H2O) is a better leaving group than hydroxide ion. The only problem with this strategy is that many nucleophiles, including cyanide, are deactivated by protonation in strong acids, effectively removing the nucleophilic co-reactant required for the substitution. The strong acids HCl, HBr and HI are not subject to this difficulty because their conjugate bases are good nucleophiles and are even weaker bases than alcohols. The following equations illustrate some substitution reactions of alcohols that may be affected by these acids. As with alkyl halides, the nucleophilic substitution of 1º-alcohols proceeds by an SN2 mechanism, whereas 3º-alcohols react by an SN1 mechanism. Reactions of 2º-alcohols may occur by both mechanisms and often produce some rearranged products. The numbers in parentheses next to the mineral acid formulas represent the weight percentage of a concentrated aqueous solution, the form in which these acids are normally used.
CH3CH2CH2CH2–OH + HBr (48%) CH3CH2CH2CH2–OH2(+) Br(–) CH3CH2CH2CH2Br + H2O SN2
(CH3)3C–OH + HCl (37%) (CH3)3C–OH2(+) Cl(–) (CH3)3C(+) Cl(–) + H2O (CH3)3C–Cl + H2O SN1
Although these reactions are sometimes referred to as "acid-catalyzed," this is not strictly correct. In the overall transformation, a strong HX acid is converted to water, a very weak acid, so at least a stoichiometric quantity of HX is required for a complete conversion of alcohol to alkyl halide. The necessity of using equivalent quantities of very strong acids in this reaction limits its usefulness to simple alcohols of the type shown above. Alcohols with acid-sensitive groups do not, of course, tolerate such treatment. Nevertheless, the idea of modifying the -OH functional group to improve its stability as a leaving anion can be pursued in other directions. The following diagram shows some modifications that have proven effective. In each case the hydroxyl group is converted to an ester of a strong acid. The first two examples show the sulfonate esters described earlier. The third and fourth examples show the formation of a phosphite ester (X represents the remaining bromines or additional alcohol substituents) and a chlorosulfite ester, respectively. All of these leaving groups (colored blue) have conjugate acids that are much stronger than water (by 13 to 16 powers of ten); thus, the leaving anion is correspondingly more stable than the hydroxide ion. The mesylate and tosylate compounds are particularly useful because they may be used in substitution reactions with a wide variety of nucleophiles. The intermediates produced in reactions of alcohols with phosphorus tribromide and thionyl chloride (last two examples) are seldom isolated, and these reactions continue to produce alkyl bromide and chloride products.
The importance of sulfonate ester intermediates in general nucleophilic substitution reactions of alcohols may be illustrated by the following conversion of 1-butanol to pentanenitrile (butyl cyanide), a reaction that does not occur with the alcohol alone. The phosphorus and thionyl halides, on the other hand, only act to convert alcohols to the corresponding alkyl halides.
CH3CH2CH2CH2–OH + CH3SO2Cl pyridine
CH3CH2CH2CH2–OSO2CH3 Na(+) CN(–)
CH3CH2CH2CH2CN + CH3SO2O(–) Na(+)
Some examples of alcohol substitution reactions using this approach to activating the hydroxyl group are shown in the following diagram. The first two cases serve to reinforce the fact that sulfonate ester derivatives of alcohols may replace alkyl halides in a variety of SN2 reactions. The next two cases demonstrate the use of phosphorus tribromide in converting alcohols to bromides. This reagent may be used without added base (e.g. pyridine) because the phosphorous acid product is a weaker acid than HBr. Phosphorus tribromide is best used with 1º-alcohols because 2º-alcohols often yield rearrangement by-products resulting from competing SN1 reactions. Note that the ether oxygen in reaction 4 is not affected by this reagent, whereas the alternative synthesis using concentrated HBr cleaves ethers. Phosphorus trichloride (PCl3) converts alcohols to alkyl chlorides in a similar manner, but thionyl chloride is usually preferred for this transformation because the inorganic products are gases (SO2 & HCl). Phosphorus triiodide is not stable but may be generated in situ from a mixture of red phosphorus and iodine and acts to convert alcohols to alkyl iodides. The last example shows the reaction of thionyl chloride with a chiral 2º-alcohol. The presence of an organic base such as pyridine is important because it provides a substantial concentration of chloride ion required for the final SN2 reaction of the chlorosufite intermediate. In the absence of a base, chlorosufites decompose upon heating to yield the expected alkyl chloride with retention of configuration
Tertiary alcohols are not commonly used for substitution reactions of the type discussed here because SN1 and E1 reaction paths are dominant and are difficult to control.
The importance of sulfonate esters as intermediates in many substitution reactions cannot be overstated. A rigorous proof of the configurational inversion that occurs at the substitution site in SN2 reactions makes use of such reactions. An example of such a proof is displayed below. Abbreviations for the more commonly used sulfonyl derivatives are given in the following table.
Sulfonyl Group CH3SO2 CH3C6H4SO2 BrC6H4SO2 CF3SO2
Name & Abbrev. Mesyl or Ms Tosyl or Ts Brosyl or Bs Trifyl or Tf
Inversion Proof
For a more complete discussion of hydroxyl substitution reactions and a description of other selective methods for this transformation, Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.03%3A_Converting_an_Alcohol_to_a_Sulfonate_Ester.txt |
The discussion of alkyl halide reactions noted that 2º and 3º-alkyl halides experience rapid E2 elimination when treated with strong bases such as hydroxide and alkoxides. Alcohols do not undergo such base-induced elimination reactions and are, in fact, often used as solvents for such reactions. This is yet another example of how leaving-group stability influences the rate of a reaction. When an alcohol is treated with sodium hydroxide, the following acid-base equilibrium occurs. Most alcohols are slightly weaker acids than water, so the left side is favored.
R–O–H + Na(+) OH(–) R–O(–) Na(+) + H–OH
The elimination of water from an alcohol is called dehydration. Recalling that water is a much better leaving group than hydroxide ion, it is sensible to use acid-catalysis rather than base-catalysis in such reactions. Four examples of this useful technique are shown below. Note that hydrohalic acids (HX) are not normally used as catalysts because their conjugate bases are good nucleophiles and may create substitution products. The conjugate bases of sulfuric and phosphoric acids are not good nucleophiles, and do not participate in substitution under typical conditions.
The first two examples (in the top row) are typical, and the more facile elimination of the 3º-alcohol suggests the predominant E1 character for the reaction. This agrees with the tendency of branched 1º and 2º-alcohols to yield rearrangement products, as shown in the last example. The last two reactions also demonstrate that the Zaitsev rule applies to alcohol dehydrations, as well as to alkyl halide eliminations. Therefore, the more highly-substituted double bond isomer is favored among the products.
It should be noted that the acid-catalyzed dehydrations discussed here are the reverse of the acid-catalyzed hydration reactions of alkenes. Indeed, for these types of reversible reactions, the laws of thermodynamics require that the mechanism in both directions proceed by the same reaction path. This is known as the principle of microscopic reversibility. To illustrate, the following diagram lists the three steps in each transformation. The dehydration reaction is shown by the blue arrows; the hydration reaction by magenta arrows. The intermediates in these reactions are common to both, and common transition states are involved. This can be seen clearly in the energy diagrams depicted by clicking the button beneath the equations.
Base-induced E2 eliminations of alcohols may be achieved from sulfonate ester derivatives. This has the advantage of avoiding strong acids, which may cause molecular rearrangement or double bond migration in some cases. Because 3º-sulfonate derivatives are sometimes unstable, this procedure is best used with 1º and 2º-mesylates or tosylates. Application of this reaction sequence is shown here for 2-butanol. The Zaitsev rule favors the formation of 2-butene (cis + trans) over 1-butene.
CH3CH2CH(CH3)–OH CH3SO2Cl
CH3CH2CH(CH3)–OSO2CH3 C2H5O(–)Na(+)
CH3CH=CHCH3 + CH3CH2CH=CH2 + CH3SO2O(–) Na(+) + C2H5OH
The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl3) in pyridine. This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols, an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. Examples of these and related reactions are given in the following figure. The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of the base at that site. The second example shows two elimination procedures applied to the same 2º-alcohol. The first uses the single step POCl3 method, which works well in this case because SN2 substitution is impeded by steric hindrance. The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case, the anionic leaving group is the conjugate base of a strong acid.
Pyrolytic syn-Eliminations
Ester derivatives of alcohols may undergo unimolecular syn-elimination upon heating. To see examples of these, Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.04%3A_Elimination_Reactions_of_Alcohols-_Dehydration.txt |
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, then the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is as follows:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Primary alcohols
Primary alcohols can be oxidized to either aldehydes or carboxylic acids, depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidized to an aldehyde, which is then oxidized further to the acid.
An aldehyde is obtained if an excess amount of the alcohol is used, and the aldehyde is distilled off as soon as it forms. An excess of the alcohol means that there is not enough oxidizing agent present to carry out the second stage, and removing the aldehyde as soon as it is formed means that it is not present to be oxidized anyway!
If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, $CH_3CHO$. The full equation for this reaction is fairly complicated, and you need to understand the electron-half-equations in order to work it out.
$3CH_3CH_2OH + Cr_2O_7^{2-} + 8H^+ \rightarrow 3CH_3CHO + 2Cr^{3+} + 7H_2O$
In organic chemistry, simplified versions are often used that concentrate on what is happening to the organic substances. To do that, oxygen from an oxidizing agent is represented as $[O]$. That would produce the much simpler equation:
It also helps in remembering what happens. You can draw simple structures to show the relationship between the primary alcohol and the aldehyde formed.
Full oxidation to carboxylic acids
An excess of the oxidizing agent must be used, and the aldehyde formed as the half-way product should remain in the mixture. The alcohol is heated under reflux with an excess of the oxidizing agent. When the reaction is complete, the carboxylic acid is distilled off. The full equation for the oxidation of ethanol to ethanoic acid is as follows:
$3CH_3CH_2OH + 2Cr_2O_7^{2-} + 16H+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O$
The more typical simplified version looks like this:
$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O$
Alternatively, you could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.
$CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O$
$CH_3CHO + [O] \rightarrow CH_3COOH$
This is what is happening in the second stage:
Secondary alcohols
Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulfuric acid, propanone is formed. Changing the reaction conditions makes no difference to the product. Folloiwng is the simple version of the equation, showing the relationship between the structures:
If you look back at the second stage of the primary alcohol reaction, you will see that an oxygen inserted between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go.
Tertiary alcohols
Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. If you look at what is happening with primary and secondary alcohols, you will see that the oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom is attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.
You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.
Using these reactions as a test for the different types of alcohols
First, the presence of an alcohol must be confirmed by testing for the -OH group. The liquid would need to be verified as neutral, free of water and that it reacted with solid phosphorus(V) chloride to produce a burst of acidic steamy hydrogen chloride fumes. A few drops of the alcohol would be added to a test tube containing potassium dichromate(VI) solution acidified with dilute sulfuric acid. The tube would be warmed in a hot water bath.
Determining the tertiary alcohol
In the case of a primary or secondary alcohol, the orange solution turns green. With a tertiary alcohol, there is no color change. After heating, the following colors are observed:
Distinguishing between the primary and secondary alcohols
A sufficient amount of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) must be produced to be able to test them. There are various reactions that aldehydes undergo that ketones do not. These include the reactions with Tollens' reagent, Fehling's solution and Benedict's solution, and these reactions are covered on a separate page.
These tests can be difficult to carry out, and the results are not always as clear-cut as the books say. A much simpler but fairly reliable test is to use Schiff's reagent. Schiff's reagent is a fuchsin dye decolorized by passing sulfur dioxide through it. In the presence of even small amounts of an aldehyde, it turns bright magenta.
It must, however, be used absolutely cold, because ketones react with it very slowly to give the same color. If you heat it, obviously the change is faster - and potentially confusing. While you are warming the reaction mixture in the hot water bath, you can pass any vapors produced through some Schiff's reagent.
• If the Schiff's reagent quickly becomes magenta, then you are producing an aldehyde from a primary alcohol.
• If there is no color change in the Schiff's reagent, or only a trace of pink color within a minute or so, then you are not producing an aldehyde; therefore, no primary alcohol is present.
Because of the color change to the acidified potassium dichromate(VI) solution, you must, therefore, have a secondary alcohol. You should check the result as soon as the potassium dichromate(VI) solution turns green - if you leave it too long, the Schiff's reagent might start to change color in the secondary alcohol case as well. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.05%3A_Oxidation_of_Alcohols.txt |
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
10.07: Nucleophilic Substitution Reactions of Epoxides
8.6A: Epoxide structure
Epoxides (also known as oxiranes) are three-membered ring structures in which one of the vertices is an oxygen and the other two are carbons.
The carbons in an epoxide group are very reactive electrophiles, due in large part to the fact that substantial ring strain is relieved when the ring opens upon nucleophilic attack.
Epoxides are very important intermediates in laboratory organic synthesis, and are also found as intermediate products in some biosynthetic pathways. The compound (3S)-2,3-oxidosqualene, for example, is an important intermediate in the biosynthesis of cholesterol (we’ll see the epoxide ring-opening step in chapter 15):
Both in the laboratory and in the cell, epoxides are usually formed by the oxidation of an alkene. These reactions will be discussed in detail in chapter 16.
Exercise 8.17: Predict the major product(s) of the ring opening reaction that occurs when the epoxide shown below is treated with:
a) ethanol and a small amount of sodium hydroxide
b) ethanol and a small amount of sulfuric acid
Hint: be sure to consider both regiochemistry and stereochemistry!
Solution
Template:ExampleEnd
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
10.08: Amines Do Not Undergo Substitution or Elimination Reactions
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.06%3A_Nucleophilic_Substitution_Reactions_of_Ethers.txt |
Amine functions seldom serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Indeed, they are even less effective in this role than are hydroxyl and alkoxyl groups. In the case of alcohols and ethers, a useful technique for enhancing the reactivity of the oxygen function was to modify the leaving group (OH(–) or OR(–)) to improve its stability as an anion (or equivalent). This stability is conveniently estimated from the strength of the corresponding conjugate acids.
As noted earlier, 1º and 2º-amines are much weaker acids than alcohols, so it is not surprising that it is difficult to force the nitrogen function to assume the role of a nucleophilic leaving group. For example, heating an amine with HBr or HI does not normally convert it to the corresponding alkyl halide, as in the case of alcohols and ethers. In this context we note that the acidity of the putative ammonium leaving group is at least ten powers of ten less than that of an analogous oxonium species. The loss of nitrogen from diazonium intermediates is a notable exception in this comparison, due to the extreme stability of this leaving group (the conjugate acid of N2 would be an extraordinarily strong acid).
One group of amine derivatives that have proven useful in SN2 and E2 reactions is that composed of the tetraalkyl (4º-) ammonium salts. Most applications involving this class of compounds are eliminations, but a few examples of SN2 substitution have been reported.
C6H5–N(CH3)3(+) Br(–) + R-S(–) Na(+)
acetone & heat
R-S-CH3 + C6H5–N(CH3)2 + NaBr
(CH3)4N(+) OH(–)
heat
CH3–OH + (CH3)3N
Hofmann Elimination
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides.
In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond.
Example #3 illustrates two important features of the Hofmann elimination:
1. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is shown again in example #4.
2. When a given alkyl group has two different sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the major product is often the alkene isomer having the less substituted double bond.
The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule, and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers).
To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated below.
The E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule.
When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.
Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path.
10.10: Phase-Transfer Catalysts
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/10%3A_Reactions_of_Alcohols_Ethers_Epoxides_Amine_and_Sulfur-_Containing_Compounds/10.09%3A_Quaternary_Ammonium_Hydroxides_Undergo_Elimination_Reactions.txt |
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. These same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to a halide anion, and the carbon bonds to the metal which has characteristics similar to a carbanion (R:-).
Formation of Organometallic Reagents
Many organometallic reagents are commercially available, however, it is often necessary to make then. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination).
• An Alkyl Lithium Reagent
$\ce{R3C-X} + \ce{2Li} \rightarrow \ce{R3C-Li} + \ce{LiX}$
• A Grignard Regent
$\ce{R3C-X} + \ce{Mg} \rightarrow \ce{R3C-MgX}$
Halide reactivity in these reactions increases in the order: Cl < Br < I and Fluorides are usually not used. The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them and received the Nobel prize in 1912 for this work. The other metals mentioned above react in a similar manner, but Grignard and Alky Lithium Reagents most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.
A suitable solvent must be used. For alkyl lithium formation pentane or hexane are usually used. Diethyl ether can also be used but the subsequent alkyl lithium reagent must be used immediately after preparation due to an interaction with the solvent. Ethyl ether or THF are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent (As pictured below). This complex helps stabilize the organometallic and increases its ability to react.
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents excellent nucleophiles and useful reactants in synthesis.
Reaction of Organometallic Reagents with Various Carbonyls
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic attack of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions.
Addition to formaldehyde gives 1° alcohols
Addition to aldehydes gives 2° alcohols
Addition to ketones gives 3° alcohols
Addition to carbon dioxide (CO2) forms a carboxylic acid
Example $1$:
Going from Reactants to Products Simplified
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown; the mechanism for an organometallic reagent is the same.
1) Nucleophilic attack
2) Protonation
Organometallic Reagents as Bases
These reagents are very strong bases (pKa's of saturated hydrocarbons range from 42 to 50). Although not usually done with Grignard reagents, organolithium reagents can be used as strong bases. Both Grignard reagents and organolithium reagents react with water to form the corresponding hydrocarbon. This is why so much care is needed to insure dry glassware and solvents when working with organometallic reagents.
In fact, the reactivity of Grignard reagents and organolithium reagents can be exploited to create a new method for the conversion of halogens to the corresponding hydrocarbon (illustrated below). The halogen is converted to an organometallic reagent and then subsequently reacted with water to from an alkane.
Limitation of Organometallic Reagents
As discussed above, Grignard and organolithium reagents are powerful bases. Because of this they cannot be used as nucleophiles on compounds which contain acidic hydrogens. If they are used they will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and attack the carbonyl. A partial list of functional groups which cannot be used are: alcohols, amides, 1o amines, 2o amines, carboxylic acids, and terminal alkynes.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please give a detailed mechanism and the final product of this reaction
4) Please show two sets of reactants which could be used to synthesize the following molecule using a Grignard reaction.
Answers
1)
2)
3)
Nucleophilic attack
Protonation
4) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/11%3A_Organometallic_Compounds/11.01%3A_Organolithium_and_Organomagnesium_Compounds.txt |
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. These same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to a halide anion, and the carbon bonds to the metal which has characteristics similar to a carbanion (R:-).
Formation of Organometallic Reagents
Many organometallic reagents are commercially available, however, it is often necessary to make then. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination).
• An Alkyl Lithium Reagent
$\ce{R3C-X} + \ce{2Li} \rightarrow \ce{R3C-Li} + \ce{LiX}$
• A Grignard Regent
$\ce{R3C-X} + \ce{Mg} \rightarrow \ce{R3C-MgX}$
Halide reactivity in these reactions increases in the order: Cl < Br < I and Fluorides are usually not used. The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them and received the Nobel prize in 1912 for this work. The other metals mentioned above react in a similar manner, but Grignard and Alky Lithium Reagents most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.
A suitable solvent must be used. For alkyl lithium formation pentane or hexane are usually used. Diethyl ether can also be used but the subsequent alkyl lithium reagent must be used immediately after preparation due to an interaction with the solvent. Ethyl ether or THF are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent (As pictured below). This complex helps stabilize the organometallic and increases its ability to react.
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents excellent nucleophiles and useful reactants in synthesis.
Reaction of Organometallic Reagents with Various Carbonyls
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic attack of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions.
Addition to formaldehyde gives 1° alcohols
Addition to aldehydes gives 2° alcohols
Addition to ketones gives 3° alcohols
Addition to carbon dioxide (CO2) forms a carboxylic acid
Example $1$:
Going from Reactants to Products Simplified
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown; the mechanism for an organometallic reagent is the same.
1) Nucleophilic attack
2) Protonation
Organometallic Reagents as Bases
These reagents are very strong bases (pKa's of saturated hydrocarbons range from 42 to 50). Although not usually done with Grignard reagents, organolithium reagents can be used as strong bases. Both Grignard reagents and organolithium reagents react with water to form the corresponding hydrocarbon. This is why so much care is needed to insure dry glassware and solvents when working with organometallic reagents.
In fact, the reactivity of Grignard reagents and organolithium reagents can be exploited to create a new method for the conversion of halogens to the corresponding hydrocarbon (illustrated below). The halogen is converted to an organometallic reagent and then subsequently reacted with water to from an alkane.
Limitation of Organometallic Reagents
As discussed above, Grignard and organolithium reagents are powerful bases. Because of this they cannot be used as nucleophiles on compounds which contain acidic hydrogens. If they are used they will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and attack the carbonyl. A partial list of functional groups which cannot be used are: alcohols, amides, 1o amines, 2o amines, carboxylic acids, and terminal alkynes.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please give a detailed mechanism and the final product of this reaction
4) Please show two sets of reactants which could be used to synthesize the following molecule using a Grignard reaction.
Answers
1)
2)
3)
Nucleophilic attack
Protonation
4)
"} }} | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/11%3A_Organometallic_Compounds/11.02%3A_The_Reaction_Organolithium_Compounds_and_Grighard_Reagents_with_Electrophiles.txt |
Learning Objectives
• To understand how Temperature, Pressure, and the presence of other solutes affect the solubility of solutes in solvents.
Solubility is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. In such an equilibrium, Le Chatelier's principle can be used to explain most of the main factors that affect solubility. Le Châtelier's principle dictates that the effect of a stress upon a system in chemical equilibrium can be predicted in that the system tends to shift in such a way as to alleviate that stress.
Solute-Solvent Interactions Affect Solubility
The relation between the solute and solvent is very important in determining solubility. Strong solute-solvent attractions equate to greater solubility while weak solute-solvent attractions equate to lesser solubility. In turn, polar solutes tend to dissolve best in polar solvents while non-polar solutes tend to dissolve best in non-polar solvents. In the case of a polar solute and non-polar solvent (or vice versa), it tends to be insoluble or only soluble to a miniscule degree. A general rule to remember is, "Like dissolves like."
Common-Ion Effect
The common-ion effect is a term that describes the decrease in solubility of an ionic compound when a salt that contains an ion that already exists in the chemical equilibrium is added to the mixture. This effect best be explained by Le Chatelier's principle. Imagine if the slightly soluble ionic compound calcium sulfate, CaSO4, is added to water. The net ionic equation for the resulting chemical equilibrium is the following:
\[ CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
Calcium sulfate is slightly soluble; at equilibrium, most of the calcium and sulfate exists in the solid form of calcium sulfate.
Suppose the soluble ionic compound copper sulfate (CuSO4) were added to the solution. Copper sulfate is soluble; therefore, its only important effect on the net ionic equation is the addition of more sulfate (SO42-) ions.
\[ CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
The sulfate ions dissociated from copper sulfate are already present (common to) in the mixture from the slight dissociation of calcium sulfate. Thus, this addition of sulfate ions places stress on the previously established equilibrium. Le Chatelier's principle dictates that the additional stress on this product side of the equilibrium results in the shift of equilibrium towards the reactants side in order to alleviate this new stress. Because of the shift toward the reactant side, the solubility of the slightly soluble calcium sulfate is reduced even further.
Temperature Affects Solubility
Temperature changes affect the solubility of solids, liquids and gases differently. However, those effects are finitely determined only for solids and gases.
Solids
The effects of temperature on the solubility of solids differ depending on whether the reaction is endothermic or exothermic. Using Le Chatelier's principle, the effects of temperature in both scenarios can be determined.
1. First, consider an endothermic reaction (\(\Delta{H_{solvation}}>0\)): Increasing the temperature results in a stress on the reactants side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the product side in order to alleviate this stress. By shifting towards the product side, more of the solid is dissociated when equilibrium is again established, resulting in increased solubility.
2. Second, consider an exothermic reaction ((\(\Delta{H_{solvation}}<0\)): Increasing the temperature results in a stress on the products side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the reactant side in order to alleviate this stress. By shifting towards the reactant's side, less of the solid is dissociated when equilibrium is again established, resulting in decreased solubility.
Liquids
In the case of liquids, there is no defined trends for the effects of temperature on the solubility of liquids.
Gases
In understanding the effects of temperature on the solubility of gases, it is first important to remember that temperature is a measure of the average kinetic energy. As temperature increases, kinetic energy increases. The greater kinetic energy results in greater molecular motion of the gas particles. As a result, the gas particles dissolved in the liquid are more likely to escape to the gas phase and the existing gas particles are less likely to be dissolved. The converse is true as well. The trend is thus as follows: increased temperatures mean lesser solubility and decreased temperatures mean higher solubility.
Le Chatelier's principle allows better conceptualization of these trends. First, note that the process of dissolving gas in liquid is usually exothermic. As such, increasing temperatures result in stress on the product side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts towards the reactant side in order to alleviate this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase increases, resulting in lowered solubility.
Conversely, decreasing temperatures result in stress on the reactant side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts toward the product side in order to compensate for this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase would decrease, resulting in greater solubility.
Pressure Affects Solubility of Gases
The effects of pressure are only significant in affecting the solubility of gases in liquids.
• Solids & Liquids: The effects of pressure changes on the solubility of solids and liquids are negligible.
• Gases: The effects of pressure on the solubility of gases in liquids can best be described through a combination of Henry's law and Le Chatelier principle. Henry's law dictates that when temperature is constant, the solubility of the gas corresponds to it's partial pressure. Consider the following formula of Henry's law:
\[ p = k_h \; c \]
where:
• \(p\) is the partial pressure of the gas above the liquid,
• \(k_h\) is Henry's law constant, and
• \(c\) is the concentrate of the gas in the liquid.
This formula indicates that (at a constant temperature) when the partial pressure decreases, the concentration of gas in the liquid decreases as well, and consequently the solubility also decreases. Conversely, when the partial pressure increases in such a situation, the concentration of gas in the liquid will increase as well; the solubility also increases. Extending the implications from Henry's law, the usefulness of Le Chatelier's principle is enhanced in predicting the effects of pressure on the solubility of gases.
Consider a system consisting of a gas that is partially dissolved in liquid. An increase in pressure would result in greater partial pressure (because the gas is being further compressed). This increased partial pressure means that more gas particles will enter the liquid (there is therefore less gas above the liquid, so the partial pressure decreases) in order to alleviate the stress created by the increase in pressure, resulting in greater solubility.
The converse case in such a system is also true, as a decrease in pressure equates to more gas particles escaping the liquid to compensate.
Example 1
Consider the following exothermic reaction that is in equilibrium
\[ CO_2 (g) + H_2O (l) \rightleftharpoons H_2CO_3 (aq) \]
What will happen to the solubility of the carbon dioxide if:
1. Temperature is increased?
2. Pressure and temperature are increased?
3. Pressure is increased but temperature is decreased?
4. Pressure is increased?
Solution
1. The reaction is exothermic, so an increase in temperature means that solubility would decrease.
2. The change in solubility cannot be determined from the given information. Increasing pressure increased solubility, but increasing temperature decreases solubility
3. An increase in pressure and an increase in temperature in this reaction results in greater solubility.
4. An increase in pressure results in more gas particles entering the liquid in order to decrease the partial pressure. Therefore, the solubility would increase.
Example 2: The Common Ion Effect
Bob is in the business of purifying silver compounds to extract the actual silver. He is extremely frugal. One day, he finds a barrel containing a saturated solution of silver chloride. Bob has a bottle of water, a jar of table salt (NaCl(s)), and a bottle of vinegar (CH3COOH). Which of the three should Bob add to the solution to maximize the amount of solid silver chloride (minimizing the solubility of the silver chloride)?
Solution
Bob should add table salt to the solution. According to the common-ion effect, the additional Cl- ions would reduce the solubility of the silver chloride, which maximizes the amount of solid silver chloride.
Example 3:
Allison has always wanted to start her own carbonated drink company. Recently, she opened a factory to produce her drinks. She wants her drink to "out-fizz" all the competitors. That is, she wants to maximize the solubility of the gas in her drink. What conditions (high/low temperature, high/low pressure) would best allow her to achieve this goal?
Solution
She would be able to maximize the solubility of the gas, (\(CO_2\) in this case, in her drink (maximize fizz) when the pressure is high and temperature is low.
Example 4
Butters is trying to increase the solubility of a solid in some water. He begins to frantically stir the mixture. Should he continue stirring? Why or why not?
Solution
He stop stop stirring. Stirring only affects how fast the system will reach equilibrium and does not affect the solubility of the solid at all.
Example 5: Outgassing Soda
With respect to Henry's law, why is it a poor ideal to open a can of soda in a low pressure environment?
Solution
The fizziness of soda originates from dissolved \(CO_2\), partially in the form of carbonic acid. The concentration of \(CO_2\) dissolved in the soda depends on the amount of ambient pressure pressing down on the liquid. Hence, the soda can will be under pressure to maintain the desired \(CO_2\) concentration. When the can is opened to a lower pressure environment (e.g., the ambient atmosphere), the soda will quickly "outgas" (\(CO_2\) will come out of solution) at a rate depending on the surrounding atmospheric pressure. If a can of soda were opened under a lower pressure environment, this outgassing will be faster and hence more explosive (and dangerous) than under a high pressure environment.
Terms
• The solubility of a solute is the concentration of the saturated solution.
• A saturated solution a solution in which the maximum amount of solute has dissolved in the solvent at a given temperature.
• An unsaturated solution a solution in which the solute has completely dissolved in the solvent.
• A supersaturated solution is a solution in which the amount of solute dissolved under given conditions exceeds it's supposed upper limit.
• Le Châtelier's principle states that when a system in chemical equilibrium is stressed, the system will shift in a way that alleviates the stress.
• Endothermic reaction: a reaction in which heat is absorbed (ΔH>0)
• Exothermic reaction: a reaction in which heat is released (ΔH < 0) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/11%3A_Organometallic_Compounds/11.03%3A_Transmetallation.txt |
Learning Objectives
• To understand how Temperature, Pressure, and the presence of other solutes affect the solubility of solutes in solvents.
Solubility is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. In such an equilibrium, Le Chatelier's principle can be used to explain most of the main factors that affect solubility. Le Châtelier's principle dictates that the effect of a stress upon a system in chemical equilibrium can be predicted in that the system tends to shift in such a way as to alleviate that stress.
Solute-Solvent Interactions Affect Solubility
The relation between the solute and solvent is very important in determining solubility. Strong solute-solvent attractions equate to greater solubility while weak solute-solvent attractions equate to lesser solubility. In turn, polar solutes tend to dissolve best in polar solvents while non-polar solutes tend to dissolve best in non-polar solvents. In the case of a polar solute and non-polar solvent (or vice versa), it tends to be insoluble or only soluble to a miniscule degree. A general rule to remember is, "Like dissolves like."
Common-Ion Effect
The common-ion effect is a term that describes the decrease in solubility of an ionic compound when a salt that contains an ion that already exists in the chemical equilibrium is added to the mixture. This effect best be explained by Le Chatelier's principle. Imagine if the slightly soluble ionic compound calcium sulfate, CaSO4, is added to water. The net ionic equation for the resulting chemical equilibrium is the following:
\[ CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
Calcium sulfate is slightly soluble; at equilibrium, most of the calcium and sulfate exists in the solid form of calcium sulfate.
Suppose the soluble ionic compound copper sulfate (CuSO4) were added to the solution. Copper sulfate is soluble; therefore, its only important effect on the net ionic equation is the addition of more sulfate (SO42-) ions.
\[ CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
The sulfate ions dissociated from copper sulfate are already present (common to) in the mixture from the slight dissociation of calcium sulfate. Thus, this addition of sulfate ions places stress on the previously established equilibrium. Le Chatelier's principle dictates that the additional stress on this product side of the equilibrium results in the shift of equilibrium towards the reactants side in order to alleviate this new stress. Because of the shift toward the reactant side, the solubility of the slightly soluble calcium sulfate is reduced even further.
Temperature Affects Solubility
Temperature changes affect the solubility of solids, liquids and gases differently. However, those effects are finitely determined only for solids and gases.
Solids
The effects of temperature on the solubility of solids differ depending on whether the reaction is endothermic or exothermic. Using Le Chatelier's principle, the effects of temperature in both scenarios can be determined.
1. First, consider an endothermic reaction (\(\Delta{H_{solvation}}>0\)): Increasing the temperature results in a stress on the reactants side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the product side in order to alleviate this stress. By shifting towards the product side, more of the solid is dissociated when equilibrium is again established, resulting in increased solubility.
2. Second, consider an exothermic reaction ((\(\Delta{H_{solvation}}<0\)): Increasing the temperature results in a stress on the products side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the reactant side in order to alleviate this stress. By shifting towards the reactant's side, less of the solid is dissociated when equilibrium is again established, resulting in decreased solubility.
Liquids
In the case of liquids, there is no defined trends for the effects of temperature on the solubility of liquids.
Gases
In understanding the effects of temperature on the solubility of gases, it is first important to remember that temperature is a measure of the average kinetic energy. As temperature increases, kinetic energy increases. The greater kinetic energy results in greater molecular motion of the gas particles. As a result, the gas particles dissolved in the liquid are more likely to escape to the gas phase and the existing gas particles are less likely to be dissolved. The converse is true as well. The trend is thus as follows: increased temperatures mean lesser solubility and decreased temperatures mean higher solubility.
Le Chatelier's principle allows better conceptualization of these trends. First, note that the process of dissolving gas in liquid is usually exothermic. As such, increasing temperatures result in stress on the product side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts towards the reactant side in order to alleviate this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase increases, resulting in lowered solubility.
Conversely, decreasing temperatures result in stress on the reactant side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts toward the product side in order to compensate for this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase would decrease, resulting in greater solubility.
Pressure Affects Solubility of Gases
The effects of pressure are only significant in affecting the solubility of gases in liquids.
• Solids & Liquids: The effects of pressure changes on the solubility of solids and liquids are negligible.
• Gases: The effects of pressure on the solubility of gases in liquids can best be described through a combination of Henry's law and Le Chatelier principle. Henry's law dictates that when temperature is constant, the solubility of the gas corresponds to it's partial pressure. Consider the following formula of Henry's law:
\[ p = k_h \; c \]
where:
• \(p\) is the partial pressure of the gas above the liquid,
• \(k_h\) is Henry's law constant, and
• \(c\) is the concentrate of the gas in the liquid.
This formula indicates that (at a constant temperature) when the partial pressure decreases, the concentration of gas in the liquid decreases as well, and consequently the solubility also decreases. Conversely, when the partial pressure increases in such a situation, the concentration of gas in the liquid will increase as well; the solubility also increases. Extending the implications from Henry's law, the usefulness of Le Chatelier's principle is enhanced in predicting the effects of pressure on the solubility of gases.
Consider a system consisting of a gas that is partially dissolved in liquid. An increase in pressure would result in greater partial pressure (because the gas is being further compressed). This increased partial pressure means that more gas particles will enter the liquid (there is therefore less gas above the liquid, so the partial pressure decreases) in order to alleviate the stress created by the increase in pressure, resulting in greater solubility.
The converse case in such a system is also true, as a decrease in pressure equates to more gas particles escaping the liquid to compensate.
Example 1
Consider the following exothermic reaction that is in equilibrium
\[ CO_2 (g) + H_2O (l) \rightleftharpoons H_2CO_3 (aq) \]
What will happen to the solubility of the carbon dioxide if:
1. Temperature is increased?
2. Pressure and temperature are increased?
3. Pressure is increased but temperature is decreased?
4. Pressure is increased?
Solution
1. The reaction is exothermic, so an increase in temperature means that solubility would decrease.
2. The change in solubility cannot be determined from the given information. Increasing pressure increased solubility, but increasing temperature decreases solubility
3. An increase in pressure and an increase in temperature in this reaction results in greater solubility.
4. An increase in pressure results in more gas particles entering the liquid in order to decrease the partial pressure. Therefore, the solubility would increase.
Example 2: The Common Ion Effect
Bob is in the business of purifying silver compounds to extract the actual silver. He is extremely frugal. One day, he finds a barrel containing a saturated solution of silver chloride. Bob has a bottle of water, a jar of table salt (NaCl(s)), and a bottle of vinegar (CH3COOH). Which of the three should Bob add to the solution to maximize the amount of solid silver chloride (minimizing the solubility of the silver chloride)?
Solution
Bob should add table salt to the solution. According to the common-ion effect, the additional Cl- ions would reduce the solubility of the silver chloride, which maximizes the amount of solid silver chloride.
Example 3:
Allison has always wanted to start her own carbonated drink company. Recently, she opened a factory to produce her drinks. She wants her drink to "out-fizz" all the competitors. That is, she wants to maximize the solubility of the gas in her drink. What conditions (high/low temperature, high/low pressure) would best allow her to achieve this goal?
Solution
She would be able to maximize the solubility of the gas, (\(CO_2\) in this case, in her drink (maximize fizz) when the pressure is high and temperature is low.
Example 4
Butters is trying to increase the solubility of a solid in some water. He begins to frantically stir the mixture. Should he continue stirring? Why or why not?
Solution
He stop stop stirring. Stirring only affects how fast the system will reach equilibrium and does not affect the solubility of the solid at all.
Example 5: Outgassing Soda
With respect to Henry's law, why is it a poor ideal to open a can of soda in a low pressure environment?
Solution
The fizziness of soda originates from dissolved \(CO_2\), partially in the form of carbonic acid. The concentration of \(CO_2\) dissolved in the soda depends on the amount of ambient pressure pressing down on the liquid. Hence, the soda can will be under pressure to maintain the desired \(CO_2\) concentration. When the can is opened to a lower pressure environment (e.g., the ambient atmosphere), the soda will quickly "outgas" (\(CO_2\) will come out of solution) at a rate depending on the surrounding atmospheric pressure. If a can of soda were opened under a lower pressure environment, this outgassing will be faster and hence more explosive (and dangerous) than under a high pressure environment.
Terms
• The solubility of a solute is the concentration of the saturated solution.
• A saturated solution a solution in which the maximum amount of solute has dissolved in the solvent at a given temperature.
• An unsaturated solution a solution in which the solute has completely dissolved in the solvent.
• A supersaturated solution is a solution in which the amount of solute dissolved under given conditions exceeds it's supposed upper limit.
• Le Châtelier's principle states that when a system in chemical equilibrium is stressed, the system will shift in a way that alleviates the stress.
• Endothermic reaction: a reaction in which heat is absorbed (ΔH>0)
• Exothermic reaction: a reaction in which heat is released (ΔH < 0) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/11%3A_Organometallic_Compounds/11.04%3A_Coupling_Reactions.txt |
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.
\begin{align*}\Delta{T_f} &= T_f(solvent) - T_f (solution) \[4pt] &= K_f \times m \end{align*}
where $\Delta{T_f}$ is the freezing point depression, $T_f$ (solution) is the freezing point of the solution, $T_f$ (solvent) is the freezing point of the solvent, $K_f$ is the freezing point depression constant, and m is the molality.
Introduction
Nonelectrolytes are substances with no ions, only molecules. Strong electrolytes, on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also, remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution.
Adding solutes to an ideal solution results in a positive $ΔS$, an increase in entropy. Because of this, the newly altered solution's chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as colligative properties. These properties are dependent on the number of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.
The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas:
$\Delta{T}_f = -K_f \times m$
$\Delta{T}_b = K_b \times m$
where $m$ is the solute molality and $K$ values are proportionality constants; ($K_f$ and $K_b$ for freezing and boiling, respectively).
Molality
Molality is defined as the number of moles of solute per kilogram solvent. Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer.
If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for $K_f$ and $K_b$ respectively, are in Table $1$:
Table $1$: Ebullioscopic and cryoscopic constants for select solvents. Note that the nature of the solute does not affect colligative property relations.
Solvent $K_f$ $K_b$
Water 1.86 .512
Acetic acid 3.90 3.07
Benzene 5.12 2.53
Phenol 7.27 3.56
The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute to the vapor pressure of the solution, and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the chemical potential of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures.
The boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease in the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus, boiling point elevation occurs with a temperature increase that is quantified using
$\Delta{T_b} = K_b m$
where $K_b$ is known as the ebullioscopic constant and $m$ is the molality of the solute.
Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the pure solid solvent. Therefore, a freezing point depression is observed.
Example $1$
2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 $^{\circ}C$. What is the molar mass of the compound?
Solution
First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved.
\begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \[4pt] &= 0.123 m \end{align*}
\begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \[4pt] &= 0.00923 \; m \; solute \end{align*}
We can now find the molecular weight of the unknown compound:
\begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \[4pt] &= 216.80 \; g/mol \end{align*}
The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water.
Exercise $1$
Benzophenone has a freezing point of 49.00oC. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59oC. Find the freezing point depression constant for the solvent.
Answer
$9.80\,^oC/m$
Applications
Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, $\ce{NaCl}$ will be ineffective. Under these conditions, $\ce{CaCl_2}$ can be used since it dissolves to make three ions instead of two for $\ce{NaCl}$. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/11%3A_Organometallic_Compounds/11.05%3A_Palladium-Catalyzed_Coupling_Reactions.txt |
Olefin metathesis, or alkene metathesis, is an important process in petroleum refining and in the synthesis of important compounds such as pharmaceuticals. The mechanism of olefin metathesis is related to pericyclic reactions like Diels Alder and [2+2] reactions. In other words, it occurs through the concerted interaction of one molecule with another.
In petroleum refining, heating alkenes over metal oxide surfaces results in the formation of longer-chain alkenes. In particular, terminal olefins (with the double bond at the end of the chain) are converted into internal olefins (with the double bond somewhere in the middle of the chain).
Figure PR4.1. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins".
What does that reaction have to do with addition reactions involving double bonds?
Clearly, the alkenes have double bonds. In addition, so do the metal oxides. Metal atoms inside the metal oxides are bridged together by oxygen atoms. The surface of the metal oxides may be covered with a mixture of hydroxyl groups as well as terminal oxides (M=O groups). The terminal oxides on the surface are the important part of the catalyst.
Figure PR4.2. A simplified structure of a chunk of metal oxide surface.
When metal alkylidene complexes were developed in the 1970's, it was found that they, too, could catalyze this reaction.
Figure PR4.3. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins"
In fact, scientists working in petroleum chemistry soon came to believe that metal oxides on the catalyst surface were converted to alkylidenes, which then carried out the work of olefin metathesis. The reaction, it turns out, involves a [2+2] cycloaddition of an alkene to a metal alkylidene (to the metal-carbon double bond). This reaction results in a four-membered ring, called a metallacyclobutane. The [2+2] cycloaddition is quickly followed by the reverse reaction, a retro-[2+2]. The metallacyclobutane pops open to form two new double bonds.
The Chauvin Mechanism
This mechanism is called the Chauvin mechanism, after its first proponent, Yves Chauvin of the French Petroleum Institute. Chauvin's proposal of this mechanism shortly after the discovery of metal alkylidenes by Dick Schrock at DuPont earned him a Nobel Prize in 2005. Chauvin and Schrock shared the prize with Bob Grubbs, who made it possible for the reaction to be adapted easily to the synthesis of complex molecules such as pharmaceuticals.
Figure PR4.4. The Chauvin mechanism for olefin metathesis.
Why does olefin metathesis lead to the formation of internal alkenes? The [2+2] addition and retro-[2+2] reactions occur in equilibrium with each other. Each time the metallacyclobutane forms, it can form two different pairs of double bonds through the retro reaction. In the presence of terminal alkenes, one of those pairs of alkenes will eventually include ethene. Since ethene is a gas, it is easily lost from the system, and equilibrium shifts to the right in the equation below.
Figure PR4.5. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins".
That leaves a longer-chain alkylidene on the metal, ready to be attached to another long chain through subsequent [2+2] addition and reversion reactions.
Figure PR4.6. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins".
In most cases, a [2+2] addition will not work unless photochemistry is involved, but it does work with metal alkylidenes. The reason for this exception is thought to involve the nature of the metal-carbon double bond. In contrast to an orbital picture for an alkene, an orbital picture for an alkylidene pi bond suggests orbital symmetry that can easily interact with the LUMO on an alkene. That's because a metal-carbon \(pi\) bond likely involves a d orbital on the metal, and the d orbital has lobes alternating in phase like a \(pi\) antibonding orbital.
Figure PR4.7. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins". | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/11%3A_Organometallic_Compounds/11.06%3A_Alkene_Metathesis.txt |
The alkanes and cycloalkanes, with the exception of cyclopropane, are probably the least chemically reactive class of organic compounds. Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar; therefore, there are no portions of the molecules that carry any significant amount of positive or negative charge that can attract other molecules or ions. Alkanes can be burned, destroying the entire molecule (Alkane Heats of Combustion), alkanes can react with some of the halogens, breaking carbon-hydrogen bonds, and alkanes can crack by breaking the carbon-carbon bonds.
12.03: Radical Stability Depends on the Number of Alkyl Groups Attached to the Carbon with the Unpaired Electron
Solutions to exercises
17.1A: The geometry and relative stability of carbon radicals
As organic chemists, we are particularly interested in radical intermediates in which the unpaired electron resides on a carbon atom. Experimental evidence indicates that the three bonds in a carbon radical have trigonal planar geometry, and therefore the carbon is considered to be sp2-hybridized with the unpaired electron occupying the perpendicular, unhybridized 2pzorbital. Contrast this picture with carbocation and carbanion intermediates, which are both also trigonal planar but whose 2pz orbitals contain zero or two electrons, respectively.
The trend in the stability of carbon radicals parallels that of carbocations (section 8.4B): tertiary radicals, for example, are more stable than secondary radicals, followed by primary and methyl radicals. This should make intuitive sense, because radicals, like carbocations, can be considered to be electron deficient, and thus are stabilized by the electron-donating effects of nearby alkyl groups. Benzylic and allylic radicals are more stable than alkyl radicals due to resonance effects - an unpaired electron can be delocalized over a system of conjugated pi bonds. An allylic radical, for example, can be pictured as a system of three parallel 2pz orbitals sharing three electrons.
Template:ExampleStart
Exercise 17.1: Draw the structure of a benzylic radical compound, and then draw a resonance form showing how the radical is stabilized.
Template:ExampleEnd
With enough resonance stabilization, radicals can be made that are quite unreactive. One example of an inert organic radical structure is shown below.
In this molecule, the already extensive resonance stabilization is further enhanced by the ability of the chlorine atoms to shield the radical center from external reagents. The radical is, in some sense, inside a protective 'cage'.
Template:ExampleStart
Exercise 17.2: Draw a resonance contributor of the structure above in which the unpaired electron is located on a chlorine atom.
Template:ExampleEnd
17.1B: The diradical character of triplet oxygen
You may be surprised to learn that molecular oxygen (O2) often reacts like a radical species - or more accurately, like a diradical, with two separate unpaired valence electrons. This puzzling phenomenon is best explained by molecular orbital theory (you may want to go back to chapter 2 at this point to review basic MO theory).
In molecular orbital energy diagram form, the configuration of O2 looks like this:
When the molecular orbitals of O2 are filled up with electrons according to Hund's rule (section 1.1C) , the HOMOs (Highest Occupied Molecular Orbitals) are the two antibonding π*2p orbitals, each holding a single, unpaired electron. This electron configuration, which describes oxygen in its lowest energy (ground) state, is referred to as the triplet state - the oxygen in the air around you istriplet oxygen. A higher energy state, in which the two highest energy electrons are paired in the same π*2p orbital, is called the singlet state of oxygen.
It is the last two unpaired electrons in triplet oxygen that react in radical-like fashion. For this reason, triplet O2 is often depicted by a Lewis structure as a diradical, with a single covalent bond, four lone pairs, and two unpaired electrons. The excited singlet state is often depicted as a doubly-bonded molecule.
These Lewis-dot depictions, while only approximations and unsatisfactory in many respects, can nonetheless be helpful in illustrating how O2 reacts. In any event, the molecular orbital picture of O2 is not simply an academic exercise: the diradical model for triplet oxygen is key to understanding its reactivity in many enzymatic and non-enzymatic contexts. We will see one example of triplet oxygen reactivity in section 17.2D.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/12%3A_Radicals_(Reactions_of_Alkanes)/12.01%3A_Alkanes_Are_Unreactive_Compounds.txt |
The radical chain reactions that we have seen so far in this chapter have all been uncatalyzed. Many enzymatic reactions also occur with single-electron mechanistic steps and free radical intermediates. Enzyme-catalyzed radical processes, however, are of course highly specific chemical events.
17.3A: Hydroxylation of alkanes
Radical mechanisms are most relevant in transformations involving functional groups that are relatively unreactive, in particular isolated alkyl carbons. None of the two-electron mechanisms that we have studied so far, for example, could be invoked for the hydroxylation of an alkane, such as this reaction in the biosynthesis of the steroid hormone cortisol.
The methylene carbon center involved is simply not reactive enough - there is no acidic proton or electronegative leaving group present, and the carbon is not in any way electrophilic. This alkane hydroxylation, and many others like it, is catalyzed by a family of monooxygenase enzymes, called the 'cytochrome P450' family, that contain a heme redox center. The heme prosthetic group, which you will learn more about in biochemistry and bioinorganic chemistry courses, is characterized by an iron atom coordinated to the nitrogen atoms of four linked pyrrole rings. Two additional sites on the octahedral iron center - at the top and bottom axial positions- are available for coordination, and in the P450 enzymes these sites are occupied by a cysteine sulfur and, after several electron-transfer steps, a neutral oxygen radical atom derived from molecular oxygen.
This radical initiator abstracts a hydrogen from an alkane substrate, leading to an alkyl radical intermediate which subsequently recombines with the activated oxygen to form an alcohol and reduced heme (the roman numerals in the figure above refer to the oxidation state of the heme iron – in this process, it gains a single electron and thus is converted from the +4 to the +3 oxidation state).
17.3B: Reductive dehydroxylation of alcohols
In the fatty acid biosynthesis pathway, an alcohol is reduced to an alkane by a two-step dehydration - hydrogenation process (we learned about these reactions in section 16.5). This reductive process somewhat of a special case, however, because the alcohol being reduced is located at the beta position relative to a carbonyl group, and thus enolate intermediates are effectively stabilized.
When alcohols which are not located two carbons down from a carbonyl are reduced to alkanes, radical mechanisms are generally involved. One of the most important examples of this type is the reaction catalyzed by ribonucleotide reductase, in which the C2'-hydroxyl group present in RNA nucleotides is removed to form the 2'-deoxynucleotides that make up DNA.
There are several different types of ribonucleotide reductases which are found in different species, but they all appear to operate by the same essential organic mechanism. A radical chain reaction is initiated when an active site cysteine is converted to an unusual thiyl radical - the exact mechanism by which this occurs depends on the type of ribonucleotide reductase enzyme, but all mechanisms are thought to involve the participation of metal redox centers (see Ann. Rev. Biochem 1998, 67, 71 for a detailed review of the various enzyme types). In the interest of clarity, arrows depicting single-electron movement in the mechanism below are colored red, while two-electron arrows are in black.
In the propagation phase of the reaction, the thiyl radical first abstracts a hydrogen from the C3' position of the ribose (step 1), generating an organic radical intermediate. After four more single-electron/two-electron rearrangement steps, the hydroxyl group at C2' is replaced by a hydrogen, a pair of cysteines on the enzyme is oxidized to a disulfide, and the starting thiyl radical is regenerated. To complete the catalytic cycle, the disulfide is reduced to free cysteines by a specific thioredoxin reductase, or by glutathione/glutathione reductase (section 16.12A), depending on the species.
17.3C: Radical mechanisms for flavin-dependent reactions
In chapter 16 we saw how flavin coenzymes, like their nicotinamide adenine dinucleotide counterparts, can act as hydride acceptors and donors. In these redox reactions, two electrons are transferred together in the form of a hydride ion. Flavin, however, is also capable of mediating chemical steps in which a single unpaired electron is transferred - in other words, radical chemistry. This is due to the ability of the flavin system to form a stabilized radical intermediate called a semiquinone, formed when FADH2 (or FMNH2) donates a single electron, or when FAD (or FMN) accepts a single electron.
This single-electron transfer capability of flavins is critical to their metabolic role as the entry point of electrons into the electron transport phase of respiration. Electrons 'harvested' from the oxidation of fuel molecules are channeled, one by one, by FMNH2 into the electron transport chain, where they eventually reduce molecular oxygen. NADH is incapable of single electron transfer - all it can do is transfer two electrons, in the form of a hydride, to FMN; the regenerated FMNH2 is then able to continue sending single electrons into the transport chain.
You will learn more details about this process in a biochemistry class.
Because flavins are capable of single-electron as well as two-electron chemistry, the relevant mechanisms of flavoenzyme-catalyzed reactions are often more difficult to determine. Recall the dehydrogenation reaction catalyzed by acyl-CoA dehydrogenase (section 16.5C) - it involves the transfer of two electrons and two protons (ie. a hydrogen molecule) to FAD. Both electrons could be transferred together, with the FAD coenzyme simply acting as a hydride acceptor (this is the mechanism we considered previously). However, because the oxidizing coenzyme being used is FAD rather than NAD+, it is also possible that the reaction could proceed by a single-electron, radical intermediate process. In the alternate radical mechanism proposed below, for example, the enolate intermediate first donates a single electron to FAD, forming a radical semiquinone intermediate (step 2). The second electron is transferred when the semiquinone intermediate abstracts a hydrogen from Cb in a homolytic fashion (step 3).
Scientists are still not sure which mechanism - the hydride transfer mechanism that we saw in section 16.5B or the single electron transfer detailed above - more accurately depicts what is going on in this reaction.
The conjugated elimination catalyzed by chorismate synthase (section 14.3B) is another example of a reaction where the participation of flavin throws doubt on the question of what is the relevant mechanism. This could simply be a conjugated E1' reaction, with formation of an allylic carbocation intermediate. The question plaguing researchers studying this enzyme, however, is why FADH2 is required. This is not a redox reaction, and correspondingly, FADH2 is not used up in the course of the transformation - it just needs to be bound in the active site in order for the reaction to proceed. Given that flavins generally participate in single-electron chemistry, this is an indication that radical intermediates may be involved. Recently an alternative mechanism, involving a flavin semiquinone intermediate, has been proposed (J. Biol. Chem 2004, 279, 9451). Notice that a single electron is transferred from substrate to coenzyme in step 2, then transferred back in step 4.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/12%3A_Radicals_(Reactions_of_Alkanes)/12.11%3A_Radical_Reactions_Occur_in_Biological_Systems.txt |
Mass spectrometry (MS) is a powerful analytical technique widely used by chemists, biologists, medical researchers, and environmental and forensic scientists, among others. With MS, we are looking at the mass of a molecule, or of different fragments of that molecule.
The basics of a mass spectrometry experiment
There are many different types of MS instruments, but they all have the same three essential components. First, there is an ionization source, where the molecule is given a positive electrical charge, either by removing an electron or by adding a proton. Depending on the ionization method used, the ionized molecule may or may not break apart into a population of smaller fragments. In the figure below, some of the sample molecules remain whole, while others fragment into smaller pieces.
Next in line there is a mass analyzer, where the cationic fragments are separated according to their mass.
Finally, there is a detector, which detects and quantifies the separated ions.
One of the more common types of MS techniques used in the organic laboratory is electron ionization. In the ionization source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of knocking a valence electron off of the molecule to form a radical cation. Because a great deal of energy is transferred by this bombardment process, the radical cation quickly begins to break up into smaller fragments, some of which are positively charged and some of which are neutral. The neutral fragments are either adsorbed onto the walls of the chamber or are removed by a vacuum source. In the mass analyzer component, the positively charged fragments and any remaining unfragmented molecular ions are accelerated down a tube by an electric field.
This tube is curved, and the ions are deflected by a strong magnetic field. Ions of different mass to charge (m/z) ratios are deflected to a different extent, resulting in a ‘sorting’ of ions by mass (virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass). A detector at the end of the curved flight tube records and quantifies the sorted ions.
Looking at mass spectra
Below is typical output for an electron-ionization MS experiment (MS data in the section is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak is at m/z = 43 - we will discuss the formation of this fragment a bit later. The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H.
Molecules with lots of oxygen atoms sometimes show a small M+2 peak (2 m/z units greater than the parent peak) in their mass spectra, due to the presence of a small amount of 18O (the most abundant isotope of oxygen is 16O). Because there are two abundant isotopes of both chlorine (about 75% 35Cl and 25% 37Cl) and bromine (about 50% 79Br and 50% 81Br), chlorinated and brominated compounds have very large and recognizable M+2 peaks. Fragments containing both isotopes of Br can be seen in the mass spectrum of ethyl bromide:
Much of the utility in electron-ionization MS comes from the fact that the radical cations generated in the electron-bombardment process tend to fragment in predictable ways. Detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but it is worthwhile to see a few representative examples, even if we don’t attempt to understand the exact process by which the fragmentation occurs. We saw, for example, that the base peak in the mass spectrum of acetone is m/z = 43. This is the result of cleavage at the ‘alpha’ position - in other words, at the carbon-carbon bond adjacent to the carbonyl. Alpha cleavage results in the formation of an acylium ion (which accounts for the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the parent peak and the base peak, the next largest peak, at a relative abundance of 23%, is at m/z = 15. This, as you might expect, is the result of formation of a methyl cation, in addition to an acyl radical (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectrometrist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
You can see many more actual examples of mass spectra in the Spectral Database for Organic Compounds
Exercise 4.1: Using the fragmentation patterns for acetone as a guide, predict the signals that you would find in the mass spectra of:
a) 2-butanone; b) 3-hexanone; c) cyclopentanone.
Exercise 4.2: Predict some signals that you would expect to see in a mass spectrum of 2-chloropropane.
Exercise 4.3: The mass spectrum of an aldehyde shows a parent peak at m/z = 58 and a base peak at m/z = 29. Propose a structure, and identify the two species whose m/z values were listed. (
Solutions
Gas Chromatography - Mass Spectrometry
Quite often, mass spectrometry is used in conjunction with a separation technique called gas chromatography (GC). The combined GC-MS procedure is very useful when dealing with a sample that is a mixture of two or more different compounds, because the various compounds are separated from one another before being subjected individually to MS analysis. We will not go into the details of gas chromatography here, although if you are taking an organic laboratory course you might well get a chance to try your hand at GC, and you will almost certainly be exposed to the conceptually analogous techniques of thin layer and column chromatography. Suffice it to say that in GC, a very small amount of a liquid sample is vaporized, injected into a long, coiled metal column, and pushed though the column by helium gas. Along the way, different compounds in the sample stick to the walls of the column to different extents, and thus travel at different speeds and emerge separately from the end of the column. In GC-MS, each purified compound is sent directly from the end of GC column into the MS instrument, so in the end we get a separate mass spectrum for each of the compounds in the original mixed sample. Because a compound's MS spectrum is a very reliable and reproducible 'fingerprint', we can instruct the instrument to search an MS database and identify each compound in the sample.
Gas chromatography-mass spectrometry (GC-MS) schematic
(Image from Wikipedia by K. Murray)
The extremely high sensitivity of modern GC-MS instrumentation makes it possible to detect and identify very small trace amounts of organic compounds. GC-MS is being used increasingly by environmental chemists to detect the presence of harmful organic contaminants in food and water samples. Airport security screeners also use high-speed GC-MS instruments to look for residue from bomb-making chemicals on checked luggage.
Mass spectrometry of proteins - applications in proteomics
Electron ionization mass spectrometry is generally not very useful for analyzing biomolecules: their high polarity makes it difficult to get them into the vapor phase, the first step in EIMS. Mass spectrometry of biomolecules has undergone a revolution over the past few decades, with many new ionization and separation techniques being developed. Generally, the strategy for biomolecule analysis involves soft ionization, in which much less energy (compared to techniques such as EIMS) is imparted to the molecule being analyzed during the ionization process. Usually, soft ionization involves adding protons rather than removing electrons: the cations formed in this way are significantly less energetic than the radical cations formed by removal of an electron. The result of soft ionization is that little or no fragmentation occurs, so the mass being measured is that of an intact molecule. Typically, large biomolecules are digested into smaller pieces using chemical or enzymatic methods, then their masses determined by 'soft' MS.
New developments in soft ionization MS technology have made it easier to detect and identify proteins that are present in very small quantities in biological samples. In electrospray ionization (ESI), the protein sample, in solution, is sprayed into a tube and the molecules are induced by an electric field to pick up extra protons from the solvent. Another common 'soft ionization' method is 'matrix-assisted laser desorption ionization' (MALDI). Here, the protein sample is adsorbed onto a solid matrix, and protonation is achieved with a laser.
Typically, both electrospray ionization and MALDI are used in conjunction with a time-of-flight (TOF) mass analyzer component.
The proteins are accelerated by an electrode through a column, and separation is achieved because lighter ions travel at greater velocity than heavier ions with the same overall charge. In this way, the many proteins in a complex biological sample (such as blood plasma, urine, etc.) can be separated and their individual masses determined very accurately. Modern protein MS is extremely sensitive – recently, scientists were even able to detect the presence of Tyrannosaurus rex protein in a fossilized skeleton! (Science 2007, 316, 277).
Soft ionization mass spectrometry has become in recent years an increasingly important tool in the field of proteomics. Traditionally, protein biochemists tend to study the structure and function of individual proteins. Proteomics researchers, in contrast, want to learn more about how large numbers of proteins in a living system interact with each other, and how they respond to changes in the state of the organism. One important subfield of proteomics is the search for protein 'biomarkers' for human disease: in other words, proteins which are present in greater quantities in the tissues of a sick person than in a healthy person. Detection in a healthy person of a known biomarker for a disease such as diabetes or cancer could provide doctors with an early warning that the patient may be especially susceptible to the disease, so that preventive measures could be taken to prevent or delay onset.
In a 2005 study, MALDI-TOF mass spectrometry was used to compare fluid samples from lung transplant recipients who had suffered from tissue rejection to samples from recipients who had not suffered rejection. Three peptides (short proteins) were found to be present at elevated levels specifically in the tissue rejection samples. It is hoped that these peptides might serve as biomarkers to identify patients who are at increased risk of rejecting their transplanted lungs. (Proteomics 2005, 5, 1705).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.01%3A_Mass_Spectrometry.txt |
Mass spectrometry (MS) is a powerful analytical technique widely used by chemists, biologists, medical researchers, and environmental and forensic scientists, among others. With MS, we are looking at the mass of a molecule, or of different fragments of that molecule.
The basics of a mass spectrometry experiment
There are many different types of MS instruments, but they all have the same three essential components. First, there is an ionization source, where the molecule is given a positive electrical charge, either by removing an electron or by adding a proton. Depending on the ionization method used, the ionized molecule may or may not break apart into a population of smaller fragments. In the figure below, some of the sample molecules remain whole, while others fragment into smaller pieces.
Next in line there is a mass analyzer, where the cationic fragments are separated according to their mass.
Finally, there is a detector, which detects and quantifies the separated ions.
One of the more common types of MS techniques used in the organic laboratory is electron ionization. In the ionization source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of knocking a valence electron off of the molecule to form a radical cation. Because a great deal of energy is transferred by this bombardment process, the radical cation quickly begins to break up into smaller fragments, some of which are positively charged and some of which are neutral. The neutral fragments are either adsorbed onto the walls of the chamber or are removed by a vacuum source. In the mass analyzer component, the positively charged fragments and any remaining unfragmented molecular ions are accelerated down a tube by an electric field.
This tube is curved, and the ions are deflected by a strong magnetic field. Ions of different mass to charge (m/z) ratios are deflected to a different extent, resulting in a ‘sorting’ of ions by mass (virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass). A detector at the end of the curved flight tube records and quantifies the sorted ions.
Looking at mass spectra
Below is typical output for an electron-ionization MS experiment (MS data in the section is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak is at m/z = 43 - we will discuss the formation of this fragment a bit later. The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H.
Molecules with lots of oxygen atoms sometimes show a small M+2 peak (2 m/z units greater than the parent peak) in their mass spectra, due to the presence of a small amount of 18O (the most abundant isotope of oxygen is 16O). Because there are two abundant isotopes of both chlorine (about 75% 35Cl and 25% 37Cl) and bromine (about 50% 79Br and 50% 81Br), chlorinated and brominated compounds have very large and recognizable M+2 peaks. Fragments containing both isotopes of Br can be seen in the mass spectrum of ethyl bromide:
Much of the utility in electron-ionization MS comes from the fact that the radical cations generated in the electron-bombardment process tend to fragment in predictable ways. Detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but it is worthwhile to see a few representative examples, even if we don’t attempt to understand the exact process by which the fragmentation occurs. We saw, for example, that the base peak in the mass spectrum of acetone is m/z = 43. This is the result of cleavage at the ‘alpha’ position - in other words, at the carbon-carbon bond adjacent to the carbonyl. Alpha cleavage results in the formation of an acylium ion (which accounts for the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the parent peak and the base peak, the next largest peak, at a relative abundance of 23%, is at m/z = 15. This, as you might expect, is the result of formation of a methyl cation, in addition to an acyl radical (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectrometrist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
You can see many more actual examples of mass spectra in the Spectral Database for Organic Compounds
Exercise 4.1: Using the fragmentation patterns for acetone as a guide, predict the signals that you would find in the mass spectra of:
a) 2-butanone; b) 3-hexanone; c) cyclopentanone.
Exercise 4.2: Predict some signals that you would expect to see in a mass spectrum of 2-chloropropane.
Exercise 4.3: The mass spectrum of an aldehyde shows a parent peak at m/z = 58 and a base peak at m/z = 29. Propose a structure, and identify the two species whose m/z values were listed. (
Solutions
Gas Chromatography - Mass Spectrometry
Quite often, mass spectrometry is used in conjunction with a separation technique called gas chromatography (GC). The combined GC-MS procedure is very useful when dealing with a sample that is a mixture of two or more different compounds, because the various compounds are separated from one another before being subjected individually to MS analysis. We will not go into the details of gas chromatography here, although if you are taking an organic laboratory course you might well get a chance to try your hand at GC, and you will almost certainly be exposed to the conceptually analogous techniques of thin layer and column chromatography. Suffice it to say that in GC, a very small amount of a liquid sample is vaporized, injected into a long, coiled metal column, and pushed though the column by helium gas. Along the way, different compounds in the sample stick to the walls of the column to different extents, and thus travel at different speeds and emerge separately from the end of the column. In GC-MS, each purified compound is sent directly from the end of GC column into the MS instrument, so in the end we get a separate mass spectrum for each of the compounds in the original mixed sample. Because a compound's MS spectrum is a very reliable and reproducible 'fingerprint', we can instruct the instrument to search an MS database and identify each compound in the sample.
Gas chromatography-mass spectrometry (GC-MS) schematic
(Image from Wikipedia by K. Murray)
The extremely high sensitivity of modern GC-MS instrumentation makes it possible to detect and identify very small trace amounts of organic compounds. GC-MS is being used increasingly by environmental chemists to detect the presence of harmful organic contaminants in food and water samples. Airport security screeners also use high-speed GC-MS instruments to look for residue from bomb-making chemicals on checked luggage.
Mass spectrometry of proteins - applications in proteomics
Electron ionization mass spectrometry is generally not very useful for analyzing biomolecules: their high polarity makes it difficult to get them into the vapor phase, the first step in EIMS. Mass spectrometry of biomolecules has undergone a revolution over the past few decades, with many new ionization and separation techniques being developed. Generally, the strategy for biomolecule analysis involves soft ionization, in which much less energy (compared to techniques such as EIMS) is imparted to the molecule being analyzed during the ionization process. Usually, soft ionization involves adding protons rather than removing electrons: the cations formed in this way are significantly less energetic than the radical cations formed by removal of an electron. The result of soft ionization is that little or no fragmentation occurs, so the mass being measured is that of an intact molecule. Typically, large biomolecules are digested into smaller pieces using chemical or enzymatic methods, then their masses determined by 'soft' MS.
New developments in soft ionization MS technology have made it easier to detect and identify proteins that are present in very small quantities in biological samples. In electrospray ionization (ESI), the protein sample, in solution, is sprayed into a tube and the molecules are induced by an electric field to pick up extra protons from the solvent. Another common 'soft ionization' method is 'matrix-assisted laser desorption ionization' (MALDI). Here, the protein sample is adsorbed onto a solid matrix, and protonation is achieved with a laser.
Typically, both electrospray ionization and MALDI are used in conjunction with a time-of-flight (TOF) mass analyzer component.
The proteins are accelerated by an electrode through a column, and separation is achieved because lighter ions travel at greater velocity than heavier ions with the same overall charge. In this way, the many proteins in a complex biological sample (such as blood plasma, urine, etc.) can be separated and their individual masses determined very accurately. Modern protein MS is extremely sensitive – recently, scientists were even able to detect the presence of Tyrannosaurus rex protein in a fossilized skeleton! (Science 2007, 316, 277).
Soft ionization mass spectrometry has become in recent years an increasingly important tool in the field of proteomics. Traditionally, protein biochemists tend to study the structure and function of individual proteins. Proteomics researchers, in contrast, want to learn more about how large numbers of proteins in a living system interact with each other, and how they respond to changes in the state of the organism. One important subfield of proteomics is the search for protein 'biomarkers' for human disease: in other words, proteins which are present in greater quantities in the tissues of a sick person than in a healthy person. Detection in a healthy person of a known biomarker for a disease such as diabetes or cancer could provide doctors with an early warning that the patient may be especially susceptible to the disease, so that preventive measures could be taken to prevent or delay onset.
In a 2005 study, MALDI-TOF mass spectrometry was used to compare fluid samples from lung transplant recipients who had suffered from tissue rejection to samples from recipients who had not suffered rejection. Three peptides (short proteins) were found to be present at elevated levels specifically in the tissue rejection samples. It is hoped that these peptides might serve as biomarkers to identify patients who are at increased risk of rejecting their transplanted lungs. (Proteomics 2005, 5, 1705).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.02%3A_The_Mass_Spectrum__Fragmentation.txt |
Mass spectrometry (MS) is a powerful analytical technique widely used by chemists, biologists, medical researchers, and environmental and forensic scientists, among others. With MS, we are looking at the mass of a molecule, or of different fragments of that molecule.
The basics of a mass spectrometry experiment
There are many different types of MS instruments, but they all have the same three essential components. First, there is an ionization source, where the molecule is given a positive electrical charge, either by removing an electron or by adding a proton. Depending on the ionization method used, the ionized molecule may or may not break apart into a population of smaller fragments. In the figure below, some of the sample molecules remain whole, while others fragment into smaller pieces.
Next in line there is a mass analyzer, where the cationic fragments are separated according to their mass.
Finally, there is a detector, which detects and quantifies the separated ions.
One of the more common types of MS techniques used in the organic laboratory is electron ionization. In the ionization source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of knocking a valence electron off of the molecule to form a radical cation. Because a great deal of energy is transferred by this bombardment process, the radical cation quickly begins to break up into smaller fragments, some of which are positively charged and some of which are neutral. The neutral fragments are either adsorbed onto the walls of the chamber or are removed by a vacuum source. In the mass analyzer component, the positively charged fragments and any remaining unfragmented molecular ions are accelerated down a tube by an electric field.
This tube is curved, and the ions are deflected by a strong magnetic field. Ions of different mass to charge (m/z) ratios are deflected to a different extent, resulting in a ‘sorting’ of ions by mass (virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass). A detector at the end of the curved flight tube records and quantifies the sorted ions.
Looking at mass spectra
Below is typical output for an electron-ionization MS experiment (MS data in the section is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak is at m/z = 43 - we will discuss the formation of this fragment a bit later. The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H.
Molecules with lots of oxygen atoms sometimes show a small M+2 peak (2 m/z units greater than the parent peak) in their mass spectra, due to the presence of a small amount of 18O (the most abundant isotope of oxygen is 16O). Because there are two abundant isotopes of both chlorine (about 75% 35Cl and 25% 37Cl) and bromine (about 50% 79Br and 50% 81Br), chlorinated and brominated compounds have very large and recognizable M+2 peaks. Fragments containing both isotopes of Br can be seen in the mass spectrum of ethyl bromide:
Much of the utility in electron-ionization MS comes from the fact that the radical cations generated in the electron-bombardment process tend to fragment in predictable ways. Detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but it is worthwhile to see a few representative examples, even if we don’t attempt to understand the exact process by which the fragmentation occurs. We saw, for example, that the base peak in the mass spectrum of acetone is m/z = 43. This is the result of cleavage at the ‘alpha’ position - in other words, at the carbon-carbon bond adjacent to the carbonyl. Alpha cleavage results in the formation of an acylium ion (which accounts for the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the parent peak and the base peak, the next largest peak, at a relative abundance of 23%, is at m/z = 15. This, as you might expect, is the result of formation of a methyl cation, in addition to an acyl radical (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectrometrist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
You can see many more actual examples of mass spectra in the Spectral Database for Organic Compounds
Exercise 4.1: Using the fragmentation patterns for acetone as a guide, predict the signals that you would find in the mass spectra of:
a) 2-butanone; b) 3-hexanone; c) cyclopentanone.
Exercise 4.2: Predict some signals that you would expect to see in a mass spectrum of 2-chloropropane.
Exercise 4.3: The mass spectrum of an aldehyde shows a parent peak at m/z = 58 and a base peak at m/z = 29. Propose a structure, and identify the two species whose m/z values were listed. (
Solutions
Gas Chromatography - Mass Spectrometry
Quite often, mass spectrometry is used in conjunction with a separation technique called gas chromatography (GC). The combined GC-MS procedure is very useful when dealing with a sample that is a mixture of two or more different compounds, because the various compounds are separated from one another before being subjected individually to MS analysis. We will not go into the details of gas chromatography here, although if you are taking an organic laboratory course you might well get a chance to try your hand at GC, and you will almost certainly be exposed to the conceptually analogous techniques of thin layer and column chromatography. Suffice it to say that in GC, a very small amount of a liquid sample is vaporized, injected into a long, coiled metal column, and pushed though the column by helium gas. Along the way, different compounds in the sample stick to the walls of the column to different extents, and thus travel at different speeds and emerge separately from the end of the column. In GC-MS, each purified compound is sent directly from the end of GC column into the MS instrument, so in the end we get a separate mass spectrum for each of the compounds in the original mixed sample. Because a compound's MS spectrum is a very reliable and reproducible 'fingerprint', we can instruct the instrument to search an MS database and identify each compound in the sample.
Gas chromatography-mass spectrometry (GC-MS) schematic
(Image from Wikipedia by K. Murray)
The extremely high sensitivity of modern GC-MS instrumentation makes it possible to detect and identify very small trace amounts of organic compounds. GC-MS is being used increasingly by environmental chemists to detect the presence of harmful organic contaminants in food and water samples. Airport security screeners also use high-speed GC-MS instruments to look for residue from bomb-making chemicals on checked luggage.
Mass spectrometry of proteins - applications in proteomics
Electron ionization mass spectrometry is generally not very useful for analyzing biomolecules: their high polarity makes it difficult to get them into the vapor phase, the first step in EIMS. Mass spectrometry of biomolecules has undergone a revolution over the past few decades, with many new ionization and separation techniques being developed. Generally, the strategy for biomolecule analysis involves soft ionization, in which much less energy (compared to techniques such as EIMS) is imparted to the molecule being analyzed during the ionization process. Usually, soft ionization involves adding protons rather than removing electrons: the cations formed in this way are significantly less energetic than the radical cations formed by removal of an electron. The result of soft ionization is that little or no fragmentation occurs, so the mass being measured is that of an intact molecule. Typically, large biomolecules are digested into smaller pieces using chemical or enzymatic methods, then their masses determined by 'soft' MS.
New developments in soft ionization MS technology have made it easier to detect and identify proteins that are present in very small quantities in biological samples. In electrospray ionization (ESI), the protein sample, in solution, is sprayed into a tube and the molecules are induced by an electric field to pick up extra protons from the solvent. Another common 'soft ionization' method is 'matrix-assisted laser desorption ionization' (MALDI). Here, the protein sample is adsorbed onto a solid matrix, and protonation is achieved with a laser.
Typically, both electrospray ionization and MALDI are used in conjunction with a time-of-flight (TOF) mass analyzer component.
The proteins are accelerated by an electrode through a column, and separation is achieved because lighter ions travel at greater velocity than heavier ions with the same overall charge. In this way, the many proteins in a complex biological sample (such as blood plasma, urine, etc.) can be separated and their individual masses determined very accurately. Modern protein MS is extremely sensitive – recently, scientists were even able to detect the presence of Tyrannosaurus rex protein in a fossilized skeleton! (Science 2007, 316, 277).
Soft ionization mass spectrometry has become in recent years an increasingly important tool in the field of proteomics. Traditionally, protein biochemists tend to study the structure and function of individual proteins. Proteomics researchers, in contrast, want to learn more about how large numbers of proteins in a living system interact with each other, and how they respond to changes in the state of the organism. One important subfield of proteomics is the search for protein 'biomarkers' for human disease: in other words, proteins which are present in greater quantities in the tissues of a sick person than in a healthy person. Detection in a healthy person of a known biomarker for a disease such as diabetes or cancer could provide doctors with an early warning that the patient may be especially susceptible to the disease, so that preventive measures could be taken to prevent or delay onset.
In a 2005 study, MALDI-TOF mass spectrometry was used to compare fluid samples from lung transplant recipients who had suffered from tissue rejection to samples from recipients who had not suffered rejection. Three peptides (short proteins) were found to be present at elevated levels specifically in the tissue rejection samples. It is hoped that these peptides might serve as biomarkers to identify patients who are at increased risk of rejecting their transplanted lungs. (Proteomics 2005, 5, 1705).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
13.04: High-Resolution Mass Spectrometry Can Reveal Molecular Formulas
Two common categories of mass spectrometry are high resolution mass spectrometry (HRMS) and low resolution mass spectrometry (LRMS). Not all mass spectrometers simply measure molecular weights as whole numbers. High resolution mass spectrometers can measure mass so accurately that they can detect the minute differences in mass between two compounds that, on a regular low-resolution instrument, would appear to be identical.
The reason is because atomic masses are not exact multiples of the mass of a proton, as we might usually think.
• An atom of 12C weighs 12.00000 amu.
• An atom of 16O weighs 15.9949 amu.
• An atom of 14N weighs 14.0031 amu.
• An atom of 1H weighs 1.00783 amu.
As a result, on a high resolution mass spectrometer, 2-octanone, C8H16O, has a molecular weight of 128.12018 instead of 128. Naphthalene, C10H8, has a molecular weight of 128.06264. Thus a high resolution mass spectrometer can supply an exact molecular formula for a compound because of the unique combination of masses that result.
• In LRMS, the molecular weight is determined to the nearest amu. The type of instrument used here is more common because it is less expensive and easier to maintain.
• In HRMS, the molecular weight in amu is determined to several decimal places. That precision allows the molecular formula to be narrowed down to only a few possibilities.
HRMS relies on the fact that the mass of an individual atom does not correspond to an integral number of atomic mass units.
Problem MS7.
Calculate the high-resolution molecular weights for the following formulae.
1. C12H20O and C11H16O2
2. C6H13N and C5H11N2 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.03%3A_Isotopes_in_Mass_Spectrometry.txt |
Mass spectrometry (MS) is a powerful analytical technique widely used by chemists, biologists, medical researchers, and environmental and forensic scientists, among others. With MS, we are looking at the mass of a molecule, or of different fragments of that molecule.
The basics of a mass spectrometry experiment
There are many different types of MS instruments, but they all have the same three essential components. First, there is an ionization source, where the molecule is given a positive electrical charge, either by removing an electron or by adding a proton. Depending on the ionization method used, the ionized molecule may or may not break apart into a population of smaller fragments. In the figure below, some of the sample molecules remain whole, while others fragment into smaller pieces.
Next in line there is a mass analyzer, where the cationic fragments are separated according to their mass.
Finally, there is a detector, which detects and quantifies the separated ions.
One of the more common types of MS techniques used in the organic laboratory is electron ionization. In the ionization source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of knocking a valence electron off of the molecule to form a radical cation. Because a great deal of energy is transferred by this bombardment process, the radical cation quickly begins to break up into smaller fragments, some of which are positively charged and some of which are neutral. The neutral fragments are either adsorbed onto the walls of the chamber or are removed by a vacuum source. In the mass analyzer component, the positively charged fragments and any remaining unfragmented molecular ions are accelerated down a tube by an electric field.
This tube is curved, and the ions are deflected by a strong magnetic field. Ions of different mass to charge (m/z) ratios are deflected to a different extent, resulting in a ‘sorting’ of ions by mass (virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass). A detector at the end of the curved flight tube records and quantifies the sorted ions.
Looking at mass spectra
Below is typical output for an electron-ionization MS experiment (MS data in the section is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak is at m/z = 43 - we will discuss the formation of this fragment a bit later. The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H.
Molecules with lots of oxygen atoms sometimes show a small M+2 peak (2 m/z units greater than the parent peak) in their mass spectra, due to the presence of a small amount of 18O (the most abundant isotope of oxygen is 16O). Because there are two abundant isotopes of both chlorine (about 75% 35Cl and 25% 37Cl) and bromine (about 50% 79Br and 50% 81Br), chlorinated and brominated compounds have very large and recognizable M+2 peaks. Fragments containing both isotopes of Br can be seen in the mass spectrum of ethyl bromide:
Much of the utility in electron-ionization MS comes from the fact that the radical cations generated in the electron-bombardment process tend to fragment in predictable ways. Detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but it is worthwhile to see a few representative examples, even if we don’t attempt to understand the exact process by which the fragmentation occurs. We saw, for example, that the base peak in the mass spectrum of acetone is m/z = 43. This is the result of cleavage at the ‘alpha’ position - in other words, at the carbon-carbon bond adjacent to the carbonyl. Alpha cleavage results in the formation of an acylium ion (which accounts for the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the parent peak and the base peak, the next largest peak, at a relative abundance of 23%, is at m/z = 15. This, as you might expect, is the result of formation of a methyl cation, in addition to an acyl radical (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectrometrist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
You can see many more actual examples of mass spectra in the Spectral Database for Organic Compounds
Exercise 4.1: Using the fragmentation patterns for acetone as a guide, predict the signals that you would find in the mass spectra of:
a) 2-butanone; b) 3-hexanone; c) cyclopentanone.
Exercise 4.2: Predict some signals that you would expect to see in a mass spectrum of 2-chloropropane.
Exercise 4.3: The mass spectrum of an aldehyde shows a parent peak at m/z = 58 and a base peak at m/z = 29. Propose a structure, and identify the two species whose m/z values were listed. (
Solutions
Gas Chromatography - Mass Spectrometry
Quite often, mass spectrometry is used in conjunction with a separation technique called gas chromatography (GC). The combined GC-MS procedure is very useful when dealing with a sample that is a mixture of two or more different compounds, because the various compounds are separated from one another before being subjected individually to MS analysis. We will not go into the details of gas chromatography here, although if you are taking an organic laboratory course you might well get a chance to try your hand at GC, and you will almost certainly be exposed to the conceptually analogous techniques of thin layer and column chromatography. Suffice it to say that in GC, a very small amount of a liquid sample is vaporized, injected into a long, coiled metal column, and pushed though the column by helium gas. Along the way, different compounds in the sample stick to the walls of the column to different extents, and thus travel at different speeds and emerge separately from the end of the column. In GC-MS, each purified compound is sent directly from the end of GC column into the MS instrument, so in the end we get a separate mass spectrum for each of the compounds in the original mixed sample. Because a compound's MS spectrum is a very reliable and reproducible 'fingerprint', we can instruct the instrument to search an MS database and identify each compound in the sample.
Gas chromatography-mass spectrometry (GC-MS) schematic
(Image from Wikipedia by K. Murray)
The extremely high sensitivity of modern GC-MS instrumentation makes it possible to detect and identify very small trace amounts of organic compounds. GC-MS is being used increasingly by environmental chemists to detect the presence of harmful organic contaminants in food and water samples. Airport security screeners also use high-speed GC-MS instruments to look for residue from bomb-making chemicals on checked luggage.
Mass spectrometry of proteins - applications in proteomics
Electron ionization mass spectrometry is generally not very useful for analyzing biomolecules: their high polarity makes it difficult to get them into the vapor phase, the first step in EIMS. Mass spectrometry of biomolecules has undergone a revolution over the past few decades, with many new ionization and separation techniques being developed. Generally, the strategy for biomolecule analysis involves soft ionization, in which much less energy (compared to techniques such as EIMS) is imparted to the molecule being analyzed during the ionization process. Usually, soft ionization involves adding protons rather than removing electrons: the cations formed in this way are significantly less energetic than the radical cations formed by removal of an electron. The result of soft ionization is that little or no fragmentation occurs, so the mass being measured is that of an intact molecule. Typically, large biomolecules are digested into smaller pieces using chemical or enzymatic methods, then their masses determined by 'soft' MS.
New developments in soft ionization MS technology have made it easier to detect and identify proteins that are present in very small quantities in biological samples. In electrospray ionization (ESI), the protein sample, in solution, is sprayed into a tube and the molecules are induced by an electric field to pick up extra protons from the solvent. Another common 'soft ionization' method is 'matrix-assisted laser desorption ionization' (MALDI). Here, the protein sample is adsorbed onto a solid matrix, and protonation is achieved with a laser.
Typically, both electrospray ionization and MALDI are used in conjunction with a time-of-flight (TOF) mass analyzer component.
The proteins are accelerated by an electrode through a column, and separation is achieved because lighter ions travel at greater velocity than heavier ions with the same overall charge. In this way, the many proteins in a complex biological sample (such as blood plasma, urine, etc.) can be separated and their individual masses determined very accurately. Modern protein MS is extremely sensitive – recently, scientists were even able to detect the presence of Tyrannosaurus rex protein in a fossilized skeleton! (Science 2007, 316, 277).
Soft ionization mass spectrometry has become in recent years an increasingly important tool in the field of proteomics. Traditionally, protein biochemists tend to study the structure and function of individual proteins. Proteomics researchers, in contrast, want to learn more about how large numbers of proteins in a living system interact with each other, and how they respond to changes in the state of the organism. One important subfield of proteomics is the search for protein 'biomarkers' for human disease: in other words, proteins which are present in greater quantities in the tissues of a sick person than in a healthy person. Detection in a healthy person of a known biomarker for a disease such as diabetes or cancer could provide doctors with an early warning that the patient may be especially susceptible to the disease, so that preventive measures could be taken to prevent or delay onset.
In a 2005 study, MALDI-TOF mass spectrometry was used to compare fluid samples from lung transplant recipients who had suffered from tissue rejection to samples from recipients who had not suffered rejection. Three peptides (short proteins) were found to be present at elevated levels specifically in the tissue rejection samples. It is hoped that these peptides might serve as biomarkers to identify patients who are at increased risk of rejecting their transplanted lungs. (Proteomics 2005, 5, 1705).
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.05%3A_Fragmentation_Patterns_of_Functional_Groups.txt |
Proteins and peptides have been characterized by high pressure liquid chromatography (HPLC) or SDS PAGE by generating peptide maps. These peptide maps have been used as fingerprints of protein or as a tool to know the purity of a known protein in a known sample. Mass spectrometry gives a peptide map when proteins are digested with amino end specific, carboxy end specific, or amino acid specific digestive enzymes. This peptide map can be used to search a sequence database to find a good match from the existing database. This is because the more accurately the peptide masses are known, the less chance there is of bad matches.
• Basir Syed
13.07: Spectroscopy and the Electromagnetic Spectrum
The electromagnetic spectrum
Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio.
Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: shorter wavelengths correspond to higher energy.
High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10-16 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10-9 m), while radio waves can be several hundred meters in length.
The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of particles, called photons, rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as:
$E = \dfrac{hc}{\lambda} \tag{4.1.1}$
where E is energy in kJ/mol, λ (the Greek letter lambda) is wavelength in meters, c is 3.00 x 108 m/s (the speed of light), and h is 3.99 x 10-13 kJ·s·mol-1, a number known as Planck’s constant.
Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s-1.
When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression:
$\lambda \nu = c \tag{4.1.2}$
where ν (the Greek letter ‘nu’) is frequency in s-1. Visible red light with a wavelength of 700 nm, for example, has a frequency of 4.29 x 1014 Hz, and an energy of 40.9 kcal per mole of photons. The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
(Image from Wikipedia commons)
Notice that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest.
Exercise 4.4: Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kJ/mol of photons?
Solutions
Overview of a molecular spectroscopy experiment
In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are not absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed.
Here is the key to molecular spectroscopy: a given molecule will specifically absorb only those wavelengths which have energies that correspond to the energy difference of the transition that is occurring. Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of ΔE, the molecule will specifically absorb radiation with wavelength that corresponds to ΔE, while allowing other wavelengths to pass through unabsorbed.
By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure.
These generalized ideas may all sound quite confusing at this point, but things will become much clearer as we begin to discuss specific examples.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.06%3A_Other_Ionization_Methods.txt |
Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual vibrational modes, a few of which are illustrated below.
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring that is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the stretching mode of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 1013 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 1013 Hz.
If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls.
Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you!
Exercise 4.5: : Express the wavenumber value of 3000 cm-1 in terms of wavelength (in meter units) frequency (in Hz), and associated energy (in kJ/mol).
The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm-1 (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 1013 Hz, and a ΔE value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: a strong peak in the 1650-1750 cm-1 region is a dead giveaway for the presence of a carbonyl group. Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm-1).
The jagged peak at approximately 2900-3000 cm-1 is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum.
You will notice that there are many additional peaks in this spectrum in the longer-wavelength 400 -1400 cm-1 region. This part of the spectrum is called the fingerprint region. While it is usually very difficult to pick out any specific functional group identifications from this region, it does, nevertheless, contain valuable information. The reason for this is suggested by the name: just like a human fingerprint, the pattern of absorbance peaks in the fingerprint region is unique to every molecule, meaning that the data from an unknown sample can be compared to the IR spectra of known standards in order to make a positive identification. It was the IR fingerprint region of the suspicious yellow paint that allowed for its identification as a pigment that could not possibly have been used by the purported artist, William Aiken Walker.
Now, let’s take a look at the IR spectrum for 1-hexanol.
As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1.
We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
You can see many more examples of IR spectra in the Spectral Database for Organic Compounds
Exercise 4.6: Explain how you could use the C-C and C-H stretching frequencies in IR spectra to distinguish between four constitutional isomers: 1,2-dimethylcyclohexene, 1,3-octadiene, 3-octyne, and 1-octyne.
Exercise 4.7: Using the online Spectral Database for Organic Compounds, look up IR spectra for the following compounds, and identify absorbance bands corresponding to those listed in the table above. List actual frequencies for each signal to the nearest cm-1 unit, using the information in tables provided on the site.
a) 1-methylcyclohexanol
b) 4-methylcyclohexene
c) 1-hexyne
d) 2-hexyne
e) 3-hexyne-2,5-diol
Exercise 4.8: A carbon-carbon single bond absorbs in the fingerprint region, and we have already seen the characteristic absorption wavelengths of carbon-carbon double and triple bonds. Rationalize the trend in wavelengths. (Hint - remember, we are thinking of bonds as springs, and looking at the frequency at which they 'bounce').
Solutions
It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in table 1 in the tables section at the end of the text.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol (this type of reaction is discussed in detail in chapter 16).
Kahn Academy video tutorials on infrared spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.08%3A_Infrared_Spectroscopy.txt |
Covalent bonds in organic molecules are not rigid sticks – rather, they behave more like springs. At room temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. These complex vibrations can be broken down mathematically into individual vibrational modes, a few of which are illustrated below.
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring that is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the stretching mode of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 1013 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 1013 Hz.
If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls.
Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you!
Exercise 4.5: : Express the wavenumber value of 3000 cm-1 in terms of wavelength (in meter units) frequency (in Hz), and associated energy (in kJ/mol).
The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm-1 (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 1013 Hz, and a ΔE value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: a strong peak in the 1650-1750 cm-1 region is a dead giveaway for the presence of a carbonyl group. Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm-1).
The jagged peak at approximately 2900-3000 cm-1 is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum.
You will notice that there are many additional peaks in this spectrum in the longer-wavelength 400 -1400 cm-1 region. This part of the spectrum is called the fingerprint region. While it is usually very difficult to pick out any specific functional group identifications from this region, it does, nevertheless, contain valuable information. The reason for this is suggested by the name: just like a human fingerprint, the pattern of absorbance peaks in the fingerprint region is unique to every molecule, meaning that the data from an unknown sample can be compared to the IR spectra of known standards in order to make a positive identification. It was the IR fingerprint region of the suspicious yellow paint that allowed for its identification as a pigment that could not possibly have been used by the purported artist, William Aiken Walker.
Now, let’s take a look at the IR spectrum for 1-hexanol.
As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1.
We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
You can see many more examples of IR spectra in the Spectral Database for Organic Compounds
Exercise 4.6: Explain how you could use the C-C and C-H stretching frequencies in IR spectra to distinguish between four constitutional isomers: 1,2-dimethylcyclohexene, 1,3-octadiene, 3-octyne, and 1-octyne.
Exercise 4.7: Using the online Spectral Database for Organic Compounds, look up IR spectra for the following compounds, and identify absorbance bands corresponding to those listed in the table above. List actual frequencies for each signal to the nearest cm-1 unit, using the information in tables provided on the site.
a) 1-methylcyclohexanol
b) 4-methylcyclohexene
c) 1-hexyne
d) 2-hexyne
e) 3-hexyne-2,5-diol
Exercise 4.8: A carbon-carbon single bond absorbs in the fingerprint region, and we have already seen the characteristic absorption wavelengths of carbon-carbon double and triple bonds. Rationalize the trend in wavelengths. (Hint - remember, we are thinking of bonds as springs, and looking at the frequency at which they 'bounce').
Solutions
It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in table 1 in the tables section at the end of the text.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol (this type of reaction is discussed in detail in chapter 16).
Kahn Academy video tutorials on infrared spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.09%3A_Characteristic_Infrared_Absorption_Bands.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.1.15%3A_Some_Vibrations_Are_Infrared_Inactive.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.1.16%3A_How_to_Interpret_An_Infrared_Spectrum.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.10%3A_The_Intensity_of_Absorption_Bands.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.11%3A_The_Position_of_Absorption_Bands.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.12%3A_The_Position_of_an_Absorption_Band_is_Affected_by_Electron_Delocalization_Elec.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.13%3A_The_Shape_of_Absorption_Bands.txt |
Infrared spectroscopy is the study of the interaction of infrared light with matter. The fundamental measurement obtained in infrared spectroscopy is an infrared spectrum, which is a plot of measured infrared intensity versus wavelength (or frequency) of light.
Introduction
In infrared spectroscopy, units called wavenumbers are normally used to denote different types of light. The frequency, wavelength, and wavenumber are related to each other via the following equation(1):
(1)
These equations show that light waves may be described by their frequency, wavelength or wavenumber. Here, we typically refer to light waves by their wavenumber, however it will be more convenient to refer to a light wave's frequency or wavelength. The wavenumber of several different types of light are shown in table 1.
Table 1. The Electromagnetic spectrum showing the wavenumber of several different types of light.
When a molecule absorbs infrared radiation, its chemical bonds vibrate. The bonds can stretch, contract, and bend. This is why infrared spectroscopy is a type of vibrational spectroscopy. Fortunately, the complex vibrational motion of a molecule can be broken down into a number of constituent vibrations called normal modes. For example, when a guitar string is plucked, the string vibrates at its normal mode frequency. Molecules, like guitar strings, vibrate at specfic frequencies so different molecules vibrate at different frequencies because their structures are different. This is why molecules can be distinguished using infrared spectroscopy. The first necessary condition for a molecule to absorb infrared light is that the molecule must have a vibration during which the change in dipole moment with respect to distance is non-zero. This condition can be summarized in equation(2) form as follows:
(2)
Vibrations that satisfy this equation are said to be infrared active. The H-Cl stretch of hydrogen chloride and the asymmetric stretch of CO2 are examples of infrared active vibrations. Infrared active vibrations cause the bands seen in an infrared spectrum.
The second necessary condition for infrared absorbance is that the energy of the light impinging on a molecule must equal a vibrational energy level difference within the molecule. This condition can be summarized in equation(3) form as follows:
(3)
If the energy of a photon does not meet the criterion in this equation, it will be transmitted by the sample and if the photon energy satisfies this equation, that photon will be absorbed by the molecule.(See Infrared: Theory for more detail)
As any other analytical techniques, infrared spectroscopy works well on some samples, and poorly on others. It is important to know the strengths and weaknesses of infrared spectroscopy so it can be used in the proper way. Some advantages and disadvantages of infrared spectroscopy are listed in table 2.
Advantages Disadvantages
Solids, Liquids, gases, semi-solids, powders and polymers are all analyzed
The peak positions, intensities, widths, and shapes all provide useful information
Fast and easy technique
Sensitive technique (Micrograms of materials can be detected routinely)
Inexpensive
Atoms or monatomic ions do not have infrared spectra
Homonuclear diatomic molecules do not posses infrared spectra
Complex mixture and aqueous solutions are difficult to analyze using infrared spectroscopy
Table 2. The Advantage and Disadvantage of Infrared Spectroscopy
Origin of Peak Positions, Intensities, and Widths
Peak Positions
The equation(4) gives the frequency of light that a molecule will absorb, and gives the frequency of vibration of the normal mode excited by that light.
(4)
Only two variables in equation(4) are a chemical bond's force constant and reduced mass. Here, the reduced mass refers to (M1M2)/(M1+M2) where M1 and M2 are the masses of the two atoms, respectively. These two molecular properties determine the wavenumber at which a molecule will absorb infrared light. No two chemical substances in the universe have the same force constants and atomic masses, which is why the infrared spectrum of each chemical substance is unique. To understand the effect of atomic masses and force constant on the positions of infrared bands, table 3 and 4 are shown as an example, respectively.
Table 3. An Example of an Mass Effect
Bond C-H Stretch in cm-1
C-1H ~3000
C-2D ~2120
The reduced masses of C-1H and C-2D are different, but their force constants are the same. By simply doubling the mass of the hydrogen atom, the carbon-hydrogen stretching vibration is reduced by over 800cm-1.
Table 4. An Example of an electronic Effect
Bond C-H Stretch in cm-1
C-H ~3000
H-C=O ~2750
When a hydrogen is attached to a carbon with a C=O bond, the C-H stretch band position decrease to ~2750cm-1. These two C-H bonds have the same reduced mass but different force constants. The oxygen in the second molecule pulls electron density away from the C-H bond so it makes weaken and reduce the C-H force constant. This cause the C-H stretching vibration to be reduced by ~250cm-1.
The Origin of Peak Intensities
The different vibrations of the different functional groups in the molecule give rise to bands of differing intensity. This is because $\frac{\partial \mu}{\partial x}$ is different for each of these vibrations. For example, the most intense band in the spectrum of octane shown in Figure 3 is at 2971, 2863 cm-1 and is due to stretching of the C-H bond. One of the weaker bands in the spectrum of octane is at 726cm-1, and it is due to long-chain methyl rock of the carbon-carbon bonds in octane. The change in dipole moment with respect to distance for the C-H stretching is greater than that for the C-C rock vibration, which is why the C-H stretching band is the more intense than C-C rock vibration.
Another factor that determines the peak intensity in infrared spectra is the concentration of molecules in the sample. The equation(5) that relates concentration to absorbance is Beer's law,
(5)
The absorptivity is the proportionality constant between concentration and absorbance, and is dependent on (µ/x)2. The absorptivity is an absolute measure of infrared absorbance intensity for a specific molecule at a specific wavenumber. For pure sample, concentration is at its maximum, and the peak intensities are true representations of the values of µ/x for different vibrations. However, in a mixture, two peaks may have different intensities because there are molecules present in different concentration.
The Orgins of Peak Widths
In general, the width of infrared bands for solid and liquid samples is determined by the number of chemical environments which is related to the strength of intermolecular interactions such as hydrogen bonding. Figure 1. shows hydrogen bond in water molecules and these water molecules are in different chemical environments. Because the number and strength of hydrogen bonds differs with chemical environment, the force constant varies and the wavenumber differs at which these molecules absorb infrared light.
Figure 1. Hydrogen Bonding in water molecules
In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. This is illustrated in the spectra of ethanol(Fig7) and hexanoic acid(Fig11). When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.
The Origin of Group Frequencies
An important observation made by early researchers is that many functional group absorb infrared radiation at about the same wavenumber, regardless of the structure of the rest of the molecule. For example, C-H stretching vibrations usually appear between 3200 and 2800cm-1 and carbonyl(C=O) stretching vibrations usually appear between 1800 and 1600cm-1. This makes these bands diagnostic markers for the presence of a functional group in a sample. These types of infrared bands are called group frequencies because they tell us about the presence or absence of specific functional groups in a sample.
Figure 2. Group frequency and fingerprint regions of the mid-infrared spectrum
The region of the infrared spectrum from 1200 to 700 cm-1 is called the fingerprint region. This region is notable for the large number of infrared bands that are found there. Many different vibrations, including C-O, C-C and C-N single bond stretches, C-H bending vibrations, and some bands due to benzene rings are found in this region. The fingerprint region is often the most complex and confusing region to interpret, and is usually the last section of a spectrum to be interpreted. However, the utility of the fingerprint region is that the many bands there provide a fingerprint for a molecule.
Spectral Interpretation by Application of Group Frequencies
Organic Compounds
One of the most common application of infrared spectroscopy is to the identification of organic compounds. The major classes of organic molecules are shown in this category and also linked on the bottom page for the number of collections of spectral information regarding organic molecules.
Hydrocarbons
Hydrocarbons compounds contain only C-H and C-C bonds, but there is plenty of information to be obtained from the infrared spectra arising from C-H stretching and C-H bending.
In alkanes, which have very few bands, each band in the spectrum can be assigned:
• Figure 3. shows the IR spectrum of octane. Since most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a routine IR spectrum. Note that the change in dipole moment with respect to distance for the C-H stretching is greater than that for others shown, which is why the C-H stretch band is the more intense.
In alkenes compounds, each band in the spectrum can be assigned:
• Figure 4. shows the IR spectrum of 1-octene. As alkanes compounds, these bands are not specific and are generally not noted because they are present in almost all organic molecules.
In alkynes, each band in the spectrum can be assigned:
• The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Figure 6. shows the spectrum of toluene.
Figure 6. Infrared Spectrum of Toluene
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• Figure 7. shows the spectrum of ethanol. Note the very broad, strong band of the O–H stretch.
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
• If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
• Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• Figure 10. shows the spectrum of ethyl benzoate.
The carbonyl stretch C=O of a carboxylic acid appears as an intense band from 1760-1690 cm-1. The exact position of this broad band depends on whether the carboxylic acid is saturated or unsaturated, dimerized, or has internal hydrogen bonding.
• Figure 11. shows the spectrum of hexanoic acid.
• Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• The spectrum of 1-chloro-2-methylpropane are shown below.
For more Infrared spectra Spectral database of organic molecules is introduced to use free database. Also, the infrared spectroscopy correlation tableis linked on bottom of page to find other assigned IR peaks.
Inorganic Compounds
Generally, the infrared bands for inorganic materials are broader, fewer in number and appear at lower wavenumbers than those observed for organic materials. If an inorganic compound forms covalent bonds within an ion, it can produce a characteristic infrared spectrum.
Main infrared bands of some common inorganic ions:
• Diatomic molecules produce one vibration along the chemical bond. Monatomic ligand, where metal s coordinate with atoms such as halogens, H, N or O, produce characteristic bands. These bands are summarized in below.
Chracteristic infrared bands of diatomic inorganic molecules: M(metal), X(halogen)
• The normal modes of vibration of linear and bent triatomic molecules are illustrated and some common linear and bent triatomic molecules are shown below. Note that some molecules show two bands for ?1because of Fermi resonance.
Characteristic infrared bands(cm-1) of triatomic inorganic molecules:
1388, 1286 3311 2053 714, 784 327
667 712 486, 471 380 249
2349 2049 748 2219 842
Bent Molecules H2O O3 SnCl 2
3675 1135 354
1595 716 120
3756 1089 334
Identification
There are a few general rules that can be used when using a mid-infrared spectrum for the determination of a molecular structure. The following is a suggested strategy for spectrum interpretation:2
1. Infrared spectroscopy is used to analyze a wide variety of samples, but it cannot solve every chemical analysis problem. When used in conjunction with other methods such as mass spectroscopy, nuclear magnetic resonance, and elemental analysis, infrared spectroscopy usually makes possible the positive identification of a sample.
• Spectral Database for Organic Compounds SDBS: http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access)
• Infrared Spectroscopy Correlation Table: en.Wikipedia.org/wiki/Infrared_spectroscopy_correlation_table
• FDM Reference Spectra Databases: http://www.fdmspectra.com/index.html
• Other Usuful Web Pages:
• www.cem.msu.edu/~reusch/Virtu...d/infrared.htm
• Fermi resonance : en.Wikipedia.org/wiki/Fermi_resonance | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.14%3A_The_Absence_of_Absorption_Bands.txt |
While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated $\pi$ systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an $n \rightarrow \pi^*$ transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Exercise 4.9: What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the lmax of b-carotene?
Exercise 4.10: Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solutions
Protecting yourself from sunburn
Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below.
Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA (para-aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O.
Looking at UV-vis spectra
We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.
Schematic for a UV-Vis spectrophotometer
(Image from Wikipedia Commons)
Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.
Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is $\lambda_{max}$, which is the wavelength at maximal light absorbance. As you can see, NAD+ has $\lambda_{max} = 260\;nm$. We also want to record how much light is absorbed at $\lambda_{max}$. Here we use a unitless number called absorbance, abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
$A = \log \dfrac{I_0}{I} \tag{4.3.1}$
You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.
Exercise 4.11: Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Solutions
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Applications of UV spectroscopy in organic and biological chemistry
UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance.
If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.
$\epsilon = \dfrac{A}{c} \tag{4.3.2}$
The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L* mol-1*cm-1. If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD+ the literature value of ε at 260 nm is 18,000 L* mol-1*cm-1. In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Exercise 4.12: 50 microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter?
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:
Both NAD+ and NADH absorb at 260 nm. However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 L*mol-1*cm-1. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:
By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction.
UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.
Kahn Academy video tutorials on UV-Vis spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.17%3A_Ultraviolet_and_Visible_Spectroscopy.txt |
While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated $\pi$ systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an $n \rightarrow \pi^*$ transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Exercise 4.9: What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the lmax of b-carotene?
Exercise 4.10: Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solutions
Protecting yourself from sunburn
Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below.
Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA (para-aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O.
Looking at UV-vis spectra
We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.
Schematic for a UV-Vis spectrophotometer
(Image from Wikipedia Commons)
Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.
Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is $\lambda_{max}$, which is the wavelength at maximal light absorbance. As you can see, NAD+ has $\lambda_{max} = 260\;nm$. We also want to record how much light is absorbed at $\lambda_{max}$. Here we use a unitless number called absorbance, abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
$A = \log \dfrac{I_0}{I} \tag{4.3.1}$
You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.
Exercise 4.11: Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Solutions
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Applications of UV spectroscopy in organic and biological chemistry
UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance.
If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.
$\epsilon = \dfrac{A}{c} \tag{4.3.2}$
The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L* mol-1*cm-1. If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD+ the literature value of ε at 260 nm is 18,000 L* mol-1*cm-1. In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Exercise 4.12: 50 microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter?
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:
Both NAD+ and NADH absorb at 260 nm. However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 L*mol-1*cm-1. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:
By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction.
UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.
Kahn Academy video tutorials on UV-Vis spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.18%3A_The_Beer-Lambert_Law.txt |
While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated $\pi$ systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an $n \rightarrow \pi^*$ transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Exercise 4.9: What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the lmax of b-carotene?
Exercise 4.10: Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solutions
Protecting yourself from sunburn
Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below.
Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA (para-aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O.
Looking at UV-vis spectra
We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.
Schematic for a UV-Vis spectrophotometer
(Image from Wikipedia Commons)
Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.
Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is $\lambda_{max}$, which is the wavelength at maximal light absorbance. As you can see, NAD+ has $\lambda_{max} = 260\;nm$. We also want to record how much light is absorbed at $\lambda_{max}$. Here we use a unitless number called absorbance, abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
$A = \log \dfrac{I_0}{I} \tag{4.3.1}$
You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.
Exercise 4.11: Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Solutions
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Applications of UV spectroscopy in organic and biological chemistry
UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance.
If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.
$\epsilon = \dfrac{A}{c} \tag{4.3.2}$
The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L* mol-1*cm-1. If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD+ the literature value of ε at 260 nm is 18,000 L* mol-1*cm-1. In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Exercise 4.12: 50 microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter?
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:
Both NAD+ and NADH absorb at 260 nm. However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 L*mol-1*cm-1. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:
By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction.
UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.
Kahn Academy video tutorials on UV-Vis spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.19%3A_The_Effect_of_Conjugation_on_max.txt |
This page explains what happens when organic compounds absorb UV or visible light, and why the wavelength of light absorbed varies from compound to compound.
What happens when light is absorbed by molecules?
When we were talking about the various sorts of orbitals present in organic compounds on the introductory page (see above), you will have come across this diagram showing their relative energies:
Remember that the diagram isn't intended to be to scale - it just shows the relative placing of the different orbitals. When light passes through the compound, energy from the light is used to promote an electron from a bonding or non-bonding orbital into one of the empty anti-bonding orbitals. The possible electron jumps that light might cause are:
In each possible case, an electron is excited from a full orbital into an empty anti-bonding orbital. Each jump takes energy from the light, and a big jump obviously needs more energy than a small one. Each wavelength of light has a particular energy associated with it. If that particular amount of energy is just right for making one of these energy jumps, then that wavelength will be absorbed - its energy will have been used in promoting an electron.
We need to work out what the relationship is between the energy gap and the wavelength absorbed. Does, for example, a bigger energy gap mean that light of a lower wavelength will be absorbed - or what? It is easier to start with the relationship between the frequency of light absorbed and its energy:
You can see that if you want a high energy jump, you will have to absorb light of a higher frequency. The greater the frequency, the greater the energy. That's easy - but unfortunately UV-visible absorption spectra are always given using wavelengths of light rather than frequency. That means that you need to know the relationship between wavelength and frequency.
You can see from this that the higher the frequency is, the lower the wavelength is. So, if you have a bigger energy jump, you will absorb light with a higher frequency - which is the same as saying that you will absorb light with a lower wavelength.
Important summary: The larger the energy jump, the lower the wavelength of the light absorbed.
Some jumps are more important than others for absorption spectrometry
An absorption spectrometer works in a range from about 200 nm (in the near ultra-violet) to about 800 nm (in the very near infra-red). Only a limited number of the possible electron jumps absorb light in that region. Look again at the possible jumps. This time, the important jumps are shown in black, and a less important one in grey. The grey dotted arrows show jumps which absorb light outside the region of the spectrum we are working in.
Remember that bigger jumps need more energy and so absorb light with a shorter wavelength. The jumps shown with grey dotted arrows absorb UV light of wavelength less that 200 nm. The important jumps are:
• from pi bonding orbitals to pi anti-bonding orbitals;
• from non-bonding orbitals to pi anti-bonding orbitals;
• from non-bonding orbitals to sigma anti-bonding orbitals.
That means that in order to absorb light in the region from 200 - 800 nm (which is where the spectra are measured), the molecule must contain either pi bonds or atoms with non-bonding orbitals. Remember that a non-bonding orbital is a lone pair on, say, oxygen, nitrogen or a halogen.
Groups in a molecule which absorb light are known as chromophores.
What does an absorption spectrum look like
The diagram below shows a simple UV-visible absorption spectrum for buta-1,3-diene - a molecule we will talk more about later. Absorbance (on the vertical axis) is just a measure of the amount of light absorbed. The higher the value, the more of a particular wavelength is being absorbed.
You will see that absorption peaks at a value of 217 nm. This is in the ultra-violet and so there would be no visible sign of any light being absorbed - buta-1,3-diene is colorless. You read the symbol on the graph as "lambda-max". In buta-1,3-diene, CH2=CH-CH=CH2, there are no non-bonding electrons. That means that the only electron jumps taking place (within the range that the spectrometer can measure) are from pi bonding to pi anti-bonding orbitals.
A chromophore producing two peaks
A chromophore such as the carbon-oxygen double bond in ethanal, for example, obviously has pi electrons as a part of the double bond, but also has lone pairs on the oxygen atom. That means that both of the important absorptions from the last energy diagram are possible. You can get an electron excited from a pi bonding to a pi anti-bonding orbital, or you can get one excited from an oxygen lone pair (a non-bonding orbital) into a pi anti-bonding orbital.
The non-bonding orbital has a higher energy than a pi bonding orbital. That means that the jump from an oxygen lone pair into a pi anti-bonding orbital needs less energy. That means it absorbs light of a lower frequency and therefore a higher wavelength. Ethanal can therefore absorb light of two different wavelengths:
• the pi bonding to pi anti-bonding absorption peaks at 180 nm;
• the non-bonding to pi anti-bonding absorption peaks at 290 nm.
Both of these absorptions are in the ultra-violet, but most spectrometers won't pick up the one at 180 nm because they work in the range from 200 - 800 nm.
The importance of conjugation and delocalisation
Consider these three molecules:
Ethene contains a simple isolated carbon-carbon double bond, but the other two have conjugated double bonds. In these cases, there is delocalization of the pi bonding orbitals over the whole molecule. Now look at the wavelengths of the light which each of these molecules absorbs.
molecule wavelength of maximum absorption (nm)
ethene 171
buta-1,3-diene 217
hexa-1,3,5-triene 258
All of the molecules give similar UV-visible absorption spectra - the only difference being that the absorptions move to longer and longer wavelengths as the amount of delocalization in the molecule increases.
Why is this? You can actually work out what must be happening.
• The maximum absorption is moving to longer wavelengths as the amount of delocalization increases.
• Therefore maximum absorption is moving to shorter frequencies as the amount of delocalization increases.
• Therefore absorption needs less energy as the amount of delocalization increases.
• Therefore there must be less energy gap between the bonding and anti-bonding orbitals as the amount of delocalization increases.
. . . and that's what is happening.
Compare ethene with buta-1,3-diene. In ethene, there is one pi bonding orbital and one pi anti-bonding orbital. In buta-1,3-diene, there are two pi bonding orbitals and two pi anti-bonding orbitals. This is all discussed in detail on the introductory page that you should have read.
The highest occupied molecular orbital is often referred to as the HOMO - in these cases, it is a pi bonding orbital. The lowest unoccupied molecular orbital (the LUMO) is a pi anti-bonding orbital. Notice that the gap between these has fallen. It takes less energy to excite an electron in the buta-1,3-diene case than with ethene.
In the hexa-1,3,5-triene case, it is less still.
If you extend this to compounds with really massive delocalisation, the wavelength absorbed will eventually be high enough to be in the visible region of the spectrum, and the compound will then be seen as colored. A good example of this is the orange plant pigment, beta-carotene - present in carrots, for example.
Why is beta-carotene orange?
Beta-carotene has the sort of delocalization that we've just been looking at, but on a much greater scale with 11 carbon-carbon double bonds conjugated together. The diagram shows the structure of beta-carotene with the alternating double and single bonds shown in red.
The more delocalization there is, the smaller the gap between the highest energy pi bonding orbital and the lowest energy pi anti-bonding orbital. To promote an electron therefore takes less energy in beta-carotene than in the cases we've looked at so far - because the gap between the levels is less.
Remember that less energy means a lower frequency of light gets absorbed - and that's equivalent to a longer wavelength. Beta-carotene absorbs throughout the ultra-violet region into the violet - but particularly strongly in the visible region between about 400 and 500 nm with a peak about 470 nm. If you have read the page in this section about electromagnetic radiation, you might remember that the wavelengths associated with the various colors are approximately:
color region wavelength (nm)
violet 380 - 435
blue 435 - 500
cyan 500 - 520
green 520 - 565
yellow 565 - 590
orange 590 - 625
red 625 - 740
So if the absorption is strongest in the violet to cyan region, what color will you actually see? It is tempting to think that you can work it out from the colors that are left - and in this particular case, you wouldn't be far wrong. Unfortunately, it isn't as simple as that!
Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors.
Complementary colors
If you arrange some colors in a circle, you get a "color wheel". The diagram shows one possible version of this. An internet search will throw up many different versions!
colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light.
What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. In the beta-carotene case, the situation is more confused because you are absorbing such a range of wavelengths. However, if you think of the peak absorption running from the blue into the cyan, it would be reasonable to think of the color you would see as being opposite that where yellow runs into red - in other words, orange.
Phenolphthalein
You have probably used phenolphthalein as an acid-base indicator, and will know that it is colorless in acidic conditions and magenta (bright pink) in an alkaline solution. How is this color change related to changes in the molecule? The structures of the two differently colored forms are:
Both of these absorb light in the ultra-violet, but the one on the right also absorbs in the visible with a peak at 553 nm. The molecule in acid solution is colorless because our eyes can't detect the fact that some light is being absorbed in the ultra-violet. However, our eyes do detect the absorption at 553 nm produced by the form in alkaline solution.
553 nm is in the green region of the spectrum. If you look back at the color wheel, you will find that the complementary color of green is magenta - and that's the color you see.
So why does the color change as the structure changes? What we have is a shift to absorption at a higher wavelength in alkaline solution. As we've already seen, a shift to higher wavelength is associated with a greater degree of delocalisation.
Here is a modified diagram of the structure of the form in acidic solution - the colorless form. The extent of the delocalization is shown in red.
Notice that there is delocalization over each of the three rings - extending out over the carbon-oxygen double bond, and to the various oxygen atoms because of their lone pairs.
But the delocalization doesn't extend over the whole molecule. The carbon atom in the centre with its four single bonds prevents the three delocalized regions interacting with each other.
Now compare that with the magenta form:
The rearrangement now lets the delocalization extend over the entire ion. This greater delocalization lowers the energy gap between the highest occupied molecular orbital and the lowest unoccupied pi anti-bonding orbital. It needs less energy to make the jump and so a longer wavelength of light is absorbed.
Increasing the amount of delocalization shifts the absorption peak to a higher wavelength.
Methyl orange
You will know that methyl orange is yellow in alkaline solutions and red in acidic ones. The structure in alkaline solution is:
In acid solution, a hydrogen ion is (perhaps unexpectedly) picked up on one of the nitrogens in the nitrogen-nitrogen double bond.
This now gets a lot more complicated! The positive charge on the nitrogen is delocalized (spread around over the structure) - especially out towards the right-hand end of the molecule as we've written it. The normally drawn structure for the red form of methyl orange is . . .
But this can be seriously misleading as regards the amount of delocalization in the structure for reasons discussed below (after the red warning box) if you are interested.
Which is the more delocalized structure?
Let's work backwards from the absorption spectra to see if that helps. The yellow form has an absorption peak at about 440 nm. That's in the blue region of the spectrum, and the complementary color of blue is yellow. That's exactly what you would expect. The red form has an absorption peak at about 520 nm. That's at the edge of the cyan region of the spectrum, and the complementary color of cyan is red. Again, there's nothing unexpected here.
Notice that the change from the yellow form to the red form has produced an increase in the wavelength absorbed. An increase in wavelength suggests an increase in delocalisation. That means that there must be more delocalization in the red form than in the yellow one. Here again is the structure of the yellow form:
delocalization will extend over most of the structure - out as far as the lone pair on the right-hand nitrogen atom.
If you use the normally written structure for the red form, the delocalization seems to be broken in the middle - the pattern of alternating single and double bonds seems to be lost.
But that is to misunderstand what this last structure represents.
Canonical forms
If you draw the two possible Kekulé structures for benzene, you will know that the real structure of benzene isn't like either of them. The real structure is somewhere between the two - all the bonds are identical and somewhere between single and double in character. That's because of the delocalization in benzene.
The two structures are known as canonical forms, and they can each be thought of as adding some knowledge to the real structure. For example, the bond drawn at the top right of the molecule is neither truly single or double, but somewhere in between. Similarly with all the other bonds.
The two structures we've previously drawn for the red form of methyl orange are also canonical forms - two out of lots of forms that could be drawn for this structure. We could represent the delocalized structure by:
These two forms can be thought of as the result of electron movements in the structure, and curly arrows are often used to show how one structure can lead to the other.
In reality, the electrons haven't shifted fully either one way or the other. Just as in the benzene case, the actual structure lies somewhere in between these.
You must also realize that drawing canonical forms has no effect on the underlying geometry of the structure. Bond types or lengths or angles don't change in the real structure.
For example, the lone pairs on the nitrogen atoms shown in the last diagram are both involved with the delocalisation. For this to happen all the bonds around these nitrogens must be in the same plane, with the lone pair sticking up so that it can overlap sideways with orbitals on the next-door atoms. The fact that in each of the two canonical forms one of these nitrogens is shown as if it had an ammonia-like arrangement of the bonds is potentially misleading - and makes it look as if the delocalization is broken.
The problem is that there is no easy way of representing a complex delocalized structure in simple structural diagrams. It is bad enough with benzene - with something as complicated as methyl orange any method just leads to possible confusion if you aren't used to working with canonical forms.
It gets even more complicated! If you were doing this properly there would be a host of other canonical forms with different arrangements of double and single bonds and with the positive charge located at various places around the rings and on the other nitrogen atom.
The real structure can't be represented properly by any one of this multitude of canonical forms, but each gives a hint of how the delocalization works.
If we take the two forms we have written as perhaps the two most important ones, it suggests that there is delocalization of the electrons over the whole structure, but that electron density is a bit low around the two nitrogens carrying the positive charge on one canonical form or the other.
Why is the red form more delocalized
Finally, we get around to an attempt at an explanation as to why the delocalization is greater in the red form of methyl orange in acid solution than in the yellow one in alkaline solution. The answer may lie in the fact that the lone pair on the nitrogen at the right-hand end of the structure as we've drawn it is more fully involved in the delocalization in the red form. The canonical form with the positive charge on that nitrogen suggests a significant movement of that lone pair towards the rest of the molecule.
Doesn't the same thing happen to the lone pair on the same nitrogen in the yellow form of methyl orange? Not to the same extent.
Any canonical form that you draw in which that happens produces another negatively charged atom somewhere in the rest of the structure. Separating negative and positive charges like this is energetically unfavourable. In the red form, we aren't producing a new separation of charge - just shifting a positive charge around the structure. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.20%3A_The_Visible_Spectrum_and_Color.txt |
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13.21: Some Uses of UV
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13.21.01: Some Uses of UV
While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated $\pi$ systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an $n \rightarrow \pi^*$ transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Exercise 4.9: What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the lmax of b-carotene?
Exercise 4.10: Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solutions
Protecting yourself from sunburn
Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below.
Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA (para-aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O.
Looking at UV-vis spectra
We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred.
Schematic for a UV-Vis spectrophotometer
(Image from Wikipedia Commons)
Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.
Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is $\lambda_{max}$, which is the wavelength at maximal light absorbance. As you can see, NAD+ has $\lambda_{max} = 260\;nm$. We also want to record how much light is absorbed at $\lambda_{max}$. Here we use a unitless number called absorbance, abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
$A = \log \dfrac{I_0}{I} \tag{4.3.1}$
You can see that the absorbance value at 260 nm (A260) is about 1.0 in this spectrum.
Exercise 4.11: Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Solutions
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Applications of UV spectroscopy in organic and biological chemistry
UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance.
If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.
$\epsilon = \dfrac{A}{c} \tag{4.3.2}$
The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L* mol-1*cm-1. If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD+ the literature value of ε at 260 nm is 18,000 L* mol-1*cm-1. In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Exercise 4.12: 50 microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter?
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:
Both NAD+ and NADH absorb at 260 nm. However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 L*mol-1*cm-1. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:
By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction.
UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.
Kahn Academy video tutorials on UV-Vis spectroscopy
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/13%3A_Mass_Spectrometry_Infrared_Spectroscopy_and_Ultraviolet_Visible_Spectroscopy/13.21%3A_Some_Uses_of_UV/13.21.1.01%3A_Some_Uses_of_UV_Vis_.txt |
5.1A: NMR-active nuclei
The basis for nuclear magnetic resonance is the observation that many atomic nuclei spin about an axis and generate their own magnetic field, or magnetic moment. For reasons that are outside the scope of this text, only those nuclei with an odd number of protons and/or neutrons have a magnetic moment. Fortunately for chemists, several common nuclei, including hydrogen (1H), the 13C isotope of carbon, the 19F isotope of fluorine, and the 31P isotope of phosphorus, all have magnetic moments and therefore can be observed by NMR – they are, in other words, NMR-active. Other nuclei - such as the common 12C and 16O isotopes of carbon and oxygen - do not have magnetic moments, and are essentially invisible in NMR. Other nuclei such as deuterium (2H) and nitrogen (14N) have magnetic moments and are NMR-active, but the nature of their magnetic moments is such that these nuclei are more difficult to analyze by NMR. In practice it is 1H, 13C, 19F, and 31P that are most often observed by NMR spectroscopy. In this chapter, we will develop our understanding of the principles behind NMR spectroscopy by focusing our attention first on the detection of protons in 1H-NMR experiments (in discussion about NMR, the terms 'hydrogen' and 'proton' are used interchangeably). Much of what we learn, however, will also apply to the detection and analysis of other NMR-active nuclei, and later in the chapter we will shift our attention to NMR experiments involving 13C and 31P atoms.
5.1B: Nuclear precession, spin states, and the resonance condition
When a sample of an organic compound is sitting in a flask on a laboratory benchtop, the magnetic moments of its hydrogen atoms are randomly oriented. When the same sample is placed within the field of a very strong magnet in an NMR instrument (this field is referred to by NMR spectroscopists as the applied field, abbreviated B0 ) each hydrogen will assume one of two possible spin states. In what is referred to as the +½ spin state, the hydrogen's magnetic moment is aligned with the direction of B0, while in the -½ spin state it is aligned opposed to the direction of B0.
Because the +½ spin state is slightly lower in energy, in a large population of organic molecules slightly more than half of the hydrogen atoms will occupy this state, while slightly less than half will occupy the –½ state. The difference in energy between the two spin states increases with increasing strength of B0.This last statement is in italics because it is one of the key ideas in NMR spectroscopy, as we shall soon see.
At this point, we need to look a little more closely at how a proton spins in an applied magnetic field. You may recall playing with spinning tops as a child. When a top slows down a little and the spin axis is no longer completely vertical, it begins to exhibit precessional motion, as the spin axis rotates slowly around the vertical. In the same way, hydrogen atoms spinning in an applied magnetic field also exhibit precessional motion about a vertical axis. It is this axis (which is either parallel or antiparallel to B0) that defines the proton’s magnetic moment. In the figure below, the proton is in the +1/2 spin state.
The frequency of precession (also called the Larmour frequency, abbreviated ωL) is simply the number of times per second that the proton precesses in a complete circle. A proton`s precessional frequency increases with the strength of B0.
If a proton that is precessing in an applied magnetic field is exposed to electromagnetic radiation of a frequency ν that matches its precessional frequency ωL, we have a condition called resonance. In the resonance condition, a proton in the lower-energy +½ spin state (aligned with B0) will transition (flip) to the higher energy –½ spin state (opposed to B0). In doing so, it will absorb radiation at this resonance frequency ν = ωL. This frequency, as you might have already guessed, corresponds to the energy difference between the proton’s two spin states. With the strong magnetic fields generated by the superconducting magnets used in modern NMR instruments, the resonance frequency for protons falls within the radio-wave range, anywhere from 100 MHz to 800 MHz depending on the strength of the magnet.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.01%3A_14.1____An_Introduction_to_NMR_Spectroscopy.txt |
Nuclear spin may be related to the nucleon composition of a nucleus in the following manner:
• Odd mass nuclei (i.e. those having an odd number of nucleons) have fractional spins.
• Examples are I = 1/2 ( 1H, 13C, 19F ), I = 3/2 ( 11B ) & I = 5/2 ( 17O ).
• Even mass nuclei composed of odd numbers of protons and neutrons have integral spins. Examples are I = 1 ( 2H, 14N ).
• Even mass nuclei composed of even numbers of protons and neutrons have zero spin ( I = 0 ). Examples are 12C, and 16O.
Spin Properties of Nuclei
Spin 1/2 nuclei have a spherical charge distribution, and their NMR behavior is the easiest to understand. Other spin nuclei have nonspherical charge distributions and may be analyzed as prolate or oblate spinning bodies. All nuclei with non-zero spins have magnetic moments (μ), but the nonspherical nuclei also have an electric quadrupole moment (eQ). Some characteristic properties of selected nuclei are given in the following table.
Isotope
Natural %
Abundance
Spin (I)
Magnetic
Moment (μ)*
Magnetogyric
Ratio (γ)
1H
99.9844 1/2 2.7927 26.753
2H
0.0156 1 0.8574 4,107
11B
81.17 3/2 2.6880 --
13C
1.108 1/2 0.7022 6,728
17O
0.037 5/2 -1.8930 -3,628
19F
100.0 1/2 2.6273 25,179
29Si
4.700 1/2 -0.5555 -5,319
31P
100.0 1/2 1.1305 10,840
* μ in units of nuclear magnetons = 5.05078•10-27 JT-1
γ in units of 107rad T-1 sec-1
A Model for NMR Spectroscopy
The model of a spinning nuclear magnet aligned with or against an external magnetic field (for I = 1/2 nuclei) must be refined for effective interpretation of NMR phenomena. Just as a spinning mass will precess in a gravitational field (a gyroscope), the magnetic moment μ associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The frequency of precession is proportional to the strength of the magnetic field, as noted by the equation: ωo = γBo. The frequency ωo is called the Larmor frequency and has units of radians per second. The proportionality constant γ is known as the gyromagnetic ratio and is proportional to the magnetic moment (γ = 2pm/hI). Some characteristic γ's were listed in the preceding table of nuclear properties.
A Spinning Gyroscope in a Gravity Field
magnetic moment
μ
A Spinning Charge in a Magnetic Field
If rf energy having a frequency matching the Larmor frequency is introduced at a right angle to the external field (e.g. along the x-axis), the precessing nucleus will absorb energy and the magnetic moment will flip to its I = _1/2 state. This excitation is shown in the following diagram. Note that frequencies in radians per second may be converted to Hz (cps) by dividing by 2π.
The energy difference between nuclear spin states is small compared with the average kinetic energy of room temperature samples, and the +1/2 and _1/2 states are nearly equally populated. Indeed, in a field of 2.34 T the excess population of the lower energy state is only six nuclei per million. Although this is a very small difference , when we consider the number of atoms in a practical sample (remember the size of Avogadro's number), the numerical excess in the lower energy state is sufficient for selective and sensitive spectroscopic measurements. The diagram on the left below illustrates the macroscopic magnetization of a sample containing large numbers of spin 1/2 nuclei at equilibrium in a strong external magnetic field (Bo). A slight excess of +1/2 spin states precess randomly in alignment with the external field and a smaller population of _1/2 spin states precess randomly in an opposite alignment. An overall net magnetization therefore lies along the z-axis.
Net Macroscopic Magnetization of a Sample in an External Magnetic Field B0
Excitation by RF Energy and Subsequent Relaxation
The diagram and animation on the right show the changes in net macroscopic magnetization that occur as energy is introduced by rf irradiation at right angles to the external field. It is convenient to show the rf transmitter on the x-axis and the receiver-detector coil on the y-axis. On clicking the "Introduce RF Energy" button the animation will begin, and will repeat five times.
• First, the net magnetization shifts away from the z-axis and toward the y-axis. This occurs because some of the +1/2 nuclei are excited to the _1/2 state, and the precession about the z-axis becomes coherent (non-random), generating a significant y component to the net magnetization (M). The animation pauses at this stage.
• After irradiation the nuclear spins return to equilibrium in a process called relaxation. As the xy coherence disappears and the population of the +1/2 state increases, energy is released and detected by the receiver. The net magnetization spirals back, and eventually the equilibrium state is reestablished.
An inherent problem of the NMR experiment must be pointed out here. We have noted that the population difference between the spin states is proportionally very small. A fundamental requirement for absorption spectroscopy is a population imbalance between a lower energy ground state and a higher energy excited state. This can be expressed by the following equation, where A is a proportionality constant. If the mole fractions of the spin states are equal (η+ = η- ) then the population difference is zero and no absorption will occur. If the rf energy used in an NMR experiment is too high this saturation of the higher spin state will result and useful signals will disappear.
Relaxation Mechanisms
For NMR spectroscopy to be practical, an efficient mechanism for nuclei in the higher energy _1/2 spin state to return to the lower energy +1/2 state must exist. In other words, the spin population imbalance existing at equilibrium must be restored if spectroscopic observations are to continue. Now an isolated spinning nucleus will not spontaneouly change its spin state in the absence of external perturbation. Indeed, hydrogen gas (H2) exists as two stable spin isomers: ortho (parallel proton spins) and para (antiparallel spins). NMR spectroscopy is normally carried out in a liquid phase (solution or neat) so that there is close contact of sample molecules with a rapidly shifting crowd of other molecules (Brownian motion). This thermal motion of atoms and molecules generates local fluctuating electromagnetic fields, having components that match the Larmor frequency of the nucleus being studied. These local fields stimulate emission/absorption events that establish spin equilibrium, the excess spin energy being detected as it is released. This relaxation mechanism is called Spin-Lattice Relaxation (or Longitudinal Relaxation). The efficiency of spin-lattice relaxation depends on factors that influence molecular movement in the lattice, such as viscosity and temperature. The relaxation process is kinetically first order, and the reciprocal of the rate constant is a characteristic variable designated T1, the spin-lattice relaxation time. In non-viscous liquids at room temperature T1 ranges from 0.1 to 20 sec. A larger T1 indicates a slower or more inefficient spin relaxation. Another relaxation mechanism called spin-spin relaxation (or transverse relaxation) is characterized by a relaxation time T2. This process, which is actually a spin exchange, will not be discussed here.
Pulsed Fourier Transform Spectroscopy
In a given strong external magnetic field, each structurally distinct set of hydrogens in a molecule has a characteristic resonance frequency, just as each tubular chime in percussion instrument has a characteristic frequency. To discover the frequency of a chime we can strike it with a mallet and measure the sound emitted. This procedure can be repeated for each chime in the group so that all the characteristic frequencies are identified.
An alternative means of aquiring the same information is to strike all the chimes simultaneously, and to subject the complex collection of frequencies produced to mathematical analysis. In the following diagram the four frequencies assigned to our set of chimes are added together to give a complex summation wave. This is a straightforward conversion; and the reverse transformation, while not as simple, is readily accomplished, provided the combination signal is adequately examined and characterized.
A CW NMR spectrometer functions by irradiating each set of distinct nuclei in turn, a process analagous to striking each chime independently. For a high resolution spectrum this must be done slowly, and a 12 ppm sweep of the proton region takes from 5 to 10 minutes. It has proven much more efficient to excite all the proton nuclei in a molecule at the same time, followed by mathematical analysis of the complex rf resonance frequencies emitted as they relax back to the equilibrium state. This is the principle on which a pulse Fourier transform spectrometer operates. By exposing the sample to a very short (10 to 100 μsec), relatively strong (about 10,000 times that used for a CW spectrometer) burst of rf energy along the x-axis, as described above, all of the protons in the sample are excited simultaneously. The macroscopic magnetization model remains useful if we recognize it is a combination of megnetization vectors for all the nuclei that have been excited.
The overlapping resonance signals generated as the excited protons relax are collected by a computer and subjected to a Fourier transform mathematical analysis. As shown in the diagram on the left, the Fourier transform analysis, abbreviated FT, converts the complex time domain signal emitted by the sample into the frequency (or field) domain spectrum we are accustomed to seeing. In this fashion a complete spectrum can be acquired in a few seconds. Because the relaxation mechanism is a first order process, the rf signal emitted by the sample decays exponentially. This is called a free induction decay signal, abbreviated FID.
Free Induction Decay Signal
Since, the FID signal collected after one pulse, may be stored and averaged with the FID's from many other identical pulses prior to the Fourier transform, the NMR signal strength from a small sample may be enhanced to provide a useable spectrum. This has been essential to acquiring spectra from low abundance isotopes, such as 13C. In practice, the pulse FT experiment has proven so versatile that many variations of the technique, suited to special purposes, have been devised and used effectively.
Examples of Anisotropy Influences on Chemical Shift
The compound on the left has a chain of ten methylene groups linking para carbons of a benzene ring. Such bridged benzenes are called paracyclophanes. The meta analogs are also known. The structural constraints of the bridging chain require the middle two methylene groups to lie over the face of the benzene ring, which is a NMR shielding region. The four hydrogen atoms that are part of these groups display resonance signals that are more than two ppm higher field than the two methylene groups bonded to the edge of the ring (a deshielding region).
The 14 π-electron bridged annulene on the right is an aromatic (4n + 2) system, and has the same anisotropy as benzene. Nuclei located over the face of the ring are shielded, and those on the periphery are deshielded. The ring hydrogens give resonance signals in the range 8.0 to 8.7 δ, as expected from their deshielded location (note that there are three structurally different hydrogens on the ring). The two propyl groups are structurally equivalent (homotopic), and are free to rotate over the faces of the ring system (one above and one below). On average all the propyl hydrogens are shielded, with the innermost methylene being the most affected. The negative chemical shifts noted here indicate that the resonances occurs at a higher field than the TMS reference signal.
A remarkable characteristic of annulenes is that antiaromatic 4n π-electron systems are anisotropic in the opposite sense as their aromatic counterparts. A dramatic illustration of this fact is provided by the dianion derivative of the above bridged annulene. This dianion, formed by the addition of two electrons, is a 16 π-electron (4n) system. In the NMR spectrum of the dianion, the ring hydrogens resonate at high field (they are shielded), and the hydrogens of the propyl group are all shifted downfield (deshielded). The innermost methylene protons (magenta) give an NMR signal at +22.2 ppm, and the signals from the adjacent methylene and methyl hydrogens also have unexpectedly large chemical shifts.
Compounds in which two or more benzene rings are fused together include examples such as naphthalene, anthracene and phenanthrene, shown in the following diagram, present interesting insights into aromaticity and reactivity. The resonance stabilization of these compounds, calculated from heats of hydrogenation or combustion, is given beneath each structure.
Unlike benzene, the structures of these compounds show measurable double bond localization, which is reflected in their increased reactivity both in substitution and addition reactions. However, the 1HNMR spectra of these aromatic hydrocarbons do not provide much insight into the distribution of their pi-electrons. As expected, naphthalene displays two equally intense signals at δ 7.46 & 7.83 ppm. Likewise, anthracene shows three signals, two equal intensity multiplets at δ 7.44 & 7.98 ppm and a signal half as intense at δ 8.4 ppm. Thus, the influence of double bond localization or competition between benzene and higher annulene stabilization cannot be discerned.
The much larger C48H24 fused benzene ring cycle, named "kekulene" by Heinz Staab and sometimes called "superbenzene" by others, serves to probe the relative importance of benzenoid versus annulenoid aromaticity. A generic structure of this remarkable compound is drawn on the left below, together with two representative Kekule contributing structures on its right. There are some 200 Kekule structures that can be drawn for kekulene, but these two canonical forms represent extremes in aromaticity. The central formula has two [4n+2] annulenes, an inner [18]annulene and an outer [30]annulene (colored pink and blue respectively). The formula on the right has six benzene rings (colored green) joined in a ring by meta bonds, and held in a planar configuration by six cis-double bond bridges.
The coupled annulene contributor in the center has an energetically equivalent canonical form in which the single and double bonds making up the annulenes are exchanged. If these contributors dominate the aromatic character of kekulene, the 6 inside hydrogens should be shielded by the ring currents, and the 18 hydrogens on the periphery should be deshielded. Furthermore, the C:C bonds composing each annulene ring should have roughly equal lengths.
If the benzene contributor on the right (and its equivalent Kekule form) dominate the aromaticity of kekulene, all the benzene hydrogens will be deshielded, and the six double bond links on the periphery will have bond lengths characteristic of fixed single and double bonds
The extreme insolubility of kekulene made it difficult to grow suitable crystals for X-ray analysis or obtain solution NMR spectra. These problems were eventually solved by using high boiling solvents, the 1HNMR spectrum being taken at 150 to 200° C in deuterated tetrachlorobenzene solution. The experimental evidence demonstrates clearly that the hexa-benzene ring structure on the right most accurately represents kekulene. This evidence is shown below. The extremely low field resonance of the inside hydrogens is assigned from similar downfield shifts in model compounds.
It is important to understand that the shielding and deshielding terms used throughout our discussion of relative chemical shifts are themselves relative. Indeed, compared to a hypothetical isolated proton, all the protons in a covalent compound are shielded by the electrons in nearby sigma and pi-bonds. Consequently, it would be more accurate to describe chemical shift differences in terms of the absolute shielding experienced by different groups of hydrogens. There is, in fact, good evidence that the anisotropy of neighboring C-H and C-C sigma bonds, together with that of the bond to the observed hydrogen, are the dominate shielding factors influencing chemical shifts. The anisotropy of pi-electron systems augments this sigma skeletal shielding. Nevertheless, the deceptive focus on anisotropic pi-electron influences is so widely and commonly used that this view has been retained and employed in these pages.
Hydrogen Bonding Influences
Hydrogen bonding of hydroxyl and amino groups not only causes large variations in the chemical shift of the proton of the hydrogen bond, but also influences its coupling with adjacent C-H groups. As shown on the right, the 60 MHz proton NMR spectrum of pure (neat) methanol exhibits two signals, as expected. At 30° C these signals are sharp singlets located at δ 3.35 and 4.80 ppm, the higher-field methyl signal (magenta) being three times as strong as the OH signal (orange) at lower field. When cooled to -45 ° C, the larger higher-field signal changes to a doublet (J = 5.2 Hz) having the same chemical shift. The smaller signal moves downfield to δ 5.5 ppm and splits into a quartet (J = 5.2 Hz). The relative intensities of the two groups of signals remains unchanged. This interesting change in the NMR spectrum, which is shown in the two spectra below, is due to increased stability of hydrogen bonded species at lower temperature. Since hydrogen bonding not only causes a resonance shift to lower field, but also decreases the rate of intermolecular proton exchange, the hydroxyl proton remains bonded to the alkoxy group for a sufficient time to exert its spin coupling influence.
Under routine conditions, rapid intermolecular exchange of the OH protons of alcohols often prevents their coupling with adjacent hydrogens from being observed. Intermediate rates of proton exchange lead to a broadening of the OH and coupled hydrogen signals, a characteristic that is useful in identifying these functions. Since traces of acid or base catalyze this hydrogen exchange, pure compounds and clean sample tubes must be used for experiments of the kind described here.
Methanol 1H NMR at approximately Room Temperature
Methanol 1H NMR at -45 oC
Another way of increasing the concentration of hydrogen bonded methanol species is to change the solvent from chloroform-d to a solvent that is a stronger hydrogen bond acceptor. Examples of such solvents are given in the following table. In contrast to the neat methanol experiment described above, very dilute solutions are used for this study. Since chloroform is a poor hydrogen bond acceptor and the dilute solution reduces the concentration of methanol clusters, the hydroxyl proton of methanol generates a resonance signal at a much higher field than that observed for the pure alcohol. Indeed, the OH resonance signal from simple alcohols in dilute chloroform solution is normally found near δ 1.0 ppm.
The exceptionally strong hydrogen bond acceptor quality of DMSO is demonstrated here by the large downfield shift of the methanol hydroxyl proton, compared with a slight upfield shift of the methyl signal. The expected spin coupling patterns shown above are also observed in this solvent. Although acetone and acetonitrile are better hydrogen-bond acceptors than chloroform, they are not as effective as DMSO.
1H Chemical Shifts of Methanol in Selected Solvents
Solvent CDCl3 CD3COCD3 CD3SOCD3 CD3C≡N
CH3–O–H
CH3
O–H
3.40
1.10
3.31
3.12
3.16
4.01
3.28
2.16
The solvent effect shown above suggests a useful diagnostic procedure for characterizing the OH resonance signals from alcohol samples. For example, a solution of ethanol in chloroform-d displays the spectrum shown on the left below, especially if traces of HCl are present (otherwise broadening of the OH and CH2 signals occurs). Note that the chemical shift of the OH signal (red) is less than that of the methylene group (blue), and no coupling of the OH proton is apparent. The vicinal coupling (J = 7 Hz) of the methyl and methylene hydrogens is typical of ethyl groups. In DMSO-d6 solution small changes of chemical shift are seen for the methyl and methylene group hydrogens, but a dramatic downfield shift of the hydroxyl signal takes place because of hydrogen bonding. Coupling of the OH proton to the adjacent methylene group is evident, and both the coupling constants can be measured. Because the coupling constants are different, the methylene signal pattern is an overlapping doublet of quartets (eight distinct lines) rather than a quintet. Note that residual hydrogens in the solvent give a small broad signal near δ 2.5 ppm.
For many alcohols in dilute chloroform-d solution, the hydroxyl resonance signal is often broad and obscured by other signals in the δ 1.5 to 3.0 region. The simple technique of using DMSO-d6 as a solvent, not only shifts this signal to a lower field, but permits 1°-, 2 °- & 3 °-alcohols to be distinguished. Thus, the hydroxyl proton of 2-propanol generates a doublet at δ 4.35 ppm, and the corresponding signal from 2-methyl-2-propanol is a singlet at δ 4.2 ppm. The more acidic OH protons of phenols are similarly shifted – from δ 4 to 7 in chloroform-d to δ 8.5 to 9.5 in DMSO-d6.
Spin-Spin Coupling
vicinal (joined by three sigma bonds). In this case a neighboring proton having a +1/2 spin shifts the resonance frequency of the proton being observed to a slightly higher value (up to 7 Hz), and a _1/2 neighboring spin shifts it to a lower frequency. Remember that the total population of these two spin states is roughly equal, differing by only a few parts per million in a strong magnetic field. If several neighboring spins are present, their effect is additive.
In the spectrum of 1,1-dichloroethane shown on the right, it is clear that the three methyl hydrogens (red) are coupled with the single methyne hydrogen (orange) in a manner that causes the former to appear as a doublet and the latter as a quartet. The light gray arrow points to the unperturbed chemical shift location for each proton set. By clicking on one of these signals, the spin relationship that leads to the coupling pattern will be displayed. Clicking elsewhere in the picture will return the original spectrum.
Full Spectrum of 1,1-dichloroethane
2.06 ppm Signal Explained
5.89 ppm Signal Explained
The statistical distribution of spins within each set explains both the n+1 rule and the relative intensities of the lines within a splitting pattern. The action of a single neighboring proton is easily deduced from the fact that it must have one of two possible spins. Interaction of these two spin states with the nuclei under observation leads to a doublet located at the expected chemical shift. The corresponding action of the three protons of the methyl group requires a more detailed analysis. In the display of this interaction four possible arrays of their spins are shown. The mixed spin states are three times as possible as the all +1/2 or all _1/2 collection. Consequently, we expect four signals, two above the chemical shift and two below it. This spin analysis also suggests that the intensity ratio of these signals will be 1:3:3:1. The line separations in splitting patterns are measured in Hz, and are characteristic of the efficiency of the spin interaction; they are referred to as coupling constants (symbol J). In the above example, the common coupling constant is 6.0 Hz.
Multipliticy Relative Line Intensity (Starting at the top: Singlet, Doublet, Triplet, Quartet, and Quintet)
A simple way of estimating the relative intensities of the lines in a first-order coupling pattern is shown on the right. This array of numbers is known as Pascal's triangle, and is easily extended to predict higher multiplicities. The number appearing at any given site is the sum of the numbers linked to it from above by the light blue lines. Thus, the central number of the five quintet values is 3 + 3 = 6. Of course, a complete analysis of the spin distributions, as shown for the case of 1,1-dichloroethane above, leads to the same relative intensities.
Coupling constants are independent of the external magnetic field, and reflect the unique spin interaction characteristics of coupled sets of nuclei in a specific structure. As noted earlier, coupling constants may vary from a fraction of a Hz to nearly 20 Hz, important factors being the nature and spatial orientation of the bonds joining the coupled nuclei. In simple, freely rotating alkane units such as CH3CH2X or YCH2CH2X the coupling constant reflects an average of all significant conformers, and usually lies in a range of 6 to 8 Hz. This conformational mobility may be restricted by incorporating the carbon atoms in a rigid ring, and in this way the influence of the dihedral orientation of the coupled hydrogens may be studied.
The structures of cis and trans-4-tert-butyl-1-chlorocyclohexane, shown above, illustrate how the coupling constant changes with the dihedral angle (φ) between coupled hydrogens. The inductive effect of chlorine shifts the resonance frequency of the red colored hydrogen to a lower field (δ ca. 4.0), allowing it to be studied apart from the other hydrogens in the molecule. The preferred equatorial orientation of the large tert-butyl group holds the six-membered ring in the chair conformation depicted in the drawing. In the trans isomer this fixes the red hydrogen in an axial orientation; whereas for the cis isomer it is equatorial. The listed values for the dihedral angles and the corresponding coupling constants suggest a relationship, which has been confirmed and clarified by numerous experiments. This relationship is expressed by the Karplus equation shown below.
Geminal couplings are most commonly observed in cyclic structures, but are also evident when methylene groups have diastereomeric hydrogens.
Spin Decoupling
We have noted that rapidly exchanging hydroxyl hydrogens are not spin-coupled to adjacent C-H groups. The reason for this should be clear. As each exchange occurs, there will be an equal chance of the new proton having a +1/2 or a _1/2 spin (remember that the overall populations of the two spin states are nearly identical). Over time, therefore, the hydroxyl hydrogen behaves as though it is rapidly changing its spin, and the adjacent nuclei see only a zero spin average from it. If we could cause other protons in a molecule to undergo a similar spin averaging, their spin-coupling influence on adjacent nuclei would cease. Such NMR experiments are possible, and are called spin decoupling.
When a given set of nuclei is irradiated with strong rf energy at its characteristic Larmor frequency, spin saturation and rapid interconversion of the spin states occurs. Neighboring nuclei with different Larmor frequencies are no longer influenced by specific long-lived spins, so spin-spin signal splitting of the neighbors vanishes. The following spectrum of 1-nitropropane may be used to illustrate this technique. The three distinct sets of hydrogens in this molecule generate three resonance signals (two triplets and a broad sextet). A carefully tuned decoupling signal may be broadcast into the sample while the remaining spectrum is scanned. The region of the decoupling signal is obscured, but resonance signals more than 60 Hz away may still be seen. By clicking on one of the three signals in the spectrum, the results of decoupling at that frequency will be displayed.
The Influence of Magnetic Field Strength
The presence of symmetrical, easily recognized first-order splitting patterns in a NMR spectrum depends on the relative chemical shifts of the spin-coupled nuclei and the magnitude of the coupling constant. If the chemical shift difference (i.e. Δδ in Hz) is large compared to J the splitting patterns will be nearly first order. If, on the other hand, the difference is relatively small (less than 10 J) second order distortion of the signal splitting will be observed. One important advantage in using very high field magnets for NMR is that the separation (or dipersion) of different sets of protons is proportional to field strength, whereas coupling constants do not change.
It is important to remember that structurally different sets of nuclei do not always produce distinctly different signals in an NMR spectrum. For example, the hydrocarbon octane has four different sets of protons, as shown in the following formula:
CH3CH2CH2CH2CH2CH2CH2CH3
Now methyl hydrogens have a smaller chemical shift than methylene hydrogens, so methyl groups (colored black here) can usually be distinguished. However, the chemical shifts of the different methylene groups (blue, red & green) are so similar that many NMR spectrometers will not resolve them. Consequently, a 90 MHz proton spectrum of octane shows a distorted triplet at δ 0.9 ppm, produced by the six methyl protons, and a strong broad singlet at δ 1.2 ppm coming from all twelve methylene protons. A similar failure to resolve structurally different hydrogen atoms occurs in the case of alkyl substituted benzene rings. The chemical shift difference between ortho, meta and para hydrogens in such compounds is often so small that they are seen as a single resonance signal in an NMR spectrum. The 90 MHz spectrum of benzyl alcohol in chloroform-d solution provides an instructive example, shown below. A broad strong signal at δ 7.24 ppm is characteristic of the aromatic protons on alkylbenzenes. Since the chemical shifts of these hydrogens are nearly identical, no spin coupling is observed. If the magnetic field strength is increased to 400 Mz (lower spectrum) the aromatic protons are more dispersed (orange, magenta and green signals), and the spin coupling of adjacent hydrogens (J = 7.6 Hz) causes overlap of the signals (gray shaded enlargement).
1H NMR of Benzyl alcohol
1H NMR of Anisole
Anisole, an isomer of benzyl alcohol, has a more dispersed set of aromatic signals, thanks to the electron donating influence of the methoxy substituent. The 90 MHz spectrum of anisole shows this greater dispersion, but the spin coupling of adjacent hydrogens still results in signal overlap. The 400 Mz spectrum at the bottom illustrates the greater dispersion of the chemical shifts, and since the coupling constants remain unchanged, the splitting patterns no longer overlap. In all these examples a very small meta-hydrogen coupling has been ignored.
Example \(1\)
Not all simple compounds have simple proton NMR spectra. The following example not only illustrates this point, but also demonstrates how a careful structural analysis can rationalize an initially complex spectrum.
The 100 MHz 1H NMR spectrum of a C3H5ClO compound is initially displayed. This spectrum is obviously complex and not easily interpreted, except for concluding that no olefinic C-H protons are present.
With a higher field spectrum it is clear that each of the five hydrogen atoms in the molecule is structurally unique, and is producing a separate signal. Also, it is clear there is considerable spin coupling of all the hydrogens. To see the coupling patterns more clearly it is necessary to expand and enhance the spectrum in these regions. For purposes of our demonstration, this can be done by clicking on any one of the signal multiplets. Clicking in an open area should return the original 500 MHz display. In some of the expanded displays two adjacent groups of signals are shown. Once an enlarged pattern is displayed, the line separations in Hz can be measured (remember that for a 500 MHz spectrum 1 ppm is 500 Hz). The middle signal at 3.2 ppm is the most complex, and overlap of some multiplet lines has occurred.
Solution
This spectrum has several interesting features. First, hydrogens A, B & C are clearly different, and are spin-coupled to each other. Hydrogens B & C are geminally related, whereas A is oriented to B & C in a vicinal manner. Since JAB and JBC are similar, the HB signal is a broad triplet. Although hydrogens D & E might seem identical at first glance, they are diastereotopic, and should therefore have different chemical shifts. The DE geminal coupling constant is 11.7 Hz, so each of these hydrogens appears as a doublet of doublets.
The splitting of the HA signal is complex and not immediately obvious. The diagram shows the consequences of the four operating coupling constants.
Additional Information from 13C NMR Spectroscopy
Broad band decoupling of the hydrogen atoms in a molecule was an essential operation for obtaining simple (single line) carbon NMR spectra. The chemical shifts of the carbon signals provide useful information, but it would also be very helpful to know how many hydrogen atoms are bonded to each carbon. Aside from the fact that carbons having no bonded hydrogens generally give weak resonance signals, this information is not present in a completely decoupled spectrum.
Clever methods of retaining the hydrogen information while still enjoying the benefits of proton decoupling have been devised. The techniques involved are beyond the scope of this discussion, but the overall results can still be appreciated. The 13C NMR spectrum of camphor shown below will serve as an illustration. It will be helpful to view an expanded section of this spectrum from δ 0.0 to 50.0 ppm, and this will be presented in the High Field Expansion spectrum. The two lowest field signals are missing in the expanded display.
High Field Expansion of Camphor
Even though the expanded display now shows the distinct carbon signals clearly, the origin of each is ambiguous. An early method of regaining coupling information was by off-resonance decoupling. In this approach a weaker and more focused proton decoupling frequency is applied as the carbon spectrum is acquired. Vestiges of the C-H coupling remain in the carbon signals, but the apparent coupling constants are greatly reduced. View the Off-Resonance Decouple spectrum. The results of such an experiment will be displayed. Notice that all the methyl groups are quartets (three coupled hydrogens), the methylene groups are triplets and methine carbons are doublets. Overlap of two quartets near δ 19 ppm and the doublet and triplet near δ 43 ppm are complicating factors.
Off-Resonance Decouple Spectrum of Camphor
A better way for classifying the carbon signals is by a technique called INEPT (insensitive nuclear enhancement by polarization transfer). This method takes advantage of the influence of hydrogen on 13C relaxation times, and can be applied in several modes. One of the most common applications of INEPT separates the signals of methyl and methine carbons from those of methylene carbons by their sign. Carbons having no hydrogen substituents have a zero signal.
INEPT Spectrum of Camphor
Properties of Some Deuterated NMR Solvents
Solvent B.P. °C Residual
1H signal (δ)
Residual
13C signal (δ)
acetone-d6 55.5 2.05 ppm 206 & 29.8 ppm
acetonitrile-d3 80.7 1.95 ppm 118 & 1.3 ppm
benzene-d6 79.1 7.16 ppm 128 ppm
chloroform-d 60.9 7.27 ppm 77.2 ppm
cyclohexane-d12 78.0 1.38 ppm 26.4 ppm
dichloromethane-d2 40.0 5.32 ppm 53.8 ppm
dimethylsulfoxide-d6 190 2.50 ppm 39.5 ppm
nitromethane-d3 100 4.33 ppm 62.8 ppm
pyridine-d5 114 7.19, 7.55 & 8.71 ppm 150, 135.5 & 123.5 ppm
tetrahydrofuran-d8 65.0 1.73 & 3.58 ppm 67.4 & 25.2 ppm
Contributors
• Layne Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.02%3A_Fourier_Transform_NMR.txt |
5.4A: Diamagnetic shielding and deshielding
We come now to the question of why nonequivalent protons have different chemical shifts. The chemical shift of a given proton is determined primarily by its immediate electronic environment. Consider the methane molecule (CH4), in which the protons have a chemical shift of 0.23 ppm. The valence electrons around the methyl carbon, when subjected to B0, are induced to circulate and thus generate their own very small magnetic field that opposes B0. This induced field, to a small but significant degree, shields the nearby protons from experiencing the full force of B0, an effect known as local diamagnetic shielding. The methane protons therefore do not experience the full force of B0 - what they experience is called Beff, or the effective field, which is slightly weaker than B0.
Therefore, their resonance frequency is slightly lower than what it would be if they did not have electrons nearby to shield them.
Now consider methyl fluoride, CH3F, in which the protons have a chemical shift of 4.26 ppm, significantly higher than that of methane. This is caused by something called the deshielding effect. Because fluorine is more electronegative than carbon, it pulls valence electrons away from the carbon, effectively decreasing the electron density around each of the protons. For the protons, lower electron density means less diamagnetic shielding, which in turn means a greater overall exposure to B0, a stronger Beff, and a higher resonance frequency. Put another way, the fluorine, by pulling electron density away from the protons, is deshielding them, leaving them more exposed to B0. As the electronegativity of the substituent increases, so does the extent of deshielding, and so does the chemical shift. This is evident when we look at the chemical shifts of methane and three halomethane compounds (remember that electronegativity increases as we move up a column in the periodic table).
To a large extent, then, we can predict trends in chemical shift by considering how much deshielding is taking place near a proton. The chemical shift of trichloromethane is, as expected, higher than that of dichloromethane, which is in turn higher than that of chloromethane.
The deshielding effect of an electronegative substituent diminishes sharply with increasing distance:
The presence of an electronegative oxygen, nitrogen, sulfur, or sp2-hybridized carbon also tends to shift the NMR signals of nearby protons slightly downfield:
Table 2 lists typical chemical shift values for protons in different chemical environments.
Armed with this information, we can finally assign the two peaks in the the 1H-NMR spectrum of methyl acetate that we saw a few pages back. The signal at 3.65 ppm corresponds to the methyl ester protons (Hb), which are deshielded by the adjacent oxygen atom. The upfield signal at 2.05 ppm corresponds to the acetate protons (Ha), which is deshielded - but to a lesser extent - by the adjacent carbonyl group.
Finally, a note on the use of TMS as a standard in NMR spectroscopy: one of the main reasons why the TMS proton signal was chosen as a zero-point is that the TMS protons are highly shielded: silicon is slightly less electronegative than carbon, and therefore donates some additional shielding electron density. Very few organic molecules contain protons with chemical shifts that are negative relative to TMS.
5.4B: The chemical shifts of aromatic and vinylic protons
Some protons resonate much further downfield than can be accounted for simply by the deshielding effect of nearby electronegative atoms. Vinylic protons (those directly bonded to an alkene carbon) and aromatic (benzylic) protons are dramatic examples.
We'll consider the aromatic proton first. Recall that in benzene and many other aromatic structures, a sextet of pelectrons is delocalized around the ring. When the molecule is exposed to B0, these pelectrons begin to circulate in a ring current, generating their own induced magnetic field that opposes B0. In this case, however, the induced field of the pelectrons does not shield the benzylic protons from B0 as you might expect– rather, it causes the protons to experience a stronger magnetic field in the direction of B0 – in other words, it adds to B0 rather than subtracting from it.
To understand how this happens, we need to understand the concept of diamagnetic anisotropy (anisotropy means `non-uniformity`). So far, we have been picturing magnetic fields as being oriented in a uniform direction. This is only true over a small area. If we step back and take a wider view, however, we see that the lines of force in a magnetic field are actually anisotropic. They start in the 'north' direction, then loop around like a snake biting its own tail.
If we are at point A in the figure above, we feel a magnetic field pointing in a northerly direction. If we are at point B, however, we feel a field pointing to the south.
In the induced field generated by the aromatic ring current, the benzylic protons are at the equivalent of ‘point B’ – this means that the induced current in this region of space is oriented in the same direction as B0.
In total, the benzylic protons are subjected to three magnetic fields: the applied field (B0) and the induced field from the pelectrons pointing in one direction, and the induced field of the non-aromatic electrons pointing in the opposite (shielding) direction. The end result is that benzylic protons, due to the anisotropy of the induced field generated by the pring current, appear to be highly deshielded. Their chemical shift is far downfield, in the 6.5-8 ppm region.
Example 5.5
The 1H-NMR spectrum of [18] annulene has two peaks, at 8.9 ppm and -1.8 ppm (upfield of TMS!) with an integration ratio of 2:1. Explain the unusual chemical shift of the latter peak.
Solution
Diamagnetic anisotropy is also responsible for the downfield chemical shifts of vinylic protons and aldehyde protons (4.5-6.5 ppm and 9-10 ppm, respectively). These groups are not aromatic and thus do not generate ring currents as does benzene, but the pelectrons circulate in such a way as to generate a magnetic field that adds to B0in the regions of space occupied by the protons.
5.4C: Hydrogen-bonded protons
Protons that are directly bonded to oxygen and nitrogen have chemical shifts that can vary widely depending on solvent and concentration. This is because these protons can participate to varying degrees in hydrogen-bonding interactions, and hydrogen bonding greatly influences the electron density around the proton. Signals for hydrogen-bonding protons also tend to be broader than those of hydrogens bonded to carbon, a phenomenon that is also due to hydrogen bonding.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.03%3A_Shielding_Causes_Different_Hydrogens_to_Show_Signals_at_Different_Frequencies.txt |
If all protons in all organic molecules had the same resonance frequency in an external magnetic field of a given strength, the information in the previous paragraph would be interesting from a theoretical standpoint, but would not be terribly useful to organic chemists. Fortunately for us, however, resonance frequencies are not uniform for all protons in a molecule. In an external magnetic field of a given strength, protons in different locations in a molecule have different resonance frequencies, because they are in non-identical electronic environments. In methyl acetate, for example, there are two ‘sets’ of protons. The three protons labeled Ha have a different - and easily distinguishable – resonance frequency than the three Hb protons, because the two sets of protons are in non-identical environments: they are, in other words, chemically nonequivalent.
On the other hand, the three Ha protons are all in the same electronic environment, and are chemically equivalent to one another. They have identical resonance frequencies. The same can be said for the three Hb protons.
The ability to recognize chemical equivalancy and nonequivalency among atoms in a molecule will be central to understanding NMR. In each of the molecules below, all protons are chemically equivalent, and therefore will have the same resonance frequency in an NMR experiment.
You might expect that the equitorial and axial hydrogens in cyclohexane would be non-equivalent, and would have different resonance frequencies. In fact, an axial hydrogen is in a different electronic environment than an equitorial hydrogen. Remember, though, that the molecule rotates rapidly between its two chair conformations, meaning that any given hydrogen is rapidly moving back and forth between equitorial and axial positions. It turns out that, except at extremely low temperatures, this rotational motion occurs on a time scale that is much faster than the time scale of an NMR experiment.
In this sense, NMR is like a camera that takes photographs of a rapidly moving object with a slow shutter speed - the result is a blurred image. In NMR terms, this means that all 12 protons in cyclohexane are equivalent.
Each the molecules in the next figure contains two sets of protons, just like our previous example of methyl acetate, and again in each case the resonance frequency of the Ha protons will be different from that of the Hb protons.
Notice how the symmetry of para-xylene results in there being only two different sets of protons.
Most organic molecules have several sets of protons in different chemical environments, and each set, in theory, will have a different resonance frequency in 1H-NMR spectroscopy.
When stereochemistry is taken into account, the issue of equivalence vs nonequivalence in NMR starts to get a little more complicated. It should be fairly intuitive that hydrogens on different sides of asymmetric ring structures and double bonds are in different electronic environments, and thus are non-equivalent and have different resonance frequencies. In the alkene and cyclohexene structures below, for example, Ha is trans to the chlorine substituent, while Hb is cis to chlorine.
What is not so intuitive is that diastereotopic hydrogens (section 3.10) on chiral molecules are also non-equivalent:
However, enantiotopic and homotopic hydrogens are chemically equivalent.
Example 5.1
How many different sets of protons do the following molecules contain? (count diastereotopic protons as non-equivalent).
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.04%3A_The_Number_of_Signals_in_an_%281H%29_NMR_Spectrum.txt |
5.3A: The basics of an NMR experiment
Given that chemically nonequivalent protons have different resonance frequencies in the same applied magnetic field, we can see how NMR spectroscopy can provide us with useful information about the structure of an organic molecule. A full explanation of how a modern NMR instrument functions is beyond the scope of this text, but in very simple terms, here is what happens. First, a sample compound (we'll use methyl acetate) is placed inside a very strong applied magnetic field (B0).
All of the protons begin to precess: the Ha protons at precessional frequency ωa, the Hb protons at ω b.At first, the magnetic moments of (slightly more than) half of the protons are aligned with B0, and half are aligned against B0. Then, the sample is hit with electromagnetic radiation in the radio frequency range. The two specific frequencies which match ωaandωb(i.e. the resonance frequencies) cause those Ha and Hb protons which are aligned with B0 to 'flip' so that they are now aligned against B0. In doing so, the protons absorb radiation at the two resonance frequencies. The NMR instrument records which frequencies were absorbed, as well as the intensity of each absorbance.
In most cases, a sample being analyzed by NMR is in solution. If we use a common laboratory solvent (diethyl ether, acetone, dichloromethane, ethanol, water, etc.) to dissolve our NMR sample, however, we run into a problem – there many more solvent protons in solution than there are sample protons, so the signals from the sample protons will be overwhelmed. To get around this problem, we use special NMR solvents in which all protons have been replaced by deuterium. Recall that deuterium is NMR-active, but its resonance frequency is very different from that of protons, and thus it is `invisible` in 1H-NMR. Some common NMR solvents are shown below.
5.3B: The chemical shift
Let's look at an actual 1H-NMR plot for methyl acetate. Just as in IR and UV-vis spectroscopy, the vertical axis corresponds to intensity of absorbance, the horizontal axis to frequency (typically the vertical axis is not shown in an NMR spectrum).
We see three absorbance signals: two of these correspond to Ha and Hb, while the peak at the far right of the spectrum corresponds to the 12 chemically equivalent protons in tetramethylsilane (TMS), a standard reference compound that was added to our sample.
You may be wondering about a few things at this point - why is TMS necessary, and what is the meaning of the `ppm (δ)` label on the horizontal axis? Shouldn't the frequency units be in Hz? Keep in mind that NMR instruments of many different applied field strengths are used in organic chemistry laboratories, and that the proton's resonance frequency range depends on the strength of the applied field. The spectrum above was generated on an instrument with an applied field of approximately 7.1 Tesla, at which strength protons resonate in the neighborhood of 300 million Hz (chemists refer to this as a 300 MHz instrument). If our colleague in another lab takes the NMR spectrum of the same molecule using an instrument with a 2.4 Tesla magnet, the protons will resonate at around 100 million Hz (so we’d call this a 100 MHz instrument). It would be inconvenient and confusing to always have to convert NMR data according to the field strength of the instrument used. Therefore, chemists report resonance frequencies not as absolute values in Hz, but rather as values relative to a common standard, generally the signal generated by the protons in TMS. This is where the ppm – parts per million – term comes in. Regardless of the magnetic field strength of the instrument being used, the resonance frequency of the 12 equivalent protons in TMS is defined as a zero point. The resonance frequencies of protons in the sample molecule are then reported in terms of how much higher they are, in ppm, relative to the TMS signal (almost all protons in organic molecules have a higher resonance frequency than those in TMS, for reasons we shall explore quite soon).
The two proton groups in our methyl acetate sample are recorded as resonating at frequencies 2.05 and 3.67 ppm higher than TMS. One-millionth (1.0 ppm) of 300 MHz is 300 Hz. Thus 2.05 ppm, on this instrument, corresponds to 615 Hz, and 3.67 ppm corresponds to 1101 Hz. If the TMS protons observed by our 7.1 Tesla instrument resonate at exactly 300,000,000 Hz, this means that the protons in our ethyl acetate samples are resonating at 300,000,615 and 300,001,101 Hz, respectively. Likewise, if the TMS protons in our colleague's 2.4 Tesla instrument resonate at exactly 100 MHz, the methyl acetate protons in her sample resonate at 100,000,205 and 100,000,367 Hz (on the 100 MHz instrument, 1.0 ppm corresponds to 100 Hz). The absolute frequency values in each case are not very useful – they will vary according to the instrument used – but the difference in resonance frequency from the TMS standard, expressed in parts per million, should be the same regardless of the instrument.
Expressed this way, the resonance frequency for a given proton in a molecule is called its chemical shift. A frequently used symbolic designation for chemical shift in ppm is the lower-case Greek letter delta (δ). Most protons in organic compounds have chemical shift values between 0 and 12 ppm from TMS, although values below zero and above 12 are occasionally observed. By convention, the left-hand side of an NMR spectrum (higher chemical shift) is called downfield, and the right-hand direction is called upfield.
In our methyl acetate example we included for illustrative purposes a small amount of TMS standard directly in the sample, as was the common procedure for determining the zero point with older NMR instruments．That practice is generally no longer necessary, as modern NMR instruments are designed to use the deuterium signal from the solvent as a standard reference point, then to extrapolate the 0 ppm baseline that corresponds to the TMS proton signal (in an applied field of 7.1 Tesla, the deuterium atom in CDCl3 resonates at 32 MHz, compared to 300 MHz for the protons in TMS). In the remaining NMR spectra that we will see in this text we will not see an actual TMS signal, but we can always assume that the 0 ppm point corresponds to where the TMS protons would resonate if they were present.
Example 5.2
A proton has a chemical shift (relative to TMS) of 4.56 ppm.
1. What is its chemical shift, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
2. What is its resonance frequency, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
(Assume that in these instruments, the TMS protons resonate at exactly 300 or 200 MHz, respectively)
Solution
5.3C: Signal integration
The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals. This is very useful, because in 1H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds. The two signals in the methyl acetate spectrum, for example, integrate to approximately the same area, because they both correspond to a set of three equivalent protons.
Take a look next at the spectrum of para-xylene (IUPAC name 1,4-dimethylbenzene):
This molecule has two sets of protons: the six methyl (Ha) protons and the four aromatic (Hb) protons. When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. This (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal.
The integration function can also be used to determine the relative amounts of two or more compounds in a mixed sample. If we have a sample that is a 50:50 (mole/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons. If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons.
Example 5.3
You take a 1H-NMR spectrum of a mixed sample of acetone (CH3(CO)CH3) and dichloromethane (CH2Cl2). The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1. What is the molar ratio of the two compounds in the sample?
Solution
Example 5.4
You take the 1H-NMR spectrum of a mixed sample of 36% para-xylene and 64% acetone in CDCl3 solvent (structures are shown earlier in this chapter). How many peaks do you expect to see? What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0)
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.05%3A_The_Chemical_Shift_Tells_How_Far_the_Signal_Is_from_the_Reference_Signal.txt |
5.4A: Diamagnetic shielding and deshielding
We come now to the question of why nonequivalent protons have different chemical shifts. The chemical shift of a given proton is determined primarily by its immediate electronic environment. Consider the methane molecule (CH4), in which the protons have a chemical shift of 0.23 ppm. The valence electrons around the methyl carbon, when subjected to B0, are induced to circulate and thus generate their own very small magnetic field that opposes B0. This induced field, to a small but significant degree, shields the nearby protons from experiencing the full force of B0, an effect known as local diamagnetic shielding. The methane protons therefore do not experience the full force of B0 - what they experience is called Beff, or the effective field, which is slightly weaker than B0.
Therefore, their resonance frequency is slightly lower than what it would be if they did not have electrons nearby to shield them.
Now consider methyl fluoride, CH3F, in which the protons have a chemical shift of 4.26 ppm, significantly higher than that of methane. This is caused by something called the deshielding effect. Because fluorine is more electronegative than carbon, it pulls valence electrons away from the carbon, effectively decreasing the electron density around each of the protons. For the protons, lower electron density means less diamagnetic shielding, which in turn means a greater overall exposure to B0, a stronger Beff, and a higher resonance frequency. Put another way, the fluorine, by pulling electron density away from the protons, is deshielding them, leaving them more exposed to B0. As the electronegativity of the substituent increases, so does the extent of deshielding, and so does the chemical shift. This is evident when we look at the chemical shifts of methane and three halomethane compounds (remember that electronegativity increases as we move up a column in the periodic table).
To a large extent, then, we can predict trends in chemical shift by considering how much deshielding is taking place near a proton. The chemical shift of trichloromethane is, as expected, higher than that of dichloromethane, which is in turn higher than that of chloromethane.
The deshielding effect of an electronegative substituent diminishes sharply with increasing distance:
The presence of an electronegative oxygen, nitrogen, sulfur, or sp2-hybridized carbon also tends to shift the NMR signals of nearby protons slightly downfield:
Table 2 lists typical chemical shift values for protons in different chemical environments.
Armed with this information, we can finally assign the two peaks in the the 1H-NMR spectrum of methyl acetate that we saw a few pages back. The signal at 3.65 ppm corresponds to the methyl ester protons (Hb), which are deshielded by the adjacent oxygen atom. The upfield signal at 2.05 ppm corresponds to the acetate protons (Ha), which is deshielded - but to a lesser extent - by the adjacent carbonyl group.
Finally, a note on the use of TMS as a standard in NMR spectroscopy: one of the main reasons why the TMS proton signal was chosen as a zero-point is that the TMS protons are highly shielded: silicon is slightly less electronegative than carbon, and therefore donates some additional shielding electron density. Very few organic molecules contain protons with chemical shifts that are negative relative to TMS.
5.4B: The chemical shifts of aromatic and vinylic protons
Some protons resonate much further downfield than can be accounted for simply by the deshielding effect of nearby electronegative atoms. Vinylic protons (those directly bonded to an alkene carbon) and aromatic (benzylic) protons are dramatic examples.
We'll consider the aromatic proton first. Recall that in benzene and many other aromatic structures, a sextet of pelectrons is delocalized around the ring. When the molecule is exposed to B0, these pelectrons begin to circulate in a ring current, generating their own induced magnetic field that opposes B0. In this case, however, the induced field of the pelectrons does not shield the benzylic protons from B0 as you might expect– rather, it causes the protons to experience a stronger magnetic field in the direction of B0 – in other words, it adds to B0 rather than subtracting from it.
To understand how this happens, we need to understand the concept of diamagnetic anisotropy (anisotropy means `non-uniformity`). So far, we have been picturing magnetic fields as being oriented in a uniform direction. This is only true over a small area. If we step back and take a wider view, however, we see that the lines of force in a magnetic field are actually anisotropic. They start in the 'north' direction, then loop around like a snake biting its own tail.
If we are at point A in the figure above, we feel a magnetic field pointing in a northerly direction. If we are at point B, however, we feel a field pointing to the south.
In the induced field generated by the aromatic ring current, the benzylic protons are at the equivalent of ‘point B’ – this means that the induced current in this region of space is oriented in the same direction as B0.
In total, the benzylic protons are subjected to three magnetic fields: the applied field (B0) and the induced field from the pelectrons pointing in one direction, and the induced field of the non-aromatic electrons pointing in the opposite (shielding) direction. The end result is that benzylic protons, due to the anisotropy of the induced field generated by the pring current, appear to be highly deshielded. Their chemical shift is far downfield, in the 6.5-8 ppm region.
Example 5.5
The 1H-NMR spectrum of [18] annulene has two peaks, at 8.9 ppm and -1.8 ppm (upfield of TMS!) with an integration ratio of 2:1. Explain the unusual chemical shift of the latter peak.
Solution
Diamagnetic anisotropy is also responsible for the downfield chemical shifts of vinylic protons and aldehyde protons (4.5-6.5 ppm and 9-10 ppm, respectively). These groups are not aromatic and thus do not generate ring currents as does benzene, but the pelectrons circulate in such a way as to generate a magnetic field that adds to B0in the regions of space occupied by the protons.
5.4C: Hydrogen-bonded protons
Protons that are directly bonded to oxygen and nitrogen have chemical shifts that can vary widely depending on solvent and concentration. This is because these protons can participate to varying degrees in hydrogen-bonding interactions, and hydrogen bonding greatly influences the electron density around the proton. Signals for hydrogen-bonding protons also tend to be broader than those of hydrogens bonded to carbon, a phenomenon that is also due to hydrogen bonding.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
14.07: Characteristic Values of Chemical Shifts
Chemical shift values are in parts per million (ppm) relative to tetramethylsilane.
Hydrogen type
Chemical shift (ppm)
RCH3
0.9 - 1.0
RCH2R
1.2 - 1.7
R3CH
1.5 – 2.0
2.0 – 2.3
1.5 – 1.8
RNH2
1 - 3
ArCH3
2.2 – 2.4
2.3 – 3.0
ROCH3
3.7 – 3.9
3.7 – 3.9
ROH
1 - 5
3.7 – 6.5
5 - 9
ArH
6.0 – 8.7
9.5 – 10.0
10 - 13
Contributors
• Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.06%3A_The_Relative_Positions_of_%281H%29_NMR_Signals.txt |
Nuclear spin may be related to the nucleon composition of a nucleus in the following manner:
• Odd mass nuclei (i.e. those having an odd number of nucleons) have fractional spins.
• Examples are I = 1/2 ( 1H, 13C, 19F ), I = 3/2 ( 11B ) & I = 5/2 ( 17O ).
• Even mass nuclei composed of odd numbers of protons and neutrons have integral spins. Examples are I = 1 ( 2H, 14N ).
• Even mass nuclei composed of even numbers of protons and neutrons have zero spin ( I = 0 ). Examples are 12C, and 16O.
Spin Properties of Nuclei
Spin 1/2 nuclei have a spherical charge distribution, and their NMR behavior is the easiest to understand. Other spin nuclei have nonspherical charge distributions and may be analyzed as prolate or oblate spinning bodies. All nuclei with non-zero spins have magnetic moments (μ), but the nonspherical nuclei also have an electric quadrupole moment (eQ). Some characteristic properties of selected nuclei are given in the following table.
Isotope
Natural %
Abundance
Spin (I)
Magnetic
Moment (μ)*
Magnetogyric
Ratio (γ)
1H
99.9844 1/2 2.7927 26.753
2H
0.0156 1 0.8574 4,107
11B
81.17 3/2 2.6880 --
13C
1.108 1/2 0.7022 6,728
17O
0.037 5/2 -1.8930 -3,628
19F
100.0 1/2 2.6273 25,179
29Si
4.700 1/2 -0.5555 -5,319
31P
100.0 1/2 1.1305 10,840
* μ in units of nuclear magnetons = 5.05078•10-27 JT-1
γ in units of 107rad T-1 sec-1
A Model for NMR Spectroscopy
The model of a spinning nuclear magnet aligned with or against an external magnetic field (for I = 1/2 nuclei) must be refined for effective interpretation of NMR phenomena. Just as a spinning mass will precess in a gravitational field (a gyroscope), the magnetic moment μ associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The frequency of precession is proportional to the strength of the magnetic field, as noted by the equation: ωo = γBo. The frequency ωo is called the Larmor frequency and has units of radians per second. The proportionality constant γ is known as the gyromagnetic ratio and is proportional to the magnetic moment (γ = 2pm/hI). Some characteristic γ's were listed in the preceding table of nuclear properties.
A Spinning Gyroscope in a Gravity Field
magnetic moment
μ
A Spinning Charge in a Magnetic Field
If rf energy having a frequency matching the Larmor frequency is introduced at a right angle to the external field (e.g. along the x-axis), the precessing nucleus will absorb energy and the magnetic moment will flip to its I = _1/2 state. This excitation is shown in the following diagram. Note that frequencies in radians per second may be converted to Hz (cps) by dividing by 2π.
The energy difference between nuclear spin states is small compared with the average kinetic energy of room temperature samples, and the +1/2 and _1/2 states are nearly equally populated. Indeed, in a field of 2.34 T the excess population of the lower energy state is only six nuclei per million. Although this is a very small difference , when we consider the number of atoms in a practical sample (remember the size of Avogadro's number), the numerical excess in the lower energy state is sufficient for selective and sensitive spectroscopic measurements. The diagram on the left below illustrates the macroscopic magnetization of a sample containing large numbers of spin 1/2 nuclei at equilibrium in a strong external magnetic field (Bo). A slight excess of +1/2 spin states precess randomly in alignment with the external field and a smaller population of _1/2 spin states precess randomly in an opposite alignment. An overall net magnetization therefore lies along the z-axis.
Net Macroscopic Magnetization of a Sample in an External Magnetic Field B0
Excitation by RF Energy and Subsequent Relaxation
The diagram and animation on the right show the changes in net macroscopic magnetization that occur as energy is introduced by rf irradiation at right angles to the external field. It is convenient to show the rf transmitter on the x-axis and the receiver-detector coil on the y-axis. On clicking the "Introduce RF Energy" button the animation will begin, and will repeat five times.
• First, the net magnetization shifts away from the z-axis and toward the y-axis. This occurs because some of the +1/2 nuclei are excited to the _1/2 state, and the precession about the z-axis becomes coherent (non-random), generating a significant y component to the net magnetization (M). The animation pauses at this stage.
• After irradiation the nuclear spins return to equilibrium in a process called relaxation. As the xy coherence disappears and the population of the +1/2 state increases, energy is released and detected by the receiver. The net magnetization spirals back, and eventually the equilibrium state is reestablished.
An inherent problem of the NMR experiment must be pointed out here. We have noted that the population difference between the spin states is proportionally very small. A fundamental requirement for absorption spectroscopy is a population imbalance between a lower energy ground state and a higher energy excited state. This can be expressed by the following equation, where A is a proportionality constant. If the mole fractions of the spin states are equal (η+ = η- ) then the population difference is zero and no absorption will occur. If the rf energy used in an NMR experiment is too high this saturation of the higher spin state will result and useful signals will disappear.
Relaxation Mechanisms
For NMR spectroscopy to be practical, an efficient mechanism for nuclei in the higher energy _1/2 spin state to return to the lower energy +1/2 state must exist. In other words, the spin population imbalance existing at equilibrium must be restored if spectroscopic observations are to continue. Now an isolated spinning nucleus will not spontaneouly change its spin state in the absence of external perturbation. Indeed, hydrogen gas (H2) exists as two stable spin isomers: ortho (parallel proton spins) and para (antiparallel spins). NMR spectroscopy is normally carried out in a liquid phase (solution or neat) so that there is close contact of sample molecules with a rapidly shifting crowd of other molecules (Brownian motion). This thermal motion of atoms and molecules generates local fluctuating electromagnetic fields, having components that match the Larmor frequency of the nucleus being studied. These local fields stimulate emission/absorption events that establish spin equilibrium, the excess spin energy being detected as it is released. This relaxation mechanism is called Spin-Lattice Relaxation (or Longitudinal Relaxation). The efficiency of spin-lattice relaxation depends on factors that influence molecular movement in the lattice, such as viscosity and temperature. The relaxation process is kinetically first order, and the reciprocal of the rate constant is a characteristic variable designated T1, the spin-lattice relaxation time. In non-viscous liquids at room temperature T1 ranges from 0.1 to 20 sec. A larger T1 indicates a slower or more inefficient spin relaxation. Another relaxation mechanism called spin-spin relaxation (or transverse relaxation) is characterized by a relaxation time T2. This process, which is actually a spin exchange, will not be discussed here.
Pulsed Fourier Transform Spectroscopy
In a given strong external magnetic field, each structurally distinct set of hydrogens in a molecule has a characteristic resonance frequency, just as each tubular chime in percussion instrument has a characteristic frequency. To discover the frequency of a chime we can strike it with a mallet and measure the sound emitted. This procedure can be repeated for each chime in the group so that all the characteristic frequencies are identified.
An alternative means of aquiring the same information is to strike all the chimes simultaneously, and to subject the complex collection of frequencies produced to mathematical analysis. In the following diagram the four frequencies assigned to our set of chimes are added together to give a complex summation wave. This is a straightforward conversion; and the reverse transformation, while not as simple, is readily accomplished, provided the combination signal is adequately examined and characterized.
A CW NMR spectrometer functions by irradiating each set of distinct nuclei in turn, a process analagous to striking each chime independently. For a high resolution spectrum this must be done slowly, and a 12 ppm sweep of the proton region takes from 5 to 10 minutes. It has proven much more efficient to excite all the proton nuclei in a molecule at the same time, followed by mathematical analysis of the complex rf resonance frequencies emitted as they relax back to the equilibrium state. This is the principle on which a pulse Fourier transform spectrometer operates. By exposing the sample to a very short (10 to 100 μsec), relatively strong (about 10,000 times that used for a CW spectrometer) burst of rf energy along the x-axis, as described above, all of the protons in the sample are excited simultaneously. The macroscopic magnetization model remains useful if we recognize it is a combination of megnetization vectors for all the nuclei that have been excited.
The overlapping resonance signals generated as the excited protons relax are collected by a computer and subjected to a Fourier transform mathematical analysis. As shown in the diagram on the left, the Fourier transform analysis, abbreviated FT, converts the complex time domain signal emitted by the sample into the frequency (or field) domain spectrum we are accustomed to seeing. In this fashion a complete spectrum can be acquired in a few seconds. Because the relaxation mechanism is a first order process, the rf signal emitted by the sample decays exponentially. This is called a free induction decay signal, abbreviated FID.
Free Induction Decay Signal
Since, the FID signal collected after one pulse, may be stored and averaged with the FID's from many other identical pulses prior to the Fourier transform, the NMR signal strength from a small sample may be enhanced to provide a useable spectrum. This has been essential to acquiring spectra from low abundance isotopes, such as 13C. In practice, the pulse FT experiment has proven so versatile that many variations of the technique, suited to special purposes, have been devised and used effectively.
Examples of Anisotropy Influences on Chemical Shift
The compound on the left has a chain of ten methylene groups linking para carbons of a benzene ring. Such bridged benzenes are called paracyclophanes. The meta analogs are also known. The structural constraints of the bridging chain require the middle two methylene groups to lie over the face of the benzene ring, which is a NMR shielding region. The four hydrogen atoms that are part of these groups display resonance signals that are more than two ppm higher field than the two methylene groups bonded to the edge of the ring (a deshielding region).
The 14 π-electron bridged annulene on the right is an aromatic (4n + 2) system, and has the same anisotropy as benzene. Nuclei located over the face of the ring are shielded, and those on the periphery are deshielded. The ring hydrogens give resonance signals in the range 8.0 to 8.7 δ, as expected from their deshielded location (note that there are three structurally different hydrogens on the ring). The two propyl groups are structurally equivalent (homotopic), and are free to rotate over the faces of the ring system (one above and one below). On average all the propyl hydrogens are shielded, with the innermost methylene being the most affected. The negative chemical shifts noted here indicate that the resonances occurs at a higher field than the TMS reference signal.
A remarkable characteristic of annulenes is that antiaromatic 4n π-electron systems are anisotropic in the opposite sense as their aromatic counterparts. A dramatic illustration of this fact is provided by the dianion derivative of the above bridged annulene. This dianion, formed by the addition of two electrons, is a 16 π-electron (4n) system. In the NMR spectrum of the dianion, the ring hydrogens resonate at high field (they are shielded), and the hydrogens of the propyl group are all shifted downfield (deshielded). The innermost methylene protons (magenta) give an NMR signal at +22.2 ppm, and the signals from the adjacent methylene and methyl hydrogens also have unexpectedly large chemical shifts.
Compounds in which two or more benzene rings are fused together include examples such as naphthalene, anthracene and phenanthrene, shown in the following diagram, present interesting insights into aromaticity and reactivity. The resonance stabilization of these compounds, calculated from heats of hydrogenation or combustion, is given beneath each structure.
Unlike benzene, the structures of these compounds show measurable double bond localization, which is reflected in their increased reactivity both in substitution and addition reactions. However, the 1HNMR spectra of these aromatic hydrocarbons do not provide much insight into the distribution of their pi-electrons. As expected, naphthalene displays two equally intense signals at δ 7.46 & 7.83 ppm. Likewise, anthracene shows three signals, two equal intensity multiplets at δ 7.44 & 7.98 ppm and a signal half as intense at δ 8.4 ppm. Thus, the influence of double bond localization or competition between benzene and higher annulene stabilization cannot be discerned.
The much larger C48H24 fused benzene ring cycle, named "kekulene" by Heinz Staab and sometimes called "superbenzene" by others, serves to probe the relative importance of benzenoid versus annulenoid aromaticity. A generic structure of this remarkable compound is drawn on the left below, together with two representative Kekule contributing structures on its right. There are some 200 Kekule structures that can be drawn for kekulene, but these two canonical forms represent extremes in aromaticity. The central formula has two [4n+2] annulenes, an inner [18]annulene and an outer [30]annulene (colored pink and blue respectively). The formula on the right has six benzene rings (colored green) joined in a ring by meta bonds, and held in a planar configuration by six cis-double bond bridges.
The coupled annulene contributor in the center has an energetically equivalent canonical form in which the single and double bonds making up the annulenes are exchanged. If these contributors dominate the aromatic character of kekulene, the 6 inside hydrogens should be shielded by the ring currents, and the 18 hydrogens on the periphery should be deshielded. Furthermore, the C:C bonds composing each annulene ring should have roughly equal lengths.
If the benzene contributor on the right (and its equivalent Kekule form) dominate the aromaticity of kekulene, all the benzene hydrogens will be deshielded, and the six double bond links on the periphery will have bond lengths characteristic of fixed single and double bonds
The extreme insolubility of kekulene made it difficult to grow suitable crystals for X-ray analysis or obtain solution NMR spectra. These problems were eventually solved by using high boiling solvents, the 1HNMR spectrum being taken at 150 to 200° C in deuterated tetrachlorobenzene solution. The experimental evidence demonstrates clearly that the hexa-benzene ring structure on the right most accurately represents kekulene. This evidence is shown below. The extremely low field resonance of the inside hydrogens is assigned from similar downfield shifts in model compounds.
It is important to understand that the shielding and deshielding terms used throughout our discussion of relative chemical shifts are themselves relative. Indeed, compared to a hypothetical isolated proton, all the protons in a covalent compound are shielded by the electrons in nearby sigma and pi-bonds. Consequently, it would be more accurate to describe chemical shift differences in terms of the absolute shielding experienced by different groups of hydrogens. There is, in fact, good evidence that the anisotropy of neighboring C-H and C-C sigma bonds, together with that of the bond to the observed hydrogen, are the dominate shielding factors influencing chemical shifts. The anisotropy of pi-electron systems augments this sigma skeletal shielding. Nevertheless, the deceptive focus on anisotropic pi-electron influences is so widely and commonly used that this view has been retained and employed in these pages.
Hydrogen Bonding Influences
Hydrogen bonding of hydroxyl and amino groups not only causes large variations in the chemical shift of the proton of the hydrogen bond, but also influences its coupling with adjacent C-H groups. As shown on the right, the 60 MHz proton NMR spectrum of pure (neat) methanol exhibits two signals, as expected. At 30° C these signals are sharp singlets located at δ 3.35 and 4.80 ppm, the higher-field methyl signal (magenta) being three times as strong as the OH signal (orange) at lower field. When cooled to -45 ° C, the larger higher-field signal changes to a doublet (J = 5.2 Hz) having the same chemical shift. The smaller signal moves downfield to δ 5.5 ppm and splits into a quartet (J = 5.2 Hz). The relative intensities of the two groups of signals remains unchanged. This interesting change in the NMR spectrum, which is shown in the two spectra below, is due to increased stability of hydrogen bonded species at lower temperature. Since hydrogen bonding not only causes a resonance shift to lower field, but also decreases the rate of intermolecular proton exchange, the hydroxyl proton remains bonded to the alkoxy group for a sufficient time to exert its spin coupling influence.
Under routine conditions, rapid intermolecular exchange of the OH protons of alcohols often prevents their coupling with adjacent hydrogens from being observed. Intermediate rates of proton exchange lead to a broadening of the OH and coupled hydrogen signals, a characteristic that is useful in identifying these functions. Since traces of acid or base catalyze this hydrogen exchange, pure compounds and clean sample tubes must be used for experiments of the kind described here.
Methanol 1H NMR at approximately Room Temperature
Methanol 1H NMR at -45 oC
Another way of increasing the concentration of hydrogen bonded methanol species is to change the solvent from chloroform-d to a solvent that is a stronger hydrogen bond acceptor. Examples of such solvents are given in the following table. In contrast to the neat methanol experiment described above, very dilute solutions are used for this study. Since chloroform is a poor hydrogen bond acceptor and the dilute solution reduces the concentration of methanol clusters, the hydroxyl proton of methanol generates a resonance signal at a much higher field than that observed for the pure alcohol. Indeed, the OH resonance signal from simple alcohols in dilute chloroform solution is normally found near δ 1.0 ppm.
The exceptionally strong hydrogen bond acceptor quality of DMSO is demonstrated here by the large downfield shift of the methanol hydroxyl proton, compared with a slight upfield shift of the methyl signal. The expected spin coupling patterns shown above are also observed in this solvent. Although acetone and acetonitrile are better hydrogen-bond acceptors than chloroform, they are not as effective as DMSO.
1H Chemical Shifts of Methanol in Selected Solvents
Solvent CDCl3 CD3COCD3 CD3SOCD3 CD3C≡N
CH3–O–H
CH3
O–H
3.40
1.10
3.31
3.12
3.16
4.01
3.28
2.16
The solvent effect shown above suggests a useful diagnostic procedure for characterizing the OH resonance signals from alcohol samples. For example, a solution of ethanol in chloroform-d displays the spectrum shown on the left below, especially if traces of HCl are present (otherwise broadening of the OH and CH2 signals occurs). Note that the chemical shift of the OH signal (red) is less than that of the methylene group (blue), and no coupling of the OH proton is apparent. The vicinal coupling (J = 7 Hz) of the methyl and methylene hydrogens is typical of ethyl groups. In DMSO-d6 solution small changes of chemical shift are seen for the methyl and methylene group hydrogens, but a dramatic downfield shift of the hydroxyl signal takes place because of hydrogen bonding. Coupling of the OH proton to the adjacent methylene group is evident, and both the coupling constants can be measured. Because the coupling constants are different, the methylene signal pattern is an overlapping doublet of quartets (eight distinct lines) rather than a quintet. Note that residual hydrogens in the solvent give a small broad signal near δ 2.5 ppm.
For many alcohols in dilute chloroform-d solution, the hydroxyl resonance signal is often broad and obscured by other signals in the δ 1.5 to 3.0 region. The simple technique of using DMSO-d6 as a solvent, not only shifts this signal to a lower field, but permits 1°-, 2 °- & 3 °-alcohols to be distinguished. Thus, the hydroxyl proton of 2-propanol generates a doublet at δ 4.35 ppm, and the corresponding signal from 2-methyl-2-propanol is a singlet at δ 4.2 ppm. The more acidic OH protons of phenols are similarly shifted – from δ 4 to 7 in chloroform-d to δ 8.5 to 9.5 in DMSO-d6.
Spin-Spin Coupling
vicinal (joined by three sigma bonds). In this case a neighboring proton having a +1/2 spin shifts the resonance frequency of the proton being observed to a slightly higher value (up to 7 Hz), and a _1/2 neighboring spin shifts it to a lower frequency. Remember that the total population of these two spin states is roughly equal, differing by only a few parts per million in a strong magnetic field. If several neighboring spins are present, their effect is additive.
In the spectrum of 1,1-dichloroethane shown on the right, it is clear that the three methyl hydrogens (red) are coupled with the single methyne hydrogen (orange) in a manner that causes the former to appear as a doublet and the latter as a quartet. The light gray arrow points to the unperturbed chemical shift location for each proton set. By clicking on one of these signals, the spin relationship that leads to the coupling pattern will be displayed. Clicking elsewhere in the picture will return the original spectrum.
Full Spectrum of 1,1-dichloroethane
2.06 ppm Signal Explained
5.89 ppm Signal Explained
The statistical distribution of spins within each set explains both the n+1 rule and the relative intensities of the lines within a splitting pattern. The action of a single neighboring proton is easily deduced from the fact that it must have one of two possible spins. Interaction of these two spin states with the nuclei under observation leads to a doublet located at the expected chemical shift. The corresponding action of the three protons of the methyl group requires a more detailed analysis. In the display of this interaction four possible arrays of their spins are shown. The mixed spin states are three times as possible as the all +1/2 or all _1/2 collection. Consequently, we expect four signals, two above the chemical shift and two below it. This spin analysis also suggests that the intensity ratio of these signals will be 1:3:3:1. The line separations in splitting patterns are measured in Hz, and are characteristic of the efficiency of the spin interaction; they are referred to as coupling constants (symbol J). In the above example, the common coupling constant is 6.0 Hz.
Multipliticy Relative Line Intensity (Starting at the top: Singlet, Doublet, Triplet, Quartet, and Quintet)
A simple way of estimating the relative intensities of the lines in a first-order coupling pattern is shown on the right. This array of numbers is known as Pascal's triangle, and is easily extended to predict higher multiplicities. The number appearing at any given site is the sum of the numbers linked to it from above by the light blue lines. Thus, the central number of the five quintet values is 3 + 3 = 6. Of course, a complete analysis of the spin distributions, as shown for the case of 1,1-dichloroethane above, leads to the same relative intensities.
Coupling constants are independent of the external magnetic field, and reflect the unique spin interaction characteristics of coupled sets of nuclei in a specific structure. As noted earlier, coupling constants may vary from a fraction of a Hz to nearly 20 Hz, important factors being the nature and spatial orientation of the bonds joining the coupled nuclei. In simple, freely rotating alkane units such as CH3CH2X or YCH2CH2X the coupling constant reflects an average of all significant conformers, and usually lies in a range of 6 to 8 Hz. This conformational mobility may be restricted by incorporating the carbon atoms in a rigid ring, and in this way the influence of the dihedral orientation of the coupled hydrogens may be studied.
The structures of cis and trans-4-tert-butyl-1-chlorocyclohexane, shown above, illustrate how the coupling constant changes with the dihedral angle (φ) between coupled hydrogens. The inductive effect of chlorine shifts the resonance frequency of the red colored hydrogen to a lower field (δ ca. 4.0), allowing it to be studied apart from the other hydrogens in the molecule. The preferred equatorial orientation of the large tert-butyl group holds the six-membered ring in the chair conformation depicted in the drawing. In the trans isomer this fixes the red hydrogen in an axial orientation; whereas for the cis isomer it is equatorial. The listed values for the dihedral angles and the corresponding coupling constants suggest a relationship, which has been confirmed and clarified by numerous experiments. This relationship is expressed by the Karplus equation shown below.
Geminal couplings are most commonly observed in cyclic structures, but are also evident when methylene groups have diastereomeric hydrogens.
Spin Decoupling
We have noted that rapidly exchanging hydroxyl hydrogens are not spin-coupled to adjacent C-H groups. The reason for this should be clear. As each exchange occurs, there will be an equal chance of the new proton having a +1/2 or a _1/2 spin (remember that the overall populations of the two spin states are nearly identical). Over time, therefore, the hydroxyl hydrogen behaves as though it is rapidly changing its spin, and the adjacent nuclei see only a zero spin average from it. If we could cause other protons in a molecule to undergo a similar spin averaging, their spin-coupling influence on adjacent nuclei would cease. Such NMR experiments are possible, and are called spin decoupling.
When a given set of nuclei is irradiated with strong rf energy at its characteristic Larmor frequency, spin saturation and rapid interconversion of the spin states occurs. Neighboring nuclei with different Larmor frequencies are no longer influenced by specific long-lived spins, so spin-spin signal splitting of the neighbors vanishes. The following spectrum of 1-nitropropane may be used to illustrate this technique. The three distinct sets of hydrogens in this molecule generate three resonance signals (two triplets and a broad sextet). A carefully tuned decoupling signal may be broadcast into the sample while the remaining spectrum is scanned. The region of the decoupling signal is obscured, but resonance signals more than 60 Hz away may still be seen. By clicking on one of the three signals in the spectrum, the results of decoupling at that frequency will be displayed.
The Influence of Magnetic Field Strength
The presence of symmetrical, easily recognized first-order splitting patterns in a NMR spectrum depends on the relative chemical shifts of the spin-coupled nuclei and the magnitude of the coupling constant. If the chemical shift difference (i.e. Δδ in Hz) is large compared to J the splitting patterns will be nearly first order. If, on the other hand, the difference is relatively small (less than 10 J) second order distortion of the signal splitting will be observed. One important advantage in using very high field magnets for NMR is that the separation (or dipersion) of different sets of protons is proportional to field strength, whereas coupling constants do not change.
It is important to remember that structurally different sets of nuclei do not always produce distinctly different signals in an NMR spectrum. For example, the hydrocarbon octane has four different sets of protons, as shown in the following formula:
CH3CH2CH2CH2CH2CH2CH2CH3
Now methyl hydrogens have a smaller chemical shift than methylene hydrogens, so methyl groups (colored black here) can usually be distinguished. However, the chemical shifts of the different methylene groups (blue, red & green) are so similar that many NMR spectrometers will not resolve them. Consequently, a 90 MHz proton spectrum of octane shows a distorted triplet at δ 0.9 ppm, produced by the six methyl protons, and a strong broad singlet at δ 1.2 ppm coming from all twelve methylene protons. A similar failure to resolve structurally different hydrogen atoms occurs in the case of alkyl substituted benzene rings. The chemical shift difference between ortho, meta and para hydrogens in such compounds is often so small that they are seen as a single resonance signal in an NMR spectrum. The 90 MHz spectrum of benzyl alcohol in chloroform-d solution provides an instructive example, shown below. A broad strong signal at δ 7.24 ppm is characteristic of the aromatic protons on alkylbenzenes. Since the chemical shifts of these hydrogens are nearly identical, no spin coupling is observed. If the magnetic field strength is increased to 400 Mz (lower spectrum) the aromatic protons are more dispersed (orange, magenta and green signals), and the spin coupling of adjacent hydrogens (J = 7.6 Hz) causes overlap of the signals (gray shaded enlargement).
1H NMR of Benzyl alcohol
1H NMR of Anisole
Anisole, an isomer of benzyl alcohol, has a more dispersed set of aromatic signals, thanks to the electron donating influence of the methoxy substituent. The 90 MHz spectrum of anisole shows this greater dispersion, but the spin coupling of adjacent hydrogens still results in signal overlap. The 400 Mz spectrum at the bottom illustrates the greater dispersion of the chemical shifts, and since the coupling constants remain unchanged, the splitting patterns no longer overlap. In all these examples a very small meta-hydrogen coupling has been ignored.
Example \(1\)
Not all simple compounds have simple proton NMR spectra. The following example not only illustrates this point, but also demonstrates how a careful structural analysis can rationalize an initially complex spectrum.
The 100 MHz 1H NMR spectrum of a C3H5ClO compound is initially displayed. This spectrum is obviously complex and not easily interpreted, except for concluding that no olefinic C-H protons are present.
With a higher field spectrum it is clear that each of the five hydrogen atoms in the molecule is structurally unique, and is producing a separate signal. Also, it is clear there is considerable spin coupling of all the hydrogens. To see the coupling patterns more clearly it is necessary to expand and enhance the spectrum in these regions. For purposes of our demonstration, this can be done by clicking on any one of the signal multiplets. Clicking in an open area should return the original 500 MHz display. In some of the expanded displays two adjacent groups of signals are shown. Once an enlarged pattern is displayed, the line separations in Hz can be measured (remember that for a 500 MHz spectrum 1 ppm is 500 Hz). The middle signal at 3.2 ppm is the most complex, and overlap of some multiplet lines has occurred.
Solution
This spectrum has several interesting features. First, hydrogens A, B & C are clearly different, and are spin-coupled to each other. Hydrogens B & C are geminally related, whereas A is oriented to B & C in a vicinal manner. Since JAB and JBC are similar, the HB signal is a broad triplet. Although hydrogens D & E might seem identical at first glance, they are diastereotopic, and should therefore have different chemical shifts. The DE geminal coupling constant is 11.7 Hz, so each of these hydrogens appears as a doublet of doublets.
The splitting of the HA signal is complex and not immediately obvious. The diagram shows the consequences of the four operating coupling constants.
Additional Information from 13C NMR Spectroscopy
Broad band decoupling of the hydrogen atoms in a molecule was an essential operation for obtaining simple (single line) carbon NMR spectra. The chemical shifts of the carbon signals provide useful information, but it would also be very helpful to know how many hydrogen atoms are bonded to each carbon. Aside from the fact that carbons having no bonded hydrogens generally give weak resonance signals, this information is not present in a completely decoupled spectrum.
Clever methods of retaining the hydrogen information while still enjoying the benefits of proton decoupling have been devised. The techniques involved are beyond the scope of this discussion, but the overall results can still be appreciated. The 13C NMR spectrum of camphor shown below will serve as an illustration. It will be helpful to view an expanded section of this spectrum from δ 0.0 to 50.0 ppm, and this will be presented in the High Field Expansion spectrum. The two lowest field signals are missing in the expanded display.
High Field Expansion of Camphor
Even though the expanded display now shows the distinct carbon signals clearly, the origin of each is ambiguous. An early method of regaining coupling information was by off-resonance decoupling. In this approach a weaker and more focused proton decoupling frequency is applied as the carbon spectrum is acquired. Vestiges of the C-H coupling remain in the carbon signals, but the apparent coupling constants are greatly reduced. View the Off-Resonance Decouple spectrum. The results of such an experiment will be displayed. Notice that all the methyl groups are quartets (three coupled hydrogens), the methylene groups are triplets and methine carbons are doublets. Overlap of two quartets near δ 19 ppm and the doublet and triplet near δ 43 ppm are complicating factors.
Off-Resonance Decouple Spectrum of Camphor
A better way for classifying the carbon signals is by a technique called INEPT (insensitive nuclear enhancement by polarization transfer). This method takes advantage of the influence of hydrogen on 13C relaxation times, and can be applied in several modes. One of the most common applications of INEPT separates the signals of methyl and methine carbons from those of methylene carbons by their sign. Carbons having no hydrogen substituents have a zero signal.
INEPT Spectrum of Camphor
Properties of Some Deuterated NMR Solvents
Solvent B.P. °C Residual
1H signal (δ)
Residual
13C signal (δ)
acetone-d6 55.5 2.05 ppm 206 & 29.8 ppm
acetonitrile-d3 80.7 1.95 ppm 118 & 1.3 ppm
benzene-d6 79.1 7.16 ppm 128 ppm
chloroform-d 60.9 7.27 ppm 77.2 ppm
cyclohexane-d12 78.0 1.38 ppm 26.4 ppm
dichloromethane-d2 40.0 5.32 ppm 53.8 ppm
dimethylsulfoxide-d6 190 2.50 ppm 39.5 ppm
nitromethane-d3 100 4.33 ppm 62.8 ppm
pyridine-d5 114 7.19, 7.55 & 8.71 ppm 150, 135.5 & 123.5 ppm
tetrahydrofuran-d8 65.0 1.73 & 3.58 ppm 67.4 & 25.2 ppm
Contributors
• Layne Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.08%3A_Diamagnetic_Anisotropy.txt |
5.3A: The basics of an NMR experiment
Given that chemically nonequivalent protons have different resonance frequencies in the same applied magnetic field, we can see how NMR spectroscopy can provide us with useful information about the structure of an organic molecule. A full explanation of how a modern NMR instrument functions is beyond the scope of this text, but in very simple terms, here is what happens. First, a sample compound (we'll use methyl acetate) is placed inside a very strong applied magnetic field (B0).
All of the protons begin to precess: the Ha protons at precessional frequency ωa, the Hb protons at ω b.At first, the magnetic moments of (slightly more than) half of the protons are aligned with B0, and half are aligned against B0. Then, the sample is hit with electromagnetic radiation in the radio frequency range. The two specific frequencies which match ωaandωb(i.e. the resonance frequencies) cause those Ha and Hb protons which are aligned with B0 to 'flip' so that they are now aligned against B0. In doing so, the protons absorb radiation at the two resonance frequencies. The NMR instrument records which frequencies were absorbed, as well as the intensity of each absorbance.
In most cases, a sample being analyzed by NMR is in solution. If we use a common laboratory solvent (diethyl ether, acetone, dichloromethane, ethanol, water, etc.) to dissolve our NMR sample, however, we run into a problem – there many more solvent protons in solution than there are sample protons, so the signals from the sample protons will be overwhelmed. To get around this problem, we use special NMR solvents in which all protons have been replaced by deuterium. Recall that deuterium is NMR-active, but its resonance frequency is very different from that of protons, and thus it is `invisible` in 1H-NMR. Some common NMR solvents are shown below.
5.3B: The chemical shift
Let's look at an actual 1H-NMR plot for methyl acetate. Just as in IR and UV-vis spectroscopy, the vertical axis corresponds to intensity of absorbance, the horizontal axis to frequency (typically the vertical axis is not shown in an NMR spectrum).
We see three absorbance signals: two of these correspond to Ha and Hb, while the peak at the far right of the spectrum corresponds to the 12 chemically equivalent protons in tetramethylsilane (TMS), a standard reference compound that was added to our sample.
You may be wondering about a few things at this point - why is TMS necessary, and what is the meaning of the `ppm (δ)` label on the horizontal axis? Shouldn't the frequency units be in Hz? Keep in mind that NMR instruments of many different applied field strengths are used in organic chemistry laboratories, and that the proton's resonance frequency range depends on the strength of the applied field. The spectrum above was generated on an instrument with an applied field of approximately 7.1 Tesla, at which strength protons resonate in the neighborhood of 300 million Hz (chemists refer to this as a 300 MHz instrument). If our colleague in another lab takes the NMR spectrum of the same molecule using an instrument with a 2.4 Tesla magnet, the protons will resonate at around 100 million Hz (so we’d call this a 100 MHz instrument). It would be inconvenient and confusing to always have to convert NMR data according to the field strength of the instrument used. Therefore, chemists report resonance frequencies not as absolute values in Hz, but rather as values relative to a common standard, generally the signal generated by the protons in TMS. This is where the ppm – parts per million – term comes in. Regardless of the magnetic field strength of the instrument being used, the resonance frequency of the 12 equivalent protons in TMS is defined as a zero point. The resonance frequencies of protons in the sample molecule are then reported in terms of how much higher they are, in ppm, relative to the TMS signal (almost all protons in organic molecules have a higher resonance frequency than those in TMS, for reasons we shall explore quite soon).
The two proton groups in our methyl acetate sample are recorded as resonating at frequencies 2.05 and 3.67 ppm higher than TMS. One-millionth (1.0 ppm) of 300 MHz is 300 Hz. Thus 2.05 ppm, on this instrument, corresponds to 615 Hz, and 3.67 ppm corresponds to 1101 Hz. If the TMS protons observed by our 7.1 Tesla instrument resonate at exactly 300,000,000 Hz, this means that the protons in our ethyl acetate samples are resonating at 300,000,615 and 300,001,101 Hz, respectively. Likewise, if the TMS protons in our colleague's 2.4 Tesla instrument resonate at exactly 100 MHz, the methyl acetate protons in her sample resonate at 100,000,205 and 100,000,367 Hz (on the 100 MHz instrument, 1.0 ppm corresponds to 100 Hz). The absolute frequency values in each case are not very useful – they will vary according to the instrument used – but the difference in resonance frequency from the TMS standard, expressed in parts per million, should be the same regardless of the instrument.
Expressed this way, the resonance frequency for a given proton in a molecule is called its chemical shift. A frequently used symbolic designation for chemical shift in ppm is the lower-case Greek letter delta (δ). Most protons in organic compounds have chemical shift values between 0 and 12 ppm from TMS, although values below zero and above 12 are occasionally observed. By convention, the left-hand side of an NMR spectrum (higher chemical shift) is called downfield, and the right-hand direction is called upfield.
In our methyl acetate example we included for illustrative purposes a small amount of TMS standard directly in the sample, as was the common procedure for determining the zero point with older NMR instruments．That practice is generally no longer necessary, as modern NMR instruments are designed to use the deuterium signal from the solvent as a standard reference point, then to extrapolate the 0 ppm baseline that corresponds to the TMS proton signal (in an applied field of 7.1 Tesla, the deuterium atom in CDCl3 resonates at 32 MHz, compared to 300 MHz for the protons in TMS). In the remaining NMR spectra that we will see in this text we will not see an actual TMS signal, but we can always assume that the 0 ppm point corresponds to where the TMS protons would resonate if they were present.
Example 5.2
A proton has a chemical shift (relative to TMS) of 4.56 ppm.
1. What is its chemical shift, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
2. What is its resonance frequency, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
(Assume that in these instruments, the TMS protons resonate at exactly 300 or 200 MHz, respectively)
Solution
5.3C: Signal integration
The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals. This is very useful, because in 1H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds. The two signals in the methyl acetate spectrum, for example, integrate to approximately the same area, because they both correspond to a set of three equivalent protons.
Take a look next at the spectrum of para-xylene (IUPAC name 1,4-dimethylbenzene):
This molecule has two sets of protons: the six methyl (Ha) protons and the four aromatic (Hb) protons. When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. This (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal.
The integration function can also be used to determine the relative amounts of two or more compounds in a mixed sample. If we have a sample that is a 50:50 (mole/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons. If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons.
Example 5.3
You take a 1H-NMR spectrum of a mixed sample of acetone (CH3(CO)CH3) and dichloromethane (CH2Cl2). The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1. What is the molar ratio of the two compounds in the sample?
Solution
Example 5.4
You take the 1H-NMR spectrum of a mixed sample of 36% para-xylene and 64% acetone in CDCl3 solvent (structures are shown earlier in this chapter). How many peaks do you expect to see? What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0)
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.09%3A_The_Integration_of_NMR_Signals_Reveals_the_Relative_Number_of_Protons_Causing_the_Signal.txt |
5.5A: The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hbhydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns.
Example 5.6
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
Example 5.7
1. How many proton signals would you expect to see in the 1H-NMR spectrum of triclosan (a common antimicrobial agent found in detergents)? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Solution
Example 5.8
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solution
5.5B: Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
5.5C: Complex coupling
In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to just one neighboring set of hydrogens. When a set of hydrogens is coupled to two or more sets of nonequivalent neighbors, the result is a phenomenon called complex coupling. A good illustration is provided by the 1H-NMR spectrum of methyl acrylate:
First, let's first consider the Hc signal, which is centered at 6.21 ppm. Here is a closer look:
With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? Hc is coupled to both Ha and Hb , but with two different coupling constants. Once again, a splitting diagram can help us to understand what we are seeing. Ha is trans to Hc across the double bond, and splits the Hc signal into a doublet with a coupling constant of 3Jac = 17.4 Hz. In addition, each of these Hc doublet sub-peaks is split again by Hb (geminal coupling) into two more doublets, each with a much smaller coupling constant of 2Jbc = 1.5 Hz.
The result of this `double splitting` is a pattern referred to as a doublet of doublets, abbreviated `dd`.
The signal for Ha at 5.95 ppm is also a doublet of doublets, with coupling constants 3Jac= 17.4 Hz and 3Jab = 10.5 Hz.
The signal for Hb at 5.64 ppm is split into a doublet by Ha, a cis coupling with 3Jab = 10.4 Hz. Each of the resulting sub-peaks is split again by Hc, with the same geminal coupling constant 2Jbc = 1.5 Hz that we saw previously when we looked at the Hc signal. The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the cis interaction. Here is a blow-up of the actual Hbsignal:
Example 5.9
Construct a splitting diagram for the Hb signal in the 1H-NMR spectrum of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).
Solution
When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).
When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `n + 1 rule` of non-complex splitting. In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for Hb to be split into a triplet by Ha, and again into doublets by Hc, resulting in a 'triplet of doublets'.
Ha and Hc are not equivalent (their chemical shifts are different), but it turns out that 3Jab is very close to 3Jbc. If we perform a splitting diagram analysis for Hb, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the n + 1 rule.
For similar reasons, the Hc peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined Hb and Hd protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Example 5.10
What splitting pattern would you expect for the signal coresponding to Hb in the molecule below? Assume that Jab ~ Jbc. Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.
Solution
In many cases, it is difficult to fully analyze a complex splitting pattern. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Hb to be a doublet, Hd a triplet, and Hc a triplet.
In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult. In this case, we would refer to the aromatic part of the spectrum as a multiplet.
When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments. Quite simply, the stronger our magnet is, the more resolution we get in our spectrum. In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide. In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider. In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.10%3A_The_Splitting_of_the_Signals_is_Described_by_the_N__1_Rule.txt |
Link to Solution Manual
P5.1: For each molecule, predict the number of signals in the 1H-NMR and the 13C-NMR spectra (do not count split peaks - eg. a quartet counts as only one signal). Assume that diastereotopic groups are non-equivalent.
P5.2: For each of the 20 common amino acids, predict the number of signals in the proton-decoupled 13C-NMR spectrum.
P5.3: Calculate the chemical shift value (expressed in Hz, to one decimal place) of each sub-peak on the 1H-NMR doublet signal below. Do this for:
a) a spectrum obtained on a 300 MHz instrument
b) a spectrum obtained on a 100 MHz instrument
P5.4: Consider a quartet signal in an 1H-NMR spectrum obtained on a 300 MHz instrument. The chemical shift is recorded as 1.7562 ppm, and the coupling constant is J = 7.6 Hz. What is the chemical shift, expressed to the nearest 0.1 Hz, of the furthest downfield sub-peak in the quartet? What is the resonance frequency (again expressed in Hz) of this sub-peak?)
P5.5: One easily recognizable splitting pattern for the aromatic proton signals from disubstituted benzene structures is a pair of doublets. Does this pattern indicate ortho, meta, or para substitution?
P5.6 :Match spectra below to their corresponding structures A-F.
Structures:
Spectrum 1
δ
splitting
integration
4.13
q
2
2.45
t
2
1.94
quintet
1
1.27
t
3
Spectrum 2
δ
splitting
integration
3.68
s
3
2.99
t
2
1.95
quintet
1
Spectrum 3
δ
splitting
integration
4.14
q
1
2.62
s
1
1.26
t
1.5
Spectrum 4
δ
splitting
integration
4.14
q
4
3.22
s
1
1.27
t
6
1.13
s
9
Spectrum 5
δ
splitting
integration
4.18
q
1
1.92
q
1
1.23
t
1.5
0.81
t
1.5
Spectrum 6
δ
splitting
integration
3.69
s
1.5
2.63
s
1
P5.7: Match spectra 7-12 below to their corresponding structures G-L .
Structures:
Spectrum 7:
δ
splitting
integration
9.96
d
1
5.88
d
1
2.17
s
3
1.98
s
3
Spectrum 8:
δ
splitting
integration
9.36
s
1
6.55
q
1
2.26
q
2
1.99
d
3
0.96
t
3
Spectrum 9:
δ
splitting
integration
9.57
s
1
6.30
s
1
6.00
s
1
1.84
s
3
Spectrum 10:
δ
splitting
integration
9.83
t
1
2.27
d
2
1.07
s
9
Spectrum 11:
δ
splitting
integration
9.75
t
1
2.30
dd
2
2.21
m
1
0.98
d
6
Spectrum 12:
δ
splitting
integration
8.08
s
1
4.13
t
2
1.70
m
2
0.96
t
3
P5.8: Match the 1H-NMR spectra 13-18 below to their corresponding structures M-R .
Structures:
Spectrum 13:
δ
splitting
integration
8.15
d
1
6.33
d
1
Spectrum 14: 1-723C (structure O)
δ
splitting
integration
6.05
s
1
2.24
s
3
Spectrum 15:
δ
splitting
integration
8.57
s (b)
1
7.89
d
1
6.30
d
1
2.28
s
3
Spectrum 16:
δ
splitting
integration
9.05
s (b)
1
8.03
s
1
6.34
s
1
5.68
s (b)
1
4.31
s
2
Spectrum 17:
δ
splitting
integration
7.76
d
1
7.57
s (b)
1
6.44
d
1
2.78
q
2
1.25
t
3
Spectrum 18:
δ
splitting
integration
4.03
s
1
2.51
t
1
2.02
t
1
P5.9: Match the 1H-NMR spectra 19-24 below to their corresponding structures S-X.
Structures:
Spectrum 19:
δ
splitting
integration
9.94
s
1
7.77
d
2
7.31
d
2
2.43
s
3
Spectrum 20:
δ
splitting
integration
10.14
s
2
8.38
s
1
8.17
d
2
7.75
t
1
Spectrum 21:
δ
splitting
integration
9.98
s
1
7.81
d
2
7.50
d
2
Spectrum 22:
δ
splitting
integration
7.15-7.29
m
2.5
2.86
t
1
2.73
t
1
2.12
s
1.5
Spectrum 23:
δ
splitting
integration
7.10
d
1
6.86
d
1
3.78
s
1.5
3.61
s
1
2.12
s
1.5
Spectrum 24:
δ
splitting
integration
7.23-7.30
m
1
3.53
s
1
P5.10: Match the 1H-NMR spectra 25-30 below to their corresponding structures AA-FF.
Structures:
Spectrum 25:
δ
splitting
integration
9.96
s
1
7.79
d
2
7.33
d
2
2.72
q
2
1.24
t
3
Spectrum 26:
δ
splitting
integration
9.73
s
1
7.71
d
2
6.68
d
2
3.06
s
6
Spectrum 27:
δ
splitting
integration
7.20-7.35
m
10
5.12
s
1
2.22
s
3
Spectrum 28:
δ
splitting
integration
8.08
s
1
7.29
d
2
6.87
d
2
5.11
s
2
3.78
s
3
Spectrum 29:
δ
splitting
integration
7.18
d
1
6.65
m
1.5
3.2
q
2
1.13
t
3
Spectrum 30:
δ
splitting
integration
8.32
s
1
4.19
t
2
2.83
t
2
2.40
s
3
P5.11: Match the 1H-NMR spectra 31-36 below to their corresponding structures GG-LL
Structures:
Spectrum 31:
δ
splitting
integration
6.98
d
1
6.64
d
1
6.54
s
1
4.95
s
1
2.23
s
3
2.17
s
3
Spectrum 32:
δ
splitting
integration
7.08
d
1
6.72
d
1
6.53
s
1
4.81
s
1
3.15
7-tet
1
2.24
s
3
1.22
d
6
Spectrum 33:
δ
splitting
integration
7.08
d
2
6.71
d
2
6.54
s
1
3.69
s
3
3.54
s
2
Spectrum 34:
δ
splitting
integration
9.63
s
1
7.45
d
2
6.77
d
2
3.95
q
2
2.05
s
3
1.33
t
3
Spectrum 35:
δ
splitting
integration
9.49
s
1
7.20
d
2
6.49
d
2
4.82
s
2
1.963
s
3
Spectrum 36:
δ
splitting
integration
9.58
s(b)
1
9.31
s
1
7.36
d
1
6.67
s
1
6.55
d
1
2.21
s
3
2.11
s
3
P5.12: Use the NMR data given to deduce structures.
a ) Molecular formula: C5H8O
1H-NMR:
δ
splitting
integration
9.56
s
1
6.25
d (J~1 Hz)
1
5.99
d (J~1 Hz)
1
2.27
q
2
1.18
t
3
13C-NMR
δ
DEPT
194.60
CH
151.77
C
132.99
CH2
20.91
CH2
11.92
CH3
b) Molecular formula: C7H14O2
1H-NMR:
δ
splitting
integration
3.85
d
2
2.32
q
2
1.93
m
1
1.14
t
3
0.94
d
6
13C-NMR
δ
DEPT
174.47
C
70.41
CH2
27.77
CH
27.64
CH2
19.09
CH3
9.21
CH3
c) Molecular formula: C5H12O
1H-NMR:
δ
splitting
integration
3.38
s
2H
2.17
s
1H
0.91
s
9H
13C-NMR
δ
DEPT
73.35
CH2
32.61
C
26.04
CH3
d) Molecular formula: C10H12O
1H-NMR:
δ
splitting
integration
7.18-7.35
m
2.5
3.66
s
1
2.44
q
1
1.01
t
1.5
13C-NMR
δ
DEPT
208.79
C
134.43
C
129.31
CH
128.61
CH
126.86
CH
49.77
CH2
35.16
CH2
7.75
CH3
P5.13:
13C-NMR data is given for the molecules shown below. Complete the peak assignment column of each NMR data table.
a)
δ
DEPT
carbon #
161.12
CH
65.54
CH2
21.98
CH2
10.31
CH3
b)
δ
DEPT
carbon #
194.72
C
149.10
C
146.33
CH
16.93
CH2
14.47
CH3
12.93
CH3
c)
δ
DEPT
carbon #
171.76
C
60.87
CH2
58.36
C
24.66
CH2
14.14
CH3
8.35
CH3
d)
δ
DEPT
carbon #
173.45
C
155.01
C
130.34
CH
125.34
C
115.56
CH
52.27
CH3
40.27
CH2
e)
δ
DEPT
carbon #
147.79
C
129.18
CH
115.36
CH
111.89
CH
44.29
CH2
12.57
CH3
P5.14: You obtain the following data for an unknown sample. Deduce its structure.
1H-NMR:
13C-NMR:
Mass Spectrometry:
P5.15:You take a 1H-NMR spectrum of a sample that comes from a bottle of 1-bromopropane. However, you suspect that the bottle might be contaminated with 2-bromopropane. The NMR spectrum shows the following peaks:
δ
splitting
integration
4.3
septet
0.0735
3.4
triplet
0.661
1.9
sextet
0.665
1.7
doublet
0.441
1.0
triplet
1.00
How badly is the bottle contaminated? Specifically, what percent of the molecules in the bottle are 2-bromopropane?
Challenge problems
C5.1: All of the 13C-NMR spectra shown in this chapter include a signal due to CDCl3, the solvent used in each case. Explain the splitting pattern for this signal.
C5.2: Researchers wanted to investigate a reaction which can be catalyzed by the enzyme alcohol dehydrogenase in yeast. They treated 4'-acylpyridine (1) with living yeast, and isolated the alcohol product(s) (some combination of 2A and 2B).
a) Will the products 2A and 2B have identical or different 1H-NMR spectra? Explain.
b) Suggest a 1H-NMR experiment that could be used to determine what percent of starting material (1) got turned into product (2A and 2B).
c) With purified 2A/2B, the researchers carried out the subsequent reaction shown below to make 3A and 3B, known as 'Mosher's esters'. Do 3A and 3B have identical or different 1H-NMR spectra? Explain.
d) Explain, very specifically, how the researchers could use 1H-NMR to determine the relative amounts of 2A and 2B formed in the reaction catalyzed by yeast enzyme.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.11%3A_More_Examples_of_%281H%29_NMR_Spectra.txt |
5.5A: The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hbhydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns.
Example 5.6
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
Example 5.7
1. How many proton signals would you expect to see in the 1H-NMR spectrum of triclosan (a common antimicrobial agent found in detergents)? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Solution
Example 5.8
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solution
5.5B: Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
5.5C: Complex coupling
In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to just one neighboring set of hydrogens. When a set of hydrogens is coupled to two or more sets of nonequivalent neighbors, the result is a phenomenon called complex coupling. A good illustration is provided by the 1H-NMR spectrum of methyl acrylate:
First, let's first consider the Hc signal, which is centered at 6.21 ppm. Here is a closer look:
With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? Hc is coupled to both Ha and Hb , but with two different coupling constants. Once again, a splitting diagram can help us to understand what we are seeing. Ha is trans to Hc across the double bond, and splits the Hc signal into a doublet with a coupling constant of 3Jac = 17.4 Hz. In addition, each of these Hc doublet sub-peaks is split again by Hb (geminal coupling) into two more doublets, each with a much smaller coupling constant of 2Jbc = 1.5 Hz.
The result of this `double splitting` is a pattern referred to as a doublet of doublets, abbreviated `dd`.
The signal for Ha at 5.95 ppm is also a doublet of doublets, with coupling constants 3Jac= 17.4 Hz and 3Jab = 10.5 Hz.
The signal for Hb at 5.64 ppm is split into a doublet by Ha, a cis coupling with 3Jab = 10.4 Hz. Each of the resulting sub-peaks is split again by Hc, with the same geminal coupling constant 2Jbc = 1.5 Hz that we saw previously when we looked at the Hc signal. The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the cis interaction. Here is a blow-up of the actual Hbsignal:
Example 5.9
Construct a splitting diagram for the Hb signal in the 1H-NMR spectrum of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).
Solution
When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).
When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `n + 1 rule` of non-complex splitting. In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for Hb to be split into a triplet by Ha, and again into doublets by Hc, resulting in a 'triplet of doublets'.
Ha and Hc are not equivalent (their chemical shifts are different), but it turns out that 3Jab is very close to 3Jbc. If we perform a splitting diagram analysis for Hb, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the n + 1 rule.
For similar reasons, the Hc peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined Hb and Hd protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Example 5.10
What splitting pattern would you expect for the signal coresponding to Hb in the molecule below? Assume that Jab ~ Jbc. Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.
Solution
In many cases, it is difficult to fully analyze a complex splitting pattern. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Hb to be a doublet, Hd a triplet, and Hc a triplet.
In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult. In this case, we would refer to the aromatic part of the spectrum as a multiplet.
When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments. Quite simply, the stronger our magnet is, the more resolution we get in our spectrum. In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide. In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider. In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.12%3A_Coupling_Constants_Identify_Coupled_Protons.txt |
5.5A: The source of spin-spin coupling
The 1H-NMR spectra that we have seen so far (of methyl acetate and para-xylene) are somewhat unusual in the sense that in both of these molecules, each set of protons generates a single NMR signal. In fact, the 1H-NMR spectra of most organic molecules contain proton signals that are 'split' into two or more sub-peaks. Rather than being a complication, however, this splitting behavior actually provides us with more information about our sample molecule.
Consider the spectrum for 1,1,2-trichloroethane. In this and in many spectra to follow, we show enlargements of individual signals so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two subpeaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three sub-peaks, with the middle peak higher than the two outside peaks - if we were to integrate each subpeak, we would see that the area under the middle peak is twice that of each of the outside peaks. This is called a triplet.
The source of signal splitting is a phenomenon called spin-spin coupling, a term that describes the magnetic interactions between neighboring, non-equivalent NMR-active nuclei. In our 1,1,2 trichloromethane example, the Ha and Hb protons are spin-coupled to each other. Here's how it works, looking first at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet). The magnetic moment of Hb will be aligned with B0 in (slightly more than) half of the molecules in the sample, while in the remaining half of the molecules it will be opposed to B0. The Beff ‘felt’ by Ha is a slightly weaker if Hb is aligned against B0, or slightly stronger if Hb is aligned with B0. In other words, in half of the molecules Ha is shielded by Hb (thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb(and the NMR signal shifted slightly downfield). What would otherwise be a single Ha peak has been split into two sub-peaks (a doublet), one upfield and one downfield of the original signal. These ideas an be illustrated by a splitting diagram, as shown below.
Now, let's think about the Hbsignal. The magnetic environment experienced by Hb is influenced by the fields of both neighboring Ha protons, which we will call Ha1 and Ha2. There are four possibilities here, each of which is equally probable. First, the magnetic fields of both Ha1 and Ha2 could be aligned with B0, which would deshield Hb, shifting its NMR signal slightly downfield. Second, both the Ha1 and Ha2 magnetic fields could be aligned opposed to B0, which would shield Hb, shifting its resonance signal slightly upfield. Third and fourth, Ha1 could be with B0 and Ha2 opposed, or Ha1opposed to B0 and Ha2 with B0. In each of the last two cases, the shielding effect of one Ha proton would cancel the deshielding effect of the other, and the chemical shift of Hb would be unchanged.
So in the end, the signal for Hb is a triplet, with the middle peak twice as large as the two outer peaks because there are two ways that Ha1 and Ha2 can cancel each other out.
Now, consider the spectrum for ethyl acetate:
We see an unsplit 'singlet' peak at 1.833 ppm that corresponds to the acetyl (Ha) hydrogens – this is similar to the signal for the acetate hydrogens in methyl acetate that we considered earlier. This signal is unsplit because there are no adjacent hydrogens on the molecule. The signal at 1.055 ppm for the Hc hydrogens is split into a triplet by the two Hb hydrogens next door. The explanation here is the same as the explanation for the triplet peak we saw previously for 1,1,2-trichloroethane.
The Hbhydrogens give rise to a quartet signal at 3.915 ppm – notice that the two middle peaks are taller then the two outside peaks. This splitting pattern results from the spin-coupling effect of the three Hc hydrogens next door, and can be explained by an analysis similar to that which we used to explain the doublet and triplet patterns.
Example 5.6
1. Explain, using left and right arrows to illustrate the possible combinations of nuclear spin states for the Hc hydrogens, why the Hb signal in ethyl acetate is split into a quartet.
2. The integration ratio of doublets is 1:1, and of triplets is 1:2:1. What is the integration ratio of the Hb quartet in ethyl acetate? (Hint – use the illustration that you drew in part a to answer this question.)
Solution
By now, you probably have recognized the pattern which is usually referred to as the n + 1 rule: if a set of hydrogens has n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks. Thus the two Hb hydrogens in ethyl acetate split the Hc signal into a triplet, and the three Hc hydrogens split the Hb signal into a quartet. This is very useful information if we are trying to determine the structure of an unknown molecule: if we see a triplet signal, we know that the corresponding hydrogen or set of hydrogens has two `neighbors`. When we begin to determine structures of unknown compounds using 1H-NMR spectral data, it will become more apparent how this kind of information can be used.
Three important points need to be emphasized here. First, signal splitting only occurs between non-equivalent hydrogens – in other words, Ha1 in 1,1,2-trichloroethane is not split by Ha2, and vice-versa.
Second, splitting occurs primarily between hydrogens that are separated by three bonds. This is why the Ha hydrogens in ethyl acetate form a singlet– the nearest hydrogen neighbors are five bonds away, too far for coupling to occur.
Occasionally we will see four-bond and even 5-bond splitting, but in these cases the magnetic influence of one set of hydrogens on the other set is much more subtle than what we typically see in three-bond splitting (more details about how we quantify coupling interactions is provided in section 5.5B). Finally, splitting is most noticeable with hydrogens bonded to carbon. Hydrogens that are bonded to heteroatoms (alcohol or amino hydrogens, for example) are coupled weakly - or not at all - to their neighbors. This has to do with the fact that these protons exchange rapidly with solvent or other sample molecules.
Below are a few more examples of chemical shift and splitting pattern information for some relatively simple organic molecules.
Example 5.7
1. How many proton signals would you expect to see in the 1H-NMR spectrum of triclosan (a common antimicrobial agent found in detergents)? For each of the proton signals, predict the splitting pattern. Assume that you see only 3-bond coupling.
Solution
Example 5.8
Predict the splitting pattern for the 1H-NMR signals corresponding to the protons at the locations indicated by arrows (the structure is that of the neurotransmitter serotonin).
Solution
5.5B: Coupling constants
Chemists quantify the spin-spin coupling effect using something called the coupling constant, which is abbreviated with the capital letter J. The coupling constant is simply the difference, expressed in Hz, between two adjacent sub-peaks in a split signal. For our doublet in the 1,1,2-trichloroethane spectrum, for example, the two subpeaks are separated by 6.1 Hz, and thus we write 3Ja-b = 6.1 Hz.
The superscript 3 tells us that this is a three-bond coupling interaction, and the a-b subscript tells us that we are talking about coupling between Ha and Hb. Unlike the chemical shift value, the coupling constant, expressed in Hz, is the same regardless of the applied field strength of the NMR magnet. This is because the strength of the magnetic moment of a neighboring proton, which is the source of the spin-spin coupling phenomenon, does not depend on the applied field strength.
When we look closely at the triplet signal in 1,1,2-trichloroethane, we see that the coupling constant - the `gap` between subpeaks - is 6.1 Hz, the same as for the doublet. This is an important concept! The coupling constant 3Ja-b quantifies the magnetic interaction between the Ha and Hb hydrogen sets, and this interaction is of the same magnitude in either direction. In other words, Ha influences Hb to the same extent that Hb influences Ha. When looking at more complex NMR spectra, this idea of reciprocal coupling constants can be very helpful in identifying the coupling relationships between proton sets.
Coupling constants between proton sets on neighboring sp3-hybridized carbons is typically in the region of 6-8 Hz. With protons bound to sp2-hybridized carbons, coupling constants can range from 0 Hz (no coupling at all) to 18 Hz, depending on the bonding arrangement.
For vinylic hydrogens in a trans configuration, we see coupling constants in the range of 3J = 11-18 Hz, while cis hydrogens couple in the 3J = 6-15 Hz range. The 2-bond coupling between hydrogens bound to the same alkene carbon (referred to as geminal hydrogens) is very fine, generally 5 Hz or lower. Ortho hydrogens on a benzene ring couple at 6-10 Hz, while 4-bond coupling of up to 4 Hz is sometimes seen between meta hydrogens.
Fine (2-3 Hz) coupling is often seen between an aldehyde proton and a three-bond neighbor. Table 4 lists typical constant values.
5.5C: Complex coupling
In all of the examples of spin-spin coupling that we have seen so far, the observed splitting has resulted from the coupling of one set of hydrogens to just one neighboring set of hydrogens. When a set of hydrogens is coupled to two or more sets of nonequivalent neighbors, the result is a phenomenon called complex coupling. A good illustration is provided by the 1H-NMR spectrum of methyl acrylate:
First, let's first consider the Hc signal, which is centered at 6.21 ppm. Here is a closer look:
With this enlargement, it becomes evident that the Hc signal is actually composed of four sub-peaks. Why is this? Hc is coupled to both Ha and Hb , but with two different coupling constants. Once again, a splitting diagram can help us to understand what we are seeing. Ha is trans to Hc across the double bond, and splits the Hc signal into a doublet with a coupling constant of 3Jac = 17.4 Hz. In addition, each of these Hc doublet sub-peaks is split again by Hb (geminal coupling) into two more doublets, each with a much smaller coupling constant of 2Jbc = 1.5 Hz.
The result of this `double splitting` is a pattern referred to as a doublet of doublets, abbreviated `dd`.
The signal for Ha at 5.95 ppm is also a doublet of doublets, with coupling constants 3Jac= 17.4 Hz and 3Jab = 10.5 Hz.
The signal for Hb at 5.64 ppm is split into a doublet by Ha, a cis coupling with 3Jab = 10.4 Hz. Each of the resulting sub-peaks is split again by Hc, with the same geminal coupling constant 2Jbc = 1.5 Hz that we saw previously when we looked at the Hc signal. The overall result is again a doublet of doublets, this time with the two `sub-doublets` spaced slightly closer due to the smaller coupling constant for the cis interaction. Here is a blow-up of the actual Hbsignal:
Example 5.9
Construct a splitting diagram for the Hb signal in the 1H-NMR spectrum of methyl acrylate. Show the chemical shift value for each sub-peak, expressed in Hz (assume that the resonance frequency of TMS is exactly 300 MHz).
Solution
When constructing a splitting diagram to analyze complex coupling patterns, it is usually easier to show the larger splitting first, followed by the finer splitting (although the reverse would give the same end result).
When a proton is coupled to two different neighboring proton sets with identical or very close coupling constants, the splitting pattern that emerges often appears to follow the simple `n + 1 rule` of non-complex splitting. In the spectrum of 1,1,3-trichloropropane, for example, we would expect the signal for Hb to be split into a triplet by Ha, and again into doublets by Hc, resulting in a 'triplet of doublets'.
Ha and Hc are not equivalent (their chemical shifts are different), but it turns out that 3Jab is very close to 3Jbc. If we perform a splitting diagram analysis for Hb, we see that, due to the overlap of sub-peaks, the signal appears to be a quartet, and for all intents and purposes follows the n + 1 rule.
For similar reasons, the Hc peak in the spectrum of 2-pentanone appears as a sextet, split by the five combined Hb and Hd protons. Technically, this 'sextet' could be considered to be a 'triplet of quartets' with overlapping sub-peaks.
Example 5.10
What splitting pattern would you expect for the signal coresponding to Hb in the molecule below? Assume that Jab ~ Jbc. Draw a splitting diagram for this signal, and determine the relative integration values of each subpeak.
Solution
In many cases, it is difficult to fully analyze a complex splitting pattern. In the spectrum of toluene, for example, if we consider only 3-bond coupling we would expect the signal for Hb to be a doublet, Hd a triplet, and Hc a triplet.
In practice, however, all three aromatic proton groups have very similar chemical shifts and their signals overlap substantially, making such detailed analysis difficult. In this case, we would refer to the aromatic part of the spectrum as a multiplet.
When we start trying to analyze complex splitting patterns in larger molecules, we gain an appreciation for why scientists are willing to pay large sums of money (hundreds of thousands of dollars) for higher-field NMR instruments. Quite simply, the stronger our magnet is, the more resolution we get in our spectrum. In a 100 MHz instrument (with a magnet of approximately 2.4 Tesla field strength), the 12 ppm frequency 'window' in which we can observe proton signals is 1200 Hz wide. In a 500 MHz (~12 Tesla) instrument, however, the window is 6000 Hz - five times wider. In this sense, NMR instruments are like digital cameras and HDTVs: better resolution means more information and clearer pictures (and higher price tags!)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.13%3A_Splitting_Diagrams_Explain_the_Multiplicity_of_a_Signal.txt |
Nuclear spin may be related to the nucleon composition of a nucleus in the following manner:
• Odd mass nuclei (i.e. those having an odd number of nucleons) have fractional spins.
• Examples are I = 1/2 ( 1H, 13C, 19F ), I = 3/2 ( 11B ) & I = 5/2 ( 17O ).
• Even mass nuclei composed of odd numbers of protons and neutrons have integral spins. Examples are I = 1 ( 2H, 14N ).
• Even mass nuclei composed of even numbers of protons and neutrons have zero spin ( I = 0 ). Examples are 12C, and 16O.
Spin Properties of Nuclei
Spin 1/2 nuclei have a spherical charge distribution, and their NMR behavior is the easiest to understand. Other spin nuclei have nonspherical charge distributions and may be analyzed as prolate or oblate spinning bodies. All nuclei with non-zero spins have magnetic moments (μ), but the nonspherical nuclei also have an electric quadrupole moment (eQ). Some characteristic properties of selected nuclei are given in the following table.
Isotope
Natural %
Abundance
Spin (I)
Magnetic
Moment (μ)*
Magnetogyric
Ratio (γ)
1H
99.9844 1/2 2.7927 26.753
2H
0.0156 1 0.8574 4,107
11B
81.17 3/2 2.6880 --
13C
1.108 1/2 0.7022 6,728
17O
0.037 5/2 -1.8930 -3,628
19F
100.0 1/2 2.6273 25,179
29Si
4.700 1/2 -0.5555 -5,319
31P
100.0 1/2 1.1305 10,840
* μ in units of nuclear magnetons = 5.05078•10-27 JT-1
γ in units of 107rad T-1 sec-1
A Model for NMR Spectroscopy
The model of a spinning nuclear magnet aligned with or against an external magnetic field (for I = 1/2 nuclei) must be refined for effective interpretation of NMR phenomena. Just as a spinning mass will precess in a gravitational field (a gyroscope), the magnetic moment μ associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The frequency of precession is proportional to the strength of the magnetic field, as noted by the equation: ωo = γBo. The frequency ωo is called the Larmor frequency and has units of radians per second. The proportionality constant γ is known as the gyromagnetic ratio and is proportional to the magnetic moment (γ = 2pm/hI). Some characteristic γ's were listed in the preceding table of nuclear properties.
A Spinning Gyroscope in a Gravity Field
magnetic moment
μ
A Spinning Charge in a Magnetic Field
If rf energy having a frequency matching the Larmor frequency is introduced at a right angle to the external field (e.g. along the x-axis), the precessing nucleus will absorb energy and the magnetic moment will flip to its I = _1/2 state. This excitation is shown in the following diagram. Note that frequencies in radians per second may be converted to Hz (cps) by dividing by 2π.
The energy difference between nuclear spin states is small compared with the average kinetic energy of room temperature samples, and the +1/2 and _1/2 states are nearly equally populated. Indeed, in a field of 2.34 T the excess population of the lower energy state is only six nuclei per million. Although this is a very small difference , when we consider the number of atoms in a practical sample (remember the size of Avogadro's number), the numerical excess in the lower energy state is sufficient for selective and sensitive spectroscopic measurements. The diagram on the left below illustrates the macroscopic magnetization of a sample containing large numbers of spin 1/2 nuclei at equilibrium in a strong external magnetic field (Bo). A slight excess of +1/2 spin states precess randomly in alignment with the external field and a smaller population of _1/2 spin states precess randomly in an opposite alignment. An overall net magnetization therefore lies along the z-axis.
Net Macroscopic Magnetization of a Sample in an External Magnetic Field B0
Excitation by RF Energy and Subsequent Relaxation
The diagram and animation on the right show the changes in net macroscopic magnetization that occur as energy is introduced by rf irradiation at right angles to the external field. It is convenient to show the rf transmitter on the x-axis and the receiver-detector coil on the y-axis. On clicking the "Introduce RF Energy" button the animation will begin, and will repeat five times.
• First, the net magnetization shifts away from the z-axis and toward the y-axis. This occurs because some of the +1/2 nuclei are excited to the _1/2 state, and the precession about the z-axis becomes coherent (non-random), generating a significant y component to the net magnetization (M). The animation pauses at this stage.
• After irradiation the nuclear spins return to equilibrium in a process called relaxation. As the xy coherence disappears and the population of the +1/2 state increases, energy is released and detected by the receiver. The net magnetization spirals back, and eventually the equilibrium state is reestablished.
An inherent problem of the NMR experiment must be pointed out here. We have noted that the population difference between the spin states is proportionally very small. A fundamental requirement for absorption spectroscopy is a population imbalance between a lower energy ground state and a higher energy excited state. This can be expressed by the following equation, where A is a proportionality constant. If the mole fractions of the spin states are equal (η+ = η- ) then the population difference is zero and no absorption will occur. If the rf energy used in an NMR experiment is too high this saturation of the higher spin state will result and useful signals will disappear.
Relaxation Mechanisms
For NMR spectroscopy to be practical, an efficient mechanism for nuclei in the higher energy _1/2 spin state to return to the lower energy +1/2 state must exist. In other words, the spin population imbalance existing at equilibrium must be restored if spectroscopic observations are to continue. Now an isolated spinning nucleus will not spontaneouly change its spin state in the absence of external perturbation. Indeed, hydrogen gas (H2) exists as two stable spin isomers: ortho (parallel proton spins) and para (antiparallel spins). NMR spectroscopy is normally carried out in a liquid phase (solution or neat) so that there is close contact of sample molecules with a rapidly shifting crowd of other molecules (Brownian motion). This thermal motion of atoms and molecules generates local fluctuating electromagnetic fields, having components that match the Larmor frequency of the nucleus being studied. These local fields stimulate emission/absorption events that establish spin equilibrium, the excess spin energy being detected as it is released. This relaxation mechanism is called Spin-Lattice Relaxation (or Longitudinal Relaxation). The efficiency of spin-lattice relaxation depends on factors that influence molecular movement in the lattice, such as viscosity and temperature. The relaxation process is kinetically first order, and the reciprocal of the rate constant is a characteristic variable designated T1, the spin-lattice relaxation time. In non-viscous liquids at room temperature T1 ranges from 0.1 to 20 sec. A larger T1 indicates a slower or more inefficient spin relaxation. Another relaxation mechanism called spin-spin relaxation (or transverse relaxation) is characterized by a relaxation time T2. This process, which is actually a spin exchange, will not be discussed here.
Pulsed Fourier Transform Spectroscopy
In a given strong external magnetic field, each structurally distinct set of hydrogens in a molecule has a characteristic resonance frequency, just as each tubular chime in percussion instrument has a characteristic frequency. To discover the frequency of a chime we can strike it with a mallet and measure the sound emitted. This procedure can be repeated for each chime in the group so that all the characteristic frequencies are identified.
An alternative means of aquiring the same information is to strike all the chimes simultaneously, and to subject the complex collection of frequencies produced to mathematical analysis. In the following diagram the four frequencies assigned to our set of chimes are added together to give a complex summation wave. This is a straightforward conversion; and the reverse transformation, while not as simple, is readily accomplished, provided the combination signal is adequately examined and characterized.
A CW NMR spectrometer functions by irradiating each set of distinct nuclei in turn, a process analagous to striking each chime independently. For a high resolution spectrum this must be done slowly, and a 12 ppm sweep of the proton region takes from 5 to 10 minutes. It has proven much more efficient to excite all the proton nuclei in a molecule at the same time, followed by mathematical analysis of the complex rf resonance frequencies emitted as they relax back to the equilibrium state. This is the principle on which a pulse Fourier transform spectrometer operates. By exposing the sample to a very short (10 to 100 μsec), relatively strong (about 10,000 times that used for a CW spectrometer) burst of rf energy along the x-axis, as described above, all of the protons in the sample are excited simultaneously. The macroscopic magnetization model remains useful if we recognize it is a combination of megnetization vectors for all the nuclei that have been excited.
The overlapping resonance signals generated as the excited protons relax are collected by a computer and subjected to a Fourier transform mathematical analysis. As shown in the diagram on the left, the Fourier transform analysis, abbreviated FT, converts the complex time domain signal emitted by the sample into the frequency (or field) domain spectrum we are accustomed to seeing. In this fashion a complete spectrum can be acquired in a few seconds. Because the relaxation mechanism is a first order process, the rf signal emitted by the sample decays exponentially. This is called a free induction decay signal, abbreviated FID.
Free Induction Decay Signal
Since, the FID signal collected after one pulse, may be stored and averaged with the FID's from many other identical pulses prior to the Fourier transform, the NMR signal strength from a small sample may be enhanced to provide a useable spectrum. This has been essential to acquiring spectra from low abundance isotopes, such as 13C. In practice, the pulse FT experiment has proven so versatile that many variations of the technique, suited to special purposes, have been devised and used effectively.
Examples of Anisotropy Influences on Chemical Shift
The compound on the left has a chain of ten methylene groups linking para carbons of a benzene ring. Such bridged benzenes are called paracyclophanes. The meta analogs are also known. The structural constraints of the bridging chain require the middle two methylene groups to lie over the face of the benzene ring, which is a NMR shielding region. The four hydrogen atoms that are part of these groups display resonance signals that are more than two ppm higher field than the two methylene groups bonded to the edge of the ring (a deshielding region).
The 14 π-electron bridged annulene on the right is an aromatic (4n + 2) system, and has the same anisotropy as benzene. Nuclei located over the face of the ring are shielded, and those on the periphery are deshielded. The ring hydrogens give resonance signals in the range 8.0 to 8.7 δ, as expected from their deshielded location (note that there are three structurally different hydrogens on the ring). The two propyl groups are structurally equivalent (homotopic), and are free to rotate over the faces of the ring system (one above and one below). On average all the propyl hydrogens are shielded, with the innermost methylene being the most affected. The negative chemical shifts noted here indicate that the resonances occurs at a higher field than the TMS reference signal.
A remarkable characteristic of annulenes is that antiaromatic 4n π-electron systems are anisotropic in the opposite sense as their aromatic counterparts. A dramatic illustration of this fact is provided by the dianion derivative of the above bridged annulene. This dianion, formed by the addition of two electrons, is a 16 π-electron (4n) system. In the NMR spectrum of the dianion, the ring hydrogens resonate at high field (they are shielded), and the hydrogens of the propyl group are all shifted downfield (deshielded). The innermost methylene protons (magenta) give an NMR signal at +22.2 ppm, and the signals from the adjacent methylene and methyl hydrogens also have unexpectedly large chemical shifts.
Compounds in which two or more benzene rings are fused together include examples such as naphthalene, anthracene and phenanthrene, shown in the following diagram, present interesting insights into aromaticity and reactivity. The resonance stabilization of these compounds, calculated from heats of hydrogenation or combustion, is given beneath each structure.
Unlike benzene, the structures of these compounds show measurable double bond localization, which is reflected in their increased reactivity both in substitution and addition reactions. However, the 1HNMR spectra of these aromatic hydrocarbons do not provide much insight into the distribution of their pi-electrons. As expected, naphthalene displays two equally intense signals at δ 7.46 & 7.83 ppm. Likewise, anthracene shows three signals, two equal intensity multiplets at δ 7.44 & 7.98 ppm and a signal half as intense at δ 8.4 ppm. Thus, the influence of double bond localization or competition between benzene and higher annulene stabilization cannot be discerned.
The much larger C48H24 fused benzene ring cycle, named "kekulene" by Heinz Staab and sometimes called "superbenzene" by others, serves to probe the relative importance of benzenoid versus annulenoid aromaticity. A generic structure of this remarkable compound is drawn on the left below, together with two representative Kekule contributing structures on its right. There are some 200 Kekule structures that can be drawn for kekulene, but these two canonical forms represent extremes in aromaticity. The central formula has two [4n+2] annulenes, an inner [18]annulene and an outer [30]annulene (colored pink and blue respectively). The formula on the right has six benzene rings (colored green) joined in a ring by meta bonds, and held in a planar configuration by six cis-double bond bridges.
The coupled annulene contributor in the center has an energetically equivalent canonical form in which the single and double bonds making up the annulenes are exchanged. If these contributors dominate the aromatic character of kekulene, the 6 inside hydrogens should be shielded by the ring currents, and the 18 hydrogens on the periphery should be deshielded. Furthermore, the C:C bonds composing each annulene ring should have roughly equal lengths.
If the benzene contributor on the right (and its equivalent Kekule form) dominate the aromaticity of kekulene, all the benzene hydrogens will be deshielded, and the six double bond links on the periphery will have bond lengths characteristic of fixed single and double bonds
The extreme insolubility of kekulene made it difficult to grow suitable crystals for X-ray analysis or obtain solution NMR spectra. These problems were eventually solved by using high boiling solvents, the 1HNMR spectrum being taken at 150 to 200° C in deuterated tetrachlorobenzene solution. The experimental evidence demonstrates clearly that the hexa-benzene ring structure on the right most accurately represents kekulene. This evidence is shown below. The extremely low field resonance of the inside hydrogens is assigned from similar downfield shifts in model compounds.
It is important to understand that the shielding and deshielding terms used throughout our discussion of relative chemical shifts are themselves relative. Indeed, compared to a hypothetical isolated proton, all the protons in a covalent compound are shielded by the electrons in nearby sigma and pi-bonds. Consequently, it would be more accurate to describe chemical shift differences in terms of the absolute shielding experienced by different groups of hydrogens. There is, in fact, good evidence that the anisotropy of neighboring C-H and C-C sigma bonds, together with that of the bond to the observed hydrogen, are the dominate shielding factors influencing chemical shifts. The anisotropy of pi-electron systems augments this sigma skeletal shielding. Nevertheless, the deceptive focus on anisotropic pi-electron influences is so widely and commonly used that this view has been retained and employed in these pages.
Hydrogen Bonding Influences
Hydrogen bonding of hydroxyl and amino groups not only causes large variations in the chemical shift of the proton of the hydrogen bond, but also influences its coupling with adjacent C-H groups. As shown on the right, the 60 MHz proton NMR spectrum of pure (neat) methanol exhibits two signals, as expected. At 30° C these signals are sharp singlets located at δ 3.35 and 4.80 ppm, the higher-field methyl signal (magenta) being three times as strong as the OH signal (orange) at lower field. When cooled to -45 ° C, the larger higher-field signal changes to a doublet (J = 5.2 Hz) having the same chemical shift. The smaller signal moves downfield to δ 5.5 ppm and splits into a quartet (J = 5.2 Hz). The relative intensities of the two groups of signals remains unchanged. This interesting change in the NMR spectrum, which is shown in the two spectra below, is due to increased stability of hydrogen bonded species at lower temperature. Since hydrogen bonding not only causes a resonance shift to lower field, but also decreases the rate of intermolecular proton exchange, the hydroxyl proton remains bonded to the alkoxy group for a sufficient time to exert its spin coupling influence.
Under routine conditions, rapid intermolecular exchange of the OH protons of alcohols often prevents their coupling with adjacent hydrogens from being observed. Intermediate rates of proton exchange lead to a broadening of the OH and coupled hydrogen signals, a characteristic that is useful in identifying these functions. Since traces of acid or base catalyze this hydrogen exchange, pure compounds and clean sample tubes must be used for experiments of the kind described here.
Methanol 1H NMR at approximately Room Temperature
Methanol 1H NMR at -45 oC
Another way of increasing the concentration of hydrogen bonded methanol species is to change the solvent from chloroform-d to a solvent that is a stronger hydrogen bond acceptor. Examples of such solvents are given in the following table. In contrast to the neat methanol experiment described above, very dilute solutions are used for this study. Since chloroform is a poor hydrogen bond acceptor and the dilute solution reduces the concentration of methanol clusters, the hydroxyl proton of methanol generates a resonance signal at a much higher field than that observed for the pure alcohol. Indeed, the OH resonance signal from simple alcohols in dilute chloroform solution is normally found near δ 1.0 ppm.
The exceptionally strong hydrogen bond acceptor quality of DMSO is demonstrated here by the large downfield shift of the methanol hydroxyl proton, compared with a slight upfield shift of the methyl signal. The expected spin coupling patterns shown above are also observed in this solvent. Although acetone and acetonitrile are better hydrogen-bond acceptors than chloroform, they are not as effective as DMSO.
1H Chemical Shifts of Methanol in Selected Solvents
Solvent CDCl3 CD3COCD3 CD3SOCD3 CD3C≡N
CH3–O–H
CH3
O–H
3.40
1.10
3.31
3.12
3.16
4.01
3.28
2.16
The solvent effect shown above suggests a useful diagnostic procedure for characterizing the OH resonance signals from alcohol samples. For example, a solution of ethanol in chloroform-d displays the spectrum shown on the left below, especially if traces of HCl are present (otherwise broadening of the OH and CH2 signals occurs). Note that the chemical shift of the OH signal (red) is less than that of the methylene group (blue), and no coupling of the OH proton is apparent. The vicinal coupling (J = 7 Hz) of the methyl and methylene hydrogens is typical of ethyl groups. In DMSO-d6 solution small changes of chemical shift are seen for the methyl and methylene group hydrogens, but a dramatic downfield shift of the hydroxyl signal takes place because of hydrogen bonding. Coupling of the OH proton to the adjacent methylene group is evident, and both the coupling constants can be measured. Because the coupling constants are different, the methylene signal pattern is an overlapping doublet of quartets (eight distinct lines) rather than a quintet. Note that residual hydrogens in the solvent give a small broad signal near δ 2.5 ppm.
For many alcohols in dilute chloroform-d solution, the hydroxyl resonance signal is often broad and obscured by other signals in the δ 1.5 to 3.0 region. The simple technique of using DMSO-d6 as a solvent, not only shifts this signal to a lower field, but permits 1°-, 2 °- & 3 °-alcohols to be distinguished. Thus, the hydroxyl proton of 2-propanol generates a doublet at δ 4.35 ppm, and the corresponding signal from 2-methyl-2-propanol is a singlet at δ 4.2 ppm. The more acidic OH protons of phenols are similarly shifted – from δ 4 to 7 in chloroform-d to δ 8.5 to 9.5 in DMSO-d6.
Spin-Spin Coupling
vicinal (joined by three sigma bonds). In this case a neighboring proton having a +1/2 spin shifts the resonance frequency of the proton being observed to a slightly higher value (up to 7 Hz), and a _1/2 neighboring spin shifts it to a lower frequency. Remember that the total population of these two spin states is roughly equal, differing by only a few parts per million in a strong magnetic field. If several neighboring spins are present, their effect is additive.
In the spectrum of 1,1-dichloroethane shown on the right, it is clear that the three methyl hydrogens (red) are coupled with the single methyne hydrogen (orange) in a manner that causes the former to appear as a doublet and the latter as a quartet. The light gray arrow points to the unperturbed chemical shift location for each proton set. By clicking on one of these signals, the spin relationship that leads to the coupling pattern will be displayed. Clicking elsewhere in the picture will return the original spectrum.
Full Spectrum of 1,1-dichloroethane
2.06 ppm Signal Explained
5.89 ppm Signal Explained
The statistical distribution of spins within each set explains both the n+1 rule and the relative intensities of the lines within a splitting pattern. The action of a single neighboring proton is easily deduced from the fact that it must have one of two possible spins. Interaction of these two spin states with the nuclei under observation leads to a doublet located at the expected chemical shift. The corresponding action of the three protons of the methyl group requires a more detailed analysis. In the display of this interaction four possible arrays of their spins are shown. The mixed spin states are three times as possible as the all +1/2 or all _1/2 collection. Consequently, we expect four signals, two above the chemical shift and two below it. This spin analysis also suggests that the intensity ratio of these signals will be 1:3:3:1. The line separations in splitting patterns are measured in Hz, and are characteristic of the efficiency of the spin interaction; they are referred to as coupling constants (symbol J). In the above example, the common coupling constant is 6.0 Hz.
Multipliticy Relative Line Intensity (Starting at the top: Singlet, Doublet, Triplet, Quartet, and Quintet)
A simple way of estimating the relative intensities of the lines in a first-order coupling pattern is shown on the right. This array of numbers is known as Pascal's triangle, and is easily extended to predict higher multiplicities. The number appearing at any given site is the sum of the numbers linked to it from above by the light blue lines. Thus, the central number of the five quintet values is 3 + 3 = 6. Of course, a complete analysis of the spin distributions, as shown for the case of 1,1-dichloroethane above, leads to the same relative intensities.
Coupling constants are independent of the external magnetic field, and reflect the unique spin interaction characteristics of coupled sets of nuclei in a specific structure. As noted earlier, coupling constants may vary from a fraction of a Hz to nearly 20 Hz, important factors being the nature and spatial orientation of the bonds joining the coupled nuclei. In simple, freely rotating alkane units such as CH3CH2X or YCH2CH2X the coupling constant reflects an average of all significant conformers, and usually lies in a range of 6 to 8 Hz. This conformational mobility may be restricted by incorporating the carbon atoms in a rigid ring, and in this way the influence of the dihedral orientation of the coupled hydrogens may be studied.
The structures of cis and trans-4-tert-butyl-1-chlorocyclohexane, shown above, illustrate how the coupling constant changes with the dihedral angle (φ) between coupled hydrogens. The inductive effect of chlorine shifts the resonance frequency of the red colored hydrogen to a lower field (δ ca. 4.0), allowing it to be studied apart from the other hydrogens in the molecule. The preferred equatorial orientation of the large tert-butyl group holds the six-membered ring in the chair conformation depicted in the drawing. In the trans isomer this fixes the red hydrogen in an axial orientation; whereas for the cis isomer it is equatorial. The listed values for the dihedral angles and the corresponding coupling constants suggest a relationship, which has been confirmed and clarified by numerous experiments. This relationship is expressed by the Karplus equation shown below.
Geminal couplings are most commonly observed in cyclic structures, but are also evident when methylene groups have diastereomeric hydrogens.
Spin Decoupling
We have noted that rapidly exchanging hydroxyl hydrogens are not spin-coupled to adjacent C-H groups. The reason for this should be clear. As each exchange occurs, there will be an equal chance of the new proton having a +1/2 or a _1/2 spin (remember that the overall populations of the two spin states are nearly identical). Over time, therefore, the hydroxyl hydrogen behaves as though it is rapidly changing its spin, and the adjacent nuclei see only a zero spin average from it. If we could cause other protons in a molecule to undergo a similar spin averaging, their spin-coupling influence on adjacent nuclei would cease. Such NMR experiments are possible, and are called spin decoupling.
When a given set of nuclei is irradiated with strong rf energy at its characteristic Larmor frequency, spin saturation and rapid interconversion of the spin states occurs. Neighboring nuclei with different Larmor frequencies are no longer influenced by specific long-lived spins, so spin-spin signal splitting of the neighbors vanishes. The following spectrum of 1-nitropropane may be used to illustrate this technique. The three distinct sets of hydrogens in this molecule generate three resonance signals (two triplets and a broad sextet). A carefully tuned decoupling signal may be broadcast into the sample while the remaining spectrum is scanned. The region of the decoupling signal is obscured, but resonance signals more than 60 Hz away may still be seen. By clicking on one of the three signals in the spectrum, the results of decoupling at that frequency will be displayed.
The Influence of Magnetic Field Strength
The presence of symmetrical, easily recognized first-order splitting patterns in a NMR spectrum depends on the relative chemical shifts of the spin-coupled nuclei and the magnitude of the coupling constant. If the chemical shift difference (i.e. Δδ in Hz) is large compared to J the splitting patterns will be nearly first order. If, on the other hand, the difference is relatively small (less than 10 J) second order distortion of the signal splitting will be observed. One important advantage in using very high field magnets for NMR is that the separation (or dipersion) of different sets of protons is proportional to field strength, whereas coupling constants do not change.
It is important to remember that structurally different sets of nuclei do not always produce distinctly different signals in an NMR spectrum. For example, the hydrocarbon octane has four different sets of protons, as shown in the following formula:
CH3CH2CH2CH2CH2CH2CH2CH3
Now methyl hydrogens have a smaller chemical shift than methylene hydrogens, so methyl groups (colored black here) can usually be distinguished. However, the chemical shifts of the different methylene groups (blue, red & green) are so similar that many NMR spectrometers will not resolve them. Consequently, a 90 MHz proton spectrum of octane shows a distorted triplet at δ 0.9 ppm, produced by the six methyl protons, and a strong broad singlet at δ 1.2 ppm coming from all twelve methylene protons. A similar failure to resolve structurally different hydrogen atoms occurs in the case of alkyl substituted benzene rings. The chemical shift difference between ortho, meta and para hydrogens in such compounds is often so small that they are seen as a single resonance signal in an NMR spectrum. The 90 MHz spectrum of benzyl alcohol in chloroform-d solution provides an instructive example, shown below. A broad strong signal at δ 7.24 ppm is characteristic of the aromatic protons on alkylbenzenes. Since the chemical shifts of these hydrogens are nearly identical, no spin coupling is observed. If the magnetic field strength is increased to 400 Mz (lower spectrum) the aromatic protons are more dispersed (orange, magenta and green signals), and the spin coupling of adjacent hydrogens (J = 7.6 Hz) causes overlap of the signals (gray shaded enlargement).
1H NMR of Benzyl alcohol
1H NMR of Anisole
Anisole, an isomer of benzyl alcohol, has a more dispersed set of aromatic signals, thanks to the electron donating influence of the methoxy substituent. The 90 MHz spectrum of anisole shows this greater dispersion, but the spin coupling of adjacent hydrogens still results in signal overlap. The 400 Mz spectrum at the bottom illustrates the greater dispersion of the chemical shifts, and since the coupling constants remain unchanged, the splitting patterns no longer overlap. In all these examples a very small meta-hydrogen coupling has been ignored.
Example \(1\)
Not all simple compounds have simple proton NMR spectra. The following example not only illustrates this point, but also demonstrates how a careful structural analysis can rationalize an initially complex spectrum.
The 100 MHz 1H NMR spectrum of a C3H5ClO compound is initially displayed. This spectrum is obviously complex and not easily interpreted, except for concluding that no olefinic C-H protons are present.
With a higher field spectrum it is clear that each of the five hydrogen atoms in the molecule is structurally unique, and is producing a separate signal. Also, it is clear there is considerable spin coupling of all the hydrogens. To see the coupling patterns more clearly it is necessary to expand and enhance the spectrum in these regions. For purposes of our demonstration, this can be done by clicking on any one of the signal multiplets. Clicking in an open area should return the original 500 MHz display. In some of the expanded displays two adjacent groups of signals are shown. Once an enlarged pattern is displayed, the line separations in Hz can be measured (remember that for a 500 MHz spectrum 1 ppm is 500 Hz). The middle signal at 3.2 ppm is the most complex, and overlap of some multiplet lines has occurred.
Solution
This spectrum has several interesting features. First, hydrogens A, B & C are clearly different, and are spin-coupled to each other. Hydrogens B & C are geminally related, whereas A is oriented to B & C in a vicinal manner. Since JAB and JBC are similar, the HB signal is a broad triplet. Although hydrogens D & E might seem identical at first glance, they are diastereotopic, and should therefore have different chemical shifts. The DE geminal coupling constant is 11.7 Hz, so each of these hydrogens appears as a doublet of doublets.
The splitting of the HA signal is complex and not immediately obvious. The diagram shows the consequences of the four operating coupling constants.
Additional Information from 13C NMR Spectroscopy
Broad band decoupling of the hydrogen atoms in a molecule was an essential operation for obtaining simple (single line) carbon NMR spectra. The chemical shifts of the carbon signals provide useful information, but it would also be very helpful to know how many hydrogen atoms are bonded to each carbon. Aside from the fact that carbons having no bonded hydrogens generally give weak resonance signals, this information is not present in a completely decoupled spectrum.
Clever methods of retaining the hydrogen information while still enjoying the benefits of proton decoupling have been devised. The techniques involved are beyond the scope of this discussion, but the overall results can still be appreciated. The 13C NMR spectrum of camphor shown below will serve as an illustration. It will be helpful to view an expanded section of this spectrum from δ 0.0 to 50.0 ppm, and this will be presented in the High Field Expansion spectrum. The two lowest field signals are missing in the expanded display.
High Field Expansion of Camphor
Even though the expanded display now shows the distinct carbon signals clearly, the origin of each is ambiguous. An early method of regaining coupling information was by off-resonance decoupling. In this approach a weaker and more focused proton decoupling frequency is applied as the carbon spectrum is acquired. Vestiges of the C-H coupling remain in the carbon signals, but the apparent coupling constants are greatly reduced. View the Off-Resonance Decouple spectrum. The results of such an experiment will be displayed. Notice that all the methyl groups are quartets (three coupled hydrogens), the methylene groups are triplets and methine carbons are doublets. Overlap of two quartets near δ 19 ppm and the doublet and triplet near δ 43 ppm are complicating factors.
Off-Resonance Decouple Spectrum of Camphor
A better way for classifying the carbon signals is by a technique called INEPT (insensitive nuclear enhancement by polarization transfer). This method takes advantage of the influence of hydrogen on 13C relaxation times, and can be applied in several modes. One of the most common applications of INEPT separates the signals of methyl and methine carbons from those of methylene carbons by their sign. Carbons having no hydrogen substituents have a zero signal.
INEPT Spectrum of Camphor
Properties of Some Deuterated NMR Solvents
Solvent B.P. °C Residual
1H signal (δ)
Residual
13C signal (δ)
acetone-d6 55.5 2.05 ppm 206 & 29.8 ppm
acetonitrile-d3 80.7 1.95 ppm 118 & 1.3 ppm
benzene-d6 79.1 7.16 ppm 128 ppm
chloroform-d 60.9 7.27 ppm 77.2 ppm
cyclohexane-d12 78.0 1.38 ppm 26.4 ppm
dichloromethane-d2 40.0 5.32 ppm 53.8 ppm
dimethylsulfoxide-d6 190 2.50 ppm 39.5 ppm
nitromethane-d3 100 4.33 ppm 62.8 ppm
pyridine-d5 114 7.19, 7.55 & 8.71 ppm 150, 135.5 & 123.5 ppm
tetrahydrofuran-d8 65.0 1.73 & 3.58 ppm 67.4 & 25.2 ppm
Contributors
• Layne Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.16%3A_Protons_Bonded_to_Oxygen_and_Nitrogen.txt |
This page describes how you interpret simple high resolution nuclear magnetic resonance (NMR) spectra. It assumes that you have already read the background page on NMR so that you understand what an NMR spectrum looks like and the use of the term "chemical shift". It also assumes that you know how to interpret simple low resolution spectra.
The difference between high and low resolution spectra
What a low resolution NMR spectrum tells you
• The number of peaks tells you the number of different environments the hydrogen atoms are in.
• The ratio of the areas under the peaks tells you the ratio of the numbers of hydrogen atoms in each of these environments.
• The chemical shifts give you important information about the sort of environment the hydrogen atoms are in.
High resolution NMR spectra
In a high resolution spectrum, you find that many of what looked like single peaks in the low resolution spectrum are split into clusters of peaks.
1 peak a singlet
2 peaks in the cluster a doublet
3 peaks in the cluster a triplet
4 peaks in the cluster a quartet
You can get exactly the same information from a high resolution spectrum as from a low resolution one - you simply treat each cluster of peaks as if it were a single one in a low resolution spectrum. But in addition, the amount of splitting of the peaks gives you important extra information.
Interpreting a high resolution spectrum
The n+1 rule
The amount of splitting tells you about the number of hydrogens attached to the carbon atom or atoms next door to the one you are currently interested in. The number of sub-peaks in a cluster is one more than the number of hydrogens attached to the next door carbon(s). So - on the assumption that there is only one carbon atom with hydrogens on next door to the carbon we're interested in.
singlet next door to carbon with no hydrogens attached
doublet next door to a CH group
triplet next door to a CH2 group
quartet next door to a CH3 group
Using the n+1 rule
What information can you get from this NMR spectrum?
Assume that you know that the compound above has the molecular formula C4H8O2.
Treating this as a low resolution spectrum to start with, there are three clusters of peaks and so three different environments for the hydrogens. The hydrogens in those three environments are in the ratio 2:3:3. Since there are 8 hydrogens altogether, this represents a CH2 group and two CH3 groups. What about the splitting?
• The CH2 group at about 4.1 ppm is a quartet. That tells you that it is next door to a carbon with three hydrogens attached - a CH3 group.
• The CH3 group at about 1.3 ppm is a triplet. That must be next door to a CH2 group. This combination of these two clusters of peaks - one a quartet and the other a triplet - is typical of an ethyl group, CH3CH2. It is very common.
• Finally, the CH3 group at about 2.0 ppm is a singlet. That means that the carbon next door doesn't have any hydrogens attached.
So what is this compound? You would also use chemical shift data to help to identify the environment each group was in, and eventually you would come up with:
Alcohols
Where is the -O-H peak? This is very confusing! Different sources quote totally different chemical shifts for the hydrogen atom in the -OH group in alcohols - often inconsistently. For example:
• The Nuffield Data Book quotes 2.0 - 4.0, but the Nuffield text book shows a peak at about 5.4.
• The OCR Data Sheet for use in their exams quotes 3.5 - 5.5.
• A reliable degree level organic chemistry text book quotes1.0 - 5.0, but then shows an NMR spectrum for ethanol with a peak at about 6.1.
• The SDBS database (used throughout this site) gives the -OH peak in ethanol at about 2.6.
The problem seems to be that the position of the -OH peak varies dramatically depending on the conditions - for example, what solvent is used, the concentration, and the purity of the alcohol - especially on whether or not it is totally dry.
A clever way of picking out the -OH peak
If you measure an NMR spectrum for an alcohol like ethanol, and then add a few drops of deuterium oxide, D2O, to the solution, allow it to settle and then re-measure the spectrum, the -OH peak disappears! By comparing the two spectra, you can tell immediately which peak was due to the -OH group.
The reason for the loss of the peak lies in the interaction between the deuterium oxide and the alcohol. All alcohols, such as ethanol, are very, very slightly acidic. The hydrogen on the -OH group transfers to one of the lone pairs on the oxygen of the water molecule. The fact that here we've got "heavy water" makes no difference to that.
The negative ion formed is most likely to bump into a simple deuterium oxide molecule to regenerate the alcohol - except that now the -OH group has turned into an -OD group.
Deuterium atoms don't produce peaks in the same region of an NMR spectrum as ordinary hydrogen atoms, and so the peak disappears.
You might wonder what happens to the positive ion in the first equation and the OD- in the second one. These get lost into the normal equilibrium which exists wherever you have water molecules - heavy or otherwise.
The lack of splitting with -OH groups
Unless the alcohol is absolutely free of any water, the hydrogen on the -OH group and any hydrogens on the next door carbon don't interact to produce any splitting. The -OH peak is a singlet and you don't have to worry about its effect on the next door hydrogens.
The left-hand cluster of peaks is due to the CH2 group. It is a quartet because of the 3 hydrogens on the next door CH3 group. You can ignore the effect of the -OH hydrogen. Similarly, the -OH peak in the middle of the spectrum is a singlet. It hasn't turned into a triplet because of the influence of the CH2 group.
Equivalent hydrogen atoms
Hydrogen atoms attached to the same carbon atom are said to be equivalent. Equivalent hydrogen atoms have no effect on each other - so that one hydrogen atom in a CH2 group doesn't cause any splitting in the spectrum of the other one.
But hydrogen atoms on neighboring carbon atoms can also be equivalent if they are in exactly the same environment. For example:
These four hydrogens are all exactly equivalent. You would get a single peak with no splitting at all. You only have to change the molecule very slightly for this no longer to be true.
Because the molecule now contains different atoms at each end, the hydrogens are no longer all in the same environment. This compound would give two separate peaks on a low resolution NMR spectrum. The high resolution spectrum would show that both peaks subdivided into triplets - because each is next door to a differently placed CH2 group. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.17%3A_The_Use_of_Deuterium_in_%281H%29_NMR_Spectroscopy.txt |
This page describes how you interpret simple high resolution nuclear magnetic resonance (NMR) spectra. It assumes that you have already read the background page on NMR so that you understand what an NMR spectrum looks like and the use of the term "chemical shift". It also assumes that you know how to interpret simple low resolution spectra.
The difference between high and low resolution spectra
What a low resolution NMR spectrum tells you
• The number of peaks tells you the number of different environments the hydrogen atoms are in.
• The ratio of the areas under the peaks tells you the ratio of the numbers of hydrogen atoms in each of these environments.
• The chemical shifts give you important information about the sort of environment the hydrogen atoms are in.
High resolution NMR spectra
In a high resolution spectrum, you find that many of what looked like single peaks in the low resolution spectrum are split into clusters of peaks.
1 peak a singlet
2 peaks in the cluster a doublet
3 peaks in the cluster a triplet
4 peaks in the cluster a quartet
You can get exactly the same information from a high resolution spectrum as from a low resolution one - you simply treat each cluster of peaks as if it were a single one in a low resolution spectrum. But in addition, the amount of splitting of the peaks gives you important extra information.
Interpreting a high resolution spectrum
The n+1 rule
The amount of splitting tells you about the number of hydrogens attached to the carbon atom or atoms next door to the one you are currently interested in. The number of sub-peaks in a cluster is one more than the number of hydrogens attached to the next door carbon(s). So - on the assumption that there is only one carbon atom with hydrogens on next door to the carbon we're interested in.
singlet next door to carbon with no hydrogens attached
doublet next door to a CH group
triplet next door to a CH2 group
quartet next door to a CH3 group
Using the n+1 rule
What information can you get from this NMR spectrum?
Assume that you know that the compound above has the molecular formula C4H8O2.
Treating this as a low resolution spectrum to start with, there are three clusters of peaks and so three different environments for the hydrogens. The hydrogens in those three environments are in the ratio 2:3:3. Since there are 8 hydrogens altogether, this represents a CH2 group and two CH3 groups. What about the splitting?
• The CH2 group at about 4.1 ppm is a quartet. That tells you that it is next door to a carbon with three hydrogens attached - a CH3 group.
• The CH3 group at about 1.3 ppm is a triplet. That must be next door to a CH2 group. This combination of these two clusters of peaks - one a quartet and the other a triplet - is typical of an ethyl group, CH3CH2. It is very common.
• Finally, the CH3 group at about 2.0 ppm is a singlet. That means that the carbon next door doesn't have any hydrogens attached.
So what is this compound? You would also use chemical shift data to help to identify the environment each group was in, and eventually you would come up with:
Alcohols
Where is the -O-H peak? This is very confusing! Different sources quote totally different chemical shifts for the hydrogen atom in the -OH group in alcohols - often inconsistently. For example:
• The Nuffield Data Book quotes 2.0 - 4.0, but the Nuffield text book shows a peak at about 5.4.
• The OCR Data Sheet for use in their exams quotes 3.5 - 5.5.
• A reliable degree level organic chemistry text book quotes1.0 - 5.0, but then shows an NMR spectrum for ethanol with a peak at about 6.1.
• The SDBS database (used throughout this site) gives the -OH peak in ethanol at about 2.6.
The problem seems to be that the position of the -OH peak varies dramatically depending on the conditions - for example, what solvent is used, the concentration, and the purity of the alcohol - especially on whether or not it is totally dry.
A clever way of picking out the -OH peak
If you measure an NMR spectrum for an alcohol like ethanol, and then add a few drops of deuterium oxide, D2O, to the solution, allow it to settle and then re-measure the spectrum, the -OH peak disappears! By comparing the two spectra, you can tell immediately which peak was due to the -OH group.
The reason for the loss of the peak lies in the interaction between the deuterium oxide and the alcohol. All alcohols, such as ethanol, are very, very slightly acidic. The hydrogen on the -OH group transfers to one of the lone pairs on the oxygen of the water molecule. The fact that here we've got "heavy water" makes no difference to that.
The negative ion formed is most likely to bump into a simple deuterium oxide molecule to regenerate the alcohol - except that now the -OH group has turned into an -OD group.
Deuterium atoms don't produce peaks in the same region of an NMR spectrum as ordinary hydrogen atoms, and so the peak disappears.
You might wonder what happens to the positive ion in the first equation and the OD- in the second one. These get lost into the normal equilibrium which exists wherever you have water molecules - heavy or otherwise.
The lack of splitting with -OH groups
Unless the alcohol is absolutely free of any water, the hydrogen on the -OH group and any hydrogens on the next door carbon don't interact to produce any splitting. The -OH peak is a singlet and you don't have to worry about its effect on the next door hydrogens.
The left-hand cluster of peaks is due to the CH2 group. It is a quartet because of the 3 hydrogens on the next door CH3 group. You can ignore the effect of the -OH hydrogen. Similarly, the -OH peak in the middle of the spectrum is a singlet. It hasn't turned into a triplet because of the influence of the CH2 group.
Equivalent hydrogen atoms
Hydrogen atoms attached to the same carbon atom are said to be equivalent. Equivalent hydrogen atoms have no effect on each other - so that one hydrogen atom in a CH2 group doesn't cause any splitting in the spectrum of the other one.
But hydrogen atoms on neighboring carbon atoms can also be equivalent if they are in exactly the same environment. For example:
These four hydrogens are all exactly equivalent. You would get a single peak with no splitting at all. You only have to change the molecule very slightly for this no longer to be true.
Because the molecule now contains different atoms at each end, the hydrogens are no longer all in the same environment. This compound would give two separate peaks on a low resolution NMR spectrum. The high resolution spectrum would show that both peaks subdivided into triplets - because each is next door to a differently placed CH2 group. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.18%3A_The_Resolution_of_%281H%29_NMR_Spectra.txt |
The 12C isotope of carbon - which accounts for up about 99% of the carbons in organic molecules - does not have a nuclear magnetic moment, and thus is NMR-inactive. Fortunately for organic chemists, however, the 13C isotope, which accounts for most of the remaining 1% of carbon atoms in nature, has a magnetic moment just like protons. Most of what we have learned about 1H-NMR spectroscopy also applies to 13C-NMR, although there are several important differences.
5.6A: The basics of 13C-NMR spectroscopy
The magnetic moment of a 13C nucleus is much weaker than that of a proton, meaning that NMR signals from 13C nuclei are inherently much weaker than proton signals. This, combined with the low natural abundance of 13C, means that it is much more difficult to observe carbon signals: more sample is required, and often the data from hundreds of scans must be averaged in order to bring the signal-to-noise ratio down to acceptable levels. Unlike 1H-NMR signals, the area under a 13C-NMR signal cannot be used to determine the number of carbons to which it corresponds. This is because the signals for some types of carbons are inherently weaker than for other types – peaks corresponding to carbonyl carbons, for example, are much smaller than those for methyl or methylene (CH2) peaks. Peak integration is generally not useful in 13C-NMR spectroscopy, except when investigating molecules that have been enriched with 13C isotope (see section 5.6B).
The resonance frequencies of 13C nuclei are lower than those of protons in the same applied field - in a 7.05 Tesla instrument, protons resonate at about 300 MHz, while carbons resonate at about 75 MHz. This is fortunate, as it allows us to look at 13C signals using a completely separate 'window' of radio frequencies. Just like in 1H-NMR, the standard used in 13C-NMR experiments to define the 0 ppm point is tetramethylsilane (TMS), although of course in 13C-NMR it is the signal from the four equivalent carbons in TMS that serves as the standard. Chemical shifts for 13C nuclei in organic molecules are spread out over a much wider range than for protons – up to 200 ppm for 13C compared to 12 ppm for protons (see Table 3 for a list of typical 13C-NMR chemical shifts). This is also fortunate, because it means that the signal from each carbon in a compound can almost always be seen as a distinct peak, without the overlapping that often plagues 1H-NMR spectra. The chemical shift of a 13C nucleus is influenced by essentially the same factors that influence a proton's chemical shift: bonds to electronegative atoms and diamagnetic anisotropy effects tend to shift signals downfield (higher resonance frequency). In addition, sp2 hybridization results in a large downfield shift. The 13C-NMR signals for carbonyl carbons are generally the furthest downfield (170-220 ppm), due to both sp2 hybridization and to the double bond to oxygen.
Example 5.11
How many sets of non-equivalent carbons are there in each of the molecules shown in exercise 5.1?
Solution
Example 5.12
How many sets of non-equivalent carbons are there in:
1. toluene
2. 2-pentanone
3. para-xylene
4. triclosan
(all structures are shown earlier in this chapter)
Solution
Because of the low natural abundance of 13C nuclei, it is very unlikely to find two 13C atoms near each other in the same molecule, and thus we do not see spin-spin coupling between neighboring carbons in a 13C-NMR spectrum. There is, however, heteronuclear coupling between 13C carbons and the hydrogens to which they are bound. Carbon-proton coupling constants are very large, on the order of 100 – 250 Hz. For clarity, chemists generally use a technique called broadband decoupling, which essentially 'turns off' C-H coupling, resulting in a spectrum in which all carbon signals are singlets. Below is the proton-decoupled13C-NMR spectrum of ethyl acetate, showing the expected four signals, one for each of the carbons.
While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH2) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH3) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13C spectral data in examples and problems.
Below are two more examples of 13C NMR spectra of simple organic molecules, along with DEPT information.
Example 5.13
Give peak assignments for the 13C-NMR spectrum of methyl methacrylate, shown above.
One of the greatest advantages of 13C-NMR compared to 1H-NMR is the breadth of the spectrum - recall that carbons resonate from 0-220 ppm relative to the TMS standard, as opposed to only 0-12 ppm for protons. Because of this, 13C signals rarely overlap, and we can almost always distinguish separate peaks for each carbon, even in a relatively large compound containing carbons in very similar environments. In the proton spectrum of 1-heptanol, for example, only the signals for the alcohol proton (Ha) and the two protons on the adjacent carbon (Hb) are easily analyzed. The other proton signals overlap, making analysis difficult.
In the 13C spectrum of the same molecule, however, we can easily distinguish each carbon signal, and we know from this data that our sample has seven non-equivalent carbons. (Notice also that, as we would expect, the chemical shifts of the carbons get progressively smaller as they get farther away from the deshielding oxygen.)
This property of 13C-NMR makes it very helpful in the elucidation of larger, more complex structures.
Example 5.14
13C-NMR (and DEPT) data for some common biomolecules are shown below (data is from the Aldrich Library of 1H and 13C NMR). Match the NMR data to the correct structure, and make complete peak assignments.
• spectrum a: 168.10 ppm (C), 159.91 ppm (C), 144.05 ppm (CH), 95.79 ppm (CH)
• spectrum b: 207.85 ppm (C), 172.69 ppm (C), 29.29 ppm (CH3)
• spectrum c: 178.54 ppm (C), 53.25 ppm (CH), 18.95 ppm (CH3)
• spectrum d: 183.81 ppm (C), 182. 63 ppm (C), 73.06 ppm (CH), 45.35 ppm (CH2)
Solution
5.6B: \(^{13}C\)-NMR in isotopic labeling studies
Although only about 1 out of 100 carbon atoms in a naturally occurring organic molecule is the 13C isotope, chemists are often able to synthesize molecules that are artificially enriched in 13C at specific carbon positions, sometimes to the point where the 13C isotope is incorporated at a given position in over half of the molecules in the sample. This can be very useful, especially in biochemical studies, because it allows us to 'label' one or more carbons in a small precursor molecule and then trace the presence of the 13C label through a metabolic pathway all the way to a larger biomolecule product. For example, scientists were able to grow bacteria in a medium in which the only source of carbon was acetate enriched in 13C at the C1 (carbonyl) position. When they isolated a large molecule called amino-bacterio-hopanetriol (very similar in structure to cholesterol) from these bacteria and looked at its 13C-NMR spectrum, they observed that the 13C label from acetate had been incorporated at eight specific positions. They knew this because the 13C-NMR signals for these carbons were much stronger compared to the same signals in a control (unlabeled) molecule.
This result was very surprising - the scientists had expected a completely different pattern of 13C incorporation based on what they believed to be the metabolic pathway involved. This unexpected result led eventually to the discovery that bacteria synthesize these types of molecules in a very different way than yeasts, plants, and animals (Eur. J. Biochem. 1988, 175, 405). The newly discovered bacterial metabolic pathway is currently a key target for the development of new antibiotic and antimalaria drugs.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.19%3A_%2813C%29_NMR_Spectroscopy.txt |
Nuclear Magnetic Resonance (NMR) interpretation plays a pivotal role in molecular identifications. As interpreting NMR spectra, the structure of an unknown compound, as well as known structures, can be assigned by several factors such as chemical shift, spin multiplicity, coupling constants, and integration. This Module focuses on the most important 1H and 13C NMR spectra to find out structure even though there are various kinds of NMR spectra such as 14N, 19F, and 31P. NMR spectrum shows that x- axis is chemical shift in ppm. It also contains integral areas, splitting pattern, and coupling constant.
Strategy for Solving Structure
Here is the general strategy for solving structure with NMR:
1. Molecular formula is determined by chemical analysis such as elementary analysis
2. Double-bond equivalent (also known as Degree of Unsaturation) is calculated by a simple equation to estimate the number of the multiple bonds and rings. It assumes that oxygen (O) and sulfur (S) are ignored and halogen (Cl, Br) and nitrogen is replaced by CH. The resulting empirical formula is CaHb
1. Structure fragmentation is determined by chemical shift, spin multiplicity, integral (peak area), and coupling constants ($^1J$, $^2J$)
2. Molecular skeleton is built up using 2-dimensional NMR spectroscopy.
3. Relative configuration is predicted by coupling constant (3J).
1H NMR
Chemical Shift
Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilane (TMS, $\ce{(CH3)4Si}$) is generally used as an internal standard to determine chemical shift of compounds: δTMS=0 ppm. In other words, frequencies for chemicals are measured for a 1H or 13C nucleus of a sample from the 1H or 13C resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1H chemical shift play a role in identifying many functional groups. Figure $1$. indicates important example to figure out the functional groups.
Chemical equivalence
Protons with Chemical equivalence has the same chemical shift due to symmetry within molecule ($CH_3COCH_3$) or fast rotation around single bond (-CH3; methyl groups).
Spin-Spin Splitting
Spin-Spin splitting means that an absorbing peak is split by more than one “neighbor” proton. Splitting signals are separated to J Hz, where is called the coupling constant. The spitting is a very essential part to obtain exact information about the number of the neighboring protons. The maximum of distance for splitting is three bonds. Chemical equivalent protons do not result in spin-spin splitting. When a proton splits, the proton’s chemical shift is determined in the center of the splitting lines.
Spin Multiplicity (Splitting pattern)
Spin Multiplicity plays a role in determining the number of neighboring protons. Here is a multiplicity rules: In case of $A_mB_n$ system, the multiplicity rule is that Nuclei of $B$ element produce a splitting the $A$ signal into $nB+1$ lines. The general formula which applies to all nuclei is $2_nI+1$, where $I$ is the spin quantum number of the coupled element. The relative intensities of the each lines are given by the coefficients of the Pascal’s triangle (Figure $2$).
First-order splitting pattern
The chemical shift difference in Hertz between coupled protons in Hertz is much larger than the $J$ coupling constant:
$\dfrac{\Delta \nu }{J} \ge 8$
Where $\Delta \nu$ is the difference of chemical shift. In other word, the proton is only coupled to other protons that are far away in chemical shift. The spectrum is called first-order spectrum. The splitting pattern depends on the magnetic field. The second-order splitting at the lower field can be resolved into first-order splitting pattern at the high field. The first-order splitting pattern is allowed to multiplicity rule (N+1) and Pascal’s triangle to determine splitting pattern and intensity distribution.
Example $1$
The note is that structure system is A3M2X2. Ha and Hx has the triplet pattern by Hm because of N+1 rule. The signal of Hm is split into six peaks by Hx and Ha (Figure3) The First order pattern easily is predicted due to separation with equal splitting pattern.
High-order splitting pattern
High-order splitting pattern takes place when chemical shift difference in Hertz is much less or the same that order of magnitude as the j coupling.
$\frac{\Delta v}{J} \leq 10$
The second order pattern is observed as leaning of a classical pattern: the inner peaks are taller and the outer peaks are shorter in case of AB system (Figure $4$). This is called the roof effect.
Here is other system as an example: A2B2 (Figure $5$). The two triplet incline toward each other. Outer lines of the triplet are less than 1 in relative area and the inner lines are more than 1. The center lines have relative area 2.
Coupling constant (J Value)
Coupling constant is the strength of the spin-spin splitting interaction and the distance between the split lines. The value of distance is equal or different depending on the coupled nuclei. The coupling constants reflect the bonding environments of the coupled nuclei. Coupling constant is classified by the number of bonds:
Geminal proton-proton coupling (2JHH)
Germinal coupling generates through two bonds (Figure $6$). Two proton having geminal coupling are not chemically equivalent. This coupling ranges from -20 to 40 Hz. 2JHHdepends on hybridization of carbon atom and the bond angle and the substituent such as electronegative atoms. When S-character is increased, Geminal coupling constant is increased: 2Jsp1>2Jsp2>2Jsp3 The bond angle(HCH) gives rise to change 2JHH value and depend on the strain of the ring in the cyclic systems. Geminal coupling constant determines ring size. When bond angle is decreased, ring size is decreased so that geminal coupling constant is more positive. If a atom is replace to an electronegative atom, Geminal coupling constant move to positive value.
Vicinal proton-proton coupling (3JHH)
Vicinal coupling occurs though three bonds (Figure $7$.). The Vicinal coupling is the most useful information of dihedral angle, leading to stereochemistry and conformation of molecules. Vicinal coupling constant always has the positive value and is affected by the dihedral angle (?;HCCH), the valence angle (?; HCC), the bond length of carbon-carbon, and the effects of electronegative atoms. Vicinal coupling constant depending on the dihedral angle (Figure $8$) is given by the Karplus equation.
$^3 J=7.0-0.5 \cos \phi+4.5 \cos ^{2} \phi$
When ? is the 90o, vicinal coupling constant is zero. In addition, vicinal coupling constant ranges from 8 to 10 Hz at the and ?=180o, where ?=0o and ?=180o means that the coupled protons have cis and trans configuration, respectively.
The valence angle(?;Figure $8$) also causes change of 3JHH value. Valence angle is related with ring size. Typically, when the valence angle decreases, the coupling constant reduces. The distance between the carbons atoms gives influences to vicinal coupling constant
The coupling constant increases with the decrease of bond length. Electronegative atoms affect vicinal coupling constants so that electronegative atoms decrease the vicinal coupling constants.
Integral
Integral is referred to integrated peak area of 1H signals. The intensity is directly proportionally to the number of hydrogen.
13C NMR
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C- 13 C spin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Figure $10$.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
• Distortionless enhancement by polarization transfer (DEPT): DEPT is used for distinguishing between a CH3 group, a CH2 group, and a CH group. The proton pulse is set at 45o, 90o, or 135o in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Figure $11$. is an example about DEPT spectrum.
2-dimensional NMR spectroscopy (COSY)
COSY stands for COrrelation SpectroscopY. COSY spectrum is more useful information about what is being correlated.
1H-1H COSY (COrrelation SpectroscopY)
1H-1H COSY is used for clearly indicate correlation with coupled protons. A point of entry into a COSY spectrum is one of the keys to predict information from it successfully. Relation of Coupling protons is determined by cross peaks(correlation peaks) and in the COSY spectrum. In other words, Diagonal peaks by lines ar e coupled to each other. Figure $12$ indicates that there are correlation peaks between proton H1 and H2 as well as between H2 and H4. This means the H2 coupled to H1 and H4.
1H-13C COSY (HETCOR)
1H-13C COSY is the heteronuclear correlation spectroscopy. The HETCOR spectrum is correlated 13C nuclei with directly attached protons. 1H-13C coupling is one bond. The cross peaks mean correlation between a proton and a carbon (Figure $13$). If a line does not have cross peak, this means that this carbon atoms has no attached proton (e.g. a quaternary carbon atom)
• NMRShiftDB: a Free web database for NMR data : nmrshiftdb.chemie.uni-mainz.de/nmrshiftdb
• NMR database from ACD/LAbs : www.acdlabs.com/products/spec_lab/exp_spectra/spec_libraries/aldrich.html
• NMR database from John Crerar Library : http://crerar.typepad.com/crerar_lib...h_ir_nmr_.html
Problems
Draw the 1H NMR spectrum for 2-Hydroxypropane in CDCl3. Assume sufficient resolution to provide a first-order spectrum and ignore vicinal proton-proton coupling(3JHH)
Solution
1) the structure of 2-hydoroxyporpane is drawn
Figure out which protons are chemically equivalent, i.e., two methyl (-CH3) groups are chemical equivalent.
4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by Hb(#of proton=1). Hb has the septet pattern by Ha (#of proton=6). Hc has one peak.(Note that Hc has doublet pattern by Hb due to vicinal proton-proton coupling.)
• You Jin Seo | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.20%3A_DEPT_%2813C%29_NMR_Spectra.txt |
Nuclear Magnetic Resonance (NMR) interpretation plays a pivotal role in molecular identifications. As interpreting NMR spectra, the structure of an unknown compound, as well as known structures, can be assigned by several factors such as chemical shift, spin multiplicity, coupling constants, and integration. This Module focuses on the most important 1H and 13C NMR spectra to find out structure even though there are various kinds of NMR spectra such as 14N, 19F, and 31P. NMR spectrum shows that x- axis is chemical shift in ppm. It also contains integral areas, splitting pattern, and coupling constant.
Strategy for Solving Structure
Here is the general strategy for solving structure with NMR:
1. Molecular formula is determined by chemical analysis such as elementary analysis
2. Double-bond equivalent (also known as Degree of Unsaturation) is calculated by a simple equation to estimate the number of the multiple bonds and rings. It assumes that oxygen (O) and sulfur (S) are ignored and halogen (Cl, Br) and nitrogen is replaced by CH. The resulting empirical formula is CaHb
1. Structure fragmentation is determined by chemical shift, spin multiplicity, integral (peak area), and coupling constants ($^1J$, $^2J$)
2. Molecular skeleton is built up using 2-dimensional NMR spectroscopy.
3. Relative configuration is predicted by coupling constant (3J).
1H NMR
Chemical Shift
Chemical shift is associated with the Larmor frequency of a nuclear spin to its chemical environment. Tetramethylsilane (TMS, $\ce{(CH3)4Si}$) is generally used as an internal standard to determine chemical shift of compounds: δTMS=0 ppm. In other words, frequencies for chemicals are measured for a 1H or 13C nucleus of a sample from the 1H or 13C resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1H chemical shift play a role in identifying many functional groups. Figure $1$. indicates important example to figure out the functional groups.
Chemical equivalence
Protons with Chemical equivalence has the same chemical shift due to symmetry within molecule ($CH_3COCH_3$) or fast rotation around single bond (-CH3; methyl groups).
Spin-Spin Splitting
Spin-Spin splitting means that an absorbing peak is split by more than one “neighbor” proton. Splitting signals are separated to J Hz, where is called the coupling constant. The spitting is a very essential part to obtain exact information about the number of the neighboring protons. The maximum of distance for splitting is three bonds. Chemical equivalent protons do not result in spin-spin splitting. When a proton splits, the proton’s chemical shift is determined in the center of the splitting lines.
Spin Multiplicity (Splitting pattern)
Spin Multiplicity plays a role in determining the number of neighboring protons. Here is a multiplicity rules: In case of $A_mB_n$ system, the multiplicity rule is that Nuclei of $B$ element produce a splitting the $A$ signal into $nB+1$ lines. The general formula which applies to all nuclei is $2_nI+1$, where $I$ is the spin quantum number of the coupled element. The relative intensities of the each lines are given by the coefficients of the Pascal’s triangle (Figure $2$).
First-order splitting pattern
The chemical shift difference in Hertz between coupled protons in Hertz is much larger than the $J$ coupling constant:
$\dfrac{\Delta \nu }{J} \ge 8$
Where $\Delta \nu$ is the difference of chemical shift. In other word, the proton is only coupled to other protons that are far away in chemical shift. The spectrum is called first-order spectrum. The splitting pattern depends on the magnetic field. The second-order splitting at the lower field can be resolved into first-order splitting pattern at the high field. The first-order splitting pattern is allowed to multiplicity rule (N+1) and Pascal’s triangle to determine splitting pattern and intensity distribution.
Example $1$
The note is that structure system is A3M2X2. Ha and Hx has the triplet pattern by Hm because of N+1 rule. The signal of Hm is split into six peaks by Hx and Ha (Figure3) The First order pattern easily is predicted due to separation with equal splitting pattern.
High-order splitting pattern
High-order splitting pattern takes place when chemical shift difference in Hertz is much less or the same that order of magnitude as the j coupling.
$\frac{\Delta v}{J} \leq 10$
The second order pattern is observed as leaning of a classical pattern: the inner peaks are taller and the outer peaks are shorter in case of AB system (Figure $4$). This is called the roof effect.
Here is other system as an example: A2B2 (Figure $5$). The two triplet incline toward each other. Outer lines of the triplet are less than 1 in relative area and the inner lines are more than 1. The center lines have relative area 2.
Coupling constant (J Value)
Coupling constant is the strength of the spin-spin splitting interaction and the distance between the split lines. The value of distance is equal or different depending on the coupled nuclei. The coupling constants reflect the bonding environments of the coupled nuclei. Coupling constant is classified by the number of bonds:
Geminal proton-proton coupling (2JHH)
Germinal coupling generates through two bonds (Figure $6$). Two proton having geminal coupling are not chemically equivalent. This coupling ranges from -20 to 40 Hz. 2JHHdepends on hybridization of carbon atom and the bond angle and the substituent such as electronegative atoms. When S-character is increased, Geminal coupling constant is increased: 2Jsp1>2Jsp2>2Jsp3 The bond angle(HCH) gives rise to change 2JHH value and depend on the strain of the ring in the cyclic systems. Geminal coupling constant determines ring size. When bond angle is decreased, ring size is decreased so that geminal coupling constant is more positive. If a atom is replace to an electronegative atom, Geminal coupling constant move to positive value.
Vicinal proton-proton coupling (3JHH)
Vicinal coupling occurs though three bonds (Figure $7$.). The Vicinal coupling is the most useful information of dihedral angle, leading to stereochemistry and conformation of molecules. Vicinal coupling constant always has the positive value and is affected by the dihedral angle (?;HCCH), the valence angle (?; HCC), the bond length of carbon-carbon, and the effects of electronegative atoms. Vicinal coupling constant depending on the dihedral angle (Figure $8$) is given by the Karplus equation.
$^3 J=7.0-0.5 \cos \phi+4.5 \cos ^{2} \phi$
When ? is the 90o, vicinal coupling constant is zero. In addition, vicinal coupling constant ranges from 8 to 10 Hz at the and ?=180o, where ?=0o and ?=180o means that the coupled protons have cis and trans configuration, respectively.
The valence angle(?;Figure $8$) also causes change of 3JHH value. Valence angle is related with ring size. Typically, when the valence angle decreases, the coupling constant reduces. The distance between the carbons atoms gives influences to vicinal coupling constant
The coupling constant increases with the decrease of bond length. Electronegative atoms affect vicinal coupling constants so that electronegative atoms decrease the vicinal coupling constants.
Integral
Integral is referred to integrated peak area of 1H signals. The intensity is directly proportionally to the number of hydrogen.
13C NMR
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C- 13 C spin-spin splitting rarely exit between adjacent carbons because 13C is naturally lower abundant (1.1%)
• 13C-1H Spin coupling: 13C-1H Spin coupling provides useful information about the number of protons attached a carbon atom. In case of one bond coupling (1JCH), -CH, -CH2, and CH3 have respectively doublet, triplet, quartets for the 13C resonances in the spectrum. However, 13C-1H Spin coupling has an disadvantage for 13C spectrum interpretation. 13C-1H Spin coupling is hard to analyze and reveal structure due to a forest of overlapping peaks that result from 100% abundance of 1H.
• Decoupling: Decoupling is the process of removing 13C-1H coupling interaction to simplify a spectrum and identify which pair of nuclei is involved in the J coupling. The decoupling 13C spectra shows only one peak(singlet) for each unique carbon in the molecule(Figure $10$.). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
• Distortionless enhancement by polarization transfer (DEPT): DEPT is used for distinguishing between a CH3 group, a CH2 group, and a CH group. The proton pulse is set at 45o, 90o, or 135o in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Figure $11$. is an example about DEPT spectrum.
2-dimensional NMR spectroscopy (COSY)
COSY stands for COrrelation SpectroscopY. COSY spectrum is more useful information about what is being correlated.
1H-1H COSY (COrrelation SpectroscopY)
1H-1H COSY is used for clearly indicate correlation with coupled protons. A point of entry into a COSY spectrum is one of the keys to predict information from it successfully. Relation of Coupling protons is determined by cross peaks(correlation peaks) and in the COSY spectrum. In other words, Diagonal peaks by lines ar e coupled to each other. Figure $12$ indicates that there are correlation peaks between proton H1 and H2 as well as between H2 and H4. This means the H2 coupled to H1 and H4.
1H-13C COSY (HETCOR)
1H-13C COSY is the heteronuclear correlation spectroscopy. The HETCOR spectrum is correlated 13C nuclei with directly attached protons. 1H-13C coupling is one bond. The cross peaks mean correlation between a proton and a carbon (Figure $13$). If a line does not have cross peak, this means that this carbon atoms has no attached proton (e.g. a quaternary carbon atom)
• NMRShiftDB: a Free web database for NMR data : nmrshiftdb.chemie.uni-mainz.de/nmrshiftdb
• NMR database from ACD/LAbs : www.acdlabs.com/products/spec_lab/exp_spectra/spec_libraries/aldrich.html
• NMR database from John Crerar Library : http://crerar.typepad.com/crerar_lib...h_ir_nmr_.html
Problems
Draw the 1H NMR spectrum for 2-Hydroxypropane in CDCl3. Assume sufficient resolution to provide a first-order spectrum and ignore vicinal proton-proton coupling(3JHH)
Solution
1) the structure of 2-hydoroxyporpane is drawn
Figure out which protons are chemically equivalent, i.e., two methyl (-CH3) groups are chemical equivalent.
4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by Hb(#of proton=1). Hb has the septet pattern by Ha (#of proton=6). Hc has one peak.(Note that Hc has doublet pattern by Hb due to vicinal proton-proton coupling.)
• You Jin Seo
14.22: NMR Used in Medicine is Called Magnetic Resonance Imaging
Magnetic Resonance Imaging (MRI), nuclear magnetic resonance imaging (NMRI), or magnetic resonance tomography (MRT) is a medical imaging technique used in radiology to visualize internal structures of the body in detail. MRI makes use of the property of nuclear magnetic resonance (NMR) to image nuclei of atoms inside the body.
An MRI scanner is a device in which the patient lies within a large, powerful magnet where the magnetic field is used to align the magnetization of some atomic nuclei in the body, and radio frequency magnetic fields are applied to systematically alter the alignment of this magnetization.[1] This causes the nuclei to produce a rotating magnetic field detectable by the scanner—and this information is recorded to construct an image of the scanned area of the body.[2]:36 Magnetic field gradients cause nuclei at different locations to precess at different speeds, which allows spatial information to be recovered using Fourier analysis of the measured signal. By using gradients in different directions, 2D images or 3D volumes can be obtained in any arbitrary orientatation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.21%3A_Two-Dimensional_NMR_Spectroscopy.txt |
X-ray Crystallography is a scientific method used to determine the arrangement of atoms of a crystalline solid in three dimensional space. This technique takes advantage of the interatomic spacing of most crystalline solids by employing them as a diffraction gradient for x-ray light, which has wavelengths on the order of 1 angstrom (10-8 cm).
Introduction
In 1895, Wilhelm Rontgen discovered x- rays. The nature of x- rays, whether they were particles or electromagnetic radiation, was a topic of debate until 1912. If the wave idea was correct, researchers knew that the wavelength of this light would need to be on the order of 1 Angstrom (A) (10-8 cm). Diffraction and measurement of such small wavelengths would require a gradient with spacing on the same order of magnitude as the light.
In 1912, Max von Laue, at the University of Munich in Germany, postulated that atoms in a crystal lattice had a regular, periodic structure with interatomic distances on the order of 1 A. Without having any evidence to support his claim on the periodic arrangements of atoms in a lattice, he further postulated that the crystalline structure can be used to diffract x-rays, much like a gradient in an infrared spectrometer can diffract infrared light. His postulate was based on the following assumptions: the atomic lattice of a crystal is periodic, x- rays are electromagnetic radiation, and the interatomic distance of a crystal are on the same order of magnitude as x- ray light. Laue's predictions were confirmed when two researchers: Friedrich and Knipping, successfully photographed the diffraction pattern associated with the x-ray radiation of crystalline $CuSO_4 \cdot 5H_2O$. The science of x-ray crystallography was born.
The arrangement of the atoms needs to be in an ordered, periodic structure in order for them to diffract the x-ray beams. A series of mathematical calculations is then used to produce a diffraction pattern that is characteristic to the particular arrangement of atoms in that crystal. X-ray crystallography remains to this day the primary tool used by researchers in characterizing the structure and bonding of organometallic compounds.
Diffraction
Diffraction is a phenomena that occurs when light encounters an obstacle. The waves of light can either bend around the obstacle, or in the case of a slit, can travel through the slits. The resulting diffraction pattern will show areas of constructive interference, where two waves interact in phase, and destructive interference, where two waves interact out of phase. Calculation of the phase difference can be explained by examining Figure 1 below.
In the figure below, two parallel waves, BD and AH are striking a gradient at an angle $θ_o$. The incident wave BD travels farther than AH by a distance of CD before reaching the gradient. The scattered wave (depicted below the gradient) HF, travels father than the scattered wave DE by a distance of HG. So the total path difference between path AHGF and BCDE is CD - HG. To observe a wave of high intensity (one created through constructive interference), the difference CD - HG must equal to an integer number of wavelengths to be observed at the angle psi, $CD - HG = n\lambda$, where $\lambda$ is the wavelength of the light. Applying some basic trigonometric properties, the following two equations can be shown about the lines:
$CD = x \cos(θ o) \nonumber$
and
$HG = x \cos (θ) \nonumber$
where $x$ is the distance between the points where the diffraction repeats. Combining the two equations,
$x(\cos θ_o - \cos θ) = n \lambda \nonumber$
Bragg's Law
Diffraction of an x-ray beam, occurs when the light interacts with the electron cloud surrounding the atoms of the crystalline solid. Due to the periodic crystalline structure of a solid, it is possible to describe it as a series of planes with an equal interplaner distance. As an x-ray's beam hits the surface of the crystal at an angle ?, some of the light will be diffracted at that same angle away from the solid (Figure 2). The remainder of the light will travel into the crystal and some of that light will interact with the second plane of atoms. Some of the light will be diffracted at an angle $theta$, and the remainder will travel deeper into the solid. This process will repeat for the many planes in the crystal. The x-ray beams travel different pathlengths before hitting the various planes of the crystal, so after diffraction, the beams will interact constructively only if the path length difference is equal to an integer number of wavelengths (just like in the normal diffraction case above). In the figure below, the difference in path lengths of the beam striking the first plane and the beam striking the second plane is equal to BG + GF. So, the two diffracted beams will constructively interfere (be in phase) only if $BG + GF = n \lambda$. Basic trigonometry will tell us that the two segments are equal to one another with the interplaner distance times the sine of the angle $\theta$. So we get:
$BG = BC = d \sin \theta \label{1}$
Thus,
$2d \sin \theta = n \lambda \label{2}$
This equation is known as Bragg's Law, named after W. H. Bragg and his son, W. L. Bragg; who discovered this geometric relationship in 1912. {C}{C}Bragg's Law relates the distance between two planes in a crystal and the angle of reflection to the x-ray wavelength. The x-rays that are diffracted off the crystal have to be in-phase in order to signal. Only certain angles that satisfy the following condition will register:
$\sin \theta = \dfrac{n \lambda}{2d} \label{3}$
For historical reasons, the resulting diffraction spectrum is represented as intensity vs. $2θ$.
Instrument Components
The main components of an x-ray instrument are similar to those of many optical spectroscopic instruments. These include a source, a device to select and restrict the wavelengths used for measurement, a holder for the sample, a detector, and a signal converter and readout. However, for x-ray diffraction; only a source, sample holder, and signal converter/readout are required.
The Source
x-ray tubes provides a means for generating x-ray radiation in most analytical instruments. An evacuated tube houses a tungsten filament which acts as a cathode opposite to a much larger, water cooled anode made of copper with a metal plate on it. The metal plate can be made of any of the following metals: chromium, tungsten, copper, rhodium, silver, cobalt, and iron. A high voltage is passed through the filament and high energy electrons are produced. The machine needs some way of controlling the intensity and wavelength of the resulting light. The intensity of the light can be controlled by adjusting the amount of current passing through the filament; essentially acting as a temperature control. The wavelength of the light is controlled by setting the proper accelerating voltage of the electrons. The voltage placed across the system will determine the energy of the electrons traveling towards the anode. X-rays are produced when the electrons hit the target metal. Because the energy of light is inversely proportional to wavelength ($E=hc=h(1/\lambda$), controlling the energy, controls the wavelength of the x-ray beam.
X-ray Filter
Monochromators and filters are used to produce monochromatic x-ray light. This narrow wavelength range is essential for diffraction calculations. For instance, a zirconium filter can be used to cut out unwanted wavelengths from a molybdenum metal target (see figure 4). The molybdenum target will produce x-rays with two wavelengths. A zirconium filter can be used to absorb the unwanted emission with wavelength Kβ, while allowing the desired wavelength, Kα to pass through.
Needle Sample Holder
The sample holder for an x-ray diffraction unit is simply a needle that holds the crystal in place while the x-ray diffractometer takes readings.
Signal Converter
In x-ray diffraction, the detector is a transducer that counts the number of photons that collide into it. This photon counter gives a digital readout in number of photons per unit time. Below is a figure of a typical x-ray diffraction unit with all of the parts labeled.
Fourier Transform
In mathematics, a Fourier transform is an operation that converts one real function into another. In the case of FTIR, a Fourier transform is applied to a function in the time domain to convert it into the frequency domain. One way of thinking about this is to draw the example of music by writing it down on a sheet of paper. Each note is in a so-called "sheet" domain. These same notes can also be expressed by playing them. The process of playing the notes can be thought of as converting the notes from the "sheet" domain into the "sound" domain. Each note played represents exactly what is on the paper just in a different way. This is precisely what the Fourier transform process is doing to the collected data of an x-ray diffraction. This is done in order to determine the electron density around the crystalline atoms in real space. The following equations can be used to determine the electrons' position:
$p(x,y,z) = \sum_h \sum_k \sum_l F(hkl) e ^{-2\pi i (hx+ky+lz)} \label{1A}$
$\int _0^1 \int _0^1 \int _0^1 p(x,y,z) e ^{2\pi i (hx+ky+lz)} dx\;dy\;dz \label{2B}$
$F(q) = | F(q) | e^{i \phi(q)} \label{3C}$
where $p(xyz)$ is the electron density function, and $F(hkl)$ is the electron density function in real space. Equation 1 represents the Fourier expansion of the electron density function. To solve for $F(hkl)$, the equation 1 needs to be evaluated over all values of h, k, and l, resulting in Equation 2. The resulting function $F(hkl)$ is generally expressed as a complex number (as seen in equation 3 above) with $| F(q)|$ representing the magnitude of the function and $\phi$ representing the phase.
Crystallization
In order to run an x-ray diffraction experiment, one must first obtain a crystal. In organometallic chemistry, a reaction might work but when no crystals form, it is impossible to characterize the products. Crystals are grown by slowly cooling a supersaturated solution. Such a solution can be made by heating a solution to decrease the amount of solvent present and to increase the solubility of the desired compound in the solvent. Once made, the solution must be cooled gradually. Rapid temperature change will cause the compound to crash out of solution, trapping solvent and impurities within the newly formed matrix. Cooling continues as a seed crystal forms. This crystal is a point where solute can deposit out of the solution and into the solid phase. Solutions are generally placed into a freezer (-78 ºC) in order to ensure all of the compound has crystallized. One way to ensure gradual cooling in a -78 ºC freezer is to place the container housing the compound into a beaker of ethanol. The ethanol will act as a temperature buffer, ensuring a slow decrease in the temperature gradient between the flask and the freezer. Once crystals are grown, it is imperative that they remain cold as any addition of energy will cause a disruption of the crystal lattice, which will yield bad diffraction data. The result of an organometallic chromium compound crystallization can be seen below.
Mounting the Crystal
Due to the air-sensitivity of most organometallic compounds, crystals must be transported in a highly viscous organic compound called paratone oil (Figure $7$). Crystals are abstracted from their respective Schlenks by dabbing the end of a spatula with the paratone oil and then sticking the crystal onto the oil. Although there might be some exposure of the compounds to air and water, crystals can withstand more exposure than solution (of the preserved protein) before degrading. On top of serving to protect the crystal, the paratone oil also serves as the glue to bind the crystal to the needle.
Rotating Crystal Method
To describe the periodic, three dimensional nature of crystals, the Laue equations are employed:
$a(\cos \theta_o – \cos \theta) = h\lambda \label{eq1}$
$b(\cos \theta_o – \cos \theta) = k\lambda \label{eq2}$
$c(\cos \theta_o – \cos \theta) = l\lambda \label{eq3}$
where $a$, $b$, and $c$ are the three axes of the unit cell, $θ_o$, $o$, $?o$ are the angles of incident radiation, and ?, ?, ? are the angles of the diffracted radiation. A diffraction signal (constructive interference) will arise when $h$, $k$, and $l$ are integer values. The rotating crystal method employs these equations. X-ray radiation is shown onto a crystal as it rotates around one of its unit cell axis. The beam strikes the crystal at a 90 degree angle. Using equation 1 above, we see that if $\theta_o$ is 90 degrees, then $\cos \theta_o = 0$. For the equation to hold true, we can set h=0, granted that $\theta= 90^o$. The above three equations will be satisfied at various points as the crystal rotates. This gives rise to a diffraction pattern (shown in the image below as multiple h values). The cylindrical film is then unwrapped and developed. The following equation can be used to determine the length axis around which the crystal was rotated:
$a = \dfrac{ch \lambda}{\sin \tan^{-1} (y/r} \nonumber$
where $a$ is the length of the axis, y is the distance from $h=0$ to the $h$ of interest, $r$ is the radius of the firm, and ? is the wavelength of the x-ray radiation used. The first length can be determined with ease, but the other two require far more work, including remounting the crystal so that it rotates around that particular axis.
X-ray Crystallography of Proteins
The crystals that form are frozen in liquid nitrogen and taken to the synchrotron which is a highly powered tunable x-ray source. They are mounted on a goniometer and hit with a beam of x-rays. Data is collected as the crystal is rotated through a series of angles. The angle depends on the symmetry of the crystal.
Proteins are among the many biological molecules that are used for x-ray Crystallography studies. They are involved in many pathways in biology, often catalyzing reactions by increasing the reaction rate. Most scientists use x-ray Crystallography to solve the structures of protein and to determine functions of residues, interactions with substrates, and interactions with other proteins or nucleic acids. Proteins can be co - crystallized with these substrates, or they may be soaked into the crystal after crystallization.
Protein Crystallization
Proteins will solidify into crystals under certain conditions. These conditions are usually made up of salts, buffers, and precipitating agents. This is often the hardest step in x-ray crystallography. Hundreds of conditions varying the salts, pH, buffer, and precipitating agents are combined with the protein in order to crystallize the protein under the right conditions. This is done using 96 well plates; each well containing a different condition and crystals; which form over the course of days, weeks, or even months. The pictures below are crystals of APS Kinase D63N from Penicillium chrysogenum taken at the Chemistry building at UC Davis after crystals formed over a period of a week. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/14%3A_NMR_Spectroscopy/14.23%3A_X-Ray_Crystallography.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.01%3A_Aromatic_Compounds_Are_Unusually_Stable.txt |
In 1931, German chemist and physicist Erich Hückel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His rule states that if a cyclic, planar molecule has 4n+2 π electrons, it is considered aromatic. This rule would come to be known as Hückel's Rule.
Four Criteria for Aromaticity
When deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic.
1. The molecule is cyclic (a ring of atoms)
2. The molecule is planar (all atoms in the molecule lie in the same plane)
3. The molecule is fully conjugated (p orbitals at every atom in the ring)
4. The molecule has $4n+2\, π$ electrons ($n=0$ or any positive integer)
According to Hückel's Molecular Orbital Theory, a compound is particularly stable if all of its bonding molecular orbitals are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by n), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of 4n+2 π electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below. Benzene has 6 π electrons. Its first 2 π electrons fill the lowest energy orbital, and it has 4 π electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons.
To apply the 4n+2 rule, first count the number of π electrons in the molecule. Then, set this number equal to 4n+2 and solve for n. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six π electrons:
\begin{align*} 4n + 2 &= 6 \[4pt] 4n &= 4 \[4pt] n &= 1 \end{align*}
For benzene, we find that $n=1$, which is a positive integer, so the rule is met.
Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward. π electrons lie in p orbitals. Sp2 hybridized atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the π electrons. A simple way to know if an atom is sp2 hybridized is to see if it has 3 attached atoms and no lone pairs of electrons. This video provides a very nice tutorial on how to determine an atom's hybridization. In a cyclic hydrocarbon compound with alternating single and double bonds, each carbon is attached to 1 hydrogen and 2 other carbons. Therefore, each carbon is sp2 hybridized and has a p orbital. Let's look at our previous example, benzene:
Each double bond (π bond) always contributes 2 π electrons. Benzene has 3 double bonds, so it has 6 π electrons.
Aromatic Ions
Hückel's Rule also applies to ions. As long as a compound has 4n+2 π electrons, it does not matter if the molecule is neutral or has a charge. For example, cyclopentadienyl anion is an aromatic ion. How do we know that it is fully conjugated? That is, how do we know that each atom in this molecule has 1 p orbital? Let's look at the following figure. Carbons 2-5 are sp2 hybridized because they have 3 attached atoms and have no lone electron pairs. What about carbon 1? Another simple rule to determine if an atom is sp2 hybridized is if an atom has 1 or more lone pairs and is attached to an sp2 hybridized atom, then that atom is sp2 hybridized also. This video explains the rule very clearly. Therefore, carbon 1 has a p orbital. Cyclopentadienyl anion has 6 π electrons and fulfills the 4n+2 rule.
Heterocyclic Aromatic Compounds
So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain 1 or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are sp2 hybridized. But is the oxygen atom sp2 hybridized? The oxygen has at least 1 lone electron pair and is attached to an sp2 hybridized atom, so it is sp2 hybridized as well. Notice how oxygen has 2 lone pairs of electrons. How many of those electrons are π electrons? An sp2 hybridized atom only has 1 p orbital, which can only hold 2 electrons, so we know that 1 electron pair is in the p orbital, while the other pair is in an sp2 orbital. So, only 1 of oxygen's 2 lone electron pairs are π electrons. Furan has 6 π electrons and fulfills the 4n+2 rule.
Problems
Using the criteria for aromaticity, determine if the following molecules are aromatic:
Answers
1. Aromatic - only 1 of S's lone pairs counts as π electrons, so there are 6 π electrons, n=1
2. Not aromatic - not fully conjugated, top C is sp3 hybridized
3. Not aromatic - top C is sp2 hybridized, but there are 4 π electrons, n=1/2
4. Aromatic - N is using its 1 p orbital for the electrons in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
5. Aromatic - there are 6 π electrons, n=1
6. Not aromatic - all atoms are sp2 hybridized, but only 1 of S's lone pairs counts as π electrons, so there 8 π electrons, n=1.5
7. Not aromatic - there are 4 π electrons, n=1/2
8. Aromatic - only 1 of N's lone pairs counts as π electrons, so there are 6 π electrons, n=1
9. Not aromatic - not fully conjugated, top C is sp3 hybridized
10. Aromatic - O is using its 1 p orbital for the elections in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
• Ramie Hosein | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.02%3A_The_Two_Criteria_for_Aromaticity.txt |
Because of the low hydrogen to carbon ratio in aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes (aliphatic compounds). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C7H6O2 compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C6H6, which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".
If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes substitution reactions rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to resonance stabilization of a conjugated cyclic triene.
Benzene:
Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequate representation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties.
Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds.
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
The Molecular Orbitals of Benzene
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.07%3A_A_Molecular_Orbital_Description_of_Aromaticity_and_Antiaromaticity.txt |
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