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In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur.
Brønsted-Lowery Definition
J.N. Brønsted and T.M. Lowry independently developed the theory of proton donors and proton acceptors in acid-base reactions, coincidentally in the same region and during the same year. The Arrhenius theory where acids and bases are defined by whether the molecule contains hydrogen and hydroxide ion is too limiting. The main effect of the Brønsted-Lowry definition is to identify the proton (H+) transfer occurring in the acid-base reaction. This is best illustrated in the following equation:
$HA+Z \rightleftharpoons A^− + HZ^+$
Acid Base
Donates hydrogen ions Accepts hydrogen ions.
HCl+ HOH → H3O+ + Cl-
HOH+ NH3 NH4+ + OH-
The determination of a substance as a Brønsted-Lowery acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case.
To determine whether a substance is an acid or a base, count the hydrogens on each substance before and after the reaction. If the number of hydrogens has decreased that substance is the acid (donates hydrogen ions). If the number of hydrogens has increased that substance is the base (accepts hydrogen ions). These definitions are normally applied to the reactants on the left. If the reaction is viewed in reverse a new acid and base can be identified. The substances on the right side of the equation are called conjugate acid and conjugate base compared to those on the left. Also note that the original acid turns in the conjugate base after the reaction is over.
Acids are Proton Donors and Bases are Proton Acceptors
For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up of conjugate acids and conjugate bases.
$HA + Z \rightleftharpoons A^- + HZ^+$
A stands for an Acidic compound and Z stands for a Basic compound
• A Donates H to form HZ+.
• Z Accepts H from A which forms HZ+
• A- becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium
• HZ+ becomes a conjugate acid of Z and in the reverse reaction it donates a H to A- recreating Z in order to remain in equilibrium
Questions
1. Why is $HA$ an Acid?
2. Why is $Z^-$ a Base?
3. How can A- be a base when HA was and Acid?
4. How can HZ+ be an acid when Z used to be a Base?
Now that we understand the concept, let's look at an an example with actual compounds!
$HCl + H_2O \rightleftharpoons H_3O^+ + Cl^¯$
• HCL is the acid because it is donating a proton to H2O
• H2O is the base because H2O is accepting a proton from HCL
• H3O+ is the conjugate acid because it is donating an acid to CL turn into it's conjugate acid H2O
• Cl¯ is the conjugate base because it accepts an H from H3O to return to it's conjugate acid HCl
How can H2O be a base? I thought it was neutral?
Answers
1. It has a proton that can be transferred
2. It receives a proton from HA
3. A- is a conjugate base because it is in need of a H in order to remain in equilibrium and return to HA
4. HZ+ is a conjugate acid because it needs to donate or give away its proton in order to return to it's previous state of Z
5. In the Brønsted-Lowry Theory what makes a compound an element or a base is whether or not it donates or accepts protons. If the H2O was in a different problem and was instead donating an H rather than accepting an H it would be an acid!
Conjugate Acid–Base Pairs
In aqueous solutions, acids and bases can be defined in terms of the transfer of a proton from an acid to a base. Thus for every acidic species in an aqueous solution, there exists a species derived from the acid by the loss of a proton. These two species that differ by only a proton constitute a conjugate acid–base pair.
All acid–base reactions contain two conjugate acid–base pairs.
For example, in the reaction of HCl with water, HCl, the parent acid, donates a proton to a water molecule, the parent base, thereby forming Cl-. Thus HCl and Cl- constitute a conjugate acid–base pair. By convention, we always write a conjugate acid–base pair as the acid followed by its conjugate base. In the reverse reaction, the Cl- ion in solution acts as a base to accept a proton from H2O and HCl. Thus H3O+ and H2O constitute a second conjugate acid–base pair. In general, any acid–base reaction must contain two conjugate acid–base pairs, which in this case are HCl/Cl- and H3O+/H2O.
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (CH3CO2H/CH3CO2-) and the parent base and its conjugate acid (H3O+/H2O).
In the reaction of ammonia with water to give ammonium ions and hydroxide ions (Figure 16.3), ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are NH4+/NH3 and H2O/OH-.
Some common conjugate acid–base pairs are shown in Figure $4$.
The Brønsted-Lowry definition of acidity
We’ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these definitions was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry, and has come to be known as the Brønsted-Lowry definition of acids and bases. An acid, by the Brønsted-Lowry definition, is a species which is able to donate a proton (H+), while a base is a proton acceptor. We have already discussed in the previous chapter one of the most familiar examples of a Brønsted-Lowry acid-base reaction, between hydrochloric acid and hydroxide ion:
In this reaction, a proton is transferred from HCl (the acid, or proton donor) to hydroxide (the base, or proton acceptor). As we learned in the previous chapter, curved arrows depict the movement of electrons in this bond-breaking and bond-forming process.
After a Brønsted-Lowry acid donates a proton, what remains – in this case, a chloride ion – is called the conjugate base. Chloride is thus the conjugate base of hydrochloric acid. Conversely, when a Brønsted-Lowry base accepts a proton it is converted into its conjugate acid form: water is thus the conjugate acid of hydroxide.
We can also talk about conjugate acid/base pairs: the two acid/base pairs involved in our first reaction are hydrochloric acid/chloride and hydroxide/water. In this next acid-base reaction, the two pairs involved are acetate/acetic acid and methyl ammonium/methylamine:
Throughout this text, we will often use the abbreviations HA and :B in order to refer in a general way to acidic and basic reactants:
In order to act as a proton acceptor, a base must have a reactive pair of electrons. In all of the examples we shall see in this chapter, this pair of electrons is a non-bonding lone pair, usually (but not always) on an oxygen, nitrogen, sulfur, or halogen atom. When acetate acts as a base in the reaction shown above, for example, one of its oxygen lone pairs is used to form a new bond to a proton. The same can be said for an amine acting as a base. Clearly, methyl ammonium ion cannot act as a base – it does not have a reactive pair of electrons with which to accept a new bond to a proton.
Later, in chapter 15, we will see several examples where the (relatively) reactive pair of electrons in a $\pi$ bond act in a basic fashion.
In this chapter, we will concentrate on those bases with non-bonding (lone pair) electrons.
Example
Exercise 7.1: Draw structures for the missing conjugate acids or conjugate bases in the reactions below.
Contributors
You are no doubt aware that some acids are stronger than others. Sulfuric acid is strong enough to be used as a drain cleaner, as it will rapidly dissolve clogs of hair and other organic material.
Not surprisingly, concentrated sulfuric acid will also cause painful burns if it touches your skin, and permanent damage if it gets in your eyes (there’s a good reason for those safety goggles you wear in chemistry lab!). Acetic acid (vinegar), will also burn your skin and eyes, but is not nearly strong enough to make an effective drain cleaner. Water, which we know can act as a proton donor, is obviously not a very strong acid. Even hydroxide ion could theoretically act as an acid – it has, after all, a proton to donate – but this is not a reaction that we would normally consider to be relevant in anything but the most extreme conditions.
The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the dissociation constant, abbreviated Ka. The common base chosen for comparison is water.
We will consider acetic acid as our first example. When a small amount of acetic acid is added to water, a proton-transfer event (acid-base reaction) occurs to some extent.
Notice the phrase ‘to some extent’ – this reaction does not run to completion, with all of the acetic acid converted to acetate, its conjugate base. Rather, a dynamic equilibrium is reached, with proton transfer going in both directions (thus the two-way arrows) and finite concentrations of all four species in play. The nature of this equilibrium situation, as you recall from General Chemistry, is expressed by an equilibrium constant, Keq. The equilibrium constant is actually a ratio of activities (represented by the symbol $a$), but activities are rarely used in courses other than analytical or physical chemistry. To simplify the discussion for general chemistry and organic chemistry courses, the activities of all of the solutes are replaced with molarities, and the activity of the solvent (usually water) is defined as having the value of 1.
In our example, we added a small amount of acetic acid to a large amount of water: water is the solvent for this reaction. Therefore, in the course of the reaction, the concentration of water changes very little, and the water can be treated as a pure solvent, which is always assigned an activity of 1. The acetic acid, acetate ion and hydronium ion are all solutes, and so their activities are approximated with molarities. The acid dissociation constant, or Ka, for acetic acid is therefore defined as:
$K_{eq} = \dfrac{a_{CH_3COO^-}·a_{H_3O^+}}{a_{CH_3COOH}·a_{H_2O}} ≈ \dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH][1]}$
Because dividing by 1 does not change the value of the constant, the "1" is usually not written, and Ka is written as:
$K_{eq} = K_{a} = \dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]} = 1.75 \times 10^{-5}$
In more general terms, the dissociation constant for a given acid is expressed as:
$K_a = \dfrac{[A^-][H_3O^+]}{[HA]} \label{First}$
or
$K_a = \dfrac{[A][H_3O^+]}{[HA^+]} \label{Second}$
The first expression applies to a neutral acid such as like HCl or acetic acid, while the second applies to a cationic acid like ammonium (NH4+).
The value of Ka = 1.75 x 10-5 for acetic acid is very small - this means that very little dissociation actually takes place, and there is much more acetic acid in solution at equilibrium than there is acetate ion. Acetic acid is a relatively weak acid, at least when compared to sulfuric acid (Ka = 109) or hydrochloric acid (Ka = 107), both of which undergo essentially complete dissociation in water.
A number like 1.75 x 10- 5 is not very easy either to say or to remember. Chemists have therefore come up with a more convenient term to express relative acidity: the pKa value.
pKa = -log Ka
Doing the math, we find that the pKa of acetic acid is 4.8. The use of pKa values allows us to express the acidity of common compounds and functional groups on a numerical scale of about –10 (very strong acid) to 50 (not acidic at all). Table 7 at the end of the text lists exact or approximate pKa values for different types of protons that you are likely to encounter in your study of organic and biological chemistry. Looking at Table 7, you see that the pKa of carboxylic acids are in the 4-5 range, the pKa of sulfuric acid is –10, and the pKa of water is 14. Alkenes and alkanes, which are not acidic at all, have pKa values above 30. The lower the pKa value, the stronger the acid.
It is important to realize that pKa is not at all the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is the measure of the hydronium ion concentration in a particular aqueous solution:
pH = -log [H3O+]
Any particular acid will always have the same pKa (assuming that we are talking about an aqueous solution at room temperature) but different aqueous solutions of the acid could have different pH values, depending on how much acid is added to how much water.
Our table of pKa values will also allow us to compare the strengths of different bases by comparing the pKavalues of their conjugate acids. The key idea to remember is this: the stronger the conjugate acid, the weaker the conjugate base. Sulfuric acid is the strongest acid on our list with a pKa value of –10, so HSO4- is the weakest conjugate base. You can see that hydroxide ion is a stronger base than ammonia (NH3), because ammonium (NH4+, pKa = 9.2) is a stronger acid than water (pKa = 14.0).
While Table 7 provides the pKa values of only a limited number of compounds, it can be very useful as a starting point for estimating the acidity or basicity of just about any organic molecule. Here is where your familiarity with organic functional groups will come in very handy. What, for example, is the pKaof cyclohexanol? It is not on the table, but as it is an alcohol it is probably somewhere near that of ethanol (pKa = 16). Likewise, we can use Table 7 to predict that para-hydroxyphenyl acetaldehyde, an intermediate compound in the biosynthesis of morphine, has a pKa in the neighborhood of 10, close to that of our reference compound, phenol.
Notice in this example that we need to evaluate the potential acidity at four different locations on the molecule.
Aldehyde and aromatic protons are not at all acidic (pKavalues are above 40 – not on our table). The two protons on the carbon next to the carbonyl are slightly acidic, with pKa values around 19-20 according to the table. The most acidic proton is on the phenol group, so if the compound were to be subjected to a single molar equivalent of strong base, this is the proton that would be donated.
As you continue your study of organic chemistry, it will be a very good idea to commit to memory the approximate pKa ranges of some important functional groups, including water, alcohols, phenols, ammonium, thiols, phosphates, carboxylic acids and carbons next to carbonyl groups (so-called a-carbons). These are the groups that you are most likely to see acting as acids or bases in biological organic reactions.
A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
Example
Exercise 7.2: Using the pKa table, estimate pKa values for the most acidic group on the compounds below, and draw the structure of the conjugate base that results when this group donates a proton.
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.3%3A_Acids_and__Bases_Electrophiles_and__Nucleophiles.txt |
Objectives
After completing this section, you should be able to
1. explain why the properties of a given organic compound are largely dependent on the functional group or groups present in the compound.
2. identify the functional groups present in each of the following compound types: alkenes, alkynes, arenes, (alkyl and aryl) halides, alcohols, ethers, aldehydes, ketones, esters, carboxylic acids, (carboxylic) acid chlorides, amides, amines, nitriles, nitro compounds, sulfides and sulfoxides.
3. identify the functional groups present in an organic compound, given its structure.
4. Given the structure of an organic compound containing a single functional group, identify which of the compound types listed under Objective 2, above, it belongs to.
5. draw the structure of a simple example of each of the compound types listed in Objective 2.
Key Terms
Make certain that you can define, and use in context, the key term below.
• functional group
Study Notes
The concept of functional groups is a very important one. We expect that you will need to refer back to tables at the end of Section 3.1 quite frequently at first, as it is not really feasible to learn the names and structures of all the functional groups and compound types at one sitting. Gradually they will become familiar, and eventually you will recognize them automatically.
Functional groups are small groups of atoms that exhibit a characteristic reactivity. A particular functional group will almost always display its distinctive chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have specific names that often carry over in the naming of individual compounds incorporating the groups.
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups tend to behave in organic reactions.
Drawing abbreviated organic structures
Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest:
The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule.
Common Functional Groups
In the following sections, many of the common functional groups found in organic chemistry will be described. Tables of these functional groups can be found at the bottom of the page.
Hydrocarbons
The simplest functional group in organic chemistry (which is often ignored when listing functional groups) is called an alkane, characterized by single bonds between two carbons and between carbon and hydrogen. Some examples of alkanes include methane, CH4, is the natural gas you may burn in your furnace or on a stove. Octane, C8H18, is a component of gasoline.
Alkanes
Alkenes (sometimes called olefins) have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds. Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening. (If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas). Ethyne, commonly called acetylene, is used as a fuel in welding blow torches.
Alkenes and alkynes
Alkenes have trigonal planar electron geometry (due to sp2 hybrid orbitals at the alkene carbons) while alkynes have linear geometry (due to sp hybrid orbitals at the alkyne carbons). Furthermore, many alkenes can take two geometric forms: cis or trans (or Z and E which will be explained in detail in Chapter 7). The cis and trans forms of a given alkene are different molecules with different physical properties there is a very high energy barrier to rotation about a double bond. In the example below, the difference between cis and trans alkenes is readily apparent.
Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens - in other words, they are saturated with hydrogen atoms. The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. As we will see in Chapter 7, hydrogen can be added to double and triple bonds, in a type of reaction called 'hydrogenation'.
The aromatic group is exemplified by benzene (which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic), and naphthalene, a compound with a distinctive 'mothball' smell. Aromatic groups are planar (flat) ring structures, and are widespread in nature. We will learn more about the structure and reactions of aromatic groups in Chapter 15.
Functional Groups with Carbon Single Bonds to other Atoms
Halides
When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or haloalkane. The presence of a halogen atom (F, Cl, Br, or I), is often represented by X due to the similar chemistry of halogens. Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules.
Alcohols and Thiols
In the alcohol functional group, a carbon is single-bonded to an OH group (the OH group, by itself, is referred to as a hydroxyl). Except for methanol, all alcohols can be classified as primary, secondary, or tertiary. In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon. In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol.
The sulfur analog of an alcohol is called a thiol (the prefix thio, derived from the Greek, refers to sulfur).
Ethers and sulfides
In an ether functional group, a central oxygen is bonded to two carbons. Below are the line and Lewis structures of diethyl ether, a common laboratory solvent and also one of the first medical anaesthesia agents.
In sulfides, the oxygen atom of an ether has been replaced by a sulfur atom.
Amines
Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations. In the case where a nitrogen has four bonds to carbon (which is somewhat unusual in biomolecules), it is called a quaternary ammonium ion.
Caution
Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols and amines - the definitions are different. In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.
Carbonyl Containing Functional Groups
Aldehydes and Ketones
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
Carboxylic acids and acid derivatives
If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a heteroatom (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl (OH) group.
As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a carboxylate (much more about carboxylic acids in Chapter 20).
In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine.
In esters, the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
In an acid anhydride, there are two carbonyl carbons with an oxygen in between. An acid anhydride is formed from combination of two carboxylic acids with the loss of water (anhydride).
In an acyl phosphate, the carbonyl carbon is bonded to the oxygen of a phosphate, and in an acid chloride, the carbonyl carbon is bonded to a chlorine.
Nitriles and Imines
In a nitrile group, a carbon is triple-bonded to a nitrogen. Nitriles are also often referred to as cyano groups.
Molecules with carbon-nitrogen double bonds are called imines, or Schiff bases.
Phosphates
Phosphorus is a very important element in biological organic chemistry, and is found as the central atom in the phosphate group. Many biological organic molecules contain phosphate, diphosphate, and triphosphate groups, which are linked to a carbon atom by the phosphate ester functionality.
Because phosphates are so abundant in biological organic chemistry, it is convenient to depict them with the abbreviation 'P'. Notice that this 'P' abbreviation includes the oxygen atoms and negative charges associated with the phosphate groups.
Molecules with Multiple Functional Groups
A single compound may contain several different functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups.
The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table found below provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
Exercise \(1\)
Identify the functional groups in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Answer
a) carboxylate, sulfide, aromatic, two amide groups (one of which is cyclic)
b) tertiary alcohol, thioester
c) carboxylate, ketone
d) ether, primary amine, alkene
2: Draw one example each (there are many possible correct answers) of compounds fitting the descriptions below, using line structures. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule).
a) a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
b) an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
c) A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group.
Functional Group Tables
Exclusively Carbon Functional Groups
Group Formula Class Name Specific Example IUPAC Name Common Name
alkene H2C=CH2 ethene ethylene
alkyne HC≡CH ethyne acetylene
arene C6H6 benzene benzene
Functional Groups with Single Bonds to Heteroatoms
Group Formula Class Name Specific Example IUPAC Name Common Name
halide H3C-I iodomethane methyl iodide
alcohol CH3CH2OH ethanol ethyl alcohol
ether CH3CH2OCH2CH3 diethyl ether ether
amine H3C-NH2 aminomethane methylamine
nitro compound H3C-NO2 nitromethane
thiol H3C-SH methanethiol methyl mercaptan
sulfide H3C-S-CH3 dimethyl sulfide
Functional Groups with Multiple Bonds to Heteroatoms
Group Formula Class Name Specific Example IUPAC Name Common Name
nitrile H3C-CN ethanenitrile acetonitrile
aldehyde H3CCHO ethanal acetaldehyde
ketone H3CCOCH3 propanone acetone
carboxylic acid H3CCO2H ethanoic Acid acetic acid
ester H3CCO2CH2CH3 ethyl ethanoate ethyl acetate
acid halide H3CCOCl ethanoyl chloride acetyl chloride
amide H3CCON(CH3)2 N,N-dimethylethanamide N,N-dimethylacetamide
acid Anhydride (H3CCO)2O ethanoic anhydride acetic anhydride | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.4%3A_Functional_Groups%3A_Centers_of__Reactivity.txt |
Alkanes are organic compounds that consist entirely of single-bonded carbon and hydrogen atoms and lack any other functional groups. Alkanes have the general formula \(C_nH_{2n+2}\) and can be subdivided into the following three groups: the linear straight-chain alkanes, branched alkanes, and cycloalkanes. Alkanes are also saturated hydrocarbons.
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Molecular Formulas
Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only. Alkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The first six alkanes are as follows:
methane CH4
ethane C2H6
propane C3H8
butane C4H10
pentane C5H12
hexane C6H14
You can work out the formula of any of the alkanes using the general formula CnH2n+2
Isomerism
Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. For example, both of the following are the same molecule. They are not isomers. Both are butane.
All of the alkanes containing four or more carbon atoms show structural isomerism, meaning that there are two or more different structural formulae that you can draw for each molecular formula.
Example 1: Butane or MethylPropane
C4H10 could be either of these two different molecules:
These are different molecules named butane (left) and 2-methylpropane (right).
There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds. If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule. In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise.
Constitutional isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, C4H10. In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.
Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond.
You could easily see this with a model. This is the example we've already used at the top of this page.
Example 2: Constitutional Isomers in Pentane
Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models.
Classification of Carbon and Hydrogen Atoms
Carbons have a special terminology to describe how many other carbons they are attached to.
• Primary carbons (1o) attached to one other C atom
• Secondary carbons (2o) are attached to two other C’s
• Tertiary carbons (3o) are attached to theree other C’s
• Quaternary carbons (4o) are attached to four C's
For example, each of the three types of carbons are found in the 2,2 -dimethyl, 4-methylpentane molecule
Hydrogen atoms are also classified in this manner. A hydrogen atom attached to a primary carbon atom is called a primary hydrogen; thus, isobutane, has nine primary hydrogens and one tertiary hydrogen.
• Primary hydrogens (1o) are attached to carbons bonded to one other C atom
• Secondary hydrogens (2o) are attached to carbons bonded to two other C’s
• Tertiary hydrogens (3o) are attached to carbons bonded to theree other C’s
Each of the three types of carbons are found in the 2,2 -dimethyl, 4-methylpentane molecule
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.5%3A_Straight-Chain_and__Branched_Alkanes.txt |
Objectives
After completing this section, you should be able to
1. provide the correct IUPAC name for any given alkane structure (Kekulé, condensed or shorthand).
2. draw the Kekulé, condensed or shorthand structure of an alkane, given its IUPAC name.
Key Terms
Make certain that you can define, and use in context, the key term below.
• IUPAC system
Study Notes
The IUPAC system of nomenclature aims to ensure
1. that every organic compound has a unique, unambiguous name.
2. that the IUPAC name of any compound conveys the structure of that compound to a person familiar with the system.
One way of checking whether the name you have given to an alkane is reasonable is to count the number of carbon atoms implied by the chosen name. For example, if you named a compound 3‑ethyl-4‑methylheptane, you have indicated that the compound contains a total of 10 carbon atoms—seven carbon atoms in the main chain, two carbon atoms in an ethyl group, and one carbon atom in a methyl group. If you were to check the given structure and find 11 carbon atoms, you would know that you had made a mistake. Perhaps the name you should have written was 3‑ethyl-4,4‑dimethylheptane!
When naming alkanes, a common error of beginning students is a failure to pick out the longest carbon chain. For example, the correct name for the compound shown below is 3‑methylheptane, not 2‑ethylhexane.
Remember that every substituent must have a number, and do not forget the prefixes: di, tri, tetra, etc.
You must use commas to separate numbers, and hyphens to separate numbers and substituents. Notice that 3‑methylhexane is one word.
Hydrocarbons having no double or triple bond functional groups are classified as alkanes or cycloalkanes, depending on whether the carbon atoms of the molecule are arranged only in chains or also in rings. Although these hydrocarbons have no functional groups, they constitute the framework on which functional groups are located in other classes of compounds, and provide an ideal starting point for studying and naming organic compounds. The alkanes and cycloalkanes are also members of a larger class of compounds referred to as aliphatic. Simply put, aliphatic compounds are compounds that do not incorporate any aromatic rings in their molecular structure.
The following table lists the IUPAC names assigned to simple continuous-chain alkanes from C-1 to C-10. A common "ane" suffix identifies these compounds as alkanes. Longer chain alkanes are well known, and their names may be found in many reference and text books. The names methane through decane should be memorized, since they constitute the root of many IUPAC names. Fortunately, common numerical prefixes are used in naming chains of five or more carbon atoms.
Table 3.4.1 : Simple Unbranched Alkanes
Name Molecular
Formula
Structural
Formula
Isomers Name Molecular
Formula
Structural
Formula
Isomers
methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
1. The formulas and structures of these alkanes increase uniformly by a CH2 increment.
2. A uniform variation of this kind in a series of compounds is called homologous.
3. These formulas all fit the CnH2n+2 rule. This is also the highest possible H/C ratio for a stable hydrocarbon.
4. Since the H/C ratio in these compounds is at a maximum, we call them saturated (with hydrogen).
Beginning with butane (C4H10), and becoming more numerous with larger alkanes, we note the existence of alkane isomers. For example, there are five C6H14 isomers, shown below as abbreviated line formulas (A through E):
Although these distinct compounds all have the same molecular formula, only one (A) can be called hexane. How then are we to name the others?
The IUPAC system requires first that we have names for simple unbranched chains, as noted above, and second that we have names for simple alkyl groups that may be attached to the chains. Examples of some common alkyl groups are given in the following table. Note that the "ane" suffix is replaced by "yl" in naming groups. The symbol R is used to designate a generic (unspecified) alkyl group.
Table 3.4.2 : Alkyl Groups Names
Group CH3 C2H5 CH3CH2CH2 (CH3)2CH– CH3CH2CH2CH2 (CH3)2CHCH2 CH3CH2CH(CH3)– (CH3)3C– R–
Name Methyl Ethyl Propyl Isopropyl Butyl Isobutyl sec-Butyl tert-Butyl Alkyl
IUPAC Rules for Alkane Nomenclature
1. Find and name the longest continuous carbon chain.
2. Identify and name groups attached to this chain.
3. Number the chain consecutively, starting at the end nearest a substituent group.
4. Designate the location of each substituent group by an appropriate number and name.
5. Assemble the name, listing groups in alphabetical order.
6. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.
Example 3.4.1 : Alkanes
The IUPAC names of the isomers of hexane are: A hexane B 2-methylpentane C 3-methylpentane D 2,2-dimethylbutane E 2,3-dimethylbutane
Halogen Groups
Halogen substituents are easily accommodated, using the names: fluoro (F-), chloro (Cl-), bromo (Br-) and iodo (I-).
Example 3.4.2 : Halogen Substitution
For example, (CH3)2CHCH2CH2Br would be named 1-bromo-3-methylbutane. If the halogen is bonded to a simple alkyl group an alternative "alkyl halide" name may be used. Thus, C2H5Cl may be named chloroethane (no locator number is needed for a two carbon chain) or ethyl chloride.
Alkyl Groups
Alkanes can be described by the general formula CnH2n+2. An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same concept can be applied to any of the straight chain alkane names provided in the table below.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Three Rules for Naming Alkanes
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substituent present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substituents and/or functional groups are written in alphabetical order.
Example 3.4.3
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in blue and consists of eight carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
Rule 3: In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the following examples.
The name of this molecule is 2,5-dimethyloctane
Example 3.4.4
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in blue and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substituent present must have the lowest possible number. In this example, numbering the chain from either the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
The name of this molecule is: 2-bromo-6-chloroheptane
Example 3.4.5
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in blue and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Parent chain: heptane Substitutents: 2-chloro 3-methyl 6-bromo
The name of this molecule is: 6-bromo-2-chloro-3-methylheptane
Exercises
Exercise 3.4.1
Give the proper IUPAC names of the following compounds.
Answer
a) Since this structure is an unbranched alkane (all single bonds) with a 5 carbon chain length, its name would be pentane.
b) This alkane has a 7 carbon longest continuous chain length that we number from right to left to get the first methyl substituent we encounter to have the lowest possible number (3 versus being 4 if numbering from left to right). This causes it to have 2 methyl substituents at positions 3 & 4 so we would name it indicating those numbers and the prefix dimethyl which gives a proper IUPAC name of 3,4-dimethylheptane.
c) This alkane has a 5 carbon longest continuous chain length (which could be numbered from left to right or right to left due to the symmetry at C-3). It has two methyl substituents off of C-3 so the proper IUPAC name is 3,3-dimethylpentane.
Exercise 3.4.2
Give the proper IUPAC names of the following compounds.
Answer
a) This alkane has a 9 carbon longest continuous chain length that we number from left to right (structure on left numbered in blue) to make the ethyl substituent be number 4. For the structure on the right (numbered in red) going from right to left, the methyl substituent is number 4. Since ethyl is higher in the alphabetical order, you want to make it have the lower number so the structure on the left (blue numbering) takes priority and the name is 4-ethyl-6-methylnonane.
b) This alkane has a 6 carbon longest continuous chain length that we number from right to left to make the first methyl be C-2 (versus the opposite direction which would make the first methyl C-3). Since there are 3 methyl substituents at positions 2,3, & 4, this compound would have the name 2,3,4-trimethylhexane.
c) This 6 carbon alkane can be numbered along different chains (see below) as well as in the opposite directions. This shows the two different chains that can be drawn (making the first substituent in that chain the lowest number). The structure on the left (numbered in blue) is the correct choice since it causes more substituents to be on the longest continuous chain (3 vs 2 in the structure on the right). This would make the IUPAC name of the structure 3-ethyl-2,4-dimethylhexane. (Notice how ethyl takes priority over methyl and the di- is not considered for alphabetizing.)
Exercise 3.4.3
All of the following names represent a compound that has been named improperly. Draw out the structure from the name and give the proper IUPAC name for the compounds.
1. 1,3-dimethylbutane
2. 4-ethylpentane
3. 2-ethyl-3-methylpentane
Answer
a) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), the correct name should be 2-methylpentane.
b) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is 6 C’s and we start the numbering on the end to the right to make the methyl substituent come off at C-3 (instead of beong at C-4 if we numbered it the opposite direction) the correct name should be 3-methyl hexane.
c) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is 6 C’s and since this molecule is symmetrical (between carbon 3 & 4), you can start the numbers from either end. In this case, we have methyl substituents coming off of carbons 3 & 4 so the proper name is 3,4-dimethyl hexane.
Exercise 3.4.4
All of the following names represent a compound that has been named improperly. Draw out the structure from the name and give the proper IUPAC name for the compounds.
1. 2,2-diethylheptane
2. 2-propylpentane
3. 4,4-diethylbutane
Answer
a) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is now 8 C’s and you have an ethyl substituent at C-3 and a methyl substituent also at C-3 so the proper name is 3- ethyl-3-methyloctane.
b) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right), notice that the longest chain is now 7 C’s and since this molecule is symmetrical (at carbon 4), you can start the numbers from either end. There is a methyl substituent at C-4 so the proper name is 4-methylheptane.
c) The structure that can be drawn for the improper name is shown below on the left. When you renumber it properly (structure on the right) going from right to left (to make the ethyl substituent have the lowest number possible), the correct name is 3-ethylhexane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.6%3A_Naming_the__Alkanes.txt |
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.
Boiling Points
The boiling points shown are for the "straight chain" isomers of which there is more than one. The first four alkanes are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!
Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.
There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.
Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.
Example
For example, the boiling points of the three isomers of \(C_5H_{12}\) are:
• pentane: 309.2 K
• 2-methylbutane: 301.0 K
• 2,2-dimethylpropane: 282.6 K
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!
Solubility
Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.
Solubility in Water
When a molecular substance dissolves in water, the following must occur:
• break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water.
As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water. The alkane does not dissolve.
Note: This is a simplification because entropic effects are important when things dissolve.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.7%3A_Structural_and__Physical_Properties_of_Alkanes.txt |
Objectives
After completing this section, you should be able to
1. explain the concept of free rotation about a carbon-carbon single bond.
2. explain the difference between conformational isomerism and structural isomerism.
3. draw the conformers of ethane using both sawhorse representation and Newman projection.
4. sketch a graph of energy versus bond rotation for ethane, and discuss the graph in terms of torsional strain.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• conformation (conformer, conformational isomer)
• dihedral angle
• eclipsed conformation
• Newman projection
• staggered conformation
• strain energy
• torsional strain (eclipsing strain)
Study Notes
You should be prepared to sketch various conformers using both sawhorse representations and Newman projections. Each method has its own advantages, depending upon the circumstances. Notice that when drawing the Newman projection of the eclipsed conformation of ethane, you cannot clearly draw the rear hydrogens exactly behind the front ones. This is an inherent limitation associated with representing a 3-D structure in two dimensions.
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity of the atoms or geometry of bonding. Two or more structures that are categorized as conformational isomers, or conformers, are really just two of the exact same molecule that differ only in rotation of one or more sigma bonds.
Ethane Conformations
Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations.
In order to better visualize these different conformations, 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 at 120°angles, which is what the actual tetrahedral geometry looks like when viewed from this perspective and flattened into two dimensions.
Interactive Element
Figure 3.6.1: A 3D Model of Staggered Ethane.
The lowest energy conformation of ethane, shown in the figure above, is called the ‘staggered’ conformation. In the staggered conformation, all of the C-H bonds on the front carbon are positioned at an angle of 60° relative to the C-H bonds on the back carbon. This angle between a sigma bond on the front carbon compared to a sigma bond on the back carbon is called the dihedral angle. In this conformation, the distance between the bonds (and the electrons in them) is maximized. Maximizing the distance between the electrons decreases the electrostatic repulsion between the electrons and results in a more stable structure.
If we now rotate the front CH3 group 60° clockwise, the molecule is in the highest energy ‘eclipsed' conformation, and the hydrogens on the front carbon are as close as possible to the hydrogens on the back carbon.
This is the highest energy conformation because of unfavorable electrostatic repulsion between the electrons in the front and back C-H bonds. The energy of the eclipsed conformation is approximately 3 kcal/mol (12 kJ/mol) higher than that of the staggered conformation. Torsional strain (or eclipsing strain) is the name give to the energy difference caused by the increased electrostatic repulsion of eclipsing bonds.
Another 60° rotation returns the molecule to a second eclipsed 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 variations in between. We will focus on the staggered and eclipsed conformers since they are, respectively, the lowest and highest energy conformers.
Unhindered (Free) Rotations Do Not Exist in Ethane
The carbon-carbon bond is not completely free to rotate – the 3 kcal/mol torsional strain in ethane creates a barrier to rotation that must be overcome for the bond to rotate from one staggered conformation to another. This rotational barrier is not large enough to prevent rotation except at extremely cold temperatures. So at normal temperatures, the carbon-carbon bond is constantly rotating. However, at any given moment the molecule is more likely to be in a staggered conformation - one of the rotational ‘energy valleys’ - than in any other conformer. The potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown in Figure 3.6.2.
Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time. The animation below illustrates the relationship between ethane's potential energy and its dihedral angle
Exercises
1) What is the most stable rotational conformation of ethane and explain why it is preferred over the other conformation?
Solutions
1) Staggered, as there is less repulsion between the hydrogen atoms. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.8%3A_Rotation_about__Single_Bonds%3A_Conformations.txt |
Objectives
After completing this section, you should be able to
1. depict the staggered and eclipsed conformers of propane (or a similar compound) using sawhorse representations and Newman projections.
2. sketch a graph of energy versus bond rotation for propane (or a similar compound) and discuss the graph in terms of torsional strain.
3. depict the anti, gauche, eclipsed and fully eclipsed conformers of butane (or a similar compound), using sawhorse representations and Newman projections.
4. sketch a graph of energy versus C2-C3 bond rotation for butane (or a similar compound), and discuss it in terms of torsional and steric repulsion.
5. assess which of two (or more) conformers of a given compound is likely to predominate at room temperature from a semi-quantitative knowledge of the energy costs of the interactions involved.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti conformation
• gauche conformation
• eclipsed conformation
• steric repulsion (strain)
In butane, there are three rotating carbon-carbon sigma 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.
Eclipsed interaction Energy (kcal/mol) Energy (kJ/mol)
H-H 1.0 4.0
H-CH3 1.4 6.0
CH3-CH3 2.6 11.0
The CH3-CH3 groups create the significantly larger eclipsed interaction of 11.0 kJ/mol. There are also two H-H eclipsed interactions at 4.0 kJ/mol each to create a total of 2(4.0 kJ/mol) + 11.0 kJ/mol = 19.0 kJ/mol of strain. This is the highest energy conformation for butane, due to torsional strain caused by the electrostatic repulsion of electrons in the eclipsed bonds, but also because of another type of strain called ‘steric repulsion’, between the two rather bulky methyl groups. Steric strain comes about when two large groups, such as two methyl groups, try to occupy the same space. What results is a repulsive non-covalent interaction caused by their respective electron densities.
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. There is still significant steric repulsion between the two bulky groups.
A further rotation of 60° gives us a second eclipsed conformation (B) in which both methyl groups are lined up with hydrogen atoms.
Due to steric repulsion between methyl and hydrogen substituents, this eclipsed conformation B is higher in energy than the gauche conformation. However, because there is no methyl-to-methyl eclipsing, it is lower in energy than eclipsed conformation A. One more 60° rotation produces the ‘anti’ conformation, where the two methyl groups are positioned opposite each other and steric repulsion is minimized.
The anti conformation is the lowest energy conformation for butane. The diagram below summarizes the relative energies for the various eclipsed, staggered, and gauche conformations.
Interactive Element
Figure \(1\): A 3D Structure of the Anti Butane Conformer.
At room temperature, butane is most likely to be in the lowest-energy anti conformation at any given moment in time, although the energy barrier between the anti and eclipsed conformations is not high enough to prevent constant rotation except at very low temperatures. For this reason (and also simply for ease of drawing), it is conventional to draw straight-chain alkanes in a zigzag form, which implies the anti conformation at all carbon-carbon bonds. For example octane is commonly drawn as:
Drawing Newman Projections
Newman projections are a valuable method for viewing the relative positions of groups within molecule. Being able to draw the Newman projection for a given molecule is a valuable skill and will be used repeatedly throughout organic chemistry. Because organic molecules often contain multiple carbon-carbon bonds it is important to precisely know which bond and which direction is being sighted for the Newman projection. The details of the Newman projection change given the molecule but for typical alkanes a full conformational analysis involves a full 360o rotation in 60o increments. This will produce three staggered conformers and three eclipsed conformers. Typically, the staggered conformers are more stable and the eclipsed conformers are less stable. The least stable conformer will have the largest groups eclipsed while the most stable conformer will have the largest groups anti (180o) to each other.
Example
Draw the Newman projection of 2,3 dimethylbutane along the C2-C3 bond. Then determine the least stable conformation.
First draw the molecule and locate the indicated bond:
Because the question asks for the least stable conformation, focus on the three possible eclipsed Newman projections. Draw out three eclipsed Newman projections as a template. Because it is difficult to draw a true staggered Newman projection, it is common to show the bonds slightly askew.
Place the substituents attached to the second carbon (C3) on the back bonds of all three Newman projections. In this example they are 2 CH3s and an H. Place the substituents in the same position on all three Newman projections.
Then place the substituents attached to the first carbon (C2) on the front bonds of the Newman projection. In this example, the substituents are also 2 CH3s and an H. Move the substituents through two 60o rotations to create the remaining two eclipsed Newman projections. Leave the substituents on the back carbon in place. Attempting to rotate the front and back carbons simultaneously is a common mistake and often leads to incorrect Newman projections.
Compare the Newman projections by looking the eclipsed interactions. Remember that the order of torsional strain interactions are CH3-CH3 > CH3-H > H-H. The third structure has two CH3-CH3 torsional interactions which will make it the least stable conformer of 2,3 dimethyl butane.
Example \(1\)
Draw Newman projections of the eclipsed and staggered conformations of propane, as if viewed down the C1-C2 bond.
Answer
Example \(2\)
Draw a Newman projection, looking down the C2-C3 bond, of 1-butene in the conformation shown below.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/02._Structure_and_Reactivity%3A_Acids_and_Bases_Polar_and_Nonpolar_Molecules/2.9%3A_Rotation_in_Substituted_Ethanes.txt |
In chemistry, a radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell, and therefore may be seen as having one or more "dangling" covalent bonds.
With some exceptions, these "dangling" bonds make free radicals highly chemically reactive towards other substances, or even towards themselves: their molecules will often spontaneously dimerize or polymerize if they come in contact with each other. Most radicals are reasonably stable only at very low concentrations in inert media or in a vacuum.
Free radicals may be created in a number of ways, including synthesis with very dilute or rarefied reagents, reactions at very low temperatures, or breakup of larger molecules. The latter can be affected by any process that puts enough energy into the parent molecule, such as ionizing radiation, heat, electrical discharges, electrolysis, and chemical reactions. Indeed, radicals are intermediate stages in many chemical reactions.
History
The first organic free radical identified was triphenylmethyl radical. This species was discovered by Moses Gomberg in 1900 at the University of Michigan USA. Historically, the term radical in radical theory was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons; however, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.
Depiction in chemical reactions
In this chapter, we will learn about some reactions in which the key steps involve the movement of single electrons. You may recall from way back in section 6.1A that single electron movement is depicted by a single-barbed'fish-hook' arrow (as opposed to the familiar double-barbed arrows that we have been using throughout the book to show two-electron movement).
Single-electron mechanisms involve the formation and subsequent reaction of free radical species, highly unstable intermediates that contain an unpaired electron. We will learn in this chapter how free radicals are often formed from homolytic cleavage, an event where the two electrons in a breaking covalent bond move in opposite directions.
(In contrast, essentially all of the reactions we have studied up to now involve bond-breaking events in which both electrons move in the same direction: this is called heterolytic cleavage).
We will also learn that many single-electron mechanisms take the form of a radical chain reaction, in which one radical causes the formation of a second radical, which in turn causes the formation of a third radical, and so on.
The high reactivity of free radical species and their ability to initiate chain reactions is often beneficial - we will learn in this chapter about radical polymerization reactions that form useful materials such as plexiglass and polyproylene fabric. We will also learn about radical reactions that are harmful, such as the degradation of atmospheric ozone by freon, and the oxidative damage done to lipids and DNA in our bodies by free radicals species. Finally, we will see how some enzymes use bound metals to catalyze high e
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.
Allyic & Benzlic > 3o > 2o > 1o > Methyl
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'.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
The Relationship between Bond Order and Bond Energy
Triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between different atoms show a wide range of bond energies, however. Table 8.6 lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends:
Table 8.6 Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K
Single Bonds Multiple Bonds
H–H 432 C–C 346 N–N ≈167 O–O ≈142 F–F 155 C=C 602
H–C 411 C–Si 318 N–O 201 O–F 190 F–Cl 249 C≡C 835
H–Si 318 C–N 305 N–F 283 O–Cl 218 F–Br 249 C=N 615
H–N 386 C–O 358 N–Cl 313 O–Br 201 F–I 278 C≡N 887
H–P ≈322 C–S 272 N–Br 243 O–I 201 Cl–Cl 240 C=O 749
H–O 459 C–F 485 P–P 201 S–S 226 Cl–Br 216 C≡O 1072
H–S 363 C–Cl 327 S–F 284 Cl–I 208 N=N 418
H–F 565 C–Br 285 S–Cl 255 Br–Br 190 N≡N 942
H–Cl 428 C–I 213 S–Br 218 Br–I 175 N=O 607
H–Br 362 Si–Si 222 I–I 149 O=O 494
H–I 295 Si–O 452 S=O 532
Source: Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, Inorganic Chemistry, 4th ed. (1993).
1. Bonds between hydrogen and atoms in the same column of the periodic table decrease in strength as we go down the column. Thus an H–F bond is stronger than an H–I bond, H–C is stronger than H–Si, H–N is stronger than H–P, H–O is stronger than H–S, and so forth. The reason for this is that the region of space in which electrons are shared between two atoms becomes proportionally smaller as one of the atoms becomes larger (part (a) in Figure 8.11).
2. Bonds between like atoms usually become weaker as we go down a column (important exceptions are noted later). For example, the C–C single bond is stronger than the Si–Si single bond, which is stronger than the Ge–Ge bond, and so forth. As two bonded atoms become larger, the region between them occupied by bonding electrons becomes proportionally smaller, as illustrated in part (b) in Figure 8.11. Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., N, O, F), which are unusually weak compared with single bonds between their larger congeners. It is likely that the N–N, O–O, and F–F single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms. The trend in bond energies for the halogens is therefore
Cl–Cl > Br–Br > F–F > I–I
Similar effects are also seen for the O–O versus S–S and for N–N versus P–P single bonds.
Note
Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column.
3. Because elements in periods 3 and 4 rarely form multiple bonds with themselves, their multiple bond energies are not accurately known. Nonetheless, they are presumed to be significantly weaker than multiple bonds between lighter atoms of the same families. Compounds containing an Si=Si double bond, for example, have only recently been prepared, whereas compounds containing C=C double bonds are one of the best-studied and most important classes of organic compounds.
Figure 8.11 The Strength of Covalent Bonds Depends on the Overlap between the Valence Orbitals of the Bonded Atoms. The relative sizes of the region of space in which electrons are shared between (a) a hydrogen atom and lighter (smaller) vs. heavier (larger) atoms in the same periodic group; and (b) two lighter versus two heavier atoms in the same group. Although the absolute amount of shared space increases in both cases on going from a light to a heavy atom, the amount of space relative to the size of the bonded atom decreases; that is, the percentage of total orbital volume decreases with increasing size. Hence the strength of the bond decreases.
4. Multiple bonds between carbon, oxygen, or nitrogen and a period 3 element such as phosphorus or sulfur tend to be unusually strong. In fact, multiple bonds of this type dominate the chemistry of the period 3 elements of groups 15 and 16. Multiple bonds to phosphorus or sulfur occur as a result of d-orbital interactions, as we discussed for the SO42− ion in Section 8.6. In contrast, silicon in group 14 has little tendency to form discrete silicon–oxygen double bonds. Consequently, SiO2 has a three-dimensional network structure in which each silicon atom forms four Si–O single bonds, which makes the physical and chemical properties of SiO2 very different from those of CO2.
Note
Bond strengths increase as bond order increases, while bond distances decrease.
Table: Average bond energies:
Bond
(kJ/mol)
C-F
485
C-Cl
328
C-Br
276
C-I
240
C-C
348
C-N
293
C-O
358
C-F
485
C-C
348
C=C
614
C=C
839
Contributors
• Kim Song (UCD), Donald Le (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/03._Reactions_of_Alkanes%3A_Bond-Dissociation_Energies_Radical_Halogenation_and_Relative_Reactivity/3-01_Strength_of_Alkane__Bonds:__Radicals.txt |
3o > 2o > 1o > Methyl
In order to understand the reason for this ordering in radical stability we must first look at the structure of a alkyl radical.
Experimental data have shown that radicals have a trigonal planar geometry associated with sp2 hybridization.
The unpaired electron is contained in a p orbital which is perpendicular to the molecular plane.
This trigonal planar geometry allows for sigma on adjacent carbons to align and overlap with the p orbital. The alignment allows for the electrons in the sigma bond to donate electrons into the electron deficient p orbital. This process is called hyperconjugation.
The more alkyl substituents, and therefore more sigma bonds, the electron density is donated into the p orbitals. This is why the general trend in radical stability is true 3o > 2o > 1o > Methyl.
3-03 Conversion of Petroleum: Pyrolysis
Objectives
After completing this section, you should be able to
1. describe the general nature of petroleum deposits, and recognize why petroleum is such an important source of organic compounds.
2. explain, in general terms, the processes involved in the refining of petroleum.
3. define the octane number of a fuel, and relate octane number to chemical structure.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• catalytic cracking
• catalytic reforming
• fractional distillation
• octane number (octane rating)
Study Notes
The refining of petroleum into usable fractions is a very important industrial process. In the laboratory component of this course, you will have the opportunity to compare this industrial process to the distillation procedure as it is performed in the student laboratory.
Petroleum
The petroleum that is pumped out of the ground is a complex mixture of several thousand organic compounds, including straight-chain alkanes, cycloalkanes, alkenes, and aromatic hydrocarbons with four to several hundred carbon atoms. The identities and relative abundance of the components vary depending on the source - Texas crude oil is somewhat different from Saudi Arabian crude oil. In fact, the analysis of petroleum from different deposits can produce a “fingerprint” of each, which is useful in tracking down the sources of spilled crude oil. For example, Texas crude oil is “sweet,” meaning that it contains a small amount of sulfur-containing molecules, whereas Saudi Arabian crude oil is “sour,” meaning that it contains a relatively large amount of sulfur-containing molecules.
Gasoline
Petroleum is converted to useful products such as gasoline in three steps: distillation, cracking, and reforming. Recall from Chapter 1 that distillation separates compounds on the basis of their relative volatility, which is usually inversely proportional to their boiling points. Part (a) in Figure 3.8.1 shows a cutaway drawing of a column used in the petroleum industry for separating the components of crude oil. The petroleum is heated to approximately 400°C (750°F) and becomes a mixture of liquid and vapor. This mixture, called the feedstock, is introduced into the refining tower. The most volatile components (those with the lowest boiling points) condense at the top of the column where it is cooler, while the less volatile components condense nearer the bottom. Some materials are so nonvolatile that they collect at the bottom without evaporating at all. Thus the composition of the liquid condensing at each level is different. These different fractions, each of which usually consists of a mixture of compounds with similar numbers of carbon atoms, are drawn off separately. Part (b) in Figure 3.8.1 shows the typical fractions collected at refineries, the number of carbon atoms they contain, their boiling points, and their ultimate uses. These products range from gases used in natural and bottled gas to liquids used in fuels and lubricants to gummy solids used as tar on roads and roofs.
The economics of petroleum refining are complex. For example, the market demand for kerosene and lubricants is much lower than the demand for gasoline, yet all three fractions are obtained from the distillation column in comparable amounts. Furthermore, most gasolines and jet fuels are blends with very carefully controlled compositions that cannot vary as their original feedstocks did. To make petroleum refining more profitable, the less volatile, lower-value fractions are converted to more volatile, higher-value mixtures that have carefully controlled formulas. The first process used to accomplish this transformation is cracking, in which the larger and heavier hydrocarbons in the kerosene and higher-boiling-point fractions are heated to temperatures as high as 900°C. High-temperature reactions cause the carbon–carbon bonds to break, which converts the compounds to lighter molecules similar to those in the gasoline fraction. Thus in cracking, a straight-chain alkane with a number of carbon atoms corresponding to the kerosene fraction is converted to a mixture of hydrocarbons with a number of carbon atoms corresponding to the lighter gasoline fraction. The second process used to increase the amount of valuable products is called reforming; it is the chemical conversion of straight-chain alkanes to either branched-chain alkanes or mixtures of aromatic hydrocarbons. Using metals such as platinum brings about the necessary chemical reactions. The mixtures of products obtained from cracking and reforming are separated by fractional distillation.
Octane Ratings
The quality of a fuel is indicated by its octane rating, which is a measure of its ability to burn in a combustion engine without knocking or pinging. Knocking and pinging signal premature combustion (Figure 3.8.2), which can be caused either by an engine malfunction or by a fuel that burns too fast. In either case, the gasoline-air mixture detonates at the wrong point in the engine cycle, which reduces the power output and can damage valves, pistons, bearings, and other engine components. The various gasoline formulations are designed to provide the mix of hydrocarbons least likely to cause knocking or pinging in a given type of engine performing at a particular level.
The octane scale was established in 1927 using a standard test engine and two pure compounds: n-heptane and isooctane (2,2,4-trimethylpentane). n-Heptane, which causes a great deal of knocking on combustion, was assigned an octane rating of 0, whereas isooctane, a very smooth-burning fuel, was assigned an octane rating of 100. Chemists assign octane ratings to different blends of gasoline by burning a sample of each in a test engine and comparing the observed knocking with the amount of knocking caused by specific mixtures of n-heptane and isooctane. For example, the octane rating of a blend of 89% isooctane and 11% n-heptane is simply the average of the octane ratings of the components weighted by the relative amounts of each in the blend. Converting percentages to decimals, we obtain the octane rating of the mixture:
$0.89(100) + 0.11(0) = 89 \label{3.8.1}$
As shown in Table $1$, many compounds that are now available have octane ratings greater than 100, which means they are better fuels than pure isooctane. In addition, anti-knock agents, also called octane enhancers, have been developed. One of the most widely used for many years was tetraethyl lead [(C2H5)4Pb], which at approximately 3 g/gal gives a 10–15-point increase in octane rating. Since 1975, however, lead compounds have been phased out as gasoline additives because they are highly toxic. Other enhancers, such as methyl t-butyl ether (MTBE), have been developed to take their place. They combine a high octane rating with minimal corrosion to engine and fuel system parts. Unfortunately, when gasoline containing MTBE leaks from underground storage tanks, the result has been contamination of the groundwater in some locations, resulting in limitations or outright bans on the use of MTBE in certain areas. As a result, the use of alternative octane enhancers such as ethanol, which can be obtained from renewable resources such as corn, sugar cane, and, eventually, corn stalks and grasses, is increasing.
Table $1$: The Octane Ratings of Some Hydrocarbons and Common Additives
Name Condensed Structural Formula Octane Rating Name Condensed Structural Formula Octane Rating
n-heptane CH3CH2CH2CH2CH2CH2CH3 0 o-xylene skeletal structure of o-xylene.cdxml 107
n-hexane CH3CH2CH2CH2CH2CH3 25 ethanol CH3CH2OH 108
n-pentane CH3CH2CH2CH2CH3 62 t-butyl alcohol (CH3)3COH 113
isooctane (CH3)3CCH2CH(CH3)2 100 p-xylene 116
benzene 106 methyl t-butyl ether H3COC(CH3)3 116
methanol CH3OH 107 toluene 118 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/03._Reactions_of_Alkanes%3A_Bond-Dissociation_Energies_Radical_Halogenation_and_Relative_Reactivity/3-02%09Structure_of_Alkyl__Radicals%3A_Hyperconjugation.txt |
Objectives
After completing this section, you should be able to
1. explain why the radical halogenation of alkanes is not usually a particularly good method of preparing pure samples of alkyl halides.
2. use $\ce{\sf{C–H}}$ bond energies to account for the fact that in radical chlorinations, the reactivity of hydrogen atoms decreases in the order $\text{tertiary} > \text{secondary} > \text{primary}. \nonumber$
3. predict the approximate ratio of the expected products from the monochlorination of a given alkane.
Study Notes
The following terms are synonymous:
1. methyl hydrogens, primary hydrogens, and 1° hydrogens.
2. methylene hydrogens, secondary hydrogens, and 2° hydrogens.
3. methine hydrogens, tertiary hydrogens, and 3° hydrogens.
Note that in radical chlorination reactions, the reactivity of methine, methylene and methyl hydrogens decreases in the ratio of approximately 5 : 3.5 : 1. This will aid in the prediction of expected products from the monochlorination of a given alkane.
Radical Halogenation
Alkanes (the simplest of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen (Cl2 or Br2) to form a haloalkane. This reaction is very important in organic chemistry because it functionalizes alkanes which opens a gateway to further chemical reactions.
General Reaction
$CH_4 + Cl_2 + energy → CH_3Cl + HCl \nonumber$
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism which is characterized by three steps: initiation, propagation, and termination. Initiation requires an input of energy but after that the reaction is self-sustaining.
Step 1: Initiation
During the initiation step free radicals are created when ultraviolet light or heat causes the X-X halogen bond to undergo homolytic to create two halogen free radicals. It is important to note that this step is not energetically favorable and cannot occur without some external energy input. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical abstracts hydrogen atom from methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step, the methyl radical reacts with more of the chlorine starting material (Cl2). One of the chlorine atoms becomes a radical and the other combines with the methyl radical to form the alkyl halide product.
Step 3: Termination
In the three termination steps of this mechanism, radicals produced in the mechanism an undergo radical coupling to form a sigma bond. These are called termination steps because a free radical is not produced as a product, which prevents the reaction from continuing. Combining the two types of radicals produced can be combined to from three possible products. Two chlorine radicals and couple to form more halogen reactant (Cl2). A chlorine radical and a methyl radical can couple to form more product (CH3Cl). An finally, two methyl radicals can couple to form a side product of ethane (CH3CH3).
This reaction is a poor synthetic method due to the formation of polyhalogenated side products. The desired product occurs when one of the hydrogen atoms in the methane has been replaced by a chlorine atom. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms to produce a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
Energetics
Why do these reactions occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ΔH that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed} \nonumber$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Chlorination of Other Alkanes
When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.
CH3-CH2-CH3 + Cl2 → 45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3
These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3.5:1. Further experiments showed that 3º-hydrogens are about 5 times more toward halogen atoms 1º-hydrogens. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.
(CH3)3CH + Cl2 → 65% (CH3)3CCl + 35% (CH3)2CHCH2Cl
The Relative Reactivity of Hydrogens to
Radical Chlorination
This difference in reactivity can only be attributed to differences in C-H bond dissociation energies. In our previous discussion of bond energy we assumed average values for all bonds of a given kind, but now we see that this is not strictly true. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table.
R (in R–H) methyl ethyl i-propyl t-butyl
Bond Dissociation Energy
(kcal/mole)
103 98 95 93
This data shows that a tertiary C-H bond (93 kcal/mole) is easier to break than a secondary (95 kcal/mole) and primary (98 kcal/mole) C-H bond. These bond dissociation energies can be used to estimate the relative stability of the radicals formed after homolytic cleavage. Because a tertiary C-H bond requires less energy to undergo homolytic cleavage than a secondary or primary C-H bond, it can be inferred that a tertiary radical is more stable than secondary or primary.
Relative Stability of Free Radicals
Exercise $1$
Write out the complete mechanism for the chlorination of methane.
Answer
The answer to this problem is actually above in the initiation, propagation and termination descriptions.
Exercise $2$
Explain, in your own words, how the first propagation step can occur without input of energy if it is energetically unfavorable.
Answer
Since the second step in propagation is energetically favorable and fast, it drives the equilibrium toward products, even though the first step is not favorable.
Exercise $3$
Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
Answer
Initiation step requires energy which can be in the form of light or het.
Exercise $4$
Having learned how to calculate the change in enthalpy for the chlorination of methane apply your knowledge and using the table provided below calculate the change in enthalpy for the bromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Answer
To calculate the enthalpy of reaction, you subtract the BDE of the bonds formed from the BDE of the bonds broken.
Bonds broken are C-H and Br-Br.
Bonds formed are H-Br adn C-Br.
Bonds broken - bonds formed = change in enthalpy
(101 kcal/mol + 46 kcal/mol) - (87 kcal/mol + 70 kcal/mol) = change in enthalpy
-10 kcal/mol = change in enthalpy for bromination of ethane.
Exercise $5$
1) Predict the mono-substituted halogenated product(s) of chlorine gas reacting with 2-methylbutane.
2) Predict the relative amount of each mono-brominated product when 3-methylpentane is reacted with Br2. Consider 1°, 2°, 3° hydrogen.
3) For the following compounds, give all possible monochlorinated derivatives.
Answer
1)
2)
.
3) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/03._Reactions_of_Alkanes%3A_Bond-Dissociation_Energies_Radical_Halogenation_and_Relative_Reactivity/3-04_Chlorination_of__Methane%3A_The__Radical_Chain__Mec.txt |
For the radical halogenation of methane, fluorine is the most reactive and iodine is the least reactive. The reason for this can be seen in the enthalpy for the first propagation step in the different halognations of methane. For fluorine the step is exothermic, however, for chlorine, bromine, and iodine the step is endothermic. This trend comes from the relative bond strengths of the H-X bond which is formed. The H-F bond is strong which causes the high reactivity of fluorine in these reactions. Correspondingly, reactions with fluorine have a relatively small activation energy and Iodine have relatively large activation energy. In the case of Iodine the energy of activation is so high that Iodination of methane does not occur.
3-06 Keys to Success: Using the "Known" Mechanism
Alkanes (the most basic of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Introduction
While the reactions possible with alkanes are few, there are many reactions that involve haloalkanes. In order to better understand the mechanism (a detailed look at the step by step process through which a reaction occurs), we will closely examine the chlorination of methane. When methane (CH4) and chlorine (Cl2) are mixed together in the absence of light at room temperature nothing happens. However, if the conditions are changed, so that either the reaction is taking place at high temperatures (denoted by Δ) or there is ultra violet irradiation, a product is formed, chloromethane (CH3Cl).
Energetics
Why does this reaction occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ($\Delta{H}$) that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed}$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism. The radical chain mechanism is characterized by three steps: initiation, propagation and termination. Initiation requires an input of energy but after that the reaction is self-sustaining. The first propagation step uses up one of the products from initiation, and the second propagation step makes another one, thus the cycle can continue until indefinitely.
Step 1: Initiation
Initiation breaks the bond between the chlorine molecule (Cl2). For this step to occur energy must be put in, this step is not energetically favorable. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy. It is important to note that this part of the mechanism cannot occur without some external energy input, through light or heat.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical combines with a hydrogen on the methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step more of the chlorine starting material (Cl2) is used, one of the chlorine atoms becomes a radical and the other combines with the methyl radical.
The first propagation step is endothermic, meaning it takes in heat (requires 2 kcal/mol) and is not energetically favorable. In contrast the second propagation step is exothermic, releasing 27 kcal/mol. Since the second propagation step is so exothermic, it occurs very quickly. The second propagation step uses up a product from the first propagation step (the methyl radical) and following Le Chatelier's principle, when the product of the first step is removed the equilibrium is shifted towards it's products. This principle is what governs the unfavorable first propagation step's occurance.
Step 3: Termination
In the termination steps, all the remaining radicals combine (in all possible manners) to form more product (CH3Cl), more reactant (Cl2) and even combinations of the two methyl radicals to form a side product of ethane (CH3CH3).
Problems with the Chlorination of Methane
The chlorination of methane does not necessarily stop after one chlorination. It may actually be very hard to get a monosubstituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed. One way to avoid this problem is to use a much higher concentration of methane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product.
References
1. Matyjaszewski, Krzysztof, Wojciech Jakubowski, Ke Min, Wei Tang, Jinyu Huang, Wade A. Braunecker, and Nicolay V. Tsarevsky. "Diminishing Catalyst Concentration in Atom Transfer Radical Polymerization with Reducing Agents." Proceedings of the National Academy of Sciences of the United States of America 103 (2006): 15309-5314.
2. Morgan, G. T. "A State Experiment in Chemical Research." Science 72 (1930): 379-90.
3. Phillips, Francis C. "# Researches upon the Chemical Properties of Gases." Researches upon the Chemical Properties of Gases 17 (1893): 149-236.
Problems
Answers to these questions are in an attached slide
1. Write out the complete mechanism for the chlorination of methane.
2. Explain, in your own words, how the first propagation step can occur without input of energy if it is energetically unfavorable.
3. Compounds other than chlorine and methane go through halogenation with the radical chain mechanism. Write out a generalized equation for the halogenation of RH with X2including all the different steps of the mechanism.
4. Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
5. Having learned how to calculate the change in enthalpy for the chlorination of methane apply your knowledge and using the table provided below calculate the change in enthalpy for the bromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Contributors
• Kristen Kelley and Britt Farquharson
Further Reading
Wikipedia
Khan Academy
Carey 5th Ed Online
Chemtube3D
MasterOrganicChemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/03._Reactions_of_Alkanes%3A_Bond-Dissociation_Energies_Radical_Halogenation_and_Relative_Reactivity/3-05__Other__Radical__Halogenations_of__Methane.txt |
Objectives
After completing this section, you should be able to
1. explain why the radical halogenation of alkanes is not usually a particularly good method of preparing pure samples of alkyl halides.
2. use $\ce{\sf{C–H}}$ bond energies to account for the fact that in radical chlorinations, the reactivity of hydrogen atoms decreases in the order $\text{tertiary} > \text{secondary} > \text{primary}. \nonumber$
3. predict the approximate ratio of the expected products from the monochlorination of a given alkane.
Study Notes
The following terms are synonymous:
1. methyl hydrogens, primary hydrogens, and 1° hydrogens.
2. methylene hydrogens, secondary hydrogens, and 2° hydrogens.
3. methine hydrogens, tertiary hydrogens, and 3° hydrogens.
Note that in radical chlorination reactions, the reactivity of methine, methylene and methyl hydrogens decreases in the ratio of approximately 5 : 3.5 : 1. This will aid in the prediction of expected products from the monochlorination of a given alkane.
Radical Halogenation
Alkanes (the simplest of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen (Cl2 or Br2) to form a haloalkane. This reaction is very important in organic chemistry because it functionalizes alkanes which opens a gateway to further chemical reactions.
General Reaction
$CH_4 + Cl_2 + energy → CH_3Cl + HCl \nonumber$
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism which is characterized by three steps: initiation, propagation, and termination. Initiation requires an input of energy but after that the reaction is self-sustaining.
Step 1: Initiation
During the initiation step free radicals are created when ultraviolet light or heat causes the X-X halogen bond to undergo homolytic to create two halogen free radicals. It is important to note that this step is not energetically favorable and cannot occur without some external energy input. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical abstracts hydrogen atom from methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step, the methyl radical reacts with more of the chlorine starting material (Cl2). One of the chlorine atoms becomes a radical and the other combines with the methyl radical to form the alkyl halide product.
Step 3: Termination
In the three termination steps of this mechanism, radicals produced in the mechanism an undergo radical coupling to form a sigma bond. These are called termination steps because a free radical is not produced as a product, which prevents the reaction from continuing. Combining the two types of radicals produced can be combined to from three possible products. Two chlorine radicals and couple to form more halogen reactant (Cl2). A chlorine radical and a methyl radical can couple to form more product (CH3Cl). An finally, two methyl radicals can couple to form a side product of ethane (CH3CH3).
This reaction is a poor synthetic method due to the formation of polyhalogenated side products. The desired product occurs when one of the hydrogen atoms in the methane has been replaced by a chlorine atom. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms to produce a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
Energetics
Why do these reactions occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ΔH that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed} \nonumber$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Chlorination of Other Alkanes
When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.
CH3-CH2-CH3 + Cl2 → 45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3
These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3.5:1. Further experiments showed that 3º-hydrogens are about 5 times more toward halogen atoms 1º-hydrogens. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.
(CH3)3CH + Cl2 → 65% (CH3)3CCl + 35% (CH3)2CHCH2Cl
The Relative Reactivity of Hydrogens to
Radical Chlorination
This difference in reactivity can only be attributed to differences in C-H bond dissociation energies. In our previous discussion of bond energy we assumed average values for all bonds of a given kind, but now we see that this is not strictly true. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table.
R (in R–H) methyl ethyl i-propyl t-butyl
Bond Dissociation Energy
(kcal/mole)
103 98 95 93
This data shows that a tertiary C-H bond (93 kcal/mole) is easier to break than a secondary (95 kcal/mole) and primary (98 kcal/mole) C-H bond. These bond dissociation energies can be used to estimate the relative stability of the radicals formed after homolytic cleavage. Because a tertiary C-H bond requires less energy to undergo homolytic cleavage than a secondary or primary C-H bond, it can be inferred that a tertiary radical is more stable than secondary or primary.
Relative Stability of Free Radicals
Exercise $1$
Write out the complete mechanism for the chlorination of methane.
Answer
The answer to this problem is actually above in the initiation, propagation and termination descriptions.
Exercise $2$
Explain, in your own words, how the first propagation step can occur without input of energy if it is energetically unfavorable.
Answer
Since the second step in propagation is energetically favorable and fast, it drives the equilibrium toward products, even though the first step is not favorable.
Exercise $3$
Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
Answer
Initiation step requires energy which can be in the form of light or het.
Exercise $4$
Having learned how to calculate the change in enthalpy for the chlorination of methane apply your knowledge and using the table provided below calculate the change in enthalpy for the bromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Answer
To calculate the enthalpy of reaction, you subtract the BDE of the bonds formed from the BDE of the bonds broken.
Bonds broken are C-H and Br-Br.
Bonds formed are H-Br adn C-Br.
Bonds broken - bonds formed = change in enthalpy
(101 kcal/mol + 46 kcal/mol) - (87 kcal/mol + 70 kcal/mol) = change in enthalpy
-10 kcal/mol = change in enthalpy for bromination of ethane.
Exercise $5$
1) Predict the mono-substituted halogenated product(s) of chlorine gas reacting with 2-methylbutane.
2) Predict the relative amount of each mono-brominated product when 3-methylpentane is reacted with Br2. Consider 1°, 2°, 3° hydrogen.
3) For the following compounds, give all possible monochlorinated derivatives.
Answer
1)
2)
.
3)
3.10: Synthetic Chlorine Compounds and the Stratosphe
The high reactivity of free radicals and the multiplicative nature of radical chain reactions can be useful in the synthesis of materials such as polyethylene plastic - but these same factors can also result in dangerous consequences. You are probably aware of the danger posed to the earth's protective stratospheric ozone layer by the use of chlorofluorocarbons (CFCs) as refrigerants and propellants in aerosol spray cans. Freon-11, or CFCl3, is a typical CFC that was widely used until fairly recently. It can take months or years for a CFC molecule to drift up into the stratosphere from the surface of the earth, and of course the concentration of CFCs at this altitude is very low. Ozone, on the other hand, is continually being formed in the stratosphere. Why all the concern, then, about destruction of the ozone layer - how could such a small amount of CFCs possibly do significant damage? The problem lies in the fact that the process by which ozone is destroyed is a chain reaction, so that a single CFC molecule can initiate the destruction of many ozone molecules before a chain termination event occurs.
Although there are several different processes by which the ozone destruction process might occur, the most important is believed to be the chain reaction shown below.
To address the problem of ozone destruction, scientists are developing new organohalogen refrigerant compounds that are less stable than the older CFCs like Freon-11, in the hope that the new compounds will break down in the lower atmosphere before they reach an altitude where they can harm the ozone layer. Most of the new compounds contain carbon-hydrogen bonds, which are subject to homolytic cleavage initiated by hydroxide radicals present in the lower atmosphere.
This degradation occurs before the refrigerant molecules have a chance to drift up to the stratosphere where the ozone plays its important protective role. The degradation products are quite unstable and quickly degrade further, by a variety of mechanisms, into relatively harmless by-products. The hydroxide radical is sometimes referred to as an atmospheric 'detergent' due to its ability to degrade refrigerants and other volatile organic pollutants which have escaped into the atmosphere.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/03._Reactions_of_Alkanes%3A_Bond-Dissociation_Energies_Radical_Halogenation_and_Relative_Reactivity/3-08_Selectivity_in_Radical_Halogenation_with__Fluorine_.txt |
Cycloalkanes are cyclic hydrocarbons with the carbons of the molecule arranged in the form of one or more rings. Cycloalkanes are saturated hydrocarbons, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds).
04. Cycloalkanes
Objectives
After completing this section, you should be able to
1. name a substituted or unsubstituted cycloalkane, given its Kekulé structure, shorthand structure or condensed structure.
2. draw the Kekulé, shorthand or condensed structure for a substituted or unsubstituted cycloalkane, given its IUPAC name.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• cycloalkane
Study Notes
Provided that you have mastered the IUPAC system for naming alkanes, you should find that the nomenclature of cycloalkanes does not present any particular difficulties.
Many organic compounds found in nature contain rings of carbon atoms. These compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The simplest examples of this class consist of a single, un-substituted carbon ring, and these form a homologous series similar to the unbranched alkanes.
Like alkanes, cycloalkane molecules are often drawn as skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens. Cyclohexane, one of the most common cycloalkanes is shown below as an example.
Cyclic hydrocarbons have the prefix "cyclo-". The IUPAC names, molecular formulas, and skeleton structures of the cycloalkanes with 3 to 10 carbons are given in Table 4.1.1. Note that the general formula for a cycloalkane composed of n carbons is CnH2n, and not CnH2n+2 as for alkanes. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so are still considered to be saturated with hydrogen.
Table \(1\): Examples of Simple Cycloalkanes
Cycloalkane Molecular Formula Skeleton Structure
Cyclopropane C3H6
Cyclobutane C4H8
Cyclopentane C5H10
Cyclohexane C6H12
Cycloheptane C7H14
Cyclooctane C8H16
Cyclononane C9H18
Cyclodecane C10H20
IUPAC Rules for Nomenclature
The naming of substituted cycloalkanes follows the same basic steps used in naming alkanes.
1. Determine the parent chain.
2. Number the substituents of the ring beginning at one substituent so that the nearest substituent is numbered the lowest possible. If there are multiple choices that are still the same, go to the next substituent and give it the lowest number possible.
3. Name the substituents and place them in alphabetical order.
More specific rules for naming substituted cycloalkanes with examples are given below.
1. Determine the cycloalkane to use as the parent. If there is an alkyl straight chain that has a greater number of carbons than the cycloalkane, then the alkyl chain must be used as the primary parent chain. Cycloalkanes substituents have an ending "-yl". If there are two cycloalkanes in the molecule, use the cycloalkane with the higher number of carbons as the parent.
Example \(1\)
The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. The parent chain in this molecule is decane and cyclopropane is a substituent. The name of this molecule is 3-cyclopropyl-6-methyldecane.
Example \(2\)
Name the cycloalkane structure.
Solution
There are two different cycloalkanes in this molecule. Because it contains more carbons, the cyclopentane ring will be named as the parent chain. The smaller ring, cyclobutane, is named as a substituent on the parent chain. The name of this molecule is cyclobutylcyclopentane.
2) When there is only one substituent on the ring, the ring carbon attached to the substituent is automatically carbon #1. Indicating the number of the carbon with the substituent in the name is optional.
Example \(3\)
1-chlorocyclobutane or cholorocyclobutane 1-propylcyclohexane or propylcyclohexane
3) If there are multiple substituents on the ring, number the carbons of the cycloalkane so that the carbons with substituents have the lowest possible number. A carbon with multiple substituents should have a lower number than a carbon with only one substituent or functional group. One way to make sure that the lowest number possible is assigned is to number the carbons so that when the numbers corresponding to the substituents are added, their sum is the lowest possible.
4) When naming the cycloalkane, the substituents must be placed in alphabetical order. Remember the prefixes di-, tri-, etc. , are not used for alphabetization.
Example \(4\)
In this example, the ethyl or the methyl subsistent could be attached to carbon one. The ethyl group attachment is assigned carbon 1 because ethyl comes before methyl alphabetically. After assigning carbon 1 the cyclohexane ring can be numbered going clockwise or counterclockwise. When looking at the numbers produced going clockwise produces lower first substituent numbers (1,3) than when numbered counterclockwise (1,5). So the correct name is 1-ethyl-3-methylcyclohexane.
Example \(5\)
Name the following structure using IUPAC rules.
Solution
Remember when dealing with cycloalkanes with more than two substituents, finding the lowest possible 2nd substituent numbering takes precedence. Consider a numbering system with each substituent attachment point as being carbon one. Compare them and whichever produces the lowest first point of difference will be correct.
The first structure would have 1,4 for the relationship between the first two groups. The next structure would have 1,3. The final 2 structures both have 1,2 so those are preferable to the first two. Now we have to determine which is better between the final 2 structures. The 3rd substituent on structure 3 would be at the 5 position leading to 1,2,5 while in the final structure the 3rd methyl group is on carbon 4 leading to 1,2,4. This follows the rules of giving the lowest numbers at the first point of difference.
The correct name for the molecule is 4-Bromo-1,2-dimethylcyclohexane.
Example \(6\)
2-bromo-1-chloro-3-methylcyclopentane
Notice that "b" of bromo alphabetically precedes the "m" of methyl. Also, notice that the chlorine attachment point is assigned carbon 1 because it comes first alphabetically and the overall sum of numbers would be the same if the methyl attachment carbon was assigned as 1 and the chlorine attachment as 3.
Example \(7\)
(2-bromo-1,1-dimethylcyclohexane)
Although "di" alphabetically precedes "f", "di" is not used in determining the alphabetical order.
Example \(8\)
2-fluoro-1,1,-dimethylcyclohexane NOT 1,1-dimethyl-2-fluorocyclohexane
also
2-fluoro-1,1,-dimethylcyclohexane NOT 1-fluoro-2,2-dimethylcyclohexane (as that would give a larger number to the first point of difference)
Although "di" alphabetically precedes "f", "di" is not used in determining the alphabetical order of the substituents. Notice that the attachment point of the two methyl groups is assigned carbon 1 despite the fact that fluorine comes first alphabetically. This is because this assignment allows for a lower overall numbering of substituents, so assigning alphabetical priority is not necessary.
Polycyclic Compounds
Hydrocarbons having more than one ring are common, and are referred to as bicyclic (two rings), tricyclic (three rings) and in general, polycyclic compounds. The molecular formulas of such compounds have H/C ratios that decrease with the number of rings. In general, for a hydrocarbon composed of \(n\) carbon atoms associated with \(m\) rings the formula is: \(\ce{C_nH_{2n + 2 - 2m}}\). The structural relationship of rings in a polycyclic compound can vary. They may be separate and independent, or they may share one or two common atoms. Some examples of these possible arrangements are shown in the following table.
Table \(2\): Examples of Isomeric \(\ce{C_8H_{14}}\) Bicycloalkanes
Isolated Rings Spiro Rings Fused Rings Bridged Rings
No common atoms One common atom One common bond Two common atoms
Polycyclic compounds, like cholesterol shown below, are biologically important and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules, and will not be covered here.
Cholesterol (polycyclic)
Exercise \(1\)
Give the IUPAC names for the following cycloalkane structures.
Answer
a) 1,3-Dimethylcyclohexane
b) 2-Cyclopropylbutane
c) 1-Ethyl-3-methylcyclooctane
d) 1-Bromo-3-methylcyclobutane
e) 1,2,4-Triethylcycloheptane
f) 1-Chloro-2,4-dimethylcyclopentane
Exercise \(2\)
Draw the structures for the IUPAC names below.
1. 2,3-dicyclopropylpentane
2. 1,2,3-triethylcyclopentane
3. 3-cyclobutyl-2-methylhexane
4. 2-bromo-1-chloro-4-methylcyclohexane
5. 1-bromo-5-propylcyclododecane
Answer
Objectives
After completing this section, you should be able to
1. draw structural formulas that distinguish between cis and trans disubstituted cycloalkanes.
2. construct models of cis and trans disubstituted cycloalkanes using ball-and-stick molecular models.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• constitutional isomer
• stereoisomer
• cis-trans isomers
Previously, constitutional isomers have been defined as molecules that have the same molecular formula, but different atom connectivity. In this section, a new class of isomers, stereoisomers, will be introduced. Stereoisomers are molecules that have the same molecular formula, the same atom connectivity, but they differ in the relative spatial orientation of the atoms.
Cycloalkanes are similar to open-chain alkanes in many respects. They both tend to be nonpolar and relatively inert. One important difference, is that cycloalkanes have much less freedom of movement than open-chain alkanes. As discussed in Sections 3.6 and 3.7, open-chain alkanes are capable of rotation around their carbon-carbon sigma bonds. The ringed structures of cycloalkanes prevent such free rotation, causing them to be more rigid and somewhat planar.
Di-substituted cycloalkanes are one class of molecules that exhibit stereoisomerism. 1,2-dibromocyclopentane can exist as two different stereoisomers: cis-1,2-dibromocyclopentane and trans-1,2-dibromocyclopentane. The cis-1,2-dibromocyclopentane and trans-1,2-dibromocyclopentane stereoisomers of 1,2-dibromocyclopentane are shown below. Both molecules have the same molecular formula and the same atom connectivity. They differ only in the relative spatial orientation of the two bromines on the ring. In cis-1,2-dibromocyclopentane, both bromine atoms are on the same "face" of the cyclopentane ring, while in trans-1,2-dibromocyclopentane, the two bromines are on opposite faces of the ring. Stereoisomers require an additional nomenclature prefix be added to the IUPAC name in order to indicate their spatial orientation. Di-substituted cycloalkane stereoisomers are designated by the nomenclature prefixes cis (Latin, meaning on this side) and trans (Latin, meaning across).
Interactive Element
The 3D Structure of cis-1,2-dibromocyclopentane
Interactive Element
The 3D Structure of trans-1,2-dibromocyclopentane
Representing 3D Structures
By convention, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the plane of the ring (coming out of the page), a dashed line for bonds to atoms or groups located below the ring (going back into the page), and solid lines for bonds in the plane of the page.
In general, if any two sp3 carbons in a ring have two different substituent groups (not counting other ring atoms) cis/trans stereoisomerism is possible. However, the cis/trans designations are not used if both groups are on the same carbon. For example, the chlorine and the methyl group are on the same carbon in 1-chloro-1-methylcyclohexane and the trans prefix should not be used.
If more than two ring carbons have substituents, the stereochemical notation distinguishing the various isomers becomes more complex and the prefixes cis and trans cannot be used to formally name the molecule. However, the relationship of any two substituents can be informally described using cis or trans. For example, in the tri-substituted cyclohexane below, the methyl group is cis to the ethyl group, and also trans to the chlorine. However, the entire molecule cannot be designated as either a cis or trans isomer. Later sections will describe how to name these more complex molecules (5.5: Sequence Rules for Specifying Configuration)
Example \(1\)
Name the following cycloalkanes:
Solution
These two example represent the two main ways of showing spatial orientation in cycloalkanes.
a) In example "a" the cycloalkane is shown as being flat and in the plane of the page. The positioning of the substituents is shown by using dash-wedge bonds. Cis/trans positioning can be determined by looking at the type of bonds attached to the substituents. If the substituents are both on the same side of the ring (Cis) they would both have either dash bonds or wedge bonds. If the the substituents are on opposite side of the ring (Trans) one substituent would have a dash bond and the other a wedge bond. Because both bromo substituents have a wedge bond they are one the same side of the ring and are cis. The name of this molecule is cis-1,4-Dibromocyclohexane.
b) Example "b" shows the cycloalkane ring roughly perpendicular to the plane of the page. When this is done, the upper and lower face of the ring is defined and each carbon in the ring will have a bond one the upper face and a bond on the lower face. Cis substituents will either both be on the upper face or the lower face. Trans substituents will have one on the upper face and one one the lower face. In example "b", one of the methyl substituents is on the upper face of the ring and one is on the lower face which makes them trans to each other. The name of this molecule is trans-1,2-Dimethylcyclopropane.
Exercises
Exercise \(1\)
Draw the following molecules:
1. trans-1,3-dimethylcyclohexane
2. trans-1,2-dibromocyclopentane
3. cis-1,3-dichlorocyclobutane
Answer
2) Cis/Trans nomenclature can be used to describe the relative positioning of substituents on molecules with more complex ring structures. The molecule below is tesosterone, the primary male sex hormone. Is the OH and the adjacent methyl group cis or trans to each other? What can you deduce about the relative positions of the indicated hydrogens?
3) Name the following compounds:
Solutions
2) Both the OH and the methyl group have wedge bonds. This implies that they are both on the same side of the testosterone ring making them cis. Two of the hydrogens have wedge bonds while one has a wedge. This means two of the hydrogens are on one side of the testosterone ring while one is on the other side.
3)
Cis-1-Bromo-3-Chlorocyclobutane
Trans-1,4-Dimethylcyclooctane
Trans-1-Bromo-3-ethylcyclopentane | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.1%3A_Names__and__Physical_Properties_of_Cycloalkanes.txt |
Objectives
After completing this section, you should be able to
1. describe the Baeyer strain theory.
2. describe how the measurement of heats of combustion provides information about the amount of strain present in a cycloalkane ring.
3. determine the relative stability of cyclic compounds, by assessing such factors as angle strain, torsional strain and steric strain.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• angle strain
• steric strain
• torsional strain
• ring strain
• heat of combustion
Heat of Combustion as a Measure of Bond Strength
The combustion of carbon compounds, especially hydrocarbons, has been the most important source of heat energy for human civilizations throughout recorded history. The practical importance of this reaction cannot be denied, but the massive and uncontrolled chemical changes that take place in combustion make it difficult to deduce mechanistic paths. Using the combustion of propane as an example, we see from the following equation that every covalent bond in the reactants has been broken and an entirely new set of covalent bonds have formed in the products. No other common reaction involves such a profound and pervasive change, and the mechanism of combustion is so complex that chemists are just beginning to explore and understand some of its elementary features.
$\ce{CH_3CH_2CH_3 + 5O_2 -> 3CO_2 + 4H_2O + heat} \nonumber$
Since all the covalent bonds in the reactant molecules are broken, the quantity of heat evolved in this reaction, and any other combustion reaction, is related to the strength of these bonds (and, of course, the strength of the bonds formed in the products). Precise heat of combustion measurements can provide useful information about the structure of molecules and their relative stability.
For example, heat of combustion is useful in determining the relative stability of isomers. Pentane has a heat of combustion of -782 kcal/mol, while that of its isomer, 2,2-dimethylpropane (neopentane), is –777 kcal/mol. These values indicate that 2,3-dimethylpentane is 5 kcal/mol more stable than pentane, since it has a lower heat of combustion.
Ring Strain
Table $1$ lists the heat of combustion data for some simple cycloalkanes. These cycloalkanes do not have the same molecular formula, so the heat of combustion per each CH2 unit present in each molecule is calculated (the fourth column) to provide a useful comparison. From the data, cyclopropane and cyclobutane have significantly higher heats of combustion per CH2, while cyclohexane has the lowest heat of combustion. This indicates that cyclohexane is more stable than cyclopropane and cyclobutane, and in fact, that cyclohexane has a same relative stability as long chain alkanes that are not cyclic. This difference in stability is seen in nature where six membered rings are by far the most common. What causes the difference in stability or the strain in small cycloalkanes?
Table $1$: Heats of combustion of select hydrocarbons
Cycloalkane
(CH2)n
CH2 Units
n
ΔH25º
kcal/mol
ΔH25º
per CH2 Unit
Ring Strain
kcal/mol
Cyclopropane n = 3 468.7 156.2 27.6
Cyclobutane n = 4 614.3 153.6 26.4
Cyclopentane n = 5 741.5 148.3 6.5
Cyclohexane n = 6 882.1 147.0 0.0
Cycloheptane n = 7 1035.4 147.9 6.3
Cyclooctane n = 8 1186.0 148.2 9.6
Cyclononane n = 9 1335.0 148.3 11.7
Cyclodecane n = 10 1481 148.1 11.0
CH3(CH2)mCH3 m = large 147.0 0.0
The Baeyer Theory on the Strain in Cycloalkane Rings
In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane are less stable than cyclohexane, because the the smaller rings are more "strained". There are many different types of strain that contribute to the overall ring strain in cycloalkanes, including angle strain, torsional strain, and steric strain. Torsional strain and steric strain were previously defined in the discussion of conformations of butane. Angle Strain occurs when the sp3 hybridized carbons in cycloalkanes do not have the expected ideal bond angle of 109.5o, causing an increase in the potential energy. An example of angle strain can be seen in the diagram of cyclopropane below in which the bond angle is 60o between the carbons. The compressed bond angles causes poor overlap of the hybrid orbitals forming the carbon-carbon sigma bonds which in turn creates destabilization.
The C-C-C bond angles in cyclopropane (diagram above) (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o. This bond angle causes cyclopropane and cyclobutane to be less stable than molecules such as cyclohexane and cyclopentane, which have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o. Changes in chemical reactivity as a consequence of angle strain are dramatic in the case of cyclopropane, and are also evident for cyclobutane.
In addition to angle strain, there is also steric (transannular) strain and torsional strain in many cycloalkanes. Transannular strain exists when there is steric repulsion between atoms.
Because cycloalkane lack the ability to freely rotate, torsional (eclipsing) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond. Torsional strain is especially prevalent in small cycloalkanes, such as cyclopropane, whose structures are nearly planar.
The Eclipsed C-H Bonds in Cyclopropane
Larger rings like cyclohexane, deal with torsional strain by forming conformers in which the rings are not planar. A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds, which decreases the angle and torsional strain in the ring. The non-planar structures of cyclohexane are very stable compared to cyclopropane and cyclobutane, and will be discussed in more detail in the next section.
The Types of Strain Which Contribute to Ring Strain in Cycloalkanes
Angle Strain The strain caused by the increase or reduction of bond angles
Torsional Strain The strain caused by eclipsing bonds on adjacent atoms
Steric Strain The strain caused by the repulsive interactions of atoms trying to occupy the same space
Exercise $1$
trans-1,2-Dimethylcyclobutane is more stable than cis-1,2-dimethylcyclobutane. Explain this observation.
Answer
The trans form does not have eclipsing methyl groups, therefore lowering the energy within the molecule. It does however have hydrogen-methyl eclipsing interactions which are not as high in energy as methyl-methyl interactions.
Exercise $2$
Cyclobutane has more torsional stain than cyclopropane. Explain this observation.
Answer
Cyclobutane has 4 CH2 groups while cyclopropane only has 3. More CH2 groups means cyclobutane has more eclipsing H-H interactions and therefore has more torsional strain.
Objectives
After completing this section, you should be able to
1. describe, and sketch the conformation of cyclopropane, cyclobutane, and cyclopentane.
2. describe the bonding in cyclopropane, and use this to account for the high reactivity of this compound.
3. analyze the stability of cyclobutane, cyclopentane and their substituted derivatives in terms of angular strain, torsional strain and steric interactions.
Study Notes
Notice that in both cyclobutane and cyclopentane, torsional strain is reduced at the cost of increasing angular (angle) strain.
Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different.
cyclopropane cyclobutane cyclopentane cyclohexane cycloheptane cyclooctane
Cyclopropane is necessarily planar (flat), with the carbon atoms at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 40 kJ/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible.
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360º, going through many different conformations. That complete rotation isn't possible in a cyclic system, because the parts that would be trying to twist away from each other would still be connected together. Thus cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation".
Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings. Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them. In nature, three- to six-membered rings are frequently encountered, so we'll focus on those.
Cyclopropane
A three membered ring has no rotational freedom whatsoever. A plane is defined by three points, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane. This lack of flexibility does not allow cyclopropane to form more stable conformers which are non-planar.
The main source of ring strain in cyclopropane is angle strain. All of the carbon atoms in cyclopropane are tetrahedral and would prefer to have a bond angle of 109.5o The angles in an equilateral triangle are actually 60o, about half as large as the optimum angle. The large deviation from the optimal bond angle means that the C-C sigma bonds forming the cyclopropane ring are bent. Maximum bonding occurs when the overlapping orbitals are pointing directly toward each other. The severely strained bond angles in cyclopropane means that the orbitals forming the C-C bonds overlap at a slight angle making them weaker. This strain is partially overcome by using so-called “banana bonds”, where the overlap between orbitals is no longer directly in a line between the two nuclei, as shown here in three representations of the bonding in cyclopropane:
The constrained nature of cyclopropane causes neighboring C-H bonds to all be held in eclipsed conformations. Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a Newman projections of cyclopropane as shown from the side.
Newman Projection of cyclopropane
Cyclopropane isn't large enough to introduce any steric strain. Steric strain does not become a factor until we reach six membered rings. Before that point, rings are not flexible enough to allow for two ring substituents to interact with each other.
Interactive Element
The 3D Structure of Cyclopropane
The combination of torsional and angle strain creates a large amount of ring strain in cyclopropane which weakens the C-C ring bonds (255 kJ/mol) when compared to C-C bonds in open-chain propane (370 kJ/mol).
Cyclobutane
Cyclobutane is a four membered ring. The larger number of ring hydrogens would cause a substantial amount of torsional strain if cyclobutane were planar.
In three dimensions, cyclobutane is flexible enough to buckle into a "puckered" shape which causes the C-H ring hydrogens to slightly deviate away from being completely eclipsed. This conformation relives some of the torsional strain but increases the angle strain because the ring bond angles decreases to 88o.
In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines.
The deviation of cyclobutane's ring C-H bonds away from being fully eclipsed can clearly be seen when viewing a Newman projections signed down one of the C-C bond.
Newman projection of cyclobutane
• With bond angles of 88o rather than 109.5o degrees, cyclobutane has significant amounts of angle strain, but less than in cyclopropane.
• Although torsional strain is still present, the neighboring C-H bonds are not exactly eclipsed in the cyclobutane's puckered conformation.
• Steric strain is very low. Cyclobutane is still not large enough that substituents can reach around to cause crowding.
• Overall the ring strain in cyclobutane (110 kJ/mol) is slightly less than cyclopropane (115 kJ/mol).
Interactive Element
The 3D Structure of Cyclobutane
Cyclopentane
Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common cycloalkane ring in nature, after cyclohexanes. Planar cyclopentane has virtually no angle strain but an immense amount of torsional strain. To reduce torsional strain, cyclopentane addops a non-planar conformation even though it slightly increases angle strain.
The lowest energy conformation of cyclopentane is known as the ‘envelope’, with four of the ring atoms in the same plane and one out of plane (notice that this shape resembles an envelope with the flap open). The out-of-plane carbon is said to be in the endo position (‘endo’ means ‘inside’). The envelope removes torsional strain along the sides and flap of the envelope. However, the neighboring carbons are eclipsed along the "bottom" of the envelope, away from the flap.
3D structure of cyclopentane (notice that the far top right carbon is the endo position).
At room temperature, cyclopentane undergoes a rapid bond rotation process in which each of the five carbons takes turns being in the endo position.
Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four. The envelope removes torsional strain along the sides and flap of the envelope by allowing the bonds to be in an almost completely staggered position. However, the neighboring bonds are eclipsed along the "bottom" of the envelope, away from the flap. Viewing a Newman projections of cyclopentane signed down one of the C-C bond show the staggered C-H bonds.
Newman projection of cyclopentane
• The angle strain in the envelope conformation of cyclopentane is low. The ideal angle in a regular pentagon is about 107o, very close to the preferred 109.5o tetrahedral bond angle.
• There is some torsional strain in cyclopentane. The envelope conformation reduces torsional strain by placing some bonds in nearly staggered positions. However, other bonds are still almost fully eclipsed.
• Cyclopentane is not large enough to allow for steric strain to be created.
• Overall, cyclopentane has very little ring strain (26 kJ/mol) when compared to cyclopropane and cyclobutane.
Interactive Element
The 3D Structure of Cyclopentane
C3-C5 Cycloalkanes in Nature
If one of the carbon-carbon bonds is broken in cyclopropane or cyclobutane, the ring will ‘spring’ open, releasing energy as the bonds reassume their preferred tetrahedral geometry. The effectiveness of two antibiotic drugs, fosfomycin and penicillin, is due in large part to the high reactivity of the three- and four-membered rings in their structures.
One of the most important five-membered rings in nature is a sugar called ribose – DNA and RNA are both constructed upon ‘backbones’ derived from ribose. Pictured below is one thymidine (T) deoxy-nucleotide from a stretch of DNA. Since the ribose has lost one of the OH groups (at carbon 2 of the ribose ring), this is part of a deoxyribonucleic acid (DNA). If the OH at carbon 2 of the ribose ring was present, this would be part of a ribonucleic acid (RNA).
The lowest-energy conformations for ribose are envelope forms in which either C3 or C2 are endo, on the same side as the C5 substituent.
Exercises
1) If cyclobutane were to be planar, how many H-H eclipsing interactions would there be? Assuming 4 kJ/mol per H-H eclipsing interaction what would the strain be on this “planar” molecule?
2) In the two conformations of trans-1,2-Dimethylcyclopentane one is more stable than the other. Explain why this is.
3) In methylcyclopentane, which carbon would most likely be in the endo position?
Solutions
1) There are 8 eclipsing interactions (two per C-C bond). The extra strain on this molecule would be 32 kJ/mol (4 kJ/mol x 8).
2) The first conformation is more stable. Even though the methyl groups are trans in both models, they are anti to one another in the first structure (which is lower energy) while they are gauche in the second structure increasing strain within the molecule.
3) The ring carbon attached to the methyl group would most likely be the endo carbon. The large methyl group would create the most torsional strain if eclipsed. Being in the endo position would place the bonds is a more staggered position which would reduce strain. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.2%3A_Ring__Strain__and__the_Structure_of_Cycloalkanes.txt |
Objectives
After completing this section, you should be able to
1. explain why cyclohexane rings are free of angular strain.
2. draw the structure of a cyclohexane ring in the chair conformation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• chair conformation
• twist-boat conformation
We will find that cyclohexanes tend to have the least angle strain and consequently are the most common cycloalkanes found in nature. A wide variety of compounds including, hormones, pharmaceuticals, and flavoring agents have substituted cyclohexane rings.
testosterone, which contains three cyclohexane rings and one cyclopentane ring
Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring). Cyclohexane has the possibility of forming multiple conformations each of which have structural differences which lead to different amounts of ring strain.
planar structure
severe angle strain (120°)
severe eclipsing strain (all bonds)
small steric strain
boat conformation
slight angle strain
eclipsing strain at two bonds
steric crowding of two hydrogens
twist boat conformation
slight angle strain
small eclipsing strain
small steric strain
chair conformation
no angle strain
no eclipsing strain
small steric strain
Conformations of Cyclohexane
A planar structure for cyclohexane is clearly improbable. The bond angles would necessarily be 120º, 10.5º larger than the ideal tetrahedral angle. Also, every carbon-hydrogen bond in such a structure would be eclipsed. The resulting angle and eclipsing strains would severely destabilize this structure. The ring strain of planar cyclohexane is in excess of 84 kJ/mol so it rarely discussed other than in theory.
Cyclohexane in the strained planar configuration showing how the hydrogens become eclipsed.
Chair Conformation of Cyclohexane
The flexibility of cyclohexane allows for a conformation which is almost free of ring strain. If two carbon atoms on opposite sides of the six-membered ring are bent out of the plane of the ring, a shape is formed that resembles a reclining beach chair. This chair conformation is the lowest energy conformation for cyclohexane with an overall ring strain of 0 kJ/mol. In this conformation, the carbon-carbon ring bonds are able to assume bonding angles of ~111o which is very near the optimal tetrahedral 109.5o so angle strain has been eliminated.
Also, the C-H ring bonds are staggered so torsional strain has also been eliminated. This is clearly seen when looking at a Newman projection of chair cyclohexane sighted down the two central C-C bonds.
Newman projection of cyclohexane
Interactive Element
The 3D Structure of Chair Cyclohexane
Boat Conformation of Cyclohexane
The Boat Conformation of cyclohexane is created when two carbon atoms on opposite sides of the six-membered ring are both lifted up out of the plane of the ring creating a shape which slightly resembles a boat. The boat conformation is less stable than the chair form for two major reasons. The boat conformation has unfavorable steric interactions between a pair of 1,4 hydrogens (the so-called "flagpole" hydrogens) that are forced to be very close together (1.83Å). This steric hindrance creates a repulsion energy of about 12 kJ/mol. An additional cause of the higher energy of the boat conformation is that adjacent hydrogen atoms on the 'bottom of the boat' are forced into eclipsed positions. For these reasons, the boat conformation about 30 kJ/mol less stable than the chair conformation.
A boat structure of cyclohexane (the interfering "flagpole" hydrogens are shown in red)
Twist-Boat Conformation of Cyclohexane
The boat form is quite flexible and by twisting it at the bottom created the twist-boat conformer. This conformation reduces the strain which characterized the boat conformer. The flagpole hydrogens move farther apart (the carbons they are attached to are shifted in opposite directions, one forward and one back) and the eight hydrogens along the sides become largely but not completely staggered. Though more stable than the boat conformation, the twist-boat (sometimes skew-boat) conformation is roughly 23 kJ/mol less stable than the chair conformation.
A twist-boat structure of cyclohexane
Half Chair Conformation of Cyclohexane
Cyclohexane can obtain a partially plane conformation called "half chair" but with only with excessive amounts of ring strain. The half chair conformation is formed by taking planar cyclohexane and lifting one carbon out of the plane of the ring. The half chair conformation has much of the same strain effects predicted by the fully planar cyclohexane. In the planar portion of half chair cyclohexane the C-C bond angles are forced to 120o which creates significant amounts of angle strain. Also, the corresponding C-H bonds are fully eclipsed which create torsional strain. The out-of-plane carbon allows for some of the ring's bond angles to reach 109.5o and for some of C-H bonds to not be fully eclipsed. Overall, the half chair conformation is roughly 45 kJ/mol less stable than the chair conformation.
Conformation Changes in Cyclohexane - "Ring Flips"
Cyclohexane is rapidly rotating between the two most stable conformations known as the chair conformations in what is called the "ring flip" shown below. The importance of the ring flip will be discussed in the next section.
"Ring flip" describes the rapid equilibrium of cyclohexane rings between the two chair conformations
It is important to note that one chair does not immediately become the other chair, rather the ring must travel through the higher energy conformations as transitions. At room temperature the energy barrier created by the half chair conformation is easily overcome allowing for equilibration between the two chair conformation on the order of 80,000 times per second. Although cyclohexane is continually converting between these different conformations, the stability of the chair conformation causes it to comprises more than 99.9% of the equilibrium mixture at room temperature.
1" id="MathJax-Element-12-Frame" role="presentation" style="position:relative;" tabindex="0">Image of energy diagram of cyclohexane conformations
1" role="presentation" style="position:relative;" tabindex="0">1
Exercises
1) Consider the conformations of cyclohexane: half chair, chair, boat, twist boat. Order them in increasing ring strain in the molecule.
Solutions
1) Chair < Twist Boat < Boat < half chair (most ring strain)
Objectives
After completing this section, you should be able to
1. Draw the chair conformation of cyclohexane, with axial and equatorial hydrogen atoms clearly shown and identified.
2. identify the axial and equatorial hydrogens in a given sketch of the cyclohexane molecule.
3. explain how chair conformations of cyclohexane and its derivatives can interconvert through the process of ring flip.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• axial position
• equatorial position
• ring flip
Axial and Equatorial Positions in Cyclohexane
Careful examination of the chair conformation of cyclohexane, shows that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termed equatorial. The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed axial because they are aligned parallel to the symmetry axis of the ring.
In the figure above, the equatorial hydrogens are colored blue, and the axial hydrogens are black. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character.
How To Draw Axial and Equatorial Bonds
How not to draw the chair:
Aside from drawing the basic chair, the key points are:
• Axial bonds alternate up and down, and are shown "vertical".
• Equatorial groups are approximately horizontal, but actually somewhat distorted from that (slightly up or slightly down), so that the angle from the axial group is a bit more than a right angle -- reflecting the common 109.5o bond angle.
• Each carbon has an axial and an equatorial bond.
• Each face of the cyclohexane ring has three axial and three equatorial bonds.
• Each face alternates between axial and equatorial bonds. Then looking at the "up" bond on each carbon in the cyclohexane ring they will alternate axial-equatorial-axial ect.
• When looking down at a cyclohexane ring:
• the equatorial bonds will form an "equator" around the ring.
• The axial bonds will either face towards you or away. These will alternate with each axial bond. The first axial bond will be coming towards with the next going away. There will be three of each type.
• Note! The terms cis and trans in regards to the stereochemistry of a ring are not directly linked to the terms axial and equatorial. It is very common to confuse the two. It typically best not to try and directly inter convert the two naming systems.
Axial vs. Equatorial Substituents
When a substituent is added to cyclohexane, the ring flip allows for two distinctly different conformations. One will have the substituent in the axial position while the other will have the substituent in the equatorial position. In the next section will discuss the energy differences between these two possible conformations. Below are the two possible chair conformations of methylcyclohexane created by a ring-flip. Although the conformation which places the methyl group in the equatorial position is more stable by 7 kJ/mol, the energy provided by ambient temperature allows the two conformations to rapidly interconvert.
The figure below illustrates how to convert a molecular model of cyclohexane between two different chair conformations - this is something that you should practice with models. Notice that a 'ring flip' causes equatorial groups to become axial, and vice-versa.
Example $1$
For the following please indicate if the substituents are in the axial or equatorial positions.
Solution
Due to the large number of bonds in cyclohexane it is common to only draw in the relevant ones (leaving off the hydrogens unless they are involved in a reaction or are important for analysis). It is still possible to determine axial and equatorial positioning with some thought. With problems such as this it is important to remember that each carbon in a cyclohexane ring has one axial and one equatorial bond. Also, remember that axial bonds are perpendicular with the ring and appear to be going either straight up or straight down. Equatorial bonds will be roughly in the plane of the cyclohexane ring (only slightly up or down). Sometimes it is valuable to draw in the additional bonds on the carbons of interest.
With this it can be concluded that the bromine and chlorine substituents are attached in equatorial positions and the CH3 substituent is attached in an axial position.
Exercises
1) Draw two conformations of cyclohexyl amine (C6H11NH2). Indicate axial and equatorial positions.
2) Draw the two isomers of 1,4-dihydroxylcyclohexane, identify which are equatorial and axial.
3) In the following molecule, label which are equatorial and which are axial, then draw the chair flip (showing labels 1,2,3).
Solutions
1)
2)
3) Original conformation: 1 = axial, 2 = equatorial, 3 = axial
Flipped chair now looks like this. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.3%3A_Cyclohexane%3A_A_Strain-Free_Cycloalkane.txt |
Objectives
After completing this section, you should be able to
1. account for the greater stability of the equatorial conformers of monosubstituted cyclohexanes compared to their axial counterparts, using the concept of 1,3‑diaxial interaction.
2. compare the gauche interactions in butane with the 1,3‑diaxial interactions in the axial conformer of methylcyclohexane.
3. arrange a given list of substituents in increasing or decreasing order of 1,3‑diaxial interactions.
Key Terms
Make certain that you can define, and use in context, the key term below.
• 1,3‑diaxial interaction
Study Notes
1,3-Diaxial interactions are steric interactions between an axial substituent located on carbon atom 1 of a cyclohexane ring and the hydrogen atoms (or other substituents) located on carbon atoms 3 and 5.
Be prepared to draw Newman-type projections for cyclohexane derivatives as the one shown for methylcyclohexane. Note that this is similar to the Newman projections from chapter 3 such as n-butane.
Newman projections of methylcyclohexane and n‑butane
When a substituent is added to a cyclohexane ring, the two possible chair conformations created during a ring flip are not equally stable. In the example of methylcyclohexane the conformation where the methyl group is in the equatorial position is more stable than the axial conformation by 7.6 kJ/mol at 25o C. The percentages of the two different conformations at equilibrium can be determined by solving the following equation for K (the equilibrium constant): ΔE = -RTlnK. In this equation ΔE is the energy difference between the two conformations, R is the gas constant (8.314 J/mol•K), T is the temperature in Kelvin, and K is the equilibrium constant for the ring flip conversion. Using this equation, we can calculate a K value of 21 which means about 95% methylcyclohexane molecules have the methyl group in the equatorial position at 25o C.
The energy difference between the two conformations comes from strain, called 1,3-diaxial interactions, created when the axial methyl group experiences steric crowding with the two axial hydrogens located on the same side of the cyclohexane ring. Because axial bonds are parallel to each other, substituents larger than hydrogen experience greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents are in the equatorial orientation. When the methyl group is in the equatorial position this strain is not present which makes the equatorial conformer more stable and favored in the ring flip equilibrium.
Actually, 1,3-diaxial steric strain is directly related to the steric strain created in the gauche conformer of butane discussed in Section: 3-7. When butane is in the gauche conformation 3.8 kJ/mol of strain was created due the steric crowding of two methyl group with a 60o dihedral angle. When looking at the a Newman projection of axial methylcyclohexane the methyl group is at a 60o dihedral angle with the ring carbon in the rear. This creates roughly the same amount of steric strain as the gauche conformer of butante. Given that there is actually two such interactions in axial methylcyclohexane, it makes sense that there is 2(3.8 kJ/mol) = 7.6 kJ/mol of steric strain in this conformation. The Newman projection of equatorial methylcyclohexane shows no such interactions and is therefore more stable.
Newman projections of methyl cyclohexane and butane showing similarity of 1,3-diaxial and gauche interactions.
Strain values for other cyclohexane substituents can also be considered. The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane determined the amount of strain generated. The strain generated can be used to evaluate the relative tendency of substituents to exist in an equatorial or axial location. Looking at the energy values in this table, it is clear that as the size of the substituent increases, the 1,3-diaxial energy tends to increase, also. Note that it is the size and not the molecular weight of the group that is important. Table 4.7.1 summarizes some of these strain values values.
Table 4.7.1: A Selection of ΔG° Values for the Change from Axial to Equatorial Orientation of Substituents for Monosubstituted Cyclohexanes
Substituent -ΔG° (kcal/mol) Substituent -ΔG° (kcal/mol)
\(\ce{CH_3\bond{-}}\) 1.7 \(\ce{O_2N\bond{-}}\) 1.1
\(\ce{CH_2H_5\bond{-}}\) 1.8 \(\ce{N#C\bond{-}}\) 0.2
\(\ce{(CH_3)_2CH\bond{-}}\) 2.2 \(\ce{CH_3O\bond{-}}\) 0.5
\(\ce{(CH_3)_3C\bond{-}}\) \(\geq 5.0\) \(\ce{HO_2C\bond{-}}\) 0.7
\(\ce{F\bond{-}}\) 0.3 \(\ce{H_2C=CH\bond{-}}\) 1.3
\(\ce{Cl\bond{-}}\) 0.5 \(\ce{C_6H_5\bond{-}}\) 3.0
\(\ce{Br\bond{-}}\) 0.5
\(\ce{I\bond{-}}\) 0.5
Exercises
1) In the molecule, cyclohexyl ethyne there is little steric strain, why?
2) Calculate the energy difference between the axial and equatorial conformations of bromocyclohexane?
3) Using your answer from Question 2) estimate the percentages of axial and equatorial conformations of bromocyclohexane at 25o C.
4) There very little in 1,3-diaxial strain when going from a methyl substituent (3.8 kJ/mol) to an ethyl substituent (4.0 kJ/mol), why? It may help to use molecular model to answer this question.
Solutions
1) The ethyne group is linear and therefore does not affect the hydrogens in the 1,3 positions to say to the extent as a bulkier or a bent group (e.g. ethene group) would. This leads to less of a strain on the molecule.
2) The equatorial conformation of bromocyclohexane will have two 1,3 diaxial interactions. The table above states that each interaction accounts for 1.2 kJ/mol of strain. The total strain in equatorial bromocyclohexane will be 2(1.2 kJ/mol) = 2.4 kJ/mol.
3) Remembering that the axial conformation is higher in energy, the energy difference between the two conformations is ΔE = (E equatorial - E axial) = (0 - 2.4 kJ/mol) = -2.4 kJ/mol. After converting oC to Kelvin and kJ/mol to J/mol we can use the equation ΔE = -RT lnK to find that -ΔE/RT = lnK or (2.4 x 103 J/mol) / (8.313 kJ/mol K • 298 K) = lnK. From this we calculate that K = 2.6. Because the ring flip reaction is an equilibrium we can say that K = [Equatorial] / [Axial]. If assumption is made that [Equatorial] = X then [Axial] must be 1-X. Plugging these values into the equilibrium expression produces K = [X] / [1-X]. After plugging in the calculated value for K, X can be solved algebraically. 2.6 = [X] / [1-X] → 2.6 - 2.6X = X → 2.6 = 3.6X → 2.6/3.6 = X = 0.72. This means that bromocyclohexane is in the equatorial position 72% of the time and in the axial position 28% of the time.
4) The fact that C-C sigma bonds can freely rotate allows the ethyl subsistent to obtain a conformation which places the bulky CH3 group away from the cyclohexane ring. This forces the ethyl substituent to have only have 1,3- diaxial interactions between hydrogens, which only provides a slight difference to a methyl group.
Exercises
Objective
After completing this section, you should be able to use conformational analysis to determine the most stable conformation of a given disubstituted cyclohexane.
Key Terms
Make certain that you can define, and use in context, the key term below.
• conformational analysis
Study Notes
When faced with the problem of trying to decide which of two conformers of a given disubstituted cyclohexane is the more stable, you may find the following generalizations helpful.
1. A conformation in which both substituents are equatorial will always be more stable than a conformation with both groups axial.
2. When one substituent is axial and the other is equatorial, the most stable conformation will be the one with the bulkiest substituent in the equatorial position. Steric bulk decreases in the order
tert-butyl > isopropyl > ethyl > methyl > hydroxyl > halogens
Monosubstituted Cyclohexanes
In the previous section, it was stated that the chair conformation in which the methyl group is equatorial is more stable because it minimizes steric repulsion, and thus the equilibrium favors the more stable conformer. This is true for all monosubstituted cyclohexanes. The chair conformation which places the substituent in the equatorial position will be the most stable and be favored in the ring flip equilibrium.
Disubstituted Cyclohexanes
Determining the more stable chair conformation becomes more complex when there are two or more substituents attached to the cyclohexane ring. To determine the stable chair conformation, the steric effects of each substituent, along with any additional steric interactions, must be taken into account for both chair conformations.
In this section, the effect of conformations on the relative stability of disubstituted cyclohexanes is examined using the two principles:
1. Substituents prefer equatorial rather than axial positions in order to minimize the steric strain created of 1,3-diaxial interactions.
2. The more stable conformation will place the larger substituent in the equatorial position.
1,1-Disubstituted Cyclohexanes
The more stable chair conformation can often be determined empirically or by using the energy values of steric interactions previously discussed in this chapter. Note, in some cases there is no discernable energy difference between the two chair conformations which means they are equally stable.
1,1-dimethylcyclohexane does not have cis or trans isomers, because both methyl groups are on the same ring carbon. Both chair conformers have one methyl group in an axial position and one methyl group in an equatorial position giving both the same relative stability. The steric strain created by the 1,3-diaxial interactions of a methyl group in an axial position (versus equatorial) is 7.6 kJ/mol (from Table 4.7.1), so both conformers will have equal amounts of steric strain. Thus, the equilibrium between the two conformers does not favor one or the other. Note, that both methyl groups cannot be equatorial at the same time without breaking bonds and creating a different molecule.
However, if the two groups are different, as in 1-tert-butyl-1-methylcyclohexane, then the equilibrium favors the conformer in which the larger group (tert-butyl in this case) is in the more stable equatorial position. The energy cost of having one tert-butyl group axial (versus equatorial) can be calculated from the values in table 4.7.1 and is approximately 22.8 kJ/mol. The conformer with the tert-butyl group axial is approximately 15.2 kJ/mol (22.8 kJ/mol - 7.6 kJ/mol) less stable then the conformer with the tert-butyl group equatorial. Solving for the equilibrium constant K shows that the equatorial is preferred about 460:1 over axial. This means that 1-tert-butyl-1-methylcyclohexane will spend the majority of its time in the more stable conformation, with the tert-butyl group in the equatorial position.
Cis and trans stereoisomers of 1,2-dimethylcyclohexane
In cis-1,2-dimethylcyclohexane, both chair conformations have one methyl group equatorial and one methyl group axial. As previously discussed, the axial methyl group creates 7.6 kJ/mol of steric strain due to 1,3-diaxial interactions. It is important to note, that both chair conformations also have an additional 3.8 kJ/mol of steric strain created by a gauche interaction between the two methyl groups. Overall, both chair conformations have 11.4 kJ/mol of steric strain and are of equal stability.
In trans-1,2-dimethylcyclohexane, one chair conformer has both methyl groups axial and the other conformer has both methyl groups equatorial. The conformer with both methyl groups equatorial has no 1,3-diaxial interactions however there is till 3.8 kJ/mol of strain created by a gauche interaction. The conformer with both methyl groups axial has four 1,3-Diaxial interactions which creates 2 x 7.6 kJ/mol (15.2 kJ/mol) of steric strain. This conformer is (15.2 kJ/mol -3.8 kJ/mol) 11.4 kJ/mol less stable than the other conformer. The equilibrium will therefore favor the conformer with both methyl groups in the equatorial position.
Cis and trans stereoisomers of 1,3-dimethylcyclohexane
A similar conformational analysis can be made for the cis and trans stereoisomers of 1,3-dimethylcyclohexane. For cis-1,3-dimethylcyclohexane one chair conformation has both methyl groups in axial positions creating 1,3-diaxial interactions. The other conformer has both methyl groups in equatorial positions thus creating no 1,3-diaxial interaction. Because the methyl groups are not on adjacent carbons in the cyclohexane rings gauche interactions are not possible. Even without energy calculations it is simple to determine that the conformer with both methyl groups in the equatorial position will be the more stable conformer.
For trans-1,3-dimethylcyclohexane both conformations have one methyl axial and one methyl group equatorial. Each conformer has one methyl group creating a 1,3-diaxial interaction so both are of equal stability.
Summary of Disubstitued Cyclohexane Chair Conformations
When considering the conformational analyses discussed above a pattern begins to form. There are only two possible relationships which can occur between ring-flip chair conformations:
1) AA/EE: One chair conformation places both substituents in axial positions creating 1,3-diaxial interactions. The other conformer places both substituents in equatorial positions creating no 1,3-diaxial interactions. This diequatorial conformer is the more stable regardless of the substituents.
2) AE/EA: Each chair conformation places one substituent in the axial position and one substituent in the equatorial position. If the substituents are the same, there will be equal 1,3-diaxial interactions in both conformers making them equal in stability. However, if the substituents are different then different 1,3-diaxial interactions will occur. The chair conformation which places the larger substituent in the equatorial position will be favored.
Substitution type Chair Conformation Relationship
cs-1,2-disubstituted cyclohexanes AE/EA
trans-1,2-disubstituted cyclohexanes AA/EE
cis-1,3-disubstituted cyclohexanes AA/EE
trans-1,3-disubstituted cyclohexanes AE/EA
cis-1,4-disubstituted cyclohexanes AE/EA
trans-1,4-disubstituted cyclohexanes AA/EE
Example \(1\)
For cis-1-chloro-4-methylcyclohexane, draw the most stable chair conformation and determine the energy difference between the two chair conformers.
Solution
Based on the table above, cis-1,4-disubstitued cyclohexanes should have two chair conformations each with one substituent axial and one equatorial. Based on this, we can surmise that the energy difference of the two chair conformations will be based on the difference in the 1,3-diaxial interactions created by the methyl and chloro substituents.
As predicted, each chair conformer places one of the substituents in the axial position. Because the methyl group is larger and has a greater 1,3-diaxial interaction than the chloro, the most stable conformer will place it the equatorial position, as shown in the structure on the right. Using the 1,3-diaxial energy values given in the previous sections we can calculate that the conformer on the right is (7.6 kJ/mol - 2.0 kJ/mol) 5.6 kJ/mol more stable than the other.
Example \(2\)
For trans-1-chloro-2-methylcyclohexane, draw the most stable chair conformation and determine the energy difference between the two chair conformers.
Solution
Based on the table above, trans-1,2-disubstitued cyclohexanes should have one chair conformation with both substituents axial and one conformation with both substituents equatorial. Based on this, we can predict that the conformer which places both substituents equatorial will be the more stable conformer. The energy difference of the two chair conformations will be based on the 1,3-diaxial interactions created by both the methyl and chloro substituents.
As predicted, one chair conformer places both substituents in the axial position and other places both substituents equatorial. The more stable conformer will place both substituents in the equatorial position, as shown in the structure on the right. Using the 1,3-diaxial energy values given in the previous sections we can calculate that the conformer on the right is (7.6 kJ/mol + 2.0 kJ/mol) 9.6 kJ/mol more stable than the other.
Conformational Analysis of Complex Six Membered Ring Structures
Cyclohexane can have more than two substituents. Also, there are multiple six membered rings which contain atoms other than carbon. All of these systems usually form chair conformations and follow the same steric constraints discussed in this section. Because the most commonly found rings in nature are six membered, conformational analysis can often help in understanding the usual shapes of some biologically important molecules. In complex six membered ring structures a direct calculation of 1,3-diaxial energy values may be difficult. In these cases a determination of the more stable chair conformer can be made by empirically applying the principles of steric interactions.
A later chapter will discuss how many sugars can exist in cyclic forms which are often six remembered rings. When in an aqueous solution the six carbon sugar, glucose, is usually a six membered ring adopting a chair conformation. When looking at the two possible ring-clip chair conformations, one has all of the substituents axial and the other has all the substutents equatorial. Even without a calculation, it is clear that the conformation with all equatorial substituents is the most stable and glucose will most commonly be found in this conformation.
Example \(3\)
The six carbon sugar, fructose, in aqueous solution is also a six-membered ring in a chair conformation. Which of the two possible chair conformations would be expected to be the most stable?
Solution
The lower energy chair conformation is the one with three of the five substituents (including the bulky –CH2OH group) in the equatorial position (pictured on the right). The left structure has 3 equatorial substituents while the structure on the right only has two equatorial substituents.
Exercises
1. Draw the two chair conformations for cis-1-ethyl-2-methylcyclohexane using bond-line structures and indicate the more energetically favored conformation.
2. Draw the most stable conformation for trans-1-ethyl-3-methylcyclohexane using bond-line structures.
3. Draw the most stable conformation for trans-1-t-butyl-4-methylcyclohexane using bond-line structures.
4. Draw the most stable conformation fo trans-1-isopropyl-3-methylcyclohexane.
5. Can a ‘ring flip’ change a cis-disubstituted cyclohexane to trans? Explain.
6. Draw the two chair conformations of the six-carbon sugar mannose, being sure to clearly show each non-hydrogen substituent as axial or equatorial. Predict which conformation is likely to be more stable, and explain why.
Solutions
4.
The bulkier isopropyl groups is in the equatorial position.
5. No. In order to change the relationship of two substituents on a ring from cis to trans, you would need to break and reform two covalent bonds. Ring flips involve only rotation of single bonds.
6. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.4%3A_Substituted_Cyclohexanes.txt |
The Baeyer strain theory suggested that the larger cycloalkanes ring are difficult to synthesize because of angle strain associated with planar rings, as calculated in Table 12-3. We now know that, except for cyclopropane, none of the cycloalkanes have planar carbon rings and that the higher cycloalkanes have normal or nearly normal bond angles. The reason that the higher cycloalkanes are generally difficult to synthesize from open-chain compounds is not so much angle strain, as Baeyer hypothesized, but the low probability of having reactive groups on the two fairly remote ends of a long hydrocarbon chain come together to effect cyclization. Usually, coupling of reactive groups on the ends of different molecules occurs in preference to cyclization, unless the reactions are carried out in very dilute solutions. This is called the high-dilution technique for achieving ring formation when the ring-forming reaction has to compete with rapid inter-molecular reactions.
With regard to conformations of the larger cycloalkanes, we first note that the chair form of cyclohexane is a “perfect” conformation for a cycloalkane. The $\ce{C-C-C}$ bond angles are close to their normal values, all the adjacent hydrogens are staggered with respect to one another, and the 1,3-axial hydrogens are not close enough together to experience nonbonded repulsions. About the only qualification one could put on the ideality of the chair form is that the trans conformation of butane is somewhat more stable than the gauche conformation (Section 5-2), and that all of the $\ce{C-C-C-C}$ segments of cyclohexane have the gauche arrangement. Arguing from this, J. Dale$^6$ has suggested that large cycloalkane rings would tend to have trans $\ce{C-C-C-C}$ segments to the degree possible and, indeed, cyclotetradecane seems to be most stable in a rectangular conformation with trans $\ce{C-C-C-C}$ bond segments (Figure 12-16). This conformation has a number of possible substituent positions, but because only single isomers of monosubstituted cyclotetradecanes have been isolated, rapid equilibration of the various conformational isomers must occur. Other evidence indicates that the barrier to interconversion of these conformations is about $7 \: \text{kcal mol}^{-1}$.
With the cycloalkanes having 7 to 10 carbons, there are problems in trying to make either trans or gauche $\ce{C-C-C-C}$ segments, because the sizes of these rings do not allow the proper bond angles or torsional angles, or else there are more or less serious nonbonded repulsions. Consequently each of these rings assumes a compromise conformation with some eclipsing, some nonbonded repulsions, and some angle distortions. Brief comments on some of these conformations follow. It will be useful to use molecular models to see the interactions involved.
Cycloheptane
Possible conformations for cycloheptane include the “comfortable” appearing chair form, $7$. However, this form has eclipsed hydrogens at $\ce{C_4}$ and $\ce{C_5}$ as well as nonbonded interactions between the axial-like hydrogens on $\ce{C_3}$ and $\ce{C_6}$. The best compromise conformation is achieved by a $30^\text{o}$-$40^\text{o}$ rotation around the $\ce{C_4-C_5}$ bond to relieve the eclipsing of the hydrogens. This spreads the interfering hydrogens at $\ce{C_3}$ and $\ce{C_6}$ and results in a somewhat less strained conformation called the twist chair. The twist chair, $8$, is very flexible and probably only about $3 \: \text{kcal mol}^{-1}$ of activation is required to interconvert the various possible monosubstituted cycloheptane conformations.
Cyclooctane
There are several more or less reasonable looking cyclooctane conformations. After much research it now is clear that the favored conformation is the boat-chair, $9$, which is in equilibrium with a few tenths percent of the crown conformation, $10$ :
The activation energy for interconversion of these two forms is about $10 \: \text{kcal mol}^{-1}$. The boat-chair conformation $9$ is quite flexible and movement of its $\ce{CH_2}$ groups between the various possible positions occurs with an activation energy of only about $5 \: \text{kcal mol}^{-1}$.
Cyclononane
Several more or less reasonable conformations of cyclononane also can be developed, but the most favorable one is called the twist-boat-chair, which has three-fold symmetry (Figure 12-17). The activation energy for inversion of the ring is about $6 \: \text{kcal mol}^{-1}$.
Cyclodecane
The stable conformation of cyclodecane (Figure 12-18) is similar to that of cyclotetradecane (Figure 12-16). However, there are relatively short $\ce{H} \cdot \cdot \cdot \cdot \ce{H}$ distances and the $\ce{C-C-C}$ bond angles are somewhat distorted because of cross-ring hydrogen-hydrogen repulsions. The most stable position for a substituent on the cyclodecane ring is the one indicated in Figure 12-18. The least stable positions are those in which a substituent replaces any of the six hydrogens shown, because nonbonded interactions are particularly strong at these positions. The activation energy for interconversion of substituent positions is about $6 \: \text{kcal mol}^{-1}$.
Most stable conformation of cyclodecane; Dale and sawhorse representations. The shaded area in the sawhorse convention indicates substantial nonbonded $\ce{H} \cdot \cdot \cdot \cdot \ce{H}$ interactions.
$^6$Pronounced Dalluh.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.5%3A_Larger__Cycloalkanes.txt |
Objective
After completing this section, you should be able to draw the structures and construct molecular models of simple polycyclic molecules.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bridgehead carbon atom
• polycyclic molecule
Study Notes
A bridgehead carbon atom is a carbon atom which is shared by at least two rings. The hydrogen atom which is attached to a bridgehead carbon may be referred to as a bridgehead hydrogen.
Note that bicyclo[2.2.1]heptane is the systematic name of norborane. You need not be concerned over the IUPAC name of norbornane. The nomenclature of compounds of this type is beyond the scope of this course.
Nomenclature of Bicyclic Ring Systems
There are many hydrocarbons and hydrocarbon derivatives with two rings having common carbon atoms. There are three main ways that the two rings can be connected. The first is called a fused bicyclic ring structure where the two rings share a covalent bond and a have two bridgehead carbons (marked in red on the structures below). A bridgehead is defined as a carbon that is part of two or more rings. Hydrogens attached to bridge head carbons are often referred to as bridge head hydrogens. The two rings can also be connected by a bridge containing one or more carbons to form a bridged bicyclic molecule. Lastly, the two rings can be joined with a singe bridge head carbon to form spiro bicyclic molecules.
Bicyclic Isomers of C10H18
Naming Fused and Bridged Compounds
Fused and bridged bicyclic compounds are follow similar naming conventions:
1. Count the total number of carbons in both rings. This is the parent name. (eg. ten carbons in the system would be decane)
2. Count the number of carbons between the bridgeheads, then place the numbers in square brackets in descending order separated by periods. Fused and bridged bycyclic compounds should have three numbers such as [2.2.0]. For fused compounds one of the numbers should be zero.
3. Place the word bicyclo at the beginning of the name.
Examples with carbons and hydrogens explicitly shown:
Naming Spiro Compounds
Spiro bicyclics are named using the same basic rules. Because there is only one bridgehead carbon only two numbers will be required in the brackets. Also, the word spiro is placed at the beginning.
Conformations in Bicyclic Ring Systems
As expected, the connection of two rings has defined effects on the possible conformations. However, the ideas previously discussed in this chapter can be used for conformational analysis. Fused rings have the possibility of two isomers where the bridgehead hydrogens are either cis or trans along the shared bond. These two isomers have significant differences in flexibility and stability as seen in bicyclo[4,4,0]decane more commonly known as decalin. If the positioning of the bridgehead hydrogens are shown in a fused ring the prefix cis or trans should be included in the name.
The trans-isomer is the easiest to describe because the fusion of the two rings creates a rigid, roughly planar, structure made up of two chair conformations. Unlike cyclohexane, the two rings cannot flip from one chair form to another. Accordingly, the orientation of the any substituents is fixed in either an axial or equatorial position in trans-decalin. This means that the C-C bonds coming away from the fused edge are held in equatorial positions relative to each ring thus preventing the possibility of any 1,3-diaxial interactions occurring between ring atoms.
Interactive Element
The 3D Structure of Trans-Decalin
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The two rings in cis-decalin are also both held in a chair conformations. In comparison, the chair-chair forms of cis-decalin are relatively flexible, and inversion of both rings at once occurs fairly easily.
Interactive Element
The 3D Structure of Cis-Decalin
The flexibility of cis-decalin allows for a substituent to interconvert between axial and equatorial conformations. In much the same fashion as cyclohexane, equatorial substituents tend to create less steric strain and create a more stable conformer.
A major difference in cis-decalin is the fact that one of C-C bonds coming away from the fused edge is held an an axial position. This is true in both ring-flip conformations. This axial C-C bond causes 1,3-diaxial interactions to occur in cis-decalin making it roughly 8.4 kJ/mol less stable than trans-decalin. This amount of 1,3-diaxial steric strain is roughly equivalent to that of an ethyl substituent attached to a cyclohexane ring (8.0 kJ/mol)
Bicyclic compounds with a bridge typically have very little flexibility and are often held in a ridged conformation. The molecule norbornane represent a cyclohexane ring connected by a single carbon bridge.
Interactive Element
The 3D Structure of Norbornane
Norbornane is estimated to have 72 kJ/mol of ring strain which can be understood when viewing the contained rings. The carbon bridge in norbornane holds the cyclohexane ring at the bottom in a boat conformation creating torsional strain from eclipsing bonds along the edge.
Also, the carbon bridge forms a cyclopentane ring (shown in red below making up the right side of the structure) with increased angle strain throughout the whole molecule.
Polycyclic Systems in Nature
Fused ring systems like decalin are very common in natural products. In fact, similar ring systems are found in steroids, which are an important class of lipids. Steroids generally have structures that include three six-membered rings and a five-membered ring connected by three fused bonds. Most natural steroids have a trans configuration at all three fusion points. This tends to give steroids a rigid and semi-flat structure.
Sex hormones are an example of steroids. The primary male hormone, testosterone, is responsible for the development of secondary sex characteristics. Two female sex hormones, progesterone and estrogen (or estradiol) control the ovulation cycle. Notice that the male and female hormones have only slight differences in structures, but yet have very different physiological effects. Testosterone promotes the normal development of male genital organs and is synthesized from cholesterol in the testes. It also promotes secondary male sexual characteristics such as deep voice, facial and body hair.
Interactive Element
The 3D Structure of Estradiol
The best known and most abundant steroid in the body is cholesterol. Cholesterol is formed in brain tissue, nerve tissue, and the blood stream. It is the major compound found in gallstones and bile salts. Cholesterol also contributes to the formation of deposits on the inner walls of blood vessels. These deposits harden and obstruct the flow of blood. This condition, known as atherosclerosis, results in various heart diseases, strokes, and high blood pressure.
Exercises
1)
i)
j)
3) The following molecule is cholic acid. Determine if the three fused bonds have a cis or trans configuration.
Solutions
1)
a) Bicyclo[2.1.1]hexane
b) Bicyclo[3.2.1]octane
c) Bicyclo[2.1.0]pentane (more commonly called "housane")
d) Bicyclo[2.2.2]octane
e) cis-Bicyclo[3.3.0]octane
f) cis-Bicyclo[1.1.0]butane
g) Bicyclo[1.1.1]pentane
h) Bicyclo[4.3.3]dodecane
i) Spiro[5.2]octane
j) Spiro[3.3]heptane
2) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.6%3A_Polycyclic_Alkanes.txt |
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring, as shown in the diagram to the right. The four rings are designated A, B, C & D as noted, and the peculiar numbering of the ring carbon atoms (shown in red) is the result of an earlier misassignment of the structure. The substituents designated by R are often alkyl groups, but may also have functionality. The R group at the A:B ring fusion is most commonly methyl or hydrogen, that at the C:D fusion is usually methyl. The substituent at C-17 varies considerably, and is usually larger than methyl if it is not a functional group. The most common locations of functional groups are C-3, C-4, C-7, C-11, C-12 & C-17. Ring A is sometimes aromatic. Since a number of tetracyclic triterpenes also have this tetracyclic structure, it cannot be considered a unique identifier.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error.
The generic steroid structure drawn above has seven chiral stereocenters (carbons 5, 8, 9, 10, 13, 14 & 17), which means that it may have as many as 128 stereoisomers. With the exception of C-5, natural steroids generally have a single common configuration. This is shown in the last of the toggled displays, along with the preferred conformations of the rings.
Chemical studies of the steroids were very important to our present understanding of the configurations and conformations of six-membered rings. Substituent groups at different sites on the tetracyclic skeleton will have axial or equatorial orientations that are fixed because of the rigid structure of the trans-fused rings. This fixed orientation influences chemical reactivity, largely due to the greater steric hindrance of axial groups versus their equatorial isomers. Thus an equatorial hydroxyl group is esterified more rapidly than its axial isomer.
It is instructive to examine a simple bicyclic system as a model for the fused rings of the steroid molecule. Decalin, short for decahydronaphthalene, exists as cis and trans isomers at the ring fusion carbon atoms. Planar representations of these isomers are drawn at the top of the following diagram, with corresponding conformational formulas displayed underneath. The numbering shown for the ring carbons follows IUPAC rules, and is different from the unusual numbering used for steroids. For purposes of discussion, the left ring is labeled A (colored blue) and the right ring B (colored red). In the conformational drawings the ring fusion and the angular hydrogens are black.
The trans-isomer is the easiest to describe because the fusion of the A & B rings creates a rigid, roughly planar, structure made up of two chair conformations. Each chair is fused to the other by equatorial bonds, leaving the angular hydrogens (Ha) axial to both rings. Note that the bonds directed above the plane of the two rings alternate from axial to equatorial and back if we proceed around the rings from C-1 to C-10 in numerical order. The bonds directed below the rings also alternate in a complementary fashion.
Conformational descriptions of cis- decalin are complicated by the fact that two energetically equivalent fusions of chair cyclohexanes are possible, and are in rapid equilibrium as the rings flip from one chair conformation to the other. In each of these all chair conformations the rings are fused by one axial and one equatorial bond, and the overall structure is bent at the ring fusion. In the conformer on the left, the red ring (B) is attached to the blue ring (A) by an axial bond to C-1 and an equatorial bond to C-6 (these terms refer to ring A substituents). In the conformer on the right, the carbon bond to C-1 is equatorial and the bond to C-6 is axial. Each of the angular hydrogens (Hae or Hea) is oriented axial to one of the rings and equatorial to the other. This relationship reverses when double ring flipping converts one cis-conformer into the other.
Cis-decalin is less stable than trans-decalin by about 2.7 kcal/mol (from heats of combustion and heats of isomerization data). This is due to steric crowding (hindrance) of the axial hydrogens in the concave region of both cis-conformers, as may be seen in the model display activated by the following button. This difference is roughly three times the energy of a gauche butane conformer relative to its anti conformer. Indeed three gauche butane interactions may be identified in each of the cis-decalin conformations, as will be displayed by clicking on the above conformational diagram. These gauche interactions are also shown in the model.
Steroids in which rings A and B are fused cis, such as the example on the right, do not have the sameconformational mobility exhibited by cis-decalin. The fusion of ring C to ring B in a trans configuration prevents ring B from undergoing a conformational flip to another chair form. If this were to occur, ring C would have to be attached to ring B by two adjacent axial bonds directed 180º apart. This is too great a distance to be bridged by the four carbon atoms making up ring C. Consequently, the steroid molecule is locked in the all chair conformation shown here. Of course, all these steroids and decalins may have one or more six-membered rings in a boat conformation. However the high energy of boat conformers relative to chairs would make such structures minor components in the overall ensemble of conformations available to these molecules.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Compounds classified as terpenes constitute what is arguably the largest and most diverse class of natural products. A majority of these compounds are found only in plants, but some of the larger and more complex terpenes (e.g. squalene & lanosterol) occur in animals. Terpenes incorporating most of the common functional groups are known, so this does not provide a useful means of classification. Instead, the number and structural organization of carbons is a definitive characteristic. Terpenes may be considered to be made up of isoprene (more accurately isopentane) units, an empirical feature known as the isoprene rule. Because of this, terpenes usually have \(5n\) carbon atoms (\(n\) is an integer), and are subdivided as follows:
Classification Isoprene Units Carbon Atoms
monoterpenes 2 C10
sesquiterpenes 3 C15
diterpenes 4 C20
sesterterpenes 5 C25
triterpenes 6 C30
Isoprene itself, a C5H8 gaseous hydrocarbon, is emitted by the leaves of various plants as a natural byproduct of plant metabolism. Next to methane it is the most common volatile organic compound found in the atmosphere. Examples of C10 and higher terpenes, representing the four most common classes are shown in the following diagrams. Most terpenes may be structurally dissected into isopentane segments. How this is done can be seen in the diagram directly below.
Figure: Monoterpenes and diterpenes
The isopentane units in most of these terpenes are easy to discern, and are defined by the shaded areas. In the case of the monoterpene camphor, the units overlap to such a degree it is easier to distinguish them by coloring the carbon chains. This is also done for alpha-pinene. In the case of the triterpene lanosterol we see an interesting deviation from the isoprene rule. This thirty carbon compound is clearly a terpene, and four of the six isopentane units can be identified. However, the ten carbons in center of the molecule cannot be dissected in this manner. Evidence exists that the two methyl groups circled in magenta and light blue have moved from their original isoprenoid locations (marked by small circles of the same color) to their present location. This rearrangement is described in the biosynthesis section. Similar alkyl group rearrangements account for other terpenes that do not strictly follow the isoprene rule.
Figure: Triterpenes
Polymeric isoprenoid hydrocarbons have also been identified. Rubber is undoubtedly the best known and most widely used compound of this kind. It occurs as a colloidal suspension called latex in a number of plants, ranging from the dandelion to the rubber tree (Hevea brasiliensis). Rubber is a polyene, and exhibits all the expected reactions of the C=C function. Bromine, hydrogen chloride and hydrogen all add with a stoichiometry of one molar equivalent per isoprene unit. Ozonolysis of rubber generates a mixture of levulinic acid ( \(CH_3COCH_2CH_2CO_2H\) ) and the corresponding aldehyde. Pyrolysis of rubber produces the diene isoprene along with other products.
The double bonds in rubber all have a Z-configuration, which causes this macromolecule to adopt a kinked or coiled conformation. This is reflected in the physical properties of rubber. Despite its high molecular weight (about one million), crude latex rubber is a soft, sticky, elastic substance. Chemical modification of this material is normal for commercial applications. Gutta-percha (structure above) is a naturally occurring E-isomer of rubber. Here the hydrocarbon chains adopt a uniform zig-zag or rod like conformation, which produces a more rigid and tough substance. Uses of gutta-percha include electrical insulation and the covering of golf balls. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/04._Cycloalkanes/4.7%3A_Carbocyclic_Products_in_Nature.txt |
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality.
05. Stereoisomers
Objectives
After completing this section, you should be able to
1. use molecular models to show that only a tetrahedral carbon atom satisfactorily accounts for the lack of isomerism in molecules of the type CH2XY, and for the existence of optical isomerism in molecules of the type CHXYZ.
2. determine whether two differently oriented wedge-and-broken-line structures are identical or represent a pair of enantiomers.
Key Terms
Make certain that you can define, and use in context, the key term below.
• enantiomer
Study Notes
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of the most interesting types of isomer is the mirror-image stereoisomer, a non-superimposable set of two molecules that are mirror images of one another. The existence of these molecules are determined by a a concept known as chirality. The word “chiral” was derived from the Greek word for hand, because our hands are a good example of chirality since they are non-superimposable mirror images of each other.
Chiral Molecules
The term chiral, from the Greek work for 'hand', refers to anything which cannot be superimposed on its own mirror image. Certain organic molecules are chiral meaning that they are not superimposable on their mirror image. Chiral molecules contain one or more chiral centers, which are almost always tetrahedral (sp3-hybridized) carbons with four different substituents. Consider the molecule A below: a tetrahedral carbon, with four different substituents denoted by balls of four different colors.
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.
A and B are called stereoisomers or optical isomers: molecules with the same molecular formula and the same bonding arrangement, but a different arrangement of atoms in space. 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.
The Many Synonyms of the Chiral Carbon
Be aware - all of the following terms can be used to describe a chiral carbon.
chiral carbon = asymmetric carbon = optically active carbon = stereo carbon = stereo center = chiral center
Let's apply our chirality discussion to real molecules.
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 substituents attached to is are different: a hydrogen (H) , a methyl (-CH3) group, an ethyl (-CH2CH3) group, and a hydroxyl (OH) group. If the bonding at C2 of 2-butanol is drawn in three dimensions and this structure called A. Then the mirror image of A can be drawn to form structure B.
When we try to superimpose A onto B, we find that we cannot do it. Because structure A and B are not superimposable on their mirror image they are both chiral molecules. Because A and B are different due only to the arrangement of atoms in space they are stereoisomers. Because A and B are mirror images of each other they are also enantiomers. When looking at simplified line structures is clear that there are two distinct ways of drawing 2-butanol which only differ in their spatial arrangement around a chiral carbon.
The 3D Structures of the Two Enantiomers of 2-Butanol
For comparison, 2-propanol, is an achiral molecule because is lacks a chiral carbon. Carbon #2 is bonded to two identical substituents (methyl groups), and so it is not a chiral carbon. Being achiral means that 2-propanol should be superimposable on its mirror image which is shown in the figure below. A more detailed explaination on why 2-propanol is achiral will be given in the next section.
2-propanol is achiral:
Stereoisomers
Stereoisomers have been defined as molecules with the same connectivity but different arrangements of the atoms in space. It is important to note that there are two types of stereoisomers: geometric and optical.
Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral. 25.7.1a" id="MathJax-Element-1-Frame" role="presentation" style="position:relative;" tabindex="0">
Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement. Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds.mers have the same connectivity, but different arrangements of atoms in space. Geometric isomers will be discussed in more detain in Sections 7.4 and 7.5.
Exercise $1$
Identify the following molecules as chiral or achiral.
Answer
a) chiral (4 different groups off C)
b) achiral (2 identical -CH3 substituents off central C)
c) achiral (2 identical -CH2CH3 substituents off central C)
d) achiral (2 identical CH3 substituents off carbon 2)
e) chiral (4 different groups off carbon 2)
f) achiral (2 identical CH3 substituents off central C)
Exercise $2$
Determine if the following sets of compounds in each group are enantiomers or the same compound.
Answer
a) enantiomers – non superimposable mirror images
b) same compound – when you rotate the molecule on the right it is identical to the one on the left
c) enantiomers – non superimposable mirror images
d) enantiomers – non superimposable mirror images
Objectives
After completing this section, you should be able to
1. determine whether or not a compound is chiral, given its Kelulé, condensed or shorthand structure, with or without the aid of molecular models.
2. label the chiral centres (carbon atoms) in a given Kelulé, condensed or shorthand structure.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• achiral
• chiral
• chiral (stereogenic) centre
• plane of symmetry
Symmetry and Chirality
Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands. Your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks 25.7.1b in the figure below. 25.7.1b.
An an important questions is why is one chiral and the other not? The answer is that the flask has a plane of symmetry and your hand does not. A plane of symmetry is a plane or a line through an object which divides the object into two halves that are mirror images of each other. When looking at the flask, a line can be drawn down the middle which separates it into two mirror image halves. However, a similar line down the middle of a hand separates it into two non-mirror image halves. This idea can be used to predict chirality. If an object or molecule has a plane of symmetry it is achiral. If if lacks a plane of symmetry it is chiral.
Symmetry can be used to explain why a carbon bonded to four different substituents is chiral. When a carbon is bonded to fewer than four different substituents it will have a plane of symmetry making it achiral. A carbon atom that is bonded to four different substituents loses all symmetry, and is often referred to as an asymmetric carbon. The lack of a plane of symmetry makes the carbon chiral. The presence of a single chiral carbon atom sufficient to render the molecule chiral, and modern terminology refers to such groupings as chiral centers or stereo centers.
An example is shown in the bromochlorofluoromethane molecule shown in part (a) of the figure below. This carbon, is attached to four different substituents making it chiral. which is often designated by an asterisk in structural drawings. If the bromine atom is replaced by another chlorine to make dichlorofluoromethane, as shown in part (b) below, the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Upon comparison, bromochlorofluoromethane lacks a plane of symmetry while dichlorofluoromethane has a plane of symmetry.
Identifying Chiral carbons
Identifying chiral carbons in a molecule is an important skill for organic chemists. The presence of a chiral carbon presents the possibility of a molecule having multiple stereoisomers. Most of the chiral centers we shall discuss in this chapter are asymmetric carbon atoms, but it should be recognized that other tetrahedral or pyramidal atoms may become chiral centers if appropriately substituted. Also, when more than one chiral center is present in a molecular structure, care must be taken to analyze their relationship before concluding that a specific molecular configuration is chiral or achiral. This aspect of stereoisomerism will be treated later. Because an carbon requires four different substituents to become asymmertric, it can be said, with few exceptions, that sp2 and sp hybridized carbons involved in multiple bonds are achiral. Also, any carbon with more than one hydrogen, such as a -CH3 or -CH2- group, are also achiral.
Looking for planes of symmetry in a molecule is useful, but often difficult in practice. It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry, where one side of the plane is the mirror image of the other. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions, the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral.
Determining if a carbon is bonded to four distinctly different substituents can often be difficult to ascertain. Remember even the slightest difference makes a substituent unique. Often these difference can be distant from the chiral carbon itself. Careful consideration and often the building of molecular models may be required. A good example is shown below. It may appear that the molecule is achiral, however, when looking at the groups directly attached to the possible chiral carbon, it is clear that they all different. The two alkyl groups are differ by a single -CH2- group which is enough to consider them different.
Example $1$
Predict if the following molecule would be chiral or achiral:
Answer
Achiral. When determining the chirality of a molecule, it best to start by locating any chiral carbons. An obvious candidate is the ring carbon attached to the methyl substituent. The question then becomes: does the ring as two different substituents making the substituted ring carbon chiral? With an uncertainty such as this, it is then helpful try to identify any planes of symmetry in the molecule. This molecule does have a plane of symmetry making the molecule achiral. The plane of symmetry would be easier see if the molecule were view from above. Typically, monosubstituted cycloalkanes have a similar plane of symmetry making them all achiral.
Exercise 5.2.1
Determine if each of the following molecules are chiral or achiral. For chiral molecules indicate any chiral carbons.
Answer
Explanation
Structures F and G are achiral. The former has a plane of symmetry passing through the chlorine atom and bisecting the opposite carbon-carbon bond. The similar structure of compound E does not have such a symmetry plane, and the carbon bonded to the chlorine is a chiral center (the two ring segments connecting this carbon are not identical). Structure G is essentially flat. All the carbons except that of the methyl group are sp2 hybridized, and therefore trigonal-planar in configuration. Compounds C, D & H have more than one chiral center, and are also chiral.
Note
In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women.
Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects.
Researchers later realized the problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms.
One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp3-hybridized carbon. These two forms of thalidomide are stereoisomers. If you make models of the two stereoisomers of thalidomide, you will see that they too are mirror images, and cannot be superimposed.
As a historical note, thalidomide was never approved for use in the United States. This was thanks in large part to the efforts of Dr. Frances Kelsey, a Food and Drug officer who, at peril to her career, blocked its approval due to her concerns about the lack of adequate safety studies, particularly with regard to the drug's ability to enter the bloodstream of a developing fetus. Unfortunately, though, at that time clinical trials for new drugs involved widespread and unregulated distribution to doctors and their patients across the country, so families in the U.S. were not spared from the damage caused.
Very recently a close derivative of thalidomide has become legal to prescribe again in the United States, with strict safety measures enforced, for the treatment of a form of blood cancer called multiple myeloma. In Brazil, thalidomide is used in the treatment of leprosy - but despite safety measures, children are still being born with thalidomide-related defects.
Example 5.2.2
Label the molecules below as chiral or achiral, and locate all stereocenters.
Answer
Exercise 5.2.2
1) For the following compounds, star (*) each chiral center, if any.
2) Explain why the following compound is chiral.
3) Determine which of the following objects is chiral.
a) A Glove.
b) A nail.
c) A pair of sunglasses.
d) The written word "Chiral".
4) Place an "*" by all of the chrial carbons in the following molecules.
a)
Erythrose, a four carbon sugar.
b) Isoflurane, an anestetic. Bright green = Chlorine, Pale green = Fluorine.
Answer
1)
2) Though the molecule does not contain a chiral carbon, it is chiral as it is non-superimposable on its mirror image due to its twisted nature (the twist comes from the structure of the double bonds needing to be at 90° angles to each other, preventing the molecule from being planar).
3)
a) Just as hands are chiral a glove must also be chiral.
b) A nail has a plane of symmetry which goes down the middle making it a achiral.
c) A pair of sunglasses has a plane of symmetry which goes through the nose making it achiral.
d) Most written words are chiral. Look one in a mirror to confirm this.
4
a)
b)
Exercise 5.2.3
Circle all of the carbon stereocenters in the molecules below.
Answer
Exercise 5.2.4
Circle all of the carbon stereocenters in the molecules below.
Answer
Here are some more examples of chiral molecules that exist as pairs of enantiomers. In each of these examples, there is a single stereocenter, indicated with an arrow. (Many molecules have more than one stereocenter, but we will get to that that a little later!)
Here are some examples of molecules that are achiral (not chiral). Notice that none of these molecules has a stereocenter.
It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry, where one side of the plane is the mirror image of the other. Chirality is tied conceptually to the idea of asymmetry, and any molecule that has a plane of symmetry cannot be chiral. When looking for a plane of symmetry, however, we must consider all possible conformations that a molecule could adopt. Even a very simple molecule like ethane, for example, is asymmetric in many of its countless potential conformations – but it has obvious symmetry in both the eclipsed and staggered conformations, and for this reason it is achiral.
Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions (see section 3.7B), the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as chiral carbons.
When evaluating a molecule for chirality, 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 stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter.
Other elements in addition to carbon can be stereocenters. The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter.
We will see in chapter 10 how researchers, in order to investigate the stereochemistry of reactions at the phosphate center, incorporated sulfur and/or 17O and 18O isotopes of oxygen (the ‘normal’ isotope is 16O) to create chiral phosphate groups. Phosphate triesters are chiral if the three substituent groups are different.
Asymmetric quaternary ammonium groups are also chiral. Amines, however, are not chiral, because they rapidly invert, or turn ‘inside out’, at room temperature.
Exercise 5.2.5
Label the molecules below as chiral or achiral, and circle all stereocenters.
a) fumarate (a citric acid cycle intermediate)
b) malate (a citric acid cycle intermediate)
b) malate (a citric acid cycle intermediate)
Answer
a) achiral (no stereocenters)
b) chiral
c) chiral
Exercise 5.2.6
Label the molecules below as chiral or achiral, and circle all stereocenters.
a) acetylsalicylic acid (aspirin)
b) acetaminophen (active ingredient in Tylenol)
c) thalidomide (drug that caused birth defects in pregnant mothers in the 1960’s)
Answer
a) achiral (no stereocenters)
b) achiral (no stereocenters)
c) chiral
Exercise 5.2.7
Draw both enantiomers of the following chiral amino acids.
a) Cysteine
b) Proline
Answer
Exercise 5.2.8
Draw both enantiomers of the following compounds from the given names.
a) 2-bromobutane
b) 2,3-dimethyl-3-pentanol
Answer
Exercise 5.2.9
Which of the following body parts are chiral?
a) Hands b) Eyes c) Feet d) Ears
Answer
a) Hands- chiral since the mirror images cannot be superimposed (think of the example in the beginning of the section)
b) Eyes- achiral since mirror images that are superimposable
c) Feet- chiral since the mirror images cannot be superimposed (Does your right foot fit in your left shoe?)
d) Ears- chiral since the mirror images cannot be superimposed
Exercise 5.2.10
Circle the chiral centers in the following compounds.
Answer
Exercise 5.2.11
Identify the chiral centers in the following compounds.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.1%3A_Chiral__Molecules.txt |
Objectives
After completing this section, you should be able to
1. describe the nature of plane-polarized light.
2. describe the features and operation of a simple polarimeter.
3. calculate the specific rotation of a compound, given the relevant experimental data.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• analyzer
• dextrorotatory
• levorotatory
• optically active
• plane-polarized light
• polarimeter
• polarizer
• specific rotation, $[\alpha]^{20}_{D}$
Study Notes
A polarizer is a device through which only light waves oscillating in a single plane may pass. A polarimeter is an instrument used to determine the angle through which plane-polarized light has been rotated by a given sample. You will have the opportunity to use a polarimeter in the laboratory component of the course. An analyzer is the component of a polarimeter that allows the angle of rotation of plane-polarized light to be determined.
Specific rotations are normally measured at 20°C, and this property may be indicated by the symbol $[\alpha]^{20}_{D}$. Sometimes the solvent is specified in parentheses behind the specific rotation value, for example,
$[\alpha]^{20}_{D} = +12^o\,(\text{chloroform}) \nonumber$
For liquids, the specific rotation may be obtained using the neat liquid rather than a solution; in such cases the formula is
$[ α ]_D^{\text{temp}} \,(\text{neat})= α \times l \times d \nonumber$
where $α$ is the observed rotation, $l$ is the path length of the cell (measured in decimetres, dm), and $d$ is the density of the liquid.
Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity.
Polarimetry
Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter (Figure $1$).
Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector.
Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α, if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission.
The observed rotations ($\alpha$) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter, meaning right, and laevus, for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined , using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously.
Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, $α$, to a specific rotation, by the following formula:
$[\alpha]_D = \dfrac{\alpha}{l c} \tag{5.3.1}$
where
• $[\alpha]_D$ is the specific rotation
• $l$ is the cell length in dm
• $c$ is the concentration in g/ml
• $D$ designates that the light used is the 589 line from a sodium lamp
Compounds that rotate the plane of polarized light are termed optically active. Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid enantiomers have the following specific rotations:
• Carvone from caraway: $[\alpha]^{20}_{D} = +62.5^o$ (this isomer may be referred to as (+)-carvone or d-carvone)
• Carvone from spearmint: $[\alpha]^{20}_{D} = -62.5^o$ (this isomer may be referred to as (–)-carvone or l-carvone)
and carvone enantiomers have the following specific rotations:
• Lactic acid from muscle tissue: $[\alpha]^{20}_{D} = +2.5^o$ (this isomer may be referred to as (+)-lactic acid or d-lactic acid)
• Lactic acid from sour milk: $[\alpha]^{20}_{D} = -2.5^o$ (this isomer may be referred to as (–)-lactic acid or l-lactic acid)
A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of $\ce{HBr}$ to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red).
$\ce{CH3CH=CHCH3 + HBr -> (±) CH3CH2}\textcolor{red}{\ce{C}} \ce{HBrCH3} \nonumber$
Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials.
There are two ways in which the condition of a chiral substance may be changed:
1. A racemate may be separated into its component enantiomers. This process is called resolution.
2. A pure enantiomer may be transformed into its racemate. This process is called racemization.
Enantiomeric Excess
The "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.
$\% \text { optical purity }=\frac{\text { specific rotation of mixture }}{\text { specific rotation of pure enantiomer }} \times 100\% \nonumber$
Because R and S enantiomers have equal but opposite optical activity, it naturally follows that a 50:50 racemic mixture of two enantiomers will have no observable optical activity. If we know the specific rotation for a chiral molecule, however, we can easily calculate the ratio of enantiomers present in a mixture of two enantiomers, based on its measured optical activity. When a mixture contains more of one enantiomer than the other, chemists often use the concept of enantiomeric excess (ee) to quantify the difference. Enantiomeric excess can be expressed as:
$e e=\frac{(\% \text { more abundant enantiomer }-50) \times 100\%}{50} \nonumber$
For example, a mixture containing 60% R enantiomer (and 40% S enantiomer) has a 20% enantiomeric excess of R: ((60-50) x 100) / 50 = 20 %.
Exercise $1$
The specific rotation of (S)-carvone is (+)61°, measured 'neat' (pure liquid sample, no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-)23°. Which enantiomer is in excess, and what is its ee? What are the percentages of (R)- and (S)-carvone in the sample?
Answer
The observed rotation of the mixture is levorotary (negative, counter-clockwise), and the specific rotation of the pure S enantiomer is given as dextrorotary (positive, clockwise), meaning that the pure R enantiomer must be levorotary, and the mixture must contain more of the R enantiomer than of the S enantiomer.
Rotation (R/S Mix) = [Fraction(S) × Rotation (S)] + [Fraction(R) × Rotation (R)]
Let Fraction (S) = x, therefore Fraction (R) = 1 – x.
Rotation (R/S Mix) = x[Rotation (S)] + (1 – x)[Rotation (R)].
–23 = x(+61) + (1 – x)(–61)
Solve for x: x = 0.3114 and (1 – x) = 0.6885
Therefore the percentages of (R)- and (S)-carvone in the sample are 68.9% and 31.1%, respectively.
ee = [(% more abundant enantiomer – 50) × 100]/50. = [68.9 – 50) × 100]/50 = 37.8%.
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration
Chiral molecules are often labeled according to the sign of their specific rotation, as in (S)-(+)-carvone and (R)-(-)-carvone, or (±)-carvone for the racemic mixture. However, there is no relationship whatsoever between a molecule's R/S designation and the sign of its specific rotation. Without performing a polarimetry experiment or looking in the literature, we would have no idea that (-)-carvone has the R configuration and (+)-carvone has the S configuration.
Separation of Chiral Compounds
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. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used of this technique is known as chiral resolution.
Exercise $2$
A 3.20 g sample of morphine ([α]D = -132) was dissolved in 10.0 mL of acetic acid ([α]D = 0). If it is put into a sample tube with a path length of 2.00 cm, what would be its observed rotation (α)?
Answer
The specific rotation, [α]D = (observed rotation, α (degrees))/ [(pathlength, l (dm)) x (concentration, c (g/cm3))] = α/(l x c)
Solving for α, α = [α]D x l x c
([α]D = -132) x (l = 2.00 cm = 0.200 dm) x (c = 3.20 g / 10.0 cm3 = 0.320 g/cm3)
α = -132 x 0.200 dm x 0.320 g/cm3 = -8.45 o
Exercise $3$
Is the morphine in the previous excercise dextrorotatory or levorotatory?
Answer
Since morphine has a (-) rotation, it indicates that it rotates light to the left (counterclockwise) and morphine is levorotatory.
Exercise $4$
Label the following compounds as dextrorotatory or levorotatory.
1. sucrose ([α]D = + 66.7)
2. cholesterol ([α]D = - 31.5)
3. cocaine ([α]D = - 16)
4. chloroform ([α]D = 0)
Answer
1. sucrose ([α]D = + 66.7) dextrorotatory
2. cholesterol ([α]D = - 31.5) levorotatory
3. cocaine ([α]D = - 16) levorotatory
4. chloroform ([α]D = 0) neither, not optically active
Exercise $\PageIndex{5a}$
The specific rotation of (S)-carvone is (+) 61o when measured neat (pure liquid sample with no solvent). The optical rotation of a neat sample of a mixture of R and S carvone is measured at (-) 23 o.
a) Which enantiomer is in excess?
Answer
Since the pure S enantiomer ((+) 61o) is dextrorotatory (positive, clockwise), the R enantiomer must be levorotatory. The observed rotation of the mixture is levorotatory since its negative (counterclockwise). This means the mixture must contain more of the R enantiomer than the S enantiomer.
Exercise $\PageIndex{5b}$
b) What are the percentages of (S)- and (R)- carvone in the sample mixture?
Answer
Optical rotation (α) of the (R/S mixture) = [fraction (S) x [α]D (S)] + [fraction (R) x [α]D (R)]
To determine the fraction of S and R, we make y = fraction (S) and 1 – y = fraction (R)
-23o = y x (61o) + (1 – y) x (-61o) solving for y: y = 0.3114 and (1-y) = 0.6885
Therefore the percentage of (S)-carvone is 31.1 % and (R)-carvone is 68.9 %
Exercise $\PageIndex{5c}$
c) What is the ee (enantiomeric excess)?
Answer
ee = [(% more abundant isomer – 50) x 100]/50 = [(68.9 – 50) x100]/50 = 37.8 % ee
Exercise $\PageIndex{6a}$
Determine the ee’s of the following from the percentages
• 95 % (R)- tartaric acid and 5.0 % (S)- tartaric acid
Answer
[(95 – 50) x 100] / 50 = 90 % ee (R)-tartaric acid
Exercise $\PageIndex{6b}$
Determine the ee’s of the following from the percentages
• 75 % (S)- limonene and 25 % (R)- limonene
Answer
[(75 – 50) x 100] / 50 = 50 % ee (S)- limonene
Exercise $\PageIndex{6c}$
Determine the ee’s of the following from the percentages
• 85 % (R) cysteine
Answer
(85 – 50) x 100] / 50 = 70 % ee (R)-cysteine
Exercise $\PageIndex{6d}$
Determine the ee’s of the following from the percentages
• 50 % (S) alanine
Answer
(50 – 50) x 100] / 50 = 0 % ee, racemic mixture | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.2%3A_Optical__Activity.txt |
Objectives
After completing this section, you should be able to
1. assign Cahn-Ingold-Prelog priorities to a given set of substituents.
2. determine whether a given wedge-and-broken-line structure corresponds to an R or an S configuration, with or without the aid of molecular models.
3. draw the wedge-and-broken-line structure for a compound, given its IUPAC name, complete with R or S designation.
4. construct a stereochemically accurate model of a given enantiomer from either a wedge-and-broken-line structure or the IUPAC name of the compound, complete with R or S designation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• absolute configuration
• R configuration
• S configuration
Study Notes
When designating a structure as R or S, you must ensure that the atom or group with the lowest priority is pointing away from you, the observer. The easiest way to show this is to use the wedge-and-broken-line representation. You can then immediately determine whether you are observing an R configuration or an S configuration.
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, as such, 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).
The Cahn-Ingold-Prelog rules of assign priorities the groups directly bonded to the chiral carbon. Having ranked the four groups attached to a chiral carbon, we describe the stereochemical configuration around the carbon by orienting the molecule so that the group with the lowest ranking (4) is given a dash bond to indicate it points directly away from us. We then look at the three remaining substituents, which now appear to radiate toward us which is shown by using wedge bonds. If a curved arrow drawn from the highest to second-highest to third-highest ranked substituent is clockwise, we say that the chirality center has the (R) configuration (Latin rectus, meaning “right”). If an arrow from is counterclockwise, the chirality center has the (S) configuration (Latin sinister, meaning “left”). To remember these assignments, think of a car’s steering wheel when making a Right (clockwise) turn. Note, the (R) or (S) configurations represent the two enantiomers of a chiral molecule. The (R) or (S) configuration is often added as a prefix, in parenthesis, to a chiral molecule's name to indicate which enantiomer is being discussed (e.g., (R)-2-bromobutane). If more than chiral carbon is present in a chiral molecule, each carbon's number is included before the (R) or (S) configuration. Ex: (2R,4S,6R)-2-bromo-6-chloro-4-methylheptane.
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 atom with a higher atomic number takes precedence over a atom with a lower atomic number. Hydrogen is the lowest possible priority atom, because it has the lowest atomic number.
1. The atom with higher atomic number has higher priority (I > Br > Cl > S > P > F > O > N > C > H).
2. When comparing isotopes, the atom with the higher mass number has higher priority [18O > 16O or 15N > 14N or 13C > 12C or T (3H) > D (2H) > H].
Rule 2
If there are two or more substituents which have the same element directly attached to chiral carbon, proceed along the substituent chains until a point of difference is found. 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.
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 two hydrogen atoms and a 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.
The "-H" (left) ranks lower than the "C-" (right) based on the relative molecular weights at the first point of difference.
Worked Exercise \(1\)
For the following pairs of substituents, determine which would have the higher and lower priority based on the Cahn-Ingold-Prelog rules. Explain your answer.
Answer
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:
The "C-" (right) ranks higher than the "H-" (left) based on
the first point of difference and their relative atomic numbers.
However:
In this case, even though the bold carbon on the right structure has two
connections to a non-hydrogen atom (C), it is the lower priority.
This is because one of the atoms attached to the bold carbon on the left molecule
ranks higher than any of the atoms attached to the bold carbon on the right structure,
since Br has a higher atomic number than C.
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.
Rule 3
For assigning priority, multiple bonds are treated as if each bond of the multiple bond is bonded to a unique atom. For example, an alkene substituent (CH2=CH-) has higher priority than an ethyl substituent (CH3CH2-). The alkene carbon priority is "two" bonds to carbon atoms and one bond to a hydrogen atom compared with the ethyl carbon that has only one bond to a carbon atom and two bonds to two hydrogen atoms. Similarly, alkyne substituent (HCC-) would have an even higher priority because the alkyne carbon is treated as if it is bonded to three carbons. This method remains the same with compounds containing a carbonyl (C=O) group. The carbon of an aldehyde substituent (O=CH-) is treated as if it is bonded to a hydrogen and two oxygen atoms.
Determining (R) or (S) Configuration Using a Molecular Model
In order to demonstrate how to determine the (R)/(S) configuration of the chiral carbon in the following molecule using molecular models, first construct a model of the bromoethanol structure:
First make a molecular model of a tetrahedral carbon with four different substituents. In many cases, this will appear as a carbon with four bonds with a different colored ball attached to each bond.
For the molecule in question, determine the location of the chiral carbon and assign CIP priorities to the substituents. In this case, Br gets the highest priority because it has the highest atomic number. The O in the OH substituent gets priority 2 and the C in CH3 gets priority 3. Lastly, H gets the lowest priority, 4, because it has the smallest atomic number.
Now take your molecular model and orientate it to match the molecule in question. Remember in the dash/wedge representation, two regular bonds are in the plane of the page. The wedge bond is coming toward you and the dashed bond is going away from you. If you were to hold a piece of paper directly in front of you, the substituents with the regular bond should both be touching the piece of paper. The dashed bond should be pointing behind the piece of paper and the wedge bond should be pointing in front.
In this structure, the bromine is going away from you, the hydrogen is coming toward you and the hydroxide and methyl groups are in the plane of the page.
Then based on the position, assign each substituent on the chiral carbon a colored ball on your molecular model. In this case, bromine is going away so it is assigned the green ball. The hydrogen is coming toward you so it is assigned the blue ball. The last two substituents are in the plane of the page, however, the CH3 is positioned higher so it is assigned the red ball which leaves OH being assigned the black ball.
Lastly, grab onto the ball for the lowest priority substitutent, in this case the blue one, and point the other three substituents towards you. The three bonds should be angled towards you as if they all have wedge bonds. Assign the original substituents and their corresponding CIP priorities to the three colored balls. The green ball was assigned to bromine which was given priority one. The OH was assigned to the black ball and given priority two. The CH3 was assigned to the red ball and given priority three. In this case the priorities are going counter clockwise so the chiral carbon has an (S) configuration.
Determining (R) or (S) Configuration Without a Molecular Model
If a molecular model cannot be used there are a couple of simple methods which can be applied if the dash/wedge bond system is being used.
After assigning CIP priorities, if the lowest priority substituent (4) is on the dash bond the configuration of substituents 1-3 can be assigned directly. As shown in the figure below, the configuration of substituents 1-3 does not change when moving to sight down the bond of substituent 4. In both cases, substituents 1-3 are ordered in a counterclockwise fashion which gives the chiral carbon an (S) configuration.
The opposite is true if the lowest priority substituent (4) is on the wedge bond. As shown in the figure below, the configuration of substituents 1-3 is inverted when moving to sight down the bond of substituent 4. When the lowest priority substituent is on the wedge bond, the configuration of substituents 1-3 can be assigned directly only if the direction is inverted. i.e. clockwise = (S) and counterclockwise = (R).
With the lowest priority group in front, drawing an arc from 1 to 2 to 3 gives the reverse of the configuration.
However, if the lowest priority substituent is on one of the regular bonds when the dash/wedge system is being used then configurations are best assigned by changing perspectives. This method can also be used if the three-dimensional configuration of the chiral carbon is represented. First, locate the chiral carbon and assign CIP priorities to its substituents. Then while perceiving the drawn molecule as a three-dimensional image, mentally change your perspective such that you are looking down the bond between the chiral carbon and the lowest CIP ranked substituent (#4). If done correctly, the bonds for substituents 1-3 should be coming towards you as wedge bonds. You can then follow the direction of the CIP priority numbers to determine the (R)/(S) configuration of the chiral carbon.
Locate the chiral carbon and assign CIP priorities to its substituents.
Mentally sight down the bond between the chiral carbon and the lowest CIP ranked substituent.
This bond is shown in magenta.
clockwise
(R)-configuration
Follow the direction of the CIP priority numbers to determine the (R)/(S) configuration.
Drawing the Structure of a Chiral Molecule from its Name
Draw the structure of (S)-2-Bromobutane:
1) Draw the basic structure of the molecule and determine the location of the chiral carbon.
2) Determine the chiral carbon's substituents and assign them a CIP priority.
-H (Priority 4)
-CH3 (Priority 3)
-CH2CH3 (Priority 2)
-Br (Priority 1)
3) Draw the chiral carbon in a dash/wedge form and add the lowest priority substituent to the dash bond. In this case, the lowest priority substituent is -H.
4) Add the remaining substituents in a clockwise fashion for (R) and a counterclockwise fashion for (S).
The molecule posed in this question has an (S) configuration so the remaining substituents are added in a counterclockwise fashion.
Exercise \(1\)
1) Orient the following so that the least priority (4) atom is paced behind, then assign stereochemistry ((R) or (S)).
2) Draw (R)-2-bromobutan-2-ol.
3) Assign (R)/(S) to the following molecule.
4) Which in the following pairs would have a higher CIP priority?
1. -H or -Cl
2. -Br or -I
3. -CH2OH or -OCH3
4. -CH2CH3 or -CH=CH2
5. -NH2 or -OH
5) Rank the following substituents in order of their CIP priority:
1. -H, -OCH3, -CH2OH, -OH
2. -OH, -CO2H, -CH=O, -CH2OH
3. -CN, -NH2, -CH=O, -NHCH3
4. -SH, -SCH3, -OH, -OOCH3
6) Determine if the chiral carbon in the following molecules have an (R) or (S) configuration. Red = Oxygen & Blue = Nitrogen.
a)
b)
Answer
1) A is (S) and B is (R).
2)
3) The stereocenter is (R).
4)
1. -Cl
2. -I
3. -OCH3
4. -CH=CH2
5. -OH
5) Rank the following substituents in order of their CIP priority:
1. -OCH3, -OH, -CH2OH, -H,
2. -OH, -CO2H, -CH=O, -CH2OH
3. -NHCH3, -NH2, -CH=O, -CN
4. -SCH3, -SH, -OOCH3, -OH
6)
a) The chiral carbon is (R). The four substituents of the chiral carbon are -OH (1), -NH2 (2), -CH3 (3), and -H (4). Then looking down the lowest priority bond, you should roughly see what appears in the picture below. The substituents with priorities 1-3 are ordered in a clockwise fashion so the chiral carbon is (R).
b) The chiral carbon is (S). The four substituents of the chiral carbon are -CO2H (1), -OH (2), -CH2CH2CH3 (3), and -H (4). Then looking down the lowest priority bond, you should roughly see what appears in the picture below. The substituents with priorities 1-3 are ordered in a counterclockwise fashion so the chiral carbon is (S).
Exercise \(2\)
Identify which substituent in the following sets has a higher ranking.
1. -H or -CH3
2. -CH2CH2CH3 or CH2CH3
3. -CH2Cl or CH2OH
Answer
1. -CH3
2. -CH2CH2CH3
3. -CH2Cl
Exercise \(3\)
Identify which substituent in the following sets has a higher ranking.
1. -NH2 or -N=NH
2. -CH2CH2OH or -CH2OH
3. -CH=CH2 or -CH2CH3
Answer
a) -N=NH
b) -CH2OH
c) -CH=CH2
Exercise \(4\)
Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)
1. -NH2, -F, -Br, -CH3
2. -SH, -NH2, -F, -H
Answer
a) -CH3 < -NH2 < -F, < -Br
b) -H < -NH2 < -F, < -SH
Exercise \(5\)
Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)
1. -CH2CH3, -CN, -CH2CH2OH, -CH2CH2CH2OH
2. -CH2NH2, -CH2SH, -C(CH3)3, -CN
Answer
a) -CH2CH3 < -CH2CH2OH < -CH2CH2CH2OH, < -CN
b) -C(CH3)3 < -CH2NH2 < -CN < -CH2SH
Exercise \(6\)
Assign the following chiral centers as (R) or (S).
Answer
a) (S): I > Br > F > H. The lowest priority substituent is going backwards so following the highest priority, it goes left (counterclockwise).
b) (R): Br > Cl > CH3 > H. Using a model kit, you need to rotate the H to the back position where the Br is. This causes the priority to go to the left (clockwise) when looking at it with the H in the back position. Alternatively, if you do not have a model kit, you can imagine the structure 3-dimensionally and since the lowest priority (H) is facing up (as drawn), if you look at it from below, starting with Br (1st priority) and moving towards Cl (2nd priority), you are moving right (clockwise) which represents (R) stereochemistry.
c) Neither (R) or (S): Since there are two identical substituents (H’s) the molecule is achiral and cannot be assigned (R) or (S).
Exercise \(7\)
Assign the following chiral centers as (R) or (S).
Answer
a) (R): OH > CN (C triple bonded to N) > CH2NH2 > H. The H needs to be moved to the back position which causes the priority to go to the right (clockwise) which indicates (R).
b) (S): COOH > CH2OH > CCH > H. Since the H is coming forward, you can assign the priority and it goes to the right (clockwise which would be (R)) but since the lowest priority is forward, you have to switch it to (S). Alternatively, you can rotate the molecule to put the lowest priority to the back and you’ll see that it rotates left (or counterclockwise) for (S).
c) (S): Br > OH > NH2 > CH3. Since the lowest priority is going back, you can follow the priority and see that it is going left (counterclockwise) and therefore (S).
Exercise \(8\)
Draw the structure of (R)-2-bromohexane.
Answer
Exercise \(9\)
Draw the structure of (S)-2-methyl-3-pentanol.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.3%3A_Absolute_Configuration%3A_R-S__Sequence_Rules.txt |
Objectives
After completing this section, you should be able to
1. draw the Fischer projection of a monosaccharide, given its wedge‑and‑broken‑line structure or a molecular model.
2. draw the wedge‑and‑broken‑line structure of a monosaccharide, given its Fischer projection or a molecular model.
3. construct a molecular model of a monosaccharide, given its Fischer projection or wedge‑and‑broken‑line structure.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Fischer projection
Study Notes
When studying this section, use your molecular model set to assist you in visualizing the structures of the compounds that are discussed. It is important that you be able to determine whether two apparently different Fischer projections represent two different structures or one single structure. Often the simplest way to check is to construct a molecular model corresponding to each projection formula, and then compare the two models.
The problem of drawing three-dimensional configurations on a two-dimensional surface, such as a piece of paper, has been a long-standing concern of chemists. The wedge and hatched line notations we have been using are effective, but can be troublesome when applied to compounds having many chiral centers. As part of his Nobel Prize-winning research on carbohydrates, the great German chemist Emil Fischer, devised a simple notation that is still widely used. In a Fischer projection drawing, the four bonds to a chiral carbon make a cross with the carbon atom at the intersection of the horizontal and vertical lines. The two horizontal bonds are directed toward the viewer (forward of the stereogenic carbon). The two vertical bonds are directed behind the central carbon (away from the viewer). Since this is not the usual way in which we have viewed such structures, the following diagram shows how a stereogenic carbon positioned in the common two-bonds-in-a-plane orientation ( x–C–y define the reference plane ) is rotated into the Fischer projection orientation (the far right formula). When writing Fischer projection formulas it is important to remember these conventions. Since the vertical bonds extend away from the viewer and the horizontal bonds toward the viewer, a Fischer structure may only be turned by 180º within the plane, thus maintaining this relationship. The structure must not be flipped over or rotated by 90º.
In the above diagram, if x = CO2H, y = CH3, a = H & b = OH, the resulting formula describes (R)-(–)-lactic acid. The mirror-image formula, where x = CO2H, y = CH3, a = OH & b = H, would, of course, represent (S)-(+)-lactic acid.
The Fischer Projection consists of both horizontal and vertical lines, where the horizontal lines represent the atoms that are pointed toward the viewer while the vertical line represents atoms that are pointed away from the viewer. The point of intersection between the horizontal and vertical lines represents the central carbon.
Using the Fischer projection notation, the stereoisomers of 2-methylamino-1-phenylpropanol are drawn in the following manner. Note that it is customary to set the longest carbon chain as the vertical bond assembly.
The usefulness of this notation to Fischer, in his carbohydrate studies, is evident in the following diagram. There are eight stereoisomers of 2,3,4,5-tetrahydroxypentanal, a group of compounds referred to as the aldopentoses. Since there are three chiral centers in this constitution, we should expect a maximum of 23 stereoisomers. These eight stereoisomers consist of four sets of enantiomers. If the configuration at C-4 is kept constant (R in the examples shown here), the four stereoisomers that result will be diastereomers. Fischer formulas for these isomers, which Fischer designated as the "D"-family, are shown in the diagram. Each of these compounds has an enantiomer, which is a member of the "L"-family so, as expected, there are eight stereoisomers in all. Determining whether a chiral carbon is R or S may seem difficult when using Fischer projections, but it is actually quite simple. If the lowest priority group (often a hydrogen) is on a vertical bond, the configuration is given directly from the relative positions of the three higher-ranked substituents. If the lowest priority group is on a horizontal bond, the positions of the remaining groups give the wrong answer (you are in looking at the configuration from the wrong side), so you simply reverse it.
The aldopentose structures drawn above are all diastereomers. A more selective term, epimer, is used to designate diastereomers that differ in configuration at only one chiral center. Thus, ribose and arabinose are epimers at C-2, and arabinose and lyxose are epimers at C-3. However, arabinose and xylose are not epimers, since their configurations differ at both C-2 and C-3.
How to make Fischer Projections
To make a Fischer Projection, it is easier to show through examples than through words. Lets start with the first example, turning a 3D structure of ethane into a 2D Fischer Projection.
Example 25.2.1
Start by mentally converting a 3D structure into a Dashed-Wedged Line Structure. Remember, the atoms that are pointed toward the viewer would be designated with a wedged lines and the ones pointed away from the viewer are designated with dashed lines.
Figure A Figure B
Notice the red balls (atoms) in Figure A above are pointed away from the screen. These atoms will be designated with dashed lines like those in Figure B by number 2 and 6. The green balls (atoms) are pointed toward the screen. These atoms will be designated with wedged lines like those in Figure B by number 3 and 5. The blue atoms are in the plane of the screen so they are designated with straight lines.
Now that we have our Dashed- Wedged Line Structure, we can convert it to a Fischer Projection. However, before we can convert this Dashed-Wedged Line Structure into a Fischer Projection, we must first convert it to a “flat” Dashed-Wedged Line Structure. Then from there we can draw our Fischer Projection. Lets start with a more simpler example. Instead of using the ethane shown in Figure A and B, we will start with a methane. The reason being is that it allows us to only focus on one central carbon, which make things a little bit easier.
Figure C Figure D
Lets start with this 3D image and work our way to a dashed-wedged image. Start by imagining yourself looking directly at the central carbon from the left side as shown in Figure C. It should look something like Figure D. Now take this Figure D and flatten it out on the surface of the paper and you should get an image of a cross.
As a reminder, the horizontal line represents atoms that are coming out of the paper and the vertical line represents atoms that are going into the paper. The cross image to the right of the arrow is a Fischer projection. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.4%3A_Fischer__Projections.txt |
Objectives
After completing this section, you should be able to
1. calculate the maximum number of stereoisomers possible for a compound containing a specified number of chiral carbon atoms.
2. draw wedge-and-broken-line structures for all possible stereoisomers of a compound containing two chiral carbon atoms, with or without the aid of molecular models.
3. assign (R)/(S) configurations to wedge-and-broken-line structures containing two chiral carbon atoms, with or without the aid of molecular models.
4. determine, with or without the aid of molecular models, whether two wedge-and-broken-line structures containing two chiral carbon atoms are identical, represent a pair of enantiomers, or represent a pair of diastereomers.
5. draw the wedge-and-broken-line structure of a specific stereoisomer of a compound containing two chiral carbon atoms, given its IUPAC name and (R)/(S) configuration.
Key Terms
Make certain that you can define, and use in context, the key term below.
• diastereomer
Diastereomers are two molecules which are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers. Unlike enantiomers which are mirror images of each other and non-superimposable, 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. In order for diastereomer stereoisomers to occur, a compound must have two or more stereocenters.
Introduction
So far, we have been analyzing compounds with a single chiral center. Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers.
We will start with a common four-carbon sugar called D-erythrose.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the (R) configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the (S) configuration. To avoid confusion, we will simply refer to the different stereoisomers by capital letters.
Now let's consider all the possible stereoisomers.
Look first at compound A below. Both chiral centers in have the (R) configuration (you should confirm this for yourself!). The mirror image of Compound A is compound B, which has the (S) configuration at both chiral centers. If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not superimposable (again, confirm this for yourself with your models!). A and B are nonsuperimposable mirror images: in other words, enantiomers.
Now, look at compound C, in which the configuration is (S) at chiral center 1 and (R) at chiral center 2. Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer). However, they are not mirror images of each other (confirm this with your models!), and so they are not enantiomers. By definition, they are diastereomers of each other.
Notice that compounds C and B also have a diastereomeric relationship, by the same definition.
So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them. Does compound C have its own enantiomer? Compound D is the mirror image of compound C, and the two are not superimposable. Therefore, C and D are a pair of enantiomers. Compound D is also a diastereomer of compounds A and B.
This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:
Stereoisomer Shortcuts
If all of the chiral centers are of opposite (R)/(S) configuration between two stereoisomers, they are enantiomers.
If at least one, but not all of the chiral centers are opposite between two stereoisomers, they are diastereomers.
These shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups: we will come to that later
Here's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue. Pairs of enantiomers are stacked together.
We know, using the shortcut above, that the enantiomer of (R,R) must be (S,S) - both chiral centers are different. We also know that (R,S) and (S,R) are diastereomers of (R,R), because in each case one - but not both - chiral centers are different.
Determining the Maximum Number of Stereoisomers for a Compound
In general, a structure with n stereocenters will have a maximum of 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
Example \(1\)
• Draw the structure of L-galactose, the enantiomer of D-galactose.
• Draw the structure of two more diastereomers of D-glucose. One should be an epimer.
Answer
Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted.
In total, there are 210 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above.
We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar.
Exercise \(1\)
Determine the number of stereoisomers a molecule can have with…
1. 3 chiral centers
2. 1 chiral center
3. 6 chiral centers
Answer
Since a molecule with n chiral centers can have 2n stereoisomers…
1. 23 = 8 possible stereoisomers
2. 21 = 2 possible stereoisomers
3. 26 = 64 possible stereoisomers
Exercise \(\PageIndex{2a}\)
What is the relationship between enantiomers?
Answer
They are mirror images of each other and when 2 or more chiral centers are present, every stereocenter is the opposite in its enantiomer.
Exercise \(\PageIndex{2b}\)
How does the stereochemistry in diastereomers differ from each other?
Answer
In diastereomers, one or more of the chiral centers is the opposite but they all can’t be the opposite or else they’d be enantiomers.
Exercise \(\PageIndex{2c}\)
What are epimers?
Answer
Epimers are when only one chiral center is the opposite (in molecules with 2 or more chiral centers) in its diastereomer.
Exercise \(\PageIndex{3a}\)
Draw the structure of (2R,3R) 2-fluoro-3-methylhexane.
Answer
Exercise \(\PageIndex{3b}\)
Draw both diastereomers of (2R,3R) 2-fluoro-3-methylhexane.
Answer
Exercise \(\PageIndex{3c}\)
Draw the enantiomer of (2R,3R) 2-fluoro-3-methylhexane.
Answer
Exercise \(\PageIndex{4a}\)
Draw the structure of L-galactose, the enantiomer of D-galactose.
Answer
Exercise \(\PageIndex{4b}\)
Draw a diastereomer of D-galactose that is an epimer.
Answer
You can draw an epimer by drawing D-galactose with 1 (and only 1) of its chiral centers reversed. Here’s an example when you switch only the first chiral center (in red). (There are 3 other epimers that could be drawn as long as you only swap a single chiral center in the diastereomer that you use.)
Exercise \(\PageIndex{4c}\)
Identify if the following diastereomer of galactose is an epimer of D- galactose or L- galactose.
Answer
Since the diastereomer above only varies from L-galactose by 1 chiral center, the above is an epimer in relationship to L-galactose. Since it varies from D-galactose by 3 chiral centers, it is not an epimer but a diastereomer. Since not all of the chiral centers are swapped, it is not an enantiomer!
Exercise \(\PageIndex{5a}\)
For the compound shown below, label each chiral center as R or S.
Answer
Exercise \(\PageIndex{5b}\)
How many stereoisomers are possible for the compound in part a)?
Answer
Since there are 3 chiral centers, 23 = 8 possible stereoisomers.
Exercise \(6\)
Consider the stereoisomers below.
1. Which is/are an enantiomer of i?
2. Which is/are a diastereomer of ii?
3. Which is/are an epimer of i?
Answer
1. iv is an enantiomer of i since both chiral centers are switched and they are non superimposable mirror images.
2. i & iv are diastereomers of ii since they are stereoisomers that are not mirror images.
3. ii and iii are epimers of i since they are diastereomers with only 1 chiral center switched and the other one the same.
Exercise \(7\)
Consider the 8 stereoisomers below.
1. Which is/are an enantiomer of i?
2. Which is/are a diastereomer of i?
3. Which is/are an epimer of i?
Answer
1. v is an enantiomer since all three chiral centers are switched and they are non superimposable mirror images.
2. ii, iii, iv, vi, vii & viii are diastereomers of i since they are stereoisomers that are not mirror images.
3. ii,iii & viii are epimers of i since they are diastereomers with only 1 chiral center switched and the other chiral centers the same. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.5%3A_Molecules__Incorporating__Several__Stereocenters%3A_Diastereomers.txt |
Objectives
After completing this section, you should be able to
1. determine whether or not a compound containing two chiral carbon atoms will have a meso form, given its Kekulé, condensed or shorthand structure, or its IUPAC name.
2. draw wedge-and-broken-line structures for the enantiomers and meso form of a compound such as tartaric acid, given its IUPAC name, or its Kekulé, condensed or shorthand structure.
3. make a general comparison of the physical properties of the enantiomers, meso form and racemic mixture of a compound such as tartaric acid.
Key Terms
Make certain that you can define, and use in context, the key term below.
• meso compound
Study Notes
You may be confused by the two sets of structures showing “rotations.” Of course in each case the two structures shown are identical, they represent the same molecule looked at from two different perspectives. In the first case, there is a 120° rotation around the single carbon-carbon bond. In the second, the whole molecule is rotated 180° top to bottom.
Introduction
A meso compound is an achiral compound that has chiral centers. A meso compound contains an internal plane of symmetry which makes it superimposable on its mirror image and is optically inactive although it contains two or more stereocenters. Remember, an internal plane of symmetry was shown to make a molecule achiral in Section 5.2.
In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of reflected stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, resulting the molecule being optically inactive.
Identification
A meso compound must have:
1. Two or more stereocenters.
2. An internal plane of symmetry, or internal mirror, that lies in the compound.
3. Stereochemistry that cancels out. This means reflected stereocenter should have the same substituents and be inverted. For instance, in a meso compound with two stereocenters one should be R and the other S.
The compounds 2,3-dichlorobutane contains two chiral carbons and therefore would be expected to provide 22 = 4 different stereoisomers. These stereoisomers should be made up of two pairs of enantiomers.
After drawing out all the possible stereoisomers of 2,3-dichlorobutane, the pair on the right in the figure below are mirror images. Also, they are non-superimposable because they have distinctly different conformation (R,R & S,S). This makes the pair enantiomers of each other. However, the pair on the left represent a meso compound, they both are identical despite being mirror images.
Upon further investigation, the meso compound has an internal plane of symmetry which is not present in the pair of enantiomers. The plane of symmetry in the meso compound comes about because there are two chiral carbons present, both chiral carbons are identically substituted (Cl, H, CH3), and one chiral carbon is R and the other is S. Despite being represented as mirror images, both structures represent the same compound. This is best proven by making molecular models of both representations and then superimposing them. Overall, 2,3-dichlorobutane only has three possible stereosiomers, the pair of enantiomers and the meso compound.
When looking for an internal plane of symmetry, it is important to remember that sigma bonds (single bonds) can rotate. Just because the immediate representation of a molecule does not have a plane of symmetry does not mean that one cannot be obtained through rotation. Often the substituents attached to a stereocenter need to be rotated to recognize the internal plane of symmetry. As the stereocenter is rotated, its configuration does not change. Building a molecular model when considering a possible meso compound is an invaluable tool because it allows for easy rotation of chiral carbons. An example of how rotation of a chiral carbon can reveal an internal plane of symmetry is shown below.
Example \(1\)
Below are the two mirror images of (meso)-2,3-Butanediol. Because it is a meso compound, the two structures are identical. Show that both mirror images can be obtained by simply rotating the three-dimensional structure provided below.
Example \(2\)
1 has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; 1 has chiral centers. Thus, 1 is a meso compound.
Example \(3\)
This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.
Other Examples of Meso Compounds
Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cycloalkanes. Although two chiral carbons must be present, meso compounds can have many more. Notice that in every case a plane of symmetry is present.
In general, a disubstituted cycloalkane is meso if the two substituents are the same and they are in a cis conformation. Trans disubstituted cycloalkanes are not meso regardless if the two substituents are the same.
Optical Activity Analysis of a Meso Compound
When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+) because a meso compound is achiral (optically inactive). Investigations of isomeric tartaric acid (2,3-dihydroxybutanedioic acid), carried out by Louis Pasteur in the mid 19th century, were instrumental in elucidating some of the subtleties of stereochemistry. Tartaric acid, has two chiral but only three stereoisomers. Two of these stereoisomers are enantiomers and the third is an achiral a meso compound. Some physical properties of these stereoisomers of tartaric acid are given in the table below. Notice that the enantiomers have the same amount of optical rotation but in different directons. Meso-tartaric acid produces no optical rotation because it is achiral and not optically active. Meso-tartaric acid is actually a diastereomer of both (-) and (+)-tartaric acid, which gives it a distinctly different melting point.
(+)-tartaric acid: [α]D = +13º m.p. 172 ºC
(–)-tartaric acid: [α]D = –13º m.p. 172 ºC
meso-tartaric acid: [α]D = 0º m.p. 140 ºC
Exercise \(1\)
1) Determine which of the following molecules are meso.
2) Explain why 2,3-dibromobutane has the possibility of being a meso compound while 2,3-dibromopentane does not.
3) Observe the following compound and determine if it is a meso compound. If so indicate the plane of symmetry. Red = oxygen. Remember sigma bonds are able to rotate.
Answer
1) A C, D, E are meso compounds.
2) One of the requirements of a meso compound is that the reflected chiral carbons have the same substituents. The compound 2,3-dibromobutane, fulfills this requirement (Br, H, CH3) and can possibly be a meso compound if the two chiral carbons have the appropriate configuration (R & S). The substituents of the two chiral carbons in 2,3-dibromopentane do not have the same substituents (Br, H, CH3 vs. Br, H, CH2CH3). This 2,3-dibromopentane cannot form a meso compound regardless of the configurations of its chiral carbons.
3) The compound is meso.
Exercise \(1\)
Which of the following are meso compounds?
Answer
a) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) between the C’s (and it has stereochemistry of S & R).
b) This is not a meso compound. No matter how you rotate the C-C bond, you do not see a plane of symmetry (and its stereochemistry is S & S)
c) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) that can be seen when you rotate the C-C bond (and it has stereochemistry of S & R).
Exercise \(2\)
Which of the following are meso compounds?
Answer
a) This is not a meso compound. There is no plane of symmetry and has stereochemistry of S & S.
b) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) and it has stereochemistry of R & S.
c) This is not a meso compound (even though it has planes of symmetry). The plane of symmetry shown in red makes it so that both chiral centers have symmetrical groups (the ring) and thus the compound is not chiral (so it can’t be a meso compound).
Exercise \(3\)
Which of the following are meso compounds?
Answer
a) This is a meso compound. If you rotate between the C-C bond, you can see that it has a mirror plane between the C’s (shown in red on the structure to the right). Notice how rotating a C-C bond doesn’t change the stereochemistry of the molecule (S & R).
b) This is a meso compound. You can see the plane of symmetry in red and the compound has stereochemistry of S & R.
c) This is not a meso compound. There is no plane of symmetry and it has stereochemistry of S & S.
Exercise \(4\)
Determine (and draw) if any of the forms of 3,4-dichlorohexane are a meso compound.
Answer
Looking at the 4 different possibilities below, i & ii are equivalent structures (with R & S stereochemistry) so it is a meso compound. iii & iv are not meso compounds but are enantiomers to each other. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.6%3A_Meso__Compounds.txt |
Objectives
After completing this section, you should be able to
1. identify a compound as being prochiral.
2. identify the Re and Si faces of prochiral sp2 centre.
3. identify atoms (or groups of atoms) as pro-R or pro-S on a prochiral sp3 centre.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• prochiral
• pro-R
• pro-S
• Re
• Si
Prochiral Carbons
When a tetrahedral carbon can be converted to a chiral center by changing only one of the attached groups, it is referred to as a ‘prochiral' carbon. The two hydrogens on the prochiral carbon can be described as 'prochiral hydrogens'.
Note that if, in a 'thought experiment', we were to change either one of the prochiral hydrogens on a prochiral carbon center to a deuterium (the 2H isotope of hydrogen), the carbon would now have four different substituents and thus would be a chiral center.
Prochirality is an important concept in biological chemistry, because enzymes can distinguish between the two ‘identical’ groups bound to a prochiral carbon center due to the fact that they occupy different regions in three-dimensional space. Consider the isomerization reaction below, which is part of the biosynthesis of isoprenoid compounds. We do not need to understand the reaction itself (it will be covered in chapter 14); all we need to recognize at this point is that the isomerase enzyme is able to distinguish between the prochiral 'red' and the 'blue' hydrogens on the isopentenyl diphosphate (IPP) substrate. In the course of the left to right reaction, IPP specifically loses the 'red' hydrogen and keeps the 'blue' one.
Prochiral hydrogens can be unambiguously designated using a variation on the R/S system for labeling chiral centers. For the sake of clarity, we'll look at a very simple molecule, ethanol, to explain this system. To name the 'red' and 'blue' prochiral hydrogens on ethanol, we need to engage in a thought experiment. If we, in our imagination, were to arbitrarily change red H to a deuterium, the molecule would now be chiral and the chiral carbon would have the R configuration (D has a higher priority than H).
For this reason, we can refer to the red H as the pro-R hydrogen of ethanol, and label it HR. Conversely, if we change the blue H to D and leave red H as a hydrogen, the configuration of the molecule would be S, so we can refer to blue H as the pro-S hydrogen of ethanol, and label it HS.
Looking back at our isoprenoid biosynthesis example, we see that it is specifically the pro-R hydrogen that the isopentenyl diphosphate substrate loses in the reaction.
Prochiral hydrogens can be designated either enantiotopic or diastereotopic. If either HR or HS on ethanol were replaced by a deuterium, the two resulting isomers would be enantiomers (because there are no other stereocenters anywhere on the molecule).
Thus, these two hydrogens are referred to as enantiotopic.
In (R)-glyceraldehyde-3-phosphate ((R)-GAP), however, we see something different:
R)-GAP already has one chiral center. If either of the prochiral hydrogens HR or HS is replaced by a deuterium, a second chiral center is created, and the two resulting molecules will be diastereomers (one is S,R, one is R,R). Thus, in this molecule, HR and HS are referred to as diastereotopic hydrogens.
Finally, hydrogens that can be designated neither enantiotopic nor diastereotopic are called homotopic. If a homotopic hydrogen is replaced by deuterium, a chiral center is not created. The three hydrogen atoms on the methyl (CH3) group of ethanol (and on any methyl group) are homotopic. An enzyme cannot distinguish among homotopic hydrogens.
Example \(1\)
Identify in the molecules below all pairs/groups of hydrogens that are homotopic, enantiotopic, or diastereotopic. When appropriate, label prochiral hydrogens as HR or HS.
Answer
Groups other than hydrogens can be considered prochiral. The alcohol below has two prochiral methyl groups - the red one is pro-R, the blue is pro-S. How do we make these designations? Simple - just arbitrarily assign the red methyl a higher priority than the blue, and the compound now has the R configuration - therefore red methyl is pro-R.
Citrate is another example. The central carbon is a prochiral center with two 'arms' that are identical except that one can be designated pro-R and the other pro-S.
In an isomerization reaction of the citric acid (Krebs) cycle, a hydroxide is shifted specifically to the pro-R arm of citrate to form isocitrate: again, the enzyme catalyzing the reaction distinguishes between the two prochiral arms of the substrate (we will study this reaction in chapter 13).
Exercise \(1\)
Assign pro-R and pro-S designations to all prochiral groups in the amino acid leucine. (Hint: there are two pairs of prochiral groups!). Are these prochiral groups diastereotopic or enantiotopic?
Answer
Prochiral Carbonyl and Imine Groups
Trigonal planar, sp2-hybridized carbons are not, as we well know, chiral centers– but they can be prochiral centers if they are bonded to three different substitutuents. We (and the enzymes that catalyze reactions for which they are substrates) can distinguish between the two planar ‘faces’ of a prochiral sp2 - hybridized group. These faces are designated by the terms re and si. To determine which is the re and which is the si face of a planar organic group, we simply use the same priority rankings that we are familiar with from the R/S system, and trace a circle: re is clockwise and si is counterclockwise.
When the two groups adjacent to a carbonyl (C=O) are not the same, we can distinguish between the re and si 'faces' of the planar structure. The concept of a trigonal planar group having two distinct faces comes into play when we consider the stereochemical outcome of a nucleophilic addition reaction. Nucleophilic additions to carbonyls will be covered in greater detail in Chapter 19. Notice that in the course of a carbonyl addition reaction, the hybridization of the carbonyl carbon changes from sp2 to sp3, meaning that the bond geometry changes from trigonal planar to tetrahedral. If the two R groups are not equivalent, then a chiral center is created upon addition of the nucleophile. The configuration of the new chiral center depends upon which side of the carbonyl plane the nucleophile attacks from. Reactions of this type often result in a 50:50 racemic mixture of stereoisomers, but it is also possible that one stereoisomer may be more abundant, depending on the structure of the reactants and the conditions under which the reaction takes place.
Below, for example, we are looking down on the re face of the ketone group in pyruvate. If we flipped the molecule over, we would be looking at the si face of the ketone group. Note that the carboxylate group does not have re and si faces, because two of the three substituents on that carbon are identical (when the two resonance forms of carboxylate are taken into account).
As we will see in chapter 10, enzymes which catalyze reactions at carbonyl carbons act specifically from one side or the other.
We need not worry about understanding the details of the reaction pictured above at this point, other than to notice the stereochemistry involved. The pro-R hydrogen (along with the two electrons in the C-H bond) is transferred to the si face of the ketone (in green), forming, in this particular example, an alcohol with the R configuration. If the transfer had taken place at the re face of the ketone, the result would have been an alcohol with the S configuration.
Exercise \(2\)
For each of the carbonyl groups in uracil, state whether we are looking at the re or the si face in the structural drawing below.
Answer
Exercise \(3\)
a) State which of the following hydrogen atoms are pro-R or pro-S.
b) Identify which side is Re or Si.
Answer
a) Left compound: Ha = pro-S and Hb = pro-R; Right compound: Ha = pro-R and Hb = pro-S
b) A – Re; B – Si; C – Re; D – Si\)
Exercise \(4\)
State whether the H's indicated below are pro-R or pro-S for the following structures.
Answer
a) Ha is pro-R; Hb is pro-S
Ha is pro-R; Hb is pro-S
Exercise \(5\)
In the structures below, determine if the H's are homotopic, enantiotopic, or diastereotopic.
Answer
In a), the CH2 is diastereotopic since there is another chiral center on the molecule. Both CH3's are homotopic since replacing one of them doesn't create a chiral center.
IN b), the CH2's are enantiotopic since it would create the only chiral center on the molecule. Both CH3's are homotopic since replacing one of them doesn't create a chiral center.
Exercise \(6\)
State whether you are looking down at the molecule from the re face or si face.
Answer
1. You are looking at the si face. The re face would be if you were facing the molecule from the back.
2. You are looking at the re face. The si face would be if you were facing the molecule from the back. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.7%3A_Stereochemistry_in_Chemical__Reactions.txt |
Objectives
After completing this section, you should be able to
1. describe a common process for separating a mixture of enantiomers.
2. explain why racemic mixtures do not rotate plane-polarized light.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• racemic mixture (or racemate)
• resolve
Study Notes
A racemic mixture is a 50:50 mixture of two enantiomers. Because they are mirror images, each enantiomer rotates plane-polarized light in an equal but opposite direction and is optically inactive. If the enantiomers are separated, the mixture is said to have been resolved. A common experiment in the laboratory component of introductory organic chemistry involves the resolution of a racemic mixture.
The dramatic biochemical consequences of chirality are illustrated by the use, in the 1950s, of the drug Thalidomide, a sedative given to pregnant women to relieve morning sickness. It was later realized that while the (+)‑form of the molecule, was a safe and effective sedative, the (−)‑form was an active teratogen. The drug caused numerous birth abnormalities when taken in the early stages of pregnancy because it contained a mixture of the two forms.
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. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as 'Moscher's esters', after Harry Stone Moscher, a chemist who pioneered the method at Stanford University.
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. Figure 5.8.1 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.
Because the physical properties of enantiomers are identical, they seldom can be separated by simple physical methods, such as fractional crystallization or distillation. It is only under the influence of another chiral substance that enantiomers behave differently, and almost all methods of resolution of enantiomers are based upon this fact. We include here a discussion of the primary methods of resolution.
Chiral Amines as Resolving Agents and Resolution of Racemic Acids
The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomers that will have different physical properties: melting point, boiling point, solubility, and so on (Section 5-5). For example, if you have a racemic or (R)/(S) mixture of enantiomers of a carboxylic acid and convert this to a salt with a chiral amine base having the (R) configuration, the salt will be a mixture of two diastereomers, ((R)-acid・(R)-base) and ((S)-acid・(R)-base). These diastereomeric salts are not identical and they are not mirror images. Therefore they will differ to some degree in their physical properties, and a separation by physical methods, such as crystallization, may be possible. If the diastereomeric salts can be completely separated, the carboxylic acid regenerated from each salt will be either exclusively the (R) or the (S) enantiomer.
Resolution of chiral acids through the formation of diastereomeric salts requires adequate supplies of suitable chiral bases. Brucine, strychnine, and quinine frequently are used for this purpose because they are readily available, naturally occurring chiral bases. Simpler amines of synthetic origin, such as 2-amino-1-butanol, amphetamine, and 1-phenylethanamine, also can be used, but first they must be resolved themselves.
Worked Example $1$
Show how (S)-1-phenylethylamine can be used to resolve a racemic mixture of lactic acid. Please draw all the structures involved.
Answer
Resolution of Racemic Bases
Chiral acids, such as (+)-tartaric acid, (-)-malic acid, (-)-mandelic acid, and (+)-camphor- 10-sulfonic acid, are used for the resolution of a racemic base.
The principle is the same as for the resolution of a racemic acid with a chiral base, and the choice of acid will depend both on the ease of separation of the diastereomeric salts and, of course, on the availability of the acid for the scale of the resolution involved. Resolution methods of this kind can be tedious, because numerous recrystallizations in different solvents may be necessary to progressively enrich the crystals in the less-soluble diastereomer. To determine when the resolution is complete, the mixture of diastereomers is recrystallized until there is no further change in the measured optical rotation of the crystals. At this stage it is hoped that the crystalline salt is a pure diastereomer from which one pure enantiomer can be recovered. The optical rotation of this enantiomer will be a maximum value if it is "optically" pure because any amount of the other enantiomer could only reduce the magnitude of the measured rotation $\alpha$.
Resolution of Racemic Alcohols
To resolve a racemic alcohol, a chiral acid can be used to convert the alcohol to a mixture of diastereomeric esters. This is not as generally useful as might be thought because esters tend to be liquids unless they are very high-molecularweight compounds. If the diastereomeric esters are not crystalline, they must be separated by some other method than fractional crystallization (for instance, by chromatography methods, Section 9-2). Two chiral acids that are useful resolving agents for alcohols are:
The most common method of resolving an alcohol is to convert it to a half-ester of a dicarboxylic acid, such as butanedioic (succinic) or 1,2-benzenedicarboxylic (phthalic) acid, with the corresponding anhydride. The resulting half-ester has a free carboxyl function and may then be resolvable with a chiral base, usually brucine:
Other Methods of Resolution
One of the major goals in the field of organic chemistry is the development of reagents with the property of "chiral recognition" such that they can affect a clean separation of enantiomers in one operation without destroying either of the enantiomers. We have not achieved that ideal yet, but it may not be far in the future. Chromatographic methods, whereby the stationary phase is a chiral reagent that adsorbs one enantiomer more strongly than the other, have been used to resolve racemic compounds, but such resolutions seldom have led to both pure enantiomers on a preparative scale. Other methods, called kinetic resolutions, are excellent when applicable. The procedure takes advantage of differences in reaction rates of enantiomers with chiral reagents. One enantiomer may react more rapidly, thereby leaving an excess of the other enantiomer behind. For example, racemic tartaric acid can be resolved with the aid of certain penicillin molds that consume the dextrorotatory enantiomer faster than the levorotatory enantiomer. As a result, almost pure (-)-tartaric acid can be recovered from the mixture:
(±)-tartaric acid + mold $\rightarrow$ (-)-tartaric acid + more mold
The crystallization procedure employed by Pasteur for his classical resolution of (±)-tartaric acid (Section 5.4) has been successful only in a very few cases. This procedure depends on the formation of individual crystals of each enantiomer. Thus if the crystallization of sodium ammonium tartrate is carried out below 27", the usual racemate salt does not form; a mixture of crystals of the (+) and (-) salts forms instead. The two different kinds of crystals, which are related as an object to its mirror image, can be separated manually with the aid of a microscope and subsequently may be converted to the tartaric acid enantiomers by strong acid. A variation on this method of resolution is the seeding of a saturated solution of a racemic mixture with crystals of one pure enantiomer in the hope of causing crystallization of just that one enantiomer, thereby leaving the other in solution. Unfortunately, very few practical resolutions have been achieved in this way.
Predicating the Chirality of the Product of a Reaction
It important to understand the changes in chirality which occur during the formation of product during a reaction. A chiral reaction product, has the possibility of forming multiple stereoisomers which all need to be considered. Changes in chirality, if possible, will be discussed with each individual reaction as this textbook moves forward. Some possible situations which can occur are:
• A new chiral carbon is formed during a reaction. This commonly occurs when an sp2 hybridized carbon in the reactant is converted to sp3 hybridized chiral carbon in the product. When this occurs, a racemic mixture of the new chiral carbon is formed.
• A chiral carbon is lost during a reaction. This commonly occurs when an sp3 hybridized chiral carbon in the reactant is converted to either a sp2 or sp hybridized carbon in the product.
• An enantiomerically pure starting material is converted to a racemic mixture in the product. This commonly occurs when a sp3 hybridized chiral carbon is temporarily converted to an sp2 hybridized carbon during a reaction's mechanism. The chiral carbon is reformed as a racemic mixture.
• Chiral carbons remain unchanged during a reaction. If a chiral carbon is not directly involved in a reaction, it will move from a reactant to a product unchanged.
Determining if a chiral carbon is involved in a given reaction is vital for determining which of these four situations is occurring.
Worked Example $2$
The following reaction involves the conversion of a carboxylic acid reacting with an alcohol to form an ester. If a pure sample of (R)-2-methylbutanoic acid is reacted with methanol to form an ester, what would be the stereochemistry of the product?
Answer
First it is important to identify the location of the chiral carbon and determine if it is directly involved in the reaction. In this case, the chiral carbon is not involved so the stereochemistry will be carried over into the product unchanged.
Exercise $1$
Indicate the reagents you could use to resolve the following compounds. Show the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 1 -phenyl-2-propanamine.
Answer
You could react the 1-phenyl-2-propanamine racemic mixture with a chiral acid such as (+)-tartaric acid (R, R). The reaction will produce a mixture of diastereomeric salts (i.e. R, R, R and S, R, R). You can separate the diastereomers through crystallization and treat the salt with a strong base (e.g. KOH) to recover the pure enantiomeric amine.
Exercise $2$
Indicate the reagents you would use to resolve the following and discuss the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 2,3-pentadienedioic acid.
Answer
You could react the 2,3-pentadienedioic acid mixture with a chiral base such as (R)‑1‑phenylethylamine. The reaction will produce a mixture of diastereomeric salts. Separate the diastereomers through crystallization and treat the resulting salt with strong acid (e.g. HCl) to recover the pure enantiomeric acid.
Exercise $3$
Indicate the reagents you would use to resolve the following and discuss the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 1 -phenylethanol.
Answer
You could react the 1-phenylethanol mixture with 1,2-benzenedicarboxylic anhydride. The reaction will produce a mixture of diastereomeric salts. You could then separate the diastereomers through crystallization and then alkaline hydrolysis treatment should recover the pure enantiomeric alcohol. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/05._Stereoisomers/5.8%3A_Resolution%3A_Separation_of_Enantiomers.txt |
Objectives
After completing this section, you should be able to
1. write the IUPAC name of a halogenated aliphatic hydrocarbon, given its Kekulé, condensed or shorthand structure.
2. draw the Kekulé, condensed or shorthand structure of a halogenated aliphatic hydrocarbon, given it IUPAC name.
3. write the IUPAC name and draw the Kekulé, condensed or shorthand structure of a simple alkyl halide, given a systematic, non-IUPAC name (e.g., sec-butyl iodide).
4. arrange a given series of carbon-halogen bonds in order of increasing or decreasing length and strength.
Study Notes
This section contains little that is new. If you mastered the IUPAC nomenclature of alkanes, you should have little difficulty in naming alkyl halides. Notice that when a group such as CH2Br must be regarded as a substituent, rather than as part of the main chain, we may use terms such as bromomethyl.
You will find it easier to understand the reactions of the alkyl halides if you keep the polarity of the C-X bond fixed permanently in your mind (see ”The Polar C-X Bond” shown in the reading below).
Alkyl halides are also known as haloalkanes. This page explains what they are and discusses their physical properties. alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). For example:
Halide Designations
Alkyl halides fall into different classes depending on how many alkyl groups are attached to the carbon which holds the halogen. There are some chemical differences between the various types. When there are no alkyl groups attached to the carbon holding the halogen, these are considered methyl halides (CH3X).
Primary alkyl halides
In a primary (1°) haloalkane, the carbon which carries the halogen atom is only attached to one other alkyl group. Some examples of primary alkyl halides include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the halogen. There is an exception to this: CH3Br and the other methyl halides are often counted as primary alkyl halides even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary alkyl halides
In a secondary (2°) haloalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary alkyl halides
In a tertiary (3°) haloalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Example $1$
Please indicate if the following haloalkanes are methyl, 1o, 2o, or 3o:
a) CH3I
b) CH3CH2Br
c)
d)
e)
f)
a) methyl
b) 1o
c) 2o
d) 3o
e) 1o
f) 2o
Nomenclature of Alkyl Halides
Alkyl halides are systematically named as alkanes (Section 3-4) where the halogen is a substituent on the parent alkane chain. To summarize the rules discussed in detail in Section 3-4, there are three basic steps to naming alkyl halides.
1. Find and name the longest carbon chain and name it as the parent. Remember is an alkene or alkyne is present, the parent chain must contain both carbons of the multiple bond.
2. Number the parent chain consecutively, starting at the end nearest a substituent group. Then assign each substituent a number. Remember the IUPAC system uses prefix to indicate the halogen followed by the suffix -ide. The prefixes are fluoro- for fluorine, chloro- for chlorine, bromo- from bromine, and iodo- for iodine. The name of a halogen is preceded by a number indicating the substituent’s location on the parent chain.
1. If there is an ambiguity in numbering the parent chain, begin on the end which is closer to the substituent which comes first alphabetically.
Common Names of Alkyl Halides
Alkyl halides with simple alkyl groups are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names. The common names of alkyl halides consist of two parts: the name of the alkyl group plus the first syllable of the name of the halogen, with the ending -ide. The names of common alkyl groups are listed in Section 3.3.
Exercise $1$
1) Give the common and IUPAC names for each compound.
1. CH3CH2CH2Br
2. (CH3)2CHCl
3. CH3CH2I
4. CH3CH2CH2CH2F
2) Give the IUPAC name for each compound.
a)
b)
c)
d)
Answer
1) a) The alkyl group (CH3CH2CH2–) is a propyl group, and the halogen is bromine (Br). The common name is therefore propyl bromide. For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane.
b) The alkyl group [(CH3)2CH–] has three carbon atoms, with a chlorine (Cl) atom attached to the middle carbon atom. The alkyl group is therefore isopropyl, and the common name of the compound is isopropyl chloride. For the IUPAC name, the Cl atom (prefix chloro-) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane.
c) The alkyl group (CH3CH2–) is a ethyl group, and the halogen is iodide (I). The common name is therefore ethyl iodide. For the IUPAC name, the prefix for Iodide (Iodod) is combined with the name for a two-carbon chain (ethane), preceded by a number identifying the carbon atom to which the I atom is attached, so the IUPAC name is 1-bromoethane
d) The alkyl group (CH3CH2CH2CH2–) is a butyl group, and the halogen is fluorine (F). The common name is therefore butyl fluoride. For the IUPAC name, the prefix for Fluorine (Fluoro) is combined with the name for a four-carbon chain (butane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-fluorobutane.
2) a) The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane.
b) The parent alkane is hexane. Methyl (CH3) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane.
c) 2-Chloro-3-methylbutane
d) 1-Bromo-2-chloro-4-methylpentane rine (F). The common name is therefore butyl fluoride. For the IUPAC name, the prefix for Fluorine (Fluoro) is combined with the name for a four-carbon chain (butane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-fluorobutane.
There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. As mentioned above, the structural differences are due to the replacement of one or more hydrogens with a halogen atom. The differences in physical properties are a result of factors such as electronegativity, bond length, bond strength, and molecular size.
Halogens and the Character of the Carbon-Halogen Bond
As discussed in Section 6. 4, halogens are more electronegative than carbon. This results in a carbon-halogen bond that is polarized with the carbon atom bearing a partial positive charge and the halogen a partial negative charge. This polarity can be distinctly seen when viewing the electrostatic potential map of a methyl halide. Electron density is shown by a red/yellow color which is almost exclusively around the halogen atom. The methyl portion of the compound lacks electron density which is shown by a blue/green color.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
Haloalkanes Have Higher Boiling Points than Alkanes
When comparing alkanes and haloalkanes, we will see that haloalkanes have higher boiling points than alkanes containing the same number of carbons. London dispersion forces are the first of two types of forces that contribute to this physical property. You might recall from general chemistry that London dispersion forces increase with molecular surface area. In comparing haloalkanes with alkanes, haloalkanes exhibit an increase in surface area due to the substitution of a halogen for hydrogen. The increase in surface area leads to an increase in London dispersion forces, which then results in a higher boiling point.
Dipole-dipole interaction is the second type of force that contributes to a higher boiling point. As you may recall, this type of interaction is a coulombic attraction between the partial positive and partial negative charges that exist between carbon-halogen bonds on separate haloalkane molecules. Similar to London dispersion forces, dipole-dipole interactions establish a higher boiling point for haloalkanes in comparison to alkanes with the same number of carbons.
The table below illustrates how boiling points are affected by some of these properties. Notice that the boiling point increases when hydrogen is replaced by a halogen, a consequence of the increase in molecular size, as well as an increase in both London dispersion forces and dipole-dipole attractions. The boiling point also increases as a result of increasing the size of the halogen, as well as increasing the size of the carbon chain.
Solubility
Solubility in water
The alkyl halides are at best only slightly soluble in water. For a haloalkane to dissolve in water you have to break attractions between the haloalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy. Energy is released when new intermolecular forces are generated between the haloalkane molecules and water molecules. These will only be dispersion forces and dipole-dipole interactions. These are not as strong as the original hydrogen bonds in the water, and so not as much energy is released as was used to separate the water molecules. The energetics of the change are sufficiently "unprofitable" that very little dissolves.
Solubility in organic solvents
Alkyl halides tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate haloalkane and solvent.
Chemical Reactivity
The pattern in strengths lies in the strength of the bond between the carbon atom and the halogen atom. Previously in this section, it was noted that the trend for bond strength increases from C-I to C-Br to C-Cl with C-F bonds being the strongest. To react with the alkyl halides, the carbon-halogen bond has got to be broken. Because that gets easier as you go from fluoride to chloride to bromide to iodide, the compounds get more reactive in that order. Iodoalkanes are the most reactive and fluoroalkanes are the least. In fact, fluoroalkanes are so unreactive that we will ignore them completely from now on in this section!
Exercise $2$
1) Give the names of the following organohalides:
2) Draw the structures of the following compounds:
a) 2-Chloro-3,3-dimethylpentane
b) 1,1-Dichloro-4-isopropylcyclohexane
c) 3-bromo-3-ethylhexane
Answer
1)
a) 5-ethyl-4-iodo-3methyl-octane
b) 1-bromo-2,3,4-trimethyl-pentane
c) 4-bromo-5-chloro-2-methyl-heptane
2) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6-01_Physical__Properties_of_Haloalkanes.txt |
Halogens and the Character of the Carbon-Halogen Bond
With respect to electronegativity, halogens are more electronegative than carbons. This results in a carbon-halogen bond that is polarized. As shown in the image below, carbon atom has a partial positive charge, while the halogen has a partial negative charge.
The following image shows the relationship between the halogens and electronegativity. Notice, as we move up the periodic table from iodine to fluorine, electronegativity increases.
The following image shows the relationships between bond length, bond strength, and molecular size. As we progress down the periodic table from fluorine to iodine, molecular size increases. As a result, we also see an increase in bond length. Conversely, as molecular size increases and we get longer bonds, the strength of those bonds decreases.
The influence of bond polarity
Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The electronegativities of carbon and iodine are equal and so there will be no separation of charge on the bond.
One of the important set of reactions of alkyl halides involves replacing the halogen by something else - substitution reactions. These reactions involve either:
• the carbon-halogen bond breaking to give positive and negative ions. The ion with the positively charged carbon atom then reacts with something either fully or slightly negatively charged.
• something either fully or negatively charged attracted to the slightly positive carbon atom and pushing off the halogen atom.
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one.
Contributors
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.
In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples.
Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. In the same way, we will see later that nucleophiles and leaving groups are sometimes protonated and sometimes not, so for now, for the sake of simplicity, we will not include protons on 'Nu' or 'X'. We will generalize the three other groups bonded on the electrophilic central carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'.
Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction:
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 on the right. 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.
In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation.
One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. If R- has no beta-hydrogens an elimination reaction is not possible, unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds.
It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Using the general reaction shown above as our reference, we can identify the following variables and observables.
Variables
R change α-carbon from 1º to 2º to 3º
if the α-carbon is a chiral center, set as (R) or (S)
X change from Cl to Br to I (F is relatively unreactive)
Nu: change from anion to neutral; change basicity; change polarizability
Solvent polar vs. non-polar; protic vs. non-protic
Observables
Products substitution, elimination, no reaction.
Stereospecificity if the α-carbon is a chiral center what happens to its configuration?
Reaction Rate measure as a function of reactant concentration.
When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: only one variable should be changed at a time, the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C2H5–CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6-02_%09Nucleophilic_Substitution.txt |
Objectives
After completing this section, you should be able to
1. explain the difference between heterolytic and homolytic bond breakage, and between heterogenic and homogenic bond formation.
2. state the two reaction types involved in symmetrical and unsymmetrical processes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• heterogenic
• heterolytic
• homogenic
• homolytic
• polar reaction
• radical reaction
• reaction mechanism
Study Notes
Upon first reading first four key terms, it is easy to be puzzled. The ending of the word tells you whether a bond is being formed (‑genic) or broken (‑lytic), while the root of the word describes the nature of that formation or decomposition. So hetero (meaning different) reactions involve asymmetrical bond making (or breaking) and homo (meaning same) involve symmetrical processes.
Because one pair of electrons constitutes a single bond, the unsymmetrical making or breaking of that bond in a hetero processes are described as polar reactions. Similarly, symmetrical homo processes of bond making and breaking are called radical reactions. Radicals (sometimes referred to as free radicals) are highly reactive neutral chemical species with one unpaired electron. In later sections we discuss radical and polar reactions in more detail.
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding (and non-bonding) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow The Equilibrium Arrow The Resonance Arrow
The following equations illustrate the proper use of these symbols:
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
Charged Intermediates Uncharged Intermediates
a carbocation
a radical
a carbanion
a carbene
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
• 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 bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined below.
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies.
Ionic Reactions
The principles and terms introduced in the previous sections can now be summarized and illustrated by the following three examples. Reactions such as these are called ionic or polar reactions, because they often involve charged species and the bonding together of electrophiles and nucleophiles. Ionic reactions normally take place in liquid solutions, where solvent molecules assist the formation of charged intermediates.
The substitution reaction shown on the left can be viewed as taking place in three steps. The first is an acid-base equilibrium, in which HCl protonates the oxygen atom of the alcohol. The resulting conjugate acid then loses water in a second step to give a carbocation intermediate. Finally, this electrophile combines with the chloride anion nucleophile to give the final product.
The addition reaction shown on the left can be viewed as taking place in two steps. The first step can again be considered an acid-base equilibrium, with the pi-electrons of the carbon-carbon double bond functioning as a base. The resulting conjugate acid is a carbocation, and this electrophile combines with the nucleophilic bromide anion.
The elimination reaction shown on the left takes place in one step. The bond breaking and making operations that take place in this step are described by the curved arrows. The initial stage may also be viewed as an acid-base interaction, with hydroxide ion serving as the base and a hydrogen atom component of the alkyl chloride as an acid.
rearrangement (tautomerism)
There are many kinds of molecular rearrangements. The examples shown on the left are from an important class called tautomerization or, more specifically, keto-enol tautomerization. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
The Brønsted-Lowry definition of acidity
We’ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these definitions was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry, and has come to be known as the Brønsted-Lowry definition of acids and bases. An acid, by the Brønsted-Lowry definition, is a species which is able to donate a proton (H+), while a base is a proton acceptor. We have already discussed in the previous chapter one of the most familiar examples of a Brønsted-Lowry acid-base reaction, between hydrochloric acid and hydroxide ion:
In this reaction, a proton is transferred from HCl (the acid, or proton donor) to hydroxide (the base, or proton acceptor). As we learned in the previous chapter, curved arrows depict the movement of electrons in this bond-breaking and bond-forming process.
After a Brønsted-Lowry acid donates a proton, what remains – in this case, a chloride ion – is called the conjugate base. Chloride is thus the conjugate base of hydrochloric acid. Conversely, when a Brønsted-Lowry base accepts a proton it is converted into its conjugate acid form: water is thus the conjugate acid of hydroxide.
We can also talk about conjugate acid/base pairs: the two acid/base pairs involved in our first reaction are hydrochloric acid/chloride and hydroxide/water. In this next acid-base reaction, the two pairs involved are acetate/acetic acid and methyl ammonium/methylamine:
Throughout this text, we will often use the abbreviations HA and :B in order to refer in a general way to acidic and basic reactants:
In order to act as a proton acceptor, a base must have a reactive pair of electrons. In all of the examples we shall see in this chapter, this pair of electrons is a non-bonding lone pair, usually (but not always) on an oxygen, nitrogen, sulfur, or halogen atom. When acetate acts as a base in the reaction shown above, for example, one of its oxygen lone pairs is used to form a new bond to a proton. The same can be said for an amine acting as a base. Clearly, methyl ammonium ion cannot act as a base – it does not have a reactive pair of electrons with which to accept a new bond to a proton.
Later, in chapter 15, we will see several examples where the (relatively) reactive pair of electrons in a $\pi$ bond act in a basic fashion.
In this chapter, we will concentrate on those bases with non-bonding (lone pair) electrons.
Example
Exercise 7.1: Draw structures for the missing conjugate acids or conjugate bases in the reactions below. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6-03_Reaction__Mechanisms_Involving__Polar__Functional_Groups%3A_Using_.txt |
Objectives
After completing this section, you should be able to
1. write an expression relating reaction rate to the concentration of reagents for a second-order reaction.
2. determine the order of a chemical reaction from experimentally obtained rate data.
3. describe the essential features of the SN2 mechanism, and draw a generalized transition state for such a reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• bimolecular
• kinetics
• rate coefficient
• rate equation
• reaction rate
• second-order reaction
• SN2
Study Notes
Most of the key terms introduced in this section should already be familiar to you from your previous general chemistry course.
Reaction rate refers to the change in concentration of a reactant or product per unit of time. Using strict SI units, reaction rates are expressed in mol · L−1 · s−1, but in some textbooks you will find this value written as M/s. In general, the reaction rate of a given reaction changes with time, as it is dependent on the concentration of one or more of the reactants.
An equation which shows the relationship between the reaction rate and the concentrations of the reactants is known as the rate equation. All rate equations contain a proportionality constant, usually given the symbol k, which is known as the rate coefficient. Some textbooks refer to this value as the “rate constant,” but this name is a little misleading as it is not a true constant. The rate coefficient of a given reaction depends on such factors as temperature and the nature of the solvent.
SN2 is short for “bimolecular nucleophilic substitution.” You will encounter abbreviations for other types of reactions later in this chapter.
If you are unclear on the point about the inversion of configuration during an SN2 reaction, construct a molecular model of a chiral alkyl halide, the transition state formed when this substance reacts with a nucleophile in an SN2 process, and the product obtained from this reaction.
Brønsted-Lowry acid-base reactions
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base. Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. We will generalize the three other groups bonded on the electrophilic central carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction.
The SN2 Mechanism
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electrophilic carbon and the halogen. 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. The mechanism starts when lone pair electrons from the nucleophile attacks the electrophilic carbon of the alkyl halide to form a C-Nu sigma bond. Simultaneously, X-C bond is broken when the electrons are pushed onto the leaving group. Overall during this mechanism, a set of lone pair electrons are transferred from the nucleophile to the leaving groups.
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. 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.
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.
Bimolecular Nucleophilic Substitution Reactions and Kinetics
In the term SN2, (as previously stated) the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics:
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.
Exercise \(1\)
1) The reaction below follows the SN2 mechanism.
a) Write the rate law for this reaction.
b) Determine the value of the rate coefficient, k, if the initial concentrations are 0.01 M CH3Cl, 0.01 M NaOH, and the initial reaction rate is 6 x 10-10 M/s.
c) Calculate the initial reaction rate if the initial reactant concentrations are changed to 0.02 M CH3Cl and 0.0005 M NaOH.
2) Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH3CO2-, Show stereochemistry.
3) Predict the structure of the product in this SN2 reaction. Be sure to specify stereochemistry.
4) Since everything is relative in chemistry, one reaction's nucleophile can be another reaction's leaving group. Some functional groups can only react as a nucleophile or electrophile, while other functional groups can react as either a nucleophile or electrophile depending on the reaction conditions. Classify the following compounds as nucleophile, electrophile, or leaving groups. More than one answer may be possible.
a) bromoethane
b) hydroxide
c) water
d) chlorocyclohexane
e) ethanol
f) bromide
Answer
1) a) rate = k [CH3Cl] [OH-]
b) substitute the data into the rate expression above and apply algebra to solve for k
k = 6 x 10-6 Lmol-1s-1
c) Using the rate law above, substitute the value for k from the previous question along with the new concentrations to determine the new initial rate.
rate = 6 x 10-10 M/s
$\begin{array}{r}2\right)\end{array}$
3)
4)
a) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
b) strong nucleophile
c) weak nucleophile and good leaving group
d) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.)
e) weak nucleophile, a poor electrophile without clever chemistry (stay tuned for future chapters), good leaving group
f) good nucleophile and a good leaving group | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6-04_A_Closer__Look__at_the__Nucleophilic_Substitution_Mechanism%3A_Kin.txt |
Objectives
After completing this section, you should be able to
1. write an equation to represent the Walden inversion.
2. write a short paragraph describing the Walden inversion.
3. describe, using equations, a series of reactions interconverting two enantiomers of 1-phenyl-2-propanol which led to the conclusion that nucleophilic substitution of primary and secondary alkyl halides proceeds with inversion of configuration.
Study Notes
The IUPAC name for malic acid is 2-hydroxybutanedioic acid. This acid is produced by apples, a fact which seems to have been appreciated by the British novelist Thomas Hardy in The Woodlanders:
Up, upward they crept, a stray beam of the sun alighting every now and then like a star on the blades of the pomace-shovels, which had been converted to steel mirrors by the action of the malic acid.
In 1896, the German chemist Paul Walden discovered that he could interconvert pure enantiomeric (+) and (-) malic acids through a series of reactions. This conversion meant that there was some kind of change in the stereochemistry made during the series of reactions. First, (−)-malic acid was reacted with phosphorus pentachloride (PCl5) to provide (+)-chlorosuccinic acid.
This was reacted with silver(I)oxide (Ag2O) to provide (+)-malic acid. These two combined steps caused an inversion of stereochemistry of (−)-malic acid to (+)-malic acid. The reaction series was then continued to convert (+)-malic acid back into (−)-malic acid by further reaction with PCl5 and Ag2O.
These results were considered astonishing. The fact that (−)-malic acid was converted into (+)-malic acid meant that the configuration of the chiral center has somehow been changed during the reaction series.
These reactions are currently referred to as nucleophilic substitution reactions because each step involves the substitution of one nucleophile by another. These are among the most common and versatile reaction types in organic chemistry.
Further investigations into these reaction were undertaken during the 1920's and 1930's to clarify the mechanism and clarify how the inversion of configurations occur. These reactions involved nucleophilic substitution of an alkyl p-toluenesulfonate (called a tosylate group). For this purpose the tosylate groups act similarly to a halogen substituent. In the series of reactions (+)-1-phenyl-2-propanol is interconverted with (-)-1-phenyl-2-propanol.
Somewhere in this three-step series of reactions the configuration at a chiral center is being inverted. In the first step the tosylate is formed without breaking the C-O bond of the chiral center, which means the configuration is unchanged. Similarly, the cleavage of the ester in step three occurs without breaking the C-O bond of the chiral center, which also means the configuration of the chiral carbon is unaffected. It was determined that the second step where acetate nucleophile undergoes a substitution with tosylate was causing the stereochemical configuration to be inverted.
Exercise \(1\)
1) Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH3CO2-, Show stereochemistry.
Answer
1)
6-07 Structure and SN2 Reactivity: The Leaving Group
Our general discussion of nucleophilic substitution reactions, we have until now been designating the leaving group simply as “X". As you may imagine, however, the nature of the leaving group is an important consideration: if the C-X bond does not break, the new bond between the nucleophile and electrophilic carbon cannot form, regardless of whether the substitution is SN1 or SN2. In this module, we are focusing on substitution reactions in which the leaving group is a halogen ion, although many reactions are known, both in the laboratory and in biochemical processes, in which the leaving group is something other than a halogen.
In order to understand the nature of the leaving group, it is important to first discuss factors that help determine whether a species will be a strong base or weak base. If you remember from general chemistry, a Lewis base is defined as a species that donates a pair of electrons to form a covalent bond. The factors that will determine whether a species wants to share its electrons or not include electronegativity, size, and resonance.
As Electronegativity Increases, Basicity Decreases: In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, electronegativity increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons.
As Size Increases, Basicity Decreases: In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons.
Resonance Decreases Basicity:The third factor to consider in determining whether or not a species will be a strong or weak base is resonance. As you may remember from general chemistry, the formation of a resonance stabilized structure results in a species that is less willing to share its electrons. Since strong bases, by definition, want to share their electrons, resonance stabilized structures are weak bases.
Now that we understand how electronegativity, size, and resonance affect basicity, we can combine these concepts with the fact that weak bases make the best leaving groups. Think about why this might be true. In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base. We will now revisit electronegativity, size, and resonance, moving our focus to the leaving group, as well providing actual examples.
Note
As the Electronegativity of the group Increases, The propensity of the Leaving Group to Leave Increases
As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
As Size Increases, The Ability of the Leaving Group to Leave Increases:Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Example 7.7.1
In each pair (A and B) below, which electrophile would be expected to react more rapidly in an SN2 reaction with the thiol group of cysteine as the common nucleophile?
Contributor
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6-06_Consequences_of_Inversion__in_SN2_Reactions.txt |
What is a nucleophile?
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates. More specifically in laboratory reactions, halide and azide (N3-) anions are commonly seen acting as nucleophiles.
Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? Enolate ions (section 7.5) are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN-) is just one example of a carbon nucleophile commonly used in the laboratory. Reactions with carbon nucleophiles will be dealt with in chapters 13 and 14, however - in this chapter and the next, we will concentrate on non-carbon nucleophiles.
When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: nucleophiles can be bases, and bases can be nucleophiles. It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity.
Protonation states and nucleophilicity
The protonation state of a nucleophilic atom has a very large effect on its nucleophilicity. This is an idea that makes intuitive sense: a hydroxide ion is much more nucleophilic (and basic) than a water molecule, because the negatively charged oxygen on the hydroxide ion carries greater electron density than the oxygen atom of a neutral water molecule. In practical terms, this means that a hydroxide nucleophile will react in an SN2 reaction with methyl bromide much faster ( about 10,000 times faster) than a water nucleophile.
There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity:
The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond.
This horizontal trends also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions.
Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity).
The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity: depending on the solvent that the reaction is taking place in, the nucleophilicity trend can go in either direction. Let's take the simple example of the SN2 reaction below:
. . .where Nu- is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile in that case where both are iodide). If this reaction is occurring in a protic solvent (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile.
Relative nucleophilicity in a protic solvent
This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the least basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile?
As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a protic solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile:
In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the cage.
The picture changes if we switch to a polar aprotic solvent, such as acetone, in which there is a molecular dipole but no hydrogens bound to oxygen or nitrogen. Now, fluoride is the best nucleophile, and iodide the weakest.
Relative nucleophilicity in a polar aprotic solvent
The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile.
A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile.
Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO).
In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a cysteine amino acid, for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS-) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do.
Resonance effects on nucleophilicity
Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance.
The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group.
Steric effects on nucleophilicity
Steric hindrance is an important consideration when evaluating nucleophility. For example, tert-butanol is less potent as a nucleophile than methanol. This is because the comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!).
It is not surprising that it is more common to observe serines acting as nucleophiles in enzymatic reactions compared to threonines - the former is a primary alcohol, while the latter is a secondary alcohol.
Example 7.8.1
Which is the better nucleophile - a cysteine side chain or a methionine side chain? Explain.
Example 7.8.2
In each of the following pairs of molecules/ions, which is the better nucleophile in a reaction with CH3Br in acetone solvent? Explain your choice.
1. phenolate ion (deprotonated phenol) or benzoate ion (deprotonated benzoic acid)
2. water and hydronium ion
3. trimethylamine and triethylamine
4. chloride anion and iodide anion
5. CH3NH- and CH3CH2NH
Contributors
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
6-09 Keys to Success: Choosing Among Multiple Mechanistic Pathways
Predicting SN1 vs. SN2 mechanisms; competition between nucleophilic substitution and elimination reactions
When considering whether a nucleophilic substitution is likely to occur via an SN1 or SN2 mechanism, we really need to consider three factors:
1. The electrophile: when the leaving group is attached to a methyl group or a primary carbon, an SN2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an SN1 mechanism is favored.
2. The nucleophile: powerful nucleophiles, especially those with negative charges, favor the SN2 mechanism. Weaker nucleophiles such as water or alcohols favor the SN1 mechanism.
3. The solvent: Polar aprotic solvents favor the SN2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the SN1 mechanism by stabilizing the carbocation intermediate. SN1 reactions are frequently solvolysis reactions.
For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we’ll assume that methanol is the solvent). Thus we’d confidently predict an SN1 reaction mechanism. Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.
In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide – with these, both SN1 and SN2 mechanisms are possible, depending on the nucleophile and the solvent. In this example, the nucleophile (a thiolate anion) is strong, and a polar protic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Example 7.17.1
Determine whether each substitution reaction shown below is likely to proceed by an SN1 or SN2 mechanism.
Solution
Contributors
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6-08_Structure_and__SN2__Reactivity:_The__Nucleophile.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.
We will be contrasting about two types of nucleophilic substitution reactions. One type is referred to as unimolecular nucleophilic substitution (SN1), whereby the rate determining step is unimolecular and bimolecular nucleophilic substitution (SN2), whereby the rate determining step is bimolecular. We will begin our discussion with SN2 reactions, and discuss SN1 reactions elsewhere.
Biomolecular Nucleophilic Substitution Reactions and Kinetics
In the term SN2, the S stands for substitution, the N stands for nucleophilic, and the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics:
Bimolecular Nucleophilic Substitution Reactions Are Concerted
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen.
The potential energy diagram for an SN2 reaction is shown below. Upon nucleophilic attack, a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed. Again, this is a single-step, concerted process with the occurrence of a single transition state.
Sterrically Hindered Substrates Will Reduce the SN2 Reaction Rate
Now that we have discussed the effects that the leaving group, nucleophile, and solvent have on biomolecular nucleophilic substitution (SN2) reactions, it's time to turn our attention to how the substrate affects the reaction. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below, we will be paying particular attention to the alkyl portion of the substrate. In other words, we are most interested in the electrophilic center that bears the leaving group.
In the section Kinetics of Nucleophilic Substitution Reactions, we learned that the SN2 transition state is very crowded. Recall that there are a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents.
If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN2 reaction to occur, the nucleophile must be able to attack the electrophilic center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile.
How does steric hindrance affect the rate at which an SN2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon, impeding nucleophilic attack.
The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked.
Substitutes on Neighboring Carbons Slow Nucleophilic Substitution Reactions
Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well.
In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction.
If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect.
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.
Contributors
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6.10:_Structure_and_SN2_Reactivity:_The_Substrate.txt |
In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon.
In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base.
In the next few sections, we are going to be discussing some general aspects of nucleophilic substitution reactions, and in doing so it will simplify things greatly if we can use some abbreviations and generalizations before we dive into real examples.
Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. In the same way, we will see later that nucleophiles and leaving groups are sometimes protonated and sometimes not, so for now, for the sake of simplicity, we will not include protons on 'Nu' or 'X'. We will generalize the three other groups bonded on the electrophilic central carbon as R1, R2, and R3: these symbols could represent hydrogens as well as alkyl groups. Finally, in order to keep figures from becoming too crowded, we will use in most cases the line structure convention in which the central, electrophilic carbon is not drawn out as a 'C'.
Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction:
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 on the right. 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.
In order to understand why some combinations of alkyl halides and nucleophiles give a substitution reaction, whereas other combinations give elimination, and still others give no observable reaction, we must investigate systematically the way in which changes in reaction variables perturb the course of the reaction. The following general equation summarizes the factors that will be important in such an investigation.
One conclusion, relating the structure of the R-group to possible products, should be immediately obvious. If R- has no beta-hydrogens an elimination reaction is not possible, unless a structural rearrangement occurs first. The first four halides shown on the left below do not give elimination reactions on treatment with base, because they have no β-hydrogens. The two halides on the right do not normally undergo such reactions because the potential elimination products have highly strained double or triple bonds.
It is also worth noting that sp2 hybridized C–X compounds, such as the three on the right, do not normally undergo nucleophilic substitution reactions, unless other functional groups perturb the double bond(s).
Using the general reaction shown above as our reference, we can identify the following variables and observables.
Variables
R change α-carbon from 1º to 2º to 3º
if the α-carbon is a chiral center, set as (R) or (S)
X change from Cl to Br to I (F is relatively unreactive)
Nu: change from anion to neutral; change basicity; change polarizability
Solvent polar vs. non-polar; protic vs. non-protic
Observables
Products substitution, elimination, no reaction.
Stereospecificity if the α-carbon is a chiral center what happens to its configuration?
Reaction Rate measure as a function of reactant concentration.
When several reaction variables may be changed, it is important to isolate the effects of each during the course of study. In other words: only one variable should be changed at a time, the others being held as constant as possible. For example, we can examine the effect of changing the halogen substituent from Cl to Br to I, using ethyl as a common R–group, cyanide anion as a common nucleophile, and ethanol as a common solvent. We would find a common substitution product, C2H5–CN, in all cases, but the speed or rate of the reaction would increase in the order: Cl < Br < I. This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/06._Properties_and_Reactions_of_Haloalkanes%3A_Bimolecular_Nucleophilic_Substitution/6.11:_The_Sn2_Reaction_at_a_Glance.txt |
Just as with SN2 reactions, the nucleophile, solvent and leaving group also affect SN1 (Unimolecular Nucleophilic Substitution) reactions. Polar protic solvents have a hydrogen atom attached to an electronegative atom so the hydrogen is highly polarized. Polar aprotic solvents have a dipole moment, but their hydrogen is not highly polarized. Polar aprotic solvents are not used in SN1 reactions because some of them can react with the carbocation intermediate and give you an unwanted product. Rather, polar protic solvents are preferred.
Introduction
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 (See SN2 Nucleophile). 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 (see example below). 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.
The figure below shows the mechanism of an SN1 reaction of an alkyl halide with water. Since water is also the solvent, this is an example of a solvolysis reaction.
Examples of polar protic solvents are: acetic acid, isopropanol, ethanol, methanol, formic acid, water, etc.
Effects of Nucleophile
The strength of the nucleophile does not affect the reaction rate of SN1 because, as stated above, the nucleophile is not involved in the rate-determining step. However, if you have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affects the distribution of products that you will get. For example, if you have (CH3)3CCl reacting in water and formic acid where the water and formic acid are competing nucleophiles, you will get two different products: (CH3)3COH and (CH3)33COCOH. The relative yields of these products depend on the concentrations and relative reactivities of the nucleophiles.
Effects of Leaving Group
An SN1 reaction speeds up with a good leaving group. This is because the leaving group is involved in the rate-determining step. A good leaving group wants to leave so it breaks the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed.
A good leaving group is a weak base because weak bases can hold the charge. They're happy to leave with both electrons and in order for the leaving group to leave, it needs to be able to accept electrons. Strong bases, on the other hand, donate electrons which is why they can't be good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whos leaving-group ability increases as you go down the column.
The two reactions below is the same reaction done with two different leaving groups. One is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster.
Other examples of good leaving groups are sulfur derivatives such as methyl sulfate ion and other sulfonate ions (See Figure below)
Methyl Sulfate Ion Mesylate Ion Triflate Ion Tosylate Ion
References
• Uggerud E. "Reactivity trends and stereospecificity in nucleophilic substitution reactions." J. Phys. Org. Chem. 2006; 19; 461-466.
• Vollhardt, K. Peter C., and Neil E. Schore. Organic Chemistry Structure and Function. New York: W. H. Freeman, 2007.
• Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Pearson Education, 2007.
• Suggs, William J. Organic Chemistry. Canada: Barron's Educational Series Inc., 2002.
Outside Links
Problems
1. Put the following leaving groups in order of decreasing leaving group ability
2. Which solvent would an SN1 reaction occur faster in? H2O or CH3CN
3. What kind of conditions disfavor SN1 reactions?
4. What are the products of the following reaction and does it proceed via SN1 or SN2?
5. How could you change the reactants in the problem 4 to favor the other substitution reaction?
6. Indicate the expected product and list why it occurs through SN1 instead of SN2?
Answers
1.
2. An SN1 reaction would occur faster in H2O because it's polar protic and would stailize the carbocation and CH3CN is polar aprotic.
3. Polar aprotic solvents, a weak leaving group and primary substrates disfavor SN1 reactions.
4.
Reaction proceeds via SN1 because a tertiary carbocation was formed, the solvent is polar protic and Br- is a good leaving group.
5. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH2- or OH-.
6.
This reaction occurs via SN1 because Cl- is a good leaving group and the solvent is polar protic. This is an example of a solvolysis reaction because the nucleophile is also the solvent.
• Ashiv Sharma | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.01%3A_Solvolysis_of_Tertiary__and_Secondary_Haloalkanes.txt |
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.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.2%3A_Unimolecular_Nucleophilic_Substitution.txt |
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.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.3%3A_Stereochemical_Consequences_of_SN1__Reactions.txt |
Objectives
After completing this section, you should be able to
1. discuss how the structure of the substrate affects the rate of a reaction occurring by the SN1 mechanism.
2. arrange a given list of carbocations (including benzyl and allyl) in order of increasing or decreasing stability.
3. explain the high stability of the allyl and benzyl carbocations.
4. arrange a given series of compounds in order of increasing or decreasing reactivity in SN1 reactions, and discuss this order in terms of the Hammond postulate.
5. discuss how the nature of the leaving group affects the rate of an SN1 reaction, and in particular, explain why SN1 reactions involving alcohols are carried out under acidic conditions.
6. explain why the nature of the nucleophile does not affect the rate of an SN1 reaction.
7. discuss the role played by the solvent in an SN1 reaction, and hence determine whether a given solvent will promote reaction by this mechanism.
8. compare the roles played by the solvent in SN1 and in SN2 reactions.
9. determine which of two SN1 reactions will occur faster, by taking into account factors such as the structure of the substrate and the polarity of the solvent.
10. determine whether a given reaction is most likely to occur by an SN1 or SN2 mechanism, based on factors such as the structure of the substrate, the solvent used, etc.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• benzylic
• dielectric constant
• polarity
The Substrate in SN1Reactions
As discussed in the previous section SN1 reactions follow first order kinetics due to a multi-step mechanism in which the rate-determining step consists of the ionization of the alkyl halide to form a carbocation.
The transition state for the rate determining step shows the transition of an alkyl halide to a carbocation. Because the rate determining step is endothermic, the Hammond postulate says that the transition state more closely resembles the carbocation intermediate. Factors which stabilize this intermediate will lower the energy of activation for the rate determining step and cause the rate of reaction to increase. In general, a more stable carbocation intermediate formed during the reaction allows for a faster the SN1 reaction rate.
In Section 7-9, the stability of alky carbocations was shown to be 3o > 2o > 1o > methyl. Two special cases of resonance-stabilized carbocations, allyl and benzyl, must be considered and added to the list. The delocalization of the positive charge over an extended p orbital system allows for allyl and benzyl carbocations to be exceptionally stable. The resonance hybrid of an allyic cation is made up of two resonance forms while the resonance hybrid of the benzylic carbocation is made up of five.
Benzyl Carbocation
Carbocation
Stability
CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) C6H5CH2(+) (CH3)3C(+)
Effects of Leaving Group
Since leaving group removal is involved in the rate-determining step of the SN1 mechanism, reaction rates increase with a good leaving group. A good leaving group can stabilize the electron pair it obtains after the breaking of the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed.
Because weak bases tend to strongly hold onto their electrons, they are lower energy molecules and they tend to make good leaving groups. Strong bases, on the other hand, donate electrons readily because they are high energy, reactive molecules. Therefore, they are not typically good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whose leaving-group ability increases as you go down the halogen column of the periodic table.
The two reactions below only vary by the different leaving groups in each reaction. The reaction with a more stable leaving group is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster.
Under acidic conditions, the -OH group of an alcohol can be converted into a neutral water leaving group through protonation. As discussed in Section 10-5, this occurs during the conversion of a tertiary alcohol to an alkyl halide through reaction with HCl or HBr. Because the -OH group itself is a poor nucleophile, the mechanism starts with protonation to form a hydronium ion. Neutral water is then lost as a leaving group to create the carbocation intermediate which then reacts with the halide ion nucleophile to provide the alkyl halide product. This reaction works best when a tertiary alcohol is used because it produces a stable carbocation intermediate.
The Nucleophile
Since nucleophiles only participate in the fast second step, their relative molar concentrations rather than their nucleophilicities should be the primary product-determining factor. If a nucleophilic solvent such as water is used, its high concentration will assure that alcohols are the major product. While recombination of the halide anion with the carbocation intermediate can occur, it simply reforms the starting compound. Also this is less likely since there are less molecules of the leaving group in solution than there are of the solvent. Note that SN1 reactions in which the nucleophile is also the solvent are commonly called solvolysis reactions. 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 strength of the nucleophile does not affect the reaction rate of SN1 because, as described previously, the nucleophile is not involved in the rate-determining step. However, if you have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affects the distribution of products that you will get. For example, if you have (CH3)3CCl reacting in water and formic acid where the water and formic acid are competing nucleophiles, you will get two different products: (CH3)3COH and (CH3)3COCOH. The relative yields of these products depend on the concentrations and relative reactivities of the nucleophiles.
Solvent Effects on the 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 reaction which depends on nucleophilic attack during the rate-determining step of the reaction. However, this 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 carbocation-like transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction.
The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for a 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.
(note that even though acetic acid is a highly polar molecule, it tends to make a dimer with itself greatly reducing its dielectric constant)
Predicting SN1 vs. SN2 mechanisms
When considering whether a nucleophilic substitution is likely to occur via an SN1 or SN2 mechanism, we really need to consider three factors:
1) The electrophile: when the leaving group is attached to a methyl group or a primary carbon, an SN2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an SN1 mechanism is favored.
2) The nucleophile: powerful nucleophiles, especially those with negative charges, favor the SN2 mechanism. Weaker nucleophiles such as water or alcohols favor the SN1 mechanism.
3) The solvent: Polar aprotic solvents favor the SN2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the SN1 mechanism by stabilizing the carbocation intermediate. SN1 reactions are frequently solvolysis reactions.
For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we’ll assume that methanol is the solvent). Thus we’d confidently predict an SN1 reaction mechanism. Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.
In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide – with these, both SN1 and SN2 mechanisms are possible, depending on the nucleophile and the solvent. In this example, the nucleophile (a thiolate anion) is strong, and a polar aprotic solvent is used – so the SN2 mechanism is heavily favored. The reaction is expected to proceed with inversion of configuration.
Exercise \(1\)
1. Put the following leaving groups in order of decreasing leaving group ability
2. Which solvent would an SN1 reaction occur faster in? H2O or CH3CN
3. What kind of conditions disfavor SN1 reactions?
4. What are the products of the following reaction and does it proceed via SN1 or SN2?
5. How could you change the reactants in the problem 4 to favor the other substitution reaction?
6. Indicate the expected product and list why it occurs through SN1 instead of SN2?
7. Rank the following by increasing reactivity in an SN1 reaction.
8. 3-bromo-1-pentene and 1-bromo-2-pentene undergo SN1 reaction at almost the same rate, but one is a secondary halide while the other is a primary halide. Explain why this is.
9. Label the following reactions as most likely occuring by an SN1 or SN2 mechanism. Suggest why.
10. Give the solvolysis product expected when the compound is heated in methanol.
a)
b)
c)
d)
11. Predict whether each compound below would be more likely to undergo a SN2 or SN1 reaction.
a)
b)
c)
d)
12. Show how each compound may be synthesized using nucleophilic substitution reactions.
a)
b)
c)
d)
e)
f)
g)
Answers
1.
2. An SN1 reaction would occur faster in H2O because it's polar protic and would stailize the carbocation and CH3CN is polar aprotic.
3. Polar aprotic solvents, a weak leaving group and primary substrates disfavor SN1 reactions.
4.
Reaction proceeds via SN1 because a tertiary carbocation was formed, the solvent is polar protic and Br- is a good leaving group.
5. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH2- or OH-.
6.
This reaction occurs via SN1 because Cl- is a good leaving group and the solvent is polar protic. This is an example of a solvolysis reaction because the nucleophile is also the solvent.
7. Consider the stability of the intermediate, the carbocation.
A < D < B < C (most reactive)
8. They have the same intermediates when you look at the resonance forms.
9.
A – SN1 *poor leaving group, protic solvent, secondary cation intermediate
B – SN2 *good leaving group, polar solvent, primary position.
10.
a)
b)
c) No reaction
d)
11.
a) SN2
b) SN2
c) SN1
d) SN1
13.
a)
b)
c)
d)
e)
f)
g) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.4%3A_Effects__of_Solvent%2C_Leaving__Group%2C_and__Nucleophi.txt |
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?
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)
Now, back to transition states. Chemists are often very interested in trying to learn about what the transition state for a given reaction looks like, but addressing this question requires an indirect approach because the transition state itself cannot be observed. In order to gain some insight into what a particular transition state looks like, chemists often invoke the Hammond postulate, which states that a transition state resembles the structure of the nearest stable species. For an exergonic reaction, therefore, the transition state resembles the reactants more than it does the products.
If we consider a hypothetical exergonic reaction between compounds A and B to form AB, the distance between A and B would be relatively large at the transition state, resembling the starting state where A and B are two isolated species. In the hypothetical endergonic reaction between C and D to form CD, however, the bond formation process would be much further along at the TS point, resembling the product.
The Hammond Postulate is a very simplistic idea, which relies on an assumption that potential energy surfaces are parabolic. Although such an assumption is not rigorously true, it is fairly reliable and allows chemists to make energetic arguments about transition states by employing arguments about the stability of a related species. Since the formation of a reactive intermediate is very reliably endergonic, arguments about the stability of reactive intermediates can serve as proxy arguments about transition state stability.
The Hammond Postulate and the SN1 Reaction
the Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate. The stability of carbocations was discussed earlier, and a qualitative relationship is given below:
Carbocation
Stability
CH3(+) < CH3CH2(+) < (CH3)2CH(+) CH2=CH-CH2(+) < C6H5CH2(+) (CH3)3C(+)
Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an SN1 mechanism. This is opposite to the reactivity order observed for the SN2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.5%3A_Effect__of__the_Alkyl__Group__on__the_SN_1_Reaction%3A_.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.
Contributors
• Satish Balasubramanian
The E1 mechanism is nearly identical to the SN1 mechanism, differing only in the course of reaction taken by the carbocation intermediate. As shown by the following equations, a carbocation bearing beta-hydrogens may function either as a Lewis acid (electrophile), as it does in the SN1 reaction, or a Brønsted acid, as in the E1 reaction.
Thus, hydrolysis of tert-butyl chloride in a mixed solvent of water and acetonitrile gives a mixture of 2-methyl-2-propanol (60%) and 2-methylpropene (40%) at a rate independent of the water concentration. The alcohol is the product of an SN1 reaction and the alkene is the product of the E1 reaction. The characteristics of these two reaction mechanisms are similar, as expected. They both show first order kinetics; neither is much influenced by a change in the nucleophile/base; and both are relatively non-stereospecific.
(CH3)3CCl + H2O ——> [ (CH3)3C(+) ] + Cl(–) + H2O ——> (CH3)3COH + (CH3)2C=CH2 + HCl + H2O
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution product.
2. The cation may transfer a beta-proton to a base, giving an alkene product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
Since the SN1 and E1 reactions proceed via the same carbocation intermediate, the product ratios are difficult to control and both substitution and elimination usually take place.
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. 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). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.6%3A_Unimolecular_Elimination%3A_E1.txt |
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. Unlike E1 reactions, E2 reactions remove two subsituents with the addition of a strong base, resulting in an alkene.
Introduction
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.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
General Reaction
Below is a mechanistic diagram of an elimination reaction by the E2 pathway:
.
In this reaction Ba represents the base and Br representents 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:
• A strong base is necessary especially necessary for primary alkyl halides. Secondary and tertirary primary halides will procede with E2 in the precesence 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 strically hindered base will result in the least substituted product.
The Leaving Group Effect in E2 Reactions
As Size Increases, The Ability of the Leaving Group to Leave Increases: Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Stereochemistry of the E2 Reaction
E2 elimination reactions of certain isomeric cycloalkyl halides show unusual rates and regioselectivity that are not explained by the principles thus far discussed. For example, trans-2-methyl-1-chlorocyclohexane reacts with alcoholic KOH at a much slower rate than does its cis-isomer. Furthermore, the product from elimination of the trans-isomer is 3-methylcyclohexene (not predicted by the Zaitsev rule), whereas the cis-isomer gives the predicted 1-methylcyclohexene as the chief product. These differences are described by the first two equations in the following diagram.
Unlike open chain structures, cyclic compounds generally restrict the spatial orientation of ring substituents to relatively few arrangements. Consequently, reactions conducted on such substrates often provide us with information about the preferred orientation of reactant species in the transition state. Stereoisomers are particularly suitable in this respect, so the results shown here contain important information about the E2 transition state.
The most sensible interpretation of the elimination reactions of 2- and 4-substituted halocyclohexanes is that this reaction prefers an anti orientation of the halogen and the beta-hydrogen which is attacked by the base. These anti orientations are colored in red in the above equations. The compounds used here all have six-membered rings, so the anti orientation of groups requires that they assume a diaxial conformation. The observed differences in rate are the result of a steric preference for equatorial orientation of large substituents, which reduces the effective concentration of conformers having an axial halogen. In the case of the 1-bromo-4-tert-butylcyclohexane isomers, the tert-butyl group is so large that it will always assume an equatorial orientation, leaving the bromine to be axial in the cis-isomer and equatorial in the trans. Because of symmetry, the two axial beta-hydrogens in the cis-isomer react equally with base, resulting in rapid elimination to the same alkene (actually a racemic mixture). This reflects the fixed anti orientation of these hydrogens to the chlorine atom. To assume a conformation having an axial bromine the trans-isomer must tolerate serious crowding distortions. Such conformers are therefore present in extremely low concentration, and the rate of elimination is very slow. Indeed, substitution by hydroxide anion predominates.
A similar analysis of the 1-chloro-2-methylcyclohexane isomers explains both the rate and regioselectivity differences. Both the chlorine and methyl groups may assume an equatorial orientation in a chair conformation of the trans-isomer, as shown in the top equation. The axial chlorine needed for the E2 elimination is present only in the less stable alternative chair conformer, but this structure has only one axial beta-hydrogen (colored red), and the resulting elimination gives 3-methylcyclohexene. In the cis-isomer the smaller chlorine atom assumes an axial position in the more stable chair conformation, and here there are two axial beta hydrogens. The more stable 1-methylcyclohexene is therefore the predominant product, and the overall rate of elimination is relatively fast.
An orbital drawing of the anti-transition state is shown on the right. Note that the base attacks the alkyl halide from the side opposite the halogen, just as in the SN2 mechanism. In this drawing the α and β carbon atoms are undergoing a rehybridization from sp3 to sp2 and the developing π-bond is drawn as dashed light blue lines. The symbol R represents an alkyl group or hydrogen. Since both the base and the alkyl halide are present in this transition state, the reaction is bimolecular and should exhibit second order kinetics. We should note in passing that a syn-transition state would also provide good orbital overlap for elimination, and in some cases where an anti-orientation is prohibited by structural constraints syn-elimination has been observed.
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. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.7%3A_Bimolecular_Elimination%3A_E2.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_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.8%3A_Keys_to_Success%3A_Substitutin_Versus_Elimination-Struc.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_(Vollhardt_and_Schore)/07._Further_Reactions_of_Haloalkanes%3A_Unimolecular_Substitution_and_Pathways_of_Elimination/7.9%3A_Summary_of_Reactivity__of_Haloalkanes.txt |
Objectives
After completing this section, you should be able to
1. identify an alcohol as being primary, secondary or tertiary, given its structure, its IUPAC name or its trivial name.
2. write the IUPAC name of an alcohol or phenol given its Kekulé, condensed or shorthand structure.
3. draw the structure of an alcohol or phenol given its IUPAC name.
4. identify a number of commonly occurring alcohols (e.g., benzyl alcohol, tert‑butyl alcohol) by their trivial names.
Study Notes
The following are common names of some alcohols (with IUPAC name).
Primary alcohols
In a primary (1°) alcohol, the carbon which carries the -OH group is only attached to one alkyl group. Some examples of primary alcohols include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage to an alkyl group from the CH2 group holding the -OH group. There is an exception to this. Methanol, CH3OH, is counted as a primary alcohol even though there are no alkyl groups attached to the carbon with the -OH group on it.
Secondary alcohols
In a secondary (2°) alcohol, the carbon with the -OH group attached is joined directly to two alkyl groups, which may be the same or different. Examples:
Tertiary alcohols
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Naming Alcohols
1. Find the longest chain containing the hydroxy group (OH). If there is a chain with more carbons than the one containing the OH group it will be named as a subsitutent.
2. Place the OH on the lowest possible number for the chain. With the exception of carbonyl groups such as ketones and aldehydes, the alcohol or hydroxy groups have first priority for naming.
3. When naming a cyclic structure, the -OH is assumed to be on the first carbon unless the carbonyl group is present, in which case the later will get priority at the first carbon.
4. When multiple -OH groups are on the cyclic structure, number the carbons on which the -OH groups reside.
5. Remove the final e from the parent alkane chain and add -ol. When multiple alcohols are present use di, tri, et.c before the ol, after the parent name. ex. 2,3-hexandiol. If a carbonyl group is present, the -OH group is named with the prefix "hydroxy," with the carbonyl group attached to the parent chain name so that it ends with -al or -one.
Examples
Ethane: CH3CH3 ----->Ethanol: (the alcohol found in beer, wine and other consumed sprits)
Secondary alcohol: 2-propanol
Other functional groups on an alcohol: 3-bromo-2-pentanol
Cyclic alcohol (two -OH groups): cyclohexan-1,4-diol
Other functional group on the cyclic structure: 3-hexeneol (the alkene is in bold and indicated by numbering the carbon closest to the alcohol)
A complex alcohol:4-ethyl-3hexanol (the parent chain is in red and the substituent is in blue)
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.
Naming phenols
Phenols are named using the rules for aromatic compounds discussed in seciton 15.1 Note! that -phenol is used rather than -benzene. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.1%3A_Naming__the_Alcohols.txt |
Objectives
After completing this section, you should be able to
1. explain why the boiling points of alcohols and phenols are much higher than those of alkanes, ethers, etc., of similar molecular mass.
2. discuss the factors that are believed to determine the acidity of alcohols and phenols.
3. list a given series of alcohols or phenols in order of increasing or decreasing acidity.
4. explain the difference in acidity between two given alcohols or phenols.
5. explain why phenols are more acidic than alcohols.
6. explain, in terms of inductive and resonance effects, why a given substituted phenol is more or less acidic than phenol itself.
7. write equations for the reactions of given alcohols and phenols with strong bases, such as sodium hydride and sodium amide.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• acid ionization constant (Ka)
• alkoxide ion (RO)
• phenoxide ion (ArO)
Study Notes
You may wish to review the concept of hydrogen bonding, which should have been discussed in your first‑year general chemistry course.
Boiling Points
The chart below shows the boiling points of the following simple primary alcohols with up to 4 carbon atoms:
These boiling points are compared with those of the equivalent alkanes (methane to butane) with the same number of carbon atoms.
Notice that:
• The boiling point of an alcohol is always significantly higher than that of the analogous alkane.
• The boiling points of the alcohols increase as the number of carbon atoms increases.
The patterns in boiling point reflect the patterns in intermolecular attractions.
Hydrogen bonding
Hydrogen bonding occurs between molecules in which a hydrogen atom is attached to a strongly electronegative element: fluorine, oxygen or nitrogen. In the case of alcohols, hydrogen bonds occur between the partially-positive hydrogen atoms and lone pairs on oxygen atoms of other molecules.
The hydrogen atoms are slightly positive because the bonding electrons are pulled toward the very electronegative oxygen atoms. In alkanes, the only intermolecular forces are van der Waals dispersion forces. Hydrogen bonds are much stronger than these, and therefore it takes more energy to separate alcohol molecules than it does to separate alkane molecules. This the main reason for higher boiling points in alcohols.
The effect of van der Waals forces
• Boiling points of the alcohols: Hydrogen bonding is not the only intermolecular force alcohols experience. There are also van der Waals dispersion forces and dipole-dipole interactions. The hydrogen bonding and dipole-dipole interactions are much the same for all alcohols, but dispersion forces increase as the alcohols get bigger. These attractions get stronger as the molecules get longer and have more electrons. This increases the sizes of the temporary dipoles formed. This is why the boiling points increase as the number of carbon atoms in the chains increases. It takes more energy to overcome the dispersion forces, and thus the boiling points rise.
• Comparison between alkanes and alcohols: Even without any hydrogen bonding or dipole-dipole interactions, the boiling point of the alcohol would be higher than the corresponding alkane with the same number of carbon atoms.
Compare ethane and ethanol:
Ethanol is a longer molecule, and the oxygen atom brings with it an extra 8 electrons. Both of these increase the size of the van der Waals dispersion forces, and subsequently the boiling point. A more accurate measurement of the effect of the hydrogen bonding on boiling point would be a comparison of ethanol with propane rather than ethane. The lengths of the two molecules are more similar, and the number of electrons is exactly the same.
Solubility of alcohols in water
Small alcohols are completely soluble in water; mixing the two in any proportion generates a single solution. However, solubility decreases as the length of the hydrocarbon chain in the alcohol increases. At four carbon atoms and beyond, the decrease in solubility is noticeable; a two-layered substance may appear in a test tube when the two are mixed.
Consider ethanol as a typical small alcohol. In both pure water and pure ethanol the main intermolecular attractions are hydrogen bonds.
In order to mix the two, the hydrogen bonds between water molecules and the hydrogen bonds between ethanol molecules must be broken. Energy is required for both of these processes. However, when the molecules are mixed, new hydrogen bonds are formed between water molecules and ethanol molecules.
The energy released when these new hydrogen bonds form approximately compensates for the energy needed to break the original interactions. In addition, there is an increase in the disorder of the system, an increase in entropy. This is another factor in deciding whether chemical processes occur. Consider a hypothetical situation involving 5-carbon alcohol molecules.
The hydrocarbon chains are forced between water molecules, breaking hydrogen bonds between those water molecules. The -OH ends of the alcohol molecules can form new hydrogen bonds with water molecules, but the hydrocarbon "tail" does not form hydrogen bonds. This means that many of the original hydrogen bonds being broken are never replaced by new ones.
In place of those original hydrogen bonds are merely van der Waals dispersion forces between the water and the hydrocarbon "tails." These attractions are much weaker, and unable to furnish enough energy to compensate for the broken hydrogen bonds. Even allowing for the increase in disorder, the process becomes less feasible. As the length of the alcohol increases, this situation becomes more pronounced, and thus the solubility decreases.
Acid/Base properties of alcohols
Several important chemical reactions of alcohols involving the O-H bond or oxygen-hydrogen bond only and leave the carbon-oxygen bond intact. An important example is salt formation with acids and bases. Alcohols, like water, are both weak bases and weak acids. The acid ionization constant (Ka) of ethanol is about 10~18, slightly less than that of water. Ethanol can be converted to its conjugate base by the conjugate base of a weaker acid such as ammonia {Ka — 10~35), or hydrogen (Ka ~ 10-38). It is convenient to employ sodium metal or sodium hydride, which react vigorously but controllably with alcohols:
The order of acidity of various liquid alcohols generally is water > primary > secondary > tertiary ROH. By this we mean that the equilibrium position for the proton-transfer reaction (Equation 15-1) lies more on the side of ROH and OHe as R is changed from primary to secondary to tertiary; therefore, tert-butyl alcohol is considered less acidic than ethanol:
\[ ROH + OH^- \rightleftharpoons RO^- + HOH\]
However, in the gas phase the order of acidity is reversed, and the equilibrium position for Equation 15-1 lies increasingly on the side of ROGas R is changed from primary to secondary to tertiary, terf-Butyl alcohol is therefore more acidic than ethanol in the gas phase. This seeming contradiction appears more reasonable when one considers what effect solvation (or the lack of it) has on equilibria expressed by Equation 15-1. In solution, the larger anions of alcohols, known as alkoxide ions, probably are less well solvated than the smaller ions, because fewer solvent molecules can be accommodated around the negatively charged oxygen in the larger ions:
Acidity of alcohols therefore decreases as the size of the conjugate base increases. However, “naked” gaseous ions are more stable the larger the associated R groups, probably because the larger R groups can stabilize the charge on the oxygen atom better than the smaller R groups. They do this by polarization of their bonding electrons, and the bigger the group, the more polarizable it is. (Also see Section 11-8A, which deals with the somewhat similar situation encountered with respect to the relative acidities of ethyne and water.)
Chemical Reactions of Alcohols involving the O-H bond of Compounds with Basic Properties
Alcohols are bases similar in strength to water and accept protons from strong acids. An example is the reaction of methanol with hydrogen bromide to give methyloxonium bromide, which is analogous to the formation of hydroxonium bromide with hydrogen bromide and water:
Acidity of Phenol
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base. For example, in solution in water:
Phenol is a very weak acid and the position of equilibrium lies well to the left. Phenol can lose a hydrogen ion because the phenoxide ion formed is stabilised to some extent. The negative charge on the oxygen atom is delocalised around the ring. The more stable the ion is, the more likely it is to form. One of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalization which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localized on the oxygen, but is spread out around the whole ion.
Spreading the charge around makes the ion more stable than it would be if all the charge remained on the oxygen. However, oxygen is the most electronegative element in the ion and the delocalized electrons will be drawn towards it. That means that there will still be a lot of charge around the oxygen which will tend to attract the hydrogen ion back again. That is why phenol is only a very weak acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
In this reaction, the hydrogen ion has been removed by the strongly basic hydroxide ion in the sodium hydroxide solution.
Acids react with the more reactive metals to give hydrogen gas. Phenol is no exception - the only difference is the slow reaction because phenol is such a weak acid. Phenol is warmed in a dry tube until it is molten, and a small piece of sodium added. There is some fizzing as hydrogen gas is given off. The mixture left in the tube will contain sodium phenoxide.
Acidity of Substituted Phenols
Substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols. This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group, as displayed in the following diagram. The alcohol cyclohexanol is shown for reference at the top left. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. The trinitro compound shown at the lower right is a very strong acid called picric acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. Formulas illustrating this electron delocalization will be displayed when the "Resonance Structures" button beneath the previous diagram is clicked. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.2%3A_Structural_and__Physical__Properties_of_Alcohols.txt |
Alcohols are very weak Brønsted acids with pKa values generally in the range of 15 - 20. Because the hydroxyl proton is the most electrophilic site, proton transfer is the most important reaction to consider with nucleophiles. There are small differences in the acidities of aliphatic alcohols in aqueous solution, which are due to differences in structure and, more importantly, solvation.
General Assessment of Acidities
When considering alcohols as organic reagents, pKas are often used because they reflect reactivity in aqueous solution. In general, alcohols in aqueous solution are slightly less acidic than water. However, the differences among the pKas of the alcohols are not large. This is not surprising because all alcohols are oxy-acids (OH), and the differences in acidities are due to the effect of substituents in the 1-position removed from the acidic site. Moreover, the more highly substituted alcohols vary only in the structure two positions removed from the acidic site. The marginal effects of additional substituents at the carbon tow positions removed from the acidic site are even evident in the gas-phase enthalpies of reaction for the reaction
$ROH \rightarrow RO^- + H^+ \tag{1}$
The pKas and gas-phase enthalpies of reaction for various alcohols, ROH, with various substituents (R) are shown in Table 1 below.
R Name pKa1 $\Delta H_{acid}\; kJ/mol^2$
Table 1: pKas and gas-phase enthalpies of reaction
H water 14.0 1633.1
CH3 methanol 15.5 1597 ± 6
CH3CH2 ethanol 15.9 1587 ± 4
(CH3)2CH propan-2-ol (isopropyl alcohol) 16.5 1569 ± 4
(CH3)3C 2-methylpropan-2-ol (t-butanol) 17 1568 ± 4
C6H5 (phenyl) phenol 9.95 1462 ± 10
Interpretation of the Relative Acidities of Alcohols
There are many sites on the internet with explanations of the relative ordering of alcohol acidities in aqueous solution. The general explanation is that the larger substituents are better electron donors, which destabilize the resulting alkoxide anions. Because hydrogen is least donating of the substituents, water is the strongest acid. Unfortunately, although this belief persists, it is incomplete because it does not account for the gas-phase results. The problem with the electron donation explanation is that it suggests that the order of acidity is due solely to the intrinsic electronic effects of the substituents. However, if that were the case, the electron donating effect should also be evident in the gas-phase data. However, the relative acidities in the gas phase are opposite to those in aqueous solution. Consequently, any interpretation of the acidities of alcohols must take the gas phase data into account.
The inversion of the acidities of alcohols between the gas phase and aqueous solution was pointed out by Brauman and Blair in 1968.3 They proposed that the ordering of acidities of alcohols in solution is predominantly due to the combination of a) polarizibility and b) solvation, and that the electron donating ability of the substituent does not play a significant role.4
Polarizibility almost completely accounts for the trend in gas-phase acidities. As the size of the substituent increases, the acid becomes stronger due to the ability for the charge to be distributed over a larger volume, thereby reducing the charge density and, consequently, the Coulombic repulsion. Therefore, in the gas-phase, t-butanol is the most acidic alcohol, more acidic than isopropanol, followed by ethanol and methanol. In the gas phase, water is much less acidic than methanol, which is consistent with the difference in polarizibility between a proton and a methyl group. As before, the fact that water is less acidic than methanol in the gas phase is not consistent with the expected electronic donating capabilities of the two substituents. Given the absence of a solvent, the gas-phase properties reflect the instrinsic effects on the acidities.
In solution, however, the ions can be stabilized by solvation, and this is what leads to the inversion of acidity ordering. Brauman and Blair3 showed that smaller ions are better stabilized by solvation, which is consistent with the Born equation. Therefore, methanol is more acidic than t-butanol because the smaller methoxide ion has a shorter radius of solvation, leading to a larger solvation energy, which overcomes the stabilization that results from polarization of the charge. Because the solvation energy of hydroxide is even larger than that of methoxide, water is more acidic than methanol.
Note: Phenol
Discussions of acidities of alcohols usually include phenol in which the enhanced acidity is generally attributed to stabilization of the phenoxide ion by resonance delocalization. In this case, the gas-phase results agree with the solution trend that phenol is a much stronger acid than the aliphatic alcohols, and the difference is certainly due to electronic effects. However, this commonly encountered explanation is incomplete because it ignores the role that inductive effects have on acidities of oxy-acids. However, it is true that the acidity of phenol is much more a result of resonance stabilization than, for example, the acidities of carboxylic acids.
Practical Considerations
With pKa values in the 15.5 - 20.0 range, useful concentrations of alkoxides cannot be formed by proton transfer with hydroxide:
$ROH + OH^- \rightarrow RO^- + H_2O \tag{2}$
The equilibrium constant for the proton transfer reaction is on the order of 10-2-10-5. Phenoxide can be formed almost completely by reaction with aqueous alkaline base because the value of the equilibrium constant is roughly 104. The acidity of alcohols also indicates that it will react by proton transfer with any base more basic than hydroxide, which includes most organic bases, such as acetylide ions, cyanide, and vinyl/phenyl/alkyl anions. Therefore, alcohols will protonate most organic nucleophiles and effectively destroy most organometallic reagents, including Grignard or organolithium reagents.
Formation of alkoxide ions
Alkoxide ions can be formed by deprotonating alcohols with an extremely strong base such as an amide ion, NH2-. However, this method is rarely used. Alkoxides are more often formed by reaction of an alkali metal such as sodium with the pure alcohol:
$2ROH + 2Na \rightarrow 2RO^- + 2Na^+ + H_2 \tag{3}$
Conclusions
Relative acidities of all acids depend on many factors, including intrinsic electronic factors such as electronegativity, inductive and resonance effects, and polarizibility, as well as extrinsic factors such as solvation. For aliphatic alcohols, the most important effects are polarizibility and solvation, not electronic donation, as is generally assumed. In systems where the intrinsic factors are large, their effects are manifested in the overall properties, regardless of the medium. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.3%3A_Alcohols__as_Acids__and__Bases.txt |
This page looks at the manufacture of alcohols by the direct hydration of alkenes, concentrating mainly on the hydration of ethene to make ethanol. It then compares that method with making ethanol by fermentation.
Manufacturing alcohols from alkenes
Ethanol is manufactured by reacting ethene with steam. The catalyst used is solid silicon dioxide coated with phosphoric(V) acid. The reaction is reversible.
Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this:
The manufacture of other alcohols from alkenes
Some - but not all - other alcohols can be made by similar reactions. The catalyst used and the reaction conditions will vary from alcohol to alcohol. The reason that there is a problem with some alcohols is well illustrated with trying to make an alcohol from propene, CH3CH=CH2. In principle, there are two different alcohols which might be formed:
You might expect to get either propan-1-ol or propan-2-ol depending on which way around the water adds to the double bond. In practice what you get is propan-2-ol. If you add a molecule H-X across a carbon-carbon double bond, the hydrogen nearly always gets attached to the carbon with the most hydrogens on it already - in this case the CH2 rather than the CH. The effect of this is that there are bound to be some alcohols which it is impossible to make by reacting alkenes with steam because the addition would be the wrong way around.
Making ethanol by fermentation
This method only applies to ethanol and you cannot make any other alcohol this way. The starting material for the process varies widely, but will normally be some form of starchy plant material such as maize (US: corn), wheat, barley or potatoes. Starch is a complex carbohydrate, and other carbohydrates can also be used - for example, in the lab sucrose (sugar) is normally used to produce ethanol. Industrially, this wouldn't make sense. It would be silly to refine sugar if all you were going to use it for was fermentation. There is no reason why you should not start from the original sugar cane, though.
The first step is to break complex carbohydrates into simpler ones. For example, if you were starting from starch in grains like wheat or barley, the grain is heated with hot water to extract the starch and then warmed with malt. Malt is germinated barley which contains enzymes which break the starch into a simpler carbohydrate called maltose, \(C_{12}H_{22}O_{11}\). Maltose has the same molecular formula as sucrose but contains two glucose units joined together, whereas sucrose contains one glucose and one fructose unit.
Yeast is then added and the mixture is kept warm (say 35°C) for perhaps several days until fermentation is complete. Air is kept out of the mixture to prevent oxidation of the ethanol produced to ethanoic acid (vinegar). Enzymes in the yeast first convert carbohydrates like maltose or sucrose into even simpler ones like glucose and fructose, both \(C_6H_{12}O_6\), and then convert these in turn into ethanol and carbon dioxide. You can show these changes as simple chemical equations, but the biochemistry of the reactions is much, much more complicated than this suggests.
\[ C_{12}H_{22}O_{11} + H_2O \longrightarrow 2C_6H_{12}O_6 \]
\[ C_6H_{12}O_6 \longrightarrow 2CH_3CH_2OH + 2CO2\]
Yeast is killed by ethanol concentrations in excess of about 15%, and that limits the purity of the ethanol that can be produced. The ethanol is separated from the mixture by fractional distillation to give 96% pure ethanol. For theoretical reasons (minimum boiling point azeotrope), it is impossible to remove the last 4% of water by fractional distillation.
Fermentation Hydration of ethene
Table 1.1.1: A comparison of fermentation with the direct hydration of ethene
Type of process A batch process. Everything is put into a container and then left until fermentation is complete. That batch is then cleared out and a new reaction set up. This is inefficient. A continuous flow process. A stream of reactants is passed continuously over a catalyst. This is a more efficient way of doing things.
Rate of reaction Very slow. Very rapid.
Quality of product Produces very impure ethanol which needs further processing Produces much purer ethanol.
Reaction conditions Uses gentle temperatures and atmospheric pressure. Uses high temperatures and pressures, needing lots of energy input.
Use of resources Uses renewable resources based on plant material. Uses finite resources based on crude oil.
8.5: Synthesis of Alcohols by Nucleophilic Substitution
Additional Resources
Carey 4th Edition On-Line Activity
Oxymercuration-Demercuration of Alkenes
Hydrolysis of Alkyl halides
Cliffs Notes
Slide Presentations | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.4%3A_Industrial__Sources__of_Alcohols%3A__Carbon__Monoxide.txt |
Objectives
After completing this section, you should be able to
1. determine whether a given reaction should be classified as an oxidation or a reduction.
2. write an equation to represent the reduction of an aldehyde or ketone using sodium borohydride or lithium aluminum hydride.
1. discuss the relative advantages and disadvantages of using sodium borohydride or lithium aluminum hydride to reduce aldehydes or ketones to alcohols.
2. identify the product formed from the reduction of a given aldehyde or ketone.
3. identify the aldehyde or ketone that should be used to produce a given alcohol in a reduction reaction.
4. identify the best reagent to carry out the reduction of a given aldehyde or ketone.
3. write an equation to represent the reduction of an ester or a carboxylic acid to an alcohol.
1. identify the product formed from the reduction of a given ester or carboxylic acid.
2. identify the esters or carboxylic acids that could be reduced to form a given alcohol.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• (organic) oxidation
• (organic) reduction
Study Notes
In your course in first‑year general chemistry, you probably discussed oxidation‑reduction reactions in terms of the transfer of electrons and changes in oxidation numbers (oxidation states). In organic chemistry, it is often more convenient to regard reduction as the gain of hydrogen or loss of oxygen, and oxidation as the gain of oxygen or the loss of hydrogen. There is no contradiction in using these various definitions. For example, when hydrogen is added across the double bond of ethene to reduce it to ethane, the oxidation number of the doubly bonded carbon atoms decreases from −II to −III. Similarly, when 2‑propanol
is oxidized to acetone
hydrogen is removed from the compound and the oxidation number of the central carbon atom increases from 0 to +II. If necessary, review the concept of oxidation number.
Reduction of Aldehydes and Ketones
The most common sources of the hydride Nucleophile are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH4 is more polar, thereby, making LiAlH4 a stronger reducing agent.
Addition of a hydride anion (H:-) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols.
In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH4 and NaBH4 are both capable of reducing aldehydes and ketones to the corresponding alcohol.
Mechanism
This mechanism is for a LiAlH4 reduction. The mechanism for a NaBH4 reduction is the same except methanol is the proton source used in the second step.
1) Nucleopilic attack by the hydride anion
2) The alkoxide is protonated
Biological Reduction
Addition to a carbonyl by a semi-anionic hydride, such as NaBH4, results in conversion of the carbonyl compound to an alcohol. The hydride from the BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
Formally, that process is referred to as a reduction. Reduction generally means a reaction in which electrons are added to a compound; the compound that gains electrons is said to be reduced. Because hydride can be thought of as a proton plus two electrons, we can think of conversion of a ketone or an aldehyde to an alcohol as a two-electron reduction. An aldehyde plus two electrons and two protons becomes an alcohol.
Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role.
NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Insetad of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD+. However, NAD+ is stabilized by the fact that its nicotinamide ring is aromatic; it was not aromatic in NADH.
Reduction of Carboxylic Acids and Esters
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4). Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (\(NaBH_4\)) is not a strong enough reducing agent to perform this reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.6%3A_Synthesis_of_Alcohols%3A_Oxidation-Reduction__Relatio.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_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.7%3A_Organometallic_Reagents%3A_Sources_of__Nucleophilic_C.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_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.8%3A_Organometallic_Reagents_in_the_Synthesis_of_Alcohols.txt |
Objective
After completing this section, you should be able to design a multistep synthesis to prepare a given product from a given starting material, using any of the reactions introduced in the textbook up to this point.
Study Notes
You should have noticed that some of the assigned problems have required that you string together a number of organic reactions to convert one organic compound to another when there is no single reaction to achieve this goal. Such a string of reactions is called an “organic synthesis.” One of the major objectives of this course is to assist you in designing such syntheses. To achieve this objective, you will need to have all of the reactions described in the course available in your memory. You will need to recall some reactions much more frequently than others, and the only way to master this objective is to practise. The examples given in this chapter will be relatively simple, but you will soon see that you can devise some quite sophisticated syntheses using a limited number of basic reactions.
Introduction
The study of organic chemistry introduces students to a wide range of interrelated reactions. Alkenes, for example, may be converted to structurally similar alkanes, alcohols, alkyl halides, epoxides, glycols and boranes; cleaved to smaller aldehydes, ketones and carboxylic acids; and enlarged by carbocation and radical additions as well as cycloadditions. Most of these reactions are shown in the Alkene Reaction Map below. All of these products may be subsequently transformed into a host of new compounds incorporating a wide variety of functional groups. Consequently, the logical conception of a multi-step synthesis for the construction of a designated compound from a specified starting material becomes one of the most challenging problems that may be posed. Functional group reaction maps like the one below for alkenes can be helpful in designing multi-step syntheses. It can be helpful to build and design your own reaction maps for each functional group studied.
Alkene Reaction Map
Please note: The reagents for each chemical transformation have been intentionally omitted so that this map can be used as a study tool. The answers are provided at the end of this section as part of the exercises.
Simple Multi-Step Syntheses
A one or two step sequence of simple reactions is not that difficult to deduce. For example, the synthesis of meso-3,4-hexanediol from 3-hexyne can occur by more than one multi-step pathway.
One approach would be to reduce the alkyne to cis or trans-3-hexene before undertaking glycol formation. Permanaganate or osmium tetroxide hydroxylation of cis-3-hexene would form the desired meso isomer.
From trans-3-hexene, it would be necessary to first epoxidize the alkene with a peracid followed by ring opening with acidic or basic hydrolysis.
Longer multi-step syntheses require careful analysis and thought, since many options need to be considered. Like an expert chess player evaluating the long range pros and cons of potential moves, the chemist must appraise the potential success of various possible reaction paths, focusing on the scope and limitations constraining each of the individual reactions being employed. The skill is acquired by practice, experience, and often trial and error.
Thinking it Through with 3 Examples
The following three examples illustrate strategies for developing multi-step syntheses from the reactions studied in the first ten chapters of this text. It is helpful to systematically look for structural changes beginning with the carbon chain and brainstorm relevant functional group conversion reactions. Retro-synthesis is the approach of working backwards from the product to the starting material.
In the first example, we are asked to synthesize 1-butanol from acetylene.
The carbon chain doubles in size indicating an acetylide SN2 reaction with an alkyl halide. Primary alcohol formation from an anti-Markovnikov alkene hydration reaction (hydroboration-oxidation) is more likely than a substitution reaction. Applying retro-synthesis, we work backwards from the alcohol to the alkene to the alkyne from an acetylide reaction that initially builds the carbon chain.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation identified from the retro-synthesis. The ethylbromide must also be derived from acetylene so multiple reaction pathways are combined as shown below.
In the second example, we are asked to synthesize 1,2-dibromobutane from acetylene.
Once again there is an increase in the carbon chain length indicating an acetylide SN2 reaction with an alkyl halide similar to the first example. The hydrohalogenation can be subtle to discern because the hydrogen atoms are not shown in bond-line structures. Comparing the chemical formulas of 1-butyne with 1,2-dibromobutane, there is a difference of two H atoms and two Br atoms indicating hydrohalogenation and not halogenation. The addition of both bromine atoms to the same carbon atom also supports the idea that hydrohalogenation occurs on an alkyne and not an alkene. The formation of the geminal dihalide also indicates hydrohalogenation instead of halogenation because halogenation produces vicinal dihalides. With this insight, the retro-synthesis indicates the following series of chemical transformations.
Retro-Synthesis
Working forwards, we specify the reagents needed for each transformation.
In the third example, we are asked to produce 6-oxoheptanal from methylcyclohexane.
Counting the carbons, the starting material and product both contain seven carbon atoms and there is a cleavage reaction of an alkene under reductive conditions. One important missing aspect of this reaction is a good leaving group (LG). Alkanes are chemically quite boring. We can burn them as fuel or perform free-radical halogenation to create alkyl halides with excellent leaving groups. With these observations, the following retro-synthesis is reasonable.
Retro-Synthesis
Working forwards, we specify the reagents needed for each reaction. For the initial free-radical halogenation of the alkane, we have the option of chlorine (Cl2) or bromine (Br2). Because methylcyclohexane has several different classifications of carbons, the selectivity of Br2 is more important than the faster reactivity of Cl2. A strong base with heat can be used for the second step to follow an E2 mechanism and form 1-methylcyclohexene. The aldehyde group on the final product indicates gentle oxidative cleavage by any of several reaction pathways. These reactions can be combined in to the following multi-step synthesis.
Reaction Maps to Build Functional Group Conversion Mastery
After working through the examples above, we can see how important it is to memorize all of the functional group reactions studied in the first ten chapters. We can apply the knowledge of these reactions to the wisdom of multi-step syntheses.
Please note: The reagents for each chemical transformation have been intentionally omitted so that these maps can be used as a study tools. The answers are provided at the end of this section as part of the exercises.
Alkyne Reaction Map
Exercise
1) Starting at 3-hexyne predict synthetic routes to achieve:
a) trans-3-hexene
b) 3,4-dibromohexane
c) 3-hexanol.
2) Starting with acetylene and any alkyl halides propose a synthesis to make
a) pentanal
b) hexane.
3) Show how you would accomplish the following synthetic transformations.
a)
b)
Answer
1)
2)
3)
a)
b) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/08._Hydroxy_of_Functional_Group%3A_Alcohols%3A_Properties_Preparation_and_Strategy_of_Synthesis/8.9%3A_Keys_to_Success%3A_An_Introduction_to_Synthetic_Strat.txt |
This page describes the reaction between alcohols and metallic sodium,and introduces the properties of the alkoxide that is formed. We will look at the reaction between sodium and ethanol as being typical, but you could substitute any other alcohol and the reaction would be the same.
The Reaction between Sodium Metal and Ethanol
If a small piece of sodium is dropped into ethanol, it reacts steadily to give off bubbles of hydrogen gas and leaves a colorless solution of sodium ethoxide: $CH_3CH_2ONa$. The anion component is an alkoxide.
$2CH_3CH_2OH_{(l)} + 2Na_{(s)} \rightarrow 2CH_3CH_2O^-_{(aq)} + 2Na^+_{(aq)} + H_{2(g)}$
If the solution is evaporated carefully to dryness, then sodium ethoxide ($CH_3CH_2ONa$) is left behind as a white solid. Although initially this appears as something new and complicated, in fact, it is exactly the same (apart from being a more gentle reaction) as the reaction between sodium and water - something you have probably known about for years.
$2H_2O_{(l)} + 2Na_{(s)} \rightarrow 2OH^-_{(aq)} + 2Na^+_{(aq)} + H_{2(g)}$
If the solution is evaporated carefully to dryness, then the sodium hydroxide ($NaOH$) is left behind as a white solid.
We normally, of course, write the sodium hydroxide formed as $NaOH$ rather than $HONa$ - but that's the only difference. Sodium ethoxide is just like sodium hydroxide, except that the hydrogen has been replaced by an ethyl group. Sodium hydroxide contains $OH^-$ ions; sodium ethoxide contains $CH_3CH_2O^-$ ions.
Note
The reason that the ethoxide formula is written with the oxygen on the right unlike the hydroxide ion is simply a matter of clarity. If you write it the other way around, it doesn't immediately look as if it comes from ethanol. You will find the same thing happens when you write formulae for organic salts like sodium ethanoate, for example.
There are two simple uses for this reaction:
• To safely dispose of small amounts of sodium: If you spill some sodium on the bench or have a small amount left over from a reaction you cannot simply dispose of it in the sink. It tends to react explosively with the water - and comes flying back out at you again! It reacts much more gently with ethanol. Ethanol is, therefore, used to dissolve small quantities of waste sodium. The solution formed can be washed away without problems (provided you remember that sodium ethoxide is strongly alkaline - see below).
• To test for the -OH group in alcohols: Because of the dangers involved in handling sodium, this is not the best test for an alcohol at this level. Because sodium reacts violently with acids to produce a salt and hydrogen, you would first have to be sure that the liquid you were testing was neutral. You would also have to be confident that there was no trace of water present because sodium reacts with the -OH group in water even better than with the one in an alcohol. With those provisos, if you add a tiny piece of sodium to a neutral liquid free of water and get bubbles of hydrogen produced, then the liquid is an alcohol.
Ethoxide Ions are Strongly Basic
If you add water to sodium ethoxide, it dissolves to give a colorless solution with a high pH. The solution is strongly alkaline because ethoxide ions are Brønsted-Lowry bases and remove hydrogen ions from water molecules to produce hydroxide ions, which increase the pH.
$CH_3CH_2O^- + H_2O \rightarrow CH_3CH_2OH + OH^-$
Ethoxide Ions are Good Nucleophiles
A nucleophile is a chemical species that carries a negative or partial negative charge that it uses to attack positive centers in other molecules or ions. Hydroxide ions are good nucleophiles, and you may have come across the reaction between a halogenoalkane (also called a haloalkane or alkyl halide) and sodium hydroxide solution. The hydroxide ions replace the halogen atom.
$CH_3CH_2CH_2Br + OH^- \rightarrow CH_3CH_2CH_2OH + Br^-$
In this case, an alcohol is formed. The ethoxide ion behaves in exactly the same way. If you knew the mechanism for the hydroxide ion reaction, you could work out exactly what happens in the reaction between a halogenoalkane and ethoxide ion.
Compare this equation with the last one.
$CH_3CH_2CH_2OH + CH_3CH_2Br \rightarrow CH_3CH_2CH_2OCH_2CH_3 + Br^-$
The only difference is that where there was a hydrogen atom at the right-hand end of the product molecule, an alkyl group is now present. Two alkyl (or other hydrocarbon) groups bridged by an oxygen atom is called an ether. This particular one is 1-ethoxypropane or ethyl propyl ether. This reaction is known as the Williamson Ether Synthesis and is a good method of synthesizing ethers in the lab.
Contributors
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9.02: Reactions of Alcohols with Strong Acids: Alkyloxonium Ions in Substitution and E
Purdue CHM 26100: Organic Chemistry for Engineers (1st Semester)
Fall 2014: Prof. Paul Wenthold | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.01%3A_Reactions_of_Alcohols__with__Base%3A__Preparation_of_Alkoxides.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_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.04%3A_Organic__and__Inorganic__Esters__from__Alcohols.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
Contributors
Objectives
After completing this section, you should be able to
1. write the normally accepted name for a crown ether, given its structure.
2. draw the structure of a crown ether, given its normally accepted name.
3. describe, briefly, the uses of crown ethers.
Key Terms
Make certain that you can define, and use in context, the key term below.
• crown ether
Study Notes
A “crown ether ” is a cyclic ether containing several (i.e., 4, 5, 6 or more) oxygen atoms. As we have indicated in the objectives above, a detailed knowledge of these compounds is not required in this course.
Crown Ethers
Crown ethers are cyclic polyethers with four or more oxygen atoms each separated by two or three carbon atoms. Crown ethers have the general formula of (OCH2CH2)n or (OCH2CH2CH2)n and are named using both the total number of atoms in the ring and the number of oxygen atoms. Thus 18-crown-6 is an 18-membered ring with six oxygen atoms. All crown ethers have a cavity in the center this is lined with oxygen atoms and can accommodate a alkali metal ion, such as K+. The cation is stabilized by interacting with lone pairs of electrons on the surrounding oxygen atoms. The negative character of center cavity of 18-crown-6 can be seen by looking at its electrostatic potential map shown below. The presence of high electron density is shown by a red color.
Crown ethers are useful for dissolving ionic substances in organic solvents, such as KMnO4 dissolving in toluene, by sequestering the cations inside a hydrophilic cavity, whereas the outer shell, consisting of C–H bonds, is hydrophobic.
The availability of crown ethers with cavities of different sizes allows specific cations to be solvated with a high degree of selectivity. Crown ethers prefer to bind alkali metal cations with sizes that match that of their binding cavity. For instance, as shown in Table-18.7.1, 14-crown-4 preferentially binds to Li+, 15-crown-5 preferentially binds to Na+, 18-crown-6 preferentially binds to K+, and 21-crown-7 preferentially binds to Cs+.
Table 18.7.1: Crown ethers preferentially bind cations with sizes that match the cavity size of their crown-shaped conformers.
Crown Ether Cavity Diameter (A˚) Preferred Cation Cation Diameter (A˚)
1.2-1.5 Li+ 1.36
1.7-2.2 Na+ 1.94
2.6-3.2 K+ 2.66
3.4-4.3 Cs+ 3.34
Cryptands
Cryptands (from the Greek kryptós, meaning “hidden”) are variations of crown ethers comprised of two nitrogens connected by three polyether strands. A common nomenclature is used where the numbers preceding the word cryptan indicate the number of oxygen atoms in each strand of the molecule.
Structures and approximate binding cavity diameters for several cryptands
Like crown ethers, cryptands are compounds that contains a central cavity that can completely surround a cation with lone pair electrons from oxygen and nitrogen atoms. Also, cryptands can be used to prepare solutions of ionic compounds in solvents that are otherwise too nonpolar to dissolve them. Similar to crown ethers, cryptands prefer to bind with alkali metal cations whose diameter matches the size of their binding cavity.
Potassium complex of 2,2,2-cryptand
Exercise \(1\)
Cryptand ligands also preferentially bind alkali metal ions with sizes that match the size of their binding cavities. The structure and approximate cavity sizes of several cryptands are shown above. Use the information in Table 18.7.1 to predict which alkali metal ion each cryptand will preferentially bind.
Answer
The cryptands might be expected to selectively bind the largest ion which fits within the cavity. Predicted selectivity of crown ethers for alkali metal cations based on the hypothesis that they will selectively bind the largest ion which fits their bindng cavity are listed below.
Cryptand Cavity Diameter of (Å) Preferred Cation Cation Diameter (Å)
2.1.1-cryptand 1.60 Li+ 1.36
2.2.1-cryptand 2.20 Na+ 1.96
2.2.2-cryptand 2.80 K+ 2.66
3.2.2-cryptand 3.60 Cs+ 3.34 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.05%3A_Names__and__Physical__Properties_of__Ethers.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the industrial preparation of simple symmetrical ethers.
2. write an equation to illustrate the Williamson synthesis of ethers.
1. identify the ether obtained from the reaction of a given alkyl halide with a given alkoxide ion.
2. identify the reagents needed to prepare a given ether through a Williamson synthesis.
3. identify the limitations of the Williamson synthesis, and make the appropriate choices when deciding how best to synthesize a given ether.
4. write an equation to describe the formation of an alkoxide from an alcohol.
5. identify silver(I) oxide as a reagent which can be used in a Williamson synthesis.
3. write an equation to show how an ether can be prepared by the alkoxymercuration-demercuration of an alkene.
1. identify the product formed from the alkoxymercuration-demercuration of a given alkene.
2. identify the alkene, the reagents, or both, needed to prepare a given ether by the alkoxymercuration-demercuration process.
3. write the detailed mechanism of the reaction between an alkene, an alcohol and mercury(II) trifluoroacetate.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alkoxymercuration
• oxymercuration
• Williamson ether synthesis
Study Notes
We studied oxymercuration as a method of converting an alkene to an alcohol in Section 8.5. “Alkoxymercuration” is a very similar process, except that we are now converting an alkene into an ether. The two processes are compared below.
Description oxymercuration alkoxymercuration
we react an alkene alkene
with water an alcohol
in the presence of Hg(O2CCH3)2 Hg(O2CCF3)2
followed by treatment with NaBH4 NaBH4
to produce an alcohol ether
Review the mechanism of the oxymercuration reaction in Section 8.5, paying particular attention to the regiochemistry and the stereochemistry of the reaction. The mechanism is identical to alkoxymercuration.
Ether Formation Though Dehydration
Acid-catalyzed dehydration of small 1º-alcohols constitutes a specialized industrial method of preparing symmetrical ethers. This reaction cannot be employed to prepare unsymmetrical ethers because a mixture of products is likely to be obtained. Also, 2o and 3o alcohols cannot be used for this reaction because they dehydrate to form alkenes by an E1 mechanism (Section 17-6).
\[\ce{2 CH_3CH_2-OH + H_2SO_4 ->[130\;^oC] CH_3CH_2\bond{-}O\bond{-}CH_2CH_3 + H_2O} \tag{18.2.1} \]
Mechanism
In the first step of the reaction mechanism, one alcohol is protonated to become a good leaving group. In the second step, a second alcohol displaces water from the protonated alcohol during an SN2 reaction yielding a protonated ether. In the final step, this intermediate is deprotonated to yield the symmetrical ether.
Williamson Ether Synthesis
One important procedure, known as the Williamson Ether Synthesis, proceeds by an SN2 reaction of an alkoxide nucleophile with a primary alkyl halide or tosylate. As previously discussed in Section 17-2, alkoxides are commonly created by deprotonating an alcohol with a strong base, such as sodium hydride (NaH). Simple alcohols can be used a solvent during a Williamson ether synthesis and with their alkoxide created through the addition of sodium metal (Na(s)).
Planning a Williamson Ether Synthesis
The Williamson ether synthesis has the same limitations as other SN2 reactions, as discussed in Section 11-3. Since alkoxide anions are strong bases, utilizing 2o or 3o halogen leaving groups could possibly produce an E2 elimination product. When considering the synthesis of an unsymmetrical ether, there are two different combinations of reactants possible and each should be carefully considered. In general, the pathway which utilizes the least sterically hindered halogen will be preferred.
The key bond cleavage in the target molecule involves a C-O bond. Because unsymmetrical ethers have two unique C-O bonds, each can be broken to provide a unique set of reactants. After cleavage, the fragment with the oxygen will become an alkoxide. The other fragment will become a halogen or tosylate.
Worked Example \(1\)
How would you prepare the following molecule using a Williamson Ether Synthesis?
Answer
Analysis: The ether is asymmetrical so each of the C-O bonds can be broken to create a different set of possible reactants. After cleavage of the C-O bond, pathway 1 shows a 3o halogen as the starting material. This reaction will most likely not be effective due to alkoxides reacting with 3o halogens to preferable form an alkenes by E2 elimination. Pathway 2 shows a 1o halogen as a starting material which is favorable for SN2 reactions.
Pathway 1
Solution 1
Pathway 2
Solution 2
Ether Synthesis Using Silver Oxide
A variation of the Williamson ether synthesis uses silver oxide (Ag2O) in the place of the strong base. The conditions of this variation are milder than the typical Willamson synthesis because a strong base and the formation of an alkoxide intermediate are not necessary. This reaction is particually useful when converting the -OH groups on a sugar into ethers.
Mechanism
During this reaction a partial positively charged silver in Ag2O gives draws electron density from the iodine in CH3I. This correspondingly removes electron density from the adjacent carbon increasing its partial positive charge which increases its electrophlicity. This allows the alcohol to act as a nucleophile in the subsequent SN2 reaction.
Ether Synthesis Using Alkoxymercuration
Alkoxymercuration, is patterned after the oxymercuration reaction discussed in Section 8-4. Reaction of an alkene with an alcohol in the presence of a trifluoroacetate mercury (II) salt [(CF3CO2)2Hg] prodcues an alkoxymercuration product. Demercuration using sodium borohydride (NaBH4) yields an ether product. Overall, this reaction allows for the Markovnikov addition of an alcohol to an alkene to create an ether. Note that the alcohol reactant is used as the solvent, and a trifluoroacetate mercury (II) salt is used in preference to the mercuric acetate (trifluoroacetate anion is a poorer nucleophile than acetate). Most 1o, 2o, 3o alcohols can be successfully used for this reaction.
Mechanism
The mechanism of alkoxymercuration is similar to that of oxymercuration, with electrophillic addition of the mercuric species to the alkene. The alcohol nucleophile attacks the more substituted carbon of the three-membered ring via a SN2 reaction. Finally, sodium borohydride (NaBH4) provides a reductive demercuration to form the ether product.
Planning the Synthesis of an Ether using Alkoxymercuration
The key bond cleavage in the target molecule involves a C-O bond. Because unsymmetrical ethers have two unique C-O bonds, each can be broken to provide a unique set of reactants. After cleavage, the fragment with the oxygen will become an alcohol. The alkyl fragment will lose a hydrogen from a adjacent carbon to form an alkene. The main point to consider when choosing a possible synthesis pathways is the ability of the alkyl fragment to form an alkene.
Worked Example \(2\)
How would you prepare the following molecule using a alkoxymercuration?
Answer
Analysis: The ether is symmetrical so each C-O bond of the ether can be cleaved to produce a set of starting materials for consideration. Pathway one shows a set of starting material which should work well for this reaction. The alcohol, methanol, can easily be used as a solvent. Although the alkene does not have a defined more and less substituted side, its symmetry will prevent a mixture of product from forming. The fragmentation for pathway 2 shows starting material which are not viable for this reaction. The alkyl fragment only has one carbon which cannot be used to form an alkene starting material. This means pathway 2 is not a viable method for the synthesis of the target molecule.
Pathway 1
Solution 1
Pathway 2
Exercises
Exercise \(1\)
When preparing ethers using the Williamson ether synthesis, what factors are important when considering the nucleophile and the electrophile?
Answer
The nucleophile ideally should be very basic, yet not sterically hindered. This will minimize any elimination reactions from occurring. The electrophile should have the characteristics of a good SN2 electrophile, preferably primary to minimize any elimination reactions from occurring.
Exercise \(2\)
How would you synthesize the following ethers? Keep in mind there are multiple ways. The Williamson ether synthesis, alkoxymercuration of alkenes, and also the acid catalyzed substitution.
(a) (b) (c) (d) (e)
Answer
The Williamson ether syntheses require added catalytic base. Also, most of the halides can be interchanged, say for example for a -Br or a -Cl. Although, typically -I is the best leaving group.
(a)
(b)
(c)
(d)
Note, there is only one ether (also called a silyl ether, and often used as an alcohol protecting group.) The other group is an ester.
(e)
Exercise \(3\)
Draw the electron arrow pushing mechanism for the formation of diethyl ether in the previous problem.
Answer
.
Exercise \(4\)
t-butoxycyclohexane can be prepared two different ways from an alkene and an alcohol, draw both possible reactions.
Answer
While both are possible, the top route is likely easier because both starting materials are a liquid.
Exercise \(5\)
Epoxides are often formed intramolecularly. Take for example this large ring, in a publication from 2016 [J. Org. Chem., 2016, 81 (20), pp 10029–10034]. If subjected to base, what epoxide would be formed? (Include stereochemistry)
Answer
Exercise \(6\)
What reagents would you use to perform the following transformations?
(a)
(b)
(c)
Answer
(a)
(b) Note the cis addition
(c)
An oxidation to an alcohol through hydroboration, and subsequent substitution with 2-bromopropane could also work, but this route provides the least likelihood of an elimination reaction occurring.
Exercise \(7\)
Predict the product of the following.
Answer
The result is the production of dioxane, a common solvent. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.06%3A_Williamson_Ether_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the industrial preparation of simple symmetrical ethers.
2. write an equation to illustrate the Williamson synthesis of ethers.
1. identify the ether obtained from the reaction of a given alkyl halide with a given alkoxide ion.
2. identify the reagents needed to prepare a given ether through a Williamson synthesis.
3. identify the limitations of the Williamson synthesis, and make the appropriate choices when deciding how best to synthesize a given ether.
4. write an equation to describe the formation of an alkoxide from an alcohol.
5. identify silver(I) oxide as a reagent which can be used in a Williamson synthesis.
3. write an equation to show how an ether can be prepared by the alkoxymercuration-demercuration of an alkene.
1. identify the product formed from the alkoxymercuration-demercuration of a given alkene.
2. identify the alkene, the reagents, or both, needed to prepare a given ether by the alkoxymercuration-demercuration process.
3. write the detailed mechanism of the reaction between an alkene, an alcohol and mercury(II) trifluoroacetate.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alkoxymercuration
• oxymercuration
• Williamson ether synthesis
Study Notes
We studied oxymercuration as a method of converting an alkene to an alcohol in Section 8.5. “Alkoxymercuration” is a very similar process, except that we are now converting an alkene into an ether. The two processes are compared below.
Description oxymercuration alkoxymercuration
we react an alkene alkene
with water an alcohol
in the presence of Hg(O2CCH3)2 Hg(O2CCF3)2
followed by treatment with NaBH4 NaBH4
to produce an alcohol ether
Review the mechanism of the oxymercuration reaction in Section 8.5, paying particular attention to the regiochemistry and the stereochemistry of the reaction. The mechanism is identical to alkoxymercuration.
Ether Formation Though Dehydration
Acid-catalyzed dehydration of small 1º-alcohols constitutes a specialized industrial method of preparing symmetrical ethers. This reaction cannot be employed to prepare unsymmetrical ethers because a mixture of products is likely to be obtained. Also, 2o and 3o alcohols cannot be used for this reaction because they dehydrate to form alkenes by an E1 mechanism (Section 17-6).
\[\ce{2 CH_3CH_2-OH + H_2SO_4 ->[130\;^oC] CH_3CH_2\bond{-}O\bond{-}CH_2CH_3 + H_2O} \tag{18.2.1} \]
Mechanism
In the first step of the reaction mechanism, one alcohol is protonated to become a good leaving group. In the second step, a second alcohol displaces water from the protonated alcohol during an SN2 reaction yielding a protonated ether. In the final step, this intermediate is deprotonated to yield the symmetrical ether.
Williamson Ether Synthesis
One important procedure, known as the Williamson Ether Synthesis, proceeds by an SN2 reaction of an alkoxide nucleophile with a primary alkyl halide or tosylate. As previously discussed in Section 17-2, alkoxides are commonly created by deprotonating an alcohol with a strong base, such as sodium hydride (NaH). Simple alcohols can be used a solvent during a Williamson ether synthesis and with their alkoxide created through the addition of sodium metal (Na(s)).
Planning a Williamson Ether Synthesis
The Williamson ether synthesis has the same limitations as other SN2 reactions, as discussed in Section 11-3. Since alkoxide anions are strong bases, utilizing 2o or 3o halogen leaving groups could possibly produce an E2 elimination product. When considering the synthesis of an unsymmetrical ether, there are two different combinations of reactants possible and each should be carefully considered. In general, the pathway which utilizes the least sterically hindered halogen will be preferred.
The key bond cleavage in the target molecule involves a C-O bond. Because unsymmetrical ethers have two unique C-O bonds, each can be broken to provide a unique set of reactants. After cleavage, the fragment with the oxygen will become an alkoxide. The other fragment will become a halogen or tosylate.
Worked Example \(1\)
How would you prepare the following molecule using a Williamson Ether Synthesis?
Answer
Analysis: The ether is asymmetrical so each of the C-O bonds can be broken to create a different set of possible reactants. After cleavage of the C-O bond, pathway 1 shows a 3o halogen as the starting material. This reaction will most likely not be effective due to alkoxides reacting with 3o halogens to preferable form an alkenes by E2 elimination. Pathway 2 shows a 1o halogen as a starting material which is favorable for SN2 reactions.
Pathway 1
Solution 1
Pathway 2
Solution 2
Ether Synthesis Using Silver Oxide
A variation of the Williamson ether synthesis uses silver oxide (Ag2O) in the place of the strong base. The conditions of this variation are milder than the typical Willamson synthesis because a strong base and the formation of an alkoxide intermediate are not necessary. This reaction is particually useful when converting the -OH groups on a sugar into ethers.
Mechanism
During this reaction a partial positively charged silver in Ag2O gives draws electron density from the iodine in CH3I. This correspondingly removes electron density from the adjacent carbon increasing its partial positive charge which increases its electrophlicity. This allows the alcohol to act as a nucleophile in the subsequent SN2 reaction.
Ether Synthesis Using Alkoxymercuration
Alkoxymercuration, is patterned after the oxymercuration reaction discussed in Section 8-4. Reaction of an alkene with an alcohol in the presence of a trifluoroacetate mercury (II) salt [(CF3CO2)2Hg] prodcues an alkoxymercuration product. Demercuration using sodium borohydride (NaBH4) yields an ether product. Overall, this reaction allows for the Markovnikov addition of an alcohol to an alkene to create an ether. Note that the alcohol reactant is used as the solvent, and a trifluoroacetate mercury (II) salt is used in preference to the mercuric acetate (trifluoroacetate anion is a poorer nucleophile than acetate). Most 1o, 2o, 3o alcohols can be successfully used for this reaction.
Mechanism
The mechanism of alkoxymercuration is similar to that of oxymercuration, with electrophillic addition of the mercuric species to the alkene. The alcohol nucleophile attacks the more substituted carbon of the three-membered ring via a SN2 reaction. Finally, sodium borohydride (NaBH4) provides a reductive demercuration to form the ether product.
Planning the Synthesis of an Ether using Alkoxymercuration
The key bond cleavage in the target molecule involves a C-O bond. Because unsymmetrical ethers have two unique C-O bonds, each can be broken to provide a unique set of reactants. After cleavage, the fragment with the oxygen will become an alcohol. The alkyl fragment will lose a hydrogen from a adjacent carbon to form an alkene. The main point to consider when choosing a possible synthesis pathways is the ability of the alkyl fragment to form an alkene.
Worked Example \(2\)
How would you prepare the following molecule using a alkoxymercuration?
Answer
Analysis: The ether is symmetrical so each C-O bond of the ether can be cleaved to produce a set of starting materials for consideration. Pathway one shows a set of starting material which should work well for this reaction. The alcohol, methanol, can easily be used as a solvent. Although the alkene does not have a defined more and less substituted side, its symmetry will prevent a mixture of product from forming. The fragmentation for pathway 2 shows starting material which are not viable for this reaction. The alkyl fragment only has one carbon which cannot be used to form an alkene starting material. This means pathway 2 is not a viable method for the synthesis of the target molecule.
Pathway 1
Solution 1
Pathway 2
Exercises
Exercise \(1\)
When preparing ethers using the Williamson ether synthesis, what factors are important when considering the nucleophile and the electrophile?
Answer
The nucleophile ideally should be very basic, yet not sterically hindered. This will minimize any elimination reactions from occurring. The electrophile should have the characteristics of a good SN2 electrophile, preferably primary to minimize any elimination reactions from occurring.
Exercise \(2\)
How would you synthesize the following ethers? Keep in mind there are multiple ways. The Williamson ether synthesis, alkoxymercuration of alkenes, and also the acid catalyzed substitution.
(a) (b) (c) (d) (e)
Answer
The Williamson ether syntheses require added catalytic base. Also, most of the halides can be interchanged, say for example for a -Br or a -Cl. Although, typically -I is the best leaving group.
(a)
(b)
(c)
(d)
Note, there is only one ether (also called a silyl ether, and often used as an alcohol protecting group.) The other group is an ester.
(e)
Exercise \(3\)
Draw the electron arrow pushing mechanism for the formation of diethyl ether in the previous problem.
Answer
.
Exercise \(4\)
t-butoxycyclohexane can be prepared two different ways from an alkene and an alcohol, draw both possible reactions.
Answer
While both are possible, the top route is likely easier because both starting materials are a liquid.
Exercise \(5\)
Epoxides are often formed intramolecularly. Take for example this large ring, in a publication from 2016 [J. Org. Chem., 2016, 81 (20), pp 10029–10034]. If subjected to base, what epoxide would be formed? (Include stereochemistry)
Answer
Exercise \(6\)
What reagents would you use to perform the following transformations?
(a)
(b)
(c)
Answer
(a)
(b) Note the cis addition
(c)
An oxidation to an alcohol through hydroboration, and subsequent substitution with 2-bromopropane could also work, but this route provides the least likelihood of an elimination reaction occurring.
Exercise \(7\)
Predict the product of the following.
Answer
The result is the production of dioxane, a common solvent. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.07%3A_Synthesis_of_Ethers%3A__Alcohols__and__Mineral__Acids.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the acidic cleavage of an ether.
2. identify the products formed when a given ether is cleaved by a strong acid.
3. identify the reagent needed to bring about cleavage of a given ether.
4. deduce the structure of an unknown ether, given the products of acidic cleavage of the ether.
5. write the detailed mechanism for the acidic cleavage of a given ether.
Study Notes
There are a number of points in this section that require additional explanations.
First, if an excess of HI (or HBr) is used in the cleavage reaction, the alcohol formed is converted by a nucleophilic substitution reaction to the appropriate alkyl halide:
ROH + HI → RI + H2O
In view of this substitution, some textbooks simplify the overall cleavage process as:
R\$\ce{-}\$O\$\ce{-}\$R′ + 2HI → RI + R′I + H2O
Second, we should consider in detail how certain ethers (those containing tertiary alkyl, benzyl or allyl groups) cleave by an SN1 mechanism:
Finally, notice that an aryl alkyl ether will always produce a phenol and an alkyl halide, never an aryl halide and an alcohol. This is because we rarely see a nucleophile attacking an aromatic ring carbon in preference to an aliphatic carbon:
As phenols do not undergo nucleophilic substitution reactions, even if an excess of HX is used, the products from the cleavage of an aryl alkyl ether are a phenol and an alkyl halide. Diaryl ethers are not cleaved by acids.
Ethers are known to be unreactive towards most reagents which makes them excellent reaction solvents. The most common reaction of ethers is cleavage of the C–O bond by using strong acids. During acidic cleavage the ether oxygen is protonated to form a good leaving groups which can be eliminated as part of an SN2, SN1, or E1 reaction mechanism. The mechanistic pathway is primarily determined by the strong acid used and the type of substituents attached to the ether.
Acidic Cleavage of Ethers
Aqueous solutions of HBr or HI (but not HCl) tend to cleave ethers into alcohol and an alkyl halide product by either an SN2 or SN1 mechanism. If the ether is attached to only primary, secondary, or methyl alkyl groups, a selective cleavage will typically take place using an SN2 mechanism. First, the strong acid protonates the ether oxygen. Then resulting halide conjugate base attacks the protonated ether at the less sterically hindered alkyl substituent forming a halogen product. The ether's more sterically hindered alkyl substituent is ejected as a leaving group and forms an alcohol product. The example below show that when ethyl isopropyl ether is cleaved with hydrobromic acid the products isopropyl alcohol and bromoethane are produced. The bromide nucleophile preferably attacks the ether's ethyl substituent because it is less hindered (1o) than the isopropyl substituent (2o).
It is important to note that a phenyl substituent on an ether is not capable of participating in the SN2 reaction of an acidic cleavage. If a phenyl group is present it will become a phenol in the product due to the halide nucleophile preferably attacking the other alkyl substituent.
When using HBr or HI, the acidic cleavage of ethers with tertiary, benzylic, or allylic substituents tend to occur by an SN1 mechanism. The ability of these substituents to produce relatively stable carbocations promotes the SN1 mechanism. The change in mechanism causes the tertiary, benzylic, or allyic group to preferably become the halogen product of the acidic cleavage. This makes the ether's other alkyl substituent become the alcohol product.
When using a strong acid whose conjugate base is a poor nucleophile, such as trifluoroacetic acid (CF3CO2H), for the the acidic cleavage of an ether with a tertiary alkyl substituent, the mechanism will often be E1. In this case the tertiary alkyl substituent will lose an adjacent hydrogen to form an alkene product. The ether's other alkyl substituent will form an alcohol product.
Worked Example \(1\)
Predict the products of the following reaction:
Answer
Analysis: When considering the acidic cleavage of ethers it is important to realize that SN2, SN1, or E1 reactions are possible depending on the conditions. First, identify if the ether has a substituent which can easily form a carbocation: tertiary, benzylic, or allylic substituents. If none of these substituents are present the reaction will most likely be SN2. However, if one of the substituents is present the reaction will most likely be SN1 or E1. Next, identify if the reaction conditions will allow for an E1 reaction. These are the presence of a tertiary alkyl substituent on the ether along with the use of a strong acid which has a poor nucleophile as a conjugate base. If this set of conditions is not present the reaction will most likely be SN1.
For a SN2 reaction: The less hindered alkyl substituent of the ether will become a halogen. The other alkyl substituent will become an alcohol.
For a SN1 reaction: The substituent which easily forms a carbocation will become a halogen and the other alkyl substituent will become an alcohol.
For an E1 reaction: The tertiary alkyl substutent will form an alkene and the other alkyl substiutent will become an alcohol.
For the reaction proposed above, the reactant contains a benzylic substituent so the reaction will most likely be SN1. Consequently, the benzylic substituent will become and Iodide product and the propyl substituent will become an alcohol.
Exercises \(1\)
1) Predict the product of the following reactions:
a)
b)
2) Please draw the mechanism for the following reaction:
3) Why is HCl less effective at cleaving ethers than HBr or HI?
Answer
1)
a)
b)
2)
3) HCl's conjugate base Cl- is a poor nucleophile when compared to Br- and I-. The chloride anion is not a strong enough nucleophile to promote the acidic cleavage of ethers.
Contributors and Attributions
Objectives
After completing this section, you should be able to:
1. explain what is meant by “protecting” a functional group during an organic synthesis.
2. describe one common method for protecting the hydroxy group of an alcohol, and give an example of its use (e.g., in the preparation of a Grignard reagent).
Often during the synthesis of complex molecules on functional group in a molecule interferes with an intended reaction on a second functional group on the same molecule. An excellent example is the fact that a Grignard reagent can't be prepared from halo alcohol because the C-Mg bond is not compatible with the acidic -OH group.
When situations like this occurs chemists circumvent eh problem by protecting the interfering functional group. Functional group protection involves three steps:
1. Blocking the interfering functionality by introducing a protecting group.
2. Performing the intended reaction.
3. Removing the protecting group and reforming the original functional group.
There are several methods for protecting an alcohol, however, the most common is the reaction with a chlorotrialkylsilane, Cl-SiR3 This reactions forms a trialkylsilyl ether, R'-O-SiR3. Chlorotrimethylsilane is often used in conjuction with a base, such as triethylamine, The base helps to form the alkoxide anion and remove the HCl produced by the reaction.
Example
The silyl ether protecting group can be removed by reaction with an aqueous acid or the fluoride ion.
By utilizing a protecting group a Grignard reagent can be formed and reacted on a halo alcohol.
1) Protect the Alcohol
2) Form the Grignard Reagent
3) Perform the Grignard Reaction
4) Deprotection
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.08%3A_Reactions_of_Ethers.txt |
Objectives
After completing this section, you should be able to
1. write an equation to describe the opening of an epoxide ring under mildly acidic conditions.
1. identify the product formed from the hydrolysis of an epoxide.
2. write the mechanism for the opening of an epoxide ring by an aqueous acid, paying particular attention to the stereochemistry of the product.
3. identify the product formed when an epoxide ring is opened by a hydrogen halide under anhydrous conditions.
2. predict the major product from the acidic cleavage of a given unsymmetrical epoxide.
3. write an equation to illustrate the cleavage of an epoxide ring by a base.
1. identify the product formed from the reaction of a given epoxide with given base.
2. explain why epoxides are susceptible to cleavage by bases, whereas other cyclic ethers are not.
Study Notes
In the discussion on base-catalyzed epoxide opening, the mechanism is essentially SN2. While oxygen is a poor leaving group, the ring strain of the epoxide really helps to drive this reaction to completion. Indeed, larger cyclic ethers would not be susceptible to either acid-catalyzed or base-catalyzed cleavage under the same conditions because the ring strain is not as great as in the three-membered epoxide ring.
Epoxide Ring-Opening by Alcoholysis
The ring-opening reactions of epoxides provide a nice overview of many of the concepts discussed in earlier chapters of this book. Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions. If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates. When an asymmetric epoxide undergoes alcoholysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon is the site of nucleophilic attack, leading to what we will refer to as product B:
Conversely, when solvolysis occurs in acidic methanol, the reaction occurs by a mechanism with substantial SN1 character, and the more substituted carbon is the site of attack. As a result, product A predominates.
These are both good examples of regioselective reactions. In a regioselective reaction, two (or more) different constitutional isomers are possible as products, but one is formed preferentially (or sometimes exclusively).
Basic Epoxide Ring-Opening by Alcoholysis
In the basic, SN2 reaction, the leaving group is an alkoxide anion, because there is no acid available to protonate the oxygen prior to ring opening. An alkoxide is a poor leaving group (Section 11-3), and thus the ring is unlikely to open without a 'push' from the nucleophile.
The nucleophile itself is potent: a deprotonated, negatively charged methoxide ion. When a nucleophilic substitution reaction involves a poor leaving group and a powerful nucleophile, it is very likely to proceed by an SN2 mechanism.
There are two electrophilic carbons in the epoxide, but the best target for the nucleophile in an SN2 reaction is the carbon that is least hindered. This accounts for the observed regiochemical outcome. Like in other SN2 reactions, nucleophilic attack takes place from the backside, resulting in inversion at the electrophilic carbon.
Acid-Catalyzed Epoxide Ring-Opening by Alcoholysis
The best way to depict the acid-catalyzed epoxide ring-opening reaction is as a hybrid, or cross, between an SN2 and SN1 mechanism. First, the oxygen is protonated, creating a good leaving group (step 1 below). Then the carbon-oxygen bond begins to break (step 2) and positive charge begins to build up on the more substituted carbon. Recall that alkyl substituents can donate electron density through hyper conjugation and stabilize a positive charge on a carbon.
Unlike in an SN1 reaction, the nucleophile attacks the electrophilic carbon (step 3) before a complete carbocation intermediate has a chance to form.
Attack takes place preferentially from the backside (like in an SN2 reaction) because the carbon-oxygen bond is still to some degree in place, and the oxygen blocks attack from the front side. Notice, however, how the regiochemical outcome is different from the base-catalyzed reaction: in the acid-catalyzed process, the nucleophile attacks the more substituted carbon because it is this carbon that holds a greater degree of positive charge.
Example 18.6.1
Predict the major product(s) of the ring opening reaction that occurs when the epoxide shown below is treated with:
1. ethanol and a small amount of sodium hydroxide
2. ethanol and a small amount of sulfuric acid
Hint: be sure to consider both regiochemistry and stereochemistry!
Answer
Epoxide Ring-Opening by Hydrolysis
Epoxides may be cleaved by hydrolysis to give trans-1,2-diols (1,2 diols are also called vicinal diols or vicinal glycols). The reaction can be preformed under acidic or basic conditions which will provide the same regioselectivity previously discussed.
Acid Catalyzed Hydrolysis
Under aqueous acidic conditions the epoxide oxygen is protonated and is subsequently attacked by a nucleophilic water. After deprotonation to reform the acid catalyst a 1,2-diol product is formed. If the epoxide is asymmetric, the incoming water nucleophile will preferably attack the more substituted epoxide carbon. The epoxide ring is opened by an SN2 like mechanism so the two -OH groups will be trans to each other in the product.
Basic Hydrolysis
Under aqueous basic conditions the epoxide is opened by the attack of hydroxide nucleophile during an SN2 reaction. The epoxide oxygen forms an alkoxide which is subsequently protonated by water forming the 1,2-diol product. If the epoxide is asymmetric the incoming hydroxide nucleophile will preferable attack the less substituted epoxide carbon. Because the reaction takes place by an SN2 mechanism the two -OH groups in the product will be trans to each other.
Epoxide Ring-Opening by HX
Epoxides can also be opened by anhydrous acids (HX) to form a trans halohydrin. When both the epoxide carbons are either primary or secondary the halogen anion will attack the less substituted carbon through an SN2 like reaction. However, if one of the epoxide carbons is tertiary, the halogen anion will primarily attack the tertiary carbon in an SN1 like reaction.
Epoxide Ring-Opening by Other Basic Nucleophiles
A wide variety of basic nucleophiles can be used for the ring opening of an epoxide including, amines, hydrides, Grignard reagents, acetylide anions, and hydride. These ring openings generally take place by an SN2 mechanism.
Reacting Grignard reagents with ethylene oxide is a particuarly useful reaction because it produces a primary alcohol containing two more carbon atoms than the original Grignard reagent.
This reaction follows the same SN2 mechanism as the opening of epoxide rings under basic conditions since Grignard reagents are both strong nucleophiles and strong bases. The first step of the mechanism of this reaction involves the SN2 attack of the Grignard reaction to open the epoxide to form an alkoxide. The second step of the mechanism involves the protonation of the alkoxide to form an alcohol.
Additional Stereochemical Considerations of Ring-Opening
During the ring-opening of an asymmetrical epoxide, the regiochemical control of the reaction usually allows for one stereoisomer to be produced. However, if the epoxide is symmetrical, each epoxide carbon has roughly the same ability to accept the incoming nucleophile. When this occurs the product typically contains a mixture of enantiomers.
Exercises
Exercise \(1\)
Given the following, predict the product assuming only the epoxide is affected. (Remember to show stereochemistry)
Answer
Note that the stereochemistry has been inverted
Exercise \(2\)
Predict the product of the following, similar to above but a different nucleophile is used and not in acidic conditions. (Remember stereochemistry)
Answer
Exercise \(3\)
Provide the structure of the product of the following reaction. Be sure to include proper stereochemistry.
Answer
The ring side of the protonated epoxide intermediate will better stabilize a partial positive charge, so would be the more likely carbon for the chloride ion to attack.
Exercise \(4\)
Predict the product of the following reaction.
Answer
Exercise \(5\)
Provide the final products of the following reactions.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.09%3A_Reactions__of_Oxacyclopropanes.txt |
Objectives
After completing this section, you should be able to
1. write the IUPAC name of a thiol, given its Kekulé, condensed or shorthand structure.
2. draw the structure of a thiol, given its IUPAC name.
3. write an equation to represent the formation of a thiol by the reaction of hydrosulfide anion with an alkyl halide.
4. write an equation to illustrate the preparation of a thiol by the reaction of thiourea with an alkyl halide.
1. write an equation to show the interconversion between thiols and disulfides.
1. write the name of a sulfide, given its structure.
2. draw the structure of a sulfide, given its name.
3. write an equation showing how a sulfide may be prepared by the reaction of a thiolate anion on an alkyl halide.
4. identify the product from the reaction of a given alkyl halide with a given thiolate anion.
5. identify the reagents necessary to prepare a given sulfide.
6. write an equation to illustrate the formation of a trialkylsulfonium salt from a sulfide and an alkyl halide.
Key Terms
Note: All of these terms are defined in the “Study Notes,” below.
• disulfide
• mercapto group
• (organic) sulfide
• sulfone
• sulfoxide
• thiol
• thiolate anion
• trialkylsulfonium ion (trialkylsulfonium salt)
Study Notes
The chemistry of sulfur-containing organic compounds is often omitted from introductory organic chemistry courses. However, we have included a short section on these compounds, not for the sake of increasing the amount of material to be digested, but because much of the chemistry of these substances can be predicted from a knowledge of their oxygen-containing analogues.
A thiol is a compound which contains an SH functional group. The SH group itself is called a mercapto group. A disulfide is a compound containing an S-S linkage.
(Organic) sulfides have the structure R-S-R′, and are therefore the sulfur analogues of ethers. The nomenclature of sulfides can be easily understood if one understands the nomenclature of the corresponding ethers. Notice that the term “thio” is also used in inorganic chemistry. For example, SO42 is the sulfate ion; while S2O32, in which one of the oxygen atoms of a sulfate ion has been replaced by a sulfur atom, is called thiosulfate.
Thiolate anions, RS-, are analogous to alkoxy anions, RO-. Thiolate anions are better nucleophiles than are alkoxy anions.
If you have trouble understanding why trialkylsulfonium ions are formed, think of them as being somewhat similar to the hydronium ions that are formed by protonating water:
Later we shall see examples of tetraalkylammonium ions, R4N+, which again may be regarded as being similar to hydronium ions.
Sulfoxides and sulfones are obtained by oxidizing organic sulfides. You need not memorize the methods used to carry out these oxidations.
Table 18.1, below, provides a quick comparison of oxygen-containing and sulfur-containing organic compounds.
Table 18.1 Comparison of compounds containing oxygen and sulfur
Oxygen-containing Compound Sulfur Analogue
ether, R-O-R′ sulfide, R-S-R′
alkoxy anion, RO- thiolate anion, RS-
alcohol, ROH thiol, RSH
hydroxy group, OH- mercapto group, SH-
peroxide, R-O-O-R′ disulfide, R-S-S-R′
Note that when we name thiols, we include the “e” of the alkane name. Thus, CH3CH2SH is called “ethanethiol,” not “ethanthiol.”
Thiols and sulfides are the "sulfur equivalent" of alcohols and ethers. You can replace the oxygen atom of an alcohol with a sulfur atom to make a thiol; similarly, you can replace the oxygen atom in an ether with S to make the corresponding alkyl sulfide. This is because thiols contain the C-S-H functional group, while sulfides contain the C-S-C group.
Oxidation States of Sulfur Compounds
Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange).
Thiols
Thiols are often called “mercaptans,” a reference to the Latin term mercurium captans (capturing mercury), since the -SH group forms strong bonds with mercury and its ions. Thiols are analogous to alcohols. Thiols are weakly acidic (pKa ~ 10) and are much stronger acids than alcohols (pKa ~ 16). However, thiols usually do not form hydrogen bonds due to the sulfur atom not have sufficient electronegativity. Thiols named using the same rules as alcohols except the parent chain is named as alkane with the suffix -thiol added. As a substituent the -SH group is called a mercapto group.
Thiols are usually prepared by using the hydrosulfide anion (-SH) as a nucleophile in an SN2 reaction with alkyl halides.
One problem with this reaction is that the thiol product can deprotonate and undergo a second SN2 reaction with an additional alkyl halide to produce a sulfide side product. This problem can be solved by using thiourea, (NH2)2C=S, as the nucleophile. The SN2 reaction first produces an alkyl isothiourea salt as an intermediate. This salt is then hydrolyzed to form the thiol by a reaction with aqueous base.
Disulfides
Thiols (R-S-H) can be oxidized to form disulfides (R-S-S-R') through reaction with Br2 or I2. Note, an equivalent oxidation of alcohols (R-O-H) to peroxides (R-O-O-R') is not normally observed. The reasons for this different behavior becomes clear when looking at bond strengths. The S–S single bond in disulfides is nearly twice as strong as the O–O bond in peroxides. Also, the O–H bond in alcohols is more than 25 kcal/mole stronger than the S–H bond in thiols. Thus, thermodynamics favors disulfide formation over peroxide. Disulfides can easily be reduced back to a thiols through reaction with zinc and acid.
Disulfide bridges in proteins
Disulfide (sulfur-sulfur) linkages between two cysteine residues are an integral component of the three-dimensional structure of many proteins. The interconversion between thiols and disulfide groups is a redox reaction: the thiol is the reduced state, and the disulfide is the oxidized state.
Notice that in the oxidized (disulfide) state, each sulfur atom has lost a bond to hydrogen and gained a bond to a sulfur - this is why the disulfide state is considered to be oxidized relative to the thiol state. The redox agent that mediates the formation and degradation of disulfide bridges in most proteins is glutathione, a versatile coenzyme. Recall that the important functional group in glutathione is the thiol, highlighted in blue in the figure below. In its reduced (free thiol) form, glutathione is abbreviated 'GSH'.
In its oxidized form, glutathione exists as a dimer of two molecules linked by a disulfide group, and is abbreviated 'GSSG'. A new disulfide in a protein forms via a 'disulfide exchange' reaction with GSSG, a process that can be described as a combination of two SN2-like attacks. The end result is that a new cysteine-cysteine disulfide forms at the expense of the disulfide in GSSG.
In its reduced (thiol) state, glutathione can reduce disulfides bridges in proteins through the reverse of the above reaction.
Sulfides
Sulfides are named using the same rules as ethers except sulfide is used in the place of ether. For more complex substances, alkylthio is used in place of alkoxy.
Sulfur analogs of ethers are called sulfides. Sulfides are less common than thiols as naturally occurring compounds. However, sulfides—especially disulfides (C-S-S-C)—have important biological functions, mainly in reducing agents (antioxidants).
Since thiols are weakly acidic their conjugate bases, called thiolate ions, can be easily formed through reaction with a strong base such as NaH. Thiolates have proven to be excellent nucleophiles and easily undergo SN2 reactions with primary and secondary alkyl halides to form sulfides. The reaction is analogous to the Williamson ether synthesis previously discussed in this chapter (Section 18-2)
The chemical behavior of sulfides contrasts with that of ethers in some important ways. Because sulfur's valence electrons involve 3P orbitals they are further away and less tightly held than the valence electrons on oxygen which involve 2P orbitals. This makes the nucleophilicity of sulfur atoms much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. For instance, sulfides, unlike ethers, easily react with primary alkyl halides through an SN2 mechanism to give ternary sulfonium ions (R3S+) in the same manner that 3º-amines can be alkylated to form quaternary ammonium salts. Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive.
Biological Sulfonium Ion Formation
In biological systems the sulfonium ion, S-adenosylmethionine (SAM), is formed by a SN2 reaction between the nucleophilic sulfur atom of the amino acid methionine and the electrophilic methene (CH2) carbon of adenosine triphosphate (ATP). This reaction is unusually in that the triphosphate ion is removed from ATP as a leaving group. In most cases biological nucleophiles attack an electrophilic phosphorus on ATP causing a diphosphate ion to be removed as the leaving group.
The sulfonium ion in SAM is an effective methylating agent. Through an SN2 reaction, biological nucleophiles can attack the electrophilic methyl group bonded to the positively charged sulfonium ion and cause the removal of a neutral sulfide as a leaving group. An example of this reaction is the catechol-O-methyltransferase catalyzed O-methylation of norepinephrin by SAM to produce adrenaline.
Sulfides can be easily oxidized. Reacting a sulfide with hydrogen peroxide, H2O2, as room temperature produces a sulfoxide (R2SO). The oxidation can be continued by reaction with a peroxyacid, such as peracetic acid (CH3CO3H), to produce the sulfone (R2SO2).
A common example of a sulfoxide is the solvent dimethyl sulfoxide (DMSO). DMSO is considered a polar aprotic solvent.
DMSO is a very polar, aprotic solvent
Exercise \(1\)
Give IUPAC names the following compounds:
a)
b)
c)
d)
e)
f)
Answer
a) 2-Pentanethiol
b) 2,2,5-Trimethyl-4-heptanethiol
c) 3-Cyclopentene-1-thiol
d) Methyl isopropyl sulfide
e) p-Di(methylthio)benzene
f) 2-(Ethylthio)cyclohexanone
Exercise \(2\)
The molecule 3-methyl-1-butanethiol has been shown to be one of the components of skunk spray. How would make the following conversions?
a)
b)
Answers
a) 1) HBr & Peroxides, 2) (NH2)2C=S, 3) NaOH & H2O
b) 1) LiAlH4, 2) PBr3, 3) (NH2)2C=S, 4) NaOH & H2O | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/09._Further_Reactions_of_Alcohols_and_the_Chemistry_of_Ethers/9.10%3A_Sulfur__Analogs__of_Alcohols__and__Ethers.txt |
This page explains what structural isomerism is, and looks at some of the various ways that structural isomers can arise.
What is structural isomerism?
Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. For example, both of the following are the same molecule. They are not isomers. Both are butane.
There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds. If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule.
In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise. The names of the various forms of structural isomerism probably don't matter all that much, but you must be aware of the different possibilities when you come to draw isomers.
Chain Isomerism
These isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, \(C_4H_{10}\). In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.
Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond.
You could easily see this with a model. This is the example we've already used at the top of this page.
Example 1: Chain Isomers in Pentane
Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models.
Position isomerism
In position isomerism, the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton.
Example 2: Positional Isomers in C5H12
For example, there are two structural isomers with the molecular formula C3H7Br. In one of them the bromine atom is on the end of the chain, whereas in the other it's attached in the middle.
If you made a model, there is no way that you could twist one molecule to turn it into the other one. You would have to break the bromine off the end and re-attach it in the middle. At the same time, you would have to move a hydrogen from the middle to the end.
Another similar example occurs in alcohols such as \(C_4H_9OH\)
These are the only two possibilities provided you keep to a four carbon chain, but there is no reason why you should do that. You can easily have a mixture of chain isomerism and position isomerism - you aren't restricted to one or the other.
So two other isomers of butanol are:
You can also get position isomers on benzene rings. Consider the molecular formula \(C_7H_7Cl\). There are four different isomers you could make depending on the position of the chlorine atom. In one case it is attached to the side-group carbon atom, and then there are three other possible positions it could have around the ring - next to the \(CH_3\) group, next-but-one to the \(CH_3\) group, or opposite the \(CH_3\) group.
Functional group isomerism
In this variety of structural isomerism, the isomers contain different functional groups - that is, they belong to different families of compounds (different homologous series).
Example 3: Isomers in C3H6O
A molecular formula \(C_3H_6O\) could be either propanal (an aldehyde) or propanone (a ketone).
There are other possibilities as well for this same molecular formula - for example, you could have a carbon-carbon double bond (an alkene) and an -OH group (an alcohol) in the same molecule.
Another common example is illustrated by the molecular formula \(C_3H_6O_2\). Amongst the several structural isomers of this are propanoic acid (a carboxylic acid) and methyl ethanoate (an ester).
Contributors
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Practice Problems | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.1%3A_Physical__and_Chemical_Tests.txt |
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_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.2%3A_Defining__Spectroscopy.txt |
Objectives
After completing this section, you should be able to
1. discuss the principles of NMR spectroscopy.
2. identify the two magnetic nuclei that are most important to an organic chemist.
Key Terms
Make certain that you can define, and use in context, the key term below.
• resonance
Study Notes
Notice that the word “resonance” has a different meaning when we are discussing nuclear magnetic resonance spectroscopy than it does when discussing molecular structures.
Introduction
Some types of atomic nuclei act as though they spin on their axis similar to the Earth. Since they are positively charged they generate an electromagnetic field just as the Earth does. So, in effect, they will act as tiny bar magnetics. Not all nuclei act this way, but fortunately both 1H and 13C do have nuclear spins and will respond to this technique.
NMR Spectrometer
In the absence of an external magnetic field the direction of the spin of the nuclei will be randomly oriented (see figure below left). However, when a sample of these nuclei is place in an external magnetic field, the nuclear spins will adopt specific orientations much as a compass needle responses to the Earth’s magnetic field and aligns with it. Two possible orientations are possible, with the external field (i.e. parallel to and in the same direction as the external field) or against the field (i.e. antiparallel to the external field) - see Figure \(1\).
If the ordered nuclei are now subjected to EM radiation of the proper frequency the nuclei aligned with the field will absorb energy and "spin-flip" to align themselves against the field, a higher energy state. When this spin-flip occurs the nuclei are said to be in "resonance" with the field, hence the name for the technique, Nuclear Magentic Resonance or NMR.
The amount of energy, and hence the exact frequency of EM radiation required for resonance to occur is dependent on both the strength of the magnetic field applied and the type of the nuclei being studied. As the strength of the magnetic field increases the energy difference between the two spin states increases and a higher frequency (more energy) EM radiation needs to be applied to achieve a spin-flip (see image below).
Superconducting magnets can be used to produce very strong magnetic field, on the order of 21 tesla (T). Lower field strengths can also be used, in the range of 4 - 7 T. At these levels the energy required to bring the nuclei into resonance is in the MHz range and corresponds to radio wavelength energies, i.e. at a field strength of 4.7 T 200 MHz bring 1H nuclei into resonance and 50 MHz bring 13C into resonance. This is considerably less energy then is required for IR spectroscopy, ~10-4 kJ/mol versus ~5 - ~50 kJ/mol.
1H and 13C are not unique in their ability to undergo NMR. All nuclei with an odd number of protons (1H, 2H, 14N, 19F, 31P ...) or nuclei with an odd number of neutrons (i.e. 13C) show the magnetic properties required for NMR. Only nuclei with even number of both protons and neutrons (12C and 16O) do not have the required magnetic properties. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.3%3A_Hydrogen__Nuclear__Magnetic__Resonance.txt |
Objectives
After completing this section, you should be able to
1. explain, in general terms, the origin of shielding effects in NMR spectroscopy.
2. explain the number of peaks occurring in the 1H or 13C NMR spectrum of a simple compound, such as methyl acetate.
3. describe, and sketch a diagram of, a simple NMR spectrometer.
4. explain the difference in time scales of NMR and infrared spectroscopy.
5. predict the number of peaks expected in the 1H or 13C NMR spectrum of a given compound.
Study Notes
Before you go on, make sure that you understand that each signal in the 1H NMR spectrum shown for methyl acetate is due to a different proton environment. The three protons on the same methyl group are equivalent and appear in the spectrum as one signal. However, the two methyl groups are in two different environments (one is more deshielded) and so we see two signals in the whole spectrum (aside from the TMS reference peak).
Methyl acetate has a very simple 1H NMR spectrum, because there is no proton-proton coupling, and therefore no splitting of the signals. In later sections, we discuss splitting patterns in 1H NMR spectra and how they help a chemist determine the structure of organic compounds.
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.
The basics of an NMR experiment
So far, you may have the impression that all 1H nuclei in a molecule would absorb the same frequency. However, this would be of little use to organic chemists if that were the case. It turns out that not all 1H nuclei absorb the same frequency and this is the same for all other NMR active nuclei. It turns out that chemically nonequivalent protons (or other nuclei) have different resonance frequencies in the same applied magnetic field. Nonequivalent protons are in different chemical environments. This allows NMR spectroscopy to 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 here is what happens. First, a sample compound (we'll use methyl acetate) is placed inside a very strong applied magnetic field (B0). There are two types of protons in methyl acetate. Ha are bonded to a C that is then bonded to a carbonyl, whereas Hb are bonded to a carbon that is then bonded to an oxygen atom. This difference in bonding leads to different types of environments for Ha and Hb. All the Ha protons are the same since they all have the same type of bonding and will be in the same chemical environment and the same is true for Hb.
The basic arrangement of an NMR spectrometer is displayed below. A sample (in a small glass tube, where the methyl acetate is in solution) is placed between the poles of a strong magnet. A radio frequency generator pulses the sample and excites the nuclei causing a spin-flip. The spin flip is detected by the detector and the signal sent to a computer where it is processed.
In the magnet, all of the protons begin to precess: the Ha protons at precessional frequency ωa, the Hb protons at ωb. At first, the magnetic moments of (slightly more than) half of the protons are aligned with B0, and half (slightly less than half) are aligned against B0. Then, the sample is hit with electromagnetic radiation in the radio frequency range. The two specific frequencies which match ωa and ωb (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 used 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, therefore 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. There are multiple deuterated solvents since molecules have different solubilities, so one molecule may dissolve in deuterated chloroform while others may not.
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
A proton has a chemical shift (relative to TMS) of 4.56 ppm.
1. a) What is its chemical shift, expressed in Hz, in a 300 MHz instrument? On a 200 MHz instrument?
2. b) 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
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. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.4%3A_Using__NMR_Spectra__to_Analyze__Molecular_Structure%3A_The__Proton__Chemical.txt |
Objectives
After completing this section, you should be able to
1. identify those protons which are equivalent in a given chemical structure.
2. use the 1H NMR spectrum of a simple organic compound to determine the number of equivalent sets of protons present.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• diastereotopic
• enantiotopic
• homotopic
Study Notes
It is important at this stage to be able to identify equivalent protons in any organic compound given the structure of that compound. Once you know the number of different groups of equivalent protons in a compound, you can predict the number (before coupling) and relative strength of signals. Look at the following examples and make sure you understand how the number and intensity ratio of signals are derived from the structure shown.
Structure Number of Signals Ratio of Signals
\(\ce{\sf{CH3OCH2CH2Br}}\) 3 A : B : C 3 : 2 : 2
1
3 A : B : C 2 : 2 : 6 (or 1 : 1 : 3)
3 A : B : C 2 : 4 : 2 (or 1 : 2 : 1)
4 A : B : C : D 3 : 2 : 2 : 3
5 A : B : C : D : E 3 : 1 : 1 : 1 : 1
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. These protons are considered to be homotopic. Homotopic protons are chemically identical, so electronically equivalent, thus show up as identical NMR absorptions.
The ability to recognize chemical equivalency 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. To determine if protons are homotopic or enantiotopic, you can do a thought experiment by replacing one H with X followed by the other H by X. In pyruvate below, if you replace any of the Hs with an X, then you would get the same molecule. These protons are homotopic. In dihydroxyacetone phosphate, this is not quite the case. If you exchange HR for X, then you would create a stereocenter with R configuration. If you exchange HS for X, then you would create a stereocenter with S configuration. The two "new molecules" would have the relationship of being enantiomers. Therefore HR and HS are enantiotopic protons. The only way these protons will show up different under NMR experimental conditions would be if you used a solvent that was chiral. Since most common solvents are all achiral, the enantiotopic protons are NMR equivalent and will show up as if they were the same proton.
Example 13.6.1
How many different sets of protons do the following molecules contain? (count diastereotopic protons as non-equivalent).
Solution
10.6: Integration
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
10.7: Spin-Spin Splitting: The Effect of Nonequivalent Neighboring Hydrogen
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University)
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.5%3A_Tests__for_Chemical___Equivalence.txt |
Objectives
After completing this section, you should be able to
1. explain how multiple coupling can give rise to complex-looking 1H NMR spectra.
2. predict the splitting pattern expected in the 1H NMR spectrum of an organic compound in which multiple coupling is possible.
3. interpret 1H NMR spectra in which multiple coupling is evident.
Key Terms
Make certain that you can define, and use in context, the key term below.
• tree diagram
Study Notes
We saw the effects of spin-spin coupling on the appearance of a 1H NMR signal. These effects can be further complicated when that signal is coupled to several different protons. For example, BrCH2CH2CH2Cl would produce three signals. The hydrogens at C1 and C3 would each be triplets because of coupling to the two hydrogens on C2. However, the hydrogen on C2 “sees” two different sets of neighboring hydrogens, and would therefore produce a triplet of triplets.
Another effect that can complicate a spectrum is the “closeness” of signals. If signals accidently overlap they can be difficult to identify. In the example above, we expected a triplet of triplets. However, if the coupling is identical (or almost identical) between the hydrogens on C2 and the hydrogens on both C1 and C3, one would observe a quintet in the 1H NMR spectrum. [You can try this yourself by drawing a tree diagram of a triplet of triplets assuming, first, different coupling constants, and then, identical coupling constants.] Keep this point in mind when interpreting real 1H NMR spectra.
Also, when multiplets are well separated, they form patterns. However, when multiplets approach each other in the spectrum they sometimes become distorted. Usually, the inner peaks become larger than the outer peaks. Note the following examples:
Aromatic ring protons quite commonly have overlapping signals and multiplet distortions. Sometimes you cannot distinguish between individual signals, and one or more messy multiplets often appear in the aromatic region.
It is much easier to rationalize the observed 1H NMR spectrum of a known compound than it is to determine the structure of an unknown compound from its 1H NMR spectrum. However, rationalizations can be a useful learning technique as you try to improve your proficiency in spectral interpretation. Remember that when a chemist tries to interpret the 1H NMR spectrum of an unknown compound, he or she usually has additional information available to make the task easier. For example, the chemist will almost certainly have an infrared spectrum of the compound and possibly a mass spectrum too. Details of how the compound was synthesized may be available, together with some indication of its chemical properties, its physical properties, or both.
In examinations, you will be given a range of information (IR, MS, UV data and empirical formulae) to aid you with your structural determination using 1H NMR spectroscopy. For example, you may be asked to determine the structure of C6H12O given the following spectra:
Infrared spectrum: 3000 cm−1 and 1720 cm−1 absorptions are both strong
1H NMR δ (ppm) Protons Multiplicity
0.87 6 doublet
1.72 1 broad multiplet
2.00 3 singlet
2.18 2 doublet
To answer this question, you note that the infrared spectrum of C6H12O shows \$\ce{\sf{C-H}}\$ stretching (3000 cm−1) and \$\ce{\sf{C-O}}\$ stretching (1720 cm−1). Now you have to piece together the information from the 1H NMR spectrum. Notice the singlet with three protons at 2.00 ppm. This signal indicates a methyl group that is not coupled to other protons. It could possibly mean the presence of a methyl ketone functional group.
The signal at 1.72 ppm is a broad multiplet, suggesting that a carbon with a single proton is beside carbons with several different protons.
The doublet signal at 2.18 ppm implies that a \$\ce{\sf{-CH2-}}\$ group is attached to a carbon having only one proton.
The six protons showing a doublet at 0.87 ppm indicate two equivalent methyl groups attached to a carbon with one proton.
Whenever you see a signal in the 0.7-1.3 ppm range that is a multiplet of three protons (3, 6, 9) it is most likely caused by equivalent methyl groups.
Using trial and error, and with the above observations, you should come up with the correct structure.
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 (or tree 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 \(1\)
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
Note
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 \(2\)
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!)
Exercises
1. Given the information below, draw the structures of compounds A through D.
1. An unknown compound A was prepared as follows:
Mass spectrum:
base peak m/e = 39
parent peak m/e = 54
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
1.0 2 triplet
5.4 1 quintet
2. Unknown compound B has the molecular formula C7H6O2.
Infrared spectrum:
3200 cm−1 (broad) and 1747 cm−1 (strong) absorptions
1H NMR spectrum:
δ (ppm) Protons
6.9 2
7.4 2
9.8 1
10.9 1
Hint: Aromatic ring currents deshield all proton signals just outside the ring.
3. Unknown compound C shows no evidence of unsaturation and contains only carbon and hydrogen.
Mass spectrum:
parent peak m/e = 68
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
1.84 3 triplet
2.45 1 septet
Hint: Think three dimensionally!
4. Unknown compound D (C15H14O) has the following spectral properties.
Infrared spectrum:
3010 cm−1 (medium)
1715 cm−1 (strong)
1610 cm−1 (strong)
1500 cm−1 (strong)
1H NMR spectrum:
δ (ppm) Relative Area Multiplicity
3.00 2 triplet
3.07 2 triplet
7.1-7.9 10 Multiplets
Answers | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.8%3A_Spin-Spin_Splitting%3A_Some__Complications.txt |
Objectives
After completing this section, you should be able to
• Determine the number of distinct C atoms in a molecule.
• Use the chemical shifts table to determine functional groups present in a molecule.
• Assign a chemical shift to each carbon in a given molecule.
The basics of 13C-NMR spectroscopy
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.
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. This means you will only see the 13C nuclei in a 13C NMR experiment like in the 1H NMR experiments we just looked at, we only saw hydrogens. 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 \(1\)
1. How many sets of non-equivalent carbons are there in each of the molecules shown in exercise 5.1?
2. How many sets of non-equivalent carbons are there in:
• toluene
• 2-pentanone
• para-xylene
• triclosan
(all structures are shown earlier in this chapter)
Solution
a
a) 8 signals (each carbon is different)
b) 11 signals (the two enantiotopic CH2CH3 groups are NMR-equivalent)
c) 6 signals (each carbon is different)
d) 16 signals (the fluorobenzene group only contributes 4 signals due to symmetry)
b
a) 5 signals
b) 5 signals
c) 3 signals
d) 6 signals
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.
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.
13C NMR Chemical Shifts
The Carbon NMR is used for determining functional groups using characteristic shift values. 13C chemical shifts are greatly affected by electronegative effects. If a H atom in an alkane is replaced by substituent X, electronegative atoms (O, N, halogen), 13C signals for nearby carbons shift downfield (left; increase in ppm) with the effect diminishing with distance from the electron withdrawing group. Figure 13.11.1 shows typical 13C chemical shift regions of the major chemical class.
Spin-Spin splitting
Comparing the 1H NMR, there is a big difference thing in the 13C NMR. The 13C-13Cspin-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 (Fig 13.11.2). Decoupling is performed by irradiating at the frequency of one proton with continuous low-power RF.
Fig \(2\): Decoupling in the 13C NMR
Objectives
After completing this section, you should be able to use data from 13C NMR spectra to distinguish between two (or more) possible structures for an unknown organic compound.
Features of a 13C NMR spectrum
Butane shows two different peaks in the 13C NMR spectrum, below. Note that: the chemical shifts of these peaks are not very different from methane. The carbons in butane are in a similar environment to the one in methane.
• there are two distinct carbons in butane: the methyl, or CH3, carbon, and the methylene, or CH2, carbon.
• the methyl carbon absorbs slightly upfield, or at lower shift, around 10 ppm.
• the methylene carbon absorbs at slightly downfield, or at higher shift, around 20 ppm.
• other factors being equal, methylene carbons show up at slightly higher shift than methyl carbons.
In the 13C NMR spectrum of pentane (below), you can see three different peaks, even though pentane just contains methyl carbons and methylene carbons like butane. As far as the NMR spectrometer is concerned, pentane contains three different kinds of carbon, in three different environments. That result comes from symmetry.
Symmetry is an important factor in spectroscopy. Nature says:
• atoms that are symmetry-inequivalent can absorb at different shifts.
• atoms that are symmetry-equivalent must absorb at the same shift.
To learn about symmetry, take a model of pentane and do the following:
• make sure the model is twisted into the most symmetric shape possible: a nice "W".
• choose one of the methyl carbons to focus on.
• rotate the model 180 degrees so that you are looking at the same "W" but from the other side.
• note that the methyl you were focusing on has simply switched places with the other methyl group. These two carbons are symmetry-equivalent via two-fold rotation.
By the same process, you can see that the second and fourth carbons along the chain are also symmetry-equivalent. However, the middle carbon is not; it never switches places with the other carbons if you rotate the model. There are three different sets of inequivalent carbons; these three groups are not the same as each other according to symmetry.
Example \(1\)
Determine how many inequivalent carbons there are in each of the following compounds. How many peaks do you expect in each 13C NMR spectrum?
Practically speaking, there is only so much room in the spectrum from one end to the other. At some point, peaks can get so crowded together that you can't distinguish one from another. You might expect to see ten different peaks in eicosane, a twenty-carbon alkane chain, but when you look at the spectrum you can only see seven different peaks. That may be frustrating, because the experiment does not seem to agree with your expectation. However, you will be using a number of methods together to minimize the problem of misleading data.
Solution
a) Three inequivalent carbons/three peaks. There is a plane of symmetry that bisects the cyclohexene horizontally. The three different carbons are one of the alkene (C1), the CH2 next to alkene (C3) and C4.
b) Six inequivalent carbons/six peaks. The two methyl groups attached to the alkene are identical.
c) Four inequivalent carbons/four peaks. This molecule has a plane of symmetry through the molecule, including the methyl group. The two carbons adjacent to the methyl group are equivalent (C2 and C5). C3 and C4 are also equivalent.
d) Five inequivalent carbons/five peaks. This molecule has a plane of symmetry that passes through the ring carbon between the two methyl groups. The two methyl carbons are identical. The two ring carbons with the methyl groups attached are identical (C1 and C3). C4 and C6 are also equivalent.
e) Six inequivalent carbons/six peaks. The three methyl groups at the end of the molecule are equivalent.
f) Ten inequivalent carbons/ten peaks. There is no symmetry for the carbons in this molecule.
The 13C NMR spectrum for ethanol
This is a simple example of a 13C NMR spectrum. Don't worry about the scale for now - we'll look at that in a minute.
Note
Note: The NMR spectra on this page have been produced from graphs taken from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.
There are two peaks because there are two different environments for the carbons. The carbon in the CH3 group is attached to 3 hydrogens and a carbon. The carbon in the CH2 group is attached to 2 hydrogens, a carbon and an oxygen. The two lines are in different places in the NMR spectrum because they need different external magnetic fields to bring them in to resonance at a particular radio frequency.
The 13C NMR spectrum for a more complicated compound
This is the 13C NMR spectrum for 1-methylethyl propanoate (also known as isopropyl propanoate or isopropyl propionate).
This time there are 5 lines in the spectrum. That means that there must be 5 different environments for the carbon atoms in the compound. Is that reasonable from the structure?
If you count the carbon atoms, there are 6 of them. So why only 5 lines? In this case, two of the carbons are in exactly the same environment. They are attached to exactly the same things. Look at the two CH3 groups on the right-hand side of the molecule.
You might reasonably ask why the carbon in the CH3 on the left is not also in the same environment. Just like the ones on the right, the carbon is attached to 3 hydrogens and another carbon. But the similarity is not exact - you have to chase the similarity along the rest of the molecule as well to be sure.
The carbon in the left-hand CH3 group is attached to a carbon atom which in turn is attached to a carbon with two oxygens on it - and so on down the molecule. That's not exactly the same environment as the carbons in the right-hand CH3 groups. They are attached to a carbon which is attached to a single oxygen - and so on down the molecule. We'll look at this spectrum again in detail on the next page - and look at some more similar examples as well. This all gets easier the more examples you look at.
For now, all you need to realize is that each line in a 13C NMR spectrum recognizes a carbon atom in one particular environment in the compound. If two (or more) carbon atoms in a compound have exactly the same environment, they will be represented by a single line.
Note
You might wonder why all this works, since only about 1% of carbon atoms are 13C. These are the only ones picked up by this form of NMR. If you had a single molecule of ethanol, then the chances are only about 1 in 50 of there being one 13C atom in it, and only about 1 in 10,000 of both being 13C.
But you have got to remember that you will be working with a sample containing huge numbers of molecules. The instrument can pick up the magnetic effect of the 13C nuclei in the carbon of the CH3 group and the carbon of the CH2 group even if they are in separate molecules. There's no need for them to be in the same one. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/10%3A_Using_Nuclear_Magnetic_Resonance_Spectroscopy_to_Deduce_Structure/10.9%3A_Carbon-13_Nuclear__Magnetic__Resonance.txt |
Objectives
After completing this section, you should be able to
1. provide the correct IUPAC name for an acyclic or cyclic alkene, given its Kekulé, condensed or shorthand structure.
2. draw the Kekulé, condensed or shorthand structure of an alkene (cyclic or acyclic), given its IUPAC name.
3. give the IUPAC equivalent of the following trivial names: ethylene, propylene, isobutylene and isoprene.
4. draw the structure of a vinyl (ethenyl) and allyl (2-propenyl) group, and use these names in alkene nomenclature.
Study Notes
This course uses IUPAC nomenclature; therefore, you need not usually memorize a large number of trivial names. However, you will encounter some trivial names so frequently in books and articles that they soon become familiar.
An alkene that can exhibit geometric isomerism has not been properly named unless its name specifies whether the double bond (or bonds) is (or are) cis or trans. The most effective way of giving this information is discussed, and more details of cis and trans follow in Section 7.4.
Alkenes contain carbon-carbon double bonds and are unsaturated hydrocarbons with the molecular formula is CnH2n; this is also the same molecular formula as cycloalkanes. The parent chain of an alkene is the longest chain containing both carbon atoms of the double bond. Alkenes are named by dropping the -ane ending of the parent and adding -ene. Also, the position of double bond in the parent chain of the alkene is indicated with a number.
The Basic Rules for Naming Alkenes
For straight chain alkenes, it is the same basic rules as nomenclature of alkanes apply except the -ane suffix is changed to -ene.
1) Find the longest carbon chain that contains both carbons of the double bond.
2) Start numbering from the end of the parent chain which gives the lowest possible number to the double bond. If the double bond is equidistant from both ends of the parent chain, number from the end which gives the substituents the lowest possible number. The double bond in cycloalkenes do not need to number because it is understood that they are in the one position.
3) Place the location number of the double bond directly before the parent name. The location number indicates the position of the first carbon of the double bond. Add substituents and their position to the alkene as prefixes. Remember substituents are written in alphabetical order.
The presence of multiple double bonds is indicated by using the appropriate suffix such as -diene, -triene, ect. Each of the multiple bonds receives a location number. Also, only -ne is removed from the parent alkane chain name leaving an "a" in the name.
Overall, the name of an alkene should look like:
(Location number of substituent)-(Name of substituent)-(Location number of double bond)-(Name of parent chain) + ene
Newer IUPAC Nomenclature
In 1993 IUPAC updated their naming recommendation to place the location number of the double bond before the -ene suffix of alkene names. The provides names such as hex-2-ene rather than 2-hexene. The newer system is slowly being accepted so it may occasionally be encountered.
Naming Cycloalkenes
Because there are no chain ends in cycloalkenes, the double bond is assumed to numbered C1 and C2 and its location number is not required in the name. The direction of the numbering is determined by which will give the substituent closest to the double bond the lowest number. If multiple double bonds are present, it may be necessary to include their location numbers in the name. One of the double bonds will be number C1 and C2 and the numbering direction is determined by which gives the remaining double bonds the lowest possible number.
Endocyclic vs. Exocyclic Alkenes
Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring.
Cyclopentene is an example of an endocyclic double bond.
Methylenecylopentane is an example of an exocyclic double bond.
Name the following compounds...
1-methylcyclobutene. The methyl group places the double bond. It is correct to also name this compound as 1-methylcyclobut-1-ene.
1-ethenylcyclohexene, the methyl group places the double bond. It is correct to also name this compound as 1-ethenylcyclohex-1-ene. A common name would be 1-vinylcyclohexene.
Try to draw structures for the following compounds...
• 3-allylcyclohex-1-ene
• 2-vinyl-1,3-cyclohexadiene
Common Names of Alkene Fragments
Some alkene containing fragments have common names which should be recognized. These common names can be used to simplify naming much the alkyl fragments discussed in Section: 3.3. Some of these fragments are the methylene group (H2C=), the vinyl group (H2C=CH-), and the allyl group (H2C=CH-CH2-).
In addition, the common name some small alkene compounds are still accepted by IUPAC. It is important to be able to identify them.
Exercise \(1\)
Name the following compounds using common fragment names.
a) b) C)
Answer
a) 2-Vinyl-1,3-cyclohexadiene
b) Methylenecylopentane
c) 3-Allylcyclohexene
Examples
Both these compounds have double bonds, making them alkenes. In example (1) the longest chain consists of six carbons, so the root name of this compound will be hexene. Three methyl substituents (colored red) are present. Numbering the six-carbon chain begins at the end nearest the double bond (the left end), so the methyl groups are located on carbons 2 & 5. The IUPAC name is therefore: 2,5,5-trimethyl-2-hexene.
In example (2) the longest chain incorporating both carbon atoms of the double bond has a length of five. There is a seven-carbon chain, but it contains only one of the double bond carbon atoms. Consequently, the root name of this compound will be pentene. There is a propyl substituent on the inside double bond carbon atom (#2), so the IUPAC name is: 2-propyl-1-pentene.
The double bond in example (3) is located in the center of a six-carbon chain. The double bond would therefore have a locator number of 3 regardless of the end chosen to begin numbering. The right hand end is selected because it gives the lowest first-substituent number (2 for the methyl as compared with 3 for the ethyl if numbering were started from the left). The IUPAC name is assigned as shown.
Example (4) is a diene (two double bonds). Both double bonds must be contained in the longest chain, which is therefore five- rather than six-carbons in length. The second and fourth carbons of this 1,4-pentadiene are both substituted, so the numbering begins at the end nearest the alphabetically first-cited substituent (the ethyl group).
These examples include rings of carbon atoms as well as some carbon-carbon triple bonds. Example (6) is best named as an alkyne bearing a cyclobutyl substituent. Example (7) is simply a ten-membered ring containing both a double and a triple bond. The double bond is cited first in the IUPAC name, so numbering begins with those two carbons in the direction that gives the triple bond carbons the lowest locator numbers. Because of the linear geometry of a triple bond, a-ten membered ring is the smallest ring in which this functional group is easily accommodated. Example (8) is a cyclooctatriene (three double bonds in an eight-membered ring). The numbering must begin with one of the end carbons of the conjugated diene moiety (adjacent double bonds), because in this way the double bond carbon atoms are assigned the smallest possible locator numbers (1, 2, 3, 4, 6 & 7). Of the two ways in which this can be done, we choose the one that gives the vinyl substituent the lower number.
Problems
Exercise \(2\)
Name the following alkenes.
Answer
a) 2-ethylhept-1-ene or 2-ethyl-1-heptene
b) 1,2-dimethylcycloheptene
c) 2,5-dimethyloct-2-ene or 2,5-dimethyl-1-octene
Exercise \(3\)
Draw structures for the following compounds from the given names.
1. 3-butylhept-2-ene (3-butyl-2-heptene)
2. 1,4-pentadiene (penta-1,4-diene)
3. 3-vinyl-1,4-cyclohexadiene (cyclohexa-1,4-diene)
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.01%3A_Naming__the_Alkenes.txt |
This is an introductory page about alkenes such as ethene, propene and the rest. It deals with their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity.
What are alkenes?
Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are:
ethene C2H4
propene C3H6
You can work out the formula of any of them using: CnH2n The table is limited to the first two, because after that there are isomers which affect the names.
Isomerism in the alkenes
Structural isomerism
All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula.
For example, with C4H8, it isn't too difficult to come up with these three structural isomers:
There is, however, another isomer. But-2-ene also exhibits geometric isomerism.
Geometric (cis-trans) isomerism
The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other.
These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides).
Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below..
Chemical Reactivity
Bonding in the alkenes
We just need to look at ethene, because what is true of C=C in ethene will be equally true of C=C in more complicated alkenes. Ethene is often modeled like this:
The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other.
One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons.
In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond.
The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other.
The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and, because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things.
Contributors
Jim Clark (Chemguide.co.uk)
11.03: Physical Properties of Alkenes
Physical state
Ethene, Propene, and Butene exists as colorless gases. Members of the 5 or more carbons such as Pentene, Hexene, and Heptene are liquid, and members of the 15 carbons or more are solids.
Density
Alkenes are lighter than water, therefore, are insoluble in water. Alkenes are only soluble in nonpolar solvent.
Solubility
Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes.
Boiling Points
The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.
Boiling points of alkenes depends on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.
Compound Boiling points (oC)
Ethene -104
Propene -47
Trans-2-Butene 0.9
Cis-2-butene 3.7
Trans 1,2-dichlorobutene 155
Cis 1,2-dichlorobutene 152
1-Pentene 30
Trans-2-Pentene 36
Cis-2-Pentene 37
1-Heptene 115
3-Octene 122
3-Nonene 147
5-Decene 170
In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.
Melting Points
Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.
Compound Melting Points (0C)
Ethene -169
Propene -185
Butene -138
1-Pentene -165
Trans-2-Pentene -135
Cis-2-Pentene -180
1-Heptene -119
3-Octene -101.9
3-Nonene -81.4
5-Decene -66.3
Polarity
Chemical structure and fuctional groups can affect the polarity of alkenes compounds. The sp2 carbon is much more electron-withdrawing than the sp3 hybridize orbitals, therefore, creates a weak dipole along the substituent weak alkenly carbon bond. The two individual dipoles together form a net molecular dipole. In trans-subsituted alkenes, the dipole cancel each other out. In cis-subsituted alkenes there is a net dipole, therefore contributing to higher boiling in cis-isomers than trans-isomers.
• Trung Nguyen | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.02%3A_Structure_and__Bonding_in_Ethene%3A_The__Pi_Bond.txt |
E2 Reaction
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.
To get a clearer picture of the interplay of these factors involved in a a reaction between a nucleophile/base and an alkyl halide, consider the reaction of a 2º-alkyl halide, isopropyl bromide, with two different nucleophiles. In one pathway, a methanethiolate nucleophile substitutes for bromine in an SN2 reaction. In the other (bottom) pathway, methoxide ion acts as a base (rather than as a nucleophile) in an elimination reaction. As we will soon see, the mechanism of this reaction is single-step, and is referred to as the E2 mechanism.
.
E1 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.
Dehydration
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond.
The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
Primary alcohol dehydrates through the E2 mechanism
Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Thereore, the trans diastereomer of the 2-butene product is most abundant.
Zaitsev's Rule
Zaitsev’s or Saytzev’s (anglicized spelling) rule is an empirical rule used to predict regioselectivity of 1,2-elimination reactions occurring via the E1 or E2 mechanisms. It states that in a regioselective E1 or E2 reaction the major product is the more stable alkene, (i.e., the alkene with the more highly substituted double bond). For example:
If two or more structurally distinct groups of beta-hydrogens are present in a given reactant, then several constitutionally isomeric alkenes may be formed by an E2 elimination. This situation is illustrated by the 2-bromobutane and 2-bromo-2,3-dimethylbutane elimination examples given below.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Objectives
After completing this section, you should be able to
1. write the mechanism of a typical E2 reaction.
2. sketch the transition state of a typical E2 reaction.
3. discuss the kinetic evidence that supports the proposed E2 mechanism.
4. discuss the stereochemistry of an E2 reaction, and explain why the anti periplanar geometry is preferred.
5. determine the structure of the alkene produced from the E2 reaction of a substrate containing two chiral carbon atoms.
6. describe the deuterium isotope effect, and discuss how it can be used to provide evidence in support of the E2 mechanism.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti periplanar
• deuterium isotope effect
• E2 reaction
• periplanar
• syn periplanar
Study Notes
An E2 reaction is a bimolecular elimination reaction; thus, two molecules are involved in the rate-limiting step. In this section, we are concerned with E2 reactions involving an alkyl halide and a base.
Use molecular models to assist you to understand the difference between syn periplanar and anti periplanar, and to appreciate why E2 eliminations are stereospecific.
Note that when deuterium is used the kinetic isotope effect (KIE) is referred to as the deuterium isotope effect. A C–H bond is about 5 kJ/mol weaker than a C–D bond. So if the rate-limiting step involves a breaking of this bond as it does at the E2 transition state there will be a substantial difference in reaction rates when comparing deuterated and non-deuterated analogues. Indeed, the reaction of 2-bromopropane with sodium ethoxide (NaOEt) is 6.7 times faster than its deuterated counterpart, providing evidence consistent with an E2 mechanism.
Introduction
E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold. E2 reactions are typically seen with secondary and tertiary alkyl halides, in the presence of a base (OH-, RO-, R2N-). For a primary halide to undergo an E2 reaction a strong sterically hindered base is usually required. The products of an E2 reaction follow Zaitsev's rule, the most substituted alkene is usually the major product.
E2 Mechanism
The E2 mechanism takes place in a single concerted step. The rate at which this mechanism occurs follows second order kinetics, and depends on the concentration of both the base and alkyl halide. The base removes a hydrogen from a carbon adjacent to the carbon with the leaving group. The electrons of the broken H-C bond move to form the pi bond of the neutral product alkene. In doing this, the C-X bond is broken causing the removal of the leaving group. Overall during this reaction an electron pair is transferred from the base to the leaving group.
E2 Reaction Transition State
Because the E2 reaction is predicted to follow a concerted mechanism only one transition state is expected in its energy coordinate diagram.
As the base begins to remove a hydrogen adjacent to the leaving group an H-B is starting to form and an H-C bond is in the process of breaking. Simultaneously, a C=C pi bond forms and the C-X bond breaks. During this reaction a set of electrons and a negative charge are transferred from the base to the leaving group. This is represented by a partial negative charge being present on both the base and the leaving group in the transition state.
Evidence for the E2 Mechanism and Transition State
Kinetic studies of E2 reactions show that they are second order overall and follow the rate law: rate = k[RX][Base]. This data is consistent with the predicted bimolecular E2 mechanism that includes both the alkyl halide and the base participating in the mechanism's rate-determining step.
Further evidence for the predicted mechanism for the E2 reaction was provided by experiments involving the deuterium isotope effect (DIE). The deuterium isotope effect (DIE) is a phenomenon associated with molecules which have had hydrogen (H) atoms isotopically substituted with deuterium (D) atoms, exhibiting different reaction rates. The bond dissociation energy for C-D bonds is about 5 kJ/mol stronger than the bond dissociation energy of C-H bonds. This difference in energy results in a reduction in the rate of reaction, if the deuterium replacement occurs at a site of bond breaking in the rate determining step of a reaction. So if the rate-limiting step involves a breaking of this bond, as it does at the E2 transition state, there will be a substantial decrease in reaction rates when comparing deuterated and non-deuterated analogues. Indeed, the reaction of 2-bromopropane with sodium ethoxide (NaOEt) is 6.7 times faster than its deuterated counterpart, providing evidence that the C-H bond is broken during the rate determining step of the E2 mechanism. This is consistent with the predicted mechanism of the E2 reaction consisting of a single concerted step. If the rate-determining step was the leaving group bond breaking first (as in an E1 reaction) the deuterium substitution would not have caused an observable change in reaction rate.
Yet another piece of evidence which supports the E2 mechanism comes when considering the stereochemistry of the elimination products. In the transition state of the E2 mechanism there are two carbons rehybridization from sp3 to sp2 while forming a π-bond. For good orbital overlap to occur during this change all the atoms involved in the reaction, the two carbons, the hydrogen, and the leaving group need to all lie in the same plane or periplanar. There are two possible ways this geometry can be achieved: syn-periplanar where the hydrogen and the leaving group are on the same side of the C-C bond and anti-periplanar where the hydrogen and the leaving group are on opposite sides of the C-C bond. To obtain a syn-periplanar geometry the substituents attached to the C-C bond must adopt an energetically unfavorable eclipsed conformation. The energy barrier to syn-orientation is such that syn-elimination is rarely observed in E2 reactions. E2 eliminations typically take place from the anti-periplanar conformation, as this is the most stable conformation due to the substituents attached to the C-C bond being staggered.
When viewing the transition state of the E2 mechanism when in the anti-periplanar conformation, the two sp3 hybridized orbitals making up the C-H and C-X sigma bonds are in the same plane. When these two carbons rehybridization from sp3 to sp2, the p orbitals forming the π-bond will also have good overlap.
The stereochemical consequences of reactant molecules obtaining anti-periplanar geometry prior to E2 reactions has been observed in numerous experiments and provides further evidence of the proposed mechanism. For example, (2S, 3R)-2-bromo-2,3-diphenylbutane only forms (Z)-2,3-diphenyl-2-butene as a product of E2 elimination. To form the Z alkene isomer, the starting material must obtain the anti-periplanar geometry preferred for E2 reactions.
No isomeric E alkene product is formed because the starting materials would need to obtain a staggard, syn-periplanar geometry. The syn-periplanar geometry would be transferred to the transition state making it higher in energy and harder to form. The fact that only the Z alkene isomer forms provides further evidence that the E2 mechanism takes place by a single bimolecular step.
Predicting the Stereochemical Product of E2 Reactions
E2 reactions have the requirements of a leaving group and a hydrogen on a carbon adjacent to the leaving group carbon. For non-ringed compounds, the C-C can usually undergo free rotation to place the C-X and C-H bonds anti-periplaner to each other. Stereochemical considerations usually come into play when both of the carbons in the C-C bond are chiral. When these carbons are achiral but contain the other requirements for an E2 reaction, products lacking stereochemistry are usually produced.
When both of the carbons in the C-C bond are chiral, one carbon will have a leaving group and the other carbon will have a single hydrogen. To consider the stereochemistry of an E2 product, start by creating a Newman projection looking down the C-C bond. Then rotate the C-C bond until the C-X and C-H bonds are in the anti position. The Newman projection now shows the relative positioning of the substituents on the C-C bond as they will appear in the double bond formed. It should be noted that if syn elimination was possible, a positional isomer of the double bond will be created. Consider the elimination products for (2R, 3R)-2-bromo-3-methylpentane.
First, locate hydrogens on carbons adjacent to the leaving group. The compound used in this example has two unique sets of hydrogens.
Because the adjacent hydrogen colored fuchsia is attached to a chiral carbon it require special consideration. Convert the structure into a Newman projection along the C-C bond that will for the alkene elimination product. The C-C bond can be view down either direction and still product the correct product.
Then rotate such the the adjacent H and the leaving group are in the anti position.
This Newman projection shows the relative positioning of the substituents on the double bond formed. Note, none of the Z isomer is formed because it would require syn orientation of the H and Br.
Lastly, consider the products made by other adjacent hydrogens and apply Zaitsev's rule to predict the preferred product.
Determining Cis and Trans Products in Elimination Reaction
Understanding that E2 reactions require an anti-periplanar geometry can explain cis and trans alkene isomers can form as products. This occurs as an important exception to the rule that stereoisomers only form as elimination products when both of the carbons in the C-C bond forming the alkene are chiral. When considering the E2 reaction products of (R)-2-butane, there are two groups of hydrogens on carbons adjacent to the leaving group a CH3 and a CH2.
The carbon in the CH2 group is achiral so both the H's appear to be equivalent. However, when the anti-periplanar geometry requirement is applied each hydrogen will produce its own unique product. Once the reactant is converted into a Newman projection it can be seen that the Br leaving group and the hydrogen colored blue are in the anti orientation and are ready to be eliminated. It should also be noted that the two CH3 groups are also in the anti position which gives them the least amount of steric interaction possible. Because the CH3 groups are on opposite sides of the Newman projection the CH3 groups will be on opposite sides of the alkene thus creating the product trans-2-butene. When the Newman projection is rotated such that the fuchsia colored hydrogen is anti to the Br leaving group, the two CH3 groups are now on the same. This orientation creates the isomer cis-2-butene. However, to obtain this orientation the two CH3 groups must overcome the steric strain associated with a gauche conformation. This makes the transition state leading the cis isomer higher in energy and more difficult to form. Because of this, the trans isomer will be the major product. It is clear that the two hydrogens of the CH2 group are not equivalent despite their not being attached to a chiral carbon.
After considering all the products made during this E2 reaction the preferred product can be determined. The cis and trans isomer both have two alkene substituents while the third product only has one. Of the cis and trans isomer, the trans isomer is expected to be easier to form. Zaitsev's rule says that trans-2-butene will be the major product of this reaction.
Exercise $1$
1) What is the product of the following molecule in an E2 reaction? What is the stereochemistry?
Answer
The stereochemistry is (Z) for the reaction. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.06%3A_Preparation_of_Alkenes_from_Haloalkanes_and_Alkyl_Sufonates%3A_Bimolecular_Elimination.txt |
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond.
Introduction
The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
The required range of reaction temperature decreases with increasing substitution of the hydroxy-containing carbon:
• 1° alcohols: 170° - 180°C
• 2° alcohols: 100°– 140 °C
• 3° alcohols: 25°– 80°C
If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers (e.g., the Williamson Ether Synthesis).
Alcohols are amphoteric; they can act both as acid or base. The lone pair of electrons on oxygen atom makes the –OH group weakly basic. Oxygen can donate two electrons to an electron-deficient proton. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion +OH2 (The pKa value of a tertiary protonated alcohol can go as low as -3.8). This basic characteristic of alcohol is essential for its dehydration reaction with an acid to form alkenes.
Mechanism for the Dehydration of Alcohol into Alkene
Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that the –OH group in the alcohol donates two electrons to H+ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the nucleophile) then attacks the hydrogen adjacent to the carbocation and form a double bond.
Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). The relative reactivity of alcohols in dehydration reaction is ranked as the following
Methanol < primary < secondary < tertiary
Primary alcohol dehydrates through the E2 mechanism
Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO4- ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and trans alkenes are more stable than cis alkenes. Therefore, the trans diastereomer of the 2-butene product is most abundant.
Dehydration reaction of secondary alcohol: The dehydration mechanism for a tertiary alcohol is analogous to that shown above for a secondary alcohol.
When more than one alkene product are possible, the favored product is usually the thermodynamically most stable alkene. More-substituted alkenes are favored over less-substituted ones; and trans-substituted alkenes are preferred compared to cis-substituted ones.
1. Since the C=C bond is not free to rotate, cis-substituted alkenes are less stable than trans-subsituted alkenes because of steric hindrance (spatial interfererence) between two bulky substituents on the same side of the double bond (as seen in the cis product in the above figure). Trans-substituted alkenes reduce this effect of spatial interference by separating the two bulky substituents on each side of the double bond (for further explanation on the rigidity of C=C bond, see Structure and Bonding in Ethene- The pi Bond).
2. Heats of hydrogenation of differently-substituted alkene isomers are lowest for more-substituted alkenes, suggesting that they are more stable than less-substituted alkenes and thus are the major products in an elimination reaction. This is partly because in more --substituted alkenes, the p orbitals of the pi bond are stabilized by neighboring alkyl substituents, a phenomenon similar to hyperconjugation.
Hydride and Alkyl Shifts
Since the dehydration reaction of alcohol has a carbocation intermediate, hydride or alkyl shifts can occur which relocates the carbocation to a more stable position. The dehydrated products therefore are a mixture of alkenes, with and without carbocation rearrangement. Tertiary cation is more stable than secondary cation, which in turn is more stable than primary cation due to a phenomenon known as hyperconjugation, where the interaction between the filled orbitals of neighboring carbons and the singly occupied p orbital in the carbocation stabilizes the positive charge in carbocation.
• In hydride shifts, a secondary or tertiary hydrogen from a carbon next to the original carbocation takes both of its electrons to the cation site, swapping place with the carbocation and renders it a more stable secondary or tertiary cation.
Similarly, when there is no hydride available for hydride shifting, an alkyl group can take its bonding electrons and swap place with an adjacent cation, a process known as alkyl shift.
Practice Problems
Test your understanding by predicting what product(s) will be formed in each of the following reactions:
1.
2.
Solutions
1. Did you notice the reaction temperature? It is only 25°, which is much lower than the required temperature of 170°C for dehydration of primary alcohol. This reaction will not produce any alkene but will form ether.
2. . Notice that the reactant is a secondary -OH group, which will form a relatively unstable secondary carbocation in the intermediate. Thus hydride shift from an adjacent hydrogen will occur to make the carbocation tertiary, which is much more stable. The products are a mixture of alkenes that are formed with or without carbocation rearrangement (A number of products are formed faster than hydride shift can occur).
• Thuy Hoang | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.07%3A_Preparation_of_Alkenes_by_Dehydration_of_Alcohols.txt |
Objectives
After completing this section, you should be able to
1. identify (by wavelength, wavenumber, or both) the region of the electromagnetic spectrum which is used in infrared (IR) spectroscopy.
2. interconvert between wavelength and wavenumber.
3. discuss, in general terms, the effect that the absorption of infrared radiation can have on a molecule.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• infrared spectrum
• wavenumber (reciprocal centimetres)
Study Notes
Notice that the scale at the bottom of the infrared spectrum for 2-hexanone shown is calibrated in wavenumbers (cm−1). A wavenumber is the reciprocal of a wavelength (1/λ); thus, a wavenumber of 1600 cm−1 corresponds to a wavelength of
$\frac{1}{1600\text{\hspace{0.17em}}{\text{cm}}^{-1}}=6.25×{10}^{-4}\text{cm or}6.25\text{\hspace{0.17em}}\mu \text{\hspace{0.17em}}\text{m}$
Organic chemists find it more convenient to deal with wavenumbers rather than wavelengths when discussing infrared spectra.
You will obtain infrared spectra for a number of the compounds you will synthesize in the laboratory component of this course.
The inverted peaks observed in the spectra correspond to molecular stretching and bending vibrations that only occur at certain quantized frequencies. When infrared radiation matching these frequencies falls on the molecule, the molecule absorbs energy and becomes excited. Eventually the molecule returns to its original (ground) state, and the energy which was absorbed is released as heat.
Infrared Spectroscopy
The full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.
Notice in the figure above that infrared light is lower energy than visible light. The wavelengths of infrared radiation are between 0.8 and 250 μm. The units that are typically used for infrared spectroscopy are wavenumbers (which is cm-1). IR spectroscopy analyzes radiation between 40 to 13,000 cm-1. But what type of excitation is occurring when infrared radiation is absorbed by a molecule?
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. This spring 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!
Objectives
After completing this section, you should be able to
1. describe how the so-called “fingerprint region” of an infrared spectrum can assist in the identification of an unknown compound.
2. identify the functional group or groups present in a compound, given a list of the most prominent absorptions in the infrared spectrum and a table of characteristic absorption frequencies.
3. identify the broad regions of the infrared spectrum in which occur absorptions caused by
1. \(\ce{\sf{N-H}}\), \(\ce{\sf{C-H}}\), and \(\ce{\sf{O-H}}\)
2. \(\ce{\sf{C#C}}\) and \(\ce{\sf{C#N}}\)
3. \(\ce{\sf{C=O}}\), \(\ce{\sf{C=N}}\$, and \$\ce{\sf{C=C}}\)
Key Terms
Make certain that you can define, and use in context, the key term below.
• fingerprint region
Study Notes
When answering assignment questions, you may use this IR table to find the characteristic infrared absorptions of the various functional groups. However, you should be able to indicate in broad terms where certain characteristic absorptions occur. You can achieve this objective by memorizing the following table.
Region of Spectrum (cm−1) Absorption
2500-4000 \$\ce{\sf{N−H}}\$, \$\ce{\sf{O−H}}\$, \$\ce{\sf{C−H}}\$
2000-2500 \$\ce{\sf{C#C}}\$, \$\ce{\sf{C#N}}\$
1500-2000 \$\ce{\sf{C=O}}\$, \$\ce{\sf{C=N}}\$, \$\ce{\sf{C=C}}\$
below 1500 Fingerprint region
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.
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.
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. In the mid-1990's, for example, several paintings were identified as forgeries because scientists were able to identify the IR footprint region of red and yellow pigment compounds that would not have been available to the artist who supposedly created the painting (for more details see Chemical and Engineering News, Sept 10, 2007, p. 28).
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.
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 this table.
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.
More examples of IR spectra
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it.
Exercises
Objective
After completing this section, you should be able to use an infrared spectrum to determine the presence of functional groups, such as alcohols, amines and carbonyl groups, in an unknown compound, given a list of infrared absorption frequencies.
Study Notes
In Chapter 12.7 you should have learned, in broad terms, where a few key absorptions occur. Otherwise, to find the characteristic infrared absorptions of the various functional groups, refer to this IR table.
Spectral Interpretation by Application of Group Frequencies
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:
• C–H stretch from 3000–2850 cm-1
• C–H bend or scissoring from 1470-1450 cm-1
• C–H rock, methyl from 1370-1350 cm-1
• C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm-1
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:
• C=C stretch from 1680-1640 cm-1
• =C–H stretch from 3100-3000 cm-1
• =C–H bend from 1000-650 cm-1
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:
• –C≡C– stretch from 2260-2100 cm-1
• –C≡C–H: C–H stretch from 3330-3270 cm-1
• –C≡C–H: C–H bend from 700-610 cm-1
The spectrum of 1-hexyne, a terminal alkyne, is shown below.
In aromatic compounds, each band in the spectrum can be assigned:
• C–H stretch from 3100-3000 cm-1
• overtones, weak, from 2000-1665 cm-1
• C–C stretch (in-ring) from 1600-1585 cm-1
• C–C stretch (in-ring) from 1500-1400 cm-1
• C–H "oop" from 900-675 cm-1
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.
Functional Groups Containing the C-O Bond
Alcohols have IR absorptions associated with both the O-H and the C-O stretching vibrations.
• O–H stretch, hydrogen bonded 3500-3200 cm-1
• C–O stretch 1260-1050 cm-1 (s)
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:
• C=O stretch - aliphatic ketones 1715 cm-1
• α, β -unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
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.
• H–C=O stretch 2830-2695 cm-1
• C=O stretch:
• aliphatic aldehydes 1740-1720 cm-1
• α, β -unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
The carbonyl stretch C=O of esters appears:
• C=O stretch
• aliphatic from 1750-1735 cm-1
• α, β -unsaturated from 1730-1715 cm-1
• C–O stretch from 1300-1000 cm-1
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.
• O–H stretch from 3300-2500 cm-1
• C=O stretch from 1760-1690 cm-1
• C–O stretch from 1320-1210 cm-1
• O–H bend from 1440-1395 and 950-910 cm-1
Figure 11. shows the spectrum of hexanoic acid.
Organic Nitrogen Compounds
• N–O asymmetric stretch from 1550-1475 cm-1
• N–O symmetric stretch from 1360-1290 cm-1
Organic Compounds Containing Halogens
Alkyl halides are compounds that have a C–X bond, where X is a halogen: bromine, chlorine, fluorene, or iodine.
• C–H wag (-CH2X) from 1300-1150 cm-1
• C–X stretches (general) from 850-515 cm-1
• C–Cl stretch 850-550 cm-1
• C–Br stretch 690-515 cm-1
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 table is linked on bottom of page to find other assigned IR peaks.
Exercises
Exercise \(1\)
Caffeine has a mass of 194.19 amu, determined by mass spectrometry, and contains C, N, H, O. What is a molecular formula for this molecule?
Answer
C8H10N4O2
C = 12 × 8 = 96
N = 14 × 4 = 56
H = 1 × 10 = 10
O = 2 × 16 = 32
96+56+10+32 = 194 g/mol
Exercise \(2\)
The following are the spectra for 2-methyl-2-hexene and 2-heptene, which spectra belongs to the correct molecule. Explain.
A:
B:
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
The (A) spectrum is 2-methyl-2-hexene and the (B) spectrum is 2-heptene. Looking at (A) the peak at 68 m/z is the fractioned molecule with just the tri-substituted alkene present. While (B) has a strong peak around the 56 m/z, which in this case is the di-substituted alkene left behind from the linear heptene.
Exercise \(3\)
What are the masses of all the components in the following fragmentations?
Answer
Exercise \(4\)
Which of the following frequencies/wavelengths are higher energy
A. λ = 2.0x10-6 m or λ = 3.0x10-9 m
B. υ = 3.0x109 Hz or υ = 3.0x10-6 Hz
Answer
A. λ = 3.0x10-9 m
B. υ = 3.0x109 Hz
Exercise \(5\)
Calculate the energies for the following;
A. Gamma Ray λ = 4.0x10-11 m
B. X-Ray λ = 4.0x10-9 m
C. UV light υ = 5.0x1015 Hz
D. Infrared Radiation λ = 3.0x10-5 m
E. Microwave Radiation υ = 3.0x1011 Hz
Answer
A. 4.965x10-15 J
B. 4.965x10-17 J
C. 3.31x10-18 J
D. 6.62x10-21 J
E. 1.99x10-22 J
Exercise \(6\)
What functional groups give the following signals in an IR spectrum?
A) 1700 cm-1
B) 1550 cm-1
C) 1700 cm-1 and 2510-3000 cm-1
Answer
Exercise \(7\)
How can you distinguish the following pairs of compounds through IR analysis?
A) CH3OH (Methanol) and CH3CH2OCH2CH3 (Diethylether)
B) Cyclopentane and 1-pentene.
C)
Answer
A) A OH peak will be present around 3300 cm-1 for methanol and will be absent in the ether.
B) 1-pentene will have a alkene peak around 1650 cm-1 for the C=C and there will be another peak around 3100 cm-1 for the sp2 C-H group on the alkene
C) Cannot distinguish these two isomers. They both have the same functional groups and therefore would have the same peaks on an IR spectra.
Exercise \(8\)
The following spectra is for the accompanying compound. What are the peaks that you can I identify in the spectrum?
Source: SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 2 December 2016)
Answer
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1610 C=C
(There is also an aromatic undertone region between 2000-1600 which describes the substitution on the phenyl ring.)
Exercise \(9\)
What absorptions would the following compounds have in an IR spectra?
Answer
A)
Frequency (cm-1) Functional Group
2900-3000 C-C-H, C=C-H
1710 C=O
1610 C=C
1100 C-O
B)
Frequency (cm-1) Functional Group
3200 C≡C-H
2900-3000 C-C-H, C=C-H
2100 C≡C
1710 C=O
C)
Frequency (cm-1) Functional Group
3300 (broad) O-H
2900-3000 C-C-H, C=C-H
2000-1800 Aromatic Overtones
1710 C=O
1610 C=C | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.08%3A_Infrared_Spectroscopy.txt |
Objectives
After completing this section, you should be able to
1. describe, briefly, how a mass spectrometer works.
2. sketch a simple diagram to show the essential features of a mass spectrometer.
3. identify peaks in a simple mass spectrum, and explain how they arise.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• base peak
• parent peak (molecular ion peak)
• cation radical
• relative abundance
• mass spectrometer
• mass spectroscopy
• mass spectrum
• molecular ion (M+·)
• mass-to-charge ratio (m/z)
Study Notes
You may remember from general first-year chemistry how mass spectroscopy has been used to establish the atomic mass and abundance of isotopes.
Mass spectrometers are large and expensive, and usually operated only by fully trained personnel, so you may not have the opportunity to use such an instrument as part of this course. Research chemists often rely quite heavily on mass spectra to assist them in the identification of compounds, and you will be required to interpret simple mass spectra both in assignments and on examinations. Note that in most attempts to identify an unknown compound, chemists do not rely exclusively on the results obtained from a single spectroscopic technique. A combination of chemical and physical properties and spectral evidence is usually employed.
The Mass Spectrometer
In order to measure the characteristics of individual molecules, a mass spectrometer converts them to ions so that they can be moved about and manipulated by external electric and magnetic fields. The three essential functions of a mass spectrometer, and the associated components, are:
1. A small sample is ionized, usually to cations by loss of an electron. The Ion Source
2. The ions are sorted and separated according to their mass and charge. The Mass Analyzer
3. The separated ions are then measured, and the results displayed on a chart. The Detector
Because ions are very reactive and short-lived, their formation and manipulation must be conducted in a vacuum. Atmospheric pressure is around 760 torr (mm of mercury). The pressure under which ions may be handled is roughly 10-5 to 10-8 torr (less than a billionth of an atmosphere). Each of the three tasks listed above may be accomplished in different ways. In one common procedure, ionization is effected by a high energy beam of electrons, and ion separation is achieved by accelerating and focusing the ions in a beam, which is then bent by an external magnetic field. The ions are then detected electronically and the resulting information is stored and analyzed in a computer. A mass spectrometer operating in this fashion is outlined in the following diagram. The heart of the spectrometer is the ion source. Here molecules of the sample (black dots) are bombarded by electrons (light blue lines) issuing from a heated filament. This is called an EI (electron-ionization) source. Gases and volatile liquid samples are allowed to leak into the ion source from a reservoir (as shown). Non-volatile solids and liquids may be introduced directly. Cations formed by the electron bombardment (red dots) are pushed away by a charged repellor plate (anions are attracted to it), and accelerated toward other electrodes, having slits through which the ions pass as a beam. Some of these ions fragment into smaller cations and neutral fragments. A perpendicular magnetic field deflects the ion beam in an arc whose radius is inversely proportional to the mass of each ion. Lighter ions are deflected more than heavier ions. By varying the strength of the magnetic field, ions of different mass can be focused progressively on a detector fixed at the end of a curved tube (also under a high vacuum).
When a high energy electron collides with a molecule it often ionizes it by knocking away one of the molecular electrons (either bonding or non-bonding). This leaves behind a molecular ion (colored red in the following diagram). Residual energy from the collision may cause the molecular ion to fragment into neutral pieces (colored green) and smaller fragment ions (colored pink and orange). The molecular ion is a radical cation, but the fragment ions may either be radical cations (pink) or carbocations (orange), depending on the nature of the neutral fragment. An animated display of this ionization process will appear if you click on the ion source of the mass spectrometer diagram.
Below is typical output for an electron-ionization MS experiment (MS data below 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, which is the mass to charge ratio (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 corresponds to a fragment with m/z = 43 - . 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. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.09%3A_Measuring_the_Molecular_Mass_of_Organic_Compounds%3A_Mass_Spectrometry.txt |
Objectives
After completing this section, you should be able to
1. suggest possible molecular formulas for a compound, given the m/z value for the molecular ion, or a mass spectrum from which this value can be obtained.
2. predict the relative heights of the M+·, (M + 1)+·, etc., peaks in the mass spectrum of a compound, given the natural abundance of the isotopes of carbon and the other elements present in the compound.
3. interpret the fragmentation pattern of the mass spectrum of a relatively simple, known compound (e.g., hexane).
4. use the fragmentation pattern in a given mass spectrum to assist in the identification of a relatively simple, unknown compound (e.g., an unknown alkane).
Study Notes
When interpreting fragmentation patterns, you may find it helpful to know that, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break. You might wish to refer to the table of bond dissociation energies when attempting problems involving the interpretation of mass spectra.
This page looks at how fragmentation patterns are formed when organic molecules are fed into a mass spectrometer, and how you can get information from the mass spectrum.
The Origin of Fragmentation Patterns
When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion and is often given the symbol M+ or . The dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process.
The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical.
The uncharged free radical will not produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump.
The ion, X+, will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this:
Note
The pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element. With an element, each line represents a different isotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up.
In the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion. The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak. This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion.
Using Fragmentation Patterns
This section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page.
Example 12.2.1: Pentane
Let's have another look at the mass spectrum for pentane:
What causes the line at m/z = 57?
How many carbon atoms are there in this ion? There cannot be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C4H9+ then?
C4H9+ would be [CH3CH2CH2CH2]+, and this would be produced by the following fragmentation:
The methyl radical produced will simply get lost in the machine.
The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion:
The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+:
The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process.
Example 12.2.2: Pentan-3-one
This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this is not produced by the same ion as the same m/z value peak in pentane.
If you remember, the m/z = 57 peak in pentane was produced by [CH3CH2CH2CH2]+. If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it.
Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH3CH2CO]+ - which is produced by this fragmentation:
You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group.
Peak Heights and Stability
The more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. We'll look at two common examples of this.
Carbocations (carbonium ions)
Summarizing the most important conclusion from the page on carbocations:
Order of stability of carbocations
primary < secondary < tertiary
Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms.
Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by:
The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by:
You would get the same ion, of course, if the left-hand CH3 group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation.
Acylium ions, [RCO]+
Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one.
The base peak, at m/z=57, is due to the [CH3CH2CO]+ ion. We've already discussed the fragmentation that produces this.
Note
The more stable an ion is, the more likely it is to form. The more of a particular ion that is formed, the higher will be its peak height.
Using mass spectra to distinguish between compounds
Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra.
pentan-2-one CH3COCH2CH2CH3
pentan-3-one CH3CH2COCH2CH3
Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this:
• [CH3CO]+
• [COCH2CH2CH3]+
That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind:
• [CH3CH2CO]+
In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one.
The two mass spectra look like this:
As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentation patterns that can occur. Provided you have a computer data base of mass spectra, any unknown spectrum can be computer analyzed and simply matched against the data base.
Exercises
Objective
After completing this section, you should be able to predict the expected fragmentation for common functional groups, such as alcohols, amines, and carbonyl compounds.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alpha (α) cleavage
• McLafferty rearrangement
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.
Click here for examples of compounds listed by functional group, which demonstrate patterns which can be seen in mass spectra of compounds ionized by electron impact ionization.
Example 12.3.1
The mass spectrum of an aldehyde gives prominent peaks at m/z = 59 (12%, highest value of m/z in the spectrum), 58 (85%), and 29 (100%), as well as others. Propose a structure, and identify the three species whose m/z values were listed.
Solution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.10%3A_Fragmentation_Patterns_of_Organic_Molecules.txt |
Objectives
After completing this section, you should be able to
1. determine the degree of unsaturation of an organic compound, given its molecular formula, and hence determine the number of double bonds, triple bonds and rings present in the compound.
2. draw all the possible isomers that correspond to a given molecular formula containing only carbon (up to a maximum of six atoms) and hydrogen.
3. draw a specified number of isomers that correspond to a given molecular formula containing carbon, hydrogen, and possibly other elements, such as oxygen, nitrogen and the halogens.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• degree of unsaturation
• saturated
• unsaturated
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 because it easily provides information about molecular structure.
Saturated and Unsaturated Molecules
Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons. The presence of a double bond causes alkenes to have less hydrogens than an alkane with the same number of carbons. Likewise, compounds containing a carbon-to-carbon triple bonds (R–C≡C–R) called alkynes (Discussed in Chapter 9), also have fewer hydrogens than the corresponding alkane. Collectively, compounds which have fewer hydrogen atoms than an alkane with the same number of carbon atoms are called unsaturated hydrocarbons. The relationship between the number of carbons (n) and hydrogens in the molecular formula for alkanes, alkenes, and alkynes are listed below.
For example, the three carbon alkane, propane has the molecular formula of C3H8. While the unsaturated compounds propene (C3H6) and propyne (C3H4) both have fewer hydrogens. Also, it is important to note that cycloalkanes with one ring have a general molecular formula of CnH2n just like alkenes. Because they also have fewer than maximum number of hydrogens possible, cyclic compounds are also considered unsaturated.
Identifying Degrees of Unsaturation
Every ring or pi bond in a compound is said to represent one degree of unsaturation. Being able to determine the degrees of unsaturation in a given compound is an important skill. Each of the following compounds are isomers of C5H7 and contain two degrees of unsaturation.
Exercise $1$
How many degrees of unsaturation do the following compounds have?
Answer
a) 0
b) 1
c) 1
d) 2
e) 2
f) 2
Calculating the Degree of Unsaturation (DoU)
As noted above, every degree of unsaturation causes the loss of two hydrogens from a compound's molecular formula when compared to an alkane with the same number of carbons. Understanding this relationship allows for the degrees of unsaturation of a compound to be calculated from its molecular formula. First, the maximum number of hydrogens possible for a given compound (2C + 2) is calculated and then the actual number of hydrogens present in the compound (H) is subtracted. If this difference is then divided by 2 the answer will be equal to the degrees of unsaturation for the compound.
For a compound which only contains carbon and hydrogen:
DoU = (2C + 2) - H / 2
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]. Because the compound only has 4 hydrogens in its molecular formula, it would have to gain 4 more hydrogens in order to be fully saturated (8-4 = 4). Degrees of unsaturation is equal to half the number of hydrogens the molecule needs to be fully saturated. This compound has 2 degrees of unsaturation (4/2 = 2).
The DoU of compounds containing elements other than carbon and hydrogen can also be calculated in a similar fashion. However, different elements can affect the formula used to calculate DoU.
For a compound which contains elements other than carbon and hydrogen:
$DoU= \dfrac{2C+2+N-X-H}{2} \tag{7.2.1}$
• $C$ 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
A halogen (X) replaces a hydrogen in a compound because both form one single bond. Therefore the DoU formula subtracts the number of halogens (X) present in a compound. For instance, 1,1-dichloroethene (C2H2Cl2) has two fewer hydrogens than ethene (C2H4) yet they both have one degree of unsaturation.
Oxygen and sulfur are not included in the DoU formula because saturation is unaffected by these elements. The inclusion of an alcohol or sulfur in a compound does not change the number of hydrogens to obtain saturation. As seen in alcohols, the number of hydrogens in cyclohexanol (C6H12O) matches the number of hydrogens in cyclohexane (C6H12) and they both have one degree of unsaturation.
When a nitrogen is present in a compound one more hydrogen is required to reach saturation. Therefore, we add the number of nitrogens (N). Propyl amine (C3H9N) has one more hydrogen compared to propane (C3H8) both of which are saturated compounds with 0 DoU.
With the degrees of unsaturation comes information about the possible number of rings and multiple bonds in a given compound. Remember, the degrees of unsaturation only gives the sum of pi bonds and/or rings.
• 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).
Example 7.3.1: 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.
References
1. Vollhardt, K. P.C. & Shore, N. (2007). Organic Chemistry (5thEd.). New York: W. H. Freeman. (473-474)
2. Shore, N. (2007). Study Guide and Solutions Manual for Organic Chemistry (5th Ed.). New York: W.H. Freeman. (201)
Exercise $2$
Determine whether the following molecules are saturated or unsaturated.
Answer
a) unsaturated (Even though the rings only contain single bonds, rings are considered unsaturated.)
b) unsaturated
c) saturated
d) unsaturated
e) unsaturated
f) saturated
Exercise $3$
Determine the degrees of unsaturation for each of the following compounds.
Answer
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 (2 rings)
b) 2 (one double bond and the double bond from the carbonyl)
c) 0 (no double bonds or rings)
d) 10 (2(10) + 2 + 4 - 0 - 6)/2 = 10
e) 1 (2(5) + 2 + 0 - 0 - 10)/2 = 1
f) 0 (2(6) + 2 + 0 - 2 - 12)/2 = 0
Exercise $4$
Calculate the degrees of unsaturation for the following molecular formulas:
a) C9H20 b) C7H8 c) C5H7Cl d) C9H9NO4
Answer
Use the formula to solve (O not involved in the formula)
(a.) 0 (2(9) + 2 - 20)/2 = 0
(b.) 4 (2(7) + 2 - 8)/2 = 4
(c.) 2 (2(5) + 2 - 1 -7)/2 = 2
(d.) 6 (2(9) + 1 - 7)/2 = 6 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/11%3A_Alkenes%3A_Infrared_Spectroscopy_and_Mass_Spectrometry/11.11%3A_Degree_of_Unsaturation%3A_Another_Aid_to_Identifying_Molecular_Structure.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the catalytic hydrogenation of an alkene.
2. identify the product obtained from the hydrogenation of a given alkene.
3. identify the alkene, the reagents, or both, required to prepare a given alkane by catalytic hydrogenation.
4. describe the mechanism of the catalytic hydrogenation of alkenes.
5. explain the difference between a heterogeneous reaction and a homogeneous reaction.
6. recognize that other types of compounds containing multiple bonds, such as ketones, esters, nitriles and aromatic compounds, do not react with hydrogen under the conditions used to hydrogenate alkenes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Adams’ catalyst
• hydrogenation
Study Notes
Chemical reactions that are heterogeneous have reactants that are in at least two different phases (e.g. gas with a solid), whereas homogeneous reactions occur in a single phase (e.g. gas with another gas).
Some confusion may arise from the description of the catalyst used in the reaction between alkenes and hydrogen. Three metals—nickel, platinum and palladium—are commonly used, but a chemist cannot simply place a piece of one of these metals in a mixture of the alkene and hydrogen and get a reaction. Each metal catalyst must be prepared in a special way:
• nickel is usually used in a finely divided form called “Raney nickel.” It is prepared by reacting a Ni-Al alloy with NaOH.
• palladium is obtained commercially “supported” on an inert substance, such as charcoal, (Pd/C). The alkene is usually dissolved in ethanol when Pd/C is used as the catalyst.
• platinum is used as PtO2, Adams’ catalyst, although it is actually platinum metal that is the catalyst. The hydrogen used to add to the carbon-carbon double bond also reduces the platinum(IV) oxide to finely divided platinum metal. Ethanol or acetic acid is used as the solvent for the alkene.
Other types of compounds containing multiple bonds, such as ketones, esters, and nitriles, do not react with hydrogen under the conditions used to hydrogenate alkenes. The examples below show reduction of an alkene, but the ketone and nitrile groups present remain intact and are not reduced.
Aromatic rings are also not reduced under the conditions used to reduce alkenes, although these rings appear to contain three carbon-carbon double bonds. As you will see later, aromatic rings do not really contain any double bonds, and many chemists prefer to represent the benzene ring as a hexagon with a circle inside it
rather than as a hexagon with three alternating double bonds.
The representation of the benzene ring will be discussed further in Section 15.2.
The reaction between carbon-carbon double bonds and hydrogen provides a method of determining the number of double bonds present in a compound. For example, one mole of cyclohexene reacts with one mole of hydrogen to produce one mole of cyclohexane:
but one mole of 1,4-cyclohexadiene reacts with two moles of hydrogen to form one mole of cyclohexane:
A chemist would say that cyclohexene reacts with one equivalent of hydrogen, and 1,4-cyclohexadiene reacts with two equivalents of hydrogen. If you take a known amount of an unknown, unsaturated hydrocarbon and determine how much hydrogen it will absorb, you can readily determine the number of double bonds present in the hydrocarbon (see question 2, below).
Addition of hydrogen to a carbon-carbon double bond is called hydrogenation. The overall effect of such an addition is the reductive removal of the double bond functional group. Regioselectivity is not an issue, since the same group (a hydrogen atom) is bonded to each of the double bond carbons. The simplest source of two hydrogen atoms is molecular hydrogen (H2), but mixing alkenes with hydrogen does not result in any discernible reaction. Although the overall hydrogenation reaction is exothermic, a high activation energy prevents it from taking place under normal conditions. This restriction may be circumvented by the use of a catalyst, as shown in the reaction coordinate diagram below.
An example of an alkene addition reaction is a process called hydrogenation. In a hydrogenation reaction, two hydrogen atoms are added across the double bond of an alkene, resulting in a saturated alkane. Hydrogenation of a double bond is a thermodynamically favorable reaction because it forms a more stable (lower energy) product. In other words, the energy of the product is lower than the energy of the reactant; thus it is exothermic (heat is released). The heat released is called the heat of hydrogenation, which is an indicator of a molecule’s stability.
Catalysts are substances that changes the rate (velocity) of a chemical reaction without being consumed or appearing as part of the product. Catalysts act by lowering the activation energy of reactions, but they do not change the relative potential energy of the reactants and products. Finely divided metals, such as platinum, palladium and nickel, are among the most widely used hydrogenation catalysts. Catalytic hydrogenation takes place in at least two stages, as depicted in the diagram. First, the alkene must be adsorbed on the surface of the catalyst along with some of the hydrogen. Next, two hydrogens shift from the metal surface to the carbons of the double bond, and the resulting saturated hydrocarbon, which is more weakly adsorbed, leaves the catalyst surface. The exact nature and timing of the last events is not well understood.
As shown in the energy diagram, the hydrogenation of alkenes is exothermic, and heat is released corresponding to the ΔH in the diagram. This heat of reaction can be used to evaluate the thermodynamic stability of alkenes having different numbers of alkyl substituents on the double bond. For example, the following table lists the heats of hydrogenation for three C5H10 alkenes which give the same alkane product (2-methylbutane). Since a larger heat of reaction indicates a higher energy reactant, these heats are inversely proportional to the stabilities of the alkene isomers. To a rough approximation, we see that each alkyl substituent on a double bond stabilizes this functional group by a bit more than 1 kcal/mole.
Alkene Isomer (CH3)2CHCH=CH2
3-methyl-1-butene
CH2=C(CH3)CH2CH3
2-methyl-1-butene
(CH3)2C=CHCH3
2-methyl-2-butene
Heat of Reaction
( ΔHº )
–30.3 kcal/mole –28.5 kcal/mole –26.9 kcal/mole
Catalytic Hydrogenation Mechanism
From the mechanism shown here we would expect the addition of hydrogen to occur with syn-stereoselectivity. This is often true, but the hydrogenation catalysts may also cause isomerization of the double bond prior to hydrogen addition, in which case stereoselectivity may be uncertain.
Exercise 1
1. In the reaction
1. 0.500 mol of ethene reacts with _______ mol of hydrogen. Thus a chemist might say that ethene reacts with one _______ of hydrogen.
2. ethene is being _______; while _______ is being oxidized.
3. the oxidation number of carbon in ethene is _______; in ethane it is _______.
Answer
1. 0.500 mol of ethene reacts with 0.500 mol of hydrogen. Thus a chemist might say that ethene reacts with one equivalent of hydrogen.
2. ethene is being reduced; while hydrogen is being oxidized.
3. the oxidation number of carbon in ethene is −2; in ethane it is −3.
Exercise 2
When 1.000 g of a certain triglyceride (fat) is treated with hydrogen gas in the presence of Adams’ catalyst, it is found that the volume of hydrogen gas consumed at 99.8 kPa and 25.0°C is 162 mL. A separate experiment indicates that the molar mass of the fat is 914 g mol−1. How many carbon-carbon double bonds does the compound contain?
Answer
Amount of hydrogen consumed
$\begin{array}{l}=n\text{\hspace{0.17em}}\text{mol}\ \text{=}\frac{PV}{RT}\ =\frac{99.8\text{\hspace{0.17em}}\text{kPa}×0.162\text{\hspace{0.17em}}\text{L}}{8.31\text{\hspace{0.17em}}\text{kPa}\cdot {\text{mol}}^{}}\end{array}$1 K1 ×298K =6.53× 103 mol H2
Amount of fat used
$\begin{array}{l}=\frac{\left(1.000\text{\hspace{0.17em}}\text{g}\right)×\left(1\text{\hspace{0.17em}}\text{mol}\right)}{\left(914\text{\hspace{0.17em}}\text{g}\right)}\ =1.09×{10}^{-3}\text{\hspace{0.17em}}\text{mol}\text{\hspace{0.17em}}\text{fat}\end{array}$
Ratio of moles of hydrogen consumed to moles of fat
$\begin{array}{l}=6.53×{10}^{-3}:1.09×{10}^{-3}\ =6:1\end{array}$
Thus, the fat contains six carbon-carbon double bonds per molecule. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.02%3A_Catalytic_Hydrogenation.txt |
Objectives
After completing this section, you should be able to
1. explain the term “electrophilic addition reaction,” using the reaction of a protic acid, HX, with an alkene as an example.
2. write the mechanism for the reaction of a protic acid, HX, with an alkene.
3. sketch a reaction energy diagram for the electrophilic addition of an acid, HX, to an alkene.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• carbocation (carbonium ion)
• electrophilic addition reaction
Study Notes
An electrophilic addition reaction is a reaction in which a substrate is initially attacked by an electrophile, and the overall result is the addition of one or more relatively simple molecules across a multiple bond.
The mechanism for the addition of hydrogen halide to propene shown in the reading is quite detailed. Normally, an organic chemist would write the reaction scheme as follows:
However, the more detailed mechanism shown in the reading does allow you to see the exact fate of all the electrons involved in the reaction.
In your previous chemistry course, you were probably taught the importance of balancing chemical equations. It may come as a surprise to you that organic chemists usually do not balance their equations, and often represent reactions using a format which is quite different from the carefully written, balanced equations encountered in general chemistry courses. In fact, organic chemists are rarely interested in the inorganic products of their reactions; furthermore, most organic reactions are non-quantitative in nature.
In many of the reactions in this course, the percentage yield is indicated beneath the products: you are not expected to memorize these figures. The question of yield is very important in organic chemistry, where two, five, ten or even twenty reactions may be needed to synthesize a desired product. For example, if a chemist wishes to prepare compound D by the following reaction sequence:
$A → B → C → D \nonumber$
and each of the individual steps gives only a 50% yield, one mole of A would give only
1 mol $×\frac{50%}{100%}×\frac{50%}{100%}×\frac{50%}{100%}=$ 0.125 mol of D
You will gain first-hand experience of such situations in the laboratory component of this course.
Introduction
One of the most important reactions for alkenes is called electrophilic addition. In this chapter several variations of the electrophilic addition reaction will be discussed. Each case will have aspects common among all electrophilic addition. In this section, the electrophilic addition reaction will be discussed in general to provide a better understanding of subsequent alkene reactions.
As discussed in Section 6-5, the double bond in alkenes is electron rich due to the prescience of 4 electrons instead of the two in a single bond. Also, the pi electrons are positioned above and below the double bond making them more accessibly for reactions. Overall, double bonds can easily donate lone pair electrons to act like a nucleophile (nucleus-loving, electron rich, a Lewis acid). During an electrophilic addition reactions double bonds donate lone pair electrons to an electrophile (Electron-loving, electron poor, a Lewis base). There are many types of electrophilic addition, but this section will focus on the addition of hydrogen halides (HX). Many of the basic ideas discussed will aplicable to subsequent electrophilic addition reactions.
General Reaction
Overall during this reaction the pi bond of the alkene is broken to form two single, sigma bonds. As shown in the reaction mechanism, one of these sigma bonds is connected to the H and the other to the X of the hydrogen halide. This reaction works well with HBr and HCl. HI can also but used but is is usually generated during the reaction by reacting potassium iodidie (KI) with phosphoric acid (H3PO4).
Addition to symmetrical alkenes
What happens?
All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other.
For example, with ethene and hydrogen chloride, you get chloroethane:
With but-2-ene you get 2-chlorobutane:
What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately.
Mechanism
Step 1) Electrophilic Attack
During the first step of the mechanism, the 2 pi electrons from the double bond attack the H in the HBr electrophile which is shown by a curved arrow. The two pi electrons form a C-H sigma bond between the hydrogen from HBr and a carbon from the double bond. Simultaneously the electrons from the H-X bond move onto the halogen to form a halide anion. The removal of pi electrons form the double bond makes one of the carbons become an electron deficient carbocation intermediate. This carbon is sp2 hybridized and the positive charge is contained in an unhybridized p orbital.
Step 2) Nucleophilic attack by halide anion
The formed carbocation now can act as an electrophile and accept an electron pair from the nucleophilic halide anion. The electron pair becomes a X-C sigma bond to create the neutral alkyl halide product of electrophilic addition.
All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s.
Reaction Energy Diagram
An energy diagram for the two-step electrophilic addition mechanism is shown below. The energy diagram has two peaks which represent the transition state for each mechanistic step. The peaks are separated by a valley which represents the high energy carbocation reaction intermediate. Because the energy of activation for the first step of the mechanism (ΔE1) is much larger than the second (ΔE1), the first step of the mechanism is the rate-determining step. Both the alkene and the hydrogen halide are reactants in the first step of the mechanism, this electrophilic addition is a second order reaction and the rate law expression can be written rate = k[Alkene][HX]. Also, any structural feature which can stabilize the transition state between the reactants the carbocation intermediate will lower ΔE1 and thereby increase the reaction rate. Overall, the alkyl halide product of this reaction more stable than the reactants making the reaction exothermic.
Reaction rates
Variation of rates when you change the halogen
Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions.
When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow.
Variation of rates when you change the alkene
This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be.
Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example:
There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions.
Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this.
Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes.
The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride.
The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction:
The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster.
Representing Organic Reactions
Organic reaction equations are often written in one of two ways. The reactant for the reaction is written to the left of the reaction arrow. The products are written to the right of the arrow. The reagent for the reaction is written above the arrow. Other reaction conditions such as the solvent or the temperature can be written above or below the reaction arrow.
Alternativley the reactant and reagent can both be written to the left of the reaction arrow. This is typically done to highlight the importance of the reactant. The solvent and reaction temperature are still written above or below the reaction arrow. The reaction products are still written to the right of the reaction arrow.
very important regarding electrophilic addition reactions is that if the starting alkene is asymmetrical, there are two possible courses that could be followed, depending on which of the two alkene carbons forms the new sigma bond in the first step.
Of course, the two reaction courses involve two different carbocation intermediates, which may have different energy levels. Two different products are possible, and in general the product which predominates will be the one that is derived from the lower-energy carbocation intermediate.
This important regiochemical principle is nicely illustrated by a simple electrophilic addition that is commonly carried out in the organic laboratory: the conversion of an alkene to an alkyl bromide by electrophilic addition of HBr to the double bond. Let's look at a hypothetical addition of HBr to 2-methyl-2-butene, pictured below. Two different regiochemical outcomes are possible:
The initial protonation step could follow two different pathways, resulting in two different carbocation intermediates: pathway 'a' gives a tertiary carbocation intermediate (Ia), while pathway 'b' gives a secondary carbocation intermediate (Ib) We know already that the tertiary carbocation is more stable (in other words, lower in energy). According to the Hammond postulate, this implies that the activation energy for pathway a is lower than in pathway b, meaning in turn that Ia forms faster.
Because the protonation step is the rate determining step for the reaction, the tertiary alkyl bromide A will form much faster than the secondary alkyl halide B, and thus A will be the predominant product observed in this reaction. This is a good example of a non-enzymatic organic reaction that is highly regiospecific.
In the example above, the difference in carbocation stability can be accounted for by the electron-donating effects of the extra methyl group on one side of the double bond. It is generally observed that, in electrophilic addition of acids (including water) to asymmetrical alkenes, the more substituted carbon is the one that ends up bonded to the heteroatom of the acid, while the less substituted carbon is protonated.
This rule of thumb is known as Markovnikov's rule, after the Russian chemist Vladimir Markovnikov who proposed it in 1869.
While it is useful in many cases, Markovikov's rule does not apply to all possible electrophilic additions. It is more accurate to use the more general principle that has already been stated above:
When an asymmetrical alkene undergoes electrophilic addition, the product that predominates is the one that results from the more stable of the two possible carbocation intermediates.
How is this different from Markovnikov's original rule? Consider the following hypothetical reaction, which is similar to the HBr addition shown above except that the six methyl hydrogens on the left side of the double bond have been replaced by highly electron-withdrawing fluorines.
Now when HBr is added, it is the less substituted carbocation that forms faster in the rate-determining protonation step, because in this intermediate the carbon bearing the positive charge is located further away from the electron-withdrawing, cation-destabilizing fluorines. As a result, the predominant product is the secondary rather than the tertiary bromoalkane. This would be referred to as an 'anti-Markovnikov' addition product, because it 'breaks' Markovnikov's rule.
Example
Predict the product of the following reaction:
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.03%3ANucleophilic_Character_of_the__Pi__Bond%3A__Electrophilic___Addition_of_Hydrogen_Halides.txt |
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane.
What Is Electrophilic Hydration?
Electrophilic hydration is the reverse dehydration of alcohols and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
The phrase "electrophilic" literally means "electron loving" (whereas "nucleophilic" means "nucleus loving"). Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see SN2 for more details).
How Does Electrophilic Hydration Work?
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
• Primary Alcohol: Less than 170ºC
• Secondary Alcohol: Less than 100ºC
• Tertiary Alcohol: Less than 25ºC
But...Why Does Electrophilic Hydration Work?
• An alkene placed in an aqueous non-nucleophilic strong acid immediately "reaches out" with its double bond and attacks one of the acid's hydrogen atoms (meanwhile, the bond between oxygen and hydrogen performs heterolytic cleavage toward the oxygen—in other words, both electrons from the oxygen/hydrogen single bond move onto the oxygen atom).
• A carbocation is formed on the original alkene (now alkane) in the more-substituted position, where the oxygen end of water attacks with its 4 non-bonded valence electrons (oxygen has 6 total valence electrons because it is found in Group 6 on the periodic table and the second row down: two electrons in a 2s-orbital and four in 2p-orbitals. Oxygen donates one valence electron to each bond it forms, leaving four 4 non-bonded valence electrons).
• After the blue oxygen atom forms its third bond with the more-substituted carbon, it develops a positive charge (3 bonds and 2 valence electrons give the blue oxygen atom a formal charge of +1).
• The bond between the green hydrogen and the blue oxygen undergoes heterolytic cleavage, and both the electrons from the bond move onto the blue oxygen. The now negatively-charged strong acid picks up the green electrophilic hydrogen.
• Now that the reaction is complete, the non-nucleophilic strong acid is regenerated as a catalyst and an alcohol forms on the most substituted carbon of the current alkane. At lower temperatures, more alcohol product can be formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the red H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction.
If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products:
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift.
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
What is Stereochemistry and How Does It Apply?
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product could look like either of the following products:
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Is There a Better Way to Add Water to Synthesize an Alcohol From an Alkene?
A more efficient pathway does exist: see Oxymercuration - Demercuration: A Special Electrophilic Addition. Oxymercuration does not allow for rearrangements, but it does require the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Problems
Predict the product of each reaction.
1)
2) How does the cyclopropane group affect the reaction?
3) (Hint: What is different about this problem?)
4) (Hint: Consider stereochemistry.)
5) Indicate any shifts as well as the major product:
Answers to Practice Problems
1) This is a basic electrophilic hydration.
2) The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
3) A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed.
4) This reaction will have poor yields due to a very unstable intermediate. For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp2 hybridization, and when the water is added on, the carbons change their hybridization to sp3. This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic.
5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes.
Contributors
• Lance Peery (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.04%3A_Alcohol__Synthesis_by_Electrophilic_Hydration%3A_Thermodynamic_Control.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the reaction of chlorine or bromine with a given alkene.
2. identify the conditions under which an addition reaction occurs between an alkene and chlorine or bromine.
3. draw the structure of the product formed when a given alkene undergoes an addition reaction with chlorine or bromine.
4. write the mechanism for the addition reaction that occurs between an alkene and chlorine or bromine, and account for the stereochemistry of the product.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anti stereochemistry
• bromonium ion
Study Notes
In the laboratory you will test a number of compounds for the presence of a carbon-carbon double bond. A common test is the decolourization of a reddish-brown bromine solution by an alkene.
The two-step mechanism shown in the LibreText pages gives you an idea of how the reaction between an alkene and a halogen occurs. Note the formation of the bridged bromonium ion intermediate and the anti stereochemistry of the final product because the two bromine atoms come from opposite faces of the double bond.
Additional evidence in support of the bromonium ion mechanism comes from the results obtained when an alkene (such as cyclopentene) reacts with bromine in the presence of sodium chloride (see Figure 8.2: Reaction of an alkene with bromine in the presence of sodium chloride, below).
Once formed, the bromonium ion is susceptible to attack by two nucleophiles—chloride ion and bromide ion—and, in fact, a mixture of two products (both produced by anti attack) is formed.
Halogens can act as electrophiles to which can be attacked by a pi bond from an alkene. Pi bonds represents a region of electron density and therefore function as a nucleophiles. How is it possible for a halogen to obtain positive charge to be an electrophile?
Introduction
A halogen molecule, for example Br2, approaches a double bond of the alkene, electrons in the double bond repel electrons in the bromine molecule causing polarization of the halogen-halogen bond. This creates a dipole moment in the halogen-halogen bond. Heterolytic bond cleavage occurs and one of the halogens obtains a positive charge and reacts as an electrophile. The reaction of the addition is not regioselective but is stereoselective. Stereochemistry of this addition can be explained by the mechanism of the reaction. In the first step, the electrophilic halogen (with the positive charge) approaches the pi bond and 2p orbitals of the halogen bond with two carbon atoms creating a cyclic ion with a halogen as the intermediate. In the second step, the remaining halide ion (halogen with the negative charge) attacks either of the two carbons in the cyclic ion from the back side of the cycle as in the SN2 reaction. Therefore stereochemistry of the product is anti addition of vicinal dihalides.
$\ce{R_2C=CR_2 + X_2 \rightarrow R_2CX-CR_2X} \tag{8.2.1}$
Step 1: In the first step of the addition the Br-Br bond polarizes, heterolytic cleavage occurs and Br with the positive charge forms a cyclic intermediate with the two carbons from the alkene.
Step 2: In the second step, bromide anion attacks either carbon of the bridged bromonium ion from the back side of the ring. The bromine atom in the bromonium ion acts as a shield in a way, forcing the bromonium anion to attack from the opposite side as it. The result of this is the ring opening up with the two halogens on opposite sides as each other. This is anti stereochemistry, which is defined as the two bromine atoms come from opposite faces of the double bond. The product is that the bromines add on trans to each other.
Halogens that are commonly used in this type of the reaction are: $Br$ and $Cl$. In thermodynamical terms $I$ is too slow for this reaction because of the size of its atom, and $F$ is too vigorous and explosive.
Because the halide ion can attack any carbon from the opposite side of the ring it creates a mixture of steric products. Optically inactive starting material produce optically inactive achiral products (meso) or a racemic mixture.
Electrophilic addition mechanism consists of two steps.
Before constructing the mechanism let us summarize conditions for this reaction. We will use Br2 in our example for halogenation of ethylene.
Nucleophile Double bond in alkene
Electrophile Br2, Cl2
Regiochemistry not relevant
Stereochemistry ANTI
Summary
Halogens can act as electrophiles due to polarizability of their covalent bond. Addition of halogens is stereospecific and produces vicinal dihalides with anti addition.
Problems
1.What is the mechanism of adding Cl2 to the cyclohexene?
2.A reaction of Br2 molecule in an inert solvent with alkene follows?
a) syn addition
b) anti addition
c) Morkovnikov rule
3)
4)
Key:
1.
2. b
3.enantiomer
4.
Contributors and Attributions
• Jim Clark (Chemguide.co.uk)
• Layne Morsch (University of Illinois Springfield)
• Dr. Krista Cunningham
• Lauren Reutenauer (Amherst College)
12.06: The Generality of Electrophilic Addition
The proton is not the only electrophilic species that initiates addition reactions to the double bond of alkenes. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to bond to the alkene pi-electrons, and the resulting complexes rearrange or are attacked by nucleophiles to give addition products.The electrophilic character of the halogens is well known. Chlorine (Cl2) and bromine(Br2) react selectively with the double bond of alkenes, and these reactions are what we will focus on. Fluorine adds uncontrollably with alkenes,and the addition of iodine is unfavorable, so these are not useful preparative methods.
The addition of chlorine and bromine to alkenes, as shown below, proceeds by an initial electrophilic attack on the pi-electrons of the double bond. Dihalo-compounds in which the halogens are bound to adjacent carbons are called vicinal, from the Latin vicinalis, meaning neighboring.
R2C=CR2 + X2 ——> R2CX-CR2X
Other halogen-containing reagents which add to double bonds include hypohalous acids, HOX, and sulfenyl chlorides, RSCl. These reagents are unsymmetrical, so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety in both of these reagents is the halogen.
(CH3)2C=CH2 + HOBr ——> (CH3)2COH-CH2Br
(CH3)2C=CH2 + C6H5SCl ——> (CH3)2CCl-CH2SC6H5
The regioselectivity of the above reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Thus, bonding of an electrophilic species to the double bond of an alkene should result in preferential formation of the more stable (more highly substituted) carbocation, and this intermediate should then combine rapidly with a nucleophilic species to produce the addition product.
To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pKa's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS(+), whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur).
Below are some examples illustrating the addition of various electrophilic halogen reagents to alkene groups. Notice the specific regiochemistry of the products, as explained above. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.05%3A_Electrophilic_Addition_of_Halogens_to_Alkenes.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the hydration of an alkene with sulfuric acid.
2. write an equation for the formation of an alcohol from an alkene by the oxymercuration-demercuration process.
3. identify the alkene, the reagents, or both, that should be used to produce a given alcohol by the oxymercuration-demercuration process.
4. write the mechanism for the reaction of an alkene with mercury(II) acetate in aqueous tetrahydrofuran (THF).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• hydration
• oxymercuration
Study Notes
Oxymercuration is the reaction of an alkene with mercury(II) acetate in aqueous THF, followed by reduction with sodium borohydride. The final product is an alcohol.
It is important that you recognize the similarity between the mechanisms of bromination and oxymercuration. Recognizing these similarities helps you to reduce the amount of factual material that you need to remember.
Mercuric acetate, or mercury(II) acetate, to give it the preferred IUPAC name, is written as Hg(OAc)2; by comparing this formula with the formula Hg(O2CCH3)2, you can equate Ac with -COCH3. In fact, Ac is an abbreviation used for the acetyl group with the structure shown below as are other similar abbreviations that you will encounter.
Ac (acetyl)
Me (methyl)
Et (ethyl)
Prn (n-propyl)
Pri (isopropyl)
But (tert-butyl)
Ph (phenyl)
What Is Electrophilic Hydration?
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane. Electrophilic hydration is the reverse dehydration of alcohols and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see SN2 for more details).
How Does Electrophilic Hydration Work?
Mechanism for 3º Alcohol (1º and 2º mechanisms are similar):
Hydration is the process where water is added to an alkene to yield an alcohol. Acid-catalyzed hydration is when a strong acid is used as a catalyst to begin the reaction, but let's look at the mechanism below and break down the steps.
Step 1: A hydrogen atom from the acid is attacked by the nucleophilic Pi-electrons in the double bond. This is similar to the other alkene reactions we have seen so far. In this process, a new C-H bond is formed to create the more stable carbocation.
Step 2: A nucleophilic water attacks or donates a lone pair to the positively charged carbon in the carbocation intermediate created in the first step. There is new C-O bond with the O having a formal charge of +1. The product is a protonated alcohol.
Step 3: To obtain the neutral alcohol product, the final step is to deprotonate the oxygen atom with the +1 formal charge using the acid. This final step regenerates the acid catalyst and yields the neutral alcohol product.
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction.
If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products:
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift.
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. After one H atom is removed from the water molecule, the alcohol is attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
What is Stereochemistry and How Does It Apply?
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product of the following reaction could look like either of the following products:
There is no stereochemical control in acid-catalyzed hydration reactions. This is due to the trigonal planar, sp2 nature of the carbocation intermediate. Water can act as a nucleophile to form a bond to either face of the carbocation, resulting in a mixture of stereochemical outcomes.
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Is There a Better Way to Add Water to Synthesize an Alcohol From an Alkene?
A more efficient pathway does exist: Oxymercuration - Demercuration, a special type of electrophilic addition. Oxymercuration does not allow for rearrangements, but it does require the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Oxymercuration is a special electrophilic addition. It is anti-stereospecific and regioselective. Regioselectivity is a process in which the substituents choses one direction it prefers to be attached to over all the other possible directions. The good thing about this reaction is that there are no carbocation rearrangement due to stabilization of the reactive intermediate. Similar stabilization is also seen in bromination addition to alkenes.
Introduction to Oxymercuration
One of the major advantages to oxymercuration is that carbocation rearrangements cannot occur under these conditions (Hg(OAc)2, H2O). Carbocation rearrangement is a process in which the carbocation intermediate can undergo a methyl or alkyl shift to form a more stable ion. Due to a possible carbocation rearrangement, the reaction below would not generate the product shown in high yields. In contrast, the oxymercuration reaction would proceed to form the desired product.
This reaction involves a mercury acting as a reagent attacking the alkene double bond to form a Mercurinium Ion Bridge. A water molecule will then attack the most substituted carbon to open the mercurium ion bridge, followed by proton transfer to solvent water molecule.
The organomercury intermediate is then reduced by sodium borohydride - the mechanism for this final step is beyond the scope of our discussion here. Notice that overall, the oxymercuration - demercuration mechanism follows Markovnikov's regioselectivity with the OH group is attached to the most substituted carbon and the H is attach to the least substituted carbon. The reaction is useful, however, because strong acids are not required, and carbocation rearrangements are avoided because no discreet carbocation intermediate forms.
It is important to note that for the mechanism shown above, the enantiomer of the product shown is also formed. This is the result of formation of the mercurium ion below the alkene in the first step.
Some Practice Problems
Practice problems
What are the end products of these reactants?
Answer
The end product to these practice problems are pretty much very similar. First, you locate where the double bond is on the reactant side. Then, you look at what substituents are attached to each side of the double bond and add the OH group to the more substituent side and the hydrogen on the less substituent side.
More Problems
Predict the product of each reaction.
1)
2) How does the cyclopropane group affect the reaction?
3) (Hint: What is different about this problem?)
4) (Hint: Consider stereochemistry.)
5) Indicate any shifts as well as the major product:
Answer
1) This is a simple electrophilic hydration.
2) The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
3) A hydride shift actually occurs from the 3 carbon of the 3-methylcyclopentene to where the carbocation had formed.
4) This reaction will have poor yields due to a very unstable intermediate. For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp2 hybridization, and when the water is added on, the carbons change their hybridization to sp3. This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic.
5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.07%3A_Oxymncuration-Demercuration%3A_A_Special_Elcctrophilic_Addition.txt |
Objectives
After completing this section, you should be able to
1. identify hydroboration (followed by oxidation) as a method for bringing about the (apparently) non-Markovnikov addition of water to an alkene.
2. write an equation for the formation of a trialkylborane from an alkene and borane.
3. write an equation for the oxidation of a trialkylborane to an alcohol.
4. draw the structure of the alcohol produced by the hydroboration, and subsequent oxidation, of a given alkene.
5. determine whether a given alcohol should be prepared by oxymercuration-demercuration or by hydroboration-oxidation, and identify the alkene and reagents required to carry out such a synthesis.
6. write the detailed mechanism for the addition of borane to an alkene, and explain the stereochemistry and regiochemistry of the reaction.
Key Terms
Make certain that you can define, and use in context, the key term below.
• hydroboration
Study Notes
The two most important factors influencing organic reactions are polar (or electronic) effects and steric effects.
Hydroboration-oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene double bond. Furthermore, the borane acts as a Lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The hydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
Introduction
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An additional feature of this reaction is that it occurs without rearrangement.
The Borane Complex
First off it is very important to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B2H6 (diborane). Additionally, the dimer B2H6 ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a Lewis acid-base complex, which allows boron to have an electron octet.BH3B2H6
The Mechanism
Step 1
• Part 1: Hydroboration of the alkene. In this first step the addition of the borane to the alkene is initiated and proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the pi bond of the nucleophile.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
• Part 2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition).
Oxidation of the Trialkylborane by Hydrogen Peroxide
Step 2
• Part 1: The first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration.
• Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
• Part 3: This is the final part of the oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate.
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well.
Stereochemistry of Hydroboration
The hydroboration reaction is among the few simple addition reactions that proceed cleanly in a syn fashion. As noted above, this is a single-step reaction. Since the bonding of the double bond carbons to boron and hydrogen is concerted, it follows that the geometry of this addition must be syn. Furthermore, rearrangements are unlikely inasmuch as a discrete carbocation intermediate is never formed. These features are illustrated for the hydroboration of α-pinene.
Since the hydroboration procedure is most commonly used to hydrate alkenes in an anti-Markovnikov fashion, we also need to know the stereoselectivity of the second oxidation reaction, which substitutes a hydroxyl group for the boron atom. Independent study has shown this reaction takes place with retention of configuration so the overall addition of water is also syn.
The hydroboration of α-pinene also provides a nice example of steric hindrance control in a chemical reaction. In the less complex alkenes used in earlier examples the plane of the double bond was often a plane of symmetry, and addition reagents could approach with equal ease from either side. In this case, one of the methyl groups bonded to C-6 (colored purple in the equation) covers one face of the double bond, blocking any approach from that side. All reagents that add to this double bond must therefore approach from the side opposite this methyl.
References
1. Vollhardt, Peter, and Neil Shore. Organic Chemistry: Structure and Function. 5th. New York: W.H. Freeman and Company, 2007.
2. Foote, S. Christopher, and William H. Brown. Organic Chemistry. 5th. Belmont, CA: Brooks/Cole Cengage Learning, 2005.
3. Bruice, Paula Yurkanis. Oragnic Chemistry. 5th. CA. Prentice Hall, 2006.
4. Bergbreiter E. David , and David P. Rainville. Stereochemistry of hydroboration-oxidation of terminal alkenes. J. Org. Chem., 1976, 41 (18), pp 3031–3033
5. Ilich, Predrag-Peter; Rickertsen, Lucas S., and Becker Erienne. Polar Addition to C=C Group: Why Is Anti-Markovnikov Hydroboration-Oxidation of Alkenes Not "Anti-"? Journal of Chemical Education., 2006, v83, n11, pg 1681-1685
Problems
What are the products of these following reactions?
1.
2.
3.
Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
4.
5. (E)-3-methyl-2-pentene
If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the alkenes.
Answer
1.
2.
3.
4.
5.
Exercises
Questions
Q8.5.1
Write out the reagents or products (A–D) shown in the following reaction schemes.
S8.5.1 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.08%3A_Hydroboration-Oxidation%3A_A_Stereospecific_Anti-Markovnikov_Hydration.txt |
Objectives
After completing this section, you should be able to
1. describe, and write the detailed mechanism for, the formation of a carbene, such as dichlorocarbene.
2. describe the structure of a carbene in terms of the hybridization of the central carbon atom.
3. write an equation for the formation of a substituted cyclopropane from an alkene and a carbene.
4. identify the reagents, the alkene, or both, needed to prepare a given substituted cyclopropane by addition of a carbene to a double bond.
5. identify the substituted cyclopropane formed from the reaction of a given alkene with the reagents necessary to form a carbene.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• carbene (R2C:)
• carbenoid
• Simmons-Smith reaction
• stereospecific
Study Notes
A carbenoid is best considered to be a reagent which, while not actually a carbene, behaves as if it were an intermediate of this type.
Dichlorocarbenes can also form cyclopropane structures and are created in situ from reagents such as chloroform and KOH.
The detailed mechanism of the formation of dichlorocarbene is given below. Note that the deprotonation of chloroform generates the trichloromethanide anion, which spontaneously expels the chloride anion.
The highly strained nature of cyclopropane compounds makes them very reactive and interesting synthetic targets. Additionally cyclopropanes are present in numerous biological compounds. One common method of cyclopropane synthesis is the reaction of carbenes with the double bond in alkenes or cycloalkenes. Methylene, H2C, is simplest carbene, and in general carbenes have the formula R2C. Other species that will also react with alkenes to form cyclopropanes but do not follow the formula of carbenes are referred to as carbenoids.
Introduction
Carbenes were once only thought of as short lived intermediates. The reactions of this section only deal with these short lived carbenes which are mostly prepared in situ, in conjunction with the main reaction. However, there do exist so called persistent carbenes. These persistent carbenes are stabilized by a variety of methods often including aromatic rings or transition metals. In general a carbene is neutral and has 6 valence electrons, 2 of which are non bonding. These electrons can either occupy the same sp2 hybridized orbital to form a singlet carbene (with paired electrons), or two different sp2 orbitals to from a triplet carbene (with unpaired electrons). The chemistry of triplet and singlet carbenes is quite different but can be oversimplified to the statement: singlet carbenes usually retain stereochemistry while triplet carbenes do not. The carbenes discussed in this section are singlet and thus retain stereochemistry.
The reactivity of a singlet carbene is concerted and similar to that of electrophilic or nucleophilic addition wheras, triplet carbenes react like biradicals, explaining why sterochemistry is not retained. The highly reactive nature of carbenes leads to very fast reactions in which the rate determining step is generally carbene formation.
Preparation of methylene
The preparation of methylene starts with the yellow gas diazomethane, CH2N2. Diazomethane can be exposed to light, heat or copper to facilitate the loss of nitrogen gas and the formation of the simplest carbene methylene. The process is driven by the formation of the nitrogen gas which is a very stable molecule.
Carbene reaction with alkenes
A carbene such as methlyene will react with an alkene which will break the double bond and result with a cyclopropane. The reaction will usually leave stereochemistry of the double bond unchanged. As stated before, carbenes are generally formed along with the main reaction; hence the starting material is diazomethane not methylene.
In the above case cis-2-butene is converted to cis-1,2-dimethylcyclopropane. Likewise, below the trans configuration is maintained. This shows that the reactions are stereospecific, only a single stereoisomer is obtained as the product.
Additional Types of Carbenes and Carbenoids
In addition to the general carbene with formula R2C there exist a number of other compounds that behave in much the same way as carbenes in the synthesis of cyclopropane. Halogenated carbenes are formed from halomethanes. An example is dicholorcarbene, Cl2C. The mechanism for the formation of dichlorocarbene is above in the study notes. These halogenated carbenes will form cyclopropanes in the same manner as methylene but with the interesting presence of two halogen atoms in place of the hydrogen atoms.
Carbenoids are substances that form cyclopropanes like carbenes but are not technically carbenes. One common example is the stereospecific Simmon-Smith reaction which utilizes the carbenoid - ICH2ZnI. The (iodomethyl) zinc iodide is formed in situ via the mixing of Zn-Cu with CH2I2. If this ICH2ZnI is in the presence of an alkene, a CH2 group is transferred to the double bond to create cyclopropane. Since this reacts as a carbene, the same methods can be applied to determine the product.
• en.Wikipedia.org/wiki/Simmons-Smith_reaction
• en.Wikipedia.org/wiki/Carbene
Problems
1. Knowing that cycloalkenes react much the same as regular alkenes what would be the expected structure of the product of cyclohexene and diazomethane facilitated by copper metal?
2. What would be the result of a Simmons-Smith reaction that used trans-2-pentene as a reagent?
3. What starting material could be used to form cis-1,2-diethylcyclopropane?
4. What would the following reaction yield?
5. Draw the product of this reaction. What type of reaction is this?
Answers
1. The product will be a bicyclic ring, Bicyclo[4.1.0]heptane.
2. The stereochemistry will be retained making a cyclopropane with trans methyl and ethyl groups. Trans-1-ethyl-2-methylcyclopropane
3. The cis configuration will be maintained from reagent to product so we would want to start with cis-3-hexene. A Simmons Smith reagent, or methylene could be used as the carbene or carbenoid.
4. The halogenated carbene will react the same as methylene yielding, cis-1,1-dichloro-2,3-dimethylcyclopropane.
5. This is a Simmons-Smith reaction which uses the carbenoid formed by the CH2I2 and Zu-Cu. The reaction results in the same product as if methylene was used and retains stereospecificity. Iodine metal and the Zn-Cu are not part of the product. The product is trans-1,2-ethyl-methylcyclopropane.
Contributors and Attributions
• Paul Tisher
• Lauren Reutenauer (Amherst College)
• Dr. Krista Cunningham | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.09%3A_Diazomethane_Carbenes_and_Cyclopropane_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the epoxidation of an alkene using meta-chloroperoxybenzoic acid.
2. identify the alkene, reagents, or both, that must be used to prepare a given epoxide.
3. write the equation for the hydroxylation of an alkene using osmium tetroxide, and draw the structure of the cyclic intermediate.
4. draw the structure of the diol formed from the reaction of a given alkene with osmium tetroxide.
5. identify the alkene, the reagents, or both, that must be used to prepare a given 1,2-diol.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• diol
• glycol
• hydroxylation
The previous section discussed the reduction of a double bond, so adding hydrogen to the the double bond. This section will discuss oxidation. In organic chemistry, this is a reaction that where the carbon atom loses electron density, which happens when new bond to a more electronegative atom occurs or when a double bond is broken between a carbon and a less electronegative atom. A simplified to say this is in organic chemistry a reduction is more bonds to hydrogen and oxidation is more bonds to oxygen often.
Oxacyclopropane Synthesis by Peroxycarboxylic Acid
One way to oxidized a double bond is to produce an oxacyclopropane ring. Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with peroxycarboxylic acid. Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Mechanism
Peroxycarboxylic acids are generally unstable. An exception is meta-chloroperoxybenzoic acid, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions.
Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example 8.7.1
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one (S) and one (R) stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of (S,S) and (R,R) enantiomers.
Epoxidation exercises
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answers
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 2-methyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane
Anti Dihydroxylation
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Introduction
The reaction with $OsO_4$ is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4 \nonumber$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Mechanism
• Electrophilic attack on the alkene
• Pi bond of the alkene acts as the nucleophile and reacts with osmium (VIII) tetroxide (OsO4)
• 2 electrons from the double bond flows toward the osmium metal
• In the process, 3 electron pairs move simultaneously
• Cyclic ester with Os (VI) is produced
• Reduction
• H2S reduces the cyclic ester
• NaHSO4 with H2O may be used
• Forms the syn-1,2-diol (glycol)
Example: Dihydroxylation of 1-ethyl-1-cycloheptene
Hydroxylation of alkenes
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
References
1. Dehestani, Ahmad et al. (2005). Ligand-assisted reduction of osmium tetroxide with molecular hydrogen via a [3+2] mechanism. Journal of the American Chemical Society, 2005, 127 (10), 3423-3432.
2. Sorrell, Thomas, N. Organic Chemistry. New York: University Science Books, 2006.
3. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Dihydroxylation Exercises
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Answers
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.10%3A_Oxacyclopropane_%28_Epoxide%29_Synthesis%3A__Epoxidation__by_Peroxycarboxylic_Acids.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the epoxidation of an alkene using meta-chloroperoxybenzoic acid.
2. identify the alkene, reagents, or both, that must be used to prepare a given epoxide.
3. write the equation for the hydroxylation of an alkene using osmium tetroxide, and draw the structure of the cyclic intermediate.
4. draw the structure of the diol formed from the reaction of a given alkene with osmium tetroxide.
5. identify the alkene, the reagents, or both, that must be used to prepare a given 1,2-diol.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• diol
• glycol
• hydroxylation
The previous section discussed the reduction of a double bond, so adding hydrogen to the the double bond. This section will discuss oxidation. In organic chemistry, this is a reaction that where the carbon atom loses electron density, which happens when new bond to a more electronegative atom occurs or when a double bond is broken between a carbon and a less electronegative atom. A simplified to say this is in organic chemistry a reduction is more bonds to hydrogen and oxidation is more bonds to oxygen often.
Oxacyclopropane Synthesis by Peroxycarboxylic Acid
One way to oxidized a double bond is to produce an oxacyclopropane ring. Oxacyclopropane rings, also called epoxide rings, are useful reagents that may be opened by further reaction to form anti vicinal diols. One way to synthesize oxacyclopropane rings is through the reaction of an alkene with peroxycarboxylic acid. Oxacyclopropane synthesis by peroxycarboxylic acid requires an alkene and a peroxycarboxylic acid as well as an appropriate solvent. The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the COOH group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the COOH group becomes COH.
Mechanism
Peroxycarboxylic acids are generally unstable. An exception is meta-chloroperoxybenzoic acid, shown in the mechanism above. Often abbreviated MCPBA, it is a stable crystalline solid. Consequently, MCPBA is popular for laboratory use. However, MCPBA can be explosive under some conditions.
Peroxycarboxylic acids are sometimes replaced in industrial applications by monoperphthalic acid, or the monoperoxyphthalate ion bound to magnesium, which gives magnesium monoperoxyphthalate (MMPP). In either case, a nonaqueous solvent such as chloroform, ether, acetone, or dioxane is used. This is because in an aqueous medium with any acid or base catalyst present, the epoxide ring is hydrolyzed to form a vicinal diol, a molecule with two OH groups on neighboring carbons. (For more explanation of how this reaction leads to vicinal diols, see below.) However, in a nonaqueous solvent, the hydrolysis is prevented and the epoxide ring can be isolated as the product. Reaction yields from this reaction are usually about 75%. The reaction rate is affected by the nature of the alkene, with more nucleophilic double bonds resulting in faster reactions.
Example 8.7.1
Since the transfer of oxygen is to the same side of the double bond, the resulting oxacyclopropane ring will have the same stereochemistry as the starting alkene. A good way to think of this is that the alkene is rotated so that some constituents are coming forward and some are behind. Then, the oxygen is inserted on top. (See the product of the above reaction.) One way the epoxide ring can be opened is by an acid catalyzed oxidation-hydrolysis. Oxidation-hydrolysis gives a vicinal diol, a molecule with OH groups on neighboring carbons. For this reaction, the dihydroxylation is anti since, due to steric hindrance, the ring is attacked from the side opposite the existing oxygen atom. Thus, if the starting alkene is trans, the resulting vicinal diol will have one (S) and one (R) stereocenter. But, if the starting alkene is cis, the resulting vicinal diol will have a racemic mixture of (S,S) and (R,R) enantiomers.
Epoxidation exercises
1. Predict the product of the reaction of cis-2-hexene with MCPBA (meta-chloroperoxybenzoic acid)
a) in acetone solvent.
b) in an aqueous medium with acid or base catalyst present.
2. Predict the product of the reaction of trans-2-pentene with magnesium monoperoxyphthalate (MMPP) in a chloroform solvent.
3. Predict the product of the reaction of trans-3-hexene with MCPBA in ether solvent.
4. Predict the reaction of propene with MCPBA.
a) in acetone solvent
b) after aqueous work-up.
5. Predict the reaction of cis-2-butene in chloroform solvent.
Answers
1. a) Cis-2-methyl-3-propyloxacyclopropane
b) Racemic (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
2. Trans-3-ethyl-2-methyloxacyclopropane.
3. Trans-3,4-diethyloxacyclopropane.
4. a) 2-methyl-oxacyclopropane
b) Racemic (2S)-1,2-propandiol and (2R)-1,2-propanediol
5. Cis-2,3-dimethyloxacyclopropane
Anti Dihydroxylation
Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is anti-hydroxylation of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups.
Syn Dihydroxylation
Osmium tetroxide oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons.
Introduction
The reaction with $OsO_4$ is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an anti dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures.
$OsO_4$ is formed slowly when osmium powder reacts with gasoues $O_2$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
$Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4 \nonumber$
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO4 was potassium permanganate, $KMnO_4$. Although syn diols will result from the reaction of KMnO4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation.
Mechanism
• Electrophilic attack on the alkene
• Pi bond of the alkene acts as the nucleophile and reacts with osmium (VIII) tetroxide (OsO4)
• 2 electrons from the double bond flows toward the osmium metal
• In the process, 3 electron pairs move simultaneously
• Cyclic ester with Os (VI) is produced
• Reduction
• H2S reduces the cyclic ester
• NaHSO4 with H2O may be used
• Forms the syn-1,2-diol (glycol)
Example: Dihydroxylation of 1-ethyl-1-cycloheptene
Hydroxylation of alkenes
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
Chemical Highlight
Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate.
References
1. Dehestani, Ahmad et al. (2005). Ligand-assisted reduction of osmium tetroxide with molecular hydrogen via a [3+2] mechanism. Journal of the American Chemical Society, 2005, 127 (10), 3423-3432.
2. Sorrell, Thomas, N. Organic Chemistry. New York: University Science Books, 2006.
3. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
Dihydroxylation Exercises
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
Answers
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.11%3A_Vicinal__SYn_Dihydroxylation_with__Osmium_Tetroxide.txt |
Objective
After completing this section, you should be able to
1. write an equation to describe the cleavage of an alkene by ozone, followed by reduction of the ozonide so formed with either sodium borohydride or zinc and acetic acid.
2. predict the products formed from the ozonolysis-reduction of a given alkene.
3. write an equation to describe the cleavage of an alkene by potassium permanganate.
4. predict the products from the oxidative cleavage of a given alkene by potassium permanganate.
5. use the results of ozonolysis-reduction, or cleavage with permanganate, to deduce the structure of an unknown alkene.
6. identify the reagents that should be used in the oxidative cleavage of an alkene to obtain a given product or products.
7. write the equation for the cleavage of a 1,2-diol by periodic acid, and draw the structure of the probable intermediate.
8. predict the product or products that will be formed from the treatment of a given 1,2-diol with periodic acid.
9. use the results of hydroxylation/1,2-diol cleavage to deduce the structure of an unknown alkene.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• molozonide
• ozonide
• ozonolysis
Study Notes
Ozonolysis, or ozonolysis-reduction, refers to the treatment of an alkene with ozone followed by a suitable reducing agent to break down complex double-bond-containing compounds into smaller, more easily identified products. From the identity of the products formed, it may be possible to deduce the structure of the original double-bond-containing substance. Ozonolysis will feature prominently in many of the road-map problems that you will encounter in this course.
A molozonide is an unstable, cyclic intermediate that is initially formed when an alkene reacts with ozone.
Alkenes can also be cleaved by other oxidizing agents such as potassium permanganate. However, KMnO4 will carry the oxidation further than ozonolysis, so products can be slightly different. Note within the summary of the following reactions that ozonolysis produces aldehydes and ketones, while KMnO4 can oxidize all the way to to carbon dioxide and carboxylic acid.
Diol cleavage is another example of a redox reaction; periodic acid, HIO4, is reduced to iodic acid, HIO3.
Ozonolysis is a method of oxidatively cleaving alkenes or alkynes using ozone (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes. by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis.
Introduction
The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead.
Reaction Mechanism
Step 1:
The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule.
Step 2:
The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. Since some ozonides are explosive, it is immediately reacted with a reductive workup to then convert the ozonide molecule into the desired carbonyl products. A typical reductive workup is zinc metal in acetic acid. A variety of carbonyl products could result depending on the starting alkene. For example, a tetrasubstituted alkene would yield two ketone products, while a trisubstituted alkene would yield one ketone product and one aldehyde product.
While there are other options for oxidative cleavage of the double bond, this is the most commonly used reaction.
References
1. Vollhardt, K., Schore, N. Organic Chemistry: Structure and Function. 5th ed. New York, NY: W. H. Freeman and Company, 2007.
2. Shore, N. Study Guide and Solutions Manual for Organic Chemistry. 5th ed. New York, NY: W.H. Freeman and Company, 2007.
Answer
Exercises
1. Draw the structure of the product or products obtained in each of the following reactions:
2. What important point did you learn from questions 1(a) and 1(b), above?
Answers:
1. Exercises 1(a) and 1(b), above, indicate that it is not possible to distinguish between cis and trans isomers of alkenes using oxidative cleavage. Both isomers give the same product or products.
Contributors and Attributions
• Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
• Lauren Reutenauer (Amherst College)
• Dr. Krista Cunningham | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.12%3A_Oxidative_Cleavage%3A_Ozonolysis.txt |
Protons and other electrophiles are not the only reactive species that initiate addition reactions to carbon-carbon double bonds. Curiously, this first became evident as a result of conflicting reports concerning the regioselectivity of HBr additions. As noted earlier, the acid-induced addition of HBr to 1-butene gave predominantly 2-bromobutane, the Markovnikov Rule product. However, in some early experiments in which peroxide contaminated reactants were used, 1-bromobutane was the chief product. Further study showed that an alternative radical chain-reaction, initiated by peroxides, was responsible for the anti-Markovnikov product. This is shown by the following equations.
The weak O–O bond of a peroxide initiator is broken homolytically by thermal or hight energy. The resulting alkoxy radical then abstracts a hydrogen atom from HBr in a strongly exothermic reaction. Once a bromine atom is formed it adds to the π-bond of the alkene in the first step of a chain reaction. This addition is regioselective, giving the more stable carbon radical as an intermediate. The second step is carbon radical abstraction of another hydrogen from HBr, generating the anti-Markovnikov alkyl bromide and a new bromine atom. Each of the steps in this chain reaction is exothermic, so once started the process continues until radicals are lost to termination events.
This free radical chain addition competes very favorably with the slower ionic addition of HBr described earlier, especially in non-polar solvents. It is important to note, however, that HBr is unique in this respect. The radical addition process is unfavorable for HCl and HI because one of the chain steps becomes endothermic (the second for HCl & the first for HI).
Other radical addition reactions to alkenes have been observed, one example being the peroxide induced addition of carbon tetrachloride shown in the following equation
RCH=CH2 + CCl4 (peroxide initiator) > RCHClCH2CCl3
The best known and most important use of free radical addition to alkenes is probably polymerization. Since the addition of carbon radicals to double bonds is energetically favorable, concentrated solutions of alkenes are prone to radical-initiated polymerization, as illustrated for propene by the following equation. The blue colored R-group represents an initiating radical species or a growing polymer chain; the propene monomers are colored maroon. The addition always occurs so that the more stable radical intermediate is formed.
RCH2(CH3)CH· + CH3CH=CH2 >RCH2(CH3)CH-CH2(CH3)CH· + CH3CH=CH2 >RCH2(CH3)CHCH2(CH3)CH-CH2(CH3)CH· > etc.
Anti-Markovnikov rule describes the regiochemistry where the substituent is bonded to a less substituted carbon, rather than the more substitued carbon. This process is quite unusual, as carboncations which are commonly formed during alkene, or alkyne reactions tend to favor the more substitued carbon. This is because substituted carbocation allow more hyperconjugation and indution to happen, making the carbocation more stable.
Introduction
This process was first explained by Morris Selig Karasch in his paper: 'The Addition of Hydrogen Bromide to Allyl Bromide' in 1933.1 Examples of Anti-Markovnikov includes Hydroboration-Oxidation and Radical Addition of HBr. A free radical is any chemical substance with unpaired electron. The more substituents the carbon is connected to, the more substituted is that carbon. For example: Tertiary carbon (most substituted), Secondary carbon (medium substituted), primary carbon (least substituted)
Anti-Markovnikov Radical Addition of Haloalkane can ONLY happen to HBr and there MUST be presence of Hydrogen Peroxide (H2O2). Hydrogen Peroxide is essential for this process, as it is the chemical which starts off the chain reaction in the initiation step. HI and HCl cannot be used in radical reactions, because in their radical reaction one of the radical reaction steps: Initiation is Endothermic, as recalled from Chem 118A, this means the reaction is unfavorable. To demonstrate the anti-Markovnikov regiochemistry, I will use 2-Methylprop-1-ene as an example below:
Initiation Steps
Hydrogen Peroxide is an unstable molecule, if we heat it, or shine it with sunlight, two free radicals of OH will be formed. These OH radicals will go on and attack HBr, which will take the Hydrogen and create a Bromine radical. Hydrogen radical do not form as they tend to be extremely unstable with only one electron, thus bromine radical which is more stable will be readily formed.
Propagation Steps
The Bromine Radical will go on and attack the LESS SUBSTITUTED carbon of the alkene. This is because after the bromine radical attacked the alkene a carbon radical will be formed. A carbon radical is more stable when it is at a more substituted carbon due to induction and hyperconjugation. Thus, the radical will be formed at the more substituted carbon, while the bromine is bonded to the less substituted carbon. After a carbon radical is formed, it will go on and attack the hydrogen of a HBr, which a bromine radical will be formed again.
Termination Steps
There are also Termination Steps, but we do not concern about the termination steps as they are just the radicals combining to create waste products. For example two bromine radical combined to give bromine. This radical addition of bromine to alkene by radical addition reaction will go on until all the alkene turns into bromoalkane, and this process will take some time to finish.
Problems
Please give the product(s) of the reactions below:
1. CH3-C(CH3)=CH-CH3 + HBr + H2O2 ==> ?
2. CH3-C(CH3)=CH-CH3 + HI + H2O2 ==> ?
3. CH3-C(CH3)=CH-CH3 + HCl + H2O2 ==> ?
4. CH3-CH=CH-CH3 + HBr + H2O2 ==> ?
5. CH3-C(CH3)=CH-CH3 + HBr ==> ?
Answers
1. CH3-CH(CH3)-CHBr-CH3 (Anti-Markovnikov)
2. CH3-C(CH3)I-CH2-CH3 (Markovnikov)
3. CH3-C(CH3)Cl-CH2-CH3 (Markovnikov)
4. CH3-CHBr-CH-CH3 or CH3-CH-CHBr-CH3 (Both molecules are the same)
5. CH3-C(CH3)Br-CH2-CH3 (Markovnikov) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.13%3ARadical_Additions%3A_Anti-Markovnikov_Product_Formation.txt |
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Virtually all of the monomers described above are subject to radical polymerization. Since this can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
By using small amounts of initiators, a wide variety of monomers can be polymerized. One example of this radical polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.14%3A_Dimerization_Oligomerization._and__Polymerization_of_Alkenes.txt |
Polymers are long chain giant organic molecules are assembled from many smaller molecules called monomers. Polymers consist of many repeating monomer units in long chains. A polymer is analogous to a necklace made from many small beads (monomers). Many monomers are alkenes or other molecules with double bonds which react by addition to their unsaturated double bonds.
Introduction
The electrons in the double bond are used to bond two monomer molecules together. This is represented by the red arrows moving from one molecule to the space between two molecules where a new bond is to form. The formation of polyethylene from ethylene (ethene) may be illustrated in the graphic on the left as follows. In the complete polymer, all of the double bonds have been turned into single bonds. No atoms have been lost and you can see that the monomers have just been joined in the process of addition. A simple representation is -[A-A-A-A-A]-. Polyethylene is used in plastic bags, bottles, toys, and electrical insulation.
• LDPE - Low Density Polyethylene: The first commercial polyethylene process used peroxide catalysts at a temperature of 500 C and 1000 atmospheres of pressure. This yields a transparent polymer with highly branched chains which do not pack together well and is low in density. LDPE makes a flexible plastic. Today most LDPE is used for blow-molding of films for packaging and trash bags and flexible snap-on lids. LDPE is recyclable plastic #4.
• HDPE - High Density Polyethylene: An alternate method is to use Ziegler-Natta aluminum titanium catalysts to make HDPE which has very little branching, allows the strands to pack closely, and thus is high density. It is three times stronger than LDPE and more opaque. About 45% of the HDPE is blow molded into milk and disposable consumer bottles. HDPE is also used for crinkly plastic bags to pack groceries at grocery stores. HDPE is recyclable plastic #2.
Other Addition Polymers
• PVC (polyvinyl chloride), which is found in plastic wrap, simulated leather, water pipes, and garden hoses, is formed from vinyl chloride (H2C=CHCl). The reaction is shown in the graphic on the left. Notice how every other carbon must have a chlorine attached.
• Polypropylene: The reaction to make polypropylene (H2C=CHCH3) is illustrated in the middle reaction of the graphic. Notice that the polymer bonds are always through the carbons of the double bond. Carbon #3 already has saturated bonds and cannot participate in any new bonds. A methyl group is on every other carbon.
• Polystyrene: The reaction is the same for polystrene where every other carbon has a benzene ring attached. Polystyrene (PS) is recyclable plastic #6. In the following illustrated example, many styrene monomers are polymerized into a long chain polystyrene molecule. The squiggly lines indicate that the polystyrene molecule extends further at both the left and right ends.
• Blowing fine gas bubbles into liquid polystyrene and letting it solidify produces expanded polystyrene, called Styrofoam by the Dow Chemical Company.
• Polystyrene with DVB: Cross-linking between polymer chains can be introduced into polystyrene by copolymerizing with p-divinylbenzene (DVB). DVB has vinyl groups (-CH=CH2) at each end of its molecule, each of which can be polymerized into a polymer chain like any other vinyl group on a styrene monomer.
Other addition polymers
Table 1: Links to various polymers with Chime molecule - Macrogalleria at U. Southern Mississippi
Monomer Polymer Name Trade Name Uses
F2C=CF2 polytetrafluoroethylene Teflon
Non-stick coating for cooking utensils, chemically-resistant specialty plastic parts, Gore-Tex
H2C=CCl2 polyvinylidene dichloride Saran Clinging food wrap
H2C=CH(CN) polyacrylonitrile Orlon, Acrilan, Creslan Fibers for textiles, carpets, upholstery
H2C=CH(OCOCH3) polyvinyl acetate Elmer's glue - Silly Putty Demo
H2C=CH(OH) polyvinyl alcohol Ghostbusters Demo
H2C=C(CH3)COOCH3 polymethyl methacrylate Plexiglass, Lucite Stiff, clear, plastic sheets, blocks, tubing, and other shapes
Addition polymers from conjugated dienes
Polymers from conjugated dienes usually give elastomer polymers having rubber-like properties.
Table 2. Addition homopolymers from conjugated dienes
Monomer Polymer name Trade name Uses
H2C=CH-C(CH3)=CH2 polyisoprene natural or some synthetic rubber applications similar to natural rubber
H2C=CH-CH=CH2 polybutadiene polybutadiene synthetic rubber select synthetic rubber applications
H2C=CH-CCl=CH2 polychloroprene Neoprene chemically-resistant rubber
All the monomers from which addition polymers are made are alkenes or functionally substituted alkenes. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions. Many of these addition reactions are known to proceed in a stepwise fashion by way of reactive intermediates, and this is the mechanism followed by most polymerizations. A general diagram illustrating this assembly of linear macromolecules, which supports the name chain growth polymers, is presented here. Since a pi-bond in the monomer is converted to a sigma-bond in the polymer, the polymerization reaction is usually exothermic by 8 to 20 kcal/mol. Indeed, cases of explosively uncontrolled polymerizations have been reported.
It is useful to distinguish four polymerization procedures fitting this general description.
• Radical Polymerization The initiator is a radical, and the propagating site of reactivity (*) is a carbon radical.
• Cationic Polymerization The initiator is an acid, and the propagating site of reactivity (*) is a carbocation.
• Anionic Polymerization The initiator is a nucleophile, and the propagating site of reactivity (*) is a carbanion.
• Coordination Catalytic Polymerization The initiator is a transition metal complex, and the propagating site of reactivity (*) is a terminal catalytic complex.
Radical Chain-Growth Polymerization
Virtually all of the monomers described above are subject to radical polymerization. Since this can be initiated by traces of oxygen or other minor impurities, pure samples of these compounds are often "stabilized" by small amounts of radical inhibitors to avoid unwanted reaction. When radical polymerization is desired, it must be started by using a radical initiator, such as a peroxide or certain azo compounds. The formulas of some common initiators, and equations showing the formation of radical species from these initiators are presented below.
By using small amounts of initiators, a wide variety of monomers can be polymerized. One example of this radical polymerization is the conversion of styrene to polystyrene, shown in the following diagram. The first two equations illustrate the initiation process, and the last two equations are examples of chain propagation. Each monomer unit adds to the growing chain in a manner that generates the most stable radical. Since carbon radicals are stabilized by substituents of many kinds, the preference for head-to-tail regioselectivity in most addition polymerizations is understandable. Because radicals are tolerant of many functional groups and solvents (including water), radical polymerizations are widely used in the chemical industry.
In principle, once started a radical polymerization might be expected to continue unchecked, producing a few extremely long chain polymers. In practice, larger numbers of moderately sized chains are formed, indicating that chain-terminating reactions must be taking place. The most common termination processes are Radical Combination and Disproportionation. These reactions are illustrated by the following equations. The growing polymer chains are colored blue and red, and the hydrogen atom transferred in disproportionation is colored green. Note that in both types of termination two reactive radical sites are removed by simultaneous conversion to stable product(s). Since the concentration of radical species in a polymerization reaction is small relative to other reactants (e.g. monomers, solvents and terminated chains), the rate at which these radical-radical termination reactions occurs is very small, and most growing chains achieve moderate length before termination.
The relative importance of these terminations varies with the nature of the monomer undergoing polymerization. For acrylonitrile and styrene combination is the major process. However, methyl methacrylate and vinyl acetate are terminated chiefly by disproportionation.
Another reaction that diverts radical chain-growth polymerizations from producing linear macromolecules is called chain transfer. As the name implies, this reaction moves a carbon radical from one location to another by an intermolecular or intramolecular hydrogen atom transfer (colored green). These possibilities are demonstrated by the following equations
Chain transfer reactions are especially prevalent in the high pressure radical polymerization of ethylene, which is the method used to make LDPE (low density polyethylene). The 1º-radical at the end of a growing chain is converted to a more stable 2º-radical by hydrogen atom transfer. Further polymerization at the new radical site generates a side chain radical, and this may in turn lead to creation of other side chains by chain transfer reactions. As a result, the morphology of LDPE is an amorphous network of highly branched macromolecules.
Chain topology
Polymers may also be classified as straight-chained or branched, leading to forms such as these:
The monomers can be joined end-to-end, and they can also be cross-linked to provide a harder material:
If the cross-links are fairly long and flexible, adjacent chains can move with respect to each other, producing an elastic polymer or
Cationic Chain-Growth Polymerization
Polymerization of isobutylene (2-methylpropene) by traces of strong acids is an example of cationic polymerization. The polyisobutylene product is a soft rubbery solid, Tg = _70º C, which is used for inner tubes. This process is similar to radical polymerization, as demonstrated by the following equations. Chain growth ceases when the terminal carbocation combines with a nucleophile or loses a proton, giving a terminal alkene (as shown here).
Monomers bearing cation stabilizing groups, such as alkyl, phenyl or vinyl can be polymerized by cationic processes. These are normally initiated at low temperature in methylene chloride solution. Strong acids, such as HClO4 , or Lewis acids containing traces of water (as shown above) serve as initiating reagents. At low temperatures, chain transfer reactions are rare in such polymerizations, so the resulting polymers are cleanly linear (unbranched).
Anionic Chain-Growth Polymerization
Treatment of a cold THF solution of styrene with 0.001 equivalents of n-butyllithium causes an immediate polymerization. This is an example of anionic polymerization, the course of which is described by the following equations. Chain growth may be terminated by water or carbon dioxide, and chain transfer seldom occurs. Only monomers having anion stabilizing substituents, such as phenyl, cyano or carbonyl are good substrates for this polymerization technique. Many of the resulting polymers are largely isotactic in configuration, and have high degrees of crystallinity.
Species that have been used to initiate anionic polymerization include alkali metals, alkali amides, alkyl lithiums and various electron sources. A practical application of anionic polymerization occurs in the use of superglue. This material is methyl 2-cyanoacrylate, CH2=C(CN)CO2CH3. When exposed to water, amines or other nucleophiles, a rapid polymerization of this monomer takes place.
Ring opening polymerization
In this kind of polymerization, molecular rings are opened in the formation of a polymer. Here epsilon-caprolactam, a 6-carbon cyclic monomer, undergoes ring opening to form a Nylon 6 homopolymer, which is somewhat similar to but not the same as Nylon 6,6 alternating copolymer.
Addition Copolymerization
Most direct copolymerizations of equimolar mixtures of different monomers give statistical copolymers, or if one monomer is much more reactive a nearly homopolymer of that monomer. The copolymerization of styrene with methyl methacrylate, for example, proceeds differently depending on the mechanism. Radical polymerization gives a statistical copolymer. However, the product of cationic polymerization is largely polystyrene, and anionic polymerization favors formation of poly(methyl methacrylate). In cases where the relative reactivities are different, the copolymer composition can sometimes be controlled by continuous introduction of a biased mixture of monomers into the reaction.
Formation of alternating copolymers is favored when the monomers have different polar substituents (e.g. one electron withdrawing and the other electron donating), and both have similar reactivities toward radicals. For example, styrene and acrylonitrile copolymerize in a largely alternating fashion.
Some Useful Copolymers
Monomer A
Monomer B
Copolymer
Uses
H2C=CHCl
H2C=CCl2
Saran
films & fibers
H2C=CHC6H5
H2C=C-CH=CH2
SBR
styrene butadiene rubber
tires
H2C=CHCN
H2C=C-CH=CH2
Nitrile Rubber
adhesives
hoses
H2C=C(CH3)2
H2C=C-CH=CH2
Butyl Rubber
inner tubes
F2C=CF(CF3)
H2C=CHF
Viton
gaskets
A terpolymer of acrylonitrile, butadiene and styrene, called ABS rubber, is used for high-impact containers, pipes and gaskets.
Block Copolymerization
Several different techniques for preparing block copolymers have been developed, many of which use condensation reactions (next section). At this point, our discussion will be limited to an application of anionic polymerization. In the anionic polymerization of styrene described above, a reactive site remains at the end of the chain until it is quenched. The unquenched polymer has been termed a living polymer, and if additional styrene or a different suitable monomer is added a block polymer will form. This is illustrated for methyl methacrylate in the following diagram.
Contributors
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
Symmetrical monomers such as ethylene and tetrafluoroethylene can join together in only one way. Monosubstituted monomers, on the other hand, may join together in two organized ways, described in the following diagram, or in a third random manner. Most monomers of this kind, including propylene, vinyl chloride, styrene, acrylonitrile and acrylic esters, prefer to join in a head-to-tail fashion, with some randomness occurring from time to time. The reasons for this regioselectivity will be discussed in the synthetic methods section.
If the polymer chain is drawn in a zig-zag fashion, as shown above, each of the substituent groups (Z) will necessarily be located above or below the plane defined by the carbon chain. Consequently we can identify three configurational isomers of such polymers. If all the substituents lie on one side of the chain the configuration is called isotactic. If the substituents alternate from one side to another in a regular manner the configuration is termed syndiotactic. Finally, a random arrangement of substituent groups is referred to as atactic. Examples of these configurations are shown here.
Many common and useful polymers, such as polystyrene, polyacrylonitrile and poly(vinyl chloride) are atactic as normally prepared. Customized catalysts that effect stereoregular polymerization of polypropylene and some other monomers have been developed, and the improved properties associated with the increased crystallinity of these products has made this an important field of investigation. The following values of Tg have been reported.
Polymer
Tg atactic
Tg isotactic
Tg syndiotactic
PP
–20 ºC
0 ºC
–8 ºC
PMMA
100 ºC
130 ºC
120 ºC
The properties of a given polymer will vary considerably with its tacticity. Thus, atactic polypropylene is useless as a solid construction material, and is employed mainly as a component of adhesives or as a soft matrix for composite materials. In contrast, isotactic polypropylene is a high-melting solid (ca. 170 ºC) which can be molded or machined into structural components.
Ziegler-Natta Catalytic Polymerization
An efficient and stereospecific catalytic polymerization procedure was developed by Karl Ziegler (Germany) and Giulio Natta (Italy) in the 1950's. Their findings permitted, for the first time, the synthesis of unbranched, high molecular weight polyethylene (HDPE), laboratory synthesis of natural rubber from isoprene, and configurational control of polymers from terminal alkenes like propene (e.g. pure isotactic and syndiotactic polymers). In the case of ethylene, rapid polymerization occurred at atmospheric pressure and moderate to low temperature, giving a stronger (more crystalline) product (HDPE) than that from radical polymerization (LDPE). For this important discovery these chemists received the 1963 Nobel Prize in chemistry.
Ziegler-Natta catalysts are prepared by reacting certain transition metal halides with organometallic reagents such as alkyl aluminum, lithium and zinc reagents. The catalyst formed by reaction of triethylaluminum with titanium tetrachloride has been widely studied, but other metals (e.g. V & Zr) have also proven effective. The following diagram presents one mechanism for this useful reaction. Others have been suggested, with changes to accommodate the heterogeneity or homogeneity of the catalyst. Polymerization of propylene through action of the titanium catalyst gives an isotactic product; whereas, a vanadium based catalyst gives a syndiotactic product. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.15%3A_Synthesis_of__Polymers.txt |
Ethene
Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking.
There is not any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C15H32 might be:
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. You will remember that during the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator.
Beta-Carotene
The long chain of alternating double bonds (conjugated) is responsible for the orange color of beta-carotene. The conjugated chain in carotenoids means that they absorb in the visible region - green/blue part of the spectrum. So β-carotene appears orange, because the red/yellow colors are reflected back to us.
Vitamin A
Vitamin A has several functions in the body. The most well known is its role in vision - hence carrots "make you able to see in the dark". The retinol is oxidized to its aldehyde, retinal, which complexes with a molecule in the eye called opsin. When a photon of light hits the complex, the retinal changes from the 11-cis form to the all-trans form, initiating a chain of events which results in the transmission of an impulse up the optic nerve. A more detailed explanation is in Photochemical Events.
Other roles of vitamin A are much less well understood. It is known to be involved in the synthesis of certain glycoproteins, and that deficiency leads to abnormal bone development, disorders of the reproductive system, xerophthalmia (a drying condition of the cornea of the eye) and ultimately death.
Vitamin A is required for healthy skin and mucus membranes, and for night vision. Its absence from diet leads to a loss in weight and failure of growth in young animals, to the eye diseases; xerophthalmia, and night blindness, and to a general susceptibility to infections. It is thought to help prevent the development of cancer. Good sources of carotene, such as green vegetables are good potential sources of vitamin A. Vitamin A is also synthetically manufactured by extraction from fish-liver oil and by synthesis from beta-ionone.
12.17: Alkenes in Nature - Insect Pheromones
Pheromones are chemicals capable of acting like hormones outside the body of the secreting individual, to impact the behavior of the receiving individuals. Pheromones of certain pest insect species, such as the Japanese beetle, acrobat ant, and the gypsy moth, can be used to trap the respective insect for monitoring purposes, to control the population by creating confusion, to disrupt mating, and to prevent further egg laying.
Figure 12.17.1: (S)-Ipsdienol is a terpene alcohol and is one of the major aggregation pheromones of the bark beetle. It was first identified from Ips confusus, in which it is believed to be a principle sex attractant
When a female bark beetle produces and secretes a sex pheromone in order to attract a mate, the pheromone can be carried on currents of air to male bark beetles, whose antennae can intercept individual pheromone molecules, completely intact, still in the same form as when they were secreted by the female, no matter how far away she is. The male recognizes the structure of the molecule; it can clearly tell if it has just intercepted the proper pheromone or some subtle imposter that also happens to have the formula C10H16O. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/12%3A_Reactions_to_Alkenes/12.16%3A_Ethene%3A_An__Important_Industrial_Feedstock.txt |
Objectives
After completing this section, you should be able to
1. provide the correct IUPAC name of an alkyne, given its Kekulé, condensed or shorthand structure.
2. provide the correct IUPAC name of a compound containing both double and triple bonds, given its Kekulé, condensed or shorthand structure.
3. draw the structure of a compound containing one or more triple bonds, and possibly one or more double bonds, given its IUPAC name.
4. name and draw the structure of simple alkynyl groups, and where appropriate, use these names as part of the IUPAC system of nomenclature.
Study Notes
Simple alkynes are named by the same rules that are used for alkenes (see Section 7.3), except that the ending is -yne instead of -ene. Alkynes cannot exhibit E,Z (cis‑trans) isomerism; hence, in this sense, their nomenclature is simpler than that of alkenes.
Alkynes are organic molecules made of the functional group carbon-carbon triple bonds and are written in the empirical formula of \(C_nH_{2n-2}\). They are unsaturated hydrocarbons. Like alkenes have the suffix –ene, alkynes use the ending –yne; this suffix is used when there is only one alkyne in the molecule.
Introduction
Here are the molecular formulas and names of the first ten carbon straight chain alkynes.
Name Molecular Formula
Ethyne C2H2
Propyne C3H4
1-Butyne C4H6
1-Pentyne C5H8
1-Hexyne C6H10
1-Heptyne C7H12
1-Octyne C8H14
1-Nonyne C9H16
1-Decyne C10H18
The more commonly used name for ethyne is acetylene, which used industrially.
Naming Alkynes
Like previously mentioned, the IUPAC rules are used for the naming of alkynes.
Rule 1
Find the longest carbon chain that includes both carbons of the triple bond.
Rule 2
Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes.
For example:
Rule 3
After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order.
For example:
If there is an alcohol present in the molecule, number the longest chain starting at the end closest to it, and follow the same rules. However, the suffix would be –ynol, because the alcohol group takes priority over the triple bond.
When there are two triple bonds in the molecule, find the longest carbon chain including both the triple bonds. Number the longest chain starting at the end closest to the triple bond that appears first. The suffix that would be used to name this molecule would be –diyne.
For example:
Rule 4
Substituents containing a triple bond are called alkynyl.
For example:
Here is a table with a few of the alkynyl substituents:
Name Molecule
Ethynyl -C≡CH
2- Propynyl -CH2C≡CH
2-Butynyl -CH3C≡CH2CH3
Rule 5
A molecule that contains both double and triple bonds is called an alkenyne. The chain can be numbered starting with the end closest to the functional group that appears first. For example:
If both functional groups are the exact same distance from the ending of the parent chain, the alkene takes precedence in the numbering.
Exercise \(1\)
1) Name the following compounds:
2) How many isomers are possible for C5H8? Draw them.
3) Draw the following compounds:
a) 4,4-dimethyl-2-pentyne
b) 3-octyne
c) 3-methyl-1-hexyne
d) trans 3-hepten-1-yne
4) Do alkynes show cis-trans isomerism? Explain.
Answer
1)
A – 3,6-diethyl-4-octyne
B – 3-methylbutyne
C – 4-ethyl-2-heptyne
D – cyclodecyne
2) 2 possible isomers
3)
4) No. A triply bonded carbon atom can form only one other bond and has linear electron geometry so there are no "sides". Allkenes have two groups attached to each inyl carbon with a trigonal planar electron geometry that creates the possibility of cis-trans isomerism.
Reference
1. Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function. 5th Edition. New York: W. H. Freeman & Company, 2007.
13-02 Properties and Bonding in the Alkynes
The simplest alkyne—a hydrocarbon with carbon-to-carbon triple bond—has the molecular formula C2H2 and is known by its common name—acetylene. Its structure is H–C≡C–H.
Terminal Alkyne: Internal Alkyne:
3-chloro-1-propyne 4,4-dichloro-2-pentyne
Bonding and Hybridization
Bond Name Location Overlap
Bond 1 s (? bond) bond Formed between 2 sp orbitals of carbon and hydrogen atoms End-on overlap
Bond 2 S (? bond) bond Formed between the 2 sp orbital of 2 unsaturated Carbon atoms. End-on overlap
Bond 3 p-bonds (? bonds) Formed between the 2 p-orbitals among the carbon atoms Side-on overlap
Orbital Name
Location
Orbital 1 sp hybrid orbitals Formed in the linear structure model of carbon atom
Orbital 2 p-orbitals Formed on each carbon
Contributors
The characteristic of the triple bond helps to explain the properties and bonding in the alkynes.
Importance of Triple Bonds
Hybridization due to triple bonds allows the uniqueness of alkyne structure. This triple bond contributes to the nonpolar bonding strength, linear, and the acidity of alkynes. Physical Properties include nonpolar due to slight solubility in polar solvents and insoluble in water. This solubility in water and polar solvents is a characteristic feature to alkenes as well. Alkynes dissolve in organic solvents.
Boiling Points
Compared to alkanes and alkenes, alkynes have a slightly higher boiling point. Ethane has a boiling point of -88.6 ?C, while Ethene is -103.7 ?C and Ethyne has a higher boiling point of -84.0 ?C.
Alkynes are High In Energy
Alkynes are involved in a high release of energy because of repulsion of electrons. The content of energy involved in the alkyne molecule contributes to this high amount of energy. The pi-bonds however, do not encompass a great amount of energy even though the concentration is small within the molecule. The combustion of Ethyne is a major contributor from CO2, water, and the ethyne molecule
??H = -311 kcal/mol
To help understand the relative stabilities of alkyne isomers, heats of hydrogenation must be used. Hydrogenation of the least energy, results in the release of the internal alkyne. With the result of the production of butane, the stability of internal versus terminal alkynes has significant relative stability due to hyperconjugation.
Problems
1. What is the carbon-carbon, carbon-hydrogen bond length for alkyne? Is it shorter or longer than alkane and alkene?
2. Which is the most acidic and most stable, alkane, alkene, or alkyne? And depends on what?
3. How many pi bonds and sigma bonds are involved in the structure of ethyne?
4. Why is the carbon-hydrogen bond so short?
5. What is the alkyne triple bond characterizes by? How is this contribute to the weakness of the pi bonds?
6. How is heat of hydrogenation effects the stability of the alkyne?
Contributors
• Bao Kha Nguyen, Garrett M. Chin | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/13%3A_Alkynes%3A_The_Carbon/13-01_Naming_the__Alkynes.txt |
Objectives
After completing this section, you should be able to
1. write an equation to describe the preparation of an alkyne by the dehydrohalogenation of a vicinal dihalide or vinylic halide.
2. identify the alkyne produced from the dehydrohalogenation of a given vicinal dihalide or vinylic halide.
3. write a reaction sequence to show how the double bond of an alkene can be transformed into a triple bond.
4. identify the vicinal dihalide (or vinylic halide) needed to synthesize a given alkyne by dehydrohalogenation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• vicinal dihalide
• vinylic halide
Alkynes can be a useful functional group to synthesize due to some of their antibacterial, antiparasitic, and antifungal properties. One simple method for alkyne synthesis is by double elimination from a dihaloalkane.
E2 Mechanism
Section 8.2 discussed that alkenes can be formed through an elimination reaction. In particular, the synthesis of alkynes will utilize the E2 elimination reaction. During the mechanism of an E2 reaction, a strong base removes a hydrogen adjacent to a halogen. The electrons from the broken C-H bond move to form the C=C double bond. Doing this causes the halogen to be ejected from the compound. Overall, a hydrogen and a halogen are eliminated from the compound to form an alkene. During this mechanism there is a stereoelectronic requirement that the adjacent hydrogen and the halogen be adjacent to each other.
E2 reaction will be discussed in greater detail in Section 11.10.
Alkyne Formation Through Dihaloalkane Elimination
Alkynes are frequently prepared through a double E2 reaction using 2 halides that are vicinal (meaning on adjacent carbons) or geminal (meaning on the same carbon). Because the E2 reaction takes place twice 2 $\pi$ bonds are formed thus creating an Alkyne. Although hydroxide and alkoxide bases could be used for the strong base required for an E2 reaction, their used opens the possibility of position rearrangement in the alkyne product. Because of this, the stronger base sodium amide in ammonia (NaNH2/NH3) is commonly used.
General Reaction
Vicinal dihalide converted to an alkyne
or
Geminal dihalide converted to an alkyne
Note! If a terminal alkyne is formed during the reaction, 3 equivalents of base are required instead of 2 as discussed below.
Mechanism
The following mechanism represents the reaction between 2,3-Dibromopentane with sodium amide in liquid ammonia to form pent-2-yne. During this mechanism an intermediate alkene is formed. Notice that in the alkene intermediate, the remaining hydrogen and halogen are anti to each other due to the stereoelectronic requirements of the E2 mechanism. The intermediate alkene is converted to an alkyne by a second E2 elimination of a hydrogen and halogen.
Terminal Alkynes
The acidity of terminal alkynes also plays a role in product determination when vicinal (or geminal) dihalides undergo base induced dielimination reactions. The following example illustrates eliminations of this kind starting from 1,2-dibromopentane, prepared from 1-pentene by addition of bromine. The initial elimination presumably forms 1-bromo-1-pentene, since base attack at the more acidic and less hindered 1º-carbon should be favored. The second elimination then produces 1-pentyne. If the very strong base such as sodium amide is used, the terminal alkyne is trapped as its sodium salt, from which it may be released by mild acid treatment. However, if the weaker base KOH with heat is used for the elimination, the terminal alkyne salt is not formed, or is formed reversibly, and the initially generated 1-pentyne rearranges to the more stable 2-pentyne via an allene intermediate. Even though terminal alkynes can be generated using sodium amide as a base, most chemists will prefer to use SN2 nucleophilic substitution instead of elimination when trying to form a terminal alkyne.
Preparation of Alkynes from Alkenes
An simple method for the preparation of alkynes utilizes alkenes as starting material. The process begins with the electrophilic addition of a halogen to the alkene bond to form the dihaloalkane. Then the double E2 elimination process is used to form the 2 $\pi$ bonds of an alkyne.
This first process is gone over in much greater detail in the page on halogenation of an alkene. In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is the reactant needed to produce the alkyne using double elimination, as covered previously on this page.
Exercise $1$
1) Why would we need 3 bases for every terminal dihaloalkane instead of 2 in order to form an alkyne?
2) What are the major products of the following reactions:
a.) 1,2-Dibromopentane with sodium amide in liquid ammonia
b.) 1-Pentene first with Br2 and chloromethane, followed by sodium ethoxide (Na+ -O-CH2CH3)
3) What would be good starting molecules for the synthesis of the following molecules:
4) Use a 6 carbon diene to synthesize a 6 carbon molecule with 2 terminal alkynes.
5) Identify the vinyl halide or halides and the vicinal dihalide or dihalides that could be used in the synthesis of:
a) 2,2,5,5-Tetramethyl-3-hexyne.
b) 4-Methyl-2-hexyne.
Answer
1) Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want.
2)
a) 1-Pentyne
b) 1-Pentyne
3)
4) Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up.
5)
a)
b) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/13%3A_Alkynes%3A_The_Carbon/13-04_Preparation_of__Alkynes_by__Double___Elimination.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the reaction that occurs between a terminal alkyne and a strong base, such as sodamide, NaNH2.
2. rank a given list of compounds, including water, acetylene and ammonia, in order of increasing or decreasing acidity.
3. rank a given list of hydrocarbons, such as acetylene, ethylene and ethane, in order of increasing or decreasing acidity.
4. describe a general method for determining which of two given compounds is the stronger acid.
5. provide an acceptable explanation of why terminal alkynes are more acidic than alkanes or alkenes.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• acetylide anion
• acidity order
Study Notes
An acetylide anion is an anion formed by removing the proton from the end carbon of a terminal alkyne:
An acidity order is a list of compounds arranged in order of increasing or decreasing acidity.
The general ideas discussed in this section should already be familiar to you from your previous exposure to chemistry and from the review in Section 2.8. A slightly different account of why terminal alkynes are stronger acids than are alkenes or alkanes is given below. However, the argument is still based on the differences between sp-, sp2- and sp3-hybrid orbitals.
The carbons of a triple bond are sp-hybridized. An sp‑hybrid orbital has a 50% s character and a 50% p character, whereas an sp2‑hybrid orbital is 33% s and 67% p, and an sp3‑hybrid orbital is 25% s and 75% p. The greater the s character of the orbital, the closer the electrons are to the nucleus. Thus in a C(sp)\$\ce{-}\$H bond, the bonding electrons are closer to the carbon nucleus than they are in a C(sp2)\$\ce{-}\$H bond. In other words, compared to a C(sp2)\$\ce{-}\$H bond (or a C(sp3)\$\ce{-}\$H bond), a C(sp)\$\ce{-}\$H bond is very slightly polar: Cδ\$\ce{-}\$Hδ+. This slight polarity makes it easier for a base to remove a proton from a terminal alkyne than from a less polar or non-polar alkene or alkane.
As you will appreciate, the reaction between sodium amide and a terminal alkyne is an acid-base reaction. The sodium acetylide product is, of course, a salt. Terminal alkynes can also form salts with certain heavy-metal cations, notably silver(I) and copper(I). In the laboratory component of this course, you will use the formation of an insoluble silver acetylide as a method for distinguishing terminal alkynes from alkenes and non-terminal alkynes:
Metal acetylides are explosive when dry. They should be destroyed while still wet by warming with dilute nitric acid:
Acidity of Terminal Alkynes: Formation of Acetylide Anions
Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the terminal proton leads to the formation of an acetylide anion, RC≡C:.
As discussed in Section 2.10, acidity typically increases with the stability of the corresponding conjugate base. The origin of the enhanced acidity of terminal alkynes can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The hybridization of an orbital affects its electronegativity. Within a shell, the s orbitals occupy the region closer to the nucleus than the p orbitals. Therefore, the spherical s orbitals are more electronegative than the lobed p orbitals. The relative electronegativity of hybridized orbitals increases as the percent s character increases and follows the order sp > sp2 > sp3. This trend indicates the sp hybridized orbitals of the acetylide anion are more electronegative and better able to stabilize a negative charge than sp2 or sp3 hybridized orbitals. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The table below shows how orbital hybridization compares with the identity of the atom when predicting relative acidity. Remember that as the pKa of a compound decreases its acidity increases.
Table 9.7.1: Akynes
Compound Conjugate Base Hybridization "s Character" pKa C-H BDE (kJ/mol)
CH3CH3 CH3CH2 sp3 25% 50 410
CH2CH2 CH2CH sp2 33% 44 473
HCCH HCC sp 50% 25 523
Acetylene, with a pKa of 25 is shown to be much more acidic than ethylene (pKa = 44) or ethane (pKa = 50). Consequently, acetylide anions can be readily formed by deprotonation of a terminal alkynes with a sufficiently strong base. The amide anion (NH2-), in the form of sodium amide (NaNH2) is commonly used for the formation of acetylide anions.
Exercise 9.7.1
Given that the pKa of water is 14.00, would you expect hydroxide ion to be capable of removing a proton from each of the substances listed below? Justify your answers, briefly.
1. ethanol (pKa = 16)
2. acetic acid (pKa = 4.72)
3. acetylene (pKa = 25)
Answer
1.
a. No, not very well. The pKa of ethanol is greater than that of water, thus the equilibrium lies to the left rather than to the right.Add texts here. Do not delete this text first.
b. Yes, very well. There is a difference of 11 pKa units between the pKa of water and the pKa of acetic acid. The equilibrium lies well to the right.
c. No, hardly at all. The hydroxide ion is too weak a base to remove a proton from acetylene. The equilibrium lies well to the left.
Objectives
After completing this section, you should be able to
1. write an equation to describe the reaction of an acetylide ion with an alkyl halide.
2. discuss the importance of the reaction between acetylide ions and alkyl halides as a method of extending a carbon chain.
3. identify the alkyne (and hence the acetylide ion) and the alkyl halide needed to synthesize a given alkyne.
4. determine whether or not the reaction of an acetylide ion with a given alkyl halide will result in substitution or elimination, and draw the structure of the product formed in either case.
Key Terms
Make certain that you can define, and use in context, the key term below.
• alkylation
Study Notes
The alkylation of acetylide ions is important in organic synthesis because it is a reaction in which a new carbon-carbon bond is formed; hence, it can be used when an organic chemist is trying to build a complicated molecule from much simpler starting materials.
The alkyl halide used in this reaction must be primary. Thus, if you were asked for a suitable synthesis of 2,2-dimethyl-3-hexyne, you would choose to attack iodoethane with the anion of 3,3- dimethyl-1-butyne
rather than to attack 2-iodo-2-methylpropane with the anion of 1-butyne.
The reasons will be made clear in Chapter 11.
Nucleophilic Substitution Reactions of Acetylides
The presence of lone pair electrons and a negative charge on a carbon, makes acetylide anions are strong bases and strong nucleophiles. Therefore, acetylide anions can attack electrophiles such as alkyl halides to cause a substitution reaction. These substitution reactions will be discussed in detail in Chapter 11.
Mechanism
The C-X bonds in 1o alkyl halides are polarized due to the high electronegativity of the halogen. The electrons of the C-X sigma bond are shifted towards the halogen giving it a partial negative charge. This also causes electrons to be shifted away from the carbon giving it a partial positive and making it electrophilic. During this reaction, the lone pair electrons on the acetylide anion attack the electrophilic carbon in the 1o alkyl halide forming a new C-C bond. The formation of this new bond causes the expulsion of the halogen as what is called a leaving group. Overall, this reaction forms a C-C bond and converts a terminal alkyne into a internal alkyne. Because a new alkyl group is added to the alkyne during this reaction, it is commonly called an alkylation.
This substitution reaction is often coupled with the acetylide formation, discussed in the previous section, and shown as a single reaction.
Example \(1\)
Terminal alkynes can be generated through the reaction of acetylene and a 1o alkyl halide.
Example \(2\)
Because the acetylide anion is a very strong base, this substitution reaction is most efficient with methyl or primary halides. Secondary, tertiary, or even bulky primary halogens will give alkenes by the E2 elimination mechanism discussed in Section 11.10. An example of this effect is seen in the reaction of bromocyclopentane with a propyne anion. The reaction produces the elimination product cyclopentene rather than the substitution product 1-propynylcyclopentane.
Nucleophilic Addition of Acetylides to Carbonyls
Acetylide anions also add to the electrophilic carbon in aldehydes and ketones to form alkoxides, which, upon protonation, give propargyl alcohols. With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers. These types of reaction will be discussed in more detail in Chapter 19.
Exercise \(1\)
1) The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by sodium amide, as shown below.
2) Give the possible reactants which could form the following molecules by an alkylation.
3) Propose a synthetic route to produce 2-pentene from propyne and an alkyl halide.
4) Using acetylene as the starting material, show how you would synthesize the following compounds
a)
b) but-2-yne
c)
d)
5) Show how you would accomplish the following synthetic transformation.
Answer
1) Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010.
2)
3)
4)
a)
b)
c)
d)
5) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/13%3A_Alkynes%3A_The_Carbon/13-05_Preparation_of__Alkynes_from_Alkynyl_Anions.txt |
Objectives
After completing this section, you should be able to
1. write equations for the catalytic hydrogenation of alkynes to alkanes and cis alkenes.
2. identify the reagent and catalyst required to produce a given alkane or cis alkene from a given alkyne.
3. identify the product formed from the reaction of a given alkyne with hydrogen and a specified catalyst.
4. identify the alkyne that must be used to produce a given alkane or cis alkene by catalytic hydrogenation.
5. write the equation for the reduction of an alkyne with an alkali metal and liquid ammonia.
6. predict the structure of the product formed when a given alkyne is reduced with an alkali metal and liquid ammonia.
7. identify the alkyne that must be used to produce a given alkene by reduction with an alkali metal and ammonia.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anion radical
• Lindlar catalyst
Study Notes
The Lindlar catalyst allows a chemist to reduce a triple bond in the presence of a double bond.
Thus
but
Hydrogenation and the Relative Stability of Hydrocarbons
Like alkenes, alkynes readily undergo catalytic hydrogenation partially to cis- or trans- alkenes or fully to alkanes depending on the reaction employed.
The catalytic addition of hydrogen to 2-butyne provides heat of reaction data that reflect the relative thermodynamic stabilities of these hydrocarbons, as shown above. From the heats of hydrogenation, shown in blue in units of kcal/mole, it would appear that alkynes are thermodynamically less stable than alkenes to a greater degree than alkenes are less stable than alkanes. The standard bond energies for carbon-carbon bonds confirm this conclusion. Thus, a double bond is stronger than a single bond, but not twice as strong. The difference ( 63 kcal/mole ) may be regarded as the strength of the π-bond component. Similarly, a triple bond is stronger than a double bond, but not 50% stronger. Here the difference ( 54 kcal/mole ) may be taken as the strength of the second π-bond. The 9 kcal/mole weakening of this second π-bond is reflected in the heat of hydrogenation numbers ( 36.7 - 28.3 = 8.4 ).
Overview of Reduction of Alkynes
Alkynes can undergo reductive hydrogenation reactions similar to alkenes. With the presence of two pi bonds within alkynes, the reduction reactions can be partial to form an alkene or complete to form an alkane. Since partial reduction of an alkyne produces an alkene, the stereochemistry provided by the reaction's mechanism determines whether a cis- or trans- alkene is formed. The three most significant alkyne reduction reactions are summarized below:
Catalytic Hydrogenation of an Alkyne
Much like alkenes, alkynes can be fully hydrogenated into alkanes with the help of a platinum, palladium, or nickel catalyst. Because the reaction is catalyzed on the surface of the metal, it is common for these catalysts to dispersed on carbon (Pd/C) or finely dispersed as nickel (Raney-Ni). The presence of two pi bonds in the alkyne cause two equivalents of H2 to be added during the reaction. During the reaction an alkene intermediate is form but not isolated.
Hydrogenation of an Alkyne to a Cis-Alkene
For catalytic hydrogenation, the Pt, Pd, or Ni catalysts are so effective in promoting addition of hydrogen to both double and triple carbon-carbon bonds that the alkene intermediate formed by hydrogen addition to an alkyne cannot be isolated. A less efficient catalyst, Lindlar's catalyst permits alkynes to be converted to alkenes without further reduction to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene because the hydrogenation reaction is occurring on the surface of the metal. Both hydrogen atoms are added to the same side of the alkyne as shown in the syn-addition mechanism for hydrogenation of alkenes in the previous chapter.
Lindlar's catalyst is prepared by deactivating (or poisoning) a conventional palladium catalyst. Lindlar’s catalyst has three components: palladium-calcium carbonate, lead acetate, and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane.
Hydrogenation of an Alkyne to a Trans-Alkene
The anti-addition of hydrogen to an alkyne pi bond occurs when reacted with sodium or lithium metal dissolved in ammonia. This reaction, also called dissolving metal reduction, involves radicals in its mechanism and produces a trans-alkene as it product.
Mechanism
Sodium metal is a powerful reducing agent due to the presence of a 3s1 electron in its valence. Sodium metal easily gives up this electron to become Na+. The mechanism start with a sodium atom donating an electron to the alkyne creating an intermediate with a negative charge and an unpaired electron called a radical anion. Next the amine solvent protonates the anion to create a vinyl radical. A second sodium atom then donates an electron to pair the radical converting it to a vinyl anion. This vinyl anion intermediate rapid interconverts between cis and trans conformations and determines the stereochemistry of the reaction. The trans-vinyl anion is more stable due to reduced steric crowding and is preferentially formed. Finally, the protonation of the trans-vinyl anion creates the trans-alkene product.
1) Electron Donation
2) Protonation
3) Electron Donation
4) Protonation
Exercise
Using any alkyne how would you prepare the following compounds: pentane, trans-4-methyl-2-pentene, cis-4-methyl-2-pentene.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/13%3A_Alkynes%3A_The_Carbon/13-06_Reduction_of__Alkynes%3A_The__Relative_Reactivity_of__the_Two__Pi__Bonds.txt |
Contributors and Attributions
Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University)
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Jim Clark (Chemguide.co.uk)
Objectives
After completing this section, you should be able to
1. write the equation for the reaction of water with an alkyne in the presence of sulfuric acid and mercury(II) sulfate.
2. describe keto-enol tautomerism.
3. predict the structure of the ketone formed when a given alkyne reacts with sulfuric acid in the presence of mercury(II) sulfate.
4. identify the reagents needed to convert a given alkyne to a given ketone.
5. identify the alkyne needed to prepare a given ketone by hydration of the triple bond.
6. write an equation for the reaction of an alkyne with borane.
7. write the equation for the reaction of a vinylic borane with basic hydrogen peroxide or hot acetic acid.
8. identify the reagents, the alkyne, or both, needed to prepare a given ketone or a given cis alkene through a vinylic borane intermediate.
9. identify the ketone produced when a given alkyne is reacted with borane followed by basic hydrogen peroxide.
10. identify the cis alkene produced when a given alkyne is reacted with borane followed by hot acetic acid.
11. explain why it is necessary to use a bulky, sterically hindered borane when preparing vinylic boranes from terminal alkynes.
12. predict the product formed when the vinylic borane produced from a terminal alkyne is treated with basic hydrogen peroxide.
13. identify the alkyne needed to prepare a given aldehyde by a vinylic borane.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• enol
• keto-enol tautomeric equilibrium
• tautomerism
• tautomers
Study Notes
Rapid interconversion between tautomers is called tautomerism; however, as the two tautomers are in equilibrium, the term tautomeric equilibrium may be used. This section demonstrates the equilibrium between a ketone and an enol; hence, the term keto-enol tautomeric equilibrium is appropriate. The term “enol” indicates the presence of a carbon-carbon double bond and a hydroxyl (i.e., alcohol) group. Later in the course, you will see the importance of keto-enol tautomerism in discussions of the reactions of ketones, carbohydrates and nucleic acids.
It is important to note that tautomerism is not restricted to keto-enol systems. Other examples include imine-enamine tautomerism
and nitroso-oxime tautomerism
However, at the moment you need only concern yourself with keto-enol tautomerism.
Notice how hydroboration complements hydration in the chemistry of both alkenes and alkynes.
Mercury(II)-Catalyzed Hydration of Alkynes
As with alkenes, hydration (addition of water) of alkynes requires a strong acid, usually sulfuric acid, and is facilitated by the mercuric ion (Hg2+). However, the hydration of alkynes gives ketone products while the hydration of alkenes gives alcohol products. Notice that the addition of oxygen in both reactions follows Markovnikov rule.
During the hydration of an alkyne, the initial product is an enol intermediate (a compound having a hydroxyl substituent attached to a double-bond), which immediately rearranges to the more stable ketone through a process called enol-keto tautomerization.
Tautomers are defined as rapidly inter-converted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The keto and enol tautomers are in equilibrium with each other and with few exceptions the keto tautomer is more thermodynamically stable and therefor favored by the equilibrium. This mechanism for tautomerization will be discussed in greater detail in Section 22-1.
General Reaction
For terminal alkynes, the addition of water follows the Markovnikov rule, and the final product is a methyl ketone. For internal alkynes the addition of water is not regioselective. Hydration of symmetrical internal alkynes produces a single ketone product. However, hydration of asymmetrical alkynes, (i.e. if R & R' are not the same ) produces two isomeric ketone products.
Mechanism
The mechanism starts with the eletrophilic addition of the mercuric ion (Hg2+) to the alkyne producing a mercury-containing vinylic carbocation intermediate. Nucleophilic attack of water on the vinylic carbocation forms a C-O bond to produce a protonated enol. Deprotonation of the enol by water then produces a organomurcury enol. The murcury is substituted with H+ to produced a neutral enol and regenerate the Hg2+ catalyst. The enol is converted to the ketone product through keto-enol tautomerization the mechanism of which is provided in Section 22-1.
1) Electrophilic addition of Hg2+
2) Nucleophilic attack by water
3) Deprotonation
4) Substitution
5) Tautomerization
Hydroboration–Oxidation of Alkynes
The hydroboration-oxidation of alkynes is analagous to the reaction with alkenes. However, where alkenes for alcohol products, alkynes for aldehyde or ketone products. In both cases the addition is anti-Markovnikov and an oxygen is placed on the less alkyl substituted carbon. With the hydroboration of an alkyne the presences of a second pi bond allows the initial product to under tautomerization to become the final aldehyde product.
Alkynes have two pi bonds both of which are capable of reacting with borane (BH3). To limit the reactivity to only one alkyne pi bond, a dialkyl borane reagent (R2BH) is used. Replacing two of the hydrogens on the borane with alkyl groups also creates steric hindrance which enhances the anti-Markovnikov regioselective of the reaction. Disiamylborane (Sia2BH) and 9-borabicyclo[3.3.1]nonane (9-BBN) are two common reagents for this hydroboration reaction. The oxidation reagents (a basic hydrogen peroxide solution) are the same for both alkenes and alkynes.
General Reaction
The hydroboration of terminal alkynes produces aldehyde products while internal alkynes produce ketone products. The hydroboration of symmetrical alkynes produces one ketone product and asymmetrical alkynes produce a mixture of product ketones.
Mechanism
The mechanism starts with the electrophilic addition of the B-H bond of the borane. The hydrogen atom and the borane all on the same side of the alkyne creating a syn addition configuration in the alkene product. Also, the addition is anti-Markovnikov regioselective which means the borane adds to the less substituted carbon of the alkyne and the hydrogen atom adds to the more substituted. The oxidative work-up replaces the borane with a hydroxy group (-OH) creating an enol intermediate. The enol immediately tautomerizes to the product aldehyde for terminal alkynes and the product ketone for internal alkynes.
Comparison of Mercury(II)-Catalyzed Hydration and Hydroboration–Oxidation of Alkynes
These two reactions are complementary for the reaction of a terminal alkyne because the produce distinctly different products. The mercury(II) catalyzed hydration of a terminal alkyne produces a methyl ketone, while the hydroboration-oxidation produces an aldehyde.
For internal alkynes, the regioslectivity of these reactions are rendered ineffective. The reactions are redundant in that they both produce the same ketone products.
Exercise \(1\)
1) Draw the structure of the product formed when each of the substances below is treated with H2O/H2SO4 in the presence of HgSO4.
a)
b)
2) Draw the structure of the keto form of the compound shown below. Which form would you expect to be the most stable?
3) What alkyne would you start with to gain the following products?
4) What alkyne would you start with to gain the following product?
5) Draw the product(s) of the following reactions:
Answer
Answers:
1)
a)
b)
2.
The keto form should be the most stable.
3)
4)
5)
For internal alkynes, there is no difference in the reaction products. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/13%3A_Alkynes%3A_The_Carbon/13-07_Electrophilic_Addition_Reactions_of__Alkynes.txt |
Objectives
After completing this section, you should be able to
1. write the equation for the reaction of water with an alkyne in the presence of sulfuric acid and mercury(II) sulfate.
2. describe keto-enol tautomerism.
3. predict the structure of the ketone formed when a given alkyne reacts with sulfuric acid in the presence of mercury(II) sulfate.
4. identify the reagents needed to convert a given alkyne to a given ketone.
5. identify the alkyne needed to prepare a given ketone by hydration of the triple bond.
6. write an equation for the reaction of an alkyne with borane.
7. write the equation for the reaction of a vinylic borane with basic hydrogen peroxide or hot acetic acid.
8. identify the reagents, the alkyne, or both, needed to prepare a given ketone or a given cis alkene through a vinylic borane intermediate.
9. identify the ketone produced when a given alkyne is reacted with borane followed by basic hydrogen peroxide.
10. identify the cis alkene produced when a given alkyne is reacted with borane followed by hot acetic acid.
11. explain why it is necessary to use a bulky, sterically hindered borane when preparing vinylic boranes from terminal alkynes.
12. predict the product formed when the vinylic borane produced from a terminal alkyne is treated with basic hydrogen peroxide.
13. identify the alkyne needed to prepare a given aldehyde by a vinylic borane.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• enol
• keto-enol tautomeric equilibrium
• tautomerism
• tautomers
Study Notes
Rapid interconversion between tautomers is called tautomerism; however, as the two tautomers are in equilibrium, the term tautomeric equilibrium may be used. This section demonstrates the equilibrium between a ketone and an enol; hence, the term keto-enol tautomeric equilibrium is appropriate. The term “enol” indicates the presence of a carbon-carbon double bond and a hydroxyl (i.e., alcohol) group. Later in the course, you will see the importance of keto-enol tautomerism in discussions of the reactions of ketones, carbohydrates and nucleic acids.
It is important to note that tautomerism is not restricted to keto-enol systems. Other examples include imine-enamine tautomerism
and nitroso-oxime tautomerism
However, at the moment you need only concern yourself with keto-enol tautomerism.
Notice how hydroboration complements hydration in the chemistry of both alkenes and alkynes.
Mercury(II)-Catalyzed Hydration of Alkynes
As with alkenes, hydration (addition of water) of alkynes requires a strong acid, usually sulfuric acid, and is facilitated by the mercuric ion (Hg2+). However, the hydration of alkynes gives ketone products while the hydration of alkenes gives alcohol products. Notice that the addition of oxygen in both reactions follows Markovnikov rule.
During the hydration of an alkyne, the initial product is an enol intermediate (a compound having a hydroxyl substituent attached to a double-bond), which immediately rearranges to the more stable ketone through a process called enol-keto tautomerization.
Tautomers are defined as rapidly inter-converted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The keto and enol tautomers are in equilibrium with each other and with few exceptions the keto tautomer is more thermodynamically stable and therefor favored by the equilibrium. This mechanism for tautomerization will be discussed in greater detail in Section 22-1.
General Reaction
For terminal alkynes, the addition of water follows the Markovnikov rule, and the final product is a methyl ketone. For internal alkynes the addition of water is not regioselective. Hydration of symmetrical internal alkynes produces a single ketone product. However, hydration of asymmetrical alkynes, (i.e. if R & R' are not the same ) produces two isomeric ketone products.
Mechanism
The mechanism starts with the eletrophilic addition of the mercuric ion (Hg2+) to the alkyne producing a mercury-containing vinylic carbocation intermediate. Nucleophilic attack of water on the vinylic carbocation forms a C-O bond to produce a protonated enol. Deprotonation of the enol by water then produces a organomurcury enol. The murcury is substituted with H+ to produced a neutral enol and regenerate the Hg2+ catalyst. The enol is converted to the ketone product through keto-enol tautomerization the mechanism of which is provided in Section 22-1.
1) Electrophilic addition of Hg2+
2) Nucleophilic attack by water
3) Deprotonation
4) Substitution
5) Tautomerization
Hydroboration–Oxidation of Alkynes
The hydroboration-oxidation of alkynes is analagous to the reaction with alkenes. However, where alkenes for alcohol products, alkynes for aldehyde or ketone products. In both cases the addition is anti-Markovnikov and an oxygen is placed on the less alkyl substituted carbon. With the hydroboration of an alkyne the presences of a second pi bond allows the initial product to under tautomerization to become the final aldehyde product.
Alkynes have two pi bonds both of which are capable of reacting with borane (BH3). To limit the reactivity to only one alkyne pi bond, a dialkyl borane reagent (R2BH) is used. Replacing two of the hydrogens on the borane with alkyl groups also creates steric hindrance which enhances the anti-Markovnikov regioselective of the reaction. Disiamylborane (Sia2BH) and 9-borabicyclo[3.3.1]nonane (9-BBN) are two common reagents for this hydroboration reaction. The oxidation reagents (a basic hydrogen peroxide solution) are the same for both alkenes and alkynes.
General Reaction
The hydroboration of terminal alkynes produces aldehyde products while internal alkynes produce ketone products. The hydroboration of symmetrical alkynes produces one ketone product and asymmetrical alkynes produce a mixture of product ketones.
Mechanism
The mechanism starts with the electrophilic addition of the B-H bond of the borane. The hydrogen atom and the borane all on the same side of the alkyne creating a syn addition configuration in the alkene product. Also, the addition is anti-Markovnikov regioselective which means the borane adds to the less substituted carbon of the alkyne and the hydrogen atom adds to the more substituted. The oxidative work-up replaces the borane with a hydroxy group (-OH) creating an enol intermediate. The enol immediately tautomerizes to the product aldehyde for terminal alkynes and the product ketone for internal alkynes.
Comparison of Mercury(II)-Catalyzed Hydration and Hydroboration–Oxidation of Alkynes
These two reactions are complementary for the reaction of a terminal alkyne because the produce distinctly different products. The mercury(II) catalyzed hydration of a terminal alkyne produces a methyl ketone, while the hydroboration-oxidation produces an aldehyde.
For internal alkynes, the regioslectivity of these reactions are rendered ineffective. The reactions are redundant in that they both produce the same ketone products.
Exercise \(1\)
1) Draw the structure of the product formed when each of the substances below is treated with H2O/H2SO4 in the presence of HgSO4.
a)
b)
2) Draw the structure of the keto form of the compound shown below. Which form would you expect to be the most stable?
3) What alkyne would you start with to gain the following products?
4) What alkyne would you start with to gain the following product?
5) Draw the product(s) of the following reactions:
Answer
Answers:
1)
a)
b)
2.
The keto form should be the most stable.
3)
4)
5)
For internal alkynes, there is no difference in the reaction products.
13.11: Naturally Occurring and Physiologically Active Alkynes
The important class of lipids called steroids are actually metabolic derivatives of terpenes, but they are customarily treated as a separate group. Steroids may be recognized by their tetracyclic skeleton, consisting of three fused six-membered and one five-membered ring.
Steroids are widely distributed in animals, where they are associated with a number of physiological processes. Examples of some important steroids are shown in the following diagram. Norethindrone is a synthetic steroid, all the other examples occur naturally. A common strategy in pharmaceutical chemistry is to take a natural compound, having certain desired biological properties together with undesired side effects, and to modify its structure to enhance the desired characteristics and diminish the undesired. This is sometimes accomplished by trial and error. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/13%3A_Alkynes%3A_The_Carbon/13-08_Anti-Markovnikov_Additions_to_Triple___Bonds.txt |
Have you ever wondered why color exists? If you know it is because of molecules, do you know what makes the different colors appear? Most of it can be explained by the presence of the π bonds, and how many there are. When photons hit the atoms and excite the electrons, the bonds that the electrons are in affect the frequency of light that is reflected when the electrons go back to th ground state and is perceived by our eyes.
Introduction
π-Bonds are formed from the overlap of two adjacent parallel p orbitals (Fig.1). π bonds are important in the addition reactions. It is vital in synthesising many products like the polymers we use in modern day society. These systems have incredibly unique thermodynamic and photochemical properties. In this section we discover the special properties that three adjacent p orbitals (a π double bond and a p orbital) can have on the reactivity of a carbon center. We call a carbon next to a double bonded carbon an allylic carbon. The electrons that are shared between all the atomic centers are commonly said to be delocalized.
Figure 1 A π Bond is the interaction between the two p orbitals of the bonded carbons.
Allylic Carbocations
Lets take a look at the energy required to homolytically cleave (bond enthalpies) a covalent bond fig.2
Figure 2: shows the decreasing energy required to a cleave a bond as we add methyl groups (for stabilization)
As you can see, the more stable the carbocation, the easier it is to form. The tertiary carbocation takes less energy to cleave than the primary carbocation. This is due to hyperconjugation which stabilizes the free electron by the slight interaction with nearby bond orbitals in the methyl groups.
Now we look at a propene molecule Figure 3.
Figure 3: propene radical formation
Even though the carbocation is a primary one, the energy required to cleave the bond is even lower than that of a seemingly stable tertiary structure. Also, the pKa of propene is about 40, compared to propane, which is about 50. This means the propene is much more willing to form the propenyl anion, about ten orders of magnitude more. Let's take a look at why:
Resonance
One way to explain the stability is in terms of resonance, the allylic effect can be shown in all carbon center forms: Cation, Radical, and Anion in figure 4. The double bond's ability to shift in between different carbons gives it extra resonance forms, which lends extra stability.
Figure 4: Each type of center is stabilized through resonance because double bond
Molecular Orbitals
It can also be clearly shown through orbitals in figure 5.
Figure 5. A conjugated ? electron system
The stability of the carbocation of propene is due to a conjugated π electron system. A "double bond" doesn't really exist. Instead, it is a group of 3 adjacent, overlapping, non-hybridized p orbitals we call a conjugated π electron system. You can clearly see the interactions between all three of the p orbitals from the three carbons resulting in a really stable cation. It all comes down to where the location of the electron-deficient carbon is.
Molecular orbital descriptions can explain allylic stability in yet another way using 2-propenyl. Fig.6
Figure 6: Shows the 3 possible Molecular orbitals of 2-propenyl
If we just take the π molecular orbital and not any of the s, we get three of them. π1 is bonding with no nodes, π2 is nonbonding (In other words, the same energy as a regular p-orbital) with a node, and π3 is antibonding with 2 nodes (none of the orbitals are interacting). The first two electrons will go into the π1 molecular orbital, regardless of whether it is a cation, radical, or anion. If it is a radical or anion, the next electron goes into the π2 molecular orbital. The last anion electron goes into the nonbonding orbital also. So no matter what kind of carbon center exists, no electron will ever go into the antibonding orbital.
The Bonding orbitals are the lowest energy orbitals and are favorable, which is why they are filled first. Even though the nonbonding orbitals can be filled, the overall energy of the system is still lower and more stable due to the filled bonding molecular orbitals.
This figure also shows that π2 is the only molecular orbital where the electrion differs, and it is also where a single node passes through the middle. Because of this, the charges of the molecule are mainly on the two terminal carbons and not the middle carbon.
This molecular orbital description can also illustrate the stability of allylic carbon centers in figure 7.
Fig.7 diagram showing how the electrons fill based on the Aufbau principle.
The π bonding orbital is lower in energy than the nonbonding p orbital. Since every carbon center shown has two electrons in the lower energy, bonding π orbitals, the energy of each system is lowered overall (and thus more stable), regardless of cation, radical, or anion.
Ultraviolet and Visible Spectroscopy
An electronic spectra can indicate the relative amounts of delocalization in a π electron system. The more conjugated π bonds there are, the longer the wavelength is reflected. Ethene has a maximum wavelength of 171 nm compared to 1,4-Pentadiene, which has a maximum wavelength of 178 nm and 1,3-Butadiene which has a maximum wavelength of 217 nm. The higher wavelength of the 1,3-Butadiene compared to only 178 nm for 1,4-Pentadiene shows the unique difference between a conjugated an non conjugated system.
Problems
1. Given 1-bromopropane and water, or 1-bromopropene, which one would react and why?
2. Order these in terms of free radical stability, 1 being most stable: tertiary alkyl, primary alkyl, secondary allylic, tertiary allylic, .CH3, primary allylic, secondary alkyl.
3. Order these molecules in terms of relative wavelength, from longest, to shortest.
4. 1,3-Cyclopentene; 2-Methyl-1,3-Pentadiene; Pyrrole; 2-Methyl-1,4-Pentadiene; trans-1,3,5-Hexatriene.
5. Draw out the resonance structures of the arenium ion during a general electrophilic aromatic substitution of a Benzene ring.
Answers
1. 1-bromopropane would have no reaction because water is a poor nucleophile, and a primary carbocation would form. 1-bromopropene is allylic and so stabilizes the carbon attached to the bromine. The water can then successfully react with it through an SN2 reaction.
2. 1 tertiary allylic, 2 secondary allylic~tertiary alkyl, 3 primary allylic~secondary alkyl, 4 primary alkyl, 5 .CH3
3. Long wavelength Pyrrole>1,3-Cyclopentadiene>trans-1,3,5-Hexatriene>2-Methyl-1,3-Pentadiene>2-Methyl-1,4-Pentadiene short wavelength
• Jeffrey Hu | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.01%3A_Overlap__of_Three__Adjacent___p_Orbitals%3A_Electron_Delocali.txt |
When halogens are in the presence of unsaturated molecules such as alkenes, the expected reaction is addition to the double bond carbons resulting in a vicinal dihalide (halogens on adjacent carbons). However, when the halogen concentration is low enough, alkenes containing allylic hydrogens undergo substitution at the allylic position rather than addition at the double bond. The product is an allylic halide (halogen on carbon next to double bond carbons), which is acquired through a radical chain mechanism.
Why Substitution of Allylic Hydrogens?
As the table below shows, the dissociation energy for the allylic C-H bond is lower than the dissociation energies for the C-H bonds at the vinylic and alkylic positions. This is because the radical formed when the allylic hydrogen is removed is resonance-stabilized. Hence, given that the halogen concentration is low, substitution at the allylic position is favored over competing reactions. However, when the halogen concentration is high, addition at the double bond is favored because a polar reaction outcompetes the radical chain reaction.
Radical Allylic Bromination (Wohl-Ziegler Reaction)
Preparation of Bromine (low concentration)
NBS (N-bromosuccinimide) is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl4), NBS reacts with trace amounts of HBr to produce a low enough concentration of bromine to facilitate the allylic bromination reaction.
Allylic Bromination Mechanism
Step 1: Initiation
Once the pre-initiation step involving NBS produces small quantities of Br2, the bromine molecules are homolytically cleaved by light to produce bromine radicals.
Step 2: Propagation
One bromine radical produced by homolytic cleavage in the initiation step removes an allylic hydrogen of the alkene molecule. A radical intermediate is generated, which is stabilized by resonance. The stability provided by delocalization of the radical in the alkene intermediate is the reason that substitution at the allylic position is favored over competing reactions such as addition at the double bond.
The intermediate radical then reacts with a Br2 molecule to generate the allylic bromide product and regenerate the bromine radical, which continues the radical chain mechanism. If the alkene reactant is asymmetric, two distinct product isomers are formed.
Step 3: Termination
The radical chain mechanism of allylic bromination can be terminated by any of the possible steps shown below.
Radical Allylic Chlorination
Like bromination, chlorination at the allylic position of an alkene is achieved when low concentrations of Cl2 are present. The reaction is run at high temperatures to achieve the desired results.
Industrial Uses
Allylic chlorination has important practical applications in industry. Since chlorine is inexpensive, allylic chlorinations of alkenes have been used in the industrial production of valuable products. For example, 3-chloropropene, which is necessary for the synthesis of products such as epoxy resin, is acquired through radical allylic chlorination (shown below).
Problems (Answers are attached as a file)
1. Cyclooctene undergoes radical allylic bromination. Write out the complete mechanism including reactants, intermediates and products.
2. Predict the two products of the allylic chlorination reaction of 1-heptene.
3. What conditions are required for allylic halogenation to occur? Why does this reaction outcompete other possible reactions such as addition when these conditions are met?
4. Predict the product of the allylic bromination reaction of 2-benzylheptane. (Hint: How are benzylic hydrogens similar to allylic hydrogens?)
5. The reactant 5-isopropyl-1-hexene generates the products 3-bromo-5-isopropyl-1-hexene and 1-bromo-5-isopropyl-2-hexene. What reagents were used in this reaction? | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.02%3A_Radical__Allylic__Halogenation.txt |
Objectives
After completing this section, you should be able to
1. write a reaction sequence to show a convenient method for preparing a given conjugated diene from an alkene, allyl halide, alkyl dihalide or alcohol (diol).
2. identify the reagents needed to prepare a given diene from one of the starting materials listed in Objective 1, above.
3. compare the stabilities of conjugated and non-conjugated dienes, using evidence obtained from hydrogenation experiments.
4. discuss the bonding in a conjugated diene, such as 1,3-butadiene, in terms of the hybridization of the carbon atoms involved.
5. discuss the bonding in 1,3-butadiene in terms of the molecular orbital theory, and draw a molecular orbital for this and similar compounds.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• delocalized electrons
• node
Study Notes
The two most frequent ways to synthesize conjugated dienes are dehydration of alcohols and dehydrohalogenation of organohalides, which were introduced in the preparation of alkenes (Section 8.1). The following scheme illustrates some of the routes to preparing a conjugated diene.
The formation of synthetic polymers from dienes such as 1,3-butadiene and isoprene is discussed in Section 14.6. Synthetic polymers are large molecules made up of smaller repeating units. You are probably somewhat familiar with a number of these polymers; for example, polyethylene, polypropylene, polystyrene and poly(vinyl chloride).
As the hydrogenation of 1,3-butadiene releases less than the predicted amount of energy, the energy content of 1,3-butadiene must be lower than we might have expected. In other words, 1,3-butadiene is more stable than its formula suggests.
Some university-level general chemistry courses do not introduce the subject of molecular orbitals. If you have taken such a course, or forgotten what is meant by the term “molecular orbital,” combine a review of Section 1.11 with your study of this section.
Conjugated vs. Non-conjugated Dienes
Dienes are compounds which contain two double bonds. These dienes can be non-conjugated (the two double bonds are separated by at least one sp3 hybridized atom. Conjugated dienes have the two double bonds separated by a single bond. Conjugated dienes have properties and reactivity which are distinctly different than non-conjugated dienes. Determining if double bonds are conjugated represents an important skill in organic chemistry.
Synthesis of Dienes
Conjugated dienes can be prepared using many of the methods used for the preparation of alkenes previously discussed (Sections 11-7, 11-8, 11-9, and 11-10). A convenient method starts with the free radical halogenation of the allylic carbon of an alkene using NBS. The resulting halide can then be reacted with a strong base to result in E2 elimination and create a diene product. The compound, 1,3-butadiene, is the simplest conjugated diene and has an important use in the synthesis of polymers which will be discussed in Section 14.6. Many simple dienes, such as 1,3-butadiene and isoprene, can be prepared industrially by the double dehydration of alcohols and the double dehydrohalogenation of organohalides.
Stability of Conjugated Dienes
Conjugated dienes are known to be more stable than non-conjugated dienes as shown by their experimentally determined heats of hydrogenation. In Section 7-6, it was shown that as alkenes become more stable, they contain less energy, and therefore release less heat during hydrogenation. Experiments have shown that conjugated dienes have a lower heat of hydrogenation (-226 kJ/mol) than non-conjugated dienes (-251 kJ/mol). Using the differences in heats of hydrogenation the stabilization energy in conjugated dienes can be estimate to be roughly 25 kJ/mol.
Here is an energy diagram comparing different types of bonds with their heats of hydrogenation to show relative stability of each molecule:
The stabilization of dienes by conjugation is less dramatic than the aromatic stabilization of benzene. Nevertheless, similar resonance and molecular orbital descriptions of conjugation may be written.
Each alkene carbon in 1,3 dienes are sp2 hybridized and therefore have one unhybridized p orbital. The four p orbitals in 1,3-butadiene overlap to form a conjugated system which can be represented by the resonance forms shown below. This delocalization of charges stabilizes conjugated diene making them more stable than non-conjugated dienes.
Evidence of conjugation in 1,3 dienes is seen in the single bond being stronger and shorter (147 pm) than an ordinary alkane C-C bond (154 pm). The resonance hybrid of a 1,3-butadiene shows this bond to contain significant double bond character with its bond length being roughly the average of a single and double bond.
The delocalized electron density of a 1,3 diene can be seen when viewing an electrostatic potential maps. In conjugated dienes, it is observed that the pi electron density overlap (shown in red) is closer together and delocalized in conjugated dienes, while in non-conjugated dienes the pi electron density is located completely on the double bonds.
Molecular Orbitals of 1,3 Dienes
According to MO theory discussed in section 1-11, when a double bond is non-conjugated, the two atomic 2pz orbitals combine to form two pi (π) molecular orbitals, one a low-energy π bonding orbital and one a high-energy π-star (π*) anti-bonding molecular orbital. These are sometimes denoted, in MO diagrams like the one below, with the Greek letter psi (Ψ) instead of π. In the bonding Ψ1 orbital, the two (+) lobes of the 2pz orbitals interact constructively with each other, as do the two (-) lobes. Therefore, there is increased electron density between the nuclei in the molecular orbital – this is why it is a bonding orbital. In the higher-energy anti-bonding Ψ2* orbital, the (+) lobes of one 2pz orbital interacts destructively with the (-) lobe of the second 2pz orbital, leading to a node between the two nuclei and overall repulsion. By the aufbau principle, the two electrons from the two atomic orbitals will be paired in the lower-energy Ψ1orbital when the molecule is in the ground state.
With a conjugated diene, such as 1,3-butadiene, the four 2p atomic orbitals combine to form four pi molecular orbitals of increasing energy. Two bonding pi orbitals and two anti-bonding pi* orbitals. The combination of four pi molecular orbitals allow for the formation of a bonding molecular orbital that is lower in energy than those created by an unconjugated alkene. The 4 pi electrons of 1,3-butadiene completely fill the bonding molecular orbitals giving is the additional stability associated with conjugated double bonds. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.05%3A_Two__Neighboring_Double__Bonds%3A__Conjugated_Dienes.txt |
Objectives
After completing this section, you should be able to
1. write an equation for the addition of one or two mole equivalents of a halogen or a hydrogen halide to a nonconjugated diene.
2. write an equation for the addition of one or two mole equivalents of a halogen or a hydrogen halide to a conjugated diene.
3. write the mechanism for the addition of one mole equivalent of hydrogen halide to a conjugated diene, and hence account for the formation of 1,2- and 1,4-addition products.
4. explain the stability of allylic carbocations in terms of resonance.
5. draw the resonance contributors for a given allylic carbocation.
6. predict the products formed from the reaction of a given conjugated diene with one mole equivalent of halogen or hydrogen halide.
7. predict which of the possible 1,2- and 1,4-addition products is likely to predominate when one mole equivalent of a hydrogen halide is reacted with a given conjugated diene.
8. use the concept of carbocation stability to explain the ratio of the products obtained when a given conjugated diene is reacted with one mole equivalent of hydrogen halide.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• 1,2-addition
• 1,4-addition
Study Notes
Notice that the numbers used in the expressions 1,2-addition and 1,4-addition do not refer to the positions of the carbon atoms in the diene molecule. Here, 1,2 indicates two neighbouring carbon atoms, while 1,4 indicates two carbon atoms which are separated in the carbon chain by two additional carbon atoms. Thus in 1,2- and 1,4-additions to 2,4-hexadiene, the additions actually occur at carbons 2 and 3, and 2 and 5, respectively.
The term “monoadduct” should be interpreted as meaning the product or products formed when one mole of reagent adds to one mole of substrate. In the objectives above, this process is referred to as the addition of one mole equivalent (or one mol equiv).
In Section 7.9 we saw that electrophilic addition to a simple alkene would follow Markovnikov’s rule, where the stability of the carbocation intermediate would increase: primary < secondary < tertiary. With conjugated dienes the allylic carbocation intermediately generated has different resonance forms. The following scheme represents the mechanism for the addition of HBr to 1,3-butadiene (at 0°C). Note the resonance contributors for the allylic carbocation intermediate and that the product resulting from the secondary cation is generated in higher yield than from the primary cation as you might expect from our discussions until now. However, in the next section you will see that the resulting product ratio can be drastically affected by a number of reaction conditions, including temperature.
In Section 7.9 we saw that electrophilic addition to a simple alkene would follow Markovnikov’s rule. Markovnikov's rule states that for the electrophilic addition of HX, the carbocation intermediate forms on the double bond carbon with the greatest number of alkyl substitutents. Because the stability of carbocation intermediates increases as it number of alky groups increases (primary < secondary < tertiary), this regioselectivity is provided by preferably forming a more stable carbocation intermediate during the reaction. During the electrophilic addition of HBr to 2-methylpropene, the more stable tertiary carbocation intermediate is preferably formed which yields the Markovnikov addition product 2-bromo-2-methylpropane. Formation of 1-bromo-2-methylpropane does not occur because it would require the formation of a less stable primary carbocation intermediate.
When an electrophilic addition is performed on a non-conjugated diene, the double bonds react in much the same manner as individual alkenes. During the addition of two equivalents of HBr to 1,4-pentadiene, a non-conjugated diene, Markovnikov's rule is still followed producing 2,4-dibromopentane as the product.
Conjugated dienes undergo the usual reactions of alkenes, such as catalytic hydrogenation, radical additions, and electrophilic additions more readily than most alkenes or dienes that have isolated double bonds. During the electrophilic addition of one equivalent of HX to a conjugated diene, the expected Markovnikov addition product is formed during a process called 1,2-addition. In addition, an unexpected product is formed from 1,4-addition, i.e. the halogen bonds at the terminal carbon atoms of a conjugated diene and the remaining double bond shifts to the 2,3-location. Note, the numbers (1 and 4) refer to which of the four carbons making up the conjugated diene the H and Br are bonded to in the products and are not used for the compounds IUPAC nomenclature. Although there are various methods for effecting the relative ratio of 1,2 and 1,4 addition products during the electrophilic addition to a conjugated diene, a mixture of these products is almost always produced. When one equivalent HBr undergoes electrophilic addition to 1,3-butadiene, 1,2 addition provides the expected Markovnikov product 3-bromo-1-butene in a 70% yield. This process is called a 1,2 addition because the hydrogen from HBr (labeled blue) bonds to the first carbon of the diene and the bromine from HBr bonds to the second carbon of the diene. The compound 1-bromo-2-butene is also produced in a 30% yield during the reaction as product of 1,4 addition. Here the hydrogen of HBr bonds to the first carbon of the diene and the Br bonds to the fourth carbon of the diene.
The Mechanism for Electrophilic Addition of HBr to a Conjugated Diene
Regardless if the 1,2 or 1,4 addition product is formed, the first step of the mechanism is the protonation of one of the double bonds. In the same manner as the electrophilic addition of HX to an alkene, the protonation occurs regioselectively to give the more stable carbocation. In the case of a conjugated diene, the more stable cation is not only secondary, but also allylic, and therefore enjoys the stabilization created from the positive charge being distributed over two carbons by resonance. The resonance hybrid of the allylic carbocation intermediate can be depicted by two resonance forms (shown below) both of which have a full positive charge.
Step 2)X−
This allylic carbocation, more properly denoted as the resonance hybrid shown below, has two carbons which have significant positive charge. The halide ion can attack either carbon. Attacking the central carbon, adjacent to the site of protonation, leads to the the 1,2-addition product. Attacking the terminal carbon, distant from the site of protonation, leads to the 1,4 addition product.
Other Electrophilic Additions
Formation of both 1,2- and 1,4-addition products occurs not only with hydrogen halides, but also with other electrophiles such as the halogens (X2). The electrophilic addition of bromine to 1,3-butadiene is an example. As shown below, a roughly 50:50 mixture of 3,4-dibromo-1-butene (the expected 1,2 addition product) and 1,4-dibromo-2-butene (the 1,4 addition product) is obtained. The double bond of the 1,4 addition product is primarily formed as the (E) isomer.
Worked Example $1$
Give the expected products from the following reaction. Show both 1,2 and 1,4 addition products.
Solution
For this example the two double bonds in the diene are not equivalent and must be considered separately. Each double bond of the reactant has the possibility of forming a 1,2 and a 1,4 addition product so the reaction has the possibility of forming four product. Protonate each double bond separately and draw out the resonance forms of the allylic carbocation intermediate created. Then react each resonance form with Br- to create the two possible products. Repeat this process with the second double bond to create the four possible product. Lastly, look for symmetry in the products to see if any are same molecule. Also, consider the stability of each carbocation created during this process. The most stable carbocation will generally product the favored product of the reaction.
Addition to the first double bond creates a resonance form with a tertiary carbocation intermediate. Due to the stability of the tertiary carbocation this product will most likely go one to be the preferred product of this reaction.
Addition to the second double bond creates a symmetrical compound so only one product is formed.
Over the reaction proposed would be expected to form three products.
Exercise $1$
1. Give the 1,2 and the 1,4 products of the addition of one equivalent of HBr to 1,3-hexa-diene.
2. Look at the previous addition reaction of HBr with a diene. Consider the transition states, predict which of them would be the major products and which will be the minor.
3. Write out the products of 1,2 addition and 1,4- addition of HBr to 1,3-cyclohexadiene.
4. What is unusual about the products of 1,2- and 1,4- addition of HX to an unsubstituted cyclic 1,3-diene?
5. Write out the products of 1,2 addition and 1,4- addition of Br2 to 1,3-cyclohexadiene.
Answer
1)
2) The products i-iii all show a secondary cation intermediate which is more stable than primary. Therefore those would be major products and the iv product would be the minor product.
3) The same product will result from 1,2 and 1,4 addition.
4) Addition of the HX to unsubstituted cycloalka-1,3-dienes in either 1,2- or 1,4- manner gives the same product because of symmetry.
5) Both 1,2 and 1,4 products will form.
Objectives
After completing this section, you should be able to
1. explain the difference between thermodynamic and kinetic control of a chemical reaction; for example, the reaction of a conjugated diene with one equivalent of hydrogen halide.
2. draw a reaction energy diagram for a reaction which can result in both a thermodynamically controlled product and a kinetically controlled product.
3. explain how reaction conditions can determine the product ratio in a reaction in which there is competition between thermodynamic and kinetic control.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• kinetic control
• thermodynamic control
Like non-conjugated dienes, conjugated dienes are subject to attack by electrophiles. In fact, conjugated dienes experience relatively greater kinetic reactivity when reacted with electrophiles than non-conjugated dienes do. The reaction mechanism is similar to other electrophilic addition reactions to alkenes (Section 7.9). During the electrophilic addition of HBr to a 1,3-butadiene, However, there are two possible outcomes once the carbocation intermediate is formed. The allyl carbocation is stabilized by resonance structures that vary in the position of the carbocation. This allows the bromide ion to add to either of these carbons resulting in the formation 1,2 and 1,4 addition products. When this reaction is run at a temperature of 0 oC or lower, the 1,2 addition product dominates. However, if the same reaction is run at 40 oC the 1,4 addition product dominates.
The reaction of one equivalent of hydrogen bromide with 1,3-butadiene yields different product ratios under different reaction conditions and is a classic example of the concept of thermodynamic versus kinetic control of a reaction.
At lower temperatures the formation of the 1,2 and 1,4 addition products are irreversible and thus do not reach equilibrium. When a reaction is irreversible the major product is determined by the relative reaction rates and not by thermodynamic stability. Of the two products, the formation of the 1,2 addition product has a higher rate of reaction and forms faster making it the major product. The reaction product which forms with a higher rate of reaction is called the Kinetic Product and when the kinetic product dominates, the reaction is said to be under Kinetic Control.
At higher temperatures the reaction to form both products becomes reversible and a reaction equilibrium is reached. When a reaction is reversible the major product is determined by thermodynamic stability. The 1,4 addition product is more stable and becomes the major product under these reaction conditions. The more stable product is called the Thermodynamic Product and when the thermodynamic product dominates, the reaction is said to be under Thermodynamic Control.
The 1,4 addition product is more stable because it has an internal, disubstituted double bond, and we know that as a general rule that the thermodynamic stability of an alkene increases with increasing substitution. So, when compared to the terminal, monosubstituted alkene of the 1,2 addition product, the 1,4 addition product is expected to be more stable.
A simple definition is that the kinetic product is the product that is formed faster, and the thermodynamic product is the product that is more stable. This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the trans isomer).
An explanation for the temperature influence is shown in the following energy diagram for the addition of HBr to 1,3-butadiene. The initial step in which a proton bonds to carbon #1 is the rate determining step, as indicated by the large activation energy (light gray arrow). The second faster step is the product determining step, and there are two reaction paths (colored blue for 1,2-addition and magenta for 1,4-addition).
At elevated temperatures, the products are more likely to have enough energy to overcome the reverse activation barrier for the second step allowing regeneration of the carbocation intermediate. Under these conditions, this step of the mechanism will be reversible and an equilibrium will be established. Since the system is no longer limited by temperature, the system will minimize its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This places the reaction under thermodynamic control and the most thermodynamically stable molecule, the 1,4 addition product, will be predominantly formed.
If the reaction temperature is kept sufficiently low, the products will not have enough energy to overcome the reverse activation barrier to regenerate the carbocation intermediate making this step of the mechanism effectively irreversible. This reaction is under kinetic control meaning the product which forms faster, the kinetic product, will predominate.
Of the two reaction pathways, the 1,2-addition has a smaller activation energy and would be expected to have a higher reaction rate than the 1,4-addition. The high reaction rate for 1,2-addition can be attributed to the formation of an ion pair during the reaction mechanism. This means that, after a double bond is protonated, the halide counterion remains in close proximity to the carbocation generated. Immediately following dissociation of HX, the chloride ion is going to be much closer to C−2 than it is to C−4, and therefore attack at C−2 is much faster. This ion pair mechanism is a pre-exponential constant effects that is attributed to the proximity and frequency of collision rather than a activation barrier effect.
Experimental Evidence for the Ion-Pair Mechanism
In 1979, Nordlander et al. carried out a similar investigation on the addition of DCl (deuterated hydrochloric acid) to a different substrate, 1,3-pentadiene. This experiment was ingenious, because it was designed to proceed via an almost symmetrical intermediate:
Resonance forms 7a and 7b are both allylic and secondary. There is a very minor difference in their stabilities arising from the different hyperconjugative ability of C-D vs C-H bonds, but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major kinetic pathway, and both 1,2- and 1,4-addition products (8 and 9) would theoretically be formed roughly equally.
Instead, it was found that the 1,2-addition product was favored over the 1,4-addition product. For example, at -78oC in the absence of solvent, there was a roughly (75:25) ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favors 1,2-addition that does not depend on the electrophilicity of the carbon being attacked! The authors attributed this effect to an ion pair mechanism. This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of DCl, the chloride ion is going to be much closer to C-2 than it is to C-4, and therefore attack at C-2 is much faster. In fact, normal electrophilic addition of HX to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms. This is reflected by the greater proportion of syn addition products to such substrates.
Exercise $1$
1) Why is the 1,4-addition product the thermodynamically more stable product?
2) Addition of 1 equivalent of bromine to 2,4-hexadiene at 0 degrees C gives 4,5-dibromo-2-hexene plus an isomer. What is the structure of that isomer?
3) The kinetically controlled product in the below reaction is:
4) For the reaction above, which product is the result of 1,4-addition?
5) What would be the major product of the addition of HBr to 2,3-dimethyl-1,3-cyclohexadiene under thermodynamic conditions?
6) Consider the reaction with 1,3-buta-diene reacting with HCl. Propose a mechanism for the reaction. Also, Predict why the 1,4 adduct is the major product in this reaction compared to the 1,2.
Answer
1) The 1,4- product is more thermodynamically stable because there are two alkyl groups on each side of the double bond and more substituted alkenes are more stable.
2) 2,5-Dibromo-3-hexene
3) 3-Chloro-1-butene
4) 1-Chloro-2-butene
5)
6)
Even though the cation would prefer to be in a secondary position in the transition state, the final product is less stable with a terminal alkene. Therefore the major product will be the 1,4 adduct. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.06%3A_Electrophilic_Attack__on_Conjugated_Dienes%3A_Kinetic__and_Th.txt |
Among the many distinctive features of benzene, its aromaticity is the major contributor to why it is so unreactive. This section will try to clarify the theory of aromaticity and why aromaticity gives unique qualities that make these conjugated alkenes inert to compounds such as Br2 and even hydrochloric acid. It will also go into detail about the unusually large resonance energy due to the six conjugated carbons of benzene.
The delocalization of the p-orbital carbons on the sp2 hybridized carbons is what gives the aromatic qualities of benzene.
This diagram shows one of the molecular orbitals containing two of the delocalized electrons, which may be found anywhere within the two "doughnuts". The other molecular orbitals are almost never drawn.
• Benzene, C6H6, is a planar molecule containing a ring of six carbon atoms, each with a hydrogen atom attached.
The six carbon atoms form a perfectly regular hexagon. All of the carbon-carbon bonds have exactly the same lengths - somewhere between single and double bonds.
• There are delocalized electrons above and below the plane of the ring.
• The presence of the delocalized electrons makes benzene particularly stable.
• Benzene resists addition reactions because those reactions would involve breaking the delocalization and losing that stability.
• Benzene is represented by this symbol, where the circle represents the delocalized electrons, and each corner of the hexagon has a carbon atom with a hydrogen attached.
Basic Structure of Benzene
Because of the aromaticity of benzene, the resulting molecule is planar in shape with each C-C bond being 1.39 Å in length and each bond angle being 120°. You might ask yourselves how it's possible to have all of the bonds to be the same length if the ring is conjugated with both single (1.47 Å) and double (1.34 Å), but it is important to note that there are no distinct single or double bonds within the benzene. Rather, the delocalization of the ring makes each count as one and a half bonds between the carbons which makes sense because experimentally we find that the actual bond length is somewhere in between a single and double bond. Finally, there are a total of six p-orbital electrons that form the stabilizing electron clouds above and below the aromatic ring.
Contributors
Jim Clark (Chemguide.co.uk)
14.08: A Special Transformation of Conjugated Dienes: Diels-Alde
Objectives
After completing this section, you should be able to
1. write an equation to represent a typical Diels-Alder reaction.
2. draw the structure of the product formed when a given conjugated diene reacts with a given dienophile in a Diels-Alder reaction.
3. identify the diene and dienophile that must be used to prepare a given compound by a Diels-Alder reaction.
4. explain the general mechanism of the Diels-Alder reaction, without necessarily being able to describe it in detail.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Diels-Alder cycloaddition
• pericyclic reaction
Study Notes
The Diels-Alder reaction is an example of an organic chemical reaction which does not proceed by either a polar or a free radical pathway, but rather a pericyclic reaction.
Although we do not expect you to be able to provide a detailed account of the mechanism of this reaction, you should learn enough about the Diels-Alder reaction to fulfillthe objectives stated above. You will find it useful to contrast the mechanism of the Diels-Alder reaction with the polar and radical mechanisms studied earlier.
The unique character of conjugated dienes manifests itself dramatically in the Diels-Alder Cycloaddition Reaction. The Diels-Alder reaction is an important and widely used synthetic method for making six-membered rings. In the Diels-Alder reaction, a conjugated diene, simply referred to as the diene, reacts with a double or triple bond co-reactant called the dienophile, because it combines with (has an affinity for) the diene. During the reaction, two pi-bonds are converted to two sigma-bonds. The Diels-Alder cycloaddition is classified as a pericyclic process. Pericyclic reactions involve the redistribution of bonding electrons in a single step mechanism and will be discussed in greater detail in Chapter 30. In particular, the Diels-alder reaction is called a [4+2] process because the diene has four pi-electrons that shift position in the reaction and the dienophile has two.
General Reaction
An example of the Diels-Alder reaction is the cycloaddition of 1,3-butadiene to cyanoethene (acrylonitrile) to form 4-cyanocyclohexene.
Mechanism
All of the electron rearrangements of the Diels-Alder reaction take place once in a single mechanistic step. During this step carbons 1 and 4 of the diene and both alkene carbons of the dienophile, rehybridize from sp2 to sp3 and electrons rearrange to create two new sigma bonds in the cyclic product. Carbons 2 and 3 of the diene remain sp2 hybridized and form a new pi bond in the product.
The mechanism occurs through a cyclic transition state in which there is head-on overlap of two p orbitals on carbons 1 and 4 of the diene with the two p orbitals from the alkene of the dienophile to form two new sigma bonds in the cyclohexene product. The remaining two p orbitals from the diene overlap to form the new pi bond.
Answers
Objectives
After completing this section, you should be able to
1. determine whether or not a given compound would behave as a reactive dienophile in a Diels-Alder reaction.
2. predict the stereochemistry of the product obtained from the reaction of a given diene with a given dienophile.
3. recognize that in order to undergo a Diels-Alder reaction, a diene must be able to assume ans-cis geometry, and determine whether or not a given diene can assume this geometry.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• dienophile
• dimerization
Study Notes
Make sure that you understand that the s-cis and s-trans forms of a diene such as 1,3-butadiene are conformers, not isomers. Note that some textbooks can confuse the issue further by referring to a compound such as (2Z, 4Z)-hexadiene as cis, cis-2,4-hexadiene, and saying that the most stable form of this compound is its s-trans conformer!
In fulfilling Objective 2, above, you must recognize that the Diels-Alder reaction is stereospecific.
Finally, note reaction B in the reading shows 1,3-cyclopentadiene reacting with another molecule of 1,3-cyclopentadiene. When the same compound acts as both diene and dienophile in a Diels-Alder reaction to couple it is a dimerization.
The Dienophile
In general, Diels-Alder reactions proceed fastest with electron-withdrawing groups on the dienophile (diene lover). Ethylene reacts slowly while propenal, ethyl propenoate, and other molecules shown below are highly reactive in a Diels-Alder reaction.
In much the same manner as electron-withdrawing substituents on a benzene ring, these are typically a double or triple bond in conjugation with the double bond in the dienophile. A resonance form can be drawn which places a positive charge in the dienophile double bond. This results in the double bond being less electron rich (greater electron density shown in Red/Orange) than ethylene. Electrostatic potential maps show that the electron-withdrawing groups pull electron density away from the double bond.
Stereochemistry of Diels-Alder (dienophile)
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 multiple new stereocenters are formed, and the reaction is inherently stereospecific. During a Diels-Alder reaction the stereochemistry of the dienophile is retained in the product. A cis dienophile will generate a cyclohexene ring with cis (syn) substitution on the two carbons from the dienophile. Likewise a trans dienophile will generate a cyclohexene ring with trans (anti) substitution on these two carbon.
Example \(1\)
The retention of stereochemistry is due to the planar nature of both reactants and that the forming process is suprafacial (i.e. to or from the same face of each plane). This stereospecificity also confirms the concerted nature of the Diels-Alder mechanism. The drawing below illustrates this fact for the reaction of 1,3-butadiene with (E)-dicyanoethene. The trans relationship of the cyano groups in the dienophile is preserved in the six-membered ring of the adduct.
Another facet of the stereochemical retention of the dienophile is that only the endo product, rather than the alternative exo product, is formed. The words endo and exo are used to indicate relative stereochemistry when referring to bicyclic structures like substituted norbornanes. 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. Diels-Alder reactions with cyclic dienes favor the formation of bicyclic structures in which substituents are in the endo position. Preference of the endo position is also a facet of the suprafacial nature of the Diels-Alder reaction. The orbital overlap required for the reaction is greater when the dienophile lies directly underneath the diene.
Alkynes can also serve as dienophiles in Diels-Alder reactions:
The Diene
In general, Diels-Alder reactions proceed fastest with electron-donating groups on the diene (eg. alkyl groups). The Diels-Alder reaction is a single step process, so the diene component must adopt an s-cis conformation in order for the end carbon atoms (#1 & #4) to bond simultaneously to the dienophile. For many acyclic dienes the s-trans conformer is more stable than the s-cis conformer (due to steric crowding of the end groups), but the two are generally in rapid equilibrium, permitting the use of all but the most hindered dienes as reactants in Diels-Alder reactions.
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:
Stereochemistry of Diels-Alder (diene)
The 1 and 4 Carbons in the diene have the possibility of forming two new stereocenters in the cyclohexene product. Similarly to the effects of dienophile stereochemistry, the positioning of substituents on the 1 and 4 carbons in the diene determine the stereochemistry in the product. The diene substituents can be thought of as being either cis (both facing in or both facing out) or trans and the stereochemistry is retained to form a cis or trans cyclohexene product.
The Essential Characteristics of the Diels-Alder Cycloaddition Reaction:
1. The reaction always creates a new six-membered ring.
2. The diene component must be able to assume an s-cis conformation.
3. Electron withdrawing groups on the dienophile facilitate reaction.
4. Electron donating groups on the diene facilitate reaction.
5. Steric hindrance at the bonding sites may inhibit or prevent reaction.
6. The reaction is stereospecific with respect to substituent configuration in both the dienophile and the diene.
link
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Predicting the Product of a Diels-Alder Reaction
Start by rotating the diene until it is in the s-cis conformation then point it towards the double bond of the dienophile. Remove the double bonds present in the diene and dieneophile. Connect carbons 1 and 4 of the the diene to a carbon in the dienophile double bond using a sigma bond to create a six-membered ring. Create a double bond between carbons 2 and 3 of diene.
Determine if any substituents attached to either the double bond of the dieneophile or carbons 1 and 4 of the diene have a cis/trans conformation. If so make sure the substituents have the same configuration in the cycloalkene product.
Predict the product of the following reaction:
The diene is locked into an s-cis configuration which will promote the reaction. The ring portion of the diene will act as electron donating groups which will also promote the reaction. Because the diene already contained a ring the product will be bicyclic. The dienophile has two nitriles attached to it both of which are electron withdrawing. Since the two nitriles in the dieneophile are cis to each other the the two nitriles will be cis to each other in the product.
Predict the product of the following reaction:
A particularly rapid Diels-Alder reaction takes place between cyclopentadiene and maleic anhydride. Cyclopentadiene is held in the required s-cis configuration so it will make a good diene for a Diels-Alder reaction. 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. Since it is part of a ring, the double bond of maleic anhydride is in a cis configuration so the cyclohexene ring will also have a cis configuration. Lastly, the product will prefer the endo position.
Exercise \(1\)
7) Of the following dienes, which are S-trans and which are s-cis? Of those that are s-trans, are they able to rotate to become s-cis?
8) Predict the product of the following reaction.
Answer
7)
A) s-trans, unable to rotate to become s-cis
B) s-cis
C) s-trans, can rotate to become s-cis.
8)
14.09: Electrocyclic Reactions
An electrocyclic reaction is the concerted cyclization of a conjugated π-electron system by converting one π-bond to a ring forming σ-bond. The reverse reaction may be called electrocyclic ring opening. Two examples are shown on the right. The electrocyclic ring closure is designated by blue arrows, and the ring opening by red arrows. Once again, the number of curved arrows that describe the bond reorganization is half the total number of electrons involved in the process.
In the first case, trans,cis,trans-2,4,6-octatriene undergoes thermal ring closure to cis-5,6-dimethyl-1,3-cyclohexadiene. The sterospecificity of this reaction is demonstrated by closure of the isomeric trans,cis,cis-triene to trans-5,6-dimethyl-1,3-cyclohexadiene, as noted in the second example.
By clicking on this diagram two examples of thermal electrocyclic opening of cyclobutenes to conjugated butadienes will be displayed. This mode of reaction is favored by relief of ring strain, and the reverse ring closure (light blue arrows) is not normally observed. Photochemical ring closure can be effected, but the stereospecificity is opposite to that of thermal ring opening.
14.10: Polymerization of Conjugated Dienes: Rubber
Conjugated dienes (alkenes with two double bonds and a single bond in between) can be polymerized to form important compounds like rubber. This takes place, in different forms, both in nature and in the laboratory. Interactions between double bonds on multiple chains leads to cross-linkage which creates elasticity within the compound.
Polymerization of 1,3-Butadiene
For rubber compounds to be synthesized, 1,3-butadiene must be polymerized. Below is a simple illustration of how this compound is formed into a chain. The 1,4 polymerization is much more useful to polymerization reactions.
Above, the green structures represent the base units of the polymers that are synthesized and the red represents the bonds between these units which form these polymers. Whether the 1,3 product or the 1,4 product is formed depends on whether the reaction is thermally or kinetically controlled.
Synthetic Rubber
The most important synthetic rubber is Neoprene which is produced by the polymerization of 2-chloro-1,3-butadiene.
In this illustration, the dashed lines represent repetition of the same base units, so both the products and reactants are polymers. The reaction proceeds with a mechanism similar to the Friedel-Crafts mechanism. Cross-linkage between the chlorine atom of one chain and the double bond of another contributes to the overall elasticity of neoprene. This cross-linkage occurs as the chains lie next to each other at random angles, and the attractions between double bonds prevent them from sliding back and forth.
Colored molecules
The counjugated double bonds in beta-carotene produce the orange color in carrots. The conjugated double bons in lycopene produce the red color in tomatoes.
ß carotene
lycopene
Problem
Draw out the mechanism for the natural synthesis of rubber from 3-methyl-3-butenyl pyrophosphate and 2-methyl-1,3-butadiene. Show the movement of electrons with arrows. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.07%3ADelocalization_among__More__than__Two__Pi__Bonds%3A__Extended_.txt |
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