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Objectives
After completing this section, you should be able to
1. identify the two possible ways in which a given enolate anion could conceivably react with an electrophile.
2. write an equation to illustrate the haloform reaction.
3. identify the products formed from the reaction of a given methyl ketone with a halogen and excess base.
4. identify the methyl ketone, the reagents, or both, needed to obtain a specified carboxylic acid through a haloform reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• haloform
• haloform reaction
Study Notes
Because the negative charge on an enolate ion is delocalized, there are two reactive sites and therefore two potential products. The α‑substituted product is much more common.
A “haloform” is any compound of the type CHX3, where X = Cl, Br or I. Of these three compounds, chloroform is the most common.
The haloform reaction describedin the reading is usually carried out with iodine. This reaction is called the “iodoform test,” and is one of the reactions carried out in the laboratory as a simple qualitative test for methyl ketones.
How Enolates React
Due to their negative charges, enolates are better and more versatile nucleophiles than enols. The increased reactivity of enolates makes them capable of a wider range of reactions than enols. Also, α-hydrogen containing compounds can be completely converted to an enolate by reaction with a strong base. Whereas enols can only be created in small amounts through manipulating their equilibrium.
Since the negative charge of an enolate anion is delocalized between the α-carbon and an oxygen, electrophiles may bond to either atom. Reactants having two or more reactive sites are called ambident, so this term applies to enolate anions. Either the C of the O reactive site in an enolate may act as a nucleophile depending on the reaction conditions. Reactions with the oxygen would create a new O-E bond and produce an enol derivative. Reactions with the α-carbon creates a new C-E bond and creates an α-substituted carbonyl compound. Although reactions with the nucleophilic oxygen are possible, reactions involving the nucleophilic α-carbon are much more common, partially due to the thermodynamic stability of the C=O bonds in the final products. Also, the enolate counter ion, such as Li+ or Na+, is more tightly associated with the negatively charged enolate oxygen which can then block incoming electrophiles, reducing their chance of reaction at the oxygen.
Stereochemical Implication of Enolate Formation
During enolate formation, an α-hydrogen is removed to form a sp2-hybridized, trigonal planar C=C bond which removes any chiral information from the original α-carbon. Because the enol alkene is planar, the incoming electrophile can attack from either the top or the bottom. If the α-carbon of the starting material has a defined stereochemistry or if a new stereocenter is formed during the reaction, the product will be a racemic mixture of enantiomers.
Base Promoted α-Halogenation
An enolate reacts rapidly with a halogen to produce α-halogenated carbonyl products. This reaction has the tendency to overreact and create polyhalogenated products. If a monohalogenated product is sought, the acid catalyzed halogenation reaction discussed in section 22.3 is preferred. Because complete formation to the enolate is not necessary, weak bases, such as the hydroxide anion, are sufficient to produce this reaction. Once a small amount of enolate is formed, it quickly reacts with the halogen. This removes the enolate and shifts the equilibrium toward forming more enolate by Le Chatelier's principle.
Overreaction During Base Promoted α-Halogenation
The α-hydrogens of halogenated carbonyl products are usually more acidic than the corresponding non-halogenated compounds. The inductive electron withdrawing effect of the electronegative halogen stabilizes the negative charge of the enolate ion. This promotes further enolate formation and also further halogenation of the α-carbon. Monohalogenated carbonyls form an enolate over 100 times faster than their non-halogenated counterparts making multiple halogenations of the α-carbon frequent. This effect is exploited to cause the haloform reaction.
The Haloform Reaction
Overall, the haloform reaction represents a method for the conversion of methyl ketones to carboxylic acids. Due to the increased reactivity of α-halogenated products, methyl ketones typically undergo base promoted halogenation three times to give a trihalo-ketone. A halomethyl ion leaving group is then substituted with a hydroxide ion during nucleophilic acyl substitution. The resulting carboxylate can then be protonated to form a carboxylic acid.
Mechanism
Note! This reaction is considered to be base promoted and not base catalyzed because an entire equivalent of base is required for each α-halogenation. Deprotonation of an α-hydrogen with hydroxide produces the nucleophilic enolate ion which subsequently reacts with the halogen. The increasing acidity of α-halogenated ketone causes this reaction to occur two more times. Once formed, the -CX3 group attached to the carbonyl can act as a leaving group. Nucleophilic acyl substitution with a hydroxide anion causes C-C bond cleavage and eventually produces a haloform (CHCl3, CHBr3, CHI3) and a carboxylate anion. The carboxylate ion is easily protonated with acid to form a carboxylic acid functional group. Often, this reaction is performed using iodine (I2) because the subsequent iodoform (CHI3) side-product is a bright yellow solid which is easily filtered off.
This reaction represents one of the few examples of a carbanion leaving group. The trihalomethyl ion (-:CX3) is particularly stabilized due to the inductive electron-withdrawing effects of the three halogens. The stability of the carbanion can be seen when considering the pKa corresponding conjugate acid. In particular, bromoform (CHBr3) has a pKa of 13.7 which is more than 1020 times more acidic than a typical alkane C-H bond.
1) Enolate formation
2) Nucleophilic attack on the halogen
3) Repeat the halogenation two more times
4) Nucleophilic attack on the electrophilic carbonyl carbon
5) Nucleophilic acyl substitution
6) Deprotonation
7) Protonation of the carboxylate
Biological Haloform Reaction
Interest in the haloform reaction has increased since the discovery that certain plants and marine algae can biosynthize chloroform, bromoform, and other small halocarbons through an analogous process. Previously it was assumed that these toxic compounds were present in the environment as a man-made pollutants.
The likely starting material of the biosynthesis are biogenic methyl ketones, a halogen anion, and oxygen. The enzyme, chloroperoxidase, catalyzes the polyhalogenation of the methyl groups. As in the haloform reaction, the final biosynthesis step involves a nucleophilic acyl substitution with a hydroxide anion to create a haloform and a carboxylate anion.
Exercises
1) Please predict the expected products of the following reactions:
a)
b)
a)
b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.06%3A_Reactivity_of_Enolate_Ions.txt |
Objectives
After completing this section, you should be able to
1. write a general mechanism for the attack of an enolate anion on an alkyl halide.
2. write a reaction sequence to illustrate the preparation of carboxylic acids via the malonic ester synthesis.
3. identify the product formed, and all the intermediates, in a given malonic ester synthesis.
4. identify all of the compounds needed to prepare a given carboxylic acid by a malonic ester synthesis.
5. write a detailed mechanism for each of the steps involved in a malonic ester synthesis.
6. write a reaction sequence to illustrate the preparation of ketones through the acetoacetic ester synthesis.
7. identify the product formed, and all the intermediates, in a given acetoacetic ester synthesis.
8. identify all of the compounds needed to prepare a given ketone by an acetoacetic ester synthesis.
9. write a detailed mechanism for each of the steps involved in an acetoacetic ester synthesis.
10. identify the product or products formed when a given lactone, ester, nitrile or ketone is treated with lithium diisopropylamide followed by an alkyl halide.
11. identify the compounds needed to prepare a given α‑substituted ketone, ester, lactone or nitrile by a method involving the alkylation of an enolate anion.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alkylation
• malonic ester synthesis
Study Notes
The two syntheses discussed in this section provide routes to a wide variety of carboxylic acids and methyl ketones. You may wish to review the factors influencing SN2 reactions (Section 11.3) in conjunction with this section.
You should try to memorize the structures of malonic ester and ethyl acetoacetate. The IUPAC names of these compounds are shown in the table below.
Structure Common name IUPAC name
malonic acid propanedioic acid
malonic ester or
diethyl malonate
diethyl propanedioate
acetoacetic acid 3‑oxobutanoic acid
ethyl acetoacetate or
acetoacetic ester
ethyl 3‑oxobutanoate
Alkylation of Enolates
Enolates can be alkylated in the alpha position through an SN2 reaction with alkyl halides. During this reaction an α-hydrogen is replaced with an alkyl group and a new C-C bond is formed. The limitations of SN2 reactions still apply. This includes preferring a good primary or secondary leaving group, X = chloride, bromide, iodide, tosylate. Tertiary leaving groups cannot be used in this reaction and typically give undesired E2 elimination products. A very strong base, such as LDA, is often used because of its ability to form the enolate completely. Removal of the carbonyl starting material from the reaction mixture makes it unavailable for nucleophilic addition by the enolate. Aldehydes are usually not directly alkylated because their enolates prefer to undergo the carbonyl condensation reactions discussed later in Section 23.1. In addition, the acidic hydrogen on carboxylic acids inhibits the formation of an enolate, and makes their direct alkylation difficult. Esters, including lactones, and symmetrical ketones readily undergo direct alkylation. However, direct alkylations, like all enolate-based reactions, can form a racemic mixture if the alkylated α-carbon produced is chiral.
Mechanism
1) Enolate formation
2) SN2 attack
Examples
When an unsymmetrical ketone with two sets of non-equivalent α-hydrogens is treated with a base, two possible enolates can form. Regioselective enolate formation is possible under the proper conditions. The main determinant is whether the reaction is under kinetic control (rate) or thermodynamic control (equilibrium). Although a predominant product can be produced, a mixture of products is usually formed causing a reduction in product yield.
Thermodynamic Enolates
The thermodynamic enolate is formed when the more substituted α-hydrogen is removed. This leads to the more alkyl substituted, therefore the more stable, enolate to be formed. The presence of additional alkyl groups causes the formation of the thermodynamic enolate to be sterically hindered and kinetically slow, especially when a bulky base like LDA is used. Thermodynamic enolates are favored by conditions which allow for equilibration between the possible enolates. When the ketone starting material is not completely deprotonated, equilibrium between the possible enolates and the α-hydrogens of the ketone can occur. During equilibrium, interconversion between the enolates allows the lower energy of the thermodynamic enolate to dominate. Other conditions can also promote the formation of the thermodynamic enolate, such as higher reaction temperatures, or the use of a smaller less sterically hindered base such as sodium hydride (NaH). Weaker bases, such as sodium ethoxide, do not completely deprotonate the ketone starting material which also allows for enolate equilibrium to occur.
Example of enolate equilibration
Kinetic Enolates
Kinetic enolates are favored under conditions which do not allow for equilibration between the enolates, such as the use of a strong bulky base, like LDA, in a molar equivalent to the ketone starting material. Kinetic enolates are formed when the less substituted α-hydrogen is deprotonated. Being less sterically hindered allows this α-hydrogen to be deprotonated faster even though it forms a less thermodynamically stable enolate. Using a molar equivalent of LDA completely converts the ketone starting material to an enolate, removing it from the reaction mixture and preventing equilibration between the possible enolates. Low reaction temperatures (-78 oC) prevent enolate equilibration and promote the formation of the kinetic enolate.
When and enolate of an asymmetric ketone is stabilized through additional resonance forms there is no competition between possible enolates despite kinetic or thermodynamics conditions. The resonance stabilized enolate will be preferentially alkylated to the point that formation of the alkylated products of other possible enolates will be minimal.
Malonic Ester Synthesis
The malonic ester synthesis is a series of reactions which converts an alkyl halide to a carboxylic acid with two additional carbons. One important use of this synthesis pathway is that it allows for the creation of α-alkylated carboxylic acids which cannot be created by direct alkylation.
The starting material of this reaction is a malonic ester: a diester derivative of malonic acid. Diethyl propanedioate, also known as diethyl malonate, is the malonic ester most commonly used in pathway. Since it is a 1,3-dicarbonyl compound, diethyl malonate has relatively acidic α-hydrogens (pKa = 12.6) and can be transformed to its enolate using sodium ethoxide as a base. Other alkoxide bases are not typically used given the possibility of a transesterification reaction.
Predicting the Product of a Malonic Ester Synthesis
The product of a Malonic Ester Synthesis can be created by simply replacing the halogen on the alkyl halide with a -CH2CO2H group.
Malonic ester synthesis takes place in four steps:
1) Enolate Formation
Reacting diethyl malonate with sodium ethoxide (NaOEt) forms a resonance-stabilized enolate.
2) Alkylation
The enolate is alkylated via an SN2 reaction to form an monoalkylmalonic ester.
3) Ester hydrolysis and protonation
After alkylation, the diester undergoes hydrolysis with sodium hydroxide to form a dicarboxylate. Subsequent protonation with acid forms a monoalkyl malonic acid.
4) Decarboxylation & Tautomerization
Monoalkyl malonic acids decarboxylate when heated, forming an α-alkyl carboxylic acid and carbon dioxide (CO2). Decarboxylation can only occur in compounds with a second carbonyl group two atoms away from carboxylic acid such as in malonic acids and β-keto acids. The mechanism occurs via a concerted mechanism involving a proton transfer between the carboxyl acid hydrogen and the nearby carbonyl group to form the enol of a carboxylic acid and CO2. The enol undergoes tautomerization to form the carboxylic acid.
Dialkylation
The presence of two α-hydrogens in malonic esters allows for a second alkylation to be performed prior to decarboxylation. This leads to dialkylated carboxylic acids. Due to the lack of stereochemical control inherent in enolate based reactions, if the two added alkyl groups are different, a racemic mixture of products will result.
Examples
In a variation of the dialkylation reaction - if one molar equivalent of malonic ester is reacted with one molar equivalent of a dihaloalkane and two molar equivalents of sodium ethoxide, a cyclization reaction occurs. By changing the dihaloalkane, three, four, five, and six-membered rings can be created.
The Acetoacetic Ester Synthesis
The acetoacetic ester synthesis is a series of reactions which converts alkyl halides into a methyl ketone with three additional carbons. This reaction creates an α-substituted methyl ketone without side-products. The starting reagent for this pathway is ethyl 3-oxobutanoate, also called ethyl acetoacetate, or acetoacetic ester. Like other 1,3-dicarbonyl compounds, ethyl acetoacetate is more acidic than ordinary esters being almost completely converted to an enolate by sodium ethoxide.
Predicting the Product of an Acetoacetic Ester Synthesis
The product of a acetoacetic ester synthesis can be created by replacing halogen on the alkyl halide with a -CH2COCH3 group.
Reaction Steps
1) Formation of the enolate
As previously described, the α-hydrogens of acetoacetic ester are rather acidic (pKa = 10.7) allowing the enolate to be easily formed when sodium ethoxide is used as a base.
2) Alkylation via an SN2 Reaction
Subsequent reaction with an alkyl halide produces a monoalkylacetoacetic ester.
3) Ester hydrolysis and decarboxylation
Hydrolysis with NaOH followed by protonation produces an alkylated beta-ketoacid. β-ketoacids are easily decoboxylated to form an α-alkyl substituted methyl ketone and carbon dioxide (CO2) using a similar mechanism as the malonic ester synthesis.
Examples
Much like the malonic ester synthesis, a second alkyl group can added before the decarboxylation step.
The reaction steps of the acetoacetic ester synthesis can also be applied to other β-keto esters with acidic α-hydrogens. Because the α-hydrogens between the two carbonyls are the most acidic, they are preferentially deprotonated allowing for a single enolate to be formed. Even cyclic beta-keto esters can be alkylated and subsequently decarboxylated to give an α-alkylated cyclic ketone.
Direct Alkylation of Nitriles
The presence of acidic α-hydrogens in nitriles gives them the ability to form an enolate equivalent which can be also be directly alkylated.
Planning a Synthesis Using Enolate Alkylations
When planning a synthesis that could involve enolates, the key is to recognize the functionality which can form an enolate. During retrosynthetic analysis a C-C bond is broken between the α-carbon and the β-carbon away from this functionality. It is also important to be able to identify specific groups of atoms which indicate if a malonic ester or an acetoacetic ester synthesis can be used. Having multiple C-C bonds which can be broken allows for multiple synthetic pathways.
After retrosynthetically breaking the C-C bond, the fragment with the functionality will gain a hydrogen and the other fragment will gain a halogen. Sometimes the fragment with the functionality will become diethyl malonate or acetoacetic ester.
Worked out example:
Plan a synthesis of the following molecule using an alkylation of an enolate. Consider multiple pathways and explain which is preferable.
The target molecule does not contain the appropriate fragments to utilize either the malonic ester or acetoacetic acid synthesis so direct alkylation of a ketone will likely be used. When analyzing this molecule, there are three α-β C-C bonds which could be cleaved to create a possible starting material. When looking at the possible starting materials, A and C are asymmetrical ketones and therefore can create multiple products during alkylation. B is a symmetrical ketone and should be the most likely to create the target molecule in high yield.
Possible Synthesis
Exercises
Questions
1) Propose a synthesis for each of the following molecules from this malonic ester.
(a)
(b)
(c)
2) Why can't you prepare tri substituted acetic acids from a malonic ester?
3) Propose a synthesis for the following molecule via a malonic ester.
4) How might you prepare the following compounds from an alkylation reaction?
(a)
(b)
(c)
(d)
(e)
(f)
Solutions
1
(a) 1) Malonic Ester, NaOEt, 2) 4-Methylbenzyl Bromide, 3) Base, 4) Acid, Heat
(b) 1) Malonic Ester, NaOEt, 2) 3-bromohexane, 3) Base, 4) Acid, Eat
(c) 1) Malonic Ester, NaOEt, 2) 1-Bromo-2,3,3-trimethylbutane, 3) Base, 4) Acid, Heat
2
Malonic esters only contain two acid protons.
3
4
(a)
(b)
(c)
(d)
(e)
(f) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.07%3A_Alkylation_of_Enolate_Ions.txt |
Concepts & Vocabulary
• Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
• Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
• The nucleophile in these reactions are new and called enols and enolates.
• In this chapter, the focus is on α substitutions reactions with aldehydes and ketones.
22.1 Keto-Enol Tautomerization
• Greek letters are used to denote the carbon atoms near carbonyls.
• The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
• Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
• Aldehyde hydrogens not given Greek leters.
• α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
• Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
• The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
• Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
• The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
• The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
• The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
• Aromaticity can also stabilize the enol tautomer over the keto tautomer.
• Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
• Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
22.2 Reactivity of Enols: The Mechanism of Alpha-Substitution Reactions
• The oxygen of the enol donates electron density to the double bond making it more electron rich and thus more reactive than a typical alkene.
• The mechanism starts with an acid-catalyzed tautomerization to form an enol.
• The double bond is then able to act as a nucleophile and attack an electrophile.
• The final product is an alpha-substituted carbonyl after the deprotonation of the carbonyl to also regenerate the acid-catalyst.
• The enol formed has planar geometry, which means the electrophile can attach on the top or bottom of the alpha-carbon.
• A racemic mixture can result if a sterocenter is created at the site of substitution.
22.3 Alpha Halogenation of Aldehydes and Ketones
• Aldehydes and ketons can undergo a substitution of an alpha hydrogen to a halogen.
• An acid-catalyzed tautomerization starts the mechanism followed by the enol attacking molecular halogens.
• The nucleophile in this reaction is the enol and the electrophile is the halogen.
• Mechanistic studies showed that the reaction was a first-order in the ketone.
• The halogen is part of a fast step after the rate-determing step.
• The formation of an enol intermediate was provided using a reaction called deuterium exchange.
• Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with the isotopic form of water, D2O.
• The mechanism for deuterium exchange is virtually the same as that of keto-enol tautomerism under acidic conditions, the difference being a deuterium is placed in the α-position.
• Both alpha halogenation and deuterium exhange reactions were found to have a common intermediate involved in the rate determining step of their mechanism, an enol.
22.4 Alpha Bromination of Carboxylic Acids
• The α-bromination of some carbonyl compounds, such as aldehydes and ketones, can be accomplished with Br2 under acidic conditions, but the reaction will generally not occur with more stable carboxylic acid derivatives.
• Carboxylic acids do not enolize to a sufficient extent since the carboxylic acid proton is preferably removed before an α-hydrogen.
• Carboxylic acids, can be brominated in the α position with a mixture of Br2 and phosphorus tribromide (PBr3) in what is called the Hell-Volhard-Zelinskii reaction.
• α-Bromo carboxylic acids are extremely useful synthetic intermediates because the halogen is highly reactive towards SN2 reaction.
• Reaction of a α-bromo carboxylic acids with an excess of ammonia provides α-amination, which is a possible route to amino acids.
22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation
• α hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the π orbitals of the adjacent carbonyl.
• The enolate has two resonance structures to contribute to the resonance hybrid.
• While α-hydrogens are weakly acidic, typical strong bases such as hydroxide or alkoxide are only capable of forming the enolate ion in very low concentrations.
• To achieve complete deprotonation of aldehyde or ketone reactants to their enolate conjugate bases, a very strong base such as LDA (lithium diisopropylamide) must be used.
• Hydrogen atoms with two or more adjacent carbonyl groups are more acidic than typical α hydrogens, such as β-diketones, β-keto-esters, and β-diesters, which create enolates that are stabilized through additional resonance forms which share the negative charge with multiple carbonyl carbons.
• The acidity of these compounds is increased to the point where typical strong bases such as hydroxide and alkoxide can be used to form the enolate.
22.6 Reactivity of Enolate Ions
• Enolates are better nucleophlies than enols.
• α-hydrogen containing compounds can be completely converted to an enolate by reaction with a strong base whereas enols can only be created in small amounts through manipulating their equilibrium.
• Either the C of the O reactive site in an enolate may act as a nucleophile depending on the reaction conditions, but reactions involving the nucleophilic α-carbon are more common, partially due to the thermodynamic stability of the C=O bonds in the final products.
• An enolate reacts rapidly with a halogen to produce α-halogenated carbonyl products.
• The α-hydrogens of halogenated carbonyl products are usually more acidic than the corresponding non-halogenated compounds, which promotes polyhalogenated products.
• The Haloform reaction represents a method for the conversion of methyl ketones to carboxylic acids.
• This reaction is considered a base promoted and not base catalyzed because an entire equivalent of base is required for each α-halogenation.
• Deprotonation of an α-hydrogen with hydroxide produces the nucleophilic enolate ion which subsequently reacts with the halogen.
• The increasing acidity of α-halogenated ketone causes this reaction to occur two more times.
• Once formed, the CX3 group attached to the carbonyl can act as a leaving group, eventually produces a haloform (CHCl3, CHBr3, CHI3) and a carboxylate anion.
22.7 Alkylation of Enolate Ions
• Enolates can be alkylated in the alpha position through an SN2 reaction with alkyl halides.
• An α hydrogen is replaced with an alkyl group and a new C-C bond is formed.
• Very strong bases like LDA are often used to fully deprotonate the carbonyl and completely form the enolate.
• Direct alkylations, like all enolate-based reactions, will form a racemic mixture if the alkylated alpha carbon is chiral.
• When an unsymmetrical ketone with two sets of nonequivalent alpha hydrogens is treated with a base, two possible enolates can form.
• The main determinant for which enolate is formed is whether the reaction is under kinetic control (rate) or thermodynamic (equilibrium) control.
• The thermodynamic enolate is formed when the more substituted alpha hydrogen is removed, yielding the more alkyl substituted, therefore the most stable, enolate.
• Formation of the thermodynamic enolate is sterically hindered and is kinetically slow, especially with a bulky base like LDA.
• Kinetic enolates are formed when the less substituted alpha hydrogen is deprotonated, being less sterically hindered allows this alpha hydrogen to be deprotonated faster even though it forms a less thermodynamically stable enolate.
• The malonic ester synthesis is a series of reactions which converts an alkyl halide to a carboxylic acid with two additional carbons.
• The importance of this synthesis pathways is that it allows for the creation of alpha alkylated carboxylic acids which cannot be created by direct alkylation.
• The acetoacetic ester synthesis is a series of reaction which converts alkyl halides into a methyl ketone with three additional carbons.
• This reaction creates an alpha substituted methyl ketone without side-products.
• In retrosynthetic analysis, a C-C bond is broken between the alpha carbon and the beta carbon away from this functionality.
Skills to Master
• Skill 22.1 Determine alpha carbons and alpha hydrogens.
• Skill 22.2 Draw the enols formed from carbonyl derivatives.
• Skill 22.3 Provide the mechanism for the keto-enol tautomerization under neutral, acidic and basic conditions.
• Skill 22.4 Provide the mechanism for the enol-keto tautomerization under neutral, acidic and basic conditions.
• Skill 22.5 Draw the products of halogenation reactions.
• Skill 22.6 Provide the mechanism for halogenation reactions.
• Skill 22.7 Write an equation to illustrate the Hell‑Volhard‑Zelinskii reaction.
• Skill 22.8 Identify the product formed from the reaction of a given carboxylic acid with bromine and phosphorus tribromide.
• Skill 22.9 Explain why the alpha hydrogen of carbonyl compounds are more acidic than a typical hydrogen.
• Skill 22.10 Explain why dicarbonyl protons are more acidic than compounds that contain a single carbonyl.
• Skill 22.11 Provide a mechanism for the haloform reaction.
• Skill 22.12 Provide a mechanism for an alkylation reaction with an enolate.
• Skill 22.13 Provide a mechanism for malonic ester synthesis.
• Skill 22.14 Propose synthesis for alkylated carboxylic acids.
• Skill 22.15 Provide a mechanism for acetoacetic acid synthesis.
• Skill 22.16 Propose a synthesis for alkylated methyl ketones.
Alkylation | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.S%3A_Carbonyl_Alpha-Substitution_Reactions_%28Summary%29.txt |
Learning Objectives
When you have completed Chapter 23, you should be able to
• fulfill all of the detailed objectives listed under each individual section.
• design multi‑step syntheses in which the reactions introduced in this unit are used in conjunction with any of the reactions described in previous units.
• solve road‑map problems that require a knowledge of carbonyl condensation reactions.
• define, and use in context, any of the key terms introduced.
In this chapter, we consider the fourth and final general type of reaction that carbonyl compounds undergo—the carbonyl condensation reaction. Carbonyl condensation reactions take place between two carbonyl‑containing reactants, one of which must possess an alpha‑hydrogen atom. The first step of the reaction involves the removal of an alpha‑hydrogen atom by a base. In the second step, the enolate anion that results from this removal attacks the carbonyl‑carbon of the second reacting molecule. In the final step of the reaction, a proton is transferred to the tetrahedral intermediate formed in the second step, although in some cases the product that results may subsequently be dehydrated.
23: Carbonyl Condensation Reactions
The importance of carbonyl condensation reactions to synthetic organic chemistry arises from the large number of combinations of carbonyl compounds that can be used in such reactions. Aldehydes or ketones can be used in a simple aldol condensation to produce β‑hydroxy aldehydes, β‑hydroxy ketones, or their dehydration products. Mixtures of aldehydes, ketones, or both, can be used in a mixed aldol condensation. Internal aldol condensations can occur in compounds containing two suitable carbonyl groups. Aldol‑like condensations can be brought about between aldehydes and a variety of compounds containing acidic alpha‑hydrogen atoms, including diethyl malonate, acetic anhydride, nitriles and nitro compounds. Esters can be used in Claisen condensations and 1,6‑ and 1,7‑diesters can give rise to internal condensations, called Dieckmann cyclizations. Related reactions include the Michael reaction, in which an α,β‑unsaturated carbonyl compound is reacted with an enolate anion; and the Stork enamine reaction, where an enamine adds to an α,β‑unsaturated ketone.
The chapter concludes with a look at how condensation reactions can be used in the synthesis of complex ring‑containing organic compounds, and at the role played by carbonyl condensation reactions in biological systems.
23.01: Carbonyl Condensations - The Aldol Reaction
Objectives
After completing this section, you should be able to
1. write a general mechanism for carbonyl condensation reactions.
2. write an equation to illustrate the aldol condensation reaction.
3. identify the product formed when an aldehyde or ketone having an alpha‑hydrogen atom is treated with base in a protic medium.
4. identify the aldehyde or ketone, and other reagents required to produce a given β‑hydroxy carbonyl compound by an aldol reaction.
5. determine whether a given aldehyde or ketone will undergo an aldol reaction.
6. write the detailed mechanism of the aldol reaction.
Study Notes
It is important that you understand the general mechanism of carbonyl condensation described in this section: once you grasp this mechanism, you will see that all the reactions that follow are very similar.
The aldol reaction is sometimes referred to as the aldol condensation. However, a condensation reaction is often regarded as a reaction in which two molecules join together with the elimination of a molecule of water (or some other compound of low molar mass). Thus, the aldol reaction described here is not a true condensation; the true aldol condensation is described later, in Section 23.3. It is perhaps unfortunate that the reactions discussed in this unit are all described as condensation reactions whether or not water is eliminated.
The term “aldol” (from "aldehyde alcohol") is used both to describe the specific compound 3‑hydroxybutanal:
and to describe β‑hydroxy aldehydes in general.
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is another example of electrophilic substitution at the alpha carbon in enolate anions. This reaction requires the formation of an enolate so at least one of the reactants must have an α-hydrogen. Due to the carbanion like nature of enolates, they can add to carbonyls through nucleophilic addition much like Grignard reagents.
The aldol reaction takes advantage of a carbonyl compound’s ability to undergo both alpha substitution and nucleophilic addition reactions. The fundamental transformation in the aldol reaction is a dimerization of an aldehyde (or ketone) to form a beta-hydroxy aldehyde (or ketone). A C-C bond is formed between the alpha carbon of one reactant molecule and the carbonyl carbon of a second reactant molecule. In the reaction’s product, the formed C-C bond links a carbon in the alpha position and a carbon in the beta position away from the carbonyl.
General Aldol reaction
A typical example involves two molecules of acetaldehyde (ethanal) reacting to form beta-hydroxybuteraldehyde (3-hydroxybutanal). This product and other beta-hydroxy aldehydes are generically called “aldols” because they contain both an aldehyde and an alcohol functional group.
An aldol reaction, like many carbonyl addition reactions, is an equilibrium reaction and is reversible. The presence of an equilibrium means weaker bases, such a hydroxides or alkoxides, can be used to perform this reaction. The reaction equilibrium favors the products when aldehydes with little steric hindrance around the carbonyl are used. However, the reaction equilibrium for ketones and sterically hindered aldehydes favors the reactants. To provide good reaction yields when using these reactants, the equilibrium must be pushed towards the products. Typically, this is done by utilizing a method to remove the product as it is formed during the reaction.
Mechanism of Aldol Reaction
1) Enolate formation
The reaction starts with a base removing an alpha hydrogen to form a nucleophilic enolate.
2) Nucleophilic attack by the enolate
Through nucleophilic addition, the enolate adds to the electrophilic carbonyl group on a second molecule. As with other nucleophilic addition reaction a tetrahedral alkoxide intermediate is formed.
3) Protonation
Protonation of the alkoxide forms the neutral aldol product and regenerates the base.
Stereochemical Ramifications of the Aldol Reaction
As previously discussed, both nucleophilic addition and alpha-substitution reactions have the possibility of creating chiral carbons. The alpha carbon and the electrophilic carbon of the reactants should be identified in the aldol product to assess their possible chirality. Most aldehydes produce chirality in both of these carbons. Most symmetrical ketones create a chiral carbon from the alpha-carbon of the reactant.
Going from reactants to products simply
Worked Example
What would be the expect product of the following aldol reaction?
Answer
Analysis:
When considering the product of an aldol reaction it is vital to consider each reactant molecule separately. Also,
Identify electrophilic carbonyl carbon and any alpha hydrogens present.
Exercises \(1\)
1) Predict the product of an aldol reaction with the following molecules:
a)
b)
c)
2) Because the aldol reaction is reversible it is possible for a beta-hydroxy carbonyl compound to undergo a retro-aldol reaction. Please draw the mechanism or the based catalyzed retro-aldol reaction shown below.
Answers
1)
a)
b)
c)
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.00%3A_Chapter_Objectives.txt |
Objectives
After completing this section, you should be able to describe the difference between a carbonyl condensation reaction and an alpha‑substitution reaction, and determine which of these two types of reaction is most likely to occur, given the appropriate experimental data.
Study Notes
So far we have discussed three of the four general reactions of carbonyl compounds: nucleophilic additions of aldehydes and ketones (Chapter 19), nucleophilic acyl substitution reactions of carboxylic acid derivatives (Chapter 21) and alpha‑substitution reactions (Chapter 22). The fourth general reaction, carbonyl condensation, is similar to the alpha‑substitution reaction, so you need to appreciate how it differs from the other three and the conditions under which it occurs.
Carbonyl condensation and alpha-substitution reactions both involve the formation of a reactive enolate ion intermediate. How is it possible to generate an enolate ion for a alpha substitution reaction without a carbonyl condensation also occurring? What reaction conditions are required to cause one reaction and not the other?
In a carbonyl condensation a catalytic amount of base is used to generate only a small amount of the the enolate ion. Most of the carbonyl compound is unreacted and can react with the enolate. During the reaction the base catalyst is regenerated which can then produce more enolate ion and continue the cycle.
If acetaldehyde was reacted with 0.05 equivalents of sodium methoxide in a methanol solvent only a small amount (~5%) of the enolate would form. The majority of the acetaldehyde would be unchanged and capable of undergoing a condensation reaction with the enolate present. The alkoxide intermediate produced is protonated by methanol to the neutral condensation product and regenerates the methoxide base catalyst. Methoxide can then deprotonate a new acetaldehyde molecule to create another enolate allowing the reaction cycle to continue.
These steps are all reversible and it should be noted that reactants and products that are close in energy level can potentially undergo the reverse reaction if conditions change enough. While from a synthetic point of view in the laboratory this may mean increasing yields by driving the reaction to completion (e.g. adding heat, removing product), in biological systems it can have more drastic consequences. Indeed, depending on metabolic conditions, retro-aldol reactions (the reverse of aldol condensations, in which carbon-carbon bonds are broken) can occur.
In contrast, the alpha-substitution reaction is often more directional by design. To reduce unwanted competition from carbonyl condensation, the enolate ion intermediate is generated all at once with a full equivalent of strong base at low temperature. This effectively removes the carbonyl from the reactive mixture making it difficult for a carbonyl condensation to occur. The reactive enolate intermediate then is quickly quenched by rapid addition of the electrophile to complete the substitution reaction. As discussed in Section 22.7, for direct alkylation, strong bases like NaNH2 and LDA were used to generate the enolate intermediate followed by addition of an alkylhalide.
An example is shown with the alpha-alkylation reaction of cyclopentanone. During this reaction, one equivalent of lithium diisopropyamide (LDA) is added to the reactant at -78 oC which completely converts cyclopenanone to the corresponding enolate ion. This leaves none of the ketone carbonyl remaining to undergo a condensation reaction. Methyliodide is quickly added to react with enolate ion forming the alpha-alkylated product. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.02%3A_Carbonyl_Condensations_versus_Alpha_Substitutions.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the formation of a conjugated enone from a β‑hydroxy aldehyde or ketone.
2. write a detailed mechanism for the basic or acidic elimination of water from a β‑hydroxy aldehyde or ketone.
3. explain why β‑hydroxy aldehydes and ketones undergo elimination reactions much more readily than most other alcohols.
4. identify the enone products from the aldol condensation of a given aldehyde or ketone.
Study Notes
Conjugated enones, like conjugated dienes, have more inherent stability compared with their non‑conjugated counterparts. You may wish to review Section 14.1 on dienes, which gives a molecular orbital description showing π electron distribution over four atomic centres.
Note that both of the elimination mechanisms described here (acidic and basic) involve either the enol form or the enolate anion of the β‑hydroxy carbonyl compound.
Aldol Condensation
Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water are termed Condensations. Hence the following examples are properly referred to as aldol condensations.
Dehydration of Aldol Products to Synthesize α, β Unsaturated carbonyl (enones)
The products of aldol reactions, with heating, often undergo a subsequent elimination of water, made up from an alpha-hydrogen and the beta-hydroxyl group. The product of this acid or base-catalyzed E1cB elimination reaction (Section 11-10) reaction is an α,β-unsaturated aldehyde or ketone (Enones). Although there may be multiple position where the alkene may form, it will always prefer to be in conjugation with the carbonyl.
Conjugated enone products are more stable than non-conjugated due to extended P orbital overlap. Conjugation of the p electrons of the alkene and carbonyl bonds provide a molecular-orbital description showing the interaction of p electrons of all four atoms. The additional stability provided by the conjugated carbonyl system of the product makes many aldol reactions thermodynamically factorable.
A representation of pi bonding molecular orbitals of the conjugated enone, propenal, are delocalized through p-orbital overlap
The elimination of water from the reaction mixture can be used to drive the equilibrium towards the products by Le Chatelier’s principal. This coupled with the thermodynamic stability of the conjugated product allow for good reaction yields when the formation of the initial aldol intermediate is unfavorable (ketones & sterically hindered aldehydes).
Stereochemical Considerations
When aldehyde starting materials are used for an aldol condensation, there is the possibility of forming both E and Z alkene isomers. When symmetrical ketones are used, the alkene formed lacks the ability to form isomers so a single product is made.
Examples
Mechanism
Base Catalyzed Mechanism
1) Form an enolate
The mechanism starts with the base removing an alpha-hydrogen to form an enolate ion.
2) Form the enone
The alkoxide reforming the carbonyl C=O bond promotes the elimination of alcohol OH as a leaving group which reforms the base catalyst. Although the base catalyzed elimination of alcohols is rare, it happens in this case in part due to the stability of the conjugated enone product.
Acidic Conditions Mechanism
1) Protonation
The mechanism starts with the two step tautomerization process to form an enol.
3) Protonation
Protonation of the alcohol OH increases its ability to act as a leaving group.
4) Elimination
Lone pair electrons from the enol reform the carbonyl C=O bond and promoted the elimination of water as a leaving group.
5) Deprotonation
Deprotonation by water in the final step create the neutral enone product and regenerates the acid catalyst.
Aldol Condensation
Whether an aldol reaction or an aldol condensation product is formed during a reaction largely depends on the reaction conditions. Typically, a reaction with a base at room temperature provides the aldol reaction product. However, if the reaction mixture is heated the aldol product is quickly converted into the aldol condensation product. If the condensation product is desired the aldol intermediate is usually not isolated.
Examples
Aldol Reaction
Aldol Condensation
Worked Example
Draw the product of an aldol condensation with the following molecule:
Answer
The overall reaction is a combination of two major steps, an aldol reaction followed by a dehydration to form the enone. In this situation it is best to consider the aldol product first (as discussed in Section 23.3, then convert it to the enone. Note! The double bond always forms in conjugation with the carbonyl. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.03%3A_Dehydration_of_Aldol_Products_-_Synthesis_of_Enones.txt |
Objectives
After completing this section, you should be able to identify the aldehyde or ketone and other necessary reagents that should be used to prepare a given enone by an aldol condensation.
Study Notes
This section stresses the importance of being able to think logically.
The experience that you have already gained through designing multi‑step syntheses and solving road‑map problems should help you to recognize when an aldol reaction may have been one of the steps in the synthesis of a given compound.
It is important that you recognize that the aldol condensation is an important part of a synthetic chemist’s repertoire, both because it involves the formation of a new carbon‑carbon bond, and also because it yields a product containing two functional groups.
Aldol reactions are excellent methods for the synthesis of many enones or beta-hydroxy carbonyls. Because of this, being able to predict when an aldol reaction might be used in a synthesis in an important skill. This can be accomplished by identifying these combinations of atoms and bonds and then, working backwards, theoretically breaking the target molecule apart into possible reactants.
Fragments which are easily formed by an aldol reaction
Determining the Reactants for an Aldol Condensation
During an aldol condensation a C-C sigma and a C-C pi bond are formed. This makes the key bond cleavage in the target molecule the C=C bond between the carbons alpha and beta away from the carbonyl. After the cleavage, the carbon that was in the alpha position (on the fragment with the carbonyl) gains two hydrogens. The carbon that was in the beta gains a =O to form a carbonyl.
Determining the Reactants for an Aldol Reaction
During an aldol condensation a C-C sigma bond is formed. This makes the key bond cleavage in the target molecule the C-C bond between the carbons alpha and beta away from the carbonyl. After the cleavage, the carbon that was in the alpha position (the fragment with the carbonyl) gains one hydrogen. The fragment loses a hydrogen from the OH and then forms a C=O carbonyl bond.
Worked Example
Show how the following molecule could be made using an aldol condensation?
Answer
Analysis: The C=C bond in the target molecule is cleaved to form two fragments. The fragment with the carbonyl gains two alpha hydrogens. The other fragment gains a =O to form a carbonyl. Both fragments end up producing the same reactant which is typical for an aldol condensation.
Additional Synthetic Considerations
The enone product of an aldol condensation is versatile because it contains two functional groups (alkene & carbonyl) which can be subject to further reactions. Among many possible reactions, an enone can undergo hydrogenation to produce an aldehyde or ketone. Also, the carbonyl group can undergo hydride reduction to produce a beta-hydroxyalkene.
These additional reactions can be applied with the consideration of using an aldol reaction in the synthesis of a target molecule. A similar analysis can be extrapolated to the other reactions possible with the alkene and carbonyl present in an enone.
Analysis for Hydrogenation
To consider a hydrogenation, remove a hydrogen from a carbon in both the alpha and beta positions relative to the carbonyl. Then connect these two carbons with a C=C double bond. This create a possible enone which can be broken apart further using the analysis described above. Target molecules can have often have multiple alkyl chains which can be used to form a double bond.
Analysis for Hydride Reduction
To consider a hydride reduction remove a hydrogen from the alcohol oxygen then a hydrogen from the adjacent carbon. Connect the oxygen and carbon with a double bond to form a C=O carbonyl. This provides an enone which can undergo further analysis.
Exercise \(1\)
Please devise a synthesis pathway for the following molecule using an aldol reaction:
Answer
Analysis: Because the target molecule has two beta-carbons with hydrogens there are two possible synthesis pathways. Both should be considered for their effectiveness. Because a carbonyl is already present fragmentations will first be performed to create two possible enone intermediates. These enones will both be broken into their aldol reactants. When looking a possible aldol reactants produced by each pathway it is clear that pathway 2 is preferred. Because both fragments are the same they will react to form on major product.
Exercises \(1\)
1) Which of the following molecules could be made using an aldol reaction or condensation. Please show the starting material for the reactions that are possible.
2) Please show the reaction steps required to make 4-methyl-2-pentanol from an aldol condensation.
3) Please design a synthesis for the following molecule: Red = Oxygen, Grey = Carbon, White = Hydrogen
Answers
1) Molecules A and C are possible.
Solution for molecule A
Solution for molecule C
2)
3) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.04%3A_Using_Aldol_Reactions_in_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate a mixed aldol reaction.
2. identify the structural features necessary to ensure that two carbonyl compounds will react together in a mixed aldol reaction to give a single product rather than a mixture of products.
3. determine whether a given mixed aldol reaction is likely to produce a single product or a mixture of products.
4. identify the product or products formed in a given mixed aldol reaction.
5. identify the carbonyl compounds needed to produce a given enone or β‑hydroxy aldehyde or ketone by a mixed aldol reaction.
Study Notes
You should satisfy yourself that you understand how the four products shown in Example 23.5.2 arise from the condensation of 2‑propanone and 2‑phenylacetaldehyde.
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolate donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this, most mixed aldol reactions are usually not performed unless one reactant has no a H’s
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
ACH2CHO + BCH2CHO + NaOH AA + BB + AB + BA
Products of a Uncontrolled Mixed Aldol Reaction
To avoid complex reaction mixtures the reactants should be chosen to favor one particular donor/acceptor reaction. Successful mixed aldol reactions usually use one of two combination of factors.
1) A reaction of an aldehyde with no alpha-hydrogens with a ketone that has alpha hydrogens: Aldehydes lacking alpha-hydrogens cannot form an enolate so they can only function as electrophilic acceptor reactants. This reduces the number of possible products by half. Although it would be possible for the ketone to react with itself it is unlikely. Aldehydes are more reactive acceptor electrophiles than ketones. This makes the preferred reaction one with the ketone as an enoloate donor and the aldehyde as an electrophilic acceptor.
The aldol condensation between an aromatic aldehyde with no α-hydrogens and an aliphatic aldehyde or ketone with α-hydrogen is called a Claisen–Schmidt condensation. The reaction product is a highly conjugated α,β‐unsaturated aldehyde or ketone which forms in the more stable (E)-alkene isomer.
Example: Claisen–Schmidt Condensation
2) One of the reactants has alpha-hydrogens which are highly acidic: The acidic compound will be preferably converted into an enolate donor which removes the possibility of its carbonyl acting as an electrophilic acceptor. An example is the aldol condensation of ethylacetoacetate and cyclopentanone. Ethylacetoacetate has α-hydrogens which are particularly acidic due to their conjugate base being stabilized by two carbonyl bonds (Section 22.5). Upon reaction with base, ethylacetoacetate is converted to the enolate donor leaving cyclopentanone to be the electrophilic acceptor. This provides one predominant aldol condensation product. In this reaction sodium ethoxide is used as a base to prevent hydrolysis side-reactions with the ester of ethylacetoacetate.
Example
Exercises \(1\)
1) Show the reactants required to make the following using a mixed aldol condensation. Indicate those which would be likely to produce a mixture of products.
a)
b)
c)
Answers
a)
b)
c) Because both reactants have α-hydrogens and neither are particularly acidic, the reaction probably will produce a mixture of condensation products. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.05%3A_Mixed_Aldol_Reactions.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate an intramolecular aldol reaction.
2. identify the product formed when a given dicarbonyl compound undergoes an intramolecular aldol condensation.
3. identify the dicarbonyl compound which, when treated with a suitable base, could be used to prepare a given cyclic enone by an intramolecular aldol condensation.
Study Notes
“Intramolecular ” means “within the same molecule.” You have already seen some examples of intramolecular reactions in previous chapters. Another term for intramolecular aldol reaction is “internal aldol reaction.”
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. The term “Intramolecular” means “within the same molecule.” In this case, it means that the enolate donor and the electrophilic acceptor of an aldol reaction are contained in the same molecule such as dialdehydes, keto aldehydes, or diketones. In these cases, the small distance between the donor and acceptor leads to faster reaction rates for intramolecular condensations making intermolecular condensations (which require two molecules to collide in solution) less favorable.
In most cases multiple sets of α-hydrogens need to be considered when determining the donor/acceptor roles for the reaction, which might lead to a mixture of products. The intramolecular aldol reaction of a 1,5-diketone, 2,6-heptanedione, could possibly yield either the six-membered ring product, 3-methyl-2-cyclohexenone, or the four-membered ring product, (2-methylcyclobutenyl)ethanone. However, the cyclohexanone product is exclusively formed.
This product selectivity is possible due to all of the steps of the mechanism being reversibly, which tends to produce the most stable product. As with most ring forming reactions, five and six membered rings are preferred due to their relative lack of ring strain compared to other sized rings (See Sections 4.4 & 4.5). Once equilibrium is reached, the relatively strain free and therefore more thermodynamically stable, cyclohexanone product will be preferably formed.
Example
Similar analysis can be used to predict the products of other intramolecular aldol reactions. In a similar reaction, 1,4-diketones, such as 2,5-hexanedione, only form the five-membered ring product, ex. 3-methyl-2-cyclopentenone, without any of the possible cyclopropane product forming.
Example
Worked Example
Please draw the expected product of an intramolecular aldol condensation with the following molecule:
Answer
Analysis: 6-Oxoheptanal has three unique sets of alpha-hydrogens which could be deprotonated to form an enolate. Selecting the correct set involves analyzing the carbonyl reactivity’s along with the possible ring sizes of the products.
Solution: Once the preferred alpha-hydrogens are determined go through the steps discussed in the previous sections to determine the aldol condensation product. Remember to form the aldol intermediate first.
Exercises \(1\)
1) Briefly explain why the molecule, 2,4-pentanedione, when reacted with a base would mostly likely not produce an intramolecular aldol condensation product.
2) Draw the product of the following aldol condensation:
Answers
1) 2,4-Pentanedione like other 1,3-Diketones have particularly acidic alpha-hydrogens between the two carbonyl. When base is applied these will be the first to deprotonate and form an enolate. For an intramolecular aldol condensation to occur an alpha-hydrogen one of the methyl groups would have to be removed which would be difficult.
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.06%3A_Intramolecular_Aldol_Reactions.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate a Claisen condensation reaction.
2. write a detailed mechanism for a Claisen condensation reaction or its reverse.
3. identify the product formed in a given Claisen condensation reaction.
4. identify the ester and other reagents needed to form a given β‑keto ester by a Claisen condensation reaction.
Key Terms
• Claisen condensation reaction
Study Notes
You have already seen that ethyl acetoacetate‑type compounds are very useful in organic syntheses. Any reaction which results in the formation of these compounds will also be of importance. In the next section, you will see how the range of β‑keto esters that can be prepared by this method is extended through the use of two different esters as starting materials.
Claisen Condensation
Because esters commonly contain both α-hydrogens and a carbonyl bond, they can undergo a reversible condensation similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol reaction, one ester acts as a Claisen enolate donor (nucleophile) while a second ester acts as the Claisen acceptor (electrophile). During the reaction a new carbon-carbon is formed to produce a β-keto ester product. This reaction is considered a condensation because it eliminates a small alcohol as an unwanted side-product.
Predicting the Product of a Claisen Condensation
Example
Although they may appear similar, there are a number of fundamental differences between an aldol and Claisen condensation. During the mechanism of the reaction, the formed tetrahedral alkoxide intermediate is not protonated to from an “aldol” type product. Rather, the alkoxide intermediate will reform the C=O carbonyl bond and eliminate a (-OR) leaving group to produce a nucleophilic acyl substitution product.
Claisen condensations cannot use a hydroxide for the reaction base due to the possibility of ester hydrolysis. Also, to prevent transesterification side products, the alkoxide base typically has the same alkyl group as alkoxy group present in the ester starting material.
Lastly, the β‑keto ester products of Claisen condensation can be acidic enough to be deprotonated by the reaction’s base during the final steps of the mechanism. This means the base is not catalytically regenerated during the reaction and a full equivalent of base is required. The β‑keto ester condensation products are removed from the equilibrium by this deprotonation, which causes the reaction to be driven forward by Le Chatelier’s principle. This concept is so important that a Claisen product will not form unless it contains an alpha hydrogen acidic enough to react completely with the reaction base. This requires that the ester starting materials have at least two alpha-hydrogens for a Claisen condensation product to form. One is removed to form an ester enolate and the second is removed to drive the reaction forward.
Mechanism
The Claisen condensation mechanism is analogous to the ester saponification reaction seen in Section 21.6.
1) Enolate formation
The mechanism starts with the alkoxide base removing an alpha-hydrogen from the ester to form a nucleophilic ester enolate ion.
2) Nucleophilic attack
The enolate nucleophile adds to carbonyl carbon of a different ester, forming a tetrahedral alkoxide intermediate.
3) Removal of leaving group
The alkoxide then reforms the carbonyl, eliminating the –OR leaving group to form a beta-ketoester.
4) Deprotonation
The acidity of beta-ketoesters (pKa ~9) is high enough to allow them to be completely deprotonated by alkoxide bases (pKa of an alcohol ~16) to form a second enolate. This make the equilibrium of this step very favorable, which is enough to drive the whole reaction towards the product.
5) Protonation
The enolate is protonated in an acid work-up to form the neutral beta-ketoester product.
Stereochemical Considerations
The alpha-carbon gains a substituent during the reaction which means it will most likely form a chiral carbon. The fact that the alpha-carbon is temporarily converted to an enolate in the last step of the mechanism means that a racemic mixture of enantiomers will form.
Worked Example
Draw the products of the following reaction:
Answer
Analysis: Remember to consider each start ester separately.
Exercise \(1\)
Please draw the products if the following molecules were to undergo a Claisen condensation.
a)
b)
c)
2) The beta-keto ester product of a claisen condensation can under hydrolysis with Sodium Hydroxide as shown in the reaction below. Please draw a curved arrow mechanism to explain how the products are formed. Also, explain why the ketone functional group preferably reacts with hydroxide instead of the ester.
Answers
a)
b)
c)
2) The carbonyl of the ketone is more likely to be attacked by the hydroxide nucleophile because is more reactive than the ester’s | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.07%3A_The_Claisen_Condensation_Reaction.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate a mixed Claisen condensation.
2. identify the structural features that should be present in the two esters if a mixed Claisen condensation is to be successful.
3. determine whether a given pair of esters is likely to produce a good yield of a single product when subjected to a mixed Claisen condensation.
4. identify the product formed when a given pair of esters is used in a mixed Claisen condensation.
5. identify the esters that should be used to produce a given β‑keto ester by a mixed Claisen condensation.
6. write an equation to illustrate the formation of a β‑diketone through a mixed Claisen‑type condensation between an ester and a ketone.
7. identify the β‑diketone formed as the result of a mixed Claisen‑type condensation between a given ester and a given ketone.
8. identify the reagents necessary to synthesize a given β‑diketone by a mixed Claisen‑type condensation between an ester and a ketone.
9. write detailed mechanisms for mixed Claisen reactions and reactions that are related to the mixed Claisen reaction, including those in which both reacting moieties are present in the same compound.
Study Notes
Just as we can react together two different aldehydes or ketones in a mixed aldol condensation, so can we react together two different esters in a mixed (or “crossed”) Claisen condensation. Again, by carefully selecting our substrates we can obtain a good yield of the desired product and minimize the number of possible by‑products. Note that even if we replace one of the esters with a ketone, the reaction is still referred to as Claisen condensation. The important thing to realize as you study these reactions is that they all take place by essentially the same mechanism—attack by an enolate anion on a carbonyl group.
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as Claisen donors and Claisen acceptors generally give complex mixtures which are difficult to separate.
General Reaction
To avoid complex mixtures most Crossed Claisen reactions are usually not successful unless one of the two esters has no alpha-hydrogens as in the following examples.
This forces the ester with alpha-hydrogens to be the Claisen donor (enolate) and the ester without alpha-hydrogens to be the Claisen acceptor (electrophile). Even with this differentiation, it is possible for the ester with alpha-hydrogens to undergo a Claisen Condensation with itself to produce an unwanted side-product. This is typically partially prevented by using the ester without alpha-hydrogens in a large excess.
Example - no alpha hydrogens
Another type of crossed Claisen condensation occurs when a ketone is reacted with an ester. The use of an ester without alpha-hydrogens is not necessary due to the greater reactivity of the ketone. The alpha-hydrogens of the ketone are much more acidic (pKa~20) than those of the ester (pKa~25). The alpha hydrogens of ketone will be preferably deprotonated to make it the enolate Claisen donor. The ester will therefore be the electrophilic Claisen acceptor and is likewise commonly used in a large excess. There is still a possibility of the ketone reacting with itself to form the product of an aldol reaction however the equilibrium is unfavorable. Also, the formation of the aldol product is reversible due to its lack of an acidic alpha hydrogen. The beta-diketone Claisen product can be irreversibly deprotonated due to its more acidic alpha-hydrogens, making it the reaction’s main product.
Example - ester with ketone
Planing a Synthesis using a Claisen or Crossed Claisen-like Reaction
A Claisen reaction should be considered for a synthesis pathway if the target molecule contains a beta-keto esters, a beta-aldo ester or a beta-dicarbonyl.
A beta-keto ester or a beta-aldo ester could possibly be made by a Claisen condensation of two esters. A beta-dicarbonyl could possibly be made by a Claisen-like condensation between a ketone and an ester. Here, the key bond cleavage is a C-C bond between one of the carbonyls and the alpha-carbon which lies between the carbonyl. In each of these situations there are two such C-C bonds so there will be two possible pathways to consider. After the C-C bond cleavage the fragment with the alpha-carbon will gain a hydrogen. The fragment with the bare carbonyl will gain an –OR group to become an ester. The –OR group added will be the same as any ester present in the target molecule. If none are present –OCH2CH3 is typically used.
Analysis for Beta-Keto and Aldo Esters
A similar analysis can be performed on a 1,3-diketone and will be presented in the worked example below.
Worked Example
Show how the following molecule can be made using a Claisen-like condensation.
Answer
Pathway 1
Solution 1
Pathway 2
Solution 2
Exercises \(1\)
Draw the product of the following reactions:
a)
b)
Answers
a)
b)
23.09: Intramolecular Claisen Condensations - The Dieckmann Cyclization
Objectives
After completing this section, you should be able to
1. write an equation to illustrate an internal Claisen condensation, that is, a Dieckmann cyclization.
2. identify the product formed when a 1,6‑ or 1,7‑diester undergoes an internal Claisen condensation.
3. identify the diester needed to prepare a given cyclic β‑keto ester by an internal Claisen condensation.
4. identify the structural features present in a diester that lead to the formation of more than one product in an internal Claisen condensation.
Key Terms
• Dieckmann cyclization
Study Notes
Essentially no new material is introduced in this section; Dieckmann cyclizations are intramolecular Claisen condensations. These reactions occur for 1,6‑ and 1,7‑diesters, as these substances result in the formation of compounds containing five‑ and six‑membered rings, respectively. You may recall that the formation of such ring systems is favoured because they are relatively free of strain.
The Dieckmann Cyclization
Diesters can undergo an intramolecular reaction, called the Dieckmann condensation, to produce cyclic beta-keto esters. This reaction works best with 1,6-diesters, which produce five-membered rings, and 1,7-diesters which produce six membered rings.
Examples
Mechanism
The mechanism of the Dieckmann condensation is the same as a Claisen condensation. An alkoxide base removes an alpha-hydrogen from one of the esters to form an ester enolate. The enolate then adds to the carbonyl carbon of the other ester to form a tetrahedral alkoxide intermediate. The alkoxide reforms the carbonyl bond which causes the elimination of the –OR leaving group and forms a cyclic beta-keto ester. The high acidity of the beta-keto ester allows it to be deprotonated by the reactions base to form a second enolate. Like the Claisen condensation, this deprotonation step drives the equilibrium towards the products and is required for the reaction to occur. A full equivalent of base is necessary during this reaction.
1) Enolate formation
2) Nucleophilic attack
3) Removal of leaving group
4) Deprotonation
5) Protonation
Further Reactions of the Dieckmann Cyclization Product
The cyclic beta-keto ester product of a Dieckmann cyclization can be modified by reaction similar to those used in the acetoacetic ester synthesis (Section 22-7). The acidic alpha-hydrogens of the beta-keto ester allow it to easily be deprotonated and alkylated in an alpha-substitution reaction. Having a carbonyl group in the beta position allows the ester to be removed through decarboxylation. The combination of these three reactions (1) Dieckmann cyclization, (2) alpha alkylation, and (3) decarboxylation provides an efficient method for preparing 2-substituted cyclopentanones and cyclohexanones.
Exercises \(1\)
1) Draw the expected product for the following reaction:
2) The Dieckmann cyclization of the following molecule is expected to give a mixture of two cyclized products. Draw the structure of the two products and briefly explain why a mixture is formed.
Answers
1)
2)
Each has the ability to act as a Claisen donor and a Claisen acceptor. The reaction should produce a roughly 50/50 mixture where each ester acts as a Claisen acceptor. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.08%3A_Mixed_Claisen_Condensations.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the Michael reaction.
2. write a detailed mechanism for a given typical Michael reaction.
3. identify the product formed in a given Michael reaction.
4. identify the reagents necessary to synthesize a given compound by a Michael reaction.
Key Terms
• Michael reaction
Study Notes
Before studying this section in detail, you should review conjugate addition to α,β‑unsaturated carbonyl compounds (Section 19.13).
Notice the wide variety of compounds that can take part in a Michael reaction. As usual, you should not feel overwhelmed by the number of different compounds that can be used, just keep in mind that in all cases the mechanism is essentially the same.
Michael Reaction
Certain nucleophiles undergo conjugate addition with the alkene of an α, β-unsaturated carbonyl compounds rather than undergo direct nucleophilic addition with the carbonyl. During conjugate addition, a nucleophile adds to the electrophilic β-alkene carbon to from a C-Nuc bond. If the starting materials contains an ester the corresponding alkoxide is used as the base in the reaction. Otherwise a hydroxide base, such as sodium or potassium hydroxide, is commonly used.
General Reaction
When enolate nucleophiles undergo conjugate addition with an α, β-unsaturated carbonyl the process is called a Michael Reaction. The Michael reaction works best with particularly acidic enolate donors such as malonic esters, β-keto esters, ect. Enolates which are weaker acids tend to undergo nucleophilic addition to the carbonyl rather than conjugate addition. For example, malonic ester’s acidic alpha-hydrogens can be easily deprotonated by sodium ethoxide to form an enolate. The enolate adds to propenal to form the conjugate addition product.
Predicting the Product of a Michael Reaction
During the is reaction the C=C pi bond of the α, β-unsaturated carbonyl is broken. An C-H bond is formed on the carbon in the α-position from the carbonyl. An acidic hydrogen is removed from the nucleophile and then it is used to form a C-C bond with the carbon in the beta position from the carbonyl.
a. In the first step, remove the double bond and add two single bonds
b. Add a hydrogen to the bond adjacent to the carbonyl
c. Add the nucleophile to the other end of the double bond (remember to remove a hydrogen from the nucleophile)
Worked Example \(1\)
What would be the product of the following conjugate addition?
Answer
1. break the double bond and add two single bonds
2. Add a hydrogen to the bond adjacent to the carbonyl
3. Remove a hydrogen from the nucleophile.
4. Add the nucleophile to the other end of the double bond.
Mechanism of Michael reaction
During a Michael reaction the enolate acts as a nucleophilic donor and the α, β-unsaturated carbonyl acts as the electrophilic acceptor. The mechanism is a mixture of an alpha-substitution for the enolate and a conjugate addition for the α, β-unsaturated carbonyl.
Notice that the Michael Reaction does not require the deprotonation of the product to push the reaction towards completion. The reaction is thermodynamically favorable because the C-C bond formed in the product is stronger than the C=C bond in the starting material. This means that, unlike the Claisen condensation, this reaction only requires a catalytic amount of reaction base.
1) Deprotonation
The alkoxide base removes an alpha-hydrogen to from an enolate nucleophile.
2) Nucleophilic attack on the carbon β to the carbonyl
The enolate nucleophile then adds to the electrophilic β-carbon of the α, β-unsaturated carbonyl. Conjugation pushes the C=C pi bond electrons onto the carbonyl oxygen, forming a new enolate enolate.
3) Protonation
The alkoxide is protonated, forming an enol.
4) Tautomerization
Tautomerization converts the enol back into a carbonyl forming the neutral conjugate addition product.
The Michael reaction can be performed with a wide variety of α, β-unsaturated carbonyl electrophilic acceptors and enolate donors. Michael acceptors such as α, β-unsaturated ketones, aldehydes, esters, amides, and nitro compounds can all participate in this reaction. Michael donors with particularly acidic α-hydrogens, such as β-diketones, β-keto esters, β-keto nitriles, α-nitro ketones, malonic esters, and nitro compounds can be used.
Table \(1\): Possible Donors and Acceptors Which can be used in a Michael Reaction
Stereochemical Consideration of Micheal Reactions
During a Michael reaction two sp2 hybridized carbons are both converted to sp3 hybridization. Both of them have the possibility of creating a chiral carbon. Michael reactions are often performed using an α, β-unsaturated carbonyl with only hydrogen substituents on the alkene to prevent the formation of chiral carbons. If either carbon in the alkene has an alkyl substituent it will likely form a chiral carbon in the Michael reaction product. The α-carbon of the enolate donor also has the possibility of forming a chiral carbon as discussed in Section 22-6.
Planning a Synthesis Using a Michael Reaction
The true utility of the Michael reaction is seen when considering it use in a synthesis. The reaction can be used to prepare aldehydes, ketones, esters, amides, nitriles, and nitro compounds. A target molecule can possibly be made using a Michael reaction if it contains one of these functional groups and an alkyl chain at least three carbons long. The key bond cleavage is a C-C bond between β and gamma carbons from a carbonyl-like group. The fragment with the Y group loses an α-hydrogen and then forms a C=C bond between the α and beta carbon. The carbon of the other fragment gains a hydrogen. This fragment should possess an acidic α-hydrogen and should be made up of Michael donor fragments such as those listed in Table \(1\). In some cases, there may be more than one fragmentation variation. Each variation should be investigated and the possible starting materials compared for viability.
Worked Example
How could the following molecule be synthesized using a Michael Reaction?
Answer
Pathway 1
Starting Materials 1
Pathway 2
Starting Material 2
Solution
When comparing the two sets of possible starting materials of pathway 1 are preferred. The Michael enolate donor is particularly acidic because its conjugate base is stabilized by two carbonyls. The Michael enolate donor for pathway 2 does not have the same acidity. Consequently, the enolate would preferably undergo nucleophilic addition rather than conjugate addition. The makes the starting materials of pathway 1 preferred for the synthesis of the starting material.
In combination with alkylations and condensations, the Michael reaction may be used to construct a wide variety of complex molecules from relatively simple starting materials. The carbon nucleophiles used in the following examples include cyanide ion, sodium diethylmalonate and the conjugate base of cyclohexane-1,3-dione. These anions are sufficiently stable that their addition reactions may be presumed reversible. If this is so, the thermodynamic argument used for hetero-nucleophile additions would apply here as well, and would indicate preferential formation of 1,4-addition products. Cyanide addition does not always follow this rule, and aldehydes often give 1,2-products (cyanohydrins). In each case the initial reaction is a Michael addition, and the new carbon-carbon bond is colored magenta. Any subsequent bonds that are formed by other reactions are colored orange.
Exercises \(1\)
1) Provide the products of the following Michael additions:
2) What would be the starting materials required to make the following molecule using a Michael reaction? Red = Oxygen, Grey = Carbon, White = Hydrogen, Blue = Nitrogen.
Answers
1)
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.10%3A_Conjugate_Carbonyl_Additions_-_The_Michael_Reaction.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the three‑step Stork enamine reaction.
2. write a detailed mechanism for each of the three steps of the Stork enamine reaction.
3. identify the product formed, and the various intermediates (i.e., the enamine, the Michael‑type adduct), in a given Stork enamine reaction.
4. identify the reagents needed to synthesize a given compound by a Stork enamine reaction.
Key Terms
• Stork enamine reaction
Study Notes
If we try to use a monoketone as a donor molecule in a Michael reaction, we will obtain a poor yield. An alternative route to the product that would be expected from such a synthesis is via the Stork enamine reaction. You may wish to review the formation of enamines (from ketones and secondary amines) before you proceed with this section; if so, review Section 19.8.
Synthesis of Enamines
As previously seen in Section 19.8, aldehydes and ketones react with 2o amines to reversibly form enamines. Enamines can add to acid halides to form 1,3-diketones and α, β-unsaturated Michael acceptors to from 1,5-dicarbonyls.
Reversibility
Typically the following 2o amines are used for enamine reactions
Enamines have resonance structures similar to enolates. The lone pair electrons on the nitrogen are conjugated with the double bond and can donate electron density to the α-carbon. This allows the α-carbon of enamines to be nucleophilic in much the same fashion as enolates. The increased electron density of the α-carbon enamine, N,N-Dimethylaminoethylene, is shown as a yellow/green color in its electrostatic map. This is analogous to the increased electron density of the α-carbon of an enolate.
Enamines act as nucleophiles in a fashion similar to enolates. Because of this enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps: 1) Formation of the enamine, 2) Reaction with an electrophile to form an iminium salt, 3) Hydrolysis of the iminium salt to reform the aldehyde or ketone. Some of the advantages of using an enamine over an enolate are enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. Also, enamine are more effective at the Michael reaction compared to a mono-carbonyl enolate.
Acylation of Enamines
Mechanism on Enamine Acylation
1) Formation of the enamine
The mechanism starts with the formation of an enamine.
2) Nucleophilic attack
The enamine adds to the electrophilic carbonyl carbon of the acid halide to form an iminium bond with tetrahedral alkoxide as an intermediate.
3) Leaving group removal
The alkoxide reforms the carbonyl bond and eliminate a halide anion as a leaving group.
4) Reform the carbonyl by hydrolysis
The iminium bond is then hydrolyzed to reform the carbonyl to create a 1,3-dicarbonyl compound as the product of a nucleophilic acyl substitution.
Michael Reaction Using Enamines: The Stork Reaction
Enamines add to α, β-unsaturated carbonyls in a Michael-like reaction. The net reaction is the addition of a ketone to a α, β-unsaturated carbonyl to product a 1,5 dicarbonyl compound as the end product. This reaction is commonly known as the Stork enamine reaction after Gilbert Stork of Columbia University who originated the work.
Mechanism of the Stork Reaction
1) Formation of the enamine
2) Nucleophilic attack on the carbon β to the carbonyl
After formation, the enamine adds to the electrophilic alkene carbon of the a α, β-unsaturated carbonyl form an iminium bond. The pi electrons of the alkene are pushed onto the oxygen through conjugation to form an alkoxide as an intermediate.
3) Protonation
Protonation of the alkoxide forms an enol.
4) Tautomerization
The enol undergoes tautomerization to reform the ketone.
4) Reform the carbonyl by hydrolysis
In the last step of the mechanism, the iminium bond is hydrolyzed to reform the carbonyl to create a 1,5-dicarbonyl compound as the product of a Michael-like reaction.
Planning a Synthesis Using the Stork Enamine Synthesis
Planning a synthesis using a Stork enamine reaction is very similar to that of a Michael reaction (Section 23-10). A target molecule can possibly be made using a Stork enamine reaction if it contains a 1,5-dicarbonyl. Like the Michael reaction, the key bond cleavage is a C-C bond between beta and gamma carbons from a carbonyl-like group. The fragment with the Y group loses an alpha-hydrogen and then forms a C=C bond between the alpha and beta carbon. The carbon of the other fragment gains a hydrogen. This fragment should possess an acidic alpha-hydrogen and should be made up of Michael donor fragments.
Worked Example
How could the following molecule be made using a Michael Reaction?
Answer
Pathway 1
Alkylation of an Enamine
Enamines undergo an SN2 reaction with reactive alkyl halides to give the iminium salt. The iminium salt can be hydrolyzed back into the carbonyl.
Individual steps
1) Formation of an enamine
2) SN2 Alkylation
3) Reform the carbonyl by hydrolysis
All three steps together:
Exercises \(1\)
1) Draw the product of the reaction with the enamine prepared from cyclopentanone and pyrrolidine, and the following molecules.
a)
b)
c)
2) What would be the starting materials necessary to make the following molecules during a Stork enamine reaction.
a)
b)
Answers
1)
a)
b)
c)
2)
a)
b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.11%3A_Carbonyl_Condensations_with_Enamines_-_The_Stork_Reaction.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the Robinson annulation reaction.
2. identify the cyclic product formed when a 1,5‑diketone is treated with base.
3. identify the carbonyl compounds and any other reagents needed to synthesize a given cyclic compound by a series of reactions, one of which is a Robinson annulation.
Key Terms
• Robinson annulation reaction
Study Notes
The building of an alicyclic compound from acyclic starting materials can present an interesting challenge to the synthetic organic chemist. One route by which such a synthesis can be achieved is through the use of the Robinson annulation reaction. This reaction, which may at first look very complex, can be readily understood once you realize that it simply involves a Michael reaction followed by an intramolecular aldol condensation. Both of these steps involve attack by enolate anions.
As in some of the other syntheses that you have studied, when you are simply given the structure of a compound and asked how it could have been prepared, it can be difficult to recognize which reactions might have been used. In this instance, keep in mind that you have studied relatively few reactions which have resulted in the formation of an alicyclic compound. Thus, when asked how such a compound might be prepared, you should keep the possibility of using a Robinson annulation reaction in mind.
The Robinson Annulation
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred.
The nucleophilic enolate donor is typically an enolate ion or enamine of a cyclic ketone, β-keto ester or β-diketone. The electrophilic acceptor is usually an α, β-unsaturated ketone. In the example below, 2-methyl-1,3-cyclohexanedione is deprotonated to form an enolate which affects a Michael reaction addition on 3-buten-2-one forming a C-C bond. The product contains a 1,5-diketone fragment which can undergo an intramolecular aldol condensation. This occurs through creation of a new enolate at the methyl ketone which undergoes an intramolecular aldol reaction. A new C-C bond is formed to one of the ring carbonyls, creating a new six-membered ring. In the last step, the resulting alcohol is eliminated to form a α, β-unsaturated ketone as the final Robinson annulation product. Because the Robinson Annulation involves an aldol reaction, a full equivalent of base is required.
Example
Worked Example
Draw the product of the following Robinson Annulation.
Answer
Analysis: When considering the product of a Robinson annulation it is usually best to consider each reaction individually. Use the steps discussed in Section 23.10 to convert the starting materials into the product of a Michael reaction, then into the product of an intramolecular aldol condensation. If the starting materials are drawn in a skeletal structure it may be helpful to convert to a condensed formula to better keep track of carbons and hydrogens.
Formation of the Michael reaction product
Formation of the intramolecular Aldol condensation product
Planning a Synthesis Using a Robinson Annulation
The presence of a cyclic, six-membered α, β-unsaturated ketone in a target molecule suggest that a Robinson Annulation may be utilized in its synthesis. The key bond cleavages are the C=C bond of the α, β-unsaturated ketone and the C-C bond between carbons in the a and g on the opposite alkyl chain of the ketone.
Worked Example
What would be the starting materials used if the following molecule was made using a Robinson annulation?
Answer
Analysis
Solution
It is necessary to use sodium ethoxide as the reaction base due to the presence of an ester.
Exercise \(1\)
1) Provide products of the following Robinson annulations.
a)
b)
c)
2) What would be the starting materials required to make the following molecule using a Robinson annulation?
Answers
1)
a)
b)
c)
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.12%3A_The_Robinson_Annulation_Reaction.txt |
Objectives
After completing this section, you should be able to
1. identify acetyl coenzyme A as an important biomolecule which undergoes carbonyl condensation reactions.
2. identify aldoases as enzymes that catalyze aldol reactions in biological systems.
3. identify the steps in which a carbonyl condensation reaction has occurred, given a general outline of a specific biosynthesis.
Study Notes
Carbonyl condensation reactions occur in biological systems; for example, in the biosynthesis of citric acid.
You first met acetyl coenzyme A in Section 21.8. Again, we stress that it is not essential that you know the detailed structure of this compound (or other large biochemical stuctures discussed), but you should know that it may be represented as
and that it has the ability to behave just as any other acetyl‑containing compound.
Aldol Reactions in Nature
So far, we have examined the non-enzymatic reaction of an aldehyde or ketone with itself (a so-called 'self-condensation' reaction, where 'condensation' means the formation of one larger molecule from two smaller ones). However, aldol reactions occur in several biological pathways, most commonly in the metabolism of carbohydrates (sugars). The enzymes that catalyze aldol reactions are called, not surprisingly, 'aldolases'. They occur in all organisms, but note that Class I aldoases are usually found in animals and higher plants, while Class II aldoases normally appear in bacteria and fungi.
Typical aldolase reactions - three variations on a theme
The first step in an aldolase reaction is the deprotonation of an alpha-carbon to generate a nucleophilic carbanion. Nature has evolved several distinct strategies to stabilize the intermediate that results. Some aldolases use a metal ion to stabilize the negative charge on an enolate intermediate, while others catalyze reactions that proceed through neutral Schiff base or enol intermediates.
Let's examine first a reaction catalyzed by a so-called 'Class II' aldolase, in which a metal cation - generally Zn2+ - bound in the active site serves to stabilize the negative charge on an enolate intermediate. Fructose 1,6-bisphosphate aldolase is an enzyme that participates in both the glycolytic (sugar burning) and gluconeogenesis (sugar building) biochemical pathways. For now, we will concentrate on its role in the gluconeogenesis pathway, but we will see it again later in its glycolytic role. The reaction catalyzed by fructose 1,6-bisphosphate aldolase is a condensation between two 3-carbon sugars, glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone phosphate (DHAP), forming a six-carbon product (which leads, after three more enzymatic steps, to glucose).
In the first step of the condensation, an alpha-carbon on DHAP is deprotonated, leading to an enolate intermediate. The strategy used to stabilize this key intermediate is to coordinate the negatively-charged enolate oxygen to an enzyme-bound zinc cation.
Next, the deprotonated a-carbon attacks the carbonyl carbon of GAP in a nucleophilic addition reaction, and protonation of the resulting alcohol leads directly to the fructose 1,6-bisphosphate product.
As with many other nucleophilic carbonyl addition reactions, a new stereocenter is created in this reaction, as a planar, achiral carbonyl group is converted to a tetrahedral, chiral alcohol. The enzyme-catalyzed reaction, not surprisingly, is completely stereospecific: the DHAP substrate is positioned in the active site so as to attack the re (front)face of the GAP carbonyl group, leading to the R configuration at the new stereocenter.
Interestingly, it appears that in bacteria, the fructose bisphosphate aldolase enzyme evolved separately from the corresponding enzyme in plants and animals. In plants and animals, the same aldol condensation reaction is carried out by a significantly different mechanism, in which the key intermediate is not a zinc-stabilized enolate but an enamine. The nucleophilic substrate (DHAP) is first linked to the enzyme through the formation of an imine (also known as a Schiff base) with a lysine residue in the active site.
The alpha-proton is then abstracted by an active site base to form an enamine.
In the next step, the alpha-carbon attacks the carbonyl carbon of GAP, and the new carbon-carbon bond is formed. In order to release the product from the enzyme active site and free the enzyme to catalyze another reaction, the imine is hydrolyzed back to a ketone group.
There are many more examples of 'Class I' aldolase reactions in which the key intermediate is a lysine-linked imine. Many bacteria are able to incorporate formaldehyde, a toxic compound, into carbohydrate metabolism by condensing it with ribulose monophosphate. The reaction proceeds through imine and enamine intermediates.
Example 23.13.1
1. Propose a complete mechanism for the condensation reaction shown above.
2. Propose a complete mechanism for the conversion of hexulose-6-phosphate (formed from the condensation of ribulose-5-phosphate and formaldehyde) into fructose-6-phosphate.
Solution
Along with aldehydes and ketones, esters and thioesters can act as the nucleophilic partner in aldol condensations. In the first step of the citric acid (Krebs) cycle, acetyl CoA condenses with oxaloacetate to form (S)-citryl CoA.
Notice that in this aldol reaction, the nucleophilic intermediate is stabilized by protonation, rather than by formation of an imine (as in the Class I aldolases) or by a metal ion (as in the Class II aldolases).
Going backwards: the retroaldol reaction
Although aldol reactions play a very important role in the formation of new carbon-carbon bonds in metabolic pathways, it is important to emphasize that they are also highly reversible: in most cases, the energy level of starting compounds and products are very close. This means that, depending on metabolic conditions, aldolases can also catalyze retro-aldol reactions (the reverse of aldol condensations, in which carbon-carbon bonds are broken). Recall that fructose 1,6-bisphosphate aldolase (section 13.3B) is active in the direction of sugar breakdown (glycolysis) as well as sugar synthesis (gluconeogenesis). In the glycolytic direction, the enzyme catalyzes - either by zinc cation or by imine/enamine mechanisms, depending on the organism - the retro-aldol cleavage of fructose bisphosphate into DHAP and GAP.
The mechanism is the exact reverse of the condensation reaction. Shown below is the mechanism for a Zn2+ - dependent (Type II) retroaldol cleavage. Notice that in the retroaldol reaction, the enolate intermediate is the leaving group, rather than the nucleophile.
The key thing to keep in mind when looking for a possible retro-aldol mechanism is that, when the carbon-carbon bond breaks, the electrons must have some place to go, where they will be stabilized by resonance. Generally, this means that there must be a carbonyl or imine group on the next carbon. If there is no adjacent carbonyl or imine group, the carbon-carbon bond is not free to break.
Here are two more examples of retro-aldol reactions. Bacterial carbohydrate metabolism involves this reversible, class I retro-aldol cleavage: (Proc. Natl. Acad. Sci 2001, 98, 3679).
Example 23.13.2
Draw the structure of the enamine intermediate in the retroaldol reaction shown above.
Another interesting example is the retro-aldol cleavage of indole-3-glycerol phosphate, a step in the biosynthesis of tryptophan.
Look carefully at this reaction - how is the leaving group stabilized? There is an imine group involved, but no participation by an enzymatic lysine. The imine is 'built into' the starting compound, available from the initial tautomerization of the cyclic enamine group in indole-3-glycerol phosphate.
Example 23.13.3
Draw the reverse (aldol condensation) direction of the reaction above.
Solution
Going both ways: transaldolase
An enzyme called transaldolase, which is part of the 'pentose phosphate pathway' of carbohydrate metabolism, catalyzes an interesting combination of class I aldol and retro-aldol reactions. The overall reaction, which can proceed in either direction depending on metabolic requirements, converts 3- and 7-carbon sugars into 6- and 4-carbon sugars. Essentially, a 3-carbon unit breaks off from a ketone sugar (ketose) and then is condensed directly with an aldehyde sugar (aldose).
Let's follow the progress of the reaction in the left-to-right direction as depicted above. Because this is a class I aldolase, the first step is the formation of an imine linkage between the ketone carbon of fructose-6-phosphate (F6P) and a lysine group from the enzyme. The enzyme-substrate adduct then undergoes a retro-aldol step to free glyceralde-3-phosphate (GAP), which leaves the active site.
The second substrate, erythrose 4-phosphate (E4P), enters the active site, and an aldol condensation occurs between E4P and the 3-carbon fragment remaining from the cleavage of fructose-6-phosphate.
The final step is hydrolysis of the imine and subsequent dissociation of sedoheptulose 7-phosphate from the active site. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.13%3A_Some_Biological_Carbonyl_Condensation_Reactions.txt |
Concepts & Vocabulary
• Design multi-step syntheses in which the reactions introduced in this unit are used in conjunction with any of the reactions described in previous units.
• Solve road-map problems that require a knowledge of carbonyl condensation reactions.
• Define, and use in context, any of the key terms introduced.
23.1 Carbonyl Condensations - The Aldol Reaction
• A carbon-carbon bond forming reaction at the alpha carbon through an enolate nucleophile.
• The Aldol Reaction proceeds by first making an enolate nucelophile on an aldehyde (or ketone).
• The enolate nucleophile then dimerizes by attacking the carbonyl of the same aldehyde (or ketone).
• The final product is a beta-hydroxy aldehyde (or ketone).
23.2 Carbonyl Condensations versus Alpha Substitutions
• Carbonyl condensation reactions are a type of alpha substitution reaction.
• Both involve an enolate nucleophile and end with an alpha substitution product.
• In carbonyl condensations, the electrophile is another carbonyl compound.
• Carbonyl condensations are reversible and use a catalytic amount of base.
• The electrophile is already present in the reaction as the enolate is formed in carbonyl condensations.
• Alpha substitution reactions are more directional by design, since a full equivalent of base is used to generate the enolate.
• The electrophile is introduced after the enolate is generated.
23.3 Dehydration of Aldol Products - Synthesis of Enones
• The aldol reaction can undergo a dehydration (loss of water) to yield an α,β-unsaturated aldehyde or ketone.
• The additional stability provided by the conjugated carbonyl system makes some of the products thermodynamically driven.
• The of small products, such as water in this case, are termed condensations, so this reaction is aldol condesation.
• Heat promotes the condensation reactions.
• Under basic conditions, the β-elimination occurs through an E1cb mechanism.
• Under acidic conditions, the β-elimination occurs through an E1 or E2 mechanism.
23.4 Using Aldol Reactions in Synthesis
• If the target contains a β-hydroxy carbonyl compound or an α,β-unsaturated carbonyl compound, then synthetically think aldol reaction or condensation.
• To reverse the aldol condensation, break the enone at the double bond.
• To reverse the aldol reaction, break the C-C bond between the alpha carbon and the carbon attached to the hydroxy group.
23.5 Mixed Aldol Reactions
• Aldol condensations between different reactants are called mixed or crossed Claisen reactions.
• Multiple products are possible, so the success of the mixed aldol reactions depends on two things:
• The electrophile (or acceptor) is an aldehyde.
• The aldehyde acceptor has no alpha protons.
• Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols.
• The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
23.6 Intramolecular Aldol Reactions
• Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction.
• In these intramolecular reactions, two sets of α-hydrogens need to be considered and most ring forming reaction five and six membered rings are preferred.
23.7 The Claisen Condensation Reaction
• Esters can contain α hydrogens, so can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation.
• One ester reacts as a nucleophile while the other reacts as an electrophile.
• A new C-C bond is formed in the reaction to form a β-keto ester product.
• An alkoxide that matches the ester group is used to help prevent side reactions in the Claisen Condensation.
• There is a thermodynamic driving step forming an enolate, which is followed by a protonation to obtain the neutral product.
23.8 Mixed Claisen Condensation Reactions
• Claisen condensations between different ester reactants are called Crossed Claisen reactions.
• Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures.
• Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens.
23.9 Intramolecular Claisen Condensation Reactions - The Dieckmann Cyclization
• A diester can undergo an intramolecular reaction called a Dieckmann condensation.
23.10 Conjugate Carbonyl Additions - The Michael Reaction
• In 1,4 additions the nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
• Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds (product of the Aldol condensation) is a process called a Michael addition.
• A new C-C bond is formed between an enolate and the 4-C of the α, β-unsaturated carbonyl compound.
• The product is a 1,5-dicarbonyl species.
23.11 Carbonyl Condensations with Enamines - The Stork Reaction
• Aldehydes and ketones react with 2o amines to reversibly form enamines.
• Enamines act as nucleophiles similar to enolates.
• This process requires a three steps:
• Formation of the enamine
• Reaction with an eletrophile to form an iminium salt
• Hydrolysis of the iminium salt to reform the aldehyde or ketone
• Advantages of using an enamine over an enolate
• Neutral
• Easier to prepare
• Prevent overreaction issue that occurs when using enolates.
• Enamines undergo an SN2 reaction with alkyl halides to yield the iminium salt.
• Enamines can react with acid halides to form β-dicarbonyls.
• Enamines can also be used as a nucleophile in a Michael reaction.
23.12 The Robinson Annulation Reaction
• Forms a cyclic product from acyclic starting materials by first creating a new C-C bond followed by ring formation.
• The Robinson Annulation reaction starts with a Michael reaction followed by an intramolecular Aldol condensation.
• The formation of 5 or 6 membered rings is preferred.
23.13 Some Biological Carbonyl Condensation Reactions
• Aldol reactions occur in several biological pathways.
• Enzymes that catalyze aldol reactions are called, not surprisingly, 'aldolases'.
• The first step in an aldolase reaction is the deprotonation of an alpha-carbon to generate a nucleophilic carbanion.
• Nature has evolved several distinct strategies to stabilize the intermediate that results.
• Some aldolases use a metal ion to stabilize the negative charge on an enolate intermediate.
• Others catalyze reactions that proceed through neutral Schiff base or enol intermediates.
Skills to Master
• Skill 23.1 Identify the product of an aldol reaction.
• Skill 23.2 Write the detailed mechanism for the aldol reaction.
• Skill 23.3 Describe the difference between a carbonyl condensation reaction and an alpha-substitution reaction.
• Skill 23.4 Know what an enone is.
• Skill 23.5 Write a detailed mechanism for the aldol condensation under basic conditions.
• Skill 23.6 Write a detailed mechanism for the aldol condensation under acidic conditions.
• Skill 23.7 Determine the products of an aldol condensation reaction.
• Skill 23.8 Provide the reactants for a target aldol condensation product.
• Skill 23.9 Write an equation to illustrate a mixed aldol reaction.
• Skill 23.10 Identify the structural features necessary to ensure that two carbonyl compounds will react together in a mixed aldol reaction to give a single product rather than a mixture of products.
• Skill 23.11 Determine whether a given mixed aldol reaction is likely to produce a single product or a mixture of products.
• Skill 23.12 Identify the carbonyl compounds needed to produce a given enone or β‑hydroxy aldehyde or ketone by a mixed aldol reaction.
• Skill 23.13 Write a detailed mechanism for the intramolecular aldol condensation.
• Skill 23.14 Write an equation to illustrate a Claisen condensation reaction.
• Skill 23.15 Write a detailed mechanism for a Claisen condensation reaction or its reverse.
• Skill 23.16 Identify the ester and other reagents needed to form a given β‑keto ester by a Claisen condensation reaction.
• Skill 23.17 Write an equation to illustrate a mixed Claisen condensation.
• Skill 23.18 Identify the structural features that should be present in the two esters if a mixed Claisen condensation is to be successful.
• Skill 23.19 Identify the product formed when a given pair of esters is used in a mixed Claisen condensation.
• Skill 23.20 Identify the esters that should be used to produce a given β‑keto ester by a mixed Claisen condensation.
• Skill 23.21 Write detailed mechanisms for mixed Claisen condensation.
• Skill 23.22 Write a detailed mechanism for an intramolecular Claisen condensation.
• Skill 23.23 Write a detailed mechanism for a given typical Michael reaction.
• Skill 23.24 Identify the product formed in a given Michael reaction.
• Skill 23.25 Identify the reagents necessary to synthesize a given compound by a Michael reaction.
• Skill 23.26 Write a detailed mechanism for each of the three steps of the Stork enamine reaction.
• Skill 23.27 Identify the reagents needed to synthesize a given compound by a Stork enamine reaction.
• Skill 23.28 Write a detailed mechanism for the Robinson Annulation reaction.
• Skill 23.29 Identify the product from an enone and an dicarbonyl under basic conditions.
• Skill 23.30 Identify the steps in which a carbonyl condensation reaction has occurred, given a general outline of a specific biosynthesis.
Summary of Reactions
Stork Enamine Reactions | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/23%3A_Carbonyl_Condensation_Reactions/23.S%3A_Carbonyl_Condensation_Reactions_%28Summary%29.txt |
Learning Objectives
When you have completed Chapter 24, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. design a multi-step synthesis that involves the use of any of the reactions described in this unit, and any of the reactions described in any previous unit.
3. solve road-map problems that require a knowledge of amine chemistry in addition to any of the chemistry discussed in previous units.
4. define, and use in context, the key terms introduced.
Amines are the first nitrogen-containing compounds that we study in detail in this course. We begin the chapter with an explanation of the differences in structure among primary, secondary and tertiary amines. We explain the nomenclature of aliphatic and arylamines, and examine the structure and bonding of these compounds, relating these features to their physical properties and basicity. We describe the use of amines to resolve racemic mixtures of chiral carboxylic acids.
Amines may be prepared by a number of different synthetic methods. We describe each of these methods and assess the relative merits of each. After a description of the reactions of aliphatic amines, we devote sections to a discussion of the use of tetraalkylammonium salts as phase-transfer agents. The chapter concludes with a summary of the spectroscopic properties of amines.
24: Amines and Heterocycles
Objectives
After completing this section, you should be able to
1. classify a given amine as being primary, secondary or tertiary.
2. explain, briefly, the difference in meaning of the terms primary, secondary and tertiary when they are applied to the structures of amines and alcohols.
3. determine whether a given structure represents a quaternary ammonium cation.
4. provide an acceptable IUPAC name for an amine, given its Kekulé, condensed or shorthand structure.
5. draw the structure of an amine, given its IUPAC name.
6. give the name and structure of one typical heterocyclic amine (e.g., pyridine).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amine
• primary amine
• secondary amine
• quaternary ammonium cation
• tertiary amine
Study Notes
You should recognize that heterocyclic amines—compounds in which the nitrogen atom occurs as part of a ring—are very common in organic chemistry. Be prepared to meet such compounds throughout this, and subsequent chapters, but do not try to memorize all of the names and structures givenin the reading. You should, however, commit the structures of pyridine and pyrrole to memory.
Classification of Amines
Amines are made up of an sp3 hybridized nitrogen and are either alkyl substituted (alkylamines) or aryl substituted (arylamines).
Amines are classified as primary, secondary, or tertiary based on the number or organic substituents directly attached to the nitrogen. An amine attached to one substituent is primary (1o), two substituents is secondary (2o), and three substituents is tertiary (3o).
Example
Amines attached to 4 substituents and at least one substituent is a hydrogen are called ammonium salts. Amines attached to 4 alkyl substituents are called quaternary ammonium salts. The nitrogen in ammonium salts lacks lone pair electrons and carries a formal positive charge. Ammonium salts are also require a negatively charged counter-ion which can vary in composition.
Nomenclature of Amines
Nomenclature of Primary Amines
Primary amines are named in two main ways using the IUPAC system. They can either be named as alkylamines or as alkanamines. Most 1o amines which are attached to linear alkanes, cycloalkanes, and alkyl groups with common names (Section 3.3), tend to be named as alkylamines. The alkyl groups is named as a substituent (prefix + yl) then the suffix -amine is added. Many amines have common names that are used by IUPAC, for example the primary arylamine (C6H5NH2) is called aniline.
Other primary amines tend to be named as alkanamines. The alkyl group is named as an alkane (prefix+ane) and –e ending is replaced with the suffix -amine. The –e ending is not removed for diamines.
Nomenclature of Symmetrical Secondary and Tertiary Amines
Symmetrical 2o and 3o amines (where all substituents are identical) are named as alkylamines and the prefix di- or tri -added to indicate the number of substituents.
Nomenclature of Unsymmetrical Secondary and Tertiary Amines
Unsymmetrical 2o and 3o amines are named with the largest chain being the base chain (prefix+yl+amine). The other alkyl groups are named as N-substituents. This notation is used to indicate that the substituent is attached to the amine nitrogen and not an alkyl carbon.
Nomenclature of Ammonium Salts
Ammonium salts and quaternary ammonium salts are named using the same rules as 2o and 3o amines except the –amine suffix is replaces with -ammonium + the name of the counter ion. The counter ion name is separated with a space.
Amines have one of the lower functional group priorities in the IUPAC nomenclature system. When present in a compound with a functional group of higher priority, the amine group is named as a substituent called “amino.”
Heterocyclic amine have one or more nitrogens as part of the ring and can be aliphatic or aromatic. Most heterocyclic amine ring systems have a common name and are numbered such that a nitrogen always gets position 1. An amine attached to a heterocyclic ring is named as an amino substituent.
Example
Exercises \(1\)
1) Name the following compounds:
a)
b)
c)
d)
e)
f)
2) Draw the structures corresponding to the following names:
a) 3-Bromo-pentan-2-amine
b) Cyclopentanamine
c) Trans-3-ethylcyclohexanamine
d) Sec-butyl-tert-butyl amine
e) N,N-Dimethyl-3-pentanamine
f) 4-Methyl-2-hexanamine
g) 6-Bromo-4-amino-2-heptanol
3) Draw the structures of the following heterocyclic compounds:
a) 4-Methoxyindole
b) 1,4-Dimethylpyrrole
c) 3-(N,N-Dimethylamino)pyridine
d) 2-Aminopyrimidine
Answers
1)
a) N-Methylpropylamine
b) Dicyclopentylamine
c) 1,4-Pentanediamine
d) 4-Methylpyridine
e) Triethylammonium Bromide
f) N,N-Dimethylcyclohexylamine
2)
a)
b)
c)
d)
e)
f)
g)
3)
a)
b)
c)
d) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.01%3A_Naming_Amines.txt |
Objectives
After completing this section, you should be able to
1. describe the geometry and bonding of simple amines.
2. explain why most chiral amines cannot be resolved into their two enantiomers.
Key Terms
Make certain that you can define, and use in context, the key term below.
Study Notes
Molecular models may help you to understand pyramidal inversion.
Bonding, Shape, and Hybridization of Amines
The nitrogen atom in most amines is sp3 hybridized. Three of the sp3 hybrid orbitals form sigma bonds with the fourth orbital carrying the lone pair electrons. Single-bonded nitrogen is trigonal pyramidal in shape, with the non-bonding lone electron pair pointing to the unoccupied corner of a tetrahedron. Due to crowding by the lone pair electrons, the C-N-C bond angle between alkyl substituents on an amine is roughly 108o which is slightly less than the 109.5o bond angle for a perfect tetrahedral geometry.
The C-N bonds (147 pm) of non-conjugated amines is shorter than C-C bond (154 pm) in alkanes, and longer than the C-O bond (143 pm) present in alcohols. This difference in bond length is expected given that the relative atom sizes are C > O > N.
Chirality of Amines
Due to their tetrahedral configuration, amines with three different substituents are chiral. The R and S enantiomeric forms of a chiral amine cannot be resolved due to their rapid interconversion by a process called pyramidal or nitrogen inversion. During the inversion, the sp3 hybridized amine momentarily rehybridizes to a sp2 hybridized, trigonal planar, transition state where the lone pair electrons occupy a p orbital. The nitrogen then returns to a tetrahedral sp3 hybridization causing the lone pair electrons to return to a sp3 orbital on the opposite side of the nitrogen. During this process substituents invert to form the enantiomer. The thermodynamic barrier for this inversion (~25 kJ/mol) is low enough to allow rapid inversion at room temperature, leading to a mixture of interconverting R and S configurations. At room temperature a nitrogen atom exists as a racemic mixture of R and S configurations at equilibrium.
Quarternary amines lack lone pair electrons and therefore do not undergo pyramidal inversions. Quarternary amines with four different substituents are chiral and are readily resolved into separate enantiomers.
Boiling Point and Water Solubility
Methyl, dimethyl, trimethyl, and ethyl amines are gases under standard conditions. Most common alkyl amines are liquids, and high molecular weight amines are, quite naturally, solids at standard temperatures.
It is instructive to compare the boiling points and water solubility of amines with those of corresponding alcohols and ethers. The dominant factor here is hydrogen bonding, and the first table below documents the powerful intermolecular attraction that results from -O-H---O- hydrogen bonding in alcohols (light blue columns). Corresponding -N-H---N- hydrogen bonding is weaker, as the lower boiling points of similarly sized amines demonstrate. Alkanes provide reference compounds in which hydrogen bonding is not possible, and the increase in boiling point for equivalent 1º-amines is roughly half the increase observed for equivalent alcohols.
A Representation of the Hydrogen Bonding in Methyamine
Compound CH3CH3 CH3OH CH3NH2 CH3CH2CH3 CH3CH2OH CH3CH2NH2
Mol.Wt. 30 32 31 44 46 45
Boiling
Point ºC
-88.6º 65º -6.0º -42º 78.5º 16.6º
The second table illustrates differences associated with isomeric 1º, 2º & 3º-amines, as well as the influence of chain branching. Since 1º-amines have two hydrogens available for hydrogen bonding, we expect them to have higher boiling points than isomeric 2º-amines, which in turn should boil higher than isomeric 3º-amines (no hydrogen bonding). Indeed, 3º-amines have boiling points similar to equivalent sized ethers; and in all but the smallest compounds, corresponding ethers, 3º-amines and alkanes have similar boiling points. In the examples shown here, it is further demonstrated that chain branching reduces boiling points by 10 to 15 ºC.
Compound CH3(CH2)2CH3 CH3(CH2)2OH CH3(CH2)2NH2 CH3CH2NHCH3 (CH3)3CH (CH3)2CHOH (CH3)2CHNH2 (CH3)3N
Mol.Wt. 58 60 59 59 58 60 59 59
Boiling
Point ºC
-0.5º 97º 48º 37º -12º 82º 34º
Most aliphatic amines display some solubility in water, reflecting their ability to form hydrogen bonds. Solubility decreases proportionally with the increase in the number of carbon atoms in the molecule – especially when the carbon atom number is greater than six. Aliphatic amines also display significant solubility in organic solvents, especially in polar organic solvents.
The water solubility of 1º and 2º-amines is similar to that of comparable alcohols. As expected, the water solubility of 3º-amines and ethers is also similar. The basicity of amines allows them to be dissolved in dilute mineral acid solutions, and this property facilitates their separation from neutral compounds such as alcohols and hydrocarbons by partitioning between the phases of non-miscible solvents.
Naturally Occurring Amines
A large and widespread class of naturally occurring amines is known as alkaloids. The structures of the plant alkaloids are extraordinarily complex, yet they are related to the simple amines in being weak nitrogen bases. In fact, the first investigator to isolate an alkaloid in pure form was F. W. A. Sertürner who, in 1816, described morphine as basic, salt-forming, and ammonia-like. He used the term "organic alkali" from which is derived the name alkaloid. Alkaloids include compounds that may be classified as antimicrobial (quinine), as analgesics (morphine, codeine), as hallucinogens (mescaline, LSD), and as topical anesthetics (cocaine).
Certain amines and ammonium compounds play key roles in the function of the central nervous system as neurotransmitters and the balance of amines in the brain is critical for normal brain functioning. The amines acetylcholine chloride, adrenalin, and serotonin play important roles in nerve function in the human body.
Most amines have “interesting” odors. The simple ones smell very much like ammonia. Higher aliphatic amines smell like decaying meat. The stench of rotting meat is due in part to two diamines: putrescine and cadaverine. They are made during the decarboxylation of the amino acids, ornithine and lysine, by bacteria. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.02%3A_Structure_and_Properties_of_Amines.txt |
Objectives
After completing this section, you should be able to
1. account for the basicity and nucleophilicity of amines.
2. explain why amines are more basic than amides, and better nucleophiles.
3. describe how an amine can be extracted from a mixture that also contains neutral compounds illustrating the reactions which take place with appropriate equations.
4. explain why primary and secondary (but not tertiary) amines may be regarded as very weak acids, and illustrate the synthetic usefulness of the strong bases that can be formed from these weak acids.
Key Terms
Make certain that you can define, and use in context, the key term below.
• amide
Study Notes
The lone pair of electrons on the nitrogen atom of amines makes these compounds not only basic, but also good nucleophiles. Indeed, we have seen in past chapters that amines react with electrophiles in several polar reactions (see for example the nucleophilic addition of amines in the formation of imines and enamines in Section 19.8).
The ammonium ions of most simple aliphatic amines have a pKa of about 10 or 11. However, these simple amines are all more basic (i.e., have a higher pKa) than ammonia. Why? Remember that, relative to hydrogen, alkyl groups are electron releasing, and that the presence of an electron‑releasing group stabilizes ions carrying a positive charge. Thus, the free energy difference between an alkylamine and an alkylammonium ion is less than the free energy difference between ammonia and an ammonium ion; consequently, an alkylamine is more easily protonated than ammonia, and therefore the former has a higher pKa than the latter.
Basicity of nitrogen groups
In this section we consider the relative basicity of amines. When evaluating the basicity of a nitrogen-containing organic functional group, the central question we need to ask ourselves is: how reactive (and thus how basic and nucleophilic) is the lone pair on the nitrogen? In other words, how much does that lone pair want to break away from the nitrogen nucleus and form a new bond with a hydrogen. The lone pair electrons makes the nitrogen in amines electron dense, which is represents by a red color in the electrostatic potential map present below left. Amine are basic and easily react with the hydrogen of acids which are electron poor as seen below.
Amines are one of the only neutral functional groups which are considered basis which is a consequence of the presence of the lone pair electrons on the nitrogen. During an acid/base reaction the lone pair electrons attack an acidic hydrogen to form a N-H bond. This gives the nitrogen in the resulting ammonium salt four single bonds and a positive charge.
Amines react with water to establish an equilibrium where a proton is transferred to the amine to produce an ammonium salt and the hydroxide ion, as shown in the following general equation:
$RNH2_{(aq)}+H_2O_{(l)} \rightleftharpoons RNH3^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb), also called the base dissociation constant:
$K_b=\dfrac{[RNH3^+][OH^−]}{[NH2]} \label{16.5.5}$
pKb = -log Kb
Just as the acid strength of a carboxylic acid can be measured by defining an acidity constant Ka (Section 2-8), the base strength of an amine can be measured by defining an analogous basicity constant Kb. The larger the value of Kb and the smaller the value of pKb, the more favorable the proton-transfer equilibrium and the stronger the base.
However, Kb values are often not used to discuss relative basicity of amines. It is common to compare basicity's of amines by using the Ka's of their conjugate acids, which is the corresponding ammonium ion. Fortunately, the Ka and Kb values for amines are directly related.
Consider the reactions for a conjugate acid-base pair, RNH3+ − RNH2:
$\ce{RNH3+}(aq)+\ce{H2O}(l)⇌\ce{RNH2}(aq)+\ce{H3O+}(aq) \hspace{20px} K_\ce{a}=\ce{\dfrac{[RNH2][H3O]}{[RNH3+]}} \nonumber$
$\ce{RNH2}(aq)+\ce{H2O}(l)⇌\ce{RNH3+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=\ce{\dfrac{[RNH3+][OH-]}{[RNH2]}} \nonumber$
Adding these two chemical equations together yields the equation for the autoionization for water:
$\cancel{\ce{RNH3+}(aq)}+\ce{H2O}(l)+\cancel{\ce{RNH2}(aq)}+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\cancel{\ce{RNH2}(aq)}+\ce{OH-}(aq)+\cancel{\ce{RNH3+}(aq)} \nonumber$
$\ce{2H2O}(l)⇌\ce{H3O+}(aq)+\ce{OH-}(aq) \nonumber$
Given that the K expression for a chemical equation formed from adding two or more other equations is the mathematical product of the input equations’ K constants.
Ka X Kb = {2 H2O} / (H3O+}{OH-} = Kw
$K_\ce{a}=\dfrac{K_\ce{w}}{K_\ce{b}} \nonumber$
pKa + pKb =14
Thus if the Ka for an ammonium ion is know the Kb for the corresponding amine can be calculated using the equation Kb = Kw / Ka. This relationship shows that as an ammonium ion becomes more acidic (Ka increases / pKa decreases) the correspond base becomes weaker (Kb decreases / pKb increases)
Weaker Base = Larger Ka and Smaller pKa of the Ammonium ion
Stronger Base = Smaller Ka and Larger pKa of the Ammonium ion
Like ammonia, most amines are Brønsted-Lowry and Lewis bases, but their base strength can be changed enormously by substituents. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their aqueous solutions are basic (have a pH of 11 to 12, depending on concentration).
Aromatic herterocyclic amines (such as pyrimidine, pyridine, imidazole, pyrrole) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In each case the heterocyclic nitrogen is sp2 hybridized. The increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton compared to sp3 hybridized nitrogens. The very low basicity of pyrrole reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Imidazole (pKa = 6.95) is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity.
Basicity of common amines (pKa of the conjugate ammonium ions)
Inductive Effects in Nitrogen Basicity
Alkyl groups donate electrons to the more electronegative nitrogen. The inductive effect makes the electron density on the alkylamine's nitrogen greater than the nitrogen of ammonia. The small amount of extra negative charge built up on the nitrogen atom makes the lone pair even more attractive towards hydrogen ions. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
Compound pKa
NH3 9.3
CH3NH2 10.66
(CH3)2NH 10.74
(CH3)3N 9.81
Comparing the Basicity of Alkylamines to Amides
The nitrogen atom is strongly basic when it is in an amine, but not significantly basic when it is part of an amide group. While the electron lone pair of an amine nitrogen is localized in one place, the lone pair on an amide nitrogen is delocalized by resonance. The electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not as available for bonding with a proton – these two electrons are too stable being part of the delocalized pi-bonding system. The electrostatic potential map shows the effect of resonance on the basicity of an amide. The map shows that the electron density, shown in red, is almost completely shifted towards the oxygen. This greatly decreases the basicity of the lone pair electrons on the nitrogen in an amide.
Amine Extraction in the Laboratory
Extraction is often employed in organic chemistry to purify compounds. Liquid-liquid extractions take advantage of the difference in solubility of a substance in two immiscible liquids (e.g. ether and water). The two immiscible liquids used in an extraction process are (1) the solvent in which the solids are dissolved, and (2) the extracting solvent. The two immiscible liquids are then easily separated using a separatory funnel. For amines one can take advantage of their basicity by forming the protonated salt (RNH2+Cl), which is soluble in water. The salt will extract into the aqueous phase leaving behind neutral compounds in the non-aqueous phase. The aqueous layer is then treated with a base (NaOH) to regenerate the amine and NaCl. A second extraction-separation is then done to isolate the amine in the non-aqueous layer and leave behind NaCl in the aqueous layer.
Important Reagent Bases
The significance of all these acid-base relationships to practical organic chemistry lies in the need for organic bases of varying strength, as reagents tailored to the requirements of specific reactions. The common base sodium hydroxide is not soluble in many organic solvents, and is therefore not widely used as a reagent in organic reactions. Most base reagents are alkoxide salts, amines or amide salts. Since alcohols are much stronger acids than amines, their conjugate bases are weaker than amine bases, and fill the gap in base strength between amines and amide salts.
Base Name Pyridine Triethyl
Amine
Hünig's Base Barton's
Base
Potassium
t-Butoxide
Sodium HMDS LDA
Formula (C2H5)3N (CH3)3CO(–) K(+) [(CH3)3Si]2N(–) Na(+) [(CH3)2CH]2N(–) Li(+)
pKa of conjugate acid 5.3 10.7 11.4 14 19 26 35.7
Basicity of common amines (pKa of the conjugate ammonium ions)
Pyridine is commonly used as an acid scavenger in reactions that produce mineral acid co-products. Its basicity and nucleophilicity may be modified by steric hindrance, as in the case of 2,6-dimethylpyridine (pKa=6.7), or resonance stabilization, as in the case of 4-dimethylaminopyridine (pKa=9.7). Hünig's base is relatively non-nucleophilic (due to steric hindrance), and is often used as the base in E2 elimination reactions conducted in non-polar solvents. Barton's base is a strong, poorly-nucleophilic, neutral base that serves in cases where electrophilic substitution of other amine bases is a problem. The alkoxides are stronger bases that are often used in the corresponding alcohol as solvent, or for greater reactivity in DMSO. Finally, the two amide bases see widespread use in generating enolate bases from carbonyl compounds and other weak carbon acids.
In addition to acting as a base, 1o and 2o amines can act as very weak acids. Their N-H proton can be removed if they are reacted with a strong enough base. An example is the formation of lithium diisopropylamide (LDA, LiN[CH(CH3)2]2) by reacting n-butyllithium with diisopropylamine (pKa 36) (Section 22-5). LDA is a very strong base and is commonly used to create enolate ions by deprotonating an alpha-hydrogen from carbonyl compounds (Section 22-7).
Exercises
Q24.3.1
Select the more basic amine from each of the following pairs of compounds.
(a)
(b)
(c)
Q24.3.2
The 4-methylbenzylammonium ion has a pKa of 9.51, and the butylammonium ion has a pKa of 10.59. Which is more basic? What's the pKb for each compound?
Solutions
S24.3.1
(a)
(b)
(c)
S24.3.2
The butylammonium is more basic. The pKb for butylammonium is 3.41, the pKb for 4-methylbenzylammonium is 4.49 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.03%3A_Basicity_of_Amines.txt |
Objectives
After completing this section, you should be able to
1. use the concept of resonance to explain why arylamines are less basic than their aliphatic counterparts.
2. arrange a given series of arylamines in order of increasing or decreasing basicity.
3. discuss, in terms of inductive and resonance effects, why a given arylamine is more or less basic than aniline.
Study Notes
With reference to the discussion of base strength, the traditional explanation for the base‑strengthening effect of electron‑releasing (I) substituents is that such substituents help to stabilize the positive charge on an arylammonium ion more than they stabilize the unprotonated compound, thereby lowering ΔG°.
The electron‑withdrawing (i.e., deactivating) substituents decrease the stability of a positively charged arylammonium ion.
Note that the arylammonium ion derived from aniline, PhNH3+, is commonly referred to as the anilinium ion.
Basicity of Aniline
Aniline is substantially less basic than methylamine, as is evident by looking at the pKa values for their respective ammonium conjugate acids (remember that the lower the pKa of the conjugate acid, the weaker the base).
This difference is basicity can be explained by the observation that, in aniline, the lone pair of electrons on the nitrogen are delocalized by the aromatic p system, making it less available for bonding to H+ and thus less basic. The lone pair electrons of aniline are involved in four resonance forms making them more stable and therefore less reactive relative to alkylamines.
The effect of delocalization can be seen when viewing the electrostatic potential maps of aniline an methyl amine. The nitrogen of methyl amine has a significant amount of electron density on its nitrogen, shown as a red color, which accounts for it basicity compared to aniline. While the electron density of aniline's nitrogen is delocalized in the aromatic ring making it less basic.
Basicity of Substituted Arylamines
The addition of substituents onto the aromatic ring can can make arylamines more or less basic. Substituents which are electron-withdrawing (-Cl, -CF3, -CN, -NO2) decrease the electron density in the aromatic ring and on the amine making the arylamine less basic. In particular, the nitro group of para-nitroaniline allows for an additional resonance form to be drawn, which further stabilizes the lone pair electrons from the nitrogen, making the substituted arylamine less basic than aniline. This effect is analogous to the one discussed for the acidity of substituted phenols in Section 17.2.
Substituents which are electron-donating (-CH3, -OCH3, -NH2) increase the electron density in the aromatic ring and on the amine making the arylamine more basic. In the case of para-methoxyaniline, the lone pair on the methoxy group donates electron density to the aromatic system, and a resonance contributor can be drawn in which a negative charge is placed on the carbon adjacent to the nitrogen, which makes the substituted arylamine more basic than aniline.
Increased Basicity of para-Methoxyaniline due to Electron-Donation
The shifting electron density of aniline, p-nitroaniline, and p-methoxyaniline are seen in their relative electrostatic potential maps. For p-Nitroaniline virtually all of the electron density, shown as a red/yellow color. is pulled toward the electron-withdrawing nitro group. In p-methoxyaninline the electron donating methoxy group donates electron density into the ring. The amine in p-methoxyaniline is shown to have more electron density, shown as a yellow color, when compared to the amine in aniline.
Exercise s\(1\)
Using the knowledge of the electron donating or withdrawing effects of subsituents gained in Section 16.6, rank the following compound in order of decreasing basicity.
1. p-Nitroaniline, methyl p-aminobenzoate, p-chloroaniline
2. p-Bromoaniline, p-Aminobenzonitrile, p-ethylaniline
3. p-(Trifluoromethyl)aniline, p-methoxyaniline, p-methylaniline
Answers
1. p-Chloroaniline, methyl p-aminobenzoate, p-nitroaniline
2. p-Ethylaniline, p-Bromoaniline, p-aminobenzonitrile
3. p-Methoxyaniline, p-methylaniline, p-(trifluoromethyl)aniline
24.05: Biological Amines and the Henderson-Hasselbalch Equation
Objectives
1. identify the form that amine bases take within living cells.
2. use the Henderson‑Hasselbalch equation to calculate the percentage of a base that is protonated in a solution, given the pKa value of the associated ion and the pH of the solution.
3. explain why organic chemists write cellular amines in their protonated form and amino acids in their ammonium carboxylate form.
The Henderson-Hasselbalch equation is very useful relating the pKa of a buffered solution to the relative amounts of an acid and its conjugate base. In Section 20-3, we used the Henderson-Hasselbalch equation to show that under physiological pH, carboxylic acids are almost completely dissociated into their carboxylate ions.
The Henderson-Hasselbalch equation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \left(\frac{\text { concentration of conjugate base }}{\text { concentration of weak acid }}\right) \nonumber$
So, what does the side chain of a lysine amino acid residue look like if it is on the surface of a protein in an aqueous solution buffered pH 7.0? Is it protonated or deprotonated? The values in the Henderson-Hasselbalch can be used for an amine with the ammonium salt written as, HA = RNH3+, and the amine as being, A- = RNH2. With an approximate pKa of 10.8 for the protonated amine HA, it should be >99% protonated, in the positively-charged, ammonium form:
7.0 = 10.8 + log ([RNH2] / [RNH3+])
([RNH2] / [RNH3+]) = 1.6 x 10-4
. . . so, [RNH3+] >> [RNH2]
So, in an aqueous solution buffered at pH 7, carboxylic acid groups can be expected to be essentially 100% deprotonated and negatively charged (ie. in the carboxylate form), and amine groups essentially 100% protonated and positively charged (i.e., in the ammonium form). Alcohols are fully protonated and neutral at pH 7, as are thiols. The imidazole group on the histidine side chain has a pKa near 7, and thus exists in physiological solutions as mixture of both protonated and deprotonated forms.
Exercise $1$
Would you expect an aromatic hetererocycle, pyrrole, to be protonated at pH = 7.3? Use the Henderson-Hasselbalch equation to determine your answer. pKa of protonated pyrrole is 0.4.
Answer
7.3 = 0.4 + log ([RNH2] / [RNH3+])
([RNH2] / [RNH3+]) = 7.9 x 106
. . . so, [RNH2] >> [RNH3+] so pyrrole would be almost completely unprotonated at pH = 7. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.04%3A_Basicity_of_Arylamines.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate the synthesis of amines by
1. reduction of nitriles or amides and nitro compounds.
2. reactions involving alkyl groups:
1. SN2 reactions of alkyl halides, ammonia and other amines.
2. nucleophilic attack by an azide ion on an alkyl halide, followed by reduction of the azide so formed.
3. alkylation of potassium phthalimide, followed by hydrolysis of the N‑alkyl phthalimide so formed (i.e., the Gabriel synthesis).
3. reductive amination of aldehydes or ketones.
4. Hofmann or Curtius rearrangements.
2. write detailed mechanisms for each of the steps involved in the synthetic routes outlined in Objective 1.
3. identify the product or products formed when
1. a given nitrile or amide is reduced using lithium aluminum hydride.
2. a given alkyl halide is reacted with ammonia or an alkylamine.
3. a given alkyl halide is reacted with azide ion and the resulting product is reduced.
4. a given alkyl halide is reacted with potassium phthalimide and the resulting product is hydrolyzed.
5. a given aldehyde or ketone is reacted with ammonia or an amine in the presence of nickel catalyst.
6. a given amide is treated with halogen and base.
7. a given acyl azide is heated and then hydrolyzed.
4. identify the starting material, the other reagents, or both, needed to synthesize a given amine by any of the routes listed in Objective 1.
5. write a general equation to illustrate the preparation of an arylamine by the reduction of a nitro compound, and balance such an equation.
6. identify the product formed from the reduction of a given aromatic nitro compound.
7. identify the organic compound, the inorganic reagents, or both, needed to prepare a given arylamine.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• azide synthesis
• Curtius rearrangement
• Hofmann rearrangement
• Gabriel synthesis
• imide
• reductive amination
Study Notes
You may wish to review the mechanism of SN2 reactions which is discussed in some detail in Sections 11.2 and 11.3.
Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3, is as shown below.
An “imide” is a compound in which an N\$\ce{-}\$H group is attached to two carbonyl groups; that is,
You should note the commonly used trivial names of the following compounds.
The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.
If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).
Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).
The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.
Reduction of Nitriles, Amides, and Nitro Compounds
Acid chlorides react with ammonia to give amides, by an addition-elimination path, and these are reduced to amines by LiAlH4 (Section 21-7).
Alkyl halides can be converted to primary amines through a two-step process. First an SN2 reaction with a cyanide anion converts the alkyl halide into a nitrile. Then the nitrile is reduced to a primary amine by LiAlH4 (Section 20-7). During this reaction sequence an additional carbon atom is added.
Arylamines are typically prepared by the reduction of a nitro group on an aromatic ring. Several methods for reducing nitro groups to amines are known. These include catalytic hydrogenation (H2 + platinum catalyst), and zinc, iron, or tin(II) chloride in dilute mineral acid. The procedures described above are sufficient for most cases. Catalytic hydrogenation can be problematic because it is known to reduce other functional groups such as alkenes, alkynes, and some carbonyl groups.
Exercise \(1\)
Propose structures of the starting materials needed to make the the following amines using the reduction of a nitrile or an amide:
a) CH3CH2CH2CH2NH2
b) (CH3CH2)2NH
c)
e) N-Propylaniline
Answer
a) CH3CH2CH2CONH2
b) CH3CONHCH2CH3
c)
e)
SN2 Reactions of Alkyl Halides
Because they posses lone pair electrons, ammonia and amines are considered good nucleophiles. Ammonia and amines can by alkylated by an SN2 reaction with alkyl halides. The product of the reaction is an ammonium salt where the negative counter-ion is the halogen of the alkyl halide. When the alkylated products are 1o, 2o, and 3o amines, they can be deprotonated with NaOH to produce the neutral amine. When a 3o amine is alkylated, a quaternary ammonium salt is produced.
Alkylation is an efficient method for the synthesis of 3o and 4o amines. However, when 1o and 2o amines are alkylated a mixture of products is typically produced. When ammonia is reacted with an alkylhalide an monoalkylammonium salt is formed.
RX + NH3 → RNH3+ + X-
It is possible for any remaining ammonia present in the reaction to deprotonate the ammonium salt to produce the neutral monoalkyl amine.
RNH3+ + NH3 → RNH2 + NH4+
The monoalkyl amine is then free to react with a second alkyl halide creating a dialkylammonium salt. This process can be repeated to to eventually to create a tetralkylammonium salt. Typically when a alkyl halide is reacted with ammonia, a mixture of mostly 1o and 2o amines, with trace amounts of 3o amines and 4o ammonium salts, is produced.
RNH2 + RX → R2NH2+ + X-
Example
To synthesize a 1o or 2o amine, specific reactions are usually employed. A more efficient method starts with an SN2 reaction between a 1o or 2o alkyl halide and the nucleophilic azide anion (N3-) to produce an alkyl azide. The alkyl azide is not nucleophilic so it cannot react with additional alkyl halide to produce overalkylation. Alkyl azide is then reduced with LiAlH4 to produce a 1o amine.
Example
Gabriel Amine Synthesis
Another common method for the synthesis of 1o amines is called the Gabriel amine synthesis. This reaction starts with the deprotonation of phthalimide by a hydroxide base such as potassium hydroxide (KOH). The N-H hydrogen of an imide functional group is acidic because its conjugate base is resonance stabilized by two carbonyl groups. The phthalimide anion is nucleophilic and easily alkylated through an SN2 reaction with an alkyl halide. The resulting N-alkylated phthalimide then undergoes base hydrolysis to produce a 1o amine product. The mechanism for the base promoted hydrolysis of the N-alkylated phthalimide is analogous to the hydrolysis of an amide (Section 21-7).
Example
Exercises \(1\)
1) Write the mechanism for the base promoted hydrolysis of an N-alkyl phthalimide to create a 1o amine and the phthalate ion.
2) Starting with any alkyl halide, show two methods to synthesize the neurotransmitter serotonin.
Answers
1)
2)
Reductive Amination
Aldehydes and ketones can be converted into 1o, 2o and 3o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of ammonia, a 1o amine, or a 2o amine to a carbonyl group to form an imine (Section 19-8). The second step is the reduction of the imine to an amine using a hydride reducing agent. Some common reducing agents used for this reaction are: sodium borohydride (NaBH4), sodium cyanoborohydride (NaBH3CN), NaBH(OAc)3, or hydrogen gas (H2) over a nickel catalyst.
General Reaction
Mechanism
Reductive amination starts with the nucleophilic addition of ammonia or a 1o amine to an aldehyde or ketone forms a cyanohydrin intermediate. Subsequent dehydration forms an imine intermediate (Section 19-8). The imine undergoes hydride reduction in a similar fashion as C=O carbonyl bonds are reduced to an alcohol.
When a 2o amine is used for reductive amination, an enamine intermediate is formed which then undergoes hydride reduction to form a 3o amine.
Predicting the Products of a Reductive Amination
Examples
Biological Reductive Aminations
Reductive amination is used in the biosynthesis of the amino acid proline. The enzyme pyrroline-5-carboxylate synthase (P5CS) catalyzes glutamate undergoing an intramolecular imine formation to produce 1-pyrrolinium 5-carboxylate. Then the enzyme pyrroline-5-carboxylate reductase (PYCR) catalyzed the nucleophilic hydride reduction of the C=N bond to produce proline. The biological hydride reducing agent for this sequence is reduced nicotinamide adenine dinucleotide (NADH).
Planning a Synthesis using a Reductive Amination
If you are trying to develop a synthesis of an amine molecule, the key bond break is a C-N bond in the target molecule. There may be multiple C-N bonds present, so each should be broken separately to produce a possible set of starting materials. Because the amine is commonly used in excess during a reductive animation, the pathway which provides the simplest amine starting material is preferred. The carbon from the broken C-N bond goes on to become a carbonyl starting material. The nitrogen gains a hydrogen to become the amine starting material.
Worked Exercise \(1\)
Show how N-methylbenzylamine can be synthesized using a reductive amination. If more than one pathway is possible, draw them both and determine which one would be preferred.
Answer
Pathway 1
Solution 1
Pathway 2
Solution 2
Because N-methylbenzylamine has two C-N bonds there are two possible synthesis pathways. Pathway 1 is most likely preferred because it uses the simplest amine starting material, methyl amine.
Exercise \(1\)
Provide starting materials which could be used to prepare the following compounds using a reductive amination. Show all possible pathways.
1)
a)
b)
c)
2) Please give the starting materials which could provide the following molecule using a reductive amination.
Answers
1)
a)
b)
c)
2)
Hofmann Rearrangement
The Hofmann rearrangement occurs when a 1o amide is reacted with bromine (Br2) and a base. The products are a 1o amine with one less carbon and carbon dioxide (CO2).
General Reaction
Example
Mechanism
The mechanism for the Hofmann rearrangement is quite complex. The mechanism starts with the deprontation of the primary amide by a base. The nitrogen is brominated during an SN2 reaction with Br2 to produce an N-bromoamide. Bromide is eliminated as a leaving group to produce a electron deficient nitrene-like nitrogen. Electrons from the adjacent C-C bond migrate to the nitrogen as part of a rearrangment to finally produce an isocyanate. The carbon in the isocyante is electrophilic and reacts with a nucleophilic hydroxide to create a C-O bond. After movement of electrons through resonance and a proton transfer a carbamate intermediate is formed. The carbamate decomposes to create an amide anion and a stable CO2 molecule. Finally, water protonates the amide anion to produce a 1o amine.
Curtius Rearrangement
The Curtius Rearrangement is another method used to synthesize a 1o amine.
General Reaction
The Curtius rearrangement involves an acyl azide reacting with water and heat to produce a 1o amine along with CO2 and N2. The acyl azide is usually made by a nucleophilic substitution of an acid chloride with sodium azide (NaN3)
Mechanism
The mechanism of the Curtius rearrangement is quite similar to the Hoffman rearrangement. The loss of N2 as a leaving group creates an electron deficient nitrogen. The adjacent -R group migrates to form an isocyanate. Reaction of the isocyanate forms a carbamate which decomposes to form a 1o amine and CO2.
Example
Planning a Synthesis using a Hoffman or Curtius Rearrangement
Because the amide and the acyl azide required for these rearrangements both are best made from an acid chloride, synthesis of amines using these reaction are best started from the corresponding carboxylic acid. For retrosynthetic analysis starting from the desired amine product - the key break is the C-NH2 bond. The amine is removed and replaced with a -CH2CO2H fragment to provide the required carboxylic acid starting material.
Worked Example \(1\)
Show how the following molecule could be prepared from a carboxylic acid starting material using both the Hoffman and Curtius rearrangements.
Answer
First determine the structure of the carboxylic acid starting material. This will be the starting material for both rearrangements. In the first step, convert the carboxylic acid into an acid halide using thionyl chloride (SOCl2). Then use the required conditions for each rearrangement.
Solution
The top solution shows the Curtius Rearrangement, while the bottom is an example using the Hofmann Rearrangement.
Exercises \(1\)
1) Using a carboxylic acid starting material explain how to prepare the following molecules using both the Hoffman and Curtius rearrangements.
a)
b)
Answers
1)
a) The top solution shows the Curtius Rearrangement, while the bottom is an example using the Hofmann Rearrangement.
b) The top solution shows the Curtius Rearrangement, while the bottom is an example using the Hofmann Rearrangement. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.06%3A_Synthesis_of_Amines.txt |
Objectives
After completing this section, you should be able to
1. write an equation to represent the reaction that takes place between ammonia, a primary or secondary amine, and an acid chloride.
2. identify the product formed when a given amine reacts with a given acid chloride.
3. identify the amine, the acid chloride, or both, needed to synthesize a given amide.
4. write a reaction sequence to illustrate the overall conversion of an amine to an alkene via a Hofmann elimination.
5. identify the alkene most likely to be formed when a given quaternary ammonium salt is heated with moist silver oxide (or silver hydroxide).
6. deduce the structure of an unknown quaternary ammonium salt, given the identity of the alkene or alkenes produced when the salt is heated with moist silver oxide.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Hofmann elimination
Study Notes
The alkylation and acylation of amines have been dealt with in previous sections: alkylation in Section 24.6 and acylation in Sections 21.4 and 21.5. Review these sections if necessary.
One explanation is given for Hofmann elimination further on in this reading. Another argument, based on the consideration of the structures of the possible transition states, has been suggested by Joseph Bunnett (Professor Emeritus, University of California, Santa Cruz) and is widely accepted. We can begin to understand Bunnett’s reasoning by considering the elimination products formed from by the dehydrohalogenation of a number of substituted hexanes in the reaction
The table below describes the proportion of each product you would expect when X in the above reaction represents a given halogen.
X % 2‑hexene % hexene
I 81 19
Br 72 28
Cl 67 33
F 30 70
As we descend this table, the electron‑withdrawing ability of X increases, and the percentage of Hofmann product also increases. Given that the (CH3)3N+ group is strongly electron‑withdrawing, it is quite consistent for a compound of the type
to give 96% Hofmann product and 4% Zaitsev product in an E2 elimination reaction. But why does the proportion of Hofmann product increase as the electron‑withdrawing ability of X increases? Bunnett suggests that the transition state in an E2 process can vary from being “carbocation‑like” on one hand to “carbanion‑like” on the other.
In a carbocation‑like transition state, the C-X bond is broken to a greater extent than the C-H bond; in the central transition state both, the C-X bond and the C-H bond are broken to an equal extent; and in the carbanion‑like transition state, the C-H bond is broken to a greater extent than the C-X bond. In the latter case, a partial negative charge will develop on the carbon atom to which the hydrogen is attached, hence the term “carbanion‑like.” As we already know, any species or transition state bearing a full or partial negative charge is stabilized by the presence of electron‑withdrawing groups and destabilized by the presence of electron‑releasing groups.
Let us now consider the two possible transition states for an E2 elimination of (CH3)3NH+ from a quaternary ammonium hydroxide,
One transition state leads to Zaitsev elimination, the other to Hofmann elimination.
In the transition state leading to the Zaitsev product, the carbon atom carrying the partial negative charge is bonded to an electron‑releasing methyl group. This is clearly a less favourable situation than the one that exists in the transition state leading to Hofmann elimination, where there are no electron‑releasing groups attached to the carbon atom bearing the partial negative charge. Bunnett’s argument is that carbanion‑like transition states are much more likely when the atom or group X in a compound such as
is strongly electron withdrawing, because the C\$\ce{-}\$X bond becomes stronger as the electron‑withdrawing ability of X increases.
Thus, as we go along the series X = I, Br, Cl, F, (CH3)3N+, the electron‑withdrawing ability of X increases, the C-X bond becomes more difficult to break, the transition state becomes more carbanion‑like in character, and Hofmann elimination becomes more pronounced.
Alkylation of Amines
It is instructive to examine these nitrogen substitution reactions using the common alkyl halide class of electrophiles. Thus, reaction of a primary alkyl bromide with a large excess of ammonia yields the corresponding 1º-amine, presumably by an SN2 mechanism. The hydrogen bromide produced in the reaction combines with some of the excess ammonia, giving ammonium bromide as a by-product. Water does not normally react with 1º-alkyl halides to give alcohols, so the enhanced nucleophilicity of nitrogen relative to oxygen is clearly demonstrated.
2 RCH2Br + NH3 (large excess) RCH2NH2 + NH4(+) Br(–)
It follows that simple amines should also be more nucleophilic than their alcohol or ether equivalents. If, for example, we wish to carry out an SN2 reaction of an alcohol with an alkyl halide to produce an ether (the Williamson synthesis), it is necessary to convert the weakly nucleophilic alcohol to its more nucleophilic conjugate base for the reaction to occur. In contrast, amines react with alkyl halides directly to give N-alkylated products. Since this reaction produces HBr as a byproduct, hydrobromide salts of the alkylated amine or unreacted starting amine (in equilibrium) will also be formed.
2 RNH2 + C2H5Br RNHC2H5 + RNH3(+) Br(–) RNH2C2H5(+) Br(–) + RNH2
Unfortunately, the direct alkylation of 1º or 2º-amines to give a more substituted product does not proceed cleanly. If a 1:1 ratio of amine to alkyl halide is used, only 50% of the amine will react because the remaining amine will be tied up as an ammonium halide salt (remember that one equivalent of the strong acid HX is produced). If a 2:1 ratio of amine to alkylating agent is used, as in the above equation, the HX issue is solved, but another problem arises. Both the starting amine and the product amine are nucleophiles. Consequently, once the reaction has started, the product amine competes with the starting material in the later stages of alkylation, and some higher alkylated products are also formed. Even 3º-amines may be alkylated to form quaternary (4º) ammonium salts. When tetraalkyl ammonium salts are desired, as shown in the following example, Hünig's base (N,N-diisopropylethylamine) may be used to scavenge the HI produced in the three SN2 reactions. Steric hindrance prevents this 3º-amine (Hünig's base) from being methylated.
C6H5NH2 + 3 CH3I + Hünig's base C6H5N(CH3)3(+) I(–) + HI salt of Hünig's base
Acylation of Amines
Ammonia, 1o amines, and 2o amines react rapidly with acid chlorides or acid anhydrides to form 1o, 2o, and 3o amides respectively (Sections 21-4 and 21.5). These reactions typically take place rapidly at room temperature and provide high reaction yields.
The reaction is commonly run with a base, such as NaOH or pyridine, to neutralize the HCl produced. Overacylation does not occur because the lone pair electrons on the amide nitrogen are conjugated with the carbonyl making it less nucleophilic than an amine starting material.
Hofmann Elimination
Amine functions seldom serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Indeed, they are even less effective in this role than are hydroxyl and alkoxyl groups. As noted earlier, 1º and 2º-amines are much weaker acids than alcohols, so it is not surprising that it is difficult to force the nitrogen function to assume the role of a leaving group. In this context we note that the acidity of the potential ammonium leaving group is at least ten powers of ten less than that of an analogous oxonium species. However, amines can be turned into a good leaving group by alkylation with an alkyl halide to form a quarternary ammonium salt. Upon elimination the quarternary ammonium salt produces a stable 3o amine as a leaving group.
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to affect the E2-like elimination of a 3º-amine. For an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, which is similar to elimination reactions of alkyl halides. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction).
General Reaction
Mechanism
During the Hofmann elimination, the hydroxide counter ion acts as a base to remove a beta-hydrogen and cause an E2 elimination to form an alkene. A trialkyl amine is eliminated as a leaving group during this reaction.
When predicting the products of an E2 elimination, the more substituted alkene is expected to be the preferred product based on Zaitsev's rule. The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule. To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state.
The reason for the difference in products in a Hofmann elimination is most likely due to steric hindrance of the base. The large size of of the quaternary ammonium group forces the base to remove a hydrogen from the least sterically hindered position. This correspondingly leads to the alkene with the least number of substituents to be preferred.
Examples
Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to remove the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path.
Eliminations analogous to the Hofmann elimination are often seen in biological systems. In these cases, ammonium ions (protonated amines) are used instead of quaternary ammonium salts. An example is seen in the biosynthesis of nucleic acids where a biological base causes a Hofmann-like elimination of protonated adenylosuccinate to produce the alkene containing fumarate and adenosine monophosphate.
Worked Exercise \(1\)
Predict the preferred product of a Hofmann elimination with the following molecule.
Answer
Predicting the product of a Hofmann elimination can be difficult so it is best done stepwise.
1) Find the least sterically hindered beta hydrogens. Beta hydrogens are on the second carbon away from the nitrogen. The order of preferred hydrogens are 1o > 2o > 3o. If the molecule is presented as a line structure it is usually beneficial to convert to a condensed structure first.
2) Break the C-N bond of the alkyl group which contains the preferred hydrogens. This will form two fragments.
3) Remove one of the original beta hydrogens in the alkyl fragment then draw a C=C between the alpha and beta carbon. This produces the alkene product of the Hofmann elimination. Sometimes the amine fragment of a Hofmann elimination is of interest. Remember that prior to elimination the amine is exhaustively alkylated with CH3I. This means any hydrogens attached to amine will starting material will be removed and the amine product will have enough methyl (-CH3) groups added to become a tertiary amine. In this case two methyl groups are added to the nitrogen. The end result are the two products of a Hofmann elimination.
Exercises
Exercise \(1\)
Draw the product for a Hofmann elimination for each of the following molecules.
(a)
(b)
(c)
(d)
Answer
(a)
(b)
(c)
(d)
Exercise \(2\)
Draw the product of a Hofmann elimination for the following molecule.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.07%3A_Reactions_of_Amines.txt |
Objectives
After completing this section, you should be able to
1. identify the product formed when a given arylamine is reacted with aqueous bromine.
2. give an appropriate example to illustrate the high reactivity of arylamines in electrophilic aromatic substitution reactions.
3. explain why arylamines cannot be used in Friedel‑Crafts reactions.
1. show how the problems associated with carrying out electrophilic aromatic substitution reactions on arylamines can be circumvented by first converting the amine to an amide, and illustrate this process with an appropriate example.
2. outline a possible synthetic route for the preparation of a given sulfa drug.
3. identify the starting material, the necessary organic reagents, inorganic reagents, or both, and the intermediate compounds formed during the synthesis of a given sulfa drug.
4. design a multi‑step synthesis for a given compound in which it is necessary to protect the amino group by acetylation.
1. write a general equation to describe the formation of an arenediazonium salt.
2. identify the product formed when a given arenediazonium salt is reacted with any of the following compounds: copper(I) chloride, copper(I) bromide, sodium iodide, copper(I) cyanide, hot aqueous acid, hypophosphorous acid.
3. identify the arenediazonium salt, the inorganic reagents, or both, needed to produce a given compound by a diazonium replacement reaction.
1. illustrate, with appropriate examples, the importance to the synthetic chemist of the overall reaction sequence A nitration, B reduction, C diazotization, and D replacement.
2. show how the removal of an amino (or nitro) group from an aromatic ring through the reaction of an arenediazonium salt with hypophosphorous acid (H3PO2) can sometimes be of use in organic synthesis.
1. write a general equation to represent a diazonium coupling reaction.
2. write the detailed mechanism for the coupling reaction which takes place between arenediazonium salts and the electon‑rich aromatic rings of phenols and arylamines.
3. identify the product formed from the reaction of a given arenediazonium salt with a given arylamine or phenol.
4. identify the arenediazonium salt and the arylamine or phenol needed to prepare a given azo compound.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• arenediazonium salt
• azo compound
• Sandmeyer reaction
• sulfa drug
Study Notes
This section contains a considerable amount of new information. To absorb all of it, you should use the three subsections indicated in the reading: electrophilic aromatic substitution and overreaction of aniline (Objectives 1 and 2), the preparation of diazonium salts and the Sandmeyer reaction (Objectives 3 and 4), and diazonium coupling reactions (Objective 5).
The general process in which an aromatic amine is reacted with acetic anhydride, substituted, and then hydrolyzed is known as “protecting the amino group.”
Sulfa drugs have the general formula
Typical examples are sulfathiazole
and sulfapyridine
The reagent used to bring about the chlorosulfonation of acetanilide is chlorosulfuric acid
Sulfanilamide itself is toxic to humans, but derivatives of this compound, such as sulfathiazole, are less harmful to humans and are effective in killing many types of bacteria. The drugs work by “deceiving” the bacteria in the following way. To survive, many micro‑organisms use p‑aminobenzoic acid to synthesize folic acid, a coenzyme in a number of biochemical processes. These micro‑organisms cannot distinguish between sulfa drugs and p‑aminobenzoic acid; so, when the drug is administered, the bacteria use it to produce a compound which has a structure similar to that of folic acid, but which is unable to act as a coenzyme in essential biochemical processes. The result is that many of the bacteria’s amino acids and nucleotides cannot be made, and the bacteria die. Amino acids are discussed in Chapter 26; nucleotides are discussed in Section 28.1.
An “arenediazonium salt” is formed by the reaction of an aromatic amine with nitrous acid at 0–5°C, and has the structure shown below.
Alkanediazonium salts are very unstable; therefore, arenediazonium salts are often simply referred to as diazonium salts.
As is mentioned in the textbook, arenediazonium salts are very useful intermediates from which a wide variety of aromatic compounds can be prepared. You should be thoroughly familiar with the use of diazonium salts to prepare each of the classes of compounds. In addition, you should be aware that fluoroarenes can also be prepared from diazonium salts, as follows:
In this case the diazonium salt is prepared using fluoroboric acid, HBF4, and sodium nitrite. The thermal decomposition of the salt, called the Schiemann reaction, can be quite hazardous.
Note that the IUPAC‑preferred name for cuprous chloride is copper(I) chloride; similarly, cuprous cyanide is called copper(I) cyanide.
Hypophosphorous acid has the structure shown below:
The presence of the letters “azo” in a compound’s name usually implies that a nitrogen‑nitrogen double bond is present in its structure. Azo compounds t in which two aryl groups are joined by an \$\ce{-}\$N\$\ce{=}\$N\$\ce{-}\$ linkage are usually very colourful.
Overreaction of Aniline
Arylamines are very reactive towards electrophilic aromomatic substitution. The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Monobromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily.
Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
In addition to overreactivity, we have previously seen (Section 16.3) that Friedel-Crafts reactions employing AlCl3 catalyst does not work with aniline. A salt complex forms and prevents electrophilic aromatic substitution.
Both this problem and the aniline overreactivity can be circumvented by first going through the corresponding amide.
Modifying the Influence of Strong Activating Groups
By acetylating the heteroatom substituent on aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
.
Sulfa Drug Synthesis
Sulfa drugs are an important group of synthetic antimicrobial agents (pharmaceuticals) that contain the sulfonamide group. The synthesis of sulfanilamide (a sulfa drug) illustrates how the reactivity of aniline can be modified to make possible an electrophilic aromatic substitution. The corresponding acetanilide undergoes chlorosulfonation. The resulting 4-acetamidobenzenesulfanyl chloride is treated with ammonia to replace the chlorine with an amino group and affords 4-acetamidobenzenesulfonamide. The subsequent hydrolysis of the sulfonamide produces the sulfanilamide.
Diazonium Salts: The Sandmeyer Reaction
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen. Some of the more commonly used substitution reactions are shown in the following diagram. Since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored. Those substitution reactions that are catalyzed by cuprous salts are known as Sandmeyer reactions. Fluoride substitution occurs on treatment with BF4(–), a reaction known as the Schiemann reaction. Stable diazonium tetrafluoroborate salts may be isolated, and on heating these lose nitrogen to give an arylfluoride product. The top reaction with hypophosphorus acid, H3PO2, is noteworthy because it achieves the reductive removal of an amino (or nitro) group. Unlike the nucleophilic substitution reactions, this reduction probably proceeds by a radical mechanism.
These aryl diazonium substitution reactions significantly expand the tactics available for the synthesis of polysubstituted benzene derivatives. Consider the following options:
1. The usual precursor to an aryl amine is the corresponding nitro compound. A nitro substituent deactivates an aromatic ring and directs electrophilic substitution to meta locations.
2. Reduction of a nitro group to an amine may be achieved in several ways. The resulting amine substituent strongly activates an aromatic ring and directs electrophilic substitution to ortho & para locations.
3. The activating character of an amine substituent may be attenuated by formation of an amide derivative (reversible), or even changed to deactivating and meta-directing by formation of a quaternary-ammonium salt (irreversible).
4. Conversion of an aryl amine to a diazonium ion intermediate allows it to be replaced by a variety of different groups (including hydrogen), which may in turn be used in subsequent reactions.
The following examples illustrate some combined applications of these options to specific cases. You should try to conceive a plausible reaction sequence for each. Once you have done so, you may check suggested answers below.
Answer 1:
It should be clear that the methyl substituent will eventually be oxidized to a carboxylic acid function. The timing is important, since a methyl substituent is ortho/para-directing and the carboxyl substituent is meta-directing. The cyano group will be introduced by a diazonium intermediate, so a nitration followed by reduction to an amine must precede this step.
Answer 2:
The hydroxyl group is a strong activating substituent and would direct aromatic ring chlorination to locations ortho & para to itself, leading to the wrong product. As an alternative, the nitro group is not only meta-directing, it can be converted to a hydroxyl group by way of a diazonium intermediate. The resulting strategy is self evident.
Answer 3:
Selective introduction of a fluorine is best achieved by treating a diazonium intermediate with boron tetrafluoride anion. To get the necessary intermediate we need to make p-nitroaniline. Since the nitro substituent on the starting material would direct a new substituent to a meta-location, we must first reduce it to an ortho/para-directing amino group. Amino groups are powerful activating substituents, so we deactivate it by acetylation before nitration. The acetyl substituent also protects the initial amine function from reaction with nitrous acid later on. It is removed in the last step.
Answer 4:
Polybromination of benzene would lead to ortho/para substitution. In order to achieve the mutual meta-relationship of three bromines, it is necessary to introduce a powerful ortho/para-directing prior to bromination, and then remove it following the tribromination. An amino group is ideal for this purpose. Reductive removal of the diazonium group may be accomplished in several ways (three are shown).
Answer 5:
The propyl substituent is best introduced by Friedel-Crafts acylation followed by reduction, and this cannot be carried out in the presence of a nitro substituent. Since an acyl substituent is a meta-director, it is logical to use this property to locate the nitro and chloro groups before reducing the carbonyl moiety. The same reduction method can be used to reduce both the nitro group (to an amine) and the carbonyl group to propyl. We have already seen the use of diazonium intermediates as precursors to phenols.
Answer 6:
Aromatic iodination can only be accomplished directly on highly activated benzene compounds, such as aniline, or indirectly by way of a diazonium intermediate. Once again, a deactivated amino group is the precursor of p-nitroaniline (prb.#3). This aniline derivative requires the more electrophilic iodine chloride (ICl) for ortho-iodination because of the presence of a deactivating nitro substituent. Finally, the third iodine is introduced by the diazonium ion procedure.
Diazonium Coupling Reactions
A resonance description of diazonium ions shows that the positive charge is delocalized over the two nitrogen atoms. It is not possible for nucleophiles to bond to the inner nitrogen, but bonding (or coupling) of negative nucleophiles to the terminal nitrogen gives neutral azo compounds. As shown in the following equation, this coupling to the terminal nitrogen should be relatively fast and reversible. The azo products may exist as E / Z stereoisomers. In practice it is found that the E-isomer predominates at equilibrium.
Unless these azo products are trapped or stabilized in some manner, reversal to the diazonium ion and slow nucleophilic substitution at carbon (with irreversible nitrogen loss) will be the ultimate course of reaction, as described in the previous section. For example, if phenyldiazonium bisufate is added rapidly to a cold solution of sodium hydroxide a relatively stable solution of sodium phenyldiazoate (the conjugate base of the initially formed diazoic acid) is obtained. Lowering the pH of this solution regenerates phenyldiazoic acid (pKa ca. 7), which disassociates back to the diazonium ion and eventually undergoes substitution, generating phenol.
C6H5N2(+) HSO4(–) + NaOH (cold solution) C6H5N2–OH + NaOH (cold) C6H5N2–O(–) Na(+)
phenyldiazonium bisulfate phenyldiazoic acid sodium phenyldiazoate
Aryl diazonium salts may be reduced to the corresponding hydrazines by mild reducing agents such as sodium bisulfite, stannous chloride or zinc dust. The bisulfite reduction may proceed by an initial sulfur-nitrogen coupling, as shown in the following equation.
Ar-N2(+) X(–)
NaHSO3
Ar-N=N-SO3H
NaHSO3
Ar-NH-NH-SO3H
H2O
Ar-NH-NH2 + H2SO4
The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles. The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes. Azobenzene (Y=Z=H) is light orange; however, the color of other azo compounds may range from red to deep blue depending on the nature of the aromatic rings and the substituents they carry. Azo compounds may exist as cis/trans isomer pairs, but most of the well-characterized and stable compounds are trans.
Some examples of azo coupling reactions are shown below. A few simple rules are helpful in predicting the course of such reactions:
1. At acid pH (< 6) an amino group is a stronger activating substituent than a hydroxyl group (i.e. a phenol). At alkaline pH (> 7.5) phenolic functions are stronger activators, due to increased phenoxide base concentration.
2. Coupling to an activated benzene ring occurs preferentially para to the activating group if that location is free. Otherwise ortho-coupling will occur.
3. Naphthalene normally undergoes electrophilic substitution at an alpha-location more rapidly than at beta-sites; however, ortho-coupling is preferred. See the diagram for examples of α / β notation in naphthalenes.
You should try to conceive a plausible product structure for each of the following couplings. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.08%3A_Reactions_of_Arylamines.txt |
Objectives
After completing this section, you should be able to
1. draw the structure of furan, pyrrole and imidazole.
2. use the Hückel 2n + 4 rule to explain the aromaticity of pyrrole.
3. predict the product formed when pyrrole is subjected to an aromatic electrophilic substitution, such as nitration, etc.
4. write the detailed mechanism for the electrophilic aromatic substitution of pyrrole to account for the fact that substitution takes place at C2 rather than C3.
5. describe the geometry of the pyridine molecule.
6. account for the difference in basicity between pyridine, pyrrole and other amines.
7. explain why pyridine undergoes electrophilic substitution much less readily than does benzene.
8. identify the presence of a fused‑ring heterocycle in a given structure.
9. make predictions about the chemical behaviour of the fused‑ring heterocycle based on what you have learned about pyrrole, imidazole, pyridine and pyrimidine.
Heterocyclic structures containing nitrogen are found in many natural products. Examples of some nitrogen compounds, known as alkaloids because of their basic properties, were given in the amine chapter.
The porphyrin aromatic heterocycle contains multiple pyrrole ring strictures. The porphyrines, heme and chlorophyll b play vital parts in the metabolism of plants and animals.
Pyrrole
Pyrrole is obtained commercially by the reaction of furan with ammonia .
As discussed in Section 15-5, pyrrole has six pi electrons contributing to its aromaticity. Each carbon in pyrrole contributes one pi electron. The nitrogen in pyrrole contributes two pi electron by becoming sp2 hybridized and placing its lone pair electrons into a p orbital. Because the nitrogen lone pair is part of the aromatic sextet, the electrons are very stable and are much less available for bonding to a proton (and if they do pick up a proton, the aromatic system is destroyed). For these reasons, pyrrole nitrogens are not strongly basic. Pyrrole is a very weak base: the conjugate acid is a strong acid with a pKa of 0.4.
The involvement of the nitrogen lone pair electrons in pyrrole's aromatic conjugation creates charge separated structures not normally found for benzene. The combination of these resonance forms creates an overall separation of charge which increases the electron density around the ring carbons and decreases the electron density on the nitrogen. This effect has a number of ramifications. First is that pyrrole has a large dipole moment (1.8 D) pointing away from the ring nitrogen. This is in sharp contrast to the non-aromatic heterocycle pyrrolidine where the dipole moment (1.5 D) points towards the ring nitrogen.
Due to the resonance forms, pyrrole has less electron density around it than the nitrogen in a typical alkyl amine. This agrees with fact that that pyrrole is a much weaker base than pyrrolidine. The electrostatic potential map of pyrrole (Shown Below) shows less electron density (shown as a red color) on the ring nitrogen when compared to its aliphatic equivalent, pyrrolidine. The map of pyrrole also shows that the resonance forms increase the electron density of pyrrole's ring carbons when compared to the ring carbons of benzene. The involvement of nitrogen's lone pair electrons in the aromaticity of pyrrole makes the ring activated towards electrophilic aromatic substitution.
Pyrrole can undergo many of the same electrophilic aromatic substitution reactions as benzene. Because of the activation of pyrrole's aromatic ring, many of these reactions are performed under a reduced temperature compared to a similar reaction with benzene. There is a clear preference for substitution at the 2-position (α) of the pyrrole ring. An explanation for the general α-selectivity of these substitution reactions is apparent from the mechanism outlined below. The intermediate formed by electrophile attack at C-2 is stabilized by charge delocalization to a greater degree than the intermediate from C-3 attack. From the Hammond postulate we may then infer that the activation energy for substitution at the former position is less than the latter substitution.
Pyridine
Pyridine is an example of a six-membered aromatic heterocycle. In the bonding picture for pyridine, the nitrogen is sp2-hybridized, with two of the three sp2 orbitals forming sigma overlaps with the sp2 orbitals of neighboring carbon atoms, and the third nitrogen sp2 orbital containing the lone pair. The unhybridized p orbital contains a single electron, which is part of the 6 pi-electron aromatic system delocalized around the ring.
Pyridine's nitrogen lone pair occupies an sp2-hybrid orbital, and is not part of the aromatic sextet. Thus, its electron pair is available for forming a bond to a proton, making pyridine's nitrogen atom (conjugate acid pKa = 5.25) more basic than pyrrole's (pKa = 0.4). However, pyridine is is less basic than a typical alkylamine (pKa ~ 10-11). The difference is due to nitrogen hybridization. The lone pair electrons on a pyridine nitrogen occupy an sp2 hybrid orbital, while the lone pair electrons on an amine nitrogen occupy an sp3 hybrid orbital. sp2 orbitals are composed of one part s and two parts p atomic orbitals, meaning that they have about 33% s character. sp3 orbitals, conversely, are only 25% s character (one part s, three parts p) . An s atomic orbital holds electrons closer to the nucleus than a p orbital, thus s orbitals are more electronegative than p orbitals. Therefore, sp2 hybrid orbitals, with their higher s-character, are more electronegative than sp3 hybrid orbitals. Lone pair electrons in the more electronegative sp2 hybrid orbitals of pyridine are held more tightly to the nitrogen nucleus, and are therefore less 'free' to break away and form a bond to a proton - in other words, they are less basic. sp3
The lack of involvement of pyrrole's nitrogen lone pair electrons in the aromatic conjugation creates charge separated structures not normally found for benzene. The combination of the resonance forms creates an overall charge separation which increases the electron density around the ring's nitrogen and decreases the electron density on the ring carbons. This causes pyridine to have a larger dipole moment (2.26 D) than its non-aromatic equivalent piperidine where the dipole moment (1.17 D). In both cases the dipole moment points towards the ring nitrogen.
Due to the resonance forms, pyridine has more electron density around its nitrogen than in a typical alkyl amine. The electrostatic potential map of pyridine (Shown Below) shows more electron density (shown as a red color) on the ring nitrogen of pyridine when compared to pyrrole. The map of pyridine also shows that the resonance forms decrease the electron density of pyridine's ring carbons (Show as a blue/green color) when compared to those of benzene. The involvement of the nitrogen's lone pair electrons in the aromaticity of pyridine makes the ring deactivated towards electrophilic aromatic substitution.
From the previous resonance description of pyridine, we expect this aromatic amine to undergo electrophilic substitution reactions far less easily than does benzene. Furthermore, the electrophilic reagents and catalysts employed in these reactions coordinate with the nitrogen electron pair, exacerbating the positive charge at positions 2,4 & 6 of the pyridine ring. When these reactions do occur then tend to produce the 3-substituted product. Three examples of the extreme conditions required for electrophilic substitution are shown below.
Example
Friedel-Crafts reactions are not feasible because alkyl halides and acid halides prefer to react with pyridine nucleophilic ring nitrogen to provide a N-substituted product. As shown below, N-alkylation and N-acylation products may be prepared as stable crystalline solids in the absence of water or other reactive nucleophiles. Pyridine is a modest base (pKa=5.2). Since the lone pair electrons on pyridine's nitrogen are not part of the aromatic sextet, the pyridinium species produced by N-substitution retains the aromaticity of pyridine.
Example
Other six-membered nitrogen contain aromatic heterocycles include pyrazine, pyrimidine, and pyridazine. The inductive effect of a second nitrogen makes all three of these heterocycles less basic than pyridine.
Imidazole
Imidazole is another important example of an aromatic heterocycle found in biomolecules - the side chain of the amino acid histidine contains an imidazole ring. The two nitrogens in imidazole are quite different. One nitrogen is pyrrole-like and donates its lone pair electrons, like pyrrole, to make imidazole aromatic. The other nitrogen is pyridine like and its lone pair electrons are contained is a sp2 hybridized orbital. These lone pair electrons are readily available for bonding which imidazole (pKa = 6.95) much more basic than pyrrole (pKa = 0.4).
Thiazole
Thiazole is a five-membered sulfur containing aromatic ring system which is found in biological systems, such as thiamine diphosphate. Thiamine diphosphate (ThDP, sometimes also abbreviated TPP or ThPP) is a coenzyme which, like PLP, acts as an electron sink to stabilize key carbanion intermediates. The most important part of the ThDP molecule from a catalytic standpoint is its thiazole ring. The presence of sulfur in thiazole's aromatic ring makes its nitrogen less basic (pKa = 2.44) than imidazole.
Exercise $1$
1) Describe how thiazole is aromatic. Use an orbital picture and include the lone pair electrons on sulfur. Assume the sulfur is sp2 hybridized.
2) Would you expect a thiazole ring to be protonated at the physiological pH of 7.3 (Section 24-5)?
Answer
1)
2)
7.3 = 2.44 + log ([RNH2] / [RNH3+])
([RNH2] / [RNH3+]) = 7.2 x 104
. . . so, [RNH2] >> [RNH3+] so thiazole would be almost completely unprotonated at pH = 7.3.
Polycyclic Heterocycles
Indole, quinoline, isoquinoline, and purine are all polycyclic aromatic heterocycles commonly found in nature. Indole, quinoline, and isoquinoline all contain a hetrocyclic ring fused to benzene. Purine is made up to two heterocyclic rings, imidazole and pyrimidine, fused together. Quinoline is found in the antimalarial drug quinine. Indole is found in the neurotransmitter serotonin. The purine ring structure is found in adenine and guanine, two important parts of DNA and in the stimulant caffeine.
The pyridine-like nitrogen atom in quinoline and isoquinoline withdraws electrons making them both less reactive to electrophilic substitution than benzene. Likewise, quinoline (pKa = 4.9) and isoquinoline (pKa = 5.4) are less basic that a typical amine (pKa ~10-11). Quinoline and isoquinoline can both undergo electrophilic aromatic substitution but substitution on the pyridine-like ring is avoided. Quinoline usually makes a roughly equal mixture of 5 and 8 substituted products. Isoquinoline favors making the 5 substituted product with a small amount of an 8 substituted side product.
Example
Indole has a ring nitrogen similar to to pyrrole. The lone pair electrons for this nitrogen are contained in a p orbital and are part of indole's 10 pi aromaticity. This makes indole relatively non-basic (pKa = -2) and activated toward electrophilic substitution. Indole undergo electrophilic substitution more easily than benzene and substitution typically occurs at the 3 position on the pyrrole ring.
Example
The purine ring system contain three pyridine-like nitrogens. This lone pair electrons of theses nitrogen are contained in sp2 hybrid orbitals, making them not not part of purine's 10 pi aromaticity, allowing them to retain base-like characteristics. The lone pair electrons for the remaining, pyrrole-like, nitrogen are contained in a p orbital and are part of purine's 10 pi aromaticity. This makes the remaining nitrogen relatively non-basic.
Exercise $1$
1) Which nitrogen atom in the neurotransmitter serotonin is expected to be the most basic. Please explain your answer.
2) Pyridine reacts with electrophiles to product a 3-substituted product rather than a 2-substituted product. Write a series of resonance forms for the cation intermediate formed during the reaction. Use these structures to explain the experimental result.
Answer
1) The
2) The resonance forms of 2-substitution places a positive charge on the ring nitrogen. The resonance forms of 3-substitution do not place a positive charge on the ring nitrogen. Having a positive charge on a nitrogen is less stable than on a carbon. Nitrogen is more electronegative than carbon making it less able to stabilize the positive charge. The 3-substituted product is preferred because it causes resonance forms which are more stable. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.09%3A_Heterocyclic_Amines.txt |
Objectives
After completing this section, you should be able to
1. identify the region of the infrared spectrum that shows absorptions resulting from the \(\ce{N-H}\) bonds of primary and secondary amines.
2. describe a characteristic change that occurs in the infrared spectrum of an amine when a small amount of mineral acid is added to the sample.
3. use 1H NMR spectra in determining the structure of an unknown amine.
4. use the “nitrogen rule” of mass spectrometry to determine whether a compound has an odd or even number of nitrogen atoms in its structure.
5. predict the prominent peaks in the mass spectrum of a given amine.
6. use the mass spectrum of an unknown amine in determining its structure.
Key Terms
Make certain that you can define, and use in context, the key term below.
• nitrogen rule
Study Notes
You should note the spectroscopic similarities between amines and alcohols: both have infrared absorptions in the 3300–3360 cm−1 region, and in both cases, the proton that is attached to the heteroatom gives rise to an often indistinct signal in the 1H NMR spectrum.
1H NMR of Amines
The hydrogens attached to an amine show up ~ 0.5-5.0 ppm. The location is dependent on the amount of hydrogen bonding and the sample's concentration. The hydrogens on carbons directly bonded to an amine typically appear ~2.3-3.0 ppm. These hydrogens are deshielded by the electron-withdrawing effects of nitrogen and appear downfield in an NMR spectra compared to alkane hydrogens.
Addition of D2O will normally cause all hydrogens on non-carbon atoms to exchange with deuterium, thus making these resonances "disappear." Addition of a few drops of D2O causing a signal to vanish can help confirm the presence of -NH.
13C NMR of Amines
Carbons directly attached to the nitrogen appear in 10-65 ppm region of a 13C NMR spectra. They are shifted slightly downfield compared to alkane carbons due to the electron-withdrawing effect of nitrogen again causing deshielding.
IR of Amines
The infrared spectra of several amines are shown beneath the following table. Some of the characteristic absorptions for C-H stretching and aromatic ring substitution are also marked, but not colored.
Amine Class Stretching Vibrations Bending Vibrations
Primary (1°)
The N-H stretching absorption is less sensitive to hydrogen bonding than are O-H absorptions. In the gas phase and in dilute CCl4 solution free N-H absorption is observed in the 3400 to 3500 cm-1 region. Primary aliphatic amines display two well-defined peaks due to asymmetric (higher frequency) and symmetric N-H stretching, separated by 80 to 100 cm-1. In aromatic amines these absorptions are usually 40 to 70 cm-1 higher in frequency. A smaller absorption near 3200 cm-1 (shaded orange in the spectra) is considered to be the result of interaction between an overtone of the 1600 cm-1 band with the symmetric N-H stretching band.
C-N stretching absorptions are found at 1200 to 1350 cm-1 for aromatic amines, and at 1000 to 1250 cm-1 for aliphatic amines.
Strong in-plane NH2 scissoring absorptions at 1550 to 1650 cm-1, and out-of-plane wagging at 650 to 900 cm-1 (usually broad) are characteristic of 1°-amines.
Secondary (2°)
Secondary amines exhibit only one absorption near 3420 cm-1. Hydrogen bonding in concentrated liquids shifts these absorptions to lower frequencies by about 100 cm-1. Again, this absorption appears at slightly higher frequency when the nitrogen atom is bonded to an aromatic ring.
The C-N absorptions are found in the same range, 1200 to 1350 cm-1(aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
A weak N-H bending absorption is sometimes visible at 1500 to 1600 cm-1. A broad wagging absorption at 650 to 900 cm-1 may be discerned in liquid film samples.
Tertiary (3°) No N-H absorptions. The C-N absorptions are found in the same range, 1200 to 1350 cm-1 (aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines. Aside from the C-N stretch noted on the left, these compounds have spectra characteristic of their alkyl and aryl substituents.
UV/Vis Spectra of Amines
Alky Amines absorb in the region around 200 nm which make them of little value. In arylamines, the lone pair electron on the nitrogen interacts with pi electron system of the aromatic ring shifting the ring 's absorption to longer wavelengths. An example is benzene's lambda max of 256 nm while aniline's lambda max of 280 nm.
Mass Spectra of Amines
Nitrogen Rule
The nitrogen rule states that a molecule that has no or even number of nitrogen atoms has an even nominal mass, whereas a molecule that has an odd number of nitrogen atoms has an odd nominal mass.
Example
Fragmentation Patterns
The molecular ion peak is an odd number any time there is an odd number of nitrogen atoms in a molecule.. The mass spectra of amines is dominated by alpha-cleavage which produces an alkyl radical on a resonance stabilized nitrogen containing cation. Secondary and tertiary amines have the possibility of multiple alpha-cleavages.
N-Butylamine (C4H11N) with MW = 73.13
Another example is a secondary amine shown below. Again, the molecular ion peak is an odd number. The base peak (m/z = 44) is from the C-C cleavage adjacent to the C-N bond. Other important peaks come from the cleavage of the N-H bond (m/z = 120) and cleavage at the benzylic position (m/z = 91).
N-Methylbenzylamine (C8H11N) with MW = 121.18 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.10%3A_Spectroscopy_of_Amines.txt |
Concepts & Vocabulary
• Design a multi‑step synthesis that involves the use of any of the reactions described in this unit.
• Solve road‑map problems that require a knowledge of amine chemistry.
• Define, and use in context, the key terms introduced.
24.1 Naming Amines
• Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group.
• Several different nomenclature systems exist for naming amines, which is complicates this topic since there is no preferred option.
• The terms primary (1º), secondary (2º) & tertiary (3º) are used to classify amines by referring to the number of alkyl (or aryl) substituents bonded to the nitrogen atom.
• A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a quaternary (4º)-ammonium cation.
• Amines are named following the IUPAC rules with -amino being added as a substitutent.
• Amines are named following the Chemical Abstract Service which the suffix -amine is attached to the root alkyl name.
• Amines are named with a common system where each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine.
24.2 Structure and Properties of Amines
• Neutral amines have three bonds and one lone pair.
• The central nitrogen has sp3 hybridization, which leads to trigonal pyramidal geometry with bond angles of 109.5o.
• Nitrogen does not lend to isolable stereisomers since it rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations.
• Hydrogen bonding is the dominant factor for amines.
• Branching of amines (less bonds to Hydrogen) lowers the boiling point of amines.
24.3 Basicity of Amines
• When evaluating the basicity of a nitrogen-containing organic functional group, the central question we need to ask ourselves is: how reactive (and thus how basic) is the lone pair on the nitrogen?
• For amines, the more alkyl groups attached, the more basic the amine is due to the electron-donating effect of the alkyl group.
• The lone pair on the nitrogen atom in an amide is in resonance with the double bond making it a more stabilized lone pair and less basic.
• When a nitrogen atom is incorporated into a ring, the bascicity depends on where the lone pair resides.
• In a ring where the lone pair on the nitrogen atom resides in a hybrid orbital, then that lone pair is basic with the more basic equating with more s character in the hybrid orbital.
• In a ring where the lone pair on the nitrogen atom resides in a p orbital, then that lone pair is delocalized and less basic.
• In extraction, one can take advantage of amines and their basicity by forming the protonated salt (RNH2+Cl), which is soluble in water, in order to separate them.
• Amines are also acidic and the same factors that decreased the basicity of amines increase their acidity.
• Amines are often utilized as bases in reactions.
24.4 Basicity of Arylamines
• Nitrogen atoms as part of aromatic rings are less basic than methylamine.
• The basic lone pair on the nitrogen is to some extent tied up in – and stabilized by – the aromatic p system.
• This effect is accentuated by the addition of an electron-withdrawing groups.
• Imines are somewhat basic, which can be explained using orbital theory and the inductive effect: the sp2 orbitals of an imine nitrogen are one part s and two parts p, meaning that they have about 67% s character.
24.5 Biological Amines and the Henderson-Hasselbalch Equation
• The chemistry of life occurs in a buffer that consists of a mixture of various phosphate and ammonium compounds.
• In an aqueous solution buffered at pH 7, carboxylic acid groups can be expected to be essentially 100% deprotonated and negatively charged (ie. in the carboxylate form), and amine groups essentially 100% protonated and positively charged (i.e., in the ammonium form).
• The imidizole group on the histidine side chain has a pKa near 7, and thus exists in physiological solutions as mixture of both protonated and deprotonated forms.
24.6 Synthesis of Amines
• Secondary amines and their salts can be synthesized from primary amines and an alkyl halide.
• Tertiary amines and their salts can be synthesized from primary amines and an alkyl halide.
• Quaternary ammonium salts can be synthesized from primary amines and an alkyl halide.
• While a primary amine can by synthesized from ammonia and an alkyl halide, there are other better options.
• The reactions that provide a more pure primary amine in high yield occur in two steps.
• First form a carbon-nitrogen bond by reacting a nitrogen nucleophile with a carbon electrophile.
• Second, any extraneous nitrogen substituents that may have facilitated this bonding are removed to give the amine product.
• Another option to synthesize amines is to reduce nitro groups, which include catalytic hydrogenation, zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution.
• Reacting a nitrile with lithium aluminum hydride will yield a primary amine.
• Amides can be reduced to primary, secondary or tertiary amines using lithium aluminum hydride.
• Aldehydes and ketones can be converted to primary, secondary, or tertiary amines using reductive amination.
• Hofmann rearrangement is the reaction of a primary amide with a halogen in strongly basic conditions to give a primary amine as a product.
• The Curtius rearrangement converts an acid chloride to an amine by the migration of an -R group form the carbonyl carbon to the the neighboring nitrogen in the acyl azide intermediate.
24.7 Reactions of Amines
• The direct alkylation of 1º or 2º-amines gives a more substituted product does not proceed cleanly.
• Acid chlorides can react with amines to form amides.
• Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations, where the products are a tertiary amine and a alkene.
• The double bond Hofmann eliminations tends to give the less-substituted double bond isomer due to the Hofmann Rule.
24.8 Reactions of Arylamines
• Arylamines are very reactive towards electrophilic aromatic substitution substitutions.
• Aniline is a strong activating group and can overreact in electrophilic aromatic substitutions.
• Changing the functional group from an amine to an amide on an aromatic ring alters the reactivity in electrophilic aromatic substitutions for more control in the reaction.
• Sulfa drugs are an important group of synthetic antimicrobial agents (pharmaceuticals) that contain the sulfonamide group.
• The synthesis of sulfanilamide (a sulfa drug) illustrates how the reactivity of aniline can be modified to make possible an electrophilic aromatic substitution.
• Aryl diazonium salts are important intermediates, since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored.
• Diazonium ions show that the positive charge is delocalized over the two nitrogen atoms while it is not possible for nucleophiles to bond to the inner nitrogen, negative nucleophiles can bond to the terminal nitrogen gives neutral azo compounds.
• The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles.
• The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes.
24.9 Heterocyclic Amines
• Heterocyclic structures are found in many natural products an example of some nitrogen compounds, known as alkaloids because of their basic properties.
• Pyrrole is obtained commercially by the reaction of furan with ammonia.
• In a pyrrole ring, the nitrogen lone pair is part of the aromatic sextet, thus not very basic.
• Substitution preference is on the 2-position of pyrrole.
• Imidazole is another five-membered heterocyclic amine, which is part of the amino acid histidine.
• Thiazole is a five-membered ring system which is found in biological systems.
• When a nitrogen atom is incorporated directly into an aromatic ring, its basicity depends on the bonding context.
• In pyridine, the nitrogen lone pair occupies an sp2-hybrid orbital, and is not part of the aromatic sextet, therefore, its electron pair is available for forming a bond to a proton, and thus the pyridine nitrogen atom is somewhat basic.
• The aromatic stabilization energy of pyridine is 21 kcal/mole based on heat of combustion measurements.
• Polycyclic heterocyclic structures are found in many natural products like caffeine.
• Derivatives of the simple fused ring heterocycle purine constitute an especially important and abundant family of natural products.
• The amino compounds adenine and guanine are two of the complementary bases that are essential components of DNA.
24.10 Spectroscopy of Amines
• The hydrogens attached to an amine show up ~ 0.5-5.0 ppm in 1H NMR.
• The broad range is due to the fact that the location is dependent on the amount of hydrogen bonding and the sample's concentration.
• The hydrogens on carbons directly bonded to an amine typically appear ~2.3-3.0 ppm in 1H NMR.
• IR for primary amines a free N-H absorption is observed in the 3400 to 3500 cm-1 region as two well-defined peaks.
• IR for amines the C-N stretching absorptions are found at 1200 to 1350 cm-1 for aromatic amines, and at 1000 to 1250 cm-1 for aliphatic amines.
• Secondary amines exhibit only one absorption near 3420 cm-1 in IR.
• For tertiary amines, there is not N-H stretch.
Skills to Master
• Skill 24.1 Name amines using IUPAC rules.
• Skill 24.2 Draw the structure of amines from the IUPAC name.
• Skill 24.3 Describe the geometries and approximate bond angles of amines.
• Skill 24.4 Explain physical properties of amines.
• Skill 24.5 Explain why one amine is more basic than another.
• Skill 24.6 Explain how the basicity of an amine when the Nitrogen atom is incorporated in a ring depends on the bonding context.
• Skill 24.7 Describe how an amine can be extracted from a mixture that also contains neutral compounds illustrating the reactions which take place with appropriate equations.
• Skill 24.8 Explain why primary and secondary (but not tertiary) amines may be regarded as very weak acids.
• Skill 24.9 Use the concept of resonance to explain why arylamines are less basic than their aliphatic counterparts.
• Skill 24.10 Arrange a given series of arylamines in order of increasing or decreasing basicity.
• Skill 24.11 Explain using inductive and resonance effects arguments as to why a given arylamine is more or less basic than aniline.
• Skill 24.12 Identify the form that amine bases take within living cells.
• Skill 24.13 Use the Henderson‑Hasselbalch equation to calculate the percentage of a base that is protonated in a solution, given the pKa value of the associated ion and the pH of the solution.
• Skill 24.14 List general ways to synthesize amines.
• Skill 24.15 Draw mechanisms for preparing amines including:
• reduction of nitriles, amides and nitro compounds.
• reactions involving alkyl groups:
1. SN2 reactions of alkyl halides, ammonia and other amines.
2. nucleophilic attack by an azide ion on an alkyl halide, followed by reduction of the azide so formed.
3. alkylation of potassium phthalimide, followed by hydrolysis of the N‑alkyl phthalimide so formed (i.e., the Gabriel synthesis).
• reductive amination of aldehydes or ketones.
• Hofmann or Curtius rearrangements.
• Skill 24.16 Write an equation to represent the reaction that takes place between ammonia, a primary or secondary amine, and an acid chloride.
• Skill 24.17 Identify the product formed when a given amine reacts with a given acid chloride.
• Skill 24.18 Draw the product for a Hoffman elimination.
• Skill 24.19 Propose a synthesis of arylamines using diazonium coupling reactions.
• Skill 24.20 Be able to draw the structure of furan, pyrrole and imidazole.
• Skill 24.21 Use the Hückel 2n + 4 rule to explain the aromaticity of pyrrole.
• Skill 24.22 Be able to predict the product formed when pyrrole is subjected to an aromatic electrophilic substitution.
• Skill 24.23 Write the detailed mechanism for the electrophilic aromatic substitution of pyrrole to account for the fact that substitution takes place at C2 rather than C3.
• Skill 24.24 Explain the difference in basicity between pyridine, pyrrole and other amines.
• Skill 24.25 Explain why pyridine undergoes electrophilic substitution much less readily than does benzene.
• Skill 24.17 Use IR, NMR and MS in combination with the nitrogen rule to identify amines.
Summary of Reactions
Amine Preparation
Amine Reactions | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/24%3A_Amines_and_Heterocycles/24.S%3A_Amines_and_Heterocycles_%28Summary%29.txt |
This chapter is designed to provide you with an overview of the biologically important group of compounds known as carbohydrates. Many of the compounds you will encounter while studying this chapter may appear to have very complex structures, but much of their chemistry can be readily understood in terms of the concepts and reactions discussed in earlier chapters of the course.
The chapter begins with an explanation of the classification schemes used to simplify the study of carbohydrates. We make extensive use of Fischer projection formulas throughout the chapter. We place considerable emphasis on gaining an appreciation of the configurations of carbohydrates, particularly of the aldoses. We describe the disadvantages of representing monosaccharides by open-chain structures, and at this point, introduce you to cyclic representations—called Haworth projections—of these substances. We describe the mutarotation of glucose, explaining it in terms of the existence of anomers. We then examine some reactions of monosaccharides, including the formation of ethers and esters, the formation of glycosides, and reduction and oxidation. We discuss the structures of some common disaccharides and polysaccharides, and conclude the chapter with a brief explanation of the role played by carbohydrates in cell recognition.
25: Biomolecules- Carbohydrates
Objectives
After completing this section, you should be able to
1. identify carbohydrates (sugars) as being polyhydroxylated aldehydes and ketones.
2. describe, briefly, the process of photosynthesis, and identify the role played by carbohydrates as an energy source for living organisms.
Key Terms
Make certain that you can define, and use in context, the key term below.
• carbohydrate
Introduction
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances. For example, because its formula is C6H12O6, glucose was once thought to be a “carbon hydrate” with the structure C6·6H2O.
Example 1
Which compounds would be classified as carbohydrates?
Solution
1. This is a carbohydrate because the molecule contains an aldehyde functional group with OH groups on the other two carbon atoms.
2. This is not a carbohydrate because the molecule does not contain an aldehyde or a ketone functional group.
3. This is a carbohydrate because the molecule contains a ketone functional group with OH groups on the other two carbon atoms.
4. This is not a carbohydrate; although it has a ketone functional group, one of the other carbons atoms does not have an OH group attached.
Exercise 1
Which compounds would be classified as carbohydrates?
Green plants are capable of synthesizing glucose (C6H12O6) from carbon dioxide (CO2) and water (H2O) by using solar energy in the process known as photosynthesis:
$\ce{6CO2 + 6H2O + 2870 kJ -> C6H12O6 + 6O2} \nonumber$
(The 2870 kJ comes from solar energy.) Plants can use the glucose for energy or convert it to larger carbohydrates, such as starch or cellulose. Starch provides energy for later use, perhaps as nourishment for a plant’s seeds, while cellulose is the structural material of plants. We can gather and eat the parts of a plant that store energy—seeds, roots, tubers, and fruits—and use some of that energy ourselves. Carbohydrates are also needed for the synthesis of nucleic acids and many proteins and lipids.
Animals, including humans, cannot synthesize carbohydrates from carbon dioxide and water and are therefore dependent on the plant kingdom to provide these vital compounds. We use carbohydrates not only for food (about 60%–65% by mass of the average diet) but also for clothing (cotton, linen, rayon), shelter (wood), fuel (wood), and paper (wood).
Contributors and Attributions
• The Basics of General, Organic, and Biological Chemistry by David W. Ball, John W. Hill, and Rhonda J. Scott. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.00%3A_Introduction.txt |
Objectives
After completing this section, you should be able to
1. classify a specific carbohydrate as being a monosaccharide, disaccharide, trisaccharide, etc., given the structure of the carbohydrate or sufficient information about its structure.
2. classify a monosaccharide according to the number of carbon atoms present and whether it contains an aldehyde or ketone group.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldose
• disaccharide
• ketose
• monosaccharide (simple sugar)
• polysaccharide
What Are Carbohydrates?
The most abundant biomolecules on earth are carbohydrates. From a chemical viewpoint, carbohydrates are primarily a combination of carbon and water, and many of them have the empirical formula (CH2O)n, where n is the number of repeated units. This view represents these molecules simply as “hydrated” carbon atom chains in which water molecules attach to each carbon atom, leading to the term “carbohydrates.” Although all carbohydrates contain carbon, hydrogen, and oxygen, there are some that also contain nitrogen, phosphorus, and/or sulfur. Carbohydrates have myriad different functions. They are abundant in terrestrial ecosystems, many forms of which we use as food sources. These molecules are also vital parts of macromolecular structures that store and transmit genetic information (i.e., DNA and RNA). They are the basis of biological polymers that impart strength to various structural components of organisms (e.g., cellulose and chitin), and they are the primary source of energy storage in the form of starch and glycogen.
Monosaccharides
In biochemistry, carbohydrates are often called saccharides, from the Greek sakcharon, meaning sugar, although not all the saccharides are sweet. The simplest carbohydrates are called monosaccharides, or simple sugars. They are the building blocks (monomers) for the synthesis of polymers or complex carbohydrates, as will be discussed further in this section. Monosaccharides are classified based on the number of carbons in the molecule. General categories are identified using a prefix that indicates the number of carbons and the suffix –ose, which indicates a saccharide; for example, triose (three carbons), tetrose (four carbons), pentose (five carbons), and hexose (six carbons). The hexose D-glucose is the most abundant monosaccharide in nature. Other very common and abundant hexose monosaccharides are galactose, used to make the disaccharide milk sugar lactose, and the fruit sugar fructose.
A second comparison can be made when looking at glucose, galactose, and fructose. All three are hexoses; however, there is a major structural difference between glucose and galactose versus fructose: the carbon that contains the carbonyl (C=O). In glucose and galactose, the carbonyl group is on the C1 carbon, forming an aldehyde group. In fructose, the carbonyl group is on the C2 carbon, forming a ketone group. The former sugars are called aldoses based on the aldehyde group that is formed; the latter is designated as a ketose based on the ketone group. Again, this difference gives fructose different chemical and structural properties from those of the related aldoses, glucose, and galactose, even though fructose, glucose, and galactose all have the same chemical composition: C6H12O6.
Complex Carbohydrates
The simple sugars form the foundation of more complex carbohydrates. The cyclic forms of two sugars can be linked together by means of a condensation reaction to form a disaccharide. Multiple sugars can be linked to form polysaccharides.
Disaccharides
Two monosaccharide molecules may chemically bond to form a disaccharide. The name given to the covalent bond between the two monosaccharides is a glycosidic bond. Glycosidic bonds form between hydroxyl groups of the two saccharide molecules, an example of the dehydration synthesis described later in this chapter.
Common disaccharides are the grain sugar maltose, made of two glucose molecules; the milk sugar lactose, made of one galactose and one glucose molecule; and the table sugar sucrose, made of one glucose and one fructose molecule.
nosaccharide—OH+HO—monosaccharidemonosaccharide—O—monosaccharidedisaccharide
Polysaccharides
Polysaccharides, also called glycans, are large polymers composed of hundreds of monosaccharide monomers. Unlike mono- and disaccharides, polysaccharides are not sweet and, in general, they are not soluble in water. Like disaccharides, the monomeric units of polysaccharides are linked together by glycosidic bonds.
Polysaccharides are very diverse in their structure. Three of the most biologically important polysaccharides—starch, glycogen, and cellulose—are all composed of repetitive glucose units, although they differ in their structure. Cellulose consists of a linear chain of glucose molecules and is a common structural component of cell walls in plants and other organisms. Glycogen and starch are branched polymers; glycogen is the primary energy-storage molecule in animals and bacteria, whereas plants primarily store energy in starch. The orientation of the glycosidic linkages in these three polymers is different as well and, as a consequence, linear and branched macromolecules have different properties.
Summary
Complexity
Simple Carbohydrates
monosaccharides
Complex Carbohydrates
disaccharides, oligosaccharides
& polysaccharides
Size
Tetrose
C4 sugars
Pentose
C5 sugars
Hexose
C6 sugars
Heptose
C7 sugars
etc.
C=O Function
Aldose
sugars having an aldehyde function or an acetal equivalent.
Ketose
sugars having a ketone function or an acetal equivalent.
Exercise $1$
1) Classify each of the following sugars.
a)
b)
c)
d)
Answer
a) Aldoterose
b) Ketopentose
c) Ketohexose
d) Aldopentose | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.01%3A_Classification_of_Carbohydrates.txt |
Objectives
After completing this section, you should be able to
1. draw the Fischer projection of a monosaccharide, given its wedge and dash structure or a molecular model.
2. draw the wedge and dash 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 dash 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.
Drawing Representations of 3D Structures
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 dash 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. Fischer Projections allow us to represent 3D molecular structures in a 2D environment without changing their properties and/or structural integrity.
Fisher projections show sugars in their open chain form. In a Fischer projection, the carbon atoms of a sugar molecule are connected vertically by solid lines, while carbon-oxygen and carbon-hydrogen bonds are shown horizontally. Stereochemical information is conveyed by a simple rule: vertical bonds point into the plane of the page, while horizontal bonds point out of the page.
Sugars can be drawn in the straight chain form as either Fisher projections or perspective structural formulas. When drawing Fischer projections, the aldehyde group is written at the top, and the H and OH groups that are attached to each chiral carbon are written to the right or left. The arrangement of the atoms distinguishes one stereoisomer from the other.
Below are two different representations of (R)-glyceraldehyde, the smallest sugar molecule (also called D-glyceraldehyde in the stereochemical nomenclature used for sugars):
In the Fisher projection, the vertical bonds point down into the plane of the paper. That's easy to visualize for 3C molecules. but more complicated for bigger molecules. For those draw a wedge and dash line drawing of the molecule. When determining the orientation of the hydroxides on each C, orient the wedge and dash drawing in your mind so that the C atoms adjacent to the one of interest are pointing down. Sighting towards the carbonyl C, if the OH is pointing to the right in the Fisher project, it should be pointing to the right in the wedge and dash drawing, as shown below for D-erthyrose and D-glucose.
Below are three representations of the open chain form of D-glucose: in the conventional Fischer projection, a wedge/dash version of a Fischer projection, and finally in the 'zigzag' style that is preferred by many organic chemists.
Fischer projections are useful when looking at many different diastereomeric sugar structures, because the eye can quickly pick out stereochemical differences according to whether a hydroxyl group is on the left or right side of the structure.
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 (aldo- since the oxidized carbon is an aldehyde and -pentose since the molecules contain 5 carbons). 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.
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.
The Fisher structures of the most common monosaccharides (other than glyceraldehyde and dihydroxyacetone), which you will encounter most frequently are shown below.
Determining R and S in Fischer Projections
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.
Worked Example \(1\)
Determine if carbon #2 in D-glucose is R or S.
Answer
When deciding whether a stereocenter in a Fischer projection is R or S, realize that the hydrogen, in a horizontal bond. Therefore, the orientation of the three remaining substituents is reversed to create the correct answer or a counterclockwise circle means R, and a clockwise circle means S. For carbon #2 in D-Glucose substituent 1, 2, and 3 form a counterclockwise circle so the carbon is R.
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.
Exercise \(1\)
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.
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.
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.
Exercise \(1\)
Draw 'zigzag' structures (using the solid/dash wedge convention to show stereochemistry) for the four sugars in the figure below. Label all stereocenters R or S.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.02%3A_Fischer_Projections.txt |
Objectives
After completing this section, you should be able to
1. identify a specific enantiomer of a monosaccharide as being D or L, given its Fischer projection.
2. identify the limitations of the D, L system of nomenclature for carbohydrates.
3. assign an R or S configuration to each of the chiral carbon atoms present in a monosaccharide, given its Fischer projection.
4. draw the Fischer projection formula for a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
5. construct a molecular model of a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• D sugar
• L sugar
Study Notes
If you find that you have forgotten the meanings of terms such as dextrorotatory and polarimeter, refer back to Section 5.3 in which the fundamentals of optical activity were introduced.
How would you set about the task of deciding whether each chiral carbon has an R or an S configuration? True, you could use molecular models, but suppose that a model set had not been available—what would you have done then?
One approach is to focus on the carbon atom of interest and sketch a three-dimensional representation of the configuration around that atom, remembering the convention used in Fischer projections: vertical lines represent bonds going into the page, and horizontal lines represent bonds coming out of the page. Thus, the configuration around carbon atom 2 in structure a can be represented as follows:
In your mind, you should be able to imagine how this molecule would look if it was rotated so that the bonds that are shown as coming out of the page are now in the plane of the page. [One possible way of doing this is to try and imagine how the molecule would look if it was viewed from a point at the bottom of the page.] What you should see in your mind is a representation similar to the one drawn below.
To determine whether the configuration about the central carbon atom is R or S, we must rotate the molecule so that the group with the lowest priority (H), is directed away from the viewer. This effect can be achieved by keeping the hydroxyl group in its present position and moving each of the other three groups one position clockwise.
The Cahn-Ingold-Prelog order of priority for the three remaining groups is OH > CHO > CH(OH)CH2OH; thus, we see that we could trace out a counterclockwise path going from the highest-priority group to the second- and third-highest, and we conclude that the central carbon atom has an S configuration.
D and L Labeling of Monosaccharide Stereochemistry
Glyceraldehyde, the simplest possible aldose, is made up of three carbons and only one these is chiral. Glyceraldehyde has two stereoisomers, an R/S pair of enantiomers. Before the R,S system for designating chiral configuration was adopted by organic chemists (R)-glyceraldehyde was called D-glyceraldehyde (Latin for right: dexter) and (S)-glyceraldehdye was called L-glyceraldehyde (Latin for left:laevus). D- and L-glyceraldehyde were then used to provide reference points for designating and drawing all other monosaccharides. Sugars whose Fischer projections have the same configuration at the chiral carbon furthest from the carbonyl group as D-glyceraldehyde are designated as D sugars; those with the same configuration as L-glyceraldehyde are designated as L sugars. D and L designations of sugars are based on the position of the hydroxyl on the chiral carbon farthest from the carbonyl group in the Fischer projection of the molecule. All D-sugars have the –OH on the right side and L-sugars have the –OH on the left side.
D-sugars predominate in nature, though L-forms of some sugars, such as fucose, do exist. The D and L designation is a bit more complicated than it would appear on the surface. The confusion about D and L arises because the L sugars of a given name (glucose, for example) are mirror images of the D sugars of the same name. This concept is most easily seen with glyceraldehyde. In the same way D- and L- glyceraldehyde represent two enantiomers, the D- and L- forms larger monosaccharides are enantiomers of one another. The figure below shows the structure of D- and L- glucose. Notice that D-glucose is not converted into L-glucose simply by flipping the configuration of the fifth carbon in the molecule. Rather all of the arrangement around all of the chiral centers (horizontal lines) in the Fischer project of D-glucose need to be opposite to make L-glucose.
.
Figure 6. Fischer projections of enantiomers of glucose (left) and fructose (right).
It is important to recognize that the sign of a compound's specific rotation of plane polarized light (+)/(−) does not correlate with its configuration (D or L). The D/L labeling does not indicate which enantiomer is (+)/(−). Rather, it says that the compound's stereochemistry is related to that of the D or L enantiomer of glyceraldehyde. It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths.
Exercise \(1\)
In the following Fischer projections, assign R and S for each chiral center and determine whether each sugar is a D or L sugar.
a)
b)
c)
Answer
a) From top to bottom, 2R, 3R, and it is a D sugar.
b) From top to bottom, 2S, 3R, 4S, and it is an L sugar.
c) From to to bottom, 3R, 4S, and it is an L sugar. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.03%3A_D_L_Sugars.txt |
Objectives
After completing this section, you should be able to
1. draw the structures of all possible aldotetroses, aldopentoses, and aldohexoses, without necessarily being able to assign names to the individual compounds.
2. draw the Fischer projection of D-glyceraldehyde, D-ribose and D-glucose from memory.
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (D)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red.
Aldotrioses: 3 carbon sugars with one chiral center. Aldotrioses have two (21) possible stereoisomers. A pair of enantiomers called D-gylceraldehyde and L-glyceraldehyde.
Aldotetroses: 4 carbon sugars with two chiral centers. Aldotetroses have four (22) possible stereoisomers. Two pairs of D/L enantiomers called erythrose and threose.
Aldopentoses: 5 carbon sugars with three chiral centers. Aldopentoses have eight (23) possible stereoisomers. Four pairs of D/L enantiomers called ribose, arabinose, xylose, and lyxose.
Aldohexoses: 6 carbon sugars with four chiral centers. Aldohexoses have sixteen (24) possible stereoisomers. five pairs of D/L enantiomers called allose, altrose, glucose, mannose, gulose, idose, galactose, and talose.
Below are the Fischer projects 3-6 carbon aldoses. Starting with the three carbon aldose D-glyceraldehyde, each additional carbon adds a new chiral center and doubles the number of possible stereoisomers of the D-aldoses. Remember that only D-aldoses are represented below. Each D-aldose has an L-aldose enantiomer which is not shown. The L-aldose versions can be draw by inverting all of the chiral centers in the D-aldose's Fischer projection as discussed in the previous section.
Worked Example \(1\)
Draw the Fisher projection of L-erythrose and L-Glucose
Solution
Use the Fischer projection provided above and reverse all of the chiral centers to provide the L-sugar. Note that in both cases the D sugars have the OH going to the right on the chiral center furthest away from the carbonyl. The L-sugars have the OH going to the left.
Worked Example \(2\)
Please draw the Fischer projection fo the following aldopentose and determine if the sugars is D or L.
Solution
First, rotate the model so that the carbonyl is at the top. This is requirement of a Fischer projection. Next rotate the model so that the H and OH of the chiral carbon just below the carbonyl are facing towards you. In this orientation, a dash/wedge model will have every other set of bonds going into the plane of the page. This is not the correct orientation of a Fischer project so they must be modified. The H and OH bonds need to be coming out of the plane of the page in a Fisher projects. When converting bonds from going into the page to going out of the page the orientation of the H and OH are reversed. Remember that the last -CH2OH of a sugar is achiral so the orientation does not need to be shown. Once the bonds are oriented correctly the wedge bonds can be converted to those of a Fischer projection.
Exercise \(1\)
For the following model of a sugar, please draw its Fischer projection and name it.
Answer
D-Mannose
Exercise \(2\)
How many heptose stereoisomers would there expected to be? How many would be D-Sugars?
Answer
There would be 25 = 32 heptose stereoisomers. Half of these would be D-sugars or 16.
Exercise \(3\)
Draw the Fischer projection of the following sugars.
1. L-Ribose
2. L-Galactose
3. L-Talose
Answer
a)
b)
c) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.04%3A_Configurations_of_Aldoses.txt |
Objectives
After completing this section, you should be able to
1. determine whether a given monosaccharide will exist as a pyranose or furanose.
2. draw the cyclic pyranose form of a monosaccharide, given its Fischer projection.
3. draw the Fischer projection of a monosaccharide, given its cyclic pyranose form.
4. draw, from memory, the cyclic pyranose form of D-glucose.
5. determine whether a given cyclic pyranose form represents the D or L form of the monosaccharide concerned.
6. describe the phenomenon known as mutarotation.
7. explain, through the use of chemical equations, exactly what happens at the molecular level during the mutarotation process.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alpha anomer
• anomer
• anomeric centre
• beta anomer
• furanose
• mutarotation
• pyranose
Study Notes
If necessary, before you attempt to study this section, review the formation of hemiacetals discussed in Section 19.10.
Cyclic Monosaccharides
In Section 19-10 it was discussed that the reaction of one equivalent of an alcohol, in the presence of an acid catalyst, adds reversibly to aldehydes and ketones to form a hydroxy ether called a hemiacetal (R2COHOR') (hemi, Greek, half).
Molecules which have both an alcohol and a carbonyl can undergo an intramolecular version of the same reaction forming a cyclic hemiacetal.
Because sugars often contain alcohol and carbonyl functional groups, intramolecular hemiacetal formation is common in carbohydrate chemistry. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown below.
Furan (5-membered ring) and pyran (6-membered ring) structures
Unlike most of the biochemical reactions you will see in this text, sugar cyclization reactions are not catalyzed by enzymes: they occur spontaneously and reversibly in aqueous solution. Sugars are often shown in their open-chain form, however, in aqueous solution, glucose, fructose, and other sugars of five or six carbons rapidly interconvert between straight-chain and cyclic forms. For most five- and six-carbon sugars, the cyclic forms predominate in equilibrium since they are more stable. The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features.
Aldohexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. At equilibrium less than 1% of glucose is in an open chain form with the rest being almost exclusively in its cyclic pyranose form. The pyranose ring is formed by attack of the hydroxyl on carbon 5 of glucose to the aldehyde carbon (carbon #1, also called the anomeric carbon in carbohydrate terminology). The cyclic form of glucose is called glucopyranose. Notice that for glucose and other aldohexoses the hydroxyl that forms the cyclic hemiacetal is also the one that determines the D/L designation of a sugar.
Pyranose rings are often drawn in a chair conformation like cyclohexane rings (Section 4-6) with substituents being either an axial or equatorial position. Pyranose rings are even capable of undergoing a ring flip to change between chair conformations. By convention the ring oxygen is placed to the right and to the rear of the structure (top right of the drawing). Groups which go to the right in a Fischer projection will be orientend 'down' of the pyranose ring while groups to the left are oriented 'up' in the chair structure. Also, the terminal -CH2OH group is oriented up on the pyranose ring for D-sugars and down for L-Sugars.
When D-glucose cyclizes it forms a 37/63 mixture of the alpha and beta anomer respectively. The beta anomer is preferred because β-D-glucopyranose is the only aldohexose which can be drawn with all its bulky substituents (-OH and -CH2OH) in equatorial positions, making it the most stable of the eight D-aldohexoses, which probably accounts for its widespread prevalence in nature.
It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium. You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then re-close to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare, meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°.
Example $1$
Fructose in aqueous solution forms a six-membered cyclic hemiketal called fructopyranose when the hydroxyl oxygen on carbon #6 attacks the ketone carbon (carbon #2, the anomeric carbon in fructose).
In this case, the β anomer is heavily favored in equilibrium by a ratio of 70:1, because in the minor α anomer the bulkier -CH2OH group occupies an axial position. Notice in the above figure that the percentages of α and β anomers present at equilibrium do not add up to 100%. Fructose also exists in solution as a five-membered cyclic hemiketal, referred to in carbohydrate nomenclature as fructofuranose. In the formation of fructofuranose from open-chain fructose, the hydroxyl group on the fifth carbon attacks the ketone.
In aqueous solution, then, fructose exists as an equilibrium mixture of 70% β-fructopyranose, 23% β-fructofuranose, and smaller percentages of the open chain and cyclic α-anomers. The β-pyranose form of fructose is one of the sweetest compounds known, and is the main component of high-fructose corn syrup. The β-furanose form is much less sweet.
Although we have been looking at specific examples for glucose and fructose, other five- and six-carbon monosaccharides also exist in solution as equilibrium mixtures of open chains and cyclic hemiacetals and hemiketals. Shorter monosaccharides are unlikely to undergo analogous ring-forming reactions, however, due to the inherent instability of three and four-membered rings.
Drawing Cyclic Structures of Monosaccharides
The cyclic forms of sugars are commonly depicted as Haworth projections. This convention, first suggested by the English chemist Walter N. Haworth, shows molecules drawn as planar rings with darkened edges representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below (bottom face) the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above (top face) the plane in a Haworth projection.
Figure: Conversion of the Fischer projection of D-glucose to the Haworth projection of ß-D-glucose.
1. When converting a Fischer projection (line) to a Haworth projection, you must first identify the type of monosaccharide involved. If the carbohydrate represents an aldohexose, the pyranose ring is typically used. A pyranose is a cyclic structure that contains five carbon atoms and an oxygen. If the carbohydrate represents a ketohexose, the furanose ring is typically used. The furanose ring contains four carbon atoms and an oxygen.
1. Indicate the arrangement of the hydroxyl group attached to the anomeric carbon to identify the sugar as an alpha or beta anomer. The α and β anomers are determined with respect to carbon 6. If the molecule represents a D-sugar, carbon 6 will be above the plane of the ring (top face) and form an L-sugar, carbon 6 will be below the plane of the ring (ring). The α anomer occurs when the OH on the anomeric carbon is trans to carbon 6 and the β anomer occurs when the OH on the anomeric carbon is cis to carbon 6. If the cyclic structure contains a furanose, since carbon 1 is not included within the ring, that carbon group would be arranged in the opposite direction of the OH group.
1. The remaining chiral centers (carbons 2, 3 and 4 of the pyranose or carbons 3 and 4 of the furanose) are arranged based on the directions of the hydroxyl from the Fischer projection structures. Groups to the left of the Fischer projection would point up (top face), while groups to the right would point down (bottom face).
Since the Fischer Projection of any given carbohydrate is always the same, the Haworth Projection is essentially always the same. The only differences between the Haworth Projection of the alpha or beta form of a single carbohydrate, is how the OH (and carbon 1 if furanose ring) is arranged around the anomeric carbon to determine whether the molecule is alpha or beta.
Stability of Chair Conformation in Pyranose Sugars
Previously, we have seen the six ring atoms of cyclic glucose drawn in two dimensions. A more accurate depiction shows that the molecule adopts, as expected, a chair conformation.
The conformation in which all substituents are equatorial is lower in energy. The two isomeric forms are referred to by the Greek letters alpha (α) and beta (β).
We have not learned about stereoisomerism quite yet, but you can still recognize that the bonding configuration on one carbon is different. On the alpha isomer, one of the hydroxyl groups is axial – this isomer is not able to adopt a chair conformation in which all non-hydrogen substituents are equatorial. The lower energy conformation is the one in which four of the five substituents are equatorial, but the presence of the one axial hydroxyl group means that the alpha isomer is, overall, less stable than the beta isomer.
The most abundant form of fructose in aqueous solution is also a six-membered ring.
The lower energy chair conformation is the one with three of the five substituents (including the bulky –CH2OH group) in the equatorial position.
Exercise $1$
Draw the following in their most stable chair conformation: α-D-galactopyranose and α-D-mannopyranose. Which is expected to be the more stable?
Answer
Because the both have two axial OH's their chair conformations should be roughly the same stability.
Exercise $2$
Draw the two chair conformations of the sugar called 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.
Answer
Exercise $3$
Draw the cyclic structure of α-D-altrose.
Answer
Exercise $4$
Draw the cyclic structure for β-D-galactose. Identify the anomeric carbon.
Answer
To identify the structure, we should first start with the Fischer projection of D-galactose. Since it is an aldohexose, we will start with the pyranose ring. The beta anomer was requested, so the OH on the anomeric carbon (C1) is cis to C6. Since C6 is top face (pointing up), the OH will be top face. Carbons 2, 3, and 4 are then arranged based on the Fischer projection arrangement at those carbons (C2 right, C3 left, and C4 left).
Exercise $5$
Given that the aldohexose D-mannose differs from D-glucose only in the configuration at the second carbon atom, draw the cyclic structure for α-D-mannose.
Answer
Exercise $6$
Draw the cyclic structure for β-D-glucose. Identify the anomeric carbon.
Answer
Exercise $7$
a) Identify the anomeric carbon of each of the sugars shown below, and specify whether the structure shown is a hemiacetal or hemiketal.
b) Draw mechanisms for cyclization of the open-chain forms to the cyclic forms shown.
Answer
Exercise $8$
Draw a mechanism for the conversion of α-glucopyranose to open-chain glucose.
Answer
6.
Exercise $9$
Identify the following monosaccharide, write its full name, and draw its open-chain form as a Fischer projection.
Answer
β-D-idopyranose. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.05%3A_Cyclic_Structures_of_Monosaccharides%3A_Anomers.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain-lengthening effect of the Kiliani-Fischer synthesis.
2. predict the product that would be produced by the Kiliani-Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani-Fischer synthesis.
1. describe the chain-shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Monosaccharides contain both alcohol and carbonyl functional groups. This allows monosaccharides to undergo many of the reactions typical for these functional groups. In particular, alcohols can be converted to esters, converted to ethers, converted to acetals, or oxidized. Carbonyls can be reacted with nucleophiles, be reduced to form alcohols, or be oxidized to form carboxylic acids.
Ester and Ether Formation
The hydroxyl groups of monosaccharides are often converted to ether or ester groups to increase solubility in organic solvents. Also, the conversion allows the sugar to be easily purified and crystallized.
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5) in the presence of a base. During these reactions all of the -OH groups are converted to esters. The monosaccharide, D-glucopyranose can be coverted to the pentacetate through reaction with acetic anhydride.
Treatment of carbohydrates with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the formation of ethers. The strongly basic conditions typically used for the Williamson ether synthesis can degrade some sugar molecules. Instead, milder bases or silver oxide (Ag2O) are used to provide ethers in high yields. The monosaccharide, D-glucopyranose can be converted to the pentamethyl ether through reaction with iodomethane and silver oxide.
Glycoside Formation
Reacting a hemiacetal with an alcohol and an acid catalyst produces an acetal in which the anomeric hydroxide has been replaced by an ether group.
Monosaccharide acetal derivatives, called glycosides, are formed when a hemiacetal reacts with an alcohol in the presence of an acid catalyst. During the reaction the -OH group from the anomeric carbon is replaced by a -OR group from the alcohol. A mixture of alpha and beta products are formed regardless of the conformation of the reactant. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. This reaction is illustrated below for D-glucopyranose and methanol which forms a mixture of alpha and beta methyl-glucopyranosides. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The acetal product is stable to base and alkaline oxidants such as Tollen's reagent. They are not in equilibrium with the open-chain form and thus do not undergo mutarotation. Since this acid-catalyzed reaction is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed below.
• Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin.
• A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins.
• Amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Koenigs–Knorr Reaction
The presence of multiple -OH groups on a sugar molecular makes the synthesis of glycosides particularly difficult. One of the oldest glycosylation reactions, called the Koenigs–Knorr reaction, is effective for preparing the beta-glycosides of glucose. The pathway starts with the reaction of glucose pentaacetate with HBr to form a pyranosyl bromide. Addition of silver oxide allows for the nucleophilic addition of the chosen alcohol. Hydrolysis of the remaning acetal groups forms the beta-glucopyranoside.
Although the final step of the reaction appears to show the inversion characteristic of an SN2 reaction, the same product forms if either the alpha or beta anomers of the glucopyranosyl bromide is used. This provides evidence that stereochemical control is lost at some point during the mechanism of the reaction.
Koenigs–Knorr Reaction Mechanism
The mechanism starts with the SN1 like removal of the bromine leaving group which is promoted by the formation of an oxonium ion intermediate. Either anomer of the glucopyranosyl bromide will produce the same oxonium intermediate. Lone pair electrons from the carbonyl oxygen of an adjacent acetate group adds to the oxonium ion in an internal ring-forming reaction. Because the adjacent acetate was on the bottom of the glucose ring the newly form C-O bond is also on the bottom. During this step a new oxonium ion is formed. The alcohol then displaces the oxonium ion as a leaving groups during an SN2 reaction. The inversion of configuration of the SN2 reaction produces a beta-glycoside. Participation of an adjacent acetate group in the mechanism of the Koenigs–Knorr reaction is called a neighboring-group effect.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP.
Carbohydrates are covalently attached to many different biomolecules, including lipids, to form glycolipids, and proteins, to form glycoproteins. Called glycoconjugates these sugar bonded molecules are often found in biological membranes, to which they are anchored through covalent bonds. These compounds are crucial to determine to how different cell types recognized one another which will be discussed further in Section 25.11.
Glycoconjugate formation begins with the formation of a phosphorylated sugar such as glucose-6-phosphate which is then reacted with uridine triphosphate (UTP), to give a glycosyl uridine diphosphate. The phosphorylation makes the anomeric -OH of the sugar a better leaving group. During nucleophilic substitutions reaction with a protein or a lipid to form a glycoconjugate the anomeric -OH does not leave as a water molecule, but rather as part of a uridine nucleotide diphosphate group.
To form a glycoprotein the anomeric carbon of a glucopyranose-UDP derivative is attacked from above by an -OH of -NH2 group from a protein. The UDP leaving group is displaced, and inversion of stereochemistry results at the anomeric carbon.
Oxidation of Monosaccharides
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as aqueous Br2 can be used for this conversion. The oxidation occurs by reaction of the open-chain form of the sugar. Because of the equilibrium between the open and ring form of the sugar, eventually the entire sample will undergo the reaction.
Historically sugars have been classified as reducing or non-reducing based on their reactivity with Tollens' (Ag+ in NH3) or Benedict's (Cu2+ and sodium citrate) reagents. If a sugar is oxidized by these reagents it is called a reducing sugar, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
Becuase aldoses contain an aldehyde group they are reducing sugars and will be oxidized by Tollen's and Benedict's reagent. Some ketoses are also reducing sugars. Despite not having an aldehyde groups, fructose is capable of isomerizing to glucose and mannose by keto-enol tautomerism under basic conditions. Once formed, these aldoses are capable of being oxidized by Tollens reagent.
Aldoses can be oxidized to a dicarboxylic acid by dilute nitric acid. Nitric acid is a strong enough oxidizing agent to cause both the aldehyde carbonyl and the -CH2OH group to be oxidized to carboxylic acids. If both ends of an aldose chain are oxidized to carboxylic acids, the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical. Such an operation will disclose any latent symmetry in the remaining molecule and possibly form an achiral meso compound. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown below.
If the -CH2OH group of an aldose is oxidized to a carboxylic acid while the aldehyde carbonyl is not affected, the monocarboxylic acid product is called a uronic acid. This selective oxidation is difficult to accomplish and can only be done enzymatically.
Exercise $1$
1) D-arabinose and D-lyxose produce the same chiral aldaric acid product when oxidized with dilute HNO3. Please explain.
2) Which two D-aldohexoses are oxidized to produce an optically inactive (meso) aldaric acid?
Answer
1) Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
2) D-Allose and D-galactose.
Reduction of Monosaccarides
Treatment of an aldose or ketose with sodium borohydride reduces it to a polyalcohol called an alditol. The reduction occurs by reaction of the open-chain form. Although only a small amount of the open-chain form is present at any given time, that small amount is reduced, more is produced by opening of the pyranose form, that additional amount is reduced, and so on, until the entire sample has undergone reaction. The reaction products can be formally named by removing the -ose ending from the open-chain sugar and replacing it with -itol. Note! The alditol products have identical end groups, HOCH2(CHOH)nCH2OH which means that forming an achiral meso compound is possible.
Exercise $2$
1) Allitol and galactitol from the reduction of allose and galactose are achiral. Explain why this occurs.
2) Altrose and talose are reduced to the same chiral alditol. Explain why this occurs.
Answer
1)
2)
Chain Lengthening: The Kiliani–Fischer Synthesis
The Kiliani–Fischer synthesis lengthens the carbon chain of an aldose by one carbon atom. In doing this two new sugar epimers are created. The synthesis starts by reacting an aldose with HCN. Nucleophilic addition adds the cyanide nucleophile to the electrophilic carbon of the aldose aldehyde forming a cyanohydrin intermediate. The cyanide nucleophile adds a carbon. Stereochemical control is lost, so a racemic mixture of two cyanohydrins differing in the stereochemistry at C2 is formed. The nitrile group of the cyanohydrin is reduced to an imine intermediate by hydrogenation over a palladium catalyst. Finally, the imine is hydrolyzed to an aldehyde to create two new aldoses with an additional carbon. For example performing the Kiliani–Fischer synthesis on D-ribose produces a mixture of D-allose and D-altrose.
Exercise $3$
1) What products would you expect from Kiliani–Fischer reaction of D-xylose?
2) What aldopentose would be expected to produce a mixture of D-xylose and D-lyxose from an Kiliani–Fischer synthesis?
Answer
1) D-Gluose and D-idose
2) D-Threose
Chain Shortening: The Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This is commonly accomplished the Wohl degradation, which is virtually the reverse of the Kiliani–Fischer synthesis. The aldose aldehyde is converted to an oxime (Section 19-8) by treatment with hydroylamine (NH2OH). Dehydration of the oxime with acetic anhydride forms a cyanohydrin. Under basic conditions, the cyanohydrine loses HCN to reform an aldehyde carbonyl. The following equation illustrates the application of this procedure to the aldopentose, D-arabinose to form the aldotetrose, D-erythrose.
Exercise $4$
Two of the four D- aldopentoses yield D-erythrose on Wohl degradation. What are their structures?
Answer
D-Ribose and D-arabinose | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.06%3A_Reactions_of_Monosaccharides.txt |
Objectives
1. identify and draw the eight essential monosaccharides
Eight monosaccharides are required for the proper functioning of human beings. Although they are typically supplied through a healthy diet, they can be biosynthesized if required. The eight monosaccharides are L-fucose, D-galactose, N-acetyl-D-glucosamine, N-acetyl-D-galactosamine, D-glucose, D-mannose, N-acetyl-D-neuraminic acid, and D-xylose.
D-Galactose, D-Glucose, and D-Mannose are commonly found aldohexoses:
D-Xylose is a common aldopentose.
D-Xylose
L-Fucose is a deoxy sugar. Fucose is is the monosaccaride L-galactose with the -OH group on C6 replaced with an -H.
N-acetylgalactosamine and N-acetylglucosamine are derivatives where the corresponding amino sugars have been converted to amides by N-acylation. Amino sugars have had their -OH groups at C2 replaced by an -NH2.
N-Acetylneuraminic acid is nine carbon sugar created by an aldol reaction between N-acetymannosamine and pyruvate. N-acetylneuaminic acid forms a pyranose ring by a intramolecular hemiacetal formation between the -OH on C6 and the ketone functional group in the pyruvate moiety.
Exercise $1$
1) Please show the mechanism of the aldol reaction between N-acetylmannosamine and pyruvate to form neuraminic acid.
Answer
1)
25.08: Disaccharides
Objectives
After completing this section, you should be able to
1. identify disaccharides as compounds consisting of two monosaccharide units joined by a glycoside link between the C1 of one sugar and one of the hydroxyl groups of a second sugar.
2. identify the two monosaccharide units in a given disaccharide.
3. identify the type of glycoside link (e.g., 1,4′‑β) present in a given disaccharide structure.
4. draw the structure of a specific disaccharide, given the structure of the monosaccharide units and the type of glycoside link involved.
Note: If α‑ or β‑D‑glucose were one of the monosaccharide units, its structure would not be provided.
5. identify the structural feature that determines whether or not a given disaccharide behaves as a reducing sugar and undergoes mutarotation, and write equations to illustrate these phenomena.
6. identify the products formed from the hydrolysis of a given disaccharide.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• 1,4′ link
• disaccharide (see Section 25.1)
• invert sugar
Study Notes
Notice that most of the disaccharides discussed in this section contain one unit of D-glucose. You are not expected to remember the detailed structures of maltose, lactose and sucrose. Similarly, we do not expect you to remember the systematic names of these substances.
Previously, you learned that monosaccharides can form cyclic structures by the reaction of the carbonyl group with an OH group. These cyclic molecules can in turn react with another alcohol to form a glycoside. When the alcohol component of a glycoside is provided by a hydroxyl function on another monosaccharide, the compound is called a disaccharide. Disaccharides (C12H22O11) are sugars composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a glycosidic linkage. This linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide. The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them. There are three common disaccharides: maltose, lactose, and sucrose. All three are white crystalline solids at room temperature and are soluble in water.
Maltose
Maltose occurs to a limited extent in sprouting grain. It is formed most often by the partial hydrolysis of starch and glycogen. In the manufacture of beer, maltose is liberated by the action of malt (germinating barley) on starch; for this reason, it is often referred to as malt sugar. Maltose is about 30% as sweet as sucrose. The human body is unable to metabolize maltose or any other disaccharide directly from the diet because the molecules are too large to pass through the cell membranes of the intestinal wall. Therefore, an ingested disaccharide must first be broken down by hydrolysis into its two constituent monosaccharide units. In the body, such hydrolysis reactions are catalyzed by enzymes such as maltase. The same reactions can be carried out in the laboratory with dilute acid as a catalyst, although in that case the rate is much slower, and high temperatures are required. Whether it occurs in the body or a glass beaker, the hydrolysis of maltose produces two molecules of D-glucose
The glucopyranose units in maltose are joined in a head-to-tail fashion through an α-linkage from the first carbon atom of one glucopyranose molecule to the fourth carbon atom of the second glucopyranose molecule (an α-1,4-glycosidic linkage). The bond from the anomeric carbon of the first monosaccharide unit is directed downward, which is why this is known as an α-glycosidic linkage. The OH group on the anomeric carbon of the second glucopyranose can be in either the α or the β position.
Cellobiose is a disaccharide made up of two D-glucopyranose molecules linked with a β-1,4-glycosidic linkage. Like maltose, the OH group on the anomeric carbon of the second glucopyranose of cellobiose can be in either the α or the β position. Although the difference between maltose and cellobiose seems small they have vastly different biological properties. Maltose can be digested by humans whereas cellobiose cannot.
Maltose is a reducing sugar. Thus, its two D-glucopyranose molecules must be linked in such a way as to leave one anomeric carbon that can open to form an aldehyde group. Although one anomeric carbon in maltose and cellobiose is used to form the 1,4-glycosidic linkage, the anomeric carbons on the second glucopyranose remain a hemiacetal. Like an individual monosaccharide, the second glucopyranose subunit undergoes mutarotation and thus is in equilibrium with its aldehyde and both anomer forms. For this reason, maltose and cellobiose contain a mixture of α and β anomers of the second glucopyranose.
Lactose
Lactose is known as milk sugar because it occurs in the milk of humans, cows, and other mammals. In fact, the natural synthesis of lactose occurs only in mammary tissue, whereas most other carbohydrates are plant products. Human milk contains about 7.5% lactose, and cow’s milk contains about 4.5%. This sugar is one of the lowest ranking in terms of sweetness, being about one-sixth as sweet as sucrose. Lactose is produced commercially from whey, a by-product in the manufacture of cheese. It is important as an infant food and in the production of penicillin.
Lactose is a disaccharide composed of one molecule of D-galactopyranose and one molecule of D-glucopyranose joined by a β-1,4-glycosidic bond between the 1 of D-galactopyranose and the C4 of glucose. Lactose is a reducing sugar and undergoes mutarotation to exhibit both anomers of the D-glucopyranose subunit. The two monosaccharides are obtained from lactose by acid hydrolysis or the catalytic action of the enzyme lactase.
Many adults and some children suffer from a deficiency of lactase. These individuals are said to be lactose intolerant because they cannot digest the lactose found in milk. A more serious problem is the genetic disease galactosemia, which results from the absence of an enzyme needed to convert galactose to glucose. Certain bacteria can metabolize lactose, forming lactic acid as one of the products. This reaction is responsible for the “souring” of milk.
Example 1
For this trisaccharide, indicate whether each glycosidic linkage is α or β.
Solution
The glycosidic linkage between sugars 1 and 2 is β because the bond is directed up from the anomeric carbon. The glycosidic linkage between sugars 2 and 3 is α because the bond is directed down from the anomeric carbon.
To Your Health: Lactose Intolerance and Galactosemia
Lactose makes up about 40% of an infant’s diet during the first year of life. Infants and small children have one form of the enzyme lactase in their small intestines and can digest the sugar easily; however, adults usually have a less active form of the enzyme, and about 70% of the world’s adult population has some deficiency in its production. As a result, many adults experience a reduction in the ability to hydrolyze lactose to galactose and glucose in their small intestine. For some people the inability to synthesize sufficient enzyme increases with age. Up to 20% of the US population suffers some degree of lactose intolerance.
In people with lactose intolerance, some of the unhydrolyzed lactose passes into the colon, where it tends to draw water from the interstitial fluid into the intestinal lumen by osmosis. At the same time, intestinal bacteria may act on the lactose to produce organic acids and gases. The buildup of water and bacterial decay products leads to abdominal distention, cramps, and diarrhea, which are symptoms of the condition.
The symptoms disappear if milk or other sources of lactose are excluded from the diet or consumed only sparingly. Alternatively, many food stores now carry special brands of milk that have been pretreated with lactase to hydrolyze the lactose. Cooking or fermenting milk causes at least partial hydrolysis of the lactose, so some people with lactose intolerance are still able to enjoy cheese, yogurt, or cooked foods containing milk. The most common treatment for lactose intolerance, however, is the use of lactase preparations (e.g., Lactaid), which are available in liquid and tablet form at drugstores and grocery stores. These are taken orally with dairy foods—or may be added to them directly—to assist in their digestion.
Galactosemia is a condition in which one of the enzymes needed to convert galactose to glucose is missing. Consequently, the blood galactose level is markedly elevated, and galactose is found in the urine. An infant with galactosemia experiences a lack of appetite, weight loss, diarrhea, and jaundice. The disease may result in impaired liver function, cataracts, mental retardation, and even death. If galactosemia is recognized in early infancy, its effects can be prevented by the exclusion of milk and all other sources of galactose from the diet. As a child with galactosemia grows older, he or she usually develops an alternate pathway for metabolizing galactose, so the need to restrict milk is not permanent. The incidence of galactosemia in the United States is 1 in every 65,000 newborn babies.
Sucrose
Sucrose, probably the largest-selling pure organic compound in the world, is known as beet sugar, cane sugar, table sugar, or simply sugar. Most of the sucrose sold commercially is obtained from sugar cane and sugar beets (whose juices are 14%–20% sucrose) by evaporation of the water and recrystallization. The dark brown liquid that remains after the recrystallization of sugar is sold as molasses.
The sucrose molecule is unique among the common disaccharides in having an α-1,β-2-glycosidic (head-to-head) linkage. Because this glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose, it ties up the anomeric carbons of both glucose and fructose.
This linkage gives sucrose certain properties that are quite different from those of maltose and lactose. As long as the sucrose molecule remains intact, neither monosaccharide is in equilibrium with its open-chain form. Thus, sucrose is incapable of mutarotation and exists in only one form both in the solid state and in solution. In addition, sucrose does not undergo reactions that are typical of aldehydes and ketones. Therefore, sucrose is a nonreducing sugar.
The hydrolysis of sucrose in dilute acid or through the action of the enzyme sucrase (also known as invertase) gives an equimolar mixture of glucose and fructose. This 1:1 mixture is referred to as invert sugar because it rotates plane-polarized light in the opposite direction than sucrose. The hydrolysis reaction has several practical applications. Sucrose readily recrystallizes from a solution, but invert sugar has a much greater tendency to remain in solution. In the manufacture of jelly and candy and in the canning of fruit, the recrystallization of sugar is undesirable. Therefore, conditions leading to the hydrolysis of sucrose are employed in these processes. Moreover, because fructose is sweeter than sucrose, the hydrolysis adds to the sweetening effect. Bees carry out this reaction when they make honey.
The average American consumes more than 100 lb of sucrose every year. About two-thirds of this amount is ingested in soft drinks, presweetened cereals, and other highly processed foods. The widespread use of sucrose is a contributing factor to obesity and tooth decay. Carbohydrates such as sucrose, are converted to fat when the caloric intake exceeds the body’s requirements, and sucrose causes tooth decay by promoting the formation of plaque that sticks to teeth.
Summary
Maltose is composed of two molecules of glucose joined by an α-1,4-glycosidic linkage. It is a reducing sugar that is found in sprouting grain. Lactose is composed of a molecule of galactose joined to a molecule of glucose by a β-1,4-glycosidic linkage. It is a reducing sugar that is found in milk. Sucrose is composed of a molecule of glucose joined to a molecule of fructose by an α-1,β-2-glycosidic linkage. It is a nonreducing sugar that is found in sugar cane and sugar beets.
Exercise \(1\)
What monosaccharides are obtained by the hydrolysis of each disaccharide?
1. sucrose
2. maltose
3. lactose
Answer
1. D-glucose and D-fructose
2. two molecules of D-glucose
3. D-glucose and D-galactose
Exercise \(2\)
For each disaccharide, indicate whether the glycosidic linkage is α or β.
a)
b)
Answer
a)
b)
Exercise \(3\)
Identify each disaccharide in Exercise 2 as a reducing or nonreducing sugar. If it is a reducing sugar, draw its structure and circle the anomeric carbon. State if the OH group at the anomeric carbon is in the α or the β position.
Answer
a) nonreducing
b) reducing
Exercise \(4\)
Melibiose is a disaccharide that occurs in some plant juices. Its structure is as follows:
1. What monosaccharide units are incorporated into melibiose?
2. What type of linkage (α or β) joins the two monosaccharide units of melibiose?
3. Melibiose has a free anomeric carbon and is thus a reducing sugar. Circle the anomeric carbon and indicate whether the OH group is α or β
Answer
a) galactose and glucose
.
b) α-glycosidic linkage
c) β.
Contributors and Attributions
• The Basics of General, Organic, and Biological Chemistry by David W. Ball, John W. Hill, and Rhonda J. Scott. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.07%3A_The_Eight_Essential_Monosaccharides.txt |
Objectives
After completing this section, you should be able to
1. identify the structural difference between cellulose and the cold-water-insoluble fraction of starch (amylose), and identify both of these substances as containing many glucose molecules joined by 1,4′-glycoside links.
2. identify the cold-water-soluble fraction of starch (amylopectin) as having a more complex structure than amylose because of the existence of 1,6′-glycoside links in addition to the 1,4′-links.
3. compare and contrast the structures and uses of starch, glycogen and cellulose.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amylopectin
• amylose
• polysaccharide
The polysaccharides are the most abundant carbohydrates in nature and serve a variety of functions, such as energy storage or as components of plant cell walls. Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages. The three most abundant polysaccharides are starch, glycogen, and cellulose. These three are referred to as homopolymers because each yields only one type of monosaccharide (glucose) after complete hydrolysis. Heteropolymers may contain sugar acids, amino sugars, or noncarbohydrate substances in addition to monosaccharides. Heteropolymers are common in nature (gums, pectins, and other substances) but will not be discussed further in this textbook. The polysaccharides are nonreducing carbohydrates, are not sweet tasting, and do not undergo mutarotation.
Starch
Starch is the most important source of carbohydrates in the human diet and accounts for more than 50% of our carbohydrate intake. It occurs in plants in the form of granules, and these are particularly abundant in seeds (especially the cereal grains) and tubers, where they serve as a storage form of carbohydrates. The breakdown of starch to glucose nourishes the plant during periods of reduced photosynthetic activity. We often think of potatoes as a “starchy” food, yet other plants contain a much greater percentage of starch (potatoes 15%, wheat 55%, corn 65%, and rice 75%). Commercial starch is a white powder.
Starch is a mixture of two polymers: amylose and amylopectin. Natural starches consist of about 10%–30% amylose and 70%–90% amylopectin. Amylose is a linear polysaccharide composed entirely of D-glucose units joined by the α-1,4-glycosidic linkages we saw in maltose (part (a) of Figure \(1\)). Experimental evidence indicates that amylose is not a straight chain of glucose units but instead is coiled like a spring, with six glucose monomers per turn (part (b) of Figure \(1\)). When coiled in this fashion, amylose has just enough room in its core to accommodate an iodine molecule. The characteristic blue-violet color that appears when starch is treated with iodine is due to the formation of the amylose-iodine complex. This color test is sensitive enough to detect even minute amounts of starch in solution.
Amylopectin is a branched-chain polysaccharide composed of glucose units linked primarily by α-1,4-glycosidic bonds but with occasional α-1,6-glycosidic bonds, which are responsible for the branching. A molecule of amylopectin may contain many thousands of glucose units with branch points occurring about every 25–30 units (Figure \(2\)). The helical structure of amylopectin is disrupted by the branching of the chain, so instead of the deep blue-violet color amylose gives with iodine, amylopectin produces a less intense reddish brown.
Dextrins are glucose polysaccharides of intermediate size. The shine and stiffness imparted to clothing by starch are due to the presence of dextrins formed when clothing is ironed. Because of their characteristic stickiness with wetting, dextrins are used as adhesives on stamps, envelopes, and labels; as binders to hold pills and tablets together; and as pastes. Dextrins are more easily digested than starch and are therefore used extensively in the commercial preparation of infant foods.
The complete hydrolysis of starch yields, in successive stages, glucose:
starch → dextrins → maltose → glucose
In the human body, several enzymes known collectively as amylases degrade starch sequentially into usable glucose units.
Glycogen
Glycogen is the energy reserve carbohydrate of animals. Practically all mammalian cells contain some stored carbohydrates in the form of glycogen, but it is especially abundant in the liver (4%–8% by weight of tissue) and in skeletal muscle cells (0.5%–1.0%). Like starch in plants, glycogen is found as granules in liver and muscle cells. When fasting, animals draw on these glycogen reserves during the first day without food to obtain the glucose needed to maintain metabolic balance.
Glycogen is structurally quite similar to amylopectin, although glycogen is more highly branched (8–12 glucose units between branches) and the branches are shorter. When treated with iodine, glycogen gives a reddish brown color. Glycogen can be broken down into its D-glucose subunits by acid hydrolysis or by the same enzymes that catalyze the breakdown of starch. In animals, the enzyme phosphorylase catalyzes the breakdown of glycogen to phosphate esters of glucose.
About 70% of the total glycogen in the body is stored in muscle cells. Although the percentage of glycogen (by weight) is higher in the liver, the much greater mass of skeletal muscle stores a greater total amount of glycogen.
Cellulose
Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%–20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of non-cellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production.
Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by β-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure \(3\)). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure \(3\)). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product.
Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the β-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes.
Polysaccaride Synthesis
The presence of multiple -OH groups in monosaccarides makes the laboratory synthesis of polysaccharides difficult. One modern method for the synthesis of polysaccharides is called the glycal assembly method. A glycal is a sugar which has been dehydrated to form a double bond. Glycals are typically prepared from the corresponding monosaccharide. As part of the synthesis pathway, the primary alcohol of the glycal is protected by forming a silyl ether (Section 17-8). Also, two adjacent secondary alcohols are protected by forming a cyclic carbonate ester. Now there are no -OH groups remanding in the protected glycal. The double bond of the protected glycal is then converted into a epoxide functional groups.
Epoxides ring can be opened by the acid-catalyzed SN2 backside attack by an alcohol (section 18-6). This reaction is exploited by reacting a protected glycal epoxide with a second glycal with an unprotected primary alcohol in the presence ZnCl2 as a Lewis acid. The disaccharide formed in this reaction is a glycal so it can be epoxidized and coupled with another glycal to form a trisaccharide etc. Once the monosaccharides are combined to form a chain of the desired length, the remaining silyl ether and cyclic carbonate protecting groups are removed by hydrolysis and the polysaccharide is formed.
Contributors and Attributions
• The Basics of General, Organic, and Biological Chemistry by David W. Ball, John W. Hill, and Rhonda J. Scott. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.09%3A_Polysaccharides_and_Their_Synthesis.txt |
Objectives
After completing this section, you should be able to identify deoxy and amino sugars, given their structures.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amino sugar
• deoxy sugar
Deoxy Sugars
As shown in Section 25-7, deoxy sugars are missing an oxygen atom. The most common deoxy sugar is 2-deoxyribose, a modified form of another sugar called ribose. When compared to ribose, 2-deoxyribose has an -OH group replaced with an -H at the 2 position. 2-deoxyribose is best known for being the sugar found in the structure of deoxyribonucleic acid (DNA).
In water, 2-deoxyribose is an equilibrium mixture of both the furanose and pyranose ring forms. The pyranose form the most stable (40% alpha anomer and 35% beta anomer), followed by the furanose structures (13% alpha anomer and 12% beta), with the uncyclized form making up the remaining 0.7%.
Pyranose, uncyclized and furanose forms of 2-deoxyribose. The more stable alpha isomer is shown for both the pyranose and furanose forms. Note the missing hydroxide at the 2 position (highlighted magenta)
Amino Sugars
An amino sugar (or more technically a 2-amino-2-deoxysugar) is a sugar molecule in which a hydroxyl group has been replaced with an amine group. More than 60 amino sugars are known, with one of the most abundant being N-acetylglucosamine, which is the main component of chitin.
Chitin is a polymer of 2-deoxy-2- N -ethanamidoglucose (N-acetyl-β-D-glucosamine) and is found in many places throughout the natural world. It is a characteristic component of the cell walls of fungi, the exoskeletons of arthropods such as crustaceans (e.g., crabs, lobsters and shrimps) and insects, the radulae of molluscs, and the beaks and internal shells of cephalopods, including squid and octopuses and on the scales and other soft tissues of fish and lissamphibians. Amino sugars are also found in antibiotics such as amikacin and tobramicin.
25.11: Cell-Surface Carbohydrates and Influenza Viruses
Objectives
After completing this section, you should be able to:
1. Explain how carbohydrates are related to blood type.
2. Explain how influenza works and how it is influenced by carbohydrates.
Oligosaccharides
An oligosaccharide is a saccharide polymer containing a small number (typically two to ten) of monosaccharides. Oligosaccharides can have many functions; for example, they are commonly found on the plasma membrane of animal cells where they can play a role in cell-cell recognition. In general, they are found attached to compatible amino acid side-chains in proteins or to lipids. Oligosaccharides are often found as a component of glycoproteins or glycolipids. They can be used as chemical markers on the outside of cells, often for cell recognition. Oligosaccharides are also responsible for determining blood type.
Glycoproteins
Carbohydrates are covalently attached to many different biomolecules, including lipids, to form glycolipids, and proteins, to form glycoproteins. Glycoproteins and glycolipids are often found in biological membranes, to which they are anchored by through nonpolar interactions. What is the function of these carbohydrates? Two are apparent. First, glycosylation of proteins helps protect the protein from degradation by enzyme catalysts within the body. However, their main functions arises from the fact that covalently attached carbohydrates that "decorate" the surface of glycoproteins or glycolipids provide new binding site interactions that allow interactions with other biomolecules. Hence glycosylation allows for cell:cell, cell:protein, or protein:protein interactions. Unfortunately, bacteria and viruses often recognize glycosylated molecules on cell membranes as well, allowing for their import into the cell.
Blood Type
Cell markers are carbohydrate chains on the surface of cells where they act as “road signs” allowing molecules to distinguish one cell from another. Blood markers are exclusively made from four monosaccharides: D-galactose, L-fucose, N-acetylgalactosamine, and N-acetylglucosamine.
Structures of monosaccharide units present in ABO blood markers
Of the four blood types, type O has the fewest types of saccharides attached to it while type AB has the most. As a result, type O blood is considered the universal donor because it doesn't have any saccharides present that will appear as foreign when transfused into blood of another type. The reverse is not true. For example:
• If type A blood is given to a patient with type O blood, it will be rejected by the body because there is an unknown species being introduced to the body. Type A blood cells contain N-acetylgalactosamine which is not present in type O blood.
• Since type AB blood has all possible saccharides, type AB blood is considered the universal acceptor.
The Rhesus factor (Rh) in blood also affects donor and acceptor properties but it does not depend on carbohydrates. The Rh factor is determined by the presence (Rh+) or absence (Rh-) of a specific protein on the surface of red blood cells.
Influenza and the Avian Flu
Three pandemics of influenza have swept the world since the "Spanish" flu of 1918.
• The "Asian" flu pandemic of 1957;
• the "Hong Kong" flu pandemic of 1968;
• the "Swine" flu pandemic that began in April of 2009.
The influenza virus is a simple yet deadly virus. It interacts with human cells through a surface protein, hemagglutinin (HA). The virus binds to host cells through interaction of HA with cell surface carbohydrates. Once bound the virus internalizes, ultimately leading to release of the RNA genome of the virus into the host cell. The HA protein is the most abundant protein on the viral surface (as surmised by antibody formation).
The influenza virus is typically classified by two kinds of glycoproteins on the surface of the virus: in addition to HA is the enzyme neuraminidase. Two viral strains which have been often discussed are H5N1 and H1N1 which stands for hemagglutinin (H: type 5 or type 1) and neuraminidase (N: type 1). 15 avian and mammalian variants have been identified (based on antibody studies). Only 3 adapted to humans in the last 100 yr, giving pandemic strains H1 (1918), H2 (957) and H3 (1968). Three recent avian variants (H5, H7, and H9) have jumped directly to humans recently but have low human to human transmissibility.
Influenza Virus binds to Cell Surface Glycoproteins with Neu5Ac - A protein on the surface of influenza virus. Hemagluttinin binds to sialic acid (Sia), which is covalently attached to many cell membrane glycoproteins. The sialic acid is usually connected through an alpha (2,3) or alpha (2,6) link to galactose on N-linked glycoproteins. The subtypes found in avian (and equine) influenza isolates bind preferentially to Sia (alpha 2,3) Gal which predominates in avian GI tract where viruses replicate. Human virus of H1, H2, and H3 subtype (cause of the 1918, 1957, and 1968 pandemics) recognize Sia (alpha 2,6) Gal, as the major form in the human respiratory tract. The swine influenza HA bind to Sia (alpha 2,6) Gal and some Sia (alpha 2,3) both of which are commonly found in swine.
The virus, before it leaves the cell, forms a bud on the intracellular side of the cell with the HA and NA in the cell membrane of the host cell. The virus in this state would not leave the cell since its HA molecules would interact with sialic acid residues in the host cell membrane, holding the virus in the membrane. Neuraminidase hydrolyzes sialic acid from cell surface glycoproteins, allowing the virus to complete the budding process and be released from the cell as new viruses. By mimicking the structure of sialic acid, the drugs Oseltamivir (Tamiflu) and zanamivir (Relenza) bind to and inhibit neuraminidase whose activity is necessary for viral release from infected cells. Tamiflu appears to work against N1 of the present H5N1 avian influenza viruses. Governments across the world are stockpiling this drug in case of a pandemic caused by the avian virus jumping directly to humans and becoming transmissible from human to human. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.10%3A_Other_Important_Carbohydrates.txt |
Concepts & Vocabulary
• fulfill all of the detailed objectives listed under each individual section.
• define, and use in context, any of the key terms introduced in this chapter.
25.1 Introduction
• Carbohydrates are composed of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes and ketones.
• Examples of carbohydrates are starch, fibers, sugar, and cellulose.
• Green plants use glucose for energy or making larger carbohydrates.
• Carbohydrates are needed for the synthesis of nucleic acids, proteins, and lipids.
• Animals rely on plants to provide them with carbohydrates, which are used for food and clothing.
25.2 Classification of Carbohydrates
• Carbohydrates originate as products from photosynthesis.
• The generic formula is Cn(H2O)n.
• Carbohydrates play a vital role in a major food source, but are also used as structural material, recognition sites on cells, and other tasks.
• Carbohydrates are called saccharides and can be classified as simple or complex.
25.3 Fischer Projections
• The Fischer projection is a type of notation often used when multiple chiral centers are present in a molecule.
• In a Fischer projection:
• 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).
• Determining whether a chiral carbon is R or S
• If the lowest priority group 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, so you reverse it.
• Epimer is used to designate diastereomers that differ in configuration at only one chiral center.
25.4 D,L Sugars
• To determine absolute configuration for carbohydrates, Emil Fischer started with an arbitrary choice that (+)-glucose would be the D-family and related aldoses.
• The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Emil Fischer as the D / L designator site.
• The L-family is the mirror image of the D-family.
• If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family.
• A left directed hydroxyl group then represented the L-family.
• It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L).
• Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths.
25.5 Configurations of Aldoses
• The aldose family is composed of three to six carbon atoms.
• Emil Fischer determined a way to determine absolute configuration of aldoses.
• In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet, proved that Fischer's choice was correct.
25.6 Anomers
• The preferred structural form of monosaccharides is a cyclic hemiacetal, especially for five and six-membered rings due to low angle strain and eclipsing strain.
• Five-membered cyclic sugars are termed furanose and six-membered cyclic sugars are pyranose.
• By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer.
• The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a Haworth formula.
• In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring.
• The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features.
• Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules.
• When a straight-chain monosaccharide forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers.
• When the OH group on the first carbon atom projected downward, this is called the alpha (α) form.
• When the OH group on the first carbon atom pointed upward, is the beta (β) form.
• These two stereoisomers of a cyclic monosaccharide are known as anomers.
• You can start with a pure crystalline sample of glucose consisting entirely of one anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer.
• The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation.
25.7 Reactions of Monosaccharides
• The -OH groups on a monosaccharide can be readily converted to esters and ethers.
• Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides.
• Phosphorylation of the -OH group using ATP.
• Oxidation can occur with the monosaccharides, which is why some sugars are classified as reducing.
• When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid.
• If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid.
• Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical like aldaric acid.
• Monosaccharides can react to have the chain length increased or decreased, which can be important in elucidation of the carbohydrate's structure.
25.8 The Eight Essential Monosaccharides
• The eight essential monosaccharides are: L-fructose, D-galactose, D-glucose, D-mannose, N-acetyl-D-glucosamine, N-acetyl-D-galacosamine, D-xylose, N-acetyl-D-neuraminic acid.
• These eight monosaccaharides are obtained through diet.
25.9 Disaccharides
• Disaccharides are sugars composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a glycosidic linkage.
• The glycosidic linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide.
• The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them.
• There are three common disaccharides: maltose, lactose, and sucrose.
• Maltose is two glucose molecules linked in a head-to-tail fashion through an α-linkage from the first carbon atom of one glucose molecule to the fourth carbon atom of the second glucose molecule.
• Lactose is composed of one molecule of D-galactose and one molecule of D-glucose joined by a β-1,4-glycosidic bond.
• Sucrose is has an α-1,β-2-glycosidic (head-to-head) linkage, which is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose, it ties up the anomeric carbons of both glucose and fructose.
25.10 Polysaccharides and Their Synthesis
• Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages.
• The three most abundant polysaccharides are starch, glycogen, and cellulose.
25.11 Other Important Carbohydrates
• The backbone of DNA is based on a repeated pattern of a sugar group and a phosphate group, the sugar present - deoxyribose.
• An amino sugar is a sugar molecule in which a hydroxyl group has been replaced with an amine group.
25.12 Cell-Surface Carbohydrates and Influenza Viruses
• Carbohydrates are covalently attached to many different biomolecules, including lipids, to form glycolipids, and proteins, to form glycoproteins.
• Glycoproteins and glycolipids are often found in biological membranes that allow other biomolecules to interact and bind to the protein.
• Bacteria and viruses often recognize glycosylated molecules on cell membranes, allowing for their import into the cell.
• This is how the influenza virus enters the body.
Skills to Master
• Skill 25.1 Determine which molecules can be classified as carbohydrates.
• Skill 25.2 Distinguish between monosaccharide, disaccharide, and trisaccharides given the structure.
• Skill 25.3 Classify a monosaccharide according to the number of carbon atoms present and whether it contains an aldehyde or ketone group.
• Skill 25.4 Draw the Fischer projection of a monosaccharide, given its wedge‑and‑broken‑line structure or a molecular model.
• Skill 25.5 Draw the wedge‑and‑broken‑line structure of a monosaccharide, given its Fischer projection or a molecular model.
• Skill 25.6 Construct a molecular model of a monosaccharide, given its Fischer projection or wedge‑and‑broken‑line structure.
• Skill 25.7 Identify a specific enantiomer of a monosaccharide as being D or L, given its Fischer projection.
• Skill 25.8 Assign an R or S configuration to each of the chiral carbon atoms present in a monosaccharide, given its Fischer projection.
• Skill 25.9 Draw the Fischer projection formula for a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
• Skill 25.10 Draw configurations of aldoses.
• Skill 25.11 Determine whether a given monosaccharide will exist as a pyranose or furanose.
• Skill 25.12 Draw the cyclic pyranose form of a monosaccharide, given its Fischer projection.
• Skill 25.13 Determine whether a given cyclic pyranose form represents the D or L form of the monosaccharide concerned.
• Skill 25.14 Describe the phenomenon known as mutarotation.
• Skill 25.15 Explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
• Skill 25.16 Predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
• Skill 25.17 Identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
• Skill 25.18 Predict the product that would be produced by the Wohl degradation of a given aldose.
• Skill 25.19 Identify the aldose that would yield a given product following Wohl degradation synthesis.
• Skill 25.20 Identify disaccharides as compounds consisting of two monosaccharide units joined by a glycoside link between the C1 of one sugar and one of the hydroxyl groups of a second sugar.
• Skill 25.21 Identify the two monosaccharide units in a given disaccharide.
• Skill 25.22 Identify the type of glycoside link (e.g., 1,4′‑β) present in a given disaccharide structure.
• Skill 25.23 Draw the structure of a specific disaccharide, given the structure of the monosaccharide units and the type of glycoside link involved.
• Skill 25.24 Identify the products formed from the hydrolysis of a given disaccharide.
Summary of Reactions
Chain Length
Oxidation - Reduction
Ester and Ether Formation | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/25%3A_Biomolecules-_Carbohydrates/25.S%3A_Biomolecules-_Carbohydrates_%28Summary%29.txt |
Learning Objectives
When you have completed Chapter 26, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. use the information provided by an amino acid analysis, an Edman degradation and a carboxypeptidase hydrolysis to determine the structure of an unknown polypeptide.
3. outline the approach that you would use to synthesize a given peptide, providing appropriate mechanistic details if requested to do so.
4. define, and use in context, the key terms introduced in this chapter.
Amino acids are important biochemicals, as they are the building blocks from which proteins and polypeptides are assembled. We begin this chapter with an examination of some of the fundamental chemistry of amino acids: their structures, stereochemistry and synthesis. We then discuss the nature of peptides and of the peptide bond, and present the complex issue of determining the order in which the various amino‑acid residues occur in a given peptide. Once a chemist knows the exact order the of the residues in a given peptide, the next challenge is to determine a method by which the same peptide can be prepared in the laboratory. Thus, two sections are devoted to the problem of protein synthesis. The final sections in the chapter deal with the classification, overall structure and denaturation of proteins.
26: Biomolecules- Amino Acids Peptides and Proteins
Objectives
After completing this section, you should be able to
1. give examples of the various biological roles played by proteins.
2. identify amino acids as being the building blocks from which all proteins are made.
3. show, in a general way, how the joining together of a number of amino acids through the formation of peptide bonds results in the formation of proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amino acid
• enzyme
• peptide bond
• protein
Study Notes
The “peptide bond” or “peptide linkage” that is formed between the amino group of one amino acid and the carboxyl group of a second amino acid is identical to the C\$\ce{-}\$N bond present in amides (see Section 21.7). We shall review the nature of such bonds in Section 26.4.
Proteins are polymers of amino acids, linked by amide groups known as peptide bonds. An amino acid can be thought of as having two components: a 'backbone', or 'main chain', composed of an ammonium group, an 'alpha-carbon', and a carboxylate, and a variable 'side chain' (in green below) bonded to the alpha-carbon.
There are twenty different side chains in naturally occurring amino acids, and it is the identity of the side chain that determines the identity of the amino acid: for example, if the side chain is a -CH3 group, the amino acid is alanine, and if the side chain is a -CH2OH group, the amino acid is serine. Many amino acid side chains contain a functional group (the side chain of serine, for example, contains a primary alcohol), while others, like alanine, lack a functional group, and contain only a simple alkane.
The two 'hooks' on an amino acid monomer are the amine and carboxylate groups. Proteins (polymers of ~50 amino acids or more) and peptides (shorter polymers) are formed when the amino group of one amino acid monomer reacts with the carboxylate carbon of another amino acid to form an amide linkage, which in protein terminology is a peptide bond. Which amino acids are linked, and in what order - the protein sequence - is what distinguishes one protein from another, and is coded for by an organism's DNA. Protein sequences are written in the amino terminal (N-terminal) to carboxylate terminal (C-terminal) direction, with either three-letter or single-letter abbreviations for the amino acids (see amino acid table). Below is a four amino acid peptide with the sequence "cysteine - histidine - glutamate - methionine". Using the single-letter code, the sequence is abbreviated CHEM.
When an amino acid is incorporated into a protein it loses a molecule of water and what remains is called a residue of the original amino acid. Thus we might refer to the 'glutamate residue' at position 3 of the CHEM peptide above.
Once a protein polymer is constructed, it in many cases folds up very specifically into a three-dimensional structure, which often includes one or more 'binding pockets' in which other molecules can be bound. It is this shape of this folded structure, and the precise arrangement of the functional groups within the structure (especially in the area of the binding pocket) that determines the function of the protein.
Enzymes are proteins which catalyze biochemical reactions. One or more reacting molecules - often called substrates - become bound in the active site pocket of an enzyme, where the actual reaction takes place. Receptors are proteins that bind specifically to one or more molecules - referred to as ligands - to initiate a biochemical process. For example, we saw in the introduction to this chapter that the TrpVI receptor in mammalian tissues binds capsaicin (from hot chili peppers) in its binding pocket and initiates a heat/pain signal which is sent to the brain.
Shown below is an image of the glycolytic enzyme fructose-1,6-bisphosphate aldolase (in grey), with the substrate molecule bound inside the active site pocket.
(x-ray crystallographic data are from Protein Science 1999, 8, 291; pdb code 4ALD. Image produced with JMol First Glance)
Intro to nucleic acids ⇒
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.01%3A_Introduction.txt |
Objectives
After completing this section, you should be able to
1. identify the structural features present in the 20 amino acids commonly found in proteins.
Note: You are not expected to remember the detailed structures of all these amino acids, but you should be prepared to draw the structures of the two simplest members, glycine and alanine.
2. draw the Fischer projection formula of a specified enantiomer of a given amino acid.
Note: To do so, you must remember that in the S enantiomer, the carboxyl group appears at the top of the projection formula and the amino group is on the left.
3. classify an amino acid as being acidic, basic or neutral, given its Kekulé, condensed or shorthand structure.
4. draw the zwitterion form of a given amino acid.
5. account for some of the typical properties of amino acids (e.g., high melting points, solubility in water) in terms of zwitterion formation.
6. write appropriate equations to illustrate the amphoteric nature of amino acids.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• α‑amino acids
• amphoteric
• essential amino acids
• zwitterion
Study Notes
This is a good point at which to review some of the principles of stereochemistry presented in Chapter 5. Be sure to make full use of molecular models when any stereochemical issues arise.
You should recognize that a three‑letter shorthand code is often used to represent individual amino acids. You need not memorize this code.
The distinction between essential and nonessential amino acids is not as clear‑cut as one might suppose. For example, arginine is often regarded as being nonessential.
Introduction to Amino Acids
Amino acids form polymers through a nucleophilic attack by the amino group of an amino acid at the electrophilic carbonyl carbon of the carboxyl group of another amino acid. The carboxyl group of the amino acid must first be activated to provide a better leaving group than OH-. (We will discuss this activation by ATP later in the course.) The resulting link between the amino acids is an amide link which biochemists call a peptide bond. In this reaction, water is released. In a reverse reaction, the peptide bond can be cleaved by water (hydrolysis).
When two amino acids link together to form an amide link, the resulting structure is called a dipeptide. Likewise, we can have tripeptides, tetrapeptides, and other polypeptides. At some point, when the structure is long enough, it is called a protein. There are many different ways to represent the structure of a polypeptide or protein, each showing differing amounts of information.
Figure: Different Representations of a Polypeptide (Heptapeptide)
(Note: above picture represents the amino acid in an unlikely protonation state with the weak acid protonated and the weak base deprotonated for simplicity in showing removal of water on peptide bond formation and the hydrolysis reaction.) Proteins are polymers of twenty naturally occurring amino acids. In contrast, nucleic acids are polymers of just 4 different monomeric nucleotides. Both the sequence of a protein and its total length differentiate one protein from another. Just for an octapeptide, there are over 25 billion different possible arrangements of amino acids. Compare this to just 65536 different oligonucleotides of 8 monomeric units (8mer). Hence the diversity of possible proteins is enormous.
Stereochemistry
The amino acids are all chiral, with the exception of glycine, whose side chain is H. As with lipids, biochemists use the L and D nomenclature. All naturally occuring proteins from all living organisms consist of L amino acids. The absolute stereochemistry is related to L-glyceraldehyde, as was the case for triacylglycerides and phospholipids. Most naturally occurring chiral amino acids are S, with the exception of cysteine. As the diagram below shows, the absolute configuration of the amino acids can be shown with the H pointed to the rear, the COOH groups pointing out to the left, the R group to the right, and the NH3 group upwards. You can remember this with the anagram CORN.
Figure: Stereochemistry of Amino Acids.
Why do biochemists still use D and L for sugars and amino acids? This explanation (taken from the link below) seems reasonable.
"In addition, however, chemists often need to define a configuration unambiguously in the absence of any reference compound, and for this purpose the alternative (R,S) system is ideal, as it uses priority rules to specify configurations. These rules sometimes lead to absurd results when they are applied to biochemical molecules. For example, as we have seen, all of the common amino acids are L, because they all have exactly the same structure, including the position of the R group if we just write the R group as R. However, they do not all have the same configuration in the (R,S) system: L-cysteine is also (R)-cysteine, but all the other L-amino acids are (S), but this just reflects the human decision to give a sulphur atom higher priority than a carbon atom, and does not reflect a real difference in configuration. Worse problems can sometimes arise in substitution reactions: sometimes inversion of configuration can result in no change in the (R) or (S) prefix; and sometimes retention of configuration can result in a change of prefix.
It follows that it is not just conservatism or failure to understand the (R,S) system that causes biochemists to continue with D and L: it is just that the DL system fulfils their needs much better. As mentioned, chemists also use D and L when they are appropriate to their needs. The explanation given above of why the (R,S) system is little used in biochemistry is thus almost the exact opposite of reality. This system is actually the only practical way of unambiguously representing the stereochemistry of complicated molecules with several asymmetric centres, but it is inconvenient with regular series of molecules like amino acids and simple sugars. "
Natural α-Amino Acids
Hydrolysis of proteins by boiling aqueous acid or base yields an assortment of small molecules identified as α-aminocarboxylic acids. More than twenty such components have been isolated, and the most common of these are listed in the following table. Those amino acids having green colored names are essential diet components, since they are not synthesized by human metabolic processes. The best food source of these nutrients is protein, but it is important to recognize that not all proteins have equal nutritional value. For example, peanuts have a higher weight content of protein than fish or eggs, but the proportion of essential amino acids in peanut protein is only a third of that from the two other sources. For reasons that will become evident when discussing the structures of proteins and peptides, each amino acid is assigned a one or three letter abbreviation.
Natural α-Amino Acids
Some common features of these amino acids should be noted. With the exception of proline, they are all 1º-amines; and with the exception of glycine, they are all chiral. The configurations of the chiral amino acids are the same when written as a Fischer projection formula, as in the drawing on the right, and this was defined as the L-configuration by Fischer. The R-substituent in this structure is the remaining structural component that varies from one amino acid to another, and in proline R is a three-carbon chain that joins the nitrogen to the alpha-carbon in a five-membered ring. Applying the Cahn-Ingold-Prelog notation, all these natural chiral amino acids, with the exception of cysteine, have an S-configuration. For the first seven compounds in the left column the R-substituent is a hydrocarbon. The last three entries in the left column have hydroxyl functional groups, and the first two amino acids in the right column incorporate thiol and sulfide groups respectively. Lysine and arginine have basic amine functions in their side-chains; histidine and tryptophan have less basic nitrogen heterocyclic rings as substituents. Finally, carboxylic acid side-chains are substituents on aspartic and glutamic acid, and the last two compounds in the right column are their corresponding amides.
The formulas for the amino acids written above are simple covalent bond representations based upon previous understanding of mono-functional analogs. The formulas are in fact incorrect. This is evident from a comparison of the physical properties listed in the following table. All four compounds in the table are roughly the same size, and all have moderate to excellent water solubility. The first two are simple carboxylic acids, and the third is an amino alcohol. All three compounds are soluble in organic solvents (e.g. ether) and have relatively low melting points. The carboxylic acids have pKa's near 4.5, and the conjugate acid of the amine has a pKa of 10. The simple amino acid alanine is the last entry. By contrast, it is very high melting (with decomposition), insoluble in organic solvents, and a million times weaker as an acid than ordinary carboxylic acids.
Physical Properties of Selected Acids and Amines
Compound Formula Mol.Wt. Solubility in Water Solubility in Ether Melting Point pKa
isobutyric acid (CH3)2CHCO2H 88 20g/100mL complete -47 ºC 5.0
lactic acid CH3CH(OH)CO2H 90 complete complete 53 ºC 3.9
3-amino-2-butanol CH3CH(NH2)CH(OH)CH3 89 complete complete 9 ºC 10.0
alanine CH3CH(NH2)CO2H 89 18g/100mL insoluble ca. 300 ºC 9.8
Zwitterion
These differences above all point to internal salt formation by a proton transfer from the acidic carboxyl function to the basic amino group. The resulting ammonium carboxylate structure, commonly referred to as a zwitterion, is also supported by the spectroscopic characteristics of alanine.
CH3CH(NH2)CO2H CH3CH(NH3)(+)CO2(–)
As expected from its ionic character, the alanine zwitterion is high melting, insoluble in nonpolar solvents and has the acid strength of a 1º-ammonium ion. Examples of a few specific amino acids may also be viewed in their favored neutral zwitterionic form. Note that in lysine the amine function farthest from the carboxyl group is more basic than the alpha-amine. Consequently, the positively charged ammonium moiety formed at the chain terminus is attracted to the negative carboxylate, resulting in a coiled conformation.
The structure of an amino acid allows it to act as both an acid and a base. An amino acid has this ability because at a certain pH value (different for each amino acid) nearly all the amino acid molecules exist as zwitterions. If acid is added to a solution containing the zwitterion, the carboxylate group captures a hydrogen (H+) ion, and the amino acid becomes positively charged. If base is added, ion removal of the H+ ion from the amino group of the zwitterion produces a negatively charged amino acid. In both circumstances, the amino acid acts to maintain the pH of the system—that is, to remove the added acid (H+) or base (OH) from solution.
Example 26.1
1. Draw the structure for the anion formed when glycine (at neutral pH) reacts with a base.
2. Draw the structure for the cation formed when glycine (at neutral pH) reacts with an acid.
Solution
1. The base removes H+ from the protonated amine group.
The acid adds H+ to the carboxylate group.
Other Natural Amino Acids
The twenty alpha-amino acids listed above are the primary components of proteins, their incorporation being governed by the genetic code. Many other naturally occurring amino acids exist, and the structures of a few of these are displayed below. Some, such as hydroxylysine and hydroxyproline, are simply functionalized derivatives of a previously described compound. These two amino acids are found only in collagen, a common structural protein. Homoserine and homocysteine are higher homologs of their namesakes. The amino group in beta-alanine has moved to the end of the three-carbon chain. It is a component of pantothenic acid, HOCH2C(CH3)2CH(OH)CONHCH2CH2CO2H, a member of the vitamin B complex and an essential nutrient. Acetyl coenzyme A is a pyrophosphorylated derivative of a pantothenic acid amide. The gamma-amino homolog GABA is a neurotransmitter inhibitor and antihypertensive agent.
Many unusual amino acids, including D-enantiomers of some common acids, are produced by microorganisms. These include ornithine, which is a component of the antibiotic bacitracin A, and statin, found as part of a pentapeptide that inhibits the action of the digestive enzyme pepsin. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.02%3A_Structures_of_Amino_Acids.txt |
Objectives
After completing this section, you should be able to
1. draw the predominant form of a given amino acid in a solution of known pH, given the isoelectric point of the amino acid.
2. describe, briefly, how a mixture of amino acids may be separated by paper electrophoresis.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• electrophoresis
• isoelectric point
Since amino acids, as well as peptides and proteins, incorporate both acidic and basic functional groups, the predominant molecular species present in an aqueous solution will depend on the pH of the solution. In order to determine the nature of the molecular and ionic species that are present in aqueous solutions at different pH's, we make use of the Henderson-Hasselbalch Equation, written below. Here, the pKa represents the acidity of a specific conjugate acid function (HA). When the pH of the solution equals pKa, the concentrations of HA and A(-) must be equal (log 1 = 0).
$pK_a = pH + \log_{10} \dfrac{[HA]}{A^-]} \nonumber$
The titration curve for alanine in Figure $2$ demonstrates this relationship. At a pH lower than 2, both the carboxylate and amine functions are protonated, so the alanine molecule has a net positive charge. At a pH greater than 10, the amine exists as a neutral base and the carboxyl as its conjugate base, so the alanine molecule has a net negative charge. At intermediate pH's the zwitterion concentration increases, and at a characteristic pH, called the isoelectric point (pI), the negatively and positively charged molecular species are present in equal concentration. This behavior is general for simple (difunctional) amino acids. Starting from a fully protonated state, the pKa's of the acidic functions range from 1.8 to 2.4 for -CO2H, and 8.8 to 9.7 for -NH3(+). The isoelectric points range from 5.5 to 6.2. Titration curves show the neutralization of these acids by added base, and the change in pH during the titration.
The distribution of charged species in a sample can be shown experimentally by observing the movement of solute molecules in an electric field, using the technique of electrophoresis (Figure $2$). For such experiments an ionic buffer solution is incorporated in a solid matrix layer, composed of paper or a crosslinked gelatin-like substance. A small amount of the amino acid, peptide or protein sample is placed near the center of the matrix strip and an electric potential is applied at the ends of the strip, as shown in the following diagram. The solid structure of the matrix retards the diffusion of the solute molecules, which will remain where they are inserted, unless acted upon by the electrostatic potential.
At pH 6.00 alanine and isoleucine exist on average as neutral zwitterionic molecules, and are not influenced by the electric field. Arginine is a basic amino acid. Both base functions exist as "onium" conjugate acids in the pH 6.00 matrix. The solute molecules of arginine therefore carry an excess positive charge, and they move toward the cathode. The two carboxyl functions in aspartic acid are both ionized at pH 6.00, and the negatively charged solute molecules move toward the anode in the electric field. Structures for all these species are shown to the right of the display.
It should be clear that the result of this experiment is critically dependent on the pH of the matrix buffer. If we were to repeat the electrophoresis of these compounds at a pH of 3.80, the aspartic acid would remain at its point of origin, and the other amino acids would move toward the cathode. Ignoring differences in molecular size and shape, the arginine would move twice as fast as the alanine and isoleucine because its solute molecules on average would carry a double positive charge.
As noted earlier, the titration curves of simple amino acids display two inflection points, one due to the strongly acidic carboxyl group (pKa1 = 1.8 to 2.4), and the other for the less acidic ammonium function (pKa2 = 8.8 to 9.7). For the 2º-amino acid proline, pKa2 is 10.6, reflecting the greater basicity of 2º-amines.
Table $1$: pKa Values of Polyfunctional Amino Acids
Amino Acid α-CO2H pKa1 α-NH3 pKa2 Side Chain pKa3 pI
Arginine 2.1 9.0 12.5 10.8
Aspartic Acid 2.1 9.8 3.9 3.0
Cysteine 1.7 10.4 8.3 5.0
Glutamic Acid 2.2 9.7 4.3 3.2
Histidine 1.8 9.2 6.0 7.6
Lysine 2.2 9.0 10.5 9.8
Tyrosine 2.2 9.1 10.1 5.7
Some amino acids have additional acidic or basic functions in their side chains. These compounds are listed in Table $1$. A third pKa, representing the acidity or basicity of the extra function, is listed in the fourth column of the table. The pI's of these amino acids (last column) are often very different from those noted above for the simpler members. As expected, such compounds display three inflection points in their titration curves, illustrated by the titrations of arginine and aspartic acid (Figure\ (3\)). For each of these compounds four possible charged species are possible, one of which has no overall charge. Formulas for these species are written to the right of the titration curves, together with the pH at which each is expected to predominate. The very high pH required to remove the last acidic proton from arginine reflects the exceptionally high basicity of the guanidine moiety at the end of the side chain.
The Isoelectric Point
The isoelectric point, pI, is the pH of an aqueous solution of an amino acid (or peptide) at which the molecules on average have no net charge. In other words, the positively charged groups are exactly balanced by the negatively charged groups. For simple amino acids such as alanine, the pI is an average of the pKa's of the carboxyl (2.34) and ammonium (9.69) groups. Thus, the pI for alanine is calculated to be: (2.34 + 9.69)/2 = 6.02, the experimentally determined value. If additional acidic or basic groups are present as side-chain functions, the pI is the average of the pKa's of the two most similar acids. To assist in determining similarity we define two classes of acids. The first consists of acids that are neutral in their protonated form (e.g. CO2H & SH). The second includes acids that are positively charged in their protonated state (e.g. -NH3+). In the case of aspartic acid, the similar acids are the alpha-carboxyl function (pKa = 2.1) and the side-chain carboxyl function (pKa = 3.9), so pI = (2.1 + 3.9)/2 = 3.0. For arginine, the similar acids are the guanidinium species on the side-chain (pKa = 12.5) and the alpha-ammonium function (pKa = 9.0), so the calculated pI = (12.5 + 9.0)/2 = 10.75 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.03%3A_Amino_Acids_the_Henderson-Hasselbalch_Equation_and_Isoelectric_Points.txt |
Objectives
After completing this section, you should be able to
1. outline, by means of equations, how a racemic mixture of given amino acid can be prepared from a carboxylic acid using reactions you studied earlier in the course.
1. outline, by means of equations, the preparation of a given amino acid by the amidomalonate synthesis.
2. identify the amino acid formed from using a given alkyl halide in an amidomalonate synthesis.
3. identify the alkyl halide needed to produce a given amino acid by the amidomalonate synthesis.
2. describe, by means of equations, how an α‑keto acid can be transformed to an amino acid by reductive amination.
1. describe a general method for resolving a racemic mixture of a given amino acid.
2. provide a brief example of how a biological method may be employed to resolve a racemic mixture of a given amino acid.
3. show the enantioselective preparation of an amino acid from the corresponding Z enamido acid.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amidomalonate synthesis
• enantioselective synthesis
• racemic mixture
Study Notes
Do not be alarmed by the number of methods to synthesize amino acids described in this section.You have seen many of these reactions in previous sections and should already be familiar with the approaches discussed here.
To fulfill the requirements of Objective 1, review the Hell‑Volhard‑Zelinskii reaction (Section 22.4) and the Gabriel phthalimide synthesis (Section 24.6).
The amidomalonate synthesis is a simple variation of the malonic ester synthesis (Section 22.7). A base abstracts a proton from the alpha carbon, which is then alkylated with an alkyl halide. Then both the hydrolysis of the esters and the amide protecting group under aqueous acidic conditions generates the α‑amino acid.
Another method of getting to the α‑amino acid is by reductive amination of the α‑keto acid which you have also previously encountered (Section 24.6).
Synthesis of α-Amino Acids
1) Amination of alpha-bromocarboxylic acids, illustrated by the following equation, provides a straightforward method for preparing alpha-aminocarboxylic acids. The bromoacids, in turn, are conveniently prepared from carboxylic acids by reaction with Br2 + PCl3. Although this direct approach gave mediocre results when used to prepare simple amines from alkyl halides, it is more effective for making amino acids, thanks to the reduced nucleophilicity of the nitrogen atom in the product. Nevertheless, more complex procedures that give good yields of pure compounds are often chosen for amino acid synthesis.
2) By modifying the nitrogen as a phthalimide salt, the propensity of amines to undergo multiple substitutions is removed, and a single clean substitution reaction of 1º- and many 2º-alkylhalides takes place. This procedure, known as the Gabriel synthesis, can be used to advantage in aminating bromomalonic esters, as shown in the upper equation of the following scheme. Since the phthalimide substituted malonic ester has an acidic hydrogen (colored orange), activated by the two ester groups, this intermediate may be converted to an ambident anion and alkylated. Finally, base catalyzed hydrolysis of the phthalimide moiety and the esters, followed by acidification and thermal decarboxylation, produces an amino acid and phthalic acid (not shown).
3) An elegant procedure, known as the Strecker synthesis, assembles an alpha-amino acid from ammonia (the amine precursor), cyanide (the carboxyl precursor), and an aldehyde. This reaction (shown below) is essentially an imino analog of cyanohydrin formation. The alpha-amino nitrile formed in this way can then be hydrolyzed to an amino acid by either acid or base catalysis.
4) Resolution The three synthetic procedures described above, and many others that can be conceived, give racemic amino acid products. If pure L or D enantiomers are desired, it is necessary to resolve these racemic mixtures. A common method of resolving racemates is by diastereomeric salt formation with a pure chiral acid or base. This is illustrated for a generic amino acid in the following diagram. Be careful to distinguish charge symbols, shown in colored circles, from optical rotation signs, shown in parenthesis.
In the initial display, the carboxylic acid function contributes to diastereomeric salt formation. The racemic amino acid is first converted to a benzamide derivative to remove the basic character of the amino group. Next, an ammonium salt is formed by combining the carboxylic acid with an optically pure amine, such as brucine (a relative of strychnine). The structure of this amine is not shown, because it is not a critical factor in the logical progression of steps. Since the amino acid moiety is racemic and the base is a single enantiomer (levorotatory in this example), an equimolar mixture of diastereomeric salts is formed (drawn in the green shaded box). Diastereomers may be separated by crystallization, chromatography or other physical manipulation, and in this way one of the isomers may be isolated for further treatment, in this illustration it is the (+):(-) diastereomer. Finally the salt is broken by acid treatment, giving the resolved (+)-amino acid derivative together with the recovered resolving agent (the optically active amine). Of course, the same procedure could be used to obtain the (-)-enantiomer of the amino acid.
Since amino acids are amphoteric, resolution could also be achieved by using the basic character of the amine function. For this approach we would need an enantiomerically pure chiral acid such as tartaric acid to use as the resolving agent. This alternative resolution strategy will be illustrated. Note that the carboxylic acid function is first esterified, so that it will not compete with the resolving acid.
Resolution of aminoacid derivatives may also be achieved by enzymatic discrimination in the hydrolysis of amides. For example, an aminoacylase enzyme from pig kidneys cleaves an amide derivative of a natural L-amino acid much faster than it does the D-enantiomer. If the racemic mixture of amides shown in the green shaded box above is treated with this enzyme, the L-enantiomer (whatever its rotation) will be rapidly converted to its free zwitterionic form, whereas the D-enantiomer will remain largely unchanged. Here, the diastereomeric species are transition states rather than isolable intermediates. This separation of enantiomers, based on very different rates of reaction, is called kinetic resolution.
Enantioselective Synthesis
Till now all of the synthetic routes to α-amino acids we have discussed yield a racemic mixture. Once produced one could resolve the mixture to obtain pure L or D enantiomers. However, enantioselective synthetic methods to produce pure compounds directly are being developed. For instance, several catalysts are now available for reduction of C=C to expose enantiopure amino acids. A good example is the industrial synthesis of L-DOPA, a drug used in the treatment of Parkinson’s disease. W.S. Knowles shared the 2001 Nobel Price with R. Noyori and K.B. Sharpless for their contributions in the area of asymmetric catalytic reductions. Knowles developed several chiral phosphine–metal catalysts for asymmetric reductions. The rhodium(I) catalyst shown, which is complexed by large organic ligands, facilitates production of almost pure L-DOPA. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.04%3A_Synthesis_of_Amino_Acids.txt |
Objectives
After completing this section, you should be able to
1. show, by means of a diagram, how two different amino acid residues can be combined to give two different dipeptides.
2. draw the structure of a relatively simple peptide, given its full or abbreviated name and the structures of the appropriate amino acids.
3. draw, or name, the six possible isomeric tripeptides that can be formed by combining three different amino acid residues (amino acid units) of given structure.
1. account for the fact that there is restricted rotation about the C\$\ce{-}\$N bonds in peptides.
2. illustrate the formation of a disulfide linkage between two cysteine residues, and show how such bonds can link together two separate peptide chains or can provide a bridge between two cysteine residues present in a single peptide molecule.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• C‑terminal amino acid
• N‑terminal amino acid
• peptides
• residues
Study Notes
If necessary, review the discussion of the delocalization of the nitrogen lone‑pair electrons in amides that was presented in Section 24.3. Similarly, you may wish to refer back to Section 18.8 to review the interconversion of thiols and disulfides.
Peptide Bond Formation or Amide Synthesis
The formation of peptides is nothing more than the application of the amide synthesis reaction. By convention, the amide bond in the peptides should be made in the order that the amino acids are written. The amine end (N terminal) of an amino acid is always on the left, while the acid end (C terminal) is on the right. The reaction of glycine with alanine to form the dipeptide glyclalanine is written as shown in the graphic on the left. Oxygen (red) from the acid and hydrogens (red) on the amine form a water molecule. The carboxyl oxygen (green) and the amine nitrogen (green) join to form the amide bond.
If the order of listing the amino acids is reversed, a different dipeptide is formed such as alaninylglycine.
Exercise \(1\)
Write the reactions for:
1. ala + gly ---> Answer graphic
2. phe + ser ----> Answer graphic
Resonance contributors for the peptide bonds
A consideration of resonance contributors is crucial to any discussion of the amide functional group. One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.
Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond.
This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly suggests that the nitrogen is sp2-hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen.
Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond.
Backbone Peptide or Protein Structure
The structure of a peptide can be written fairly easily without showing the complete amide synthesis reaction by learning the structure of the "backbone" for peptides and proteins.
The peptide backbone consists of repeating units of "N-H 2, CH, C double bond O; N-H 2, CH, C double bond O; etc. See the graphic on the left .
After the backbone is written, go back and write the specific structure for the side chains as represented by the "R" as gly-ala-leu for this example. The amine end (N terminal) of an amino acid is always on the left (gly), while the acid end (C terminal) is on the right (leu).
Exercise \(2\)
Write the tripeptide structure for val-ser-cys. First write the "backbone" and then add the specific side chains.
Solution
Answer graphic
QUES. Write the structure for the tripeptide:
2 a ) glu-cys-gly ---> Answer graphic
2 b) phe-tyr-asn ---> Answer graphic
Disulfide Bridges and Oxidation-Reduction
The amino acid cysteine undergoes oxidation and reduction reactions involving the -SH (sulfhydryl group). The oxidation of two sulfhydryl groups results in the formation of a disulfide bond by the removal of two hydrogens. The oxidation of two cysteine amino acids is shown in the graphic. An unspecified oxidizing agent (O) provides an oxygen which reacts with the hydrogen (red) on the -SH group to form water. The sulfurs (yellow) join to make the disulfide bridge. This is an important bond to recognize in protein tertiary structure. The reduction of a disulfide bond is the opposite reaction which again leads to two separate cysteine molecules. Remember that reduction is the addition of hydrogen.
Cysteine residues in the the peptide chain can form a loop buy forming the disulfide bond (—S—S—), while cysteine residues in different peptide chains can actually link what were otherwise separate chains. Insulin was the first protein whose amino acid sequence was determined. This pioneering work, completed in 1953 after some 10 years of effort, earned a Nobel Prize for British biochemist Frederick Sanger (born 1918). He found the primary structure to comprise of two chains linked by two cysteine disulfide bridges. Also note the first peptide chain possesses an internal loop.
Insulin | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.05%3A_Peptides_and_Proteins.txt |
Objectives
After completing this section, you should be able to describe, briefly, how the identity and amounts of each amino acid residue present in a peptide of unknown structure may be determined.
Key Terms
Make certain that you can define, and use in context, the key term below.
• amino acid analyzer
Study Notes
You need not memorize the reaction between ninhydrin and an α‑amino acid.
Ion-exchange chromatography
When a protein is to be analysed, it is first heated with acid to hydrolyse all the peptide bonds. When such a mixture of amino acids is to be purified and estimated quantitatively, ion-exchange chromatography is the technique of choice. Fully automated amino acid analyzers are now available, which are equipped with a solvent pump to deliver the required buffer(s) in a programmed manner. There is a column, filled with Dowex 50 resin (Fig 26.5.1). This solid support is made up of polymeric beads. Chemically speaking they are polymers bearing arylsulfonic acid groups. The cation exchange resin helps in the separation of amino acids. In a typical run (Fig 26.5.2), the eluent is a buffer. The pH value of the buffer could be varied as step elution or as gradient elution. The chromatogram shown in Fig 26.5.2 is a chromatogram run with gradient elution technique, using ninhydrin as the post column treatment. The detector is a UV detector scanning the wavelengths 570 nm and 440 nm.
Fig 26.5.1: A Cation Rasin like Dowex 50 is a polymeric bead bearing aryl sulfonic acid groups
Fig 26.5.2: Some typical chromatograms from an amino acid analyzer
The Ninhydrin Reaction
Alpha-amino acids show reactivity at their the carboxylic acid and amine sites typical of those functional groups. In addition to these common reactions of amines and carboxylic acids, common alpha-amino acids, except proline, undergo a unique reaction with the triketohydrindene hydrate known as ninhydrin. Among the products of this unusual reaction (shown on the left below) is a purple colored imino derivative, which provides as a useful color test for these amino acids, most of which are colorless. A common application of the ninhydrin test is the visualization of amino acids in paper chromatography. As shown in the graphic on the right, samples of amino acids or mixtures thereof are applied along a line near the bottom of a rectangular sheet of paper (the baseline). The bottom edge of the paper is immersed in an aqueous buffer, and this liquid climbs slowly toward the top edge. As the solvent front passes the sample spots, the compounds in each sample are carried along at a rate which is characteristic of their functionality, size and interaction with the cellulose matrix of the paper. Some compounds move rapidly up the paper, while others may scarcely move at all. The ratio of the distance a compound moves from the baseline to the distance of the solvent front from the baseline is defined as the retardation (or retention) factor Rf. Different amino acids usually have different Rf's under suitable conditions. In the example on the right, the three sample compounds (1, 2 & 3) have respective Rf values of 0.54, 0.36 & 0.78. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.06%3A_Amino_Acid_Analysis_of_Peptides.txt |
Objectives
After completing this section, you should be able to
1. describe how an Edman degradation is used to determine the sequence of the amino acid residues in peptides containing up to 20 such residues.
2. describe, briefly, how the procedure is modified to deal with peptides and proteins containing more than 20 amino acid residues.
3. write a detailed mechanism for the Edman degradation.
4. determine the structure of a peptide, given a list of the fragments that are produced by a partial acid hydrolysis.
5. determine the structure of a peptide, given a list of the fragments that are produced when the peptide is cleaved by a specific enzyme and the details of the types of bonds cleaved by that enzyme.
6. predict the fragments that would be produced when a peptide of known structure is cleaved by a specific enzyme, given sufficient information about the types of bonds that are cleaved by the enzyme in question.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Edman degradation
Study Notes
The reagent used in the Edman degradation is phenyl isothiocyanate. You may find it helpful to review the relationship between cyanates, isocyanates, thiocyanates and isothiocyanates.
You need not memorize the specific peptide bonds that are broken by the enzymes trypsin and chymotrypsin.
Edman degradation is the process of purifying protein by sequentially removing one residue at a time from the amino end of a peptide. To solve the problem of damaging the protein by hydrolyzing conditions, Pehr Edman created a new way of labeling and cleaving the peptide. Edman thought of a way of removing only one residue at a time, which did not damage the overall sequencing. This was done by adding Phenyl isothiocyanate, which creates a phenylthiocarbamoyl derivative with the N-terminal. The N-terminal is then cleaved under less harsh acidic conditions, creating a cyclic compound of phenylthiohydantoin PTH-amino acid. This does not damage the protein and leaves two constituents of the peptide. This method can be repeated for the rest of the residues, separating one residue at a time.
Edman degradation is very useful because it does not damage the protein. This allows sequencing of the protein to be done in less time. Edman sequencing is done best if the composition of the amino acid is known. As we saw in Section 26.5, to determine the composition of the amino acid, the peptide must be hydrolyzed. This can be done by denaturing the protein and heating it and adding HCl for a long time. This causes the individual amino acids to be separated, and they can be separated by ion exchange chromatography. They are then dyed with ninhydrin and the amount of amino acid can be determined by the amount of optical absorbance. This way, the composition but not the sequence can be determined
Sequencing Larger Proteins
Larger proteins cannot be sequenced by the Edman sequencing because of the less than perfect efficiency of the method. A strategy called divide and conquer successfully cleaves the larger protein into smaller, practical amino acids. This is done by using a certain chemical or enzyme which can cleave the protein at specific amino acid residues. The separated peptides can be isolated by chromatography. Then they can be sequenced using the Edman method, because of their smaller size.
In order to put together all the sequences of the different peptides, a method of overlapping peptides is used. The strategy of divide and conquer followed by Edman sequencing is used again a second time, but using a different enzyme or chemical to cleave it into different residues. This allows two different sets of amino acid sequences of the same protein, but at different points. By comparing these two sequences and examining for any overlap between the two, the sequence can be known for the original protein.
For example, trypsin can be used on the initial peptide to cleave it at the carboxyl side of arginine and lysine residues. Using trypsin to cleave the protein and sequencing them individually with Edman degradation will yield many different individual results. Although the sequence of each individual cleaved amino acid segment is known, the order is scrambled. Chymotrypsin, which cleaves on the carboxyl side of aromatic and other bulky nonpolar residues, can be used. The sequence of these segments overlap with those of the trypsin. They can be overlapped to find the original sequence of the initial protein. However, this method is limited in analyzing larger sized proteins (more than 100 amino acids) because of secondary hydrogen bond interference. Other weak intermolecular bonding such as hydrophobic interactions cannot be properly predicted. Only the linear sequence of a protein can be properly predicted assuming the sequence is small enough.
Contributors and Attributions
• Wikibooks (Structural Biopchemistry). The content on this page is licensed under a CC-SA-BY 3.0 license. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.07%3A_The_Edman_Degradation.txt |
Objectives
After completing this section, you should be able to
1. describe why it is necessary to protect certain amino and carboxyl groups during the synthesis of a peptide.
2. describe, using appropriate equations, how carboxyl groups are protected by ester formation and amino groups are protected by the formation of their tert‑butoxycarbonyl amide derivatives.
3. write a detailed mechanism for the formation of a peptide link between an amino acid with a protected amino group and an amino acid with a protected carboxyl group using dicyclohexylcarbodiimide.
4. outline the five steps required in order to form a dipeptide from two given amino acids.
In order to synthesize a peptide from its component amino acids, two obstacles must be overcome. The first of these is statistical in nature, and is illustrated by considering the dipeptide Ala-Gly as a proposed target. If we ignore the chemistry involved, a mixture of equal molar amounts of alanine and glycine would generate four different dipeptides. These are: Ala-Ala, Gly-Gly, Ala-Gly & Gly-Ala. In the case of tripeptides, the number of possible products from these two amino acids rises to eight. Clearly, some kind of selectivity must be exercised if complex mixtures are to be avoided.
The second difficulty arises from the fact that carboxylic acids and 1º or 2º-amines do not form amide bonds on mixing, but will generally react by proton transfer to give salts (the intermolecular equivalent of zwitterion formation).
From the perspective of an organic chemist, peptide synthesis requires selective acylation of a free amine. To accomplish the desired amide bond formation, we must first deactivate all extraneous amine functions so they do not compete for the acylation reagent. Then we must selectively activate the designated carboxyl function so that it will acylate the one remaining free amine. Fortunately, chemical reactions that permit us to accomplish these selections are well known.
First, the basicity and nucleophilicity of amines are substantially reduced by amide formation. Consequently, the acylation of amino acids by treatment with acyl chlorides or anhydrides at pH > 10, as described earlier, serves to protect their amino groups from further reaction.
Second, acyl halide or anhydride-like activation of a specific carboxyl reactant must occur as a prelude to peptide (amide) bond formation. This is possible, provided competing reactions involving other carboxyl functions that might be present are precluded by preliminary ester formation. Remember, esters are weaker acylating reagents than either anhydrides or acyl halides, as noted earlier.
Finally, dicyclohexylcarbodiimide (DCC) effects the dehydration of a carboxylic acid and amine mixture to the corresponding amide under relatively mild conditions. The structure of this reagent and the mechanism of its action have been described. Its application to peptide synthesis will become apparent in the following discussion.
The strategy for peptide synthesis, as outlined here, should now be apparent. The following example shows a selective synthesis of the dipeptide Ala-Gly.
An important issue remains to be addressed. Since the N-protective group is an amide, removal of this function might require conditions that would also cleave the just formed peptide bond. Furthermore, the harsh conditions often required for amide hydrolysis might cause extensive racemization of the amino acids in the resulting peptide. This problem strikes at the heart of our strategy, so it is important to give careful thought to the design of specific N-protective groups. In particular, three qualities are desired:
1. The protective amide should be easy to attach to amino acids.
2. The protected amino group should not react under peptide forming conditions.
3. The protective amide group should be easy to remove under mild conditions.
A number of protective groups that satisfy these conditions have been devised; and two of the most widely used, carbobenzoxy (Cbz) and t-butoxycarbonyl (BOC or t-BOC), are described here.
The reagents for introducing these N-protective groups are the acyl chlorides or anhydrides shown in the left portion of the above diagram. Reaction with a free amine function of an amino acid occurs rapidly to give the "protected" amino acid derivative shown in the center. This can then be used to form a peptide (amide) bond to a second amino acid. Once the desired peptide bond is created the protective group can be removed under relatively mild non-hydrolytic conditions. Equations showing the protective group removal will be displayed above by are shown above. Cleavage of the reactive benzyl or tert-butyl groups generates a common carbamic acid intermediate (HOCO-NHR) which spontaneously loses carbon dioxide, giving the corresponding amine. If the methyl ester at the C-terminus is left in place, this sequence of reactions may be repeated, using a different N-protected amino acid as the acylating reagent. Removal of the protective groups would then yield a specific tripeptide, determined by the nature of the reactants and order of the reactions.
26.09: The Merrifield Solid-Phase Technique
Objectives
After completing this section, you should be able to describe, briefly, the Merrifield solid‑phase technique for the synthesis of polypeptides.
Key Terms
Make certain that you can define, and use in context, the key term below.
• solid‑phase method (solid‑phase synthesis)
Study Notes
The solid‑phase used in this method is a polymer support. You will not be examined on the details of the Merrifield solid‑phase method; however, you should be prepared to write a couple of paragraphs describing this important process.
For his work on the synthesis of peptides, Bruce Merrifield was awarded the 1984 Nobel Prize in chemistry.
The synthesis of a peptide of significant length (e.g. ten residues) by this approach requires many steps, and the product must be carefully purified after each step to prevent unwanted cross-reactions. To facilitate the tedious and time consuming purifications, and reduce the material losses that occur in handling, a clever modification of this strategy has been developed. This procedure, known as the Merrifield Synthesis after its inventor R. Bruce Merrifield, involves attaching the C-terminus of the peptide chain to a polymeric solid, usually having the form of very small beads. Separation and purification is simply accomplished by filtering and washing the beads with appropriate solvents. The reagents for the next peptide bond addition are then added, and the purification steps repeated. The entire process can be automated, and peptide synthesis machines based on the Merrifield approach are commercially available. A series of equations illustrating the Merrifield synthesis may be viewed below. The final step, in which the completed peptide is released from the polymer support, is a simple benzyl ester cleavage. This is not shown in the display.
The Merrifield Peptide Synthesis
Two or more moderately sized peptides can be joined together by selective peptide bond formation, provided side-chain functions are protected and do not interfere. In this manner good sized peptides and small proteins may be synthesized in the laboratory. However, even if chemists assemble the primary structure of a natural protein in this or any other fashion, it may not immediately adopt its native secondary, tertiary and quaternary structure. Many factors, such as pH, temperature and inorganic ion concentration influence the conformational coiling of peptide chains. Indeed, scientists are still trying to understand how and why these higher structures are established in living organisms. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.08%3A_Peptide_Synthesis.txt |
Objectives
After completing this section, you should be able to
1. discuss, with reference to a suitable example (either given or of your own choice), the structure of proteins, paying particular attention to distinguishing between the primary, secondary, tertiary and quaternary structure.
2. describe the α‑helical secondary structure displayed by many proteins.
3. describe the β‑pleated‑sheet structure displayed by many proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• α helix
• β pleated sheet
• primary structure
• quaternary structure
• secondary structure
• tertiary structure
Study Notes
Note that in a diagram of the α‑helical structure of a protein, the C‑terminal of the protein is at the bottom of the diagram and the N‑terminal is at the top. In an α helix, such as the one shown in Figure 26.9.1, the bulky R groups are all found on the outside of the helix, where they have the most room.
The four levels of protein structure
Protein structure can be discussed at four distinct levels. A protein’s primary structure is two-dimensional - simply the sequence of amino acids in the peptide chain. Below is a Lewis structure of a short segment of a protein with the sequence CHEM (cysteine - histidine - glutamate - methionine)
Secondary structure is three-dimensional, but is a local phenomenon, confined to a relatively short stretch of amino acids. For the most part, there are three important elements of secondary structure: helices, beta-sheets, and loops. In a helix, the main chain of the protein adopts the shape of a clockwise spiral staircase, and the side chains point out laterally.
In a beta-sheet (or beta-strand) structure, two sections of protein chain are aligned side-by-side in an extended conformation. The figure below shows two different views of the same beta-sheet: in the left-side view, the two regions of protein chain are differentiated by color.
Loops are relatively disordered segments of protein chain, but often assume a very ordered structure when in contact with a second protein or a smaller organic compound.
Both helix and the beta-sheet structures are held together by very specific hydrogen-bonding interactions between the amide nitrogen on one amino acid and the carbonyl oxygen on another. The hydrogen bonding pattern in a section of a beta-strand is shown below.
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
α-Helices
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil. Such a hydrogen bond is formed exactly every 4 amino acid residues, and every complete turn of the helix is only 3.6 amino acid residues. This regular pattern gives the α-helix very definite features with regards to the thickness of the coil and the length of each complete turn along the helix axis.
The structural integrity of an α-helix is in part dependent on correct steric configuration. Amino acids whose R-groups are too large (tryptophan, tyrosine) or too small (glycine) destabilize α-helices. Proline also destabilizes α-helices because of its irregular geometry; its R-group bonds back to the nitrogen of the amide group, which causes steric hindrance. In addition, the lack of a hydrogen on Proline's nitrogen prevents it from participating in hydrogen bonding.
Another factor affecting α-helix stability is the total dipole moment of the entire helix due to individual dipoles of the C=O groups involved in hydrogen bonding. Stable α-helices typically end with a charged amino acid to neutralize the dipole moment.
BETA-PLEATED SHEETS
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement:
Or in anti-parallel arrangement:
Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain. In anti-parallel arrangement, the C-terminus end of one segment is on the same side as the N-terminus end of the other segment. In parallel arrangement, the C-terminus end and the N-terminus end are on the same sides for both segments. The "pleat" occurs because of the alternating planes of the peptide bonds between amino acids; the aligned amino and carbonyl group of each opposite segment alternate their orientation from facing towards each other to facing opposite directions.
The parallel arrangement is less stable because the geometry of the individual amino acid molecules forces the hydrogen bonds to occur at an angle, making them longer and thus weaker. Contrarily, in the anti-parallel arrangement the hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds.
Commonly, an anti-parallel beta-pleated sheet forms when a polypeptide chain sharply reverses direction. This can occur in the presence of two consecutive proline residues, which create an angled kink in the polypeptide chain and bend it back upon itself. This is not necessary for distant segments of a polypeptide chain to form beta-pleated sheets, but for proximal segments it is a definite requirement. For short distances, the two segments of a beta-pleated sheet are separated by 4+2n amino acid residues, with 4 being the minimum number of residues.
α-PLEATED SHEETS
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction. The polarization of the amino and carbonyl groups results in a net dipole moment on the α-pleated sheet. The carbonyl side acquires a net negative charge, and the amino side acquires a net positive charge.
A protein’s tertiary structure is the shape in which the entire protein chain folds together in three-dimensional space, and it is this level of structure that provides protein scientists with the most information about a protein’s specific function.
While a protein's secondary and tertiary structure is defined by how the protein chain folds together, quaternary structure is defined by how two or more folded protein chains come together to form a 'superstructure'. Many proteins consist of only one protein chain, or subunit, and thus have no quaternary structure. Many other proteins consist of two identical subunits (these are called homodimers) or two non-identical subunits (these are called heterodimers).
Quaternary structures can be quite elaborate: below we see a protein whose quaternary structure is defined by ten identical subunits arranged in two five-membered rings, forming what can be visualized as a 'double donut' shape (this is fructose 1,6-bisphosphate aldolase):
The molecular forces that hold proteins together
The question of exactly how a protein ‘finds’ its specific folded structure out of the vast number of possible folding patterns is still an active area of research. What is known, however, is that the forces that cause a protein to fold properly and to remain folded are the same basic noncovalent forces that we talked about in chapter 2: ion-ion, ion-dipole, dipole-dipole, hydrogen bonding, and hydrophobic (van der Waals) interactions. One interesting type of hydrophobic interaction is called ‘aromatic stacking’, and occurs when two or more planar aromatic rings on the side chains of phenylalanine, tryptophan, or tyrosine stack together like plates, thus maximizing surface area contact.
Hydrogen bonding networks are extensive within proteins, with both side chain and main chain atoms participating. Ionic interactions often play a role in protein structure, especially on the protein surface, as negatively charged residues such as aspartate interact with positively-charged groups on lysine or arginine.
One of the most important ideas to understand regarding tertiary structure is that a protein, when properly folded, is polar on the surface and nonpolar in the interior. It is the protein's surface that is in contact with water, and therefore the surface must be hydrophilic in order for the whole structure to be soluble. If you examine a three dimensional protein structure you will see many charged side chains (e.g. lysine, arginine, aspartate, glutamate) and hydrogen-bonding side chains (e.g. serine, threonine, glutamine, asparagine) exposed on the surface, in direct contact with water. Inside the protein, out of contact with the surrounding water, there tend to be many more hydrophobic residues such as alanine, valine, phenylalanine, etc. If a protein chain is caused to come unfolded (through exposure to heat, for example, or extremes of pH), it will usually lose its solubility and form solid precipitates, as the hydrophobic residues from the interior come into contact with water. You can see this phenomenon for yourself if you pour a little bit of vinegar (acetic acid) into milk. The solid clumps that form in the milk are proteins that have come unfolded due to the sudden acidification, and precipitated out of solution.
In recent years, scientists have become increasing interested in the proteins of so-called ‘thermophilic’ (heat-loving) microorganisms that thrive in hot water environments such as geothermal hot springs. While the proteins in most organisms (including humans) will rapidly unfold and precipitate out of solution when put in hot water, the proteins of thermophilic microbes remain completely stable, sometimes even in water that is just below the boiling point. In fact, these proteins typically only gain full biological activity when in appropriately hot water - at room-temperature they act is if they are ‘frozen’. Is the chemical structure of these thermostable proteins somehow unique and exotic? As it turns out, the answer to this question is ‘no’: the overall three-dimensional structures of thermostable proteins look very much like those of ‘normal’ proteins. The critical difference seems to be simply that thermostable proteins have more extensive networks of noncovalent interactions, particularly ion-ion interactions on their surface, that provides them with a greater stability to heat. Interestingly, the proteins of ‘psychrophilic’ (cold-loving) microbes isolated from pockets of water in arctic ice show the opposite characteristic: they have far fewer ion-ion interactions, which gives them greater flexibility in cold temperatures but leads to their rapid unfolding in room temperature water. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.10%3A_Protein_Structure.txt |
Objectives
After completing this section, you should be able to
1. describe the catalytic role of an enzyme in a biochemical reaction.
2. give an example of one fat‑soluble and one water‑soluble vitamin.
Key Terms
Make certain that you can define, and use in context, the key term below.
• coenzyme
• cofactor
• enzyme
• substrate
• vitamin
Study Notes
You should have a general knowledge of the function of enzymes, but you need not memorize specific names or the classification system.
A catalyst is any substance that increases the rate or speed of a chemical reaction without being changed or consumed in the reaction. Enzymes are biological catalysts, and nearly all of them are proteins. In addition, enzymes are highly specific in their action; that is, each enzyme catalyzes only one type of reaction in only one compound or a group of structurally related compounds. The compound or compounds on which an enzyme acts are known as its substrates. Enzymes are classified by reaction type into six categories show in Table $1$.
Table $1$: Classes of Enzymes
Class Type of Reaction Catalyzed Examples
oxidoreductases oxidation-reduction reactions Dehydrogenases catalyze oxidation-reduction reactions involving hydrogen and reductases catalyze reactions in which a substrate is reduced.
transferases transfer reactions of groups, such as methyl, amino, and acetyl Transaminases catalyze the transfer of amino group, and kinases catalyze the transfer of a phosphate group.
hydrolases hydrolysis reactions Lipases catalyze the hydrolysis of lipids, and proteases catalyze the hydrolysis of proteins
lyases reactions in which groups are removed without hydrolysis or addition of groups to a double bond Decarboxylases catalyze the removal of carboxyl groups.
isomerases reactions in which a compound is converted to its isomer Isomerases may catalyze the conversion of an aldose to a ketose, and mutases catalyze reactions in which a functional group is transferred from one atom in a substrate to another.
ligases reactions in which new bonds are formed between carbon and another atom; energy is required Synthetases catalyze reactions in which two smaller molecules are linked to form a larger one.
Enzyme-catalyzed reactions occur in at least two steps. In the first step, an enzyme molecule (E) and the substrate molecule or molecules (S) collide and react to form an intermediate compound called the enzyme-substrate (E–S) complex (Equation $\ref{step1}$). This step is reversible because the complex can break apart into the original substrate or substrates and the free enzyme. Once the E–S complex forms, the enzyme is able to catalyze the formation of product (P), which is then released from the enzyme surface (Equation $\ref{step2}$):
$S + E \rightleftharpoons E–S \label{step1}$
$E–S → P + E \label{step2}$
Hydrogen bonding and other electrostatic interactions hold the enzyme and substrate together in the complex. The structural features or functional groups on the enzyme that participate in these interactions are located in a cleft or pocket on the enzyme surface. This pocket, where the enzyme combines with the substrate and transforms the substrate to product is called the active site of the enzyme (Figure $1$).
Figure $1$: Substrate Binding to the Active Site of an Enzyme. The enzyme dihydrofolate reductase is shown with one of its substrates: NADP+ (a) unbound and (b) bound. The NADP+ (shown in red) binds to a pocket that is complementary to it in shape and ionic properties.
The active site possesses a unique conformation (including correctly positioned bonding groups) that is complementary to the structure of the substrate, so that the enzyme and substrate molecules fit together in much the same manner as a key fits into a tumbler lock. In fact, an early model describing the formation of the enzyme-substrate complex was called the lock-and-key model (Figure $2$). This model portrayed the enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site.
Working out the precise three-dimensional structures of numerous enzymes has enabled chemists to refine the original lock-and-key model of enzyme actions. They discovered that the binding of a substrate often leads to a large conformational change in the enzyme, as well as to changes in the structure of the substrate or substrates. The current theory, known as theinduced-fit model, says that enzymes can undergo a change in conformation when they bind substrate molecules, and the active site has a shape complementary to that of the substrate only after the substrate is bound, as shown for hexokinase in Figure $3$. After catalysis, the enzyme resumes its original structure.
The structural changes that occur when an enzyme and a substrate join together bring specific parts of a substrate into alignment with specific parts of the enzyme’s active site. Amino acid side chains in or near the binding site can then act as acid or base catalysts, provide binding sites for the transfer of functional groups from one substrate to another or aid in the rearrangement of a substrate. The participating amino acids, which are usually widely separated in the primary sequence of the protein, are brought close together in the active site as a result of the folding and bending of the polypeptide chain or chains when the protein acquires its tertiary and quaternary structure. Binding to enzymes brings reactants close to each other and aligns them properly, which has the same effect as increasing the concentration of the reacting compounds.
Example $1$
1. What type of interaction would occur between an OH group present on a substrate molecule and a functional group in the active site of an enzyme?
2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified.
Solution
1. An OH group would most likely engage in hydrogen bonding with an appropriate functional group present in the active site of an enzyme.
2. Several amino acid side chains would be able to engage in hydrogen bonding with an OH group. One example would be asparagine, which has an amide functional group.
Exercise $1$
1. What type of interaction would occur between an COO group present on a substrate molecule and a functional group in the active site of an enzyme?
2. Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified.
Enzyme Cofactors and Vitamins
Many enzymes are simple proteins consisting entirely of one or more amino acid chains. Other enzymes contain a nonprotein component called a cofactor that is necessary for the enzyme’s proper functioning. There are two types of cofactors: inorganic ions [e.g., zinc or Cu(I) ions] and organic molecules known as coenzymes. Most coenzymes are vitamins or are derived from vitamins.
Vitamins are organic compounds that are essential in very small (trace) amounts for the maintenance of normal metabolism. They generally cannot be synthesized at adequate levels by the body and must be obtained from the diet. The absence or shortage of a vitamin may result in a vitamin-deficiency disease. In the first half of the 20th century, a major focus of biochemistry was the identification, isolation, and characterization of vitamins. Despite accumulating evidence that people needed more than just carbohydrates, fats, and proteins in their diets for normal growth and health, it was not until the early 1900s that research established the need for trace nutrients in the diet.
Table $2$: Fat-Soluble Vitamins and Physiological Functions
Vitamin Physiological Function Effect of Deficiency
vitamin A (retinol) formation of vision pigments; differentiation of epithelial cells night blindness; continued deficiency leads to total blindness
vitamin D (cholecalciferol) increases the body’s ability to absorb calcium and phosphorus osteomalacia (softening of the bones); known as rickets in children
vitamin E (tocopherol) fat-soluble antioxidant damage to cell membranes
vitamin K (phylloquinone) formation of prothrombin, a key enzyme in the blood-clotting process increases the time required for blood to clot
Because organisms differ in their synthetic abilities, a substance that is a vitamin for one species may not be so for another. Over the past 100 years, scientists have identified and isolated 13 vitamins required in the human diet and have divided them into two broad categories: the fat-soluble vitamins (Table $2$), which include vitamins A, D, E, and K, and the water-soluble vitamins, which are the B complex vitamins and vitamin C (Table $3$). All fat-soluble vitamins contain a high proportion of hydrocarbon structural components. There are one or two oxygen atoms present, but the compounds as a whole are nonpolar. In contrast, water-soluble vitamins contain large numbers of electronegative oxygen and nitrogen atoms, which can engage in hydrogen bonding with water. Most water-soluble vitamins act as coenzymes or are required for the synthesis of coenzymes. The fat-soluble vitamins are important for a variety of physiological functions.
Table $3$: Water-Soluble Vitamins and Physiological Functions
Vitamin Coenzyme Coenzyme Function Deficiency Disease
vitamin B1 (thiamine) thiamine pyrophosphate decarboxylation reactions beri-beri
vitamin B2 (riboflavin) flavin mononucleotide or flavin adenine dinucleotide oxidation-reduction reactions involving two hydrogen atoms
vitamin B3 (niacin) nicotinamide adenine dinucleotide or nicotinamide adenine dinucleotide phosphate oxidation-reduction reactions involving the hydride ion (H) pellagra
vitamin B6 (pyridoxine) pyridoxal phosphate variety of reactions including the transfer of amino groups
vitamin B12 (cyanocobalamin) methylcobalamin or deoxyadenoxylcobalamin intramolecular rearrangement reactions pernicious anemia
biotin biotin carboxylation reactions
folic acid tetrahydrofolate carrier of one-carbon units such as the formyl group anemia
pantothenic Acid coenzyme A carrier of acyl groups
vitamin C (ascorbic acid) none antioxidant; formation of collagen, a protein found in tendons, ligaments, and bone scurvy
One characteristic that distinguishes an enzyme from all other types of catalysts is its substrate specificity. An inorganic acid such as sulfuric acid can be used to increase the reaction rates of many different reactions, such as the hydrolysis of disaccharides, polysaccharides, lipids, and proteins, with complete impartiality. In contrast, enzymes are much more specific. Some enzymes act on a single substrate, while other enzymes act on any of a group of related molecules containing a similar functional group or chemical bond. Some enzymes even distinguish between D- and L-stereoisomers, binding one stereoisomer but not the other. Urease, for example, is an enzyme that catalyzes the hydrolysis of a single substrate—urea—but not the closely related compounds methyl urea, thiourea, or biuret. The enzyme carboxypeptidase, on the other hand, is far less specific. It catalyzes the removal of nearly any amino acid from the carboxyl end of any peptide or protein.
Enzyme specificity results from the uniqueness of the active site in each different enzyme because of the identity, charge, and spatial orientation of the functional groups located there. It regulates cell chemistry so that the proper reactions occur in the proper place at the proper time. Clearly, it is crucial to the proper functioning of the living cell.
Concept Review Exercises
1. Distinguish between the lock-and-key model and induced-fit model of enzyme action.
2. Which enzyme has greater specificity—urease or carboxypeptidase? Explain.
Answers
1. The lock-and-key model portrays an enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. The induced fit model portrays the enzyme structure as more flexible and is complementary to the substrate only after the substrate is bound.
2. Urease has the greater specificity because it can bind only to a single substrate. Carboxypeptidase, on the other hand, can catalyze the removal of nearly any amino acid from the carboxyl end of a peptide or protein.
Takeaways
• A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product.
• The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions.
• The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate.
• Enzymes exhibit varying degrees of substrate specificity.
Exercises
1. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. COOH
2. NH3+
3. OH
4. CH(CH3)2
2. What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme?
1. SH
2. NH2
3. C6H5
4. COO
3. For each functional group in Exercise 1, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
4. For each functional group in Exercise 2, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified.
Answers
1. hydrogen bonding
2. ionic bonding
3. hydrogen bonding
4. dispersion forces
1. The amino acid has a polar side chain capable of engaging in hydrogen bonding; serine (answers will vary).
2. The amino acid has a negatively charged side chain; aspartic acid (answers will vary).
3. The amino acid has a polar side chain capable of engaging in hydrogen bonding; asparagine (answers will vary).
4. The amino acid has a nonpolar side chain; isoleucine (answers will vary).
26.12: How do Enzymes Work Citrate Synthase
Objectives
After completing this section, you should be able to
1. describe and explain the general function of an enzyme like citrate synthase in a reaction.
2. identify the structures of ten common coenzymes.
Oxaloacetate to Citrate Catalyzed by Citrate Synthase
Citrate synthase is a protein with 433 amino acids with various functional groups that can react with substrates. This enzyme catalyzes oxaloacetate to eventually produce citrate as part of the citric acid (Krebs) cycle. In the first step of the citric acid (Krebs) cycle, acetyl CoA condenses with oxaloacetate to form (S)-citryl CoA. The carboxylate group of an aspartic acid (B:) on citrate synthase removes the acidic alpha proton on acetyl CoA, while a histidine site (H-A) donates a proton to form the enol. Then a second histidine site (H-A) protonates the carbonyl oxygen of oxaloacetate, while the carbon of the carbonyl is attacked by the enol. Simultaeously, that first histidine (:A-) deprotonates the acetyl CoA enol. (S)-citryl CoA is generated.
The acyl group of a thioester of (S)-citryl CoA can be transferred to a water molecule in a hydrolysis reaction to converting (S)-citryl CoA to citrate. Again histidine sites on citrate synthase are an integral part of the mechanism and assist with removal and addition of protons. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/26%3A_Biomolecules-_Amino_Acids_Peptides_and_Proteins/26.11%3A_Enzymes_and_Coenzymes.txt |
Learning Objectives
When you have completed Chapter 28, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. draw the structure of a given nucleotide.
3. discuss the structure of DNA and RNA.
4. describe the processes involved in DNA replication, transcription, translation, and protein synthesis.
5. define, and use in context, the key terms introduced in this chapter.
Two types of nucleic acids are found in cells—deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). These highly complex substances are built up from a number of simpler units, called nucleotides. Each nucleotide consists of three parts: a phosphoric acid residue, a sugar and a nitrogen‑containing heterocyclic base. Thus, in order to understand the biochemistry of the nucleic acids, you must first study the chemistry of the sugars (see Chapter 25) and simple heterocyclic systems. We have already discussed certain aspects of the structure of heterocyclic ring systems during our study of aromaticity (Sections 15.5–15.6). You may find it helpful to review this chapter.
Chapter 28 examines the structure and replication of DNA and then describes the structure and synthesis of RNA. The chapter closes with a brief study of the role played by RNA in the biosynthesis of proteins.
28: Biomolecules - Nucleic Acids
Objectives
After completing this section, you should be able to
1. outline the relationship between nucleic acids, nucleotides and nucleosides.
2. identify, in general terms, the enzymatic hydrolysis products of nucleosides.
3. explain the structural difference between the sugar components of DNA and RNA.
4. identify by name the four heterocyclic amine bases found in deoxyribonucleotides.
5. identify by name the four heterocyclic amine bases found in ribonucleotides.
6. draw the general structure of a nucleotide and a nucleoside.
7. indicate the nitrogen atom by which a given purine or pyrimidine base attaches to the sugar component in nucleotides and nucleosides.
8. sketch a section of nucleic acid to show how the nucleotide units are joined together.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• deoxyribonucleic acid (DNA)
• nucleosides nucleotides
• ribonucleic acid (RNA)
Study Notes
The five bases that are found in nucleotides are often represented by their initial letter: adenine, A; guanine, G; cytosine, C; thymine, T; and uracil, U. Note that A, G, C and T occur in DNA; A, G, C and U occur in RNA. You are not required to memorize the structures of these bases, but you must know how each one bonds to the sugar unit in a nucleotide.
To fulfill Objective 6, you should be able to reproduce the figure below.
The Learning Objective of this Module is to identify the different molecules that combine to form nucleotides.
The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides. The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 109 nucleotides. Nucleotides can be further broken down to a phosphate group (PO4-3) and a nucleoside which is composed of a aldopentose sugar (a sugar with five carbon atoms), and a heterocyclic purine or pyrimidine base (a base containing nitrogen atoms).
$\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \tag{28.1.1}$
If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide, and the resulting nucleosides are used in ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide, and the nucleoside are used in deoxyribonucleic acid (DNA). The prefix -deoxy implies that there is an oxygen missing from the 2' position of ribose.
The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure $1$). The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring.
Nucleosides are formed by a bond between the anomeric C1′ of the pentose sugar and N1 position of the pyrimidine base or the N9 position of the purine base. The addition of a phospate groups at the 5' position of a nucleoside creates a corresponding nucleotide. DNA is made from four deoxyribonucleotides (Cytosine, Thymine, Adenine, Guanine) and RNA is made from four ribonucleotides Cytosine, Uracil, Adenine, Guanine. The names and structures of the major ribonucleotides and deoxyribonucleotides are given in Figure $2$.
Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure $3$, have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD+), and coenzyme A, contain adenine nucleotides as structural components.
Primary Structure of Nucleic Acids
Nucleotides are joined together through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the 3' carbon atom of the sugar unit of a second nucleotide. This unit joins to a third nucleotide, and the process is repeated to produce a long nucleic acid chain (Figure 28.1.4). The backbone of the chain consists of alternating phosphate and sugar units (2-deoxyribose in DNA and ribose in RNA). The purine and pyrimidine bases branch off this backbone.
Like proteins, nucleic acids have a primary structure that is defined as the sequence of their nucleotides. Unlike proteins, which have 20 different kinds of amino acids, there are only 4 different kinds of nucleotides in nucleic acids. For amino acid sequences in proteins, the convention is to write the amino acids in order starting with the N-terminal amino acid. In writing nucleotide sequences for nucleic acids, the convention is to write the nucleotides (usually using the one-letter abbreviations for the bases, shown in Figure 28.1.4) starting with the nucleotide having a free phosphate group, which is known as the 5′ end, and indicate the nucleotides in order. For DNA, a lowercase d is often written in front of the sequence to indicate that the monomers are deoxyribonucleotides. The final nucleotide has a free OH group on the 3′ carbon atom and is called the 3′ end. The sequence of nucleotides in the DNA segment shown in Figure 28.1.4 would be written 5′-dG-dT-dA-dC-3′, which is often further abbreviated to dGTAC or just GTAC.
Note
Each phosphate group has one acidic hydrogen atom that is ionized at physiological pH. This is why these compounds are known as nucleic acids.
Exercise $1$
Classify each compound as a pentose sugar, a purine, or a pyrimidine.
1. adenine
2. guanine
3. deoxyribose
4. thymine
5. ribose
6. cytosine
Answer
a) purine
b) purine
c) pentose sugar
d) pyrimidine
e) pentose sugar
f) pyrimidine
Exercise $2$
Identify the three molecules needed to form the nucleotides in each nucleic acid.
1. DNA
2. RNA
Answer
a)nitrogenous base (adenine, guanine, cytosine, and thymine), 2-deoxyribose, and H3PO4
b) nitrogenous base (adenine, guanine, cytosine, and uracil), ribose, and H3PO4
Exercise $3$
For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
Answer
a.
b.
Exercise $4$
For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.01%3A_Nucleotides_and_Nucleic_Acids.txt |
Objectives
After completing this section, you should be able, given the necessary Kekulé structures, to show how hydrogen bonding can occur between thymine and adenine, and between guanine and cytosine; and to explain the significance of such interactions to the primary and secondary structures of DNA.
Study Notes
Watson and Crick received the Nobel Prize in 1962 for elucidating the structure of DNA and proposing the mechanism for gene reproduction. Their work rested heavily on X-ray crystallographic work done on RNA and DNA by Franklin and Wilkins. Wilkins shared the Nobel Prize with Watson and Crick, but Franklin had been dead four years at the time of the award (you cannot be awarded the Nobel Prize posthumously).
The history of Watson and Crick’s proposed DNA model is controversial and a travesty of scientific ethics. Rosalind Franklin was deeply involved in the determination of the structure of DNA, and had collected numerous diffraction patterns. Watson attended a departmental colloquium at King’s College given by Franklin, and came into possession of an internal progress report she had written. Both departmental colloquia and progress reports are merely methods of discussion between colleagues; works presented in these fora are not considered by scientists to be “published” works, and therefore are not in the public domain. Watson and Crick not only were aware of Franklin’s work, but used her unpublished data, presented in confidence within her own college.
The final blow came about a year after the colloquium. Watson visited Wilkins at King’s College, and Wilkins inexplicably handed over Franklin’s diffraction photographs without her consent. Had Franklin’s work not been secretly taken from her, she might quite possibly have solved the DNA structure before Watson and Crick, who at the time did not yet have their own photographs. This is truly one of the sadder episodes of questionable scientific ethics and discovery that I have ever encountered.
References
Kass-Simon, G., and P. Farnes. Women of Science: Righting the Record. Bloomington, IN: Indiana University Press, 1990.
Maddox, B. Rosalind Franklin: The Dark Lady of DNA. New York: HarperCollins, 2002.
Intermolecular Forces in Nucleic Acids
The nucleic acids RNA and DNA are involved in the storage and expression of genetic information in a cell. Both are polymers of monomeric nucleotides. DNA exists in the cell as double-stranded helices while RNA typically is a single-stranded molecule which can fold in 3D space to form complex secondary (double-stranded helices) and tertiary structures in a fashion similar to proteins. The complex 3D structures formed by RNA allow it to perform functions other than simple genetic information storage, such as catalysis. Hence most scientists believe that RNA preceded both DNA and proteins in evolution as it can both store genetic information and catalyze chemical reactions.
DNA
DNA is a polymer, consisting of monomers call deoxynucleotides. The monomer contains a simple sugar (deoxyribose, shown in black below), a phosphate group (in red), and a cyclic organic R group (in blue) that is analogous to the side chain of an amino acid.
Only four bases are used in DNA (in contrast to the 20 different side chains in proteins) which we will abbreviate, for simplicity, as A, G, C and T. They are bases since they contain amine groups that can accept protons. The polymer consists of a sugar - phosphate - sugar - phosphate backbone, with one base attached to each sugar molecule. As with proteins, the DNA backbone is polar but also charged. It is a polyanion. The bases, analogous to the side chains of amino acids, are predominately polar. Given the charged nature of the backbone, you might expect that DNA does not fold to a compact globular (spherical) shape, even if positively charged cations like Mg bind to and stabilize the charge on the polymer. Instead, DNA exists usually as a double-stranded (ds) structure with the sugar-phosphate backbones of the two different strands running in opposite directions (5'-3' and the other 3'-5'). The strands are held together by hydrogen bonds between bases on complementary strands. Hence like proteins, DNA has secondary structure but in this case, the hydrogen bonds are not within the backbone but between the "side chain" bases on opposing strands. It is actually a misnomer to call dsDNA a molecule, since it really consists of two different, complementary strands held together by hydrogen bonds. A structure of ds-DNA showing the opposite polarity of the strands is shown below.
In 1950, Erwin Chargaff of Columbia University showed that the molar amount of adenine (A) in DNA was always equal to that of thymine (T). Similarly, he showed that the molar amount of guanine (G) was the same as that of cytosine (C). Chargaff's findings clearly indicate that some type of heterocyclic amine base pairing exists in the DNA structure. In double stranded DNA, the guanine (G) base on one strand can form three H-bonds with a cytosine (C) base on another strand (this is called a GC base pair). The thymine (T) base on one strand can form two H-bonds with an adenine (A) base on the other strand (this is called an AT base pair). Double-stranded DNA has a regular geometric structure with a fixed distance between the two backbones. This requires the bases pairs to consists of one base with a two-ring (bicyclic) structure (these bases are called purines) and one with a single ring structure (these bases are called pyrimidines). Hence a G and A or a T and C are not possible base pair partners.
Secondary Structure of DNA
The three-dimensional structure of DNA was the subject of an intensive research effort in the late 1940s to early 1950s. DNA exists as a double-stranded molecule that twists around its axis to form a helical structure,stabilized through Watson-Crick hydrogen bonding between purines and pyrimidines, and through pi-pi stacking interactions among the bases arranged in structure. helical column. Each strand is a complement to the other; the nucleotides on one strand hydrogen-bond with complementary nucleotides on the opposite strand—that is, side-by-side with the 5′ end of one chain next to the 3′ end of the other. The purine and pyrimidine bases face the inside of the helix, with guanine always opposite cytosine and adenine always opposite thymine. The double helical "twist" occurs because of the angular geometry of each bonded nucleotide.
Initial work revealed the DNA polymer had a regular repeating pattern X-ray diffraction data shows that a repeating helical pattern is 20 Angstrom units wide and occurs every 34 Angstrom units with 10 nucleotide subunits per turn. Each subunit occupies 3.4 Angstrom units which is the same amount of space occupied by a single nucleotide unit. The helix is Under most conditions, the two strands are slightly offset, which creates a 12 Angstrom major groove on one face of the double helix, and a 6 Angstrom minor groove on the other. The overall DNA polymer varies in length (number of sugar-phosphate units connected), base composition (how many of each set of bases) and sequence (the order of the bases in the backbone).
What do we mean when we say information is encoded in the DNA molecule? An organism’s DNA can be compared to a book containing directions for assembling a model airplane or for knitting a sweater. Letters of the alphabet are arranged into words, and these words direct the individual to perform certain operations with specific materials. If all the directions are followed correctly, a model airplane or sweater is produced.
In DNA, the particular sequences of nucleotides along the chains encode the directions for building an organism. Just as saw means one thing in English and was means another, the sequence of bases CGT means one thing, and TGC means something different. Although there are only four letters—the four nucleotides—in the genetic code of DNA, their sequencing along the DNA strands can vary so widely that information storage is essentially unlimited.
Deoxyribonucleic acid (DNA) stores genetic information, while ribonucleic acid (RNA) is responsible for transmitting or expressing genetic information by directing the synthesis of thousands of proteins found in living organisms. But how do the nucleic acids perform these functions?
Three processes are required:
1. Replication, in which new copies of DNA are made.
2. Transcription, in which a segment of DNA is used to produce RNA.
3. Translation, in which the information in RNA is translated into a protein sequence.
Exercise \(1\)
For this short DNA segment,
1. Identify the 5′ end and the 3′ end of the molecule.
2. Circle the atoms that comprise the backbone of the nucleic acid chain.
3. Write the nucleotide sequence of this DNA segment.
Answer
Exercise \(2\)
Which nitrogenous base in DNA pairs with each listed nitrogenous base?
1. Cytosine
2. Adenine
3. Guanine
4. Thymine
Answer
1. Guanine
2. Thymine
3. Cytosine
4. Adenine
Exercise \(3\)
How many hydrogen bonds can form between the two strands in the short DNA segment shown below?
5′ ATGCGACTA 3′ 3′ TACGCTGAT 5′
Answer
22 (2 between each AT base pair and 3 between each GC base pair).
Exercise \(4\)
A segment of one strand from a DNA molecule has the sequence 5′-TCCATGAGTTGA-3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain?
Answer
Knowing that the two strands are antiparallel and that T base pairs with A, while C base pairs with G, the sequence of the complementary strand will be 3′-AGGTACTCAACT-5′ (can also be written as TCAACTCATGGA). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.02%3A_Base_Pairing_in_DNA_-_The_Watson-Crick_Model.txt |
Objectives
After completing this section, you should be able to describe, very briefly, the replication of DNA.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• replication
• semiconservative replication
Study Notes
Notice that the objective for this section requires only that you be able to describe the replication process briefly.
According to the central dogma of molecular genetics, DNA is the genetically active component of the chromosomes of a cell. That is, DNA in the cell nucleus contains all the information necessary to control synthesis of the proteins, enzymes, and other molecules which are needed as that cell grows, carries on metabolism, and eventually reproduces. Thus when a cell divides, its DNA must pass on genetic information to both daughter cells. It must somehow be able to divide into duplicate copies. This process is called replication. Given the complementary double strands of DNA, it is relatively easy to see how DNA as a molecule is well structured for replication, as is show in Figure 20.21.1. Each strand serves as a template for a new strand. Thus, after DNA is replicated, each new DNA double helix will have one strand from the original DNA molecule, and one newly synthesized molecule. This is referred to as semiconservative replicatio
A rather complex mechanism exists for DNA replication, involving many different enzymes and protein factors. Let us consider some of the more important aspects of DNA replication. First, the double strand needs to be opened up to replicate each template strand. To do this, a set of proteins and enzymes bind to and open up the double helix at an origin point in the molecule. This forms replication forks, points where double stranded DNA opens up, allowing replication to occur. A helicase enzyme binds at the replication forks, with the function of further unwinding the DNA and allowing the replication fork to move along the double strand as DNA is replicated. Another enzyme, DNA gyrase, is also required to relieve stress on the duplex caused by unwinding the double strand. Further, single strand binding proteins are needed to prevent the single strands from reforming a double strand. Another essential enzyme in this initiation phase is primase, which creates an RNA primer on each single strand of DNA to begin replication from.
All of these initial functions are necessary to prepare the DNA for the main enzyme which builds then new strands, DNA polymerase. Multiple polymerase enzymes exist, but for the moment we will DNA polymerase III, the main DNA polymerase in E. coli. DNA polymerase III catalyzes the reaction by which a new nucleotide is added to a growing DNA strand. That reaction is seen in Figure 20.21.2. The DNA polymerase enzymes need a free 3' OH group in order to begin synthesizing a new strand, which explains the necessity of the RNA primer, which gives a 3'OH group for DNA polymerase III to start from.
This leads to another constraint on DNA polymerase III. One strand, the leading strand can be polymerized continuously since the new strand being created goes 5' to 3' from the replication fork, but since the original strands are anti-parallel, the other strand, the lagging strand is going in the wrong direction for polymerization. In this case, the polymerization reaction starts away from the replication fork and works back toward it. This means that the lagging strand is synthesized in disconnected segments, known as Okazaki fragments, instead of continuously. Later, another DNA polymerase, in the case of E. coli, DNA polymerase I, removes RNA primers and fills in the missing discontinuities. Then, another enzyme, DNA ligase, connects breaks between 3'OH groups and 5' phosphate groups in the newly synthesized strands that exist due to these discontinuities. While the enzymes of this process differ in eukaryotes, they fulfill similar mechanisms. Even with this complexity of this process, DNA polymerase III is able to add new nucleotides at a rate of 250-1,000 nucleotides per second.[3]
A number of advantages of the double-stranded structure held together by hydrogen bonds is evident in the process of replication. Complementary base pairing, A to T and G to C, insures that the two new DNA molecules will be the same as the original. The large number of hydrogen bonds, each of which is relatively weak, makes complete separation of the two strands unlikely, but one hydrogen bond, or even a few, can be broken rather easily. The helicase portion of the replication complex can therefore separate the two strands in much the same way that a zipper operates. Like the teeth of a zipper, hydrogen bonds provide great strength when all work together, but the proper tool can separate them one at a time.
Figure 28.3.1 A Schematic Diagram of DNA Replication. DNA replication occurs by the sequential unzipping of segments of the double helix.
The Scope of the Problem of DNA Replication
The 46 chromosomes of the human genome consists of roughly 6.5 billion base pairs of DNA if one considers the full diploid genome (i.e. if you count the DNA inherited from both parents). Considering that this DNA can be copied in just a few hours shows that the rate of DNA replication is staggering.
Although DNA replication is typically a highly accurate process and proofreading DNA polymerases help to keep the error rate low (about one per 10 billion bases), mistakes still occur. In addition to errors of replication, environmental damage may also occur to the DNA. Such uncorrected errors of replication or environmental DNA damage may lead to serious consequences. Therefore, Nature has evolved several mechanisms for detecting and repairing damaged or incorrectly synthesized DNA.
Example \(1\)
A segment of one strand from a DNA molecule has the sequence 5′-TCCATGAGTTGA-3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain?
Solution
Knowing that the two strands are antiparallel and that T base pairs with A, while C base pairs with G, the sequence of the complementary strand will be 3′-AGGTACTCAACT-5′ (can also be written as TCAACTCATGGA).
Exercise 28.3.1
A segment of one strand from a DNA molecule has the sequence 5′-CCAGTGAATTGCCTAT-3′. What is the sequence of nucleotides in the opposite, or complementary, DNA chain? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.03%3A_Replication_of_DNA.txt |
Objectives
After completing this section, you should be able to
1. describe, very briefly, how RNA is synthesized in the nucleus of the cell by transcription of DNA.
2. identify the important structural differences between DNA and DNA.
3. given the appropriate Kekulé structures, show how uracil can form strong hydrogen bonds to adenine.
4. identify the base sequence in RNA that would be complementary to a given base sequence in DNA.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• messenger RNA
• RNA polymerase
• ribosomal RNA
• transcription
• transfer RNA
Study Notes
“Messenger RNA” (mRNA) carries the genetic information from the DNA in the nucleus to the cytoplasm where protein synthesis occurs. The code carried by mRNA is read by “transfer RNA” (tRNA) in a process called translation (see Section 28.5).
“Ribosomal RNA” (rRNA) is the term used to describe the RNA molecules which, together with proteins, make up the ribosomes on which proteins are synthesized.
Three types of RNA are formed during transcription: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). These three types of RNA differ in function, size, and percentage of the total cell RNA (Table 28.4.1). mRNA makes up only a small percent of the total amount of RNA within the cell, primarily because each molecule of mRNA exists for a relatively short time; it is continuously being degraded and resynthesized. The molecular dimensions of the mRNA molecule vary according to the amount of genetic information a given molecule contains. After transcription, which takes place in the nucleus, the mRNA passes into the cytoplasm, carrying the genetic message from DNA to the ribosomes, the sites of protein synthesis. Elsewhere, we shall see how mRNA directly determines the sequence of amino acids during protein synthesis.
Table 28.4.1: Properties of Cellular RNA in Escherichia coli
Type Function Approximate Number of Nucleotides Percentage of Total Cell RNA
mRNA codes for proteins 100–6,000 ~3
rRNA component of ribosomes 120–2900 83
tRNA adapter molecule that brings the amino acid to the ribosome 75–90 14
For the hereditary information in DNA to be useful, it must be “expressed,” that is, used to direct the growth and functioning of an organism. The flow of genetic information in cells goes from DNA to mRNA to protein, by genes which specify the sequences of mRNAs, which in turn specify the sequences of proteins.
The first step in the processes that constitute DNA expression is the synthesis of RNA, by a template mechanism that is in many ways analogous to DNA replication. Because the RNA that is synthesized is a complementary copy of information contained in DNA, RNA synthesis is referred to as transcription. Transcription requires the DNA double helix to partially unwind in the region of mRNA synthesis. The region of unwinding is called a transcription bubble. The DNA sequence has patterns which indicate where RNA polymerase should start and end transcription. A DNA sequence at which the RNA polymerase binds to start transcription is called a promoter. The DNA sequence that indicates the endpoint of transcription, where the RNA polymerase should stop adding nucleotides and dissociate from the template is known as a terminator sequence. The promoter and terminator, thus, bracket the region of the DNA that is to be transcribed.
In bacteria, the promotor sequence contains two 6 bp region called consensus sequences. One sequence is centered about 10 bp upstream from the transcription start site. The second sequence at about 35 basepairs upstream from the start of transcription. The consensus sequences at -10 and -35 are necessary for recognition of the promoter region by RNA polymerase.
The DNA sequence that is transcribed to make RNA is called the antisense strand (also called template, anticoding, or transcribed strand), while the complementary sequence on the other DNA strand is called the sense strand (also called the coding or informational strand). To initiate RNA synthesis, the two DNA strands unwind at specific sites along the DNA molecule. Ribonucleotides are attracted to the uncoiling region of the DNA molecule, beginning at the 3′ end of the template strand, according to the rules of base pairing. Thymine in DNA calls for adenine in RNA, cytosine specifies guanine, guanine calls for cytosine, and adenine requires uracil. RNA polymerase—an enzyme—binds the complementary ribonucleotide and catalyzes the formation of the ester linkage between ribonucleotides, a reaction very similar to that catalyzed by DNA polymerase (Figure 28.4.1). Synthesis of the RNA strand takes place in the 5′ to 3′ direction, antiparallel to the template strand. Only a short segment of the RNA molecule is hydrogen-bonded to the template strand at any time during transcription. When transcription is completed, the RNA is released, and the DNA helix reforms. The nucleotide sequence of the RNA strand formed during transcription is identical to that of the corresponding coding strand of the DNA, except that U replaces T.
Figure 28.4.1 A Schematic Diagram of RNA Transcription from a DNA Template. The representation of RNA polymerase is proportionately much smaller than the actual molecule, which encompasses about 50 nucleotides at a time.
The genes of eukaryotes (animals and plants) usually done have continuous segments of coding DNA. Rather they have intervening sequences of DNA (introns) within a given gene that separate coding fragments of DNA (exons). In a process called splicing, a primary transcript is made from the DNA, and then splicesomes cut out the introns and join the exons to form a contiguous stretch to form messenger RNA, mRNA. Once formed the mRNA leaves the nucleus is to be translated into a protein sequence..
New findings make it even more complicated to define a gene, especially if the transcripts of a "gene region" are studied. Cheng et al studied all transcripts from 10 different human chromosomes and 8 different cell lines. They found a large number of different transcripts, many of which overlapped. Splicing often occur between nonadjacent introns. Transcripts were found from both strands and were from regions containing introns and exons. Other studies found up to 5% of transcripts continued through the end of "gene" into other genes. 63% of the entire mouse genome, which is comprised of only 2% exons, is transcribed.
Exercise \(1\)
A portion of the template strand of a gene has the sequence 5′-TCCATGAGTTGA-3′. What is the sequence of nucleotides in the RNA that is formed from this template?
Answer
Four things must be remembered in answering this question: (1) the DNA strand and the RNA strand being synthesized are antiparallel; (2) RNA is synthesized in a 5′ to 3′ direction, so transcription begins at the 3′ end of the template strand; (3) ribonucleotides are used in place of deoxyribonucleotides; and (4) thymine (T) base pairs with adenine (A), A base pairs with uracil (U; in RNA), and cytosine (C) base pairs with guanine (G). The sequence is determined to be 3′-AGGUACUCAACU-5′ (can also be written as 5′-UCAACUCAUGGA-3′).
Exercise \(2\)
What would be the DNA base sequence of the coding strand required to transcribe the following RNA sequence?
5′-AUGAGCGACUUUGCGGGAUUA-3′
Answer
5′-ATGAGCGACTTTGCGGGATTA-3′. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.04%3A_Transcription_of_DNA.txt |
Objectives
After completing this section, you should be able to describe, very briefly, the roles of messenger RNA and transfer RNA in the biosynthesis of proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• anticodon
• codon
• translation
Study Notes
As in the preceding section, you should not be too concerned about trying to memorize details. The objective requires you to have a general understanding of the roles played by mRNA and tRNA in the biosynthesis of proteins, and that you be able to describe this process.
After transcription, which takes place in the nucleus, the mRNA passes into the cytoplasm, carrying the genetic message from DNA to the ribosomes, the sites of protein synthesis. Ribosomes are cellular substructures where proteins are synthesized. They contain about 65% rRNA and 35% protein, held together by numerous noncovalent interactions, such as hydrogen bonding, in an overall structure consisting of two globular particles of unequal size. We turn now to the question of how the sequence of nucleotides in a molecule of ribonucleic acid (RNA) is translated into an amino acid sequence.
How can a molecule containing just 4 different nucleotides specify the sequence of the 20 amino acids that occur in proteins? If each nucleotide coded for 1 amino acid, then obviously the nucleic acids could code for only 4 amino acids. What if amino acids were coded for by groups of 2 nucleotides? There are 42, or 16, different combinations of 2 nucleotides (AA, AU, AC, AG, UU, and so on). Such a code is more extensive but still not adequate to code for 20 amino acids. However, if the nucleotides are arranged in groups of 3, the number of different possible combinations is 43, or 64. Here we have a code that is extensive enough to direct the synthesis of the primary structure of a protein molecule.
The genetic code can therefore be described as the identification of each group of three nucleotides and its particular amino acid. The sequence of these triplet groups in the mRNA dictates the sequence of the amino acids in the protein. Each individual three-nucleotide coding unit, as we have seen, is called a codon.
Early experimenters were faced with the task of determining which of the 64 possible codons stood for each of the 20 amino acids. The cracking of the genetic code was the joint accomplishment of several well-known geneticists—notably Har Khorana, Marshall Nirenberg, Philip Leder, and Severo Ochoa—from 1961 to 1964. The genetic dictionary they compiled, summarized in Figure 28.5.3, shows that 61 codons code for amino acids, and 3 codons serve as signals for the termination of polypeptide synthesis (much like the period at the end of a sentence). Notice that only methionine (AUG) and tryptophan (UGG) have single codons. All other amino acids have two or more codons.
Figure 28.5.1: The Genetic Code
Protein synthesis is accomplished by orderly interactions between mRNA and the other ribonucleic acids (transfer RNA [tRNA] and ribosomal RNA [rRNA]), the ribosome, and more than 100 enzymes. The mRNA formed in the nucleus during transcription is transported across the nuclear membrane into the cytoplasm to the ribosomes—carrying with it the genetic instructions. The process in which the information encoded in the mRNA is used to direct the sequencing of amino acids and thus ultimately to synthesize a protein is referred to as translation.
Before an amino acid can be incorporated into a polypeptide chain, it must be attached to its unique tRNA. The carboxylic acid group of the amino acid forms an ester linkage with with the 3' hydroxyl group on the riboses bonded at the 3' end of the tRNA. This crucial process requires an enzyme known as aminoacyl-tRNA synthetase (Figure 28.5.1). There is a specific aminoacyl-tRNA synthetase for each amino acid. This high degree of specificity is vital to the incorporation of the correct amino acid into a protein. After the amino acid molecule has been bound to its tRNA carrier, protein synthesis can take place.
Figure 28.5.2 Binding of an Amino Acid to Its tRNA
The two-dimensional structure of a tRNA molecule is reminiscent of a cloverleaf. At one end of the tRNA molecule is the acceptor stem, where the amino acid is attached. The tRNA Molecule has three distinctive loops. One of these is called the anticodon loop which holds a sequence of three nucleotides called the anticodon. Each anticodon corresponds to the amino acid each is tRNA molecule is specifically designed to carry. For example, the amino acid lysine has the codon AAG, so the anticodon is UUC. Therefore, lysine would be carried by a tRNA molecule with the anticodon UUC. Each of the 20 amino acids found in proteins has at least one corresponding kind of tRNA, and most amino acids have more than one.
Figure 28.5.3: Transfer RNA (a) In the two-dimensional structure of a yeast tRNA molecule for phenylalanine, the amino acid binds to the acceptor stem located at the 3′ end of the tRNA primary sequence. (The nucleotides that are not specifically identified here are slightly altered analogs of the four common ribonucleotides A, U, C, and G.) (b) In the three-dimensional structure of yeast phenylalanine tRNA, note that the anticodon loop is at the bottom and the acceptor stem is at the top right. (c) This shows a space-filling model of the tRNA.
During protein synthesis the codon on the mRNA determines which kind of tRNA will add its amino acid to the growing chain. Wherever the codon AAG appears in mRNA, a UUC anticodon on a tRNA temporarily binds to the codon ect. As each different tRNA brings an amino acid into position an enzyme adds it to the growing protein chain. The protein is released from the ribosome once it is completed. Figure 28.5.2 depicts a schematic stepwise representation of this all-important process.
Figure 28.5.4: The Elongation Steps in Protein Synthesis
Exercise \(1\)
What are the roles of mRNA and tRNA in protein synthesis?
Answer
mRNA provides the code that determines the order of amino acids in the protein; tRNA transports the amino acids to the ribosome to incorporate into the growing protein chain.
Exercise \(2\)
A portion of an mRNA molecule has the sequence 5′-AUGCCACGAGUUGAC-3′. What amino acid sequence does this code for?
Answer
Use Figure 28.5.1 to determine what amino acid each set of three nucleotides (codon) codes for. Remember that the sequence is read starting from the 5′ end and that a protein is synthesized starting with the N-terminal amino acid. The sequence 5′-AUGCCACGAGUUGAC-3′ codes for met-pro-arg-val-asp.
Exercise \(3\)
Write the anticodon on tRNA that would pair with each mRNA codon.
a) 5′-UUU-3′
b) 5′-CAU-3′
c) 5′-AGC-3′
d) 5′-CCG-3′
Answer
a) 3′-AAA-5′
b) 3′-GUA-5′
c) 3′-UCG-5′
d) 3′-GGC-5′
Exercise \(4\)
The peptide hormone oxytocin contains 9 amino acid units. What is the minimum number of nucleotides needed to code for this peptide?
Answer
27 nucleotides (3 nucleotides/codon)
Exercise \(5\)
Determine the amino acid sequence produced from this mRNA sequence: 5′-AUGAGCGACUUUGCGGGAUUA-3′.
Answer
met-ser-asp-phe-ala-gly-leu | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.05%3A_Translation_of_RNA_-_Protein_Biosynthesis.txt |
Objectives
After completing this section, you should be able to
• Describe briefly how DNA sequencing is carried out.
DNA sequencing determines the order of nucleotide bases within a given fragment of DNA. This information can be used to infer the RNA or protein sequence encoded by the gene, from which further inferences may be made about the gene’s function and its relationship to other genes and gene products. DNA sequence information is also useful in studying the regulation of gene expression. If DNA sequencing is applied to the study of many genes, or even a whole genome, it is considered an example of genomics.
While techniques to sequence proteins have been around since the 1950s, techniques to sequence DNA were not developed until the mid-1970s, when two distinct sequencing methods were developed almost simultaneously, one by Walter Gilbert’s group at Harvard University, the other by Frederick Sanger’s group at Cambridge University. However, until the 1990s, the sequencing of DNA was a relatively expensive and long process. Using radiolabeled nucleotides also compounded the problem through safety concerns. With currently-available technology and automated machines, the process is cheaper, safer, and can be completed in a matter of hours. The Sanger sequencing method was used for the human genome sequencing project, which was finished its sequencing phase in 2003, but today both it and the Gilbert method have been largely replaced by better methods.
Restriction Enzymes
To be able to sequence DNA, it is first necessary to cut it into smaller fragments. What is needed is a way to cleave the DNA molecule at a few precisely-located sites so that a small set of homogeneous fragments are produced. To cut DNA at known locations, researchers use restriction endonucleases enzymes that have been purified from various bacterial species, and which can be purchased from various commercial sources. REs occur naturally in bacteria, where they specifically recognize short stretches of nucleotides in DNA and catalyze double-strand breaks at or near the recognition site (also known as a restriction site). These enzymes are usually named after the bacterium from which they were first isolated. For example, EcoRI and EcoRV are both enzymes from E. coli.
Restriction enzymes like EcoRI are frequently called 6-cutters, because they recognize a 6-nucleotide sequence. Assuming a random distribution of A, C, G and Ts in DNA, probability predicts that a recognition site for a 6-cutter should occur about once for every 4096 bp (46) in DNA. Of course, the distribution of nucleotides in DNA is not random, so the actual sizes of DNA fragments produced by EcoRI range from hundreds to many thousands of base pairs, but the mean size is close to 4000 bp. DNA fragments of this length are useful in the lab, since they long enough to contain the coding sequence for proteins and are well-resolved on agarose gels.
EcoRI recognizes the sequence G A A T T C in double stranded DNA. This recognition sequence is a palindrome with a two-fold axis of symmetry, because reading from 5’ to 3’ on either strand of the helix gives the same sequence. The palindromic nature of the restriction site is more obvious in the figure below. The dot in the center of the restriction site denotes the axis of symmetry. EcoRI catalyzes the hydrolysis of the phosphodiester bonds between G and A on both DNA strands. The restriction fragments generated in the reaction have short single-stranded tails at the 5’-ends. These ends are often referred to as “sticky ends,” because of their ability to form hydrogen bonds with complementary DNA sequences.
Reading DNA Sequences
We will discuss one method of reading the sequence of DNA. This method, developed by Sanger won him a second Nobel prize. Sanger sequencing, also known as chain-termination sequencing, requires a single-stranded DNA template, a DNA primer, a DNA polymerase, normal deoxynucleotidetriphosphates (dNTPs), and modified nucleotides (dideoxyNTPs - ddNTP) that terminate DNA strand elongation. These chain-terminating nucleotides lack a 3′-OH group required for the formation of a phosphodiester bond between two nucleotides, causing DNA polymerase to cease extension of DNA when a ddNTP is incorporated.
Four reaction tubes are set up, each containing the template DNA to be sequenced, a primer of known sequence, all four of the standard deoxynucleotides (dATP, dGTP, dCTP and dTTP), and the DNA polymerase. To each reaction is added only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP, or ddTTP) which has been fluorescently labeled. Most of the time in a Sanger sequencing reaction, DNA Polymerase will add a proper dNTP to the growing strand it is synthesizing in vitro. But at random locations, it will instead add a ddNTP. When it does, that strand will be terminated at the ddNTP just added. If enough template DNAs are included in the reaction mix, each one will have the labeled ddNTP inserted at a different random location, and there will be at least one DNA terminated at each different nucleotide along its length for as long as the in vitro reaction can take place (about 900 nucleotides under optimal conditions.)
After the reactions are over, the newly synthesized strands can be denatured from the template, and then separated by capillary electrophoresis or other equivalent methods. Since each band differs in length by one nucleotide, and the identity of that nucleotide is known from its fluorescence, the DNA sequence can be read simply from the order of the colors in successive bands.
As each differently-sized fragment exits the capillary column, a laser excites the florescent tag on its terminal nucleotide. From the color of the resulting florescence, a computer can keep track of which nucleotide was present as the terminating nucleotide. The computer also keeps track of the order in which the terminating nucleotides appeared, which is the sequence of the DNA used in the original reaction. In practice, the maximum length of sequence that can be read from a single sequencing reaction is about 700 bp.
Scientists now know the sequence of all the 3 billion DNA base pairs in the entire human genome. This knowledge was attained by the Human Genome Project (HGP), a \$3 billion, international scientific research project that was formally launched in 1990. The project was completed in 2003, two years ahead of its 15-year projected deadline. Determining the sequence of the billions of base pairs that make up human DNA was the main goal of the HGP. Another goal was mapping the location and determining the function of all the genes in the human genome. There are only about 20,500 genes in human beings. If modern methods were used it might bring the cost of sequencing the human genome down from the initial billion dollar range to \$100.
Example 28.6.1
You will pretend to sequence a single stranded piece of DNA as shown below. The new nucleotides are added by the enzyme DNA polymerase to the primer, GACT, in the 5' to 3' direction. You will set up 4 reaction tubes, Each tube contains all the dXTP's. In addition, add ddATP to tube 1, ddTTP to tube 2, ddCTP to tube 3, and ddGTP to tube 4. For each separate reaction mixture, determine all the possible sequences made by writing the possible sequences on one of the unfinished complementary sequences below. Cut the completed sequences from the page, determine the size of the polynucleotide sequences made, and place them as they would migrate (based on size) in the appropriate lane of a imaginary gel which you have drawn on a piece of paper. Lane 1 will contain the nucleotides made in tube 1, etc. Then draw lines under the positions of the cutout nucleotides to represent DNA bands in the gel. Read the sequence of the complementary DNA synthesized. Then write the sequence of the ssDNA that was to be sequenced.
• 5' T C A A C G A T C T G A 3' (STAND TO SEQUENCE)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
Since the DNA fragments have no detectable color, they can not be directly visualized in the gel. Alternative methods are used. In the one described above, radiolabeled ddXTP's where used. Once the sequencing gel is run, it can be dried and the bands visualized by radioautography (also called autoradiography). A place of x-ray film is placed over the dried gel in a dark environment. The radiolabeled bands will emit radiation which will expose the x-ray film directly over the bands. The film can be developed to detect the bands. In a newer technique, the primer can be labeled with a flourescent dye. If a different dye is used for each reaction mixture, all the reaction mixtures can be run in one lane of a gel. (Actually only one reaction mix containing all the ddXTP's together need be performed.) The gel can then be scanned by a laser, which detects fluorescence from the dyes, each at a different wavelength. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.06%3A_DNA_Sequencing.txt |
Objectives
After completing this section, you should be able to
1. describe, briefly, the steps required for chemical synthesis of DNA segments.
DNA must be synthesized to study genes, the sequence of genomes, and many other studies. This occurs in two fashions, by polymerase chain reaction (PCR) and chemical synthesis. PCR is covered in Section 28.8. Here we will focus on chemical synthesis of short DNA segments, which which are called oligonucleotides. Oligonucleotide synthesis is the chemical synthesis of relatively short fragments of nucleic acids, both DNA and RNA with a defined chemical structure (sequence). The technique is extremely useful in current laboratory practice because it provides a rapid and inexpensive access to custom-made oligonucleotides of the desired sequence.
The synthesis of DNA is more difficult than peptide synthesis (Section 26-8) because of the complexity of nucleotide monomers. Commercial automated DNA synthesizers are available which allow for DNA segments to be made quickly and at a low cost. DNA synthesizers typically uses solid-phase techniques similar to the Merrifield solid-phase peptide synthesizer. Nucleotides are protected then covalently bonded to a solid support. Nucleotides are sequentially coupled to the growing oligonucleotide chain in the order required by the sequence of the product. Upon the completion of the chain assembly, the product is released from the solid phase to solution, it is then deprotected, and collected. The occurrence of side reactions sets practical limits for the length of synthetic oligonucleotides (up to about 200 nucleotide residues) because the number of errors accumulates with the length of the oligonucleotide being synthesized. Products are often isolated by HPLC to obtain the desired oligonucleotides in high purity. Typically, synthetic oligonucleotides are single-stranded DNA or RNA molecules around 15–25 bases in length.
DNA Chemical Synthesis
The synthesis of DNA involves five steps:
Step 1)
Nucleosides used for this synthesis are modified with a linking agent at the 3' hydroxyl group of deoxyribose. The 5' hydroxyl group of the nucleosides is protected with p-dimethoxytrityl (DMT) ether. The amine groups on the nucleoside's heterocyclic bases are also protected. The amines of adenine and cytosine bases are protected with benzoyl groups. Guanine's amine are protected by a isobutyryl group and thymine has no amine groups so protection is not required.
The solid phase support used for DNA synthesis is commonly silica (SO2) spheres which have been functionalized with (3-aminopropyl)triethoxysilane such that an amine group is available for reaction on the surface. A protected nucleoside is coupled to the solid phase support through an ester linkage with the 3' hydroxyl group of the nucleoside and an amide linkage with amine group from the silica surface.
Step 2)
Reaction with with dichloroacetic acid removes the DMT protecting group from the 5' hydroxyl of the nucleoside attached to the silica surface. The p-dimethoxytrityl leaving group forms a relatively stable dimethoxytrityl cation which is both tertiary and benzylic. The reaction proceeds rapidly through a SN1 mechanism.
Step 3)
The nucleoside attached to the silica surface is then reacted with a protected nucleoside which has a phosphoramidite functional group [R2NP(OR)2] attached to the 3' hydroxyl group of its deoxyribose moiety. In addition, one of the phosphorus oxygen atoms of the phosphoramidite group is protected with a beta-cyanoethyl group (-OCH2CH2CN). The two nucleosides are coupled in a reaction which uses acetonitrile as a polar aprotic solvent, tetrazole as a heterocyclic amine catalyst, and produces a product with a phosphite functional group [P(OR)3].
Step 4)
Next the phosphite product of the previous step is oxidized to a phosphate by reaction with iodine (I2) along with 2,6-dimethylpyridine in aqueous tetrahydrofuran (THF). Additional nucleosides can now be added by repeating the phosphoramidite oligodeoxynucleotide synthesis cycle of (1) DMT deprotection, (2) phosphoramidite coupling, and (3) oxidation to a phosphite.
Step 5)
After the oligonucleotide chain of the desired sequence has been made, the final step is the removal of all the protecting groups and the linkage to silica by reaction with aqueous ammonia (NH3).
Exercise $1$
Draw a mechanism which shows why dimethoxytrityl cation produced by the cleavage of a p-dimethoxytrityl protecting group is exceptionally stable.
Answer
Exercise $1$
When the beta-cyanoethyl protecting group is cleaved with aqueous ammonia, Acrylonitrile, H2CCHCN , is also produced as a by-product. Draw a mechanism for the reaction.
Answer
28.08: The Polymerase Chain Reaction
Objectives
After completing this section, you should be able to
1. describe, briefly the three steps of PCR.
The polymerase chain reaction (PCR), allows one to use the power of DNA replication to amplify DNA enormously in a short period of time. As you know, cells replicate their DNA before they divide, and in doing so, double the amount of the cell’s DNA. PCR essentially mimics cellular DNA replication in the test tube, repeatedly copying the target DNA over and over, to produce large quantities of the desired DNA. Kary B. Mullis was awarded a Nobel Prize in 1993 for his development of PCR, which is now the basis of innumerable research studies of gene structure, function and evolution as well as applications in criminal forensics, medical diagnostics and other commercial uses.
PCR requires a DNA fragment, some primers, which are short synthetic oligonucleotides whose sequence matches a region flanking the target sequence. All four deoxynucleotide triphosphates (dATP, dCTP, dGTP, dTTP), are added along with a heat stable DNA polymerase, Taq, from the organism Thermophilus aquaticus (which lives in hot springs).
PCR is run in a repeating cycle which involves three steps, template denaturing, primer annealing and primer extension.
Template Denaturing
This step is the first regular cycling event and consists of heating the reaction to 94-98°C. It causes denaturing of the DNA template by disrupting the hydrogen bonds between complementary bases, yielding single-stranded DNA molecules. The Taq polymerase is heat-stable so it is not denatured by the high temperature needed to separate the DNA template strands.
Primer Annealing
Next, the solution is cooled to 50-65°C, a temperature that favors complementary DNA sequences finding each other and making base pairs, a process called annealing. Since the primers are present in great excess, the complementary sequences they target are readily found and base-paired to the primers. These primers direct the synthesis of DNA. Only where primer anneals to a DNA strand will replication occur, since DNA polymerases require a primer to begin synthesis of a new strand.
Primer Extension
The temperature at this step depends on the DNA polymerase used; Taq polymerase has its optimum activity temperature at 75-80°C, and commonly a temperature of 72°C is used with this enzyme. At this step the DNA polymerase synthesizes a new DNA strand complementary to the DNA template strand by adding dNTPs that are complementary to the template in 5′ to 3′ direction, condensing the 5′-phosphate group of the dNTPs with the 3′-hydroxyl group at the end of the nascent (extending) DNA strand. On completion of this step there are two copies of the DNA template. The new DNA then be denatured to start the cycle again. Each cycle doubles the amount of DNA. Using automated equipment, each cycle of replication can be completed in less than 5 minutes. After 30 cycles, what began as a single molecule of DNA has been amplified into more than a billion copies. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/28%3A_Biomolecules_-_Nucleic_Acids/28.07%3A_DNA_Synthesis.txt |
• 30.1: Molecular Orbitals of Conjugated Pi Systems
HOMO and LUMO are often referred to as frontier orbitals and their energy difference is termed the HOMO–LUMO gap. One common way of thinking about reactions in this way is through the concept of frontier orbitals. This idea says that if one species is going to donate electrons to another in order to form a new bond, then the donated electrons are most likely going to come from the highest occupied energy level.
• 30.2: 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 key sigma bond must be formed at the terminus of a pi system. These reactions classified by the number of pi electrons involved.
• 30.3: Stereochemistry of Thermal Electrocyclic Reactions
Frontier orbital theory can predict the stereochemistry of electrocyclic reactions. Electrons in the HOMO are the highest energy and therefore the most easily moved during a reaction. A molecular orbital diagram can be used to determine the orbital symmetry of a conjugated polyene's HOMO. Thermal reactions utilize the HOMO from the ground-state electron configuration of the molecular orbital diagram while photochemical reactions utilize the HOMO in the excited-state electron configuration.
• 30.4: Photochemical Electrocyclic Reactions
Electron excitation changes the symmetry of the new HOMO which has a corresponding effect on the reaction stereochemistry. Under photochemical reaction conditions conjugated dienes undergo disrotatory cyclization whereas under thermal conditions they underwent conrotatory cyclization. Likewise, conjugated triene undergo conrotatory photochemical cyclization while undergoing disrotatory thermal cyclization.
• 30.5: Cycloaddition Reactions
A concerted combination of two π-electron systems to form a ring of atoms having two new σ bonds and two fewer π bonds is called a cycloaddition reaction. The number of participating π-electrons in each component is given in brackets preceding the name of the reaction. The Diels-Alder reaction is the most useful cycloaddition reaction due to the ubiquity of 6-membered rings and its ability to reliably control stereochemistry in the product.
• 30.6: Stereochemistry of Cycloadditions
Frontier orbital theory can be used to predict if a given cycloaddition will occur with suprafacial or with antarafacial geometry. In a standard Diels-Alder reaction, bonding interactions are created when the electron containing HOMO of the diene donates electrons to the electron vacant LUMO of the other the dienophile. The dienophile has one pi bond, so it will use the pi MOs for a 2 atom system.
• 30.7: Sigmatropic Rearrangements
Molecular rearrangements in which a σ-bonded atom or group, flanked by one or more π-electron systems, shifts to a new location with a corresponding reorganization of the π-bonds are called sigmatropic reactions. The reactant and product have the same number and type of bonds, just different bond locations.
• 30.8: Some Examples of Sigmatropic Rearrangements
• 30.9: A Summary of Rules for Pericyclic Reactions
Chapter 30: Orbitals and Organic Chemistry - Pericyclic Reactions
Frontier Molecular Orbital Theory
Prior to 1965, pericyclic reactions were known as "no mechanism reactions" since no one could adequately explain why reaction outcomes changed depending on whether reactants were exposed to heat or light. In 1965 Robert Burns Woodward and Roald Hoffmann used Frontier Molecular Orbital Theory, initially proposed by Kenichi Fukui, to develop their Theory of Conservation of Orbital Symmetry where outcomes of pericyclic reactions are explained by examining the Highest Occupied Molecular Orbital (HOMO) or Lowest Unoccupied Molecular Orbital (LUMO) of the reacting system. Their analysis of cycloadditions, electrocyclic reactions, and sigmatropic rearrangements is commonly referred to as the Woodward-Hoffmann Rules. A detailed analysis of three reaction types is provided in the subsequent sections of this chapter.
HOMO and LUMO are often referred to as frontier orbitals and their energy difference is termed the HOMO–LUMO gap. One common way of thinking about reactions in this way is through the concept of frontier orbitals. This idea says that if one species is going to donate electrons to another in order to form a new bond, then the donated electrons are most likely going to come from the highest occupied energy level. In this level, called the highest occupied molecular orbital (HOMO), the electrons are further from the nucleus and therefore less tightly held by the protons in the nucleus. The electrons would be donated, in turn, to the lowest empty energy level on the other species, called the lowest unoccupied molecular orbital (LUMO).
Molecular orbital interaction between frontier orbitals.
Molecular Orbitals
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 antibonding 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.
Electronic Transitions
When a double-bonded molecule such as ethene (common name ethylene) absorbs 165 nm light, it undergoes a π - π* transition. An electron is moved from the HOMO of ethene to the LUMO placing the molecule in an excited state.
Where electronic transition becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems. In these groups, the HOMO–LUMO gap energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. The MO diagram for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are pi bonding, while the upper two are pi antibonding. Comparing this MO picture to that of ethene, our isolated pi-bond example the HOMO would be psi 2 and the LUMO would be psi 3. The HOMO-LUMO energy gap is smaller for the conjugated 1,3-butadiene system which absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the HOMO–LUMO gap energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm.
Woodward-Hoffmann Rules
Much of what we have said about the electronic factors controlling pericyclic reaction was formulated in the mid 1960's by the American chemists R. B. Woodward and R. Hoffmann, in terms of what came to be called the orbital symmetry principles, or the Woodward-Hoffmann rules. This is a particularly simple approach says that many details of pericyclic reactions can explained by "conservation of orbital symmetry." This requires the symmetries of the molecular orbitals of reactants to be the same as the molecular orbitals of the products for a reaction to proceed.
The original approach of Woodward and Hoffmann involved construction of an "orbital correlation diagram" to see if the lobes of the reactant molecular orbitals match phases and allow for overlap required for bonding to occur. The symmetries of the appropriate reactant and product orbitals were matched to determine whether the transformation could proceed without a symmetry imposed conversion of bonding reactant orbitals to antibonding product orbitals. If the correlation diagram indicated that the reaction could occur without encountering such a symmetry-imposed barrier, it was termed symmetry allowed. If a symmetry barrier was present, the reaction was designated symmetry-forbidden.
Exercise \(1\)
Using the molecular orbital diagram for 1,3,5-hexatriene determine the HOMO and LUMO for both the ground and excited state.
Answer
For the ground state the HOMO is psi 3 and LUMO is psi 4.
For the excited state the HOMO is psi 4 and LUMO is psi 5. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.1%3A_Molecular_Orbitals_of_Conjugated_Pi_Systems.txt |
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 key sigma bond must be formed at the terminus of a pi system. These reactions classified by the number of pi electrons involved. Thus, 4 pi reactions form 4 membered rings, as in a conjugated diene can being converted into a cyclobutene. Also, 6 pi reactions form 6 membered rings as in a conjugated triene can be converted into a cyclohexadiene. These reactions are often reversible with the reverse reaction may be called electrocyclic ring opening. Although more more pi electrons can be used, the 4 pi and 6 pi variants are by far the most common and are illustrated below with the key sigma bond highlighted in magenta.
A striking feature of electrocyclic reactions that proceed by concerted mechanisms is their high degree of stereospecificity. For example, 2-trans, 4-cis, 6-trans-2,4,6-octatriene undergoes ring closure to cis-5,6-dimethyl-1,3-cyclohexadiene under thermal conditions i.e. when heated. Similarly the isomeric 2-trans, 4-cis, 6-cis-2,4,6-octatriene produces trans-5,6-dimethyl-1,3-cyclohexadiene, as noted below. However these results are completely reversed if the reaction is run under photochemical conditions (Irradiation with ultraviolet light). For example if 2-trans, 4-cis, 6-cis-2,4,6-octatriene is irradiated with UV light cis-5,6-dimethyl-1,3-cyclohexadiene would be produced.
Similar results are seen with the 4 pi electrocyclic reaction of cis,trans-2,4-hexadiene being heated to exclusively form cis-3,4-dimethylcyclobutene and being irradiated with UV light to exclusively form trans-3,4-dimethylcyclobutene. Likewise, trans,trans-2,4-hexadiene forms trans-3,4-dimethylcyclobutene when heated and cis-3,4-dimethylcyclobutene when being irradiated with UV light.
The stereospecificity of electrocyclic reaction can be explained by considering the terminal lobes of the molecular orbital fo the conjugated π-electron system. For electrocyclic reactions of occur, molecular orbital lobes with the same sign from the HOMO of the molecule must rotate to form/break the key ring sigma bond. If the orbital lobes involved both rotate in the same direction (both counterclockwise or both clockwise), the process is called conrotatory. This typically occurs when orbital lobes of the same sign are on opposite sides of the molecule.
If the orbitals involved rotate in opposite directions (one clockwise and one counterclockwise), the process is called disrotatory. These differences in rotation are critically important when stereocenters are formed or broken. This typically occurs when orbital lobes of the same sign are on the same side of the molecule.
30.3: Stereochemistry of Thermal Electrocyclic Reactions
Frontier orbital theory can be used to predict the stereochemistry of electrocyclic reactions. Electrons in the HOMO are the highest energy and therefore the most easily moved during a reaction. A molecular orbital diagram can be used to determine the orbital symmetry of a conjugated polyene's HOMO. Thermal reactions utilize the HOMO from the ground-state electron configuration of the molecular orbital diagram while photochemical reactions utilize the HOMO in the excited-state electron configuration.
The molecular orbital of 1,3,5-hexatriene in its ground state electron configuration has psi three as its HOMO. The terminal molecular orbital lobes of the HOMO with the same sign are on the same side which predicts disrotatory ring closure under thermal conditions.
Disrotatory cyclization is observed during the electrocyclic reaction of 2,4,6-octatriene. The trans,cis,cis-2,4,6-octatriene isomer produces cis-5,6-dimethyl-1,3-cyclohexadiene as the product of thermal cyclization while the trans,cis,cis-2,4,6-octatriene isomer produces trans-5,6-dimethyl-1,3-cyclohexadiene.
The molecular orbital of a conjugated diene, such as 1,3-butadiene has psi two as the HOMO in its ground state electron configuration. The terminal molecular orbital lobes of the HOMO with the same sign are on opposite sides which predicts conrotatory ring closure under thermal conditions. However, the equilibrium of the electrocyclic reaction only allows for the ring opening to be observed.
Thus heating cis-3,4-dimethylcyclobutene causes the conrotatory ring opening to form cis,trans-2,4-hexadiene. Likewise, trans-3,4-dimethylcyclobutene forms trans,trans-2,4-hexadiene when heated.
A pattern begins to form revealing a relationship between the number of double bonds in the conjugated polyene and the rotation during electrocyclic reactions. For thermal electrocyclic reactions, polyenes with an odd number of double bonds undergo disrotation and those with an even number of double bond undergo conrotation.
Exercise \(1\)
1) The thermal electrocyclic ring opening of trans-3,4-dimethylcyclobutene could form trans,trans-2,4-hexadiene or cis,cis-2,4-hexadiene. However, the trans,trans-2,4-hexadiene is the isomer obtained from the reaction. Explain how it is possible to get both products and why the trans,trans-2,4-hexadiene is preferred.
2) If a conjugated tetraene were to undergo a thermal electrocylic reaction would the orbital nodes undergo con or disrotation?
Answer
1) Dienes undergo conrotation during thermal electrocyclic ring opening. Conrotation means the orbital nodes both rotate in the same direction either both clock with or both counter clockwise. If the nodes both rotate counter clockwise the trans,trans-2,4-hexadiene isomers forms. If they both rotate clockwise the cis,cis-2,4-hexadiene isomer is formed. Trans double bonds are more stable than cis due to steric strain. The trans,trans-2,4-hexadiene is perferably formed because it is more stable.
2) A tetraene has an even number of double bonds so it would be expected to undergo conrotation. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.2%3A_Electrocyclic_Reactions.txt |
As discussed in Section 30.1, irradiation of a conjugated polyene with ultraviolet light causes an electron from the groud-state HOMO to be excited to the ground state LUMO. This creates a new higher energy HOMO in an electron configuration called the excited state. Electron excitation changes the symmetry of the new HOMO which has a corresponding effect on the reaction stereochemistry. Under photochemical reaction conditions conjugated dienes undergo disrotatory cyclization whereas under thermal conditions they underwent conrotatory cyclization. Likewise, conjugated triene undergo conrotatory photochemical cyclization while undergoing disrotatory thermal cyclization. For example, trans,trans-2,4-hexadiene undergoes conrotatory photochemical cyclization to form cis-3,4-dimethylcyclobutene.
The conjugated tirene, trans,cis,trans-2,4,6-octatriene undergoes conrotation to form trans-5,6-dimethyl-1,3-cyclohexadiene during photochemical cyclization.
Generalized Statement of Woodward-Hoffmann Rules for Electrocyclic Reactions
Thermal and photochemical electrocyclic reactions always produce the opposite stereochemistry products due to the difference in symmetries in their HOMO frontier orbitals. This idea can be combined with the trend of even and odd polyenes to provide simple rules to predict the stereochemistry of electrocyclic reactions.
Number of Double Bonds Thermal Photochemical
Odd Disrotatory Conrotatory
Even Conrotatory Disrotatory
Exercise \(1\)
Would the following electrocyclic reaction be con or disrotatory? Please draw the expected product.
Answer
1) An even number of double bonds in photochemical conditions predicts a disrotatory reaction.
30.5: Cycloaddition Reactions
A concerted combination of two π-electron systems to form a ring of atoms having two new σ bonds and two fewer π bonds is called a cycloaddition reaction. The number of participating π-electrons in each component is given in brackets preceding the name of the reaction. The Diels-Alder reaction (Section 14-4) is the most useful cycloaddition reaction due to the ubiquity of 6-membered rings and its ability to reliably control stereochemistry in the product. In the Diels-Alder cycloaddition reaction, a conjugated diene, simply referred to as the diene, reacts with a double or triple bond coreactant called the dienophile, because it combines with (has an affinity for) the diene. The Diels-Alder reaction is a [4+2] cycloaddition (4 pi electrons from the diene and 2 pi electrons from the dienophile) that yields a functionalized 6-membered ring product. During the Diels-Alder reaction, two pi-bonds are converted to two sigma-bonds.
Due to the concerted mechanism for cycloaddition reactions, the geometry of atoms on the dienophile or the diene maintain their orientation in the product. This is a critical point for the 4 atoms (both dienophile atoms and the terminal atoms of the diene) that become sp3 hybridized and thus are potential stereocenters in the product. (As a reminder, when chiral products are formed, we obtain a racemic mixture of enantiomers.) As highlighted below, cis dienophiles yield cis substituents in the product, while trans dienophiles yield trans product substituents. Substituents on the terminal atoms of the diene also can become stereocenters and this analysis is a little less straightforward than for dienophile substituents. The way to think about the diene substituents is whether the are pointing "outside" or "inside" the diene. These orientations are illustrated below. When groups are both pointing "outside" or "inside", we can consider them to be cis and they will end up cis in the product. When one group is pointing "outside" and one "inside", we can consider them as trans and they will be trans in the product.
Another important reaction is the [2+2] cycloaddition of two alkene containing molecules to form a 4-membered cyclobutane ring. The [2 + 2] cycloaddition of two alkenes does not occur by simply heating but can only be achieved by irradiation with ultraviolet light.
Many other cycloadditions are known, such as [2 + 2 + 2], and other types of [2 + 2], which give different size of rings. Some specific examples are shown below:
Like other pericyclic reactions, cycloadditions are determined by the orbital symmetry of the frontier orbitals of the reactants. For bonding to occur in a cycloaddition, the terminal lobes of the frontier orbitals of the two pi systems must have the correct symmetry. Correct symmetry can be obtained in two different ways. If the signs on the orbital lobes are the same on the faces of both reactants then the reaction undergoes a suprafacial cycloaddition. If the signs of the orbital lobes are the same on the face of one reactant but opposite on the other reactant then the reaction undergoes an antarafacial cycloaddition. Although both types are symmetry allowed, the fact that antarafacial cycloadditions require the twisting of a pi orbital system makes them more difficult to achieve. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.4%3A_Photochemical_Electrocyclic_Reactions.txt |
Frontier orbital theory can be used to predict if a given cycloaddition will occur with suprafacial or with antarafacial geometry. In a standard Diels-Alder reaction, bonding interactions are created when the electron containing HOMO of the diene donates electrons to the electron vacant LUMO of the other the dienophile. The dienophile has one pi bond, so it will use the pi MOs for a 2 atom system. The dienophile has 2 pi electrons which makes psi 2* its LUMO. The diene has two pi bonds, so it will use the pi MOs for a 4 atom system. The diene has has 4 pi electrons makes psi 2 its HOMO.
These MO diagrams show that the ground-state fronteir orbitals of both reactants have terminal lobes with matching signs. The symmery of these orbitals are such that bond formation will easily occur under thermal conditions with suprafacial geomerty as shown below. (Note: The dashed black lines in the figure below represent nodes in the pi molecular orbitals of the diene and dienophile.) The two new sigma bonds, shown as dashed magenta lines below, are formed from constructive overlap of the terminal dienophile orbitals with the terminal orbitals of the diene.
Photochemical [2+2] cycloadditions are excellent reactions for the synthesis of strained products containing 4-membered rings. One of the reaction partners must be conjugated so that it can absorb light and become an excited state molecule. These reactions produce strained 4-membered rings but are not reversible because the products lack conjugation and, thus, can't absorb light to facilitate a cycloreversion.
So, why won't this reaction happen thermally? How does our molecular orbital analysis help us understand the importance of this being a photochemical reaction? First, let's look at the orbital analysis if we tried to do a thermal [2+2] reaction. As shown below, we cannot get suprafacial overlap for both of the 2 pi reactants when trying to combine psi 1 HOMO with psi 2* LUMO. This means that it is not favorable to convert the two reactant pi bonds into two new product pi bonds.
What happens when we shine light on the reaction? Light creates an excited state molecule by promoting an electron in the HOMO to the LUMO, as shown below. This means the excited state HOMO is the ground state LUMO. We need to understand a few key points about photoreactions before doing our molecular orbital analysis. Excited state molecules are very short lived, relaxing back to the ground state very quickly. Therefore, it is practically impossible for two excited state molecules to find each other in a reaction. Instead, reactions occur between one excited state molecule and one ground state molecule. When only one molecule is conjugated, that is the molecule that will form the excited state. If both reactants are conjugated, either can form the excited state. The orbital analysis is shown below. First, we see the orbital picture when a ground state molecule absorbs light to form an excited state. Second, when we analyze the reaction, it is now psi 2* HOMO of the excited state molecule reacting with psi 2* LUMO of the ground state molecule. This gives suprafacial constructive overlap for both orbitals and the 2 reactant pi bonds can be converted into two product sigma bonds.
Cycloaddition reactions can be categorized based on the total number electrons involved in the rearrangement. The [4+2] Diels-Alder reaction involves 6 electrons and takes place using a suprafacial pathway under thermal conditions. The thermal [2+2] cycloaddtion of two alkenes involves 4 electrons and must take place by antarafacial pathway. However, the pathway is reversed for photochemical [2+2] cycloaddtions which take place by a supraficial pathway. These ideas can be generalized using the rules for cycloadditions below.
Generalized Statement of Woodward-Hoffmann Rules for Cycloadditions
Number of Electrons Thermal Photochemical
4n + 2 Suprafacial (Allowed) Antarafacial (Forbidden)
4n Antarafacial (Forbidden) Suprafacial (Allowed)
Exercise \(1\)
1) For the following reactions determine what type of cycloaddition is occurring, is the reaction supra or antarafacial, and would the reaction require thermal or photochemical conditions.
a)
b)
c)
Answer
1)a) [6+4], suprafacial, thermal.
b) [4+2], suprafacial, thermal.
c) [2+2], antarafacial, photochemical. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.6%3A_Stereochemistry_of_Cycloadditions.txt |
Molecular rearrangements in which a σ-bonded atom or group, flanked by one or more π-electron systems, shifts to a new location with a corresponding reorganization of the π-bonds are called sigmatropic reactions. The reactant and product have the same number and type of bonds, just different bond locations. These rearrangements are described by two numbers set in brackets, which refer to the position of the sigma bond in the reactant that is broken compared to the position of the sigma bond in the product that is formed. The most common examples include hydrogen shifts across a diene system (called a [1,5] H shift) and rearrangements of double allyl-type systems (called [3,3] rearrangements).
As shown in the examples below, the atoms on the sigma bond in the reactant that is broken (magenta bond) are both labeled "1", and the numbering of atoms on each side of that sigma bond continue until the atoms connected by the new sigma bond in the product (magenta) are reached. Thus, the H shift is [1,5] because the key sigma bond in both the reactant and product is to the H while the H moves from C-1 to C-5. For the [3,3] rearrangement, the broken sigma bond migrates across two allyl-type systems and forms between atoms "3" and "3" in the product. This particular example is a Claisen rearrangement since an allyl vinyl ether is transformed into a 1,4-carbonyl alkene.
The migration of a group during a sigmatropic rearrangement is controlled by the orbital symmetries of the alkenes involved. Sigmatropic rearrangements can occur on one face of the molecule (think top or bottom, like a syn addition to an alkene) which is called a suprafacial reaction or from one face to the other (think from top to bottom or vice versa, like an anti addition to an alkene) which is called antarafacial. We will explore this idea further in the next section.
Generalized Statement of Woodward-Hoffmann Rules for Sigmatropic Rearrangements
Suprafacial and antarafacial sigmatropic rearrangements are considered symmetry-allowed by the Woodward-Hoffmann rules. However, suprafacial reactions are much more common. Note! These rules for sigmatropic rearrangements are the same as those given for cycloaddition reactions in Section 30-3.
Number of Double Bonds Thermal Photochemical
Odd Suprafacial Antarafacial
Even Antarafacial Suprafacial
Exercise \(1\)
For the following sigmatropic hydrogen shift please use braked number to discribe the reaction. Predict if the hydrogen shift will be suprafacial and antarafacial. Lastly, draw in arrows to discribe the mechanism for this reaction.
Answer
An example of an [1,7] hydrogen shift is shown in the following diagram. The conjugated alkene has three double bonds, which is an odd number, and the reaction is occuing under thermal conditions. This means the reaction is predicted to be suprafacial. The conjugated triene assumes a nearly planar coiled conformation in which a methyl hydrogen is oriented just above the end carbon atom of the last double bond. A [1s,7a] sigmatropic hydrogen shift may then take place, as described by the four curved arrows. With reference to the approximate plane of this π-electron system (defined by the green bonds), the hydrogen atom departs from the bottom face and bonds to the top face, so the transfer is antarafacial. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.7%3A_Sigmatropic_Rearrangements.txt |
A sigmatropic rearrangement which involves a [1,5] hydrogen shift, like the one shown below illustrates, a hydrogen atom (s orbital) moving across a 5 atom system containing two conjugated double bonds. Since there are three curved arrows in the mechanism, this is a 6 pi reaction. The rules for sigmatropic rearrangements discussed in the last section predict the hydrogen atom will undergo a suprafacial shift. The C-H bond breaking and the C-H bond forming, magenta dashed lines, are both on the bottom face of the molecule's pi system.
The suprafacial nature of this reaction comes from the orbital symmetry of the HOMO from a 6 pi electron molecular orbital. The orbital lobes on the terminal ends have the same sign on the same side.
During the reaction mechanism the hydrogen atom, an s orbital, is passes from one molecular to another of the same sign on the same side of the molecule or suprafacial.
For example, 5-t-butyl-1,3-cyclopentadiene easily undergoes a [1,5] suprafacial shift of a hydrogen atom under thermal conditions to yield 1-t-butyl-1,3-cyclopentadiene as a rearranged product.
As another example, heating S-2-deuterio-6-methyl-(2E,4Z)-2,4-octadiene undergoes a thermal [1,5] suprafacial shift of a hydrogen to produce the product, R-2-deuterio-6-methyl-(2Z,4E)-3,5-octadiene with stereochemical control.
Claisen Rearrangement
A very important example of a [3,3] sigmatropic reaction is the Claisen rearrangement of an allyl aryl ether or an allyl vinyl ether (Section 18-4). Heating an allyl aryl ether to 250 oC causes an sigmatropic rearrangement to produce an o-allylphenol. This rearrangement initially produces the non-aromatic 6-allyl-2,4-cyclohexadienone intermediate which quickly undergoes a proton shift to reform the aromatic ring in the o-allylphenol product. Claisen rearrangement occurs in a six-membered, cyclic transition state involving the movement of six bonding electron.
Allyl vinyl ethers can also undergo a Claisen rearrangement when heated to form gamma, delta -unsaturated ketones or aldehydes.
Cope Rearrangement
Another important example of a [3,3] sigmatropic reaction is the cope rearrangement. This reaction converts between isomers of a 1,5-hexadiene through a cyclic transition state.
Both Cope and Claisen rearrangements involve the movement of six electrons which means they both react by suprafacial pathways under thermal conditions. These reactions can be considered to occur due to changes in overlap between orbitals around the cyclic transition state. Two orbitals which form the sigma bond being broken tilt away from each other while two orbitals that are pi bonding tilt toward each other to from a new sigma bond. After the change there are now two p orbitals parallel to each other on the left which then form new pi bonds.
Orbital rearrangement in a Cope rearrangement
Sigmatropic rearrangements are rare in biological chemistry. One example is the chorismate mutase catalyzed Claisen rearrangement of chorismate (a allylic vinyl ether) to form prephenate. Prephenate is a precursor in the biosynthetic pathway of aromatic amino acids phenylalanine and tyrosine.
.
Exercise \(1\)
1) A Claisen rearrangement performed in the presence of ortho-substituents exclusively leads to para-substituted rearrangement products. This occurs via a Claisen rearrangement followed by a Cope rearrangement. Draw out the intermediates formed during this transformation and the mechanims for both rearrangments.
2) Please draw the expected products of the following [1,5] suprafacial shifts.
a)
b)
3) Please draw the product of the following [3,3] suprafacial shifts.
a)
Answer
1)
2)
a)
b)
3)
a)
30.9: A Summary of Rules for Pericyclic Reactions
Before pericyclic reactions can be put to use in a predictable and controlled manner, a broad mechanistic understanding of the factors that influence these concerted transformations must be formulated. The simplest, albeit least rigorous, method for predicting the configurational path favored by a proposed pericyclic reaction is based upon a transition state electron count. In most of the earlier examples, pericyclic reactions were described by a cycle of curved arrows, each representing a pair of bonding electrons. The total number of electrons undergoing reorganization is always even, and is either a 4n+2 or 4n number (where n is an integer). Once this electron count is made, the following table may be used for predictions. It is important to remember that going from thermal to photochemical conditions or going from 4n to 4n+2 reaction electrons changes the outcome of the reaction.
Thermal Reactions (Ground State) Electron Count Stereochemistry
4n + 2 Suprafacial or Disrotatory
4n Antarafacial or Conrotatory
Photochemical Reactions
(Excited State)
Electron Count Stereochemistry
4n + 2 Antarafacial or Conrotatory
4n Suprafacial or Disrotatory
Exercise $1$
Predict the stereochemistry of the following reactions:
1. The photochemical cyclization of a conjugated tetraene.
2. The thermal cyclization of a conjugated tetraene
3. A thermal [4+4] cycloaddition
4. A photochemical [2+5] cycloaddition
5. A thermal [3,5] sigmatropic rearrangement
Answer
1)a) Disrotatory
b) Conrotatory
c) Antarafacial
d) Suprafacial
e) Antarafacial | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_30%3A_Orbitals_and_Organic_Chemistry_-_Pericyclic_Reactions/30.8%3A_Some_Examples_of_Sigmatropic_Rearrangements.txt |
Synthetic polymer are man-made polymer that is not a biopolymer (e..g, proteins or complex carbohydrates). Synthetic polymers are mostly non-biodegradable and often synthesized from petroleum. The eight most common types of synthetic organic polymers are: Low-density polyethylene (LDPE), High-density polyethylene (HDPE), Polypropylene (PP), Polyvinyl chloride (PVC), Polystyrene (PS), Nylon, nylon 6, nylon 6,6, Teflon (Polytetrafluoroethylene) and Thermoplastic polyurethanes (TPU).
• 31.1: Chain-Growth Polymers
Polymers resulting from additions to alkenes monomers are chain-growth polymers. In these processes each addition step results in a longer chain which ends in a reactive site. The mechanism of each addition step is the same, and each addition step adds another monomer to extend the chain by one repeating unit. The most common and thermodynamically favored chemical transformations of alkenes are addition reactions.
• 31.2: Stereochemistry of Polymerization: Ziegler-Natta Catalysts
An efficient and stereospecific catalytic polymerization procedure was developed by Karl Ziegler (Germany) and Giulio Natta (Italy) in the 1950's. Ziegler-Natta catalysts are prepared by reacting certain transition metal halides with organometallic reagents. The catalyst formed by reaction of triethylaluminum with titanium tetrachloride is commonly used. Ziegler-Natta catalysts allowed for the first time, the stereochemically controlled synthesis of polymers with virtually no branching.
• 31.3: Copolymers
Homopolymers are made with a single monomer and are made up of identical repeating units. Copolymers is made when two or more different monomers are polymerized together to create a polymer with variable repeating units. For example the monomers hexafluoropropene and vinylidene fluoride can be polymerized together to create the copolymer vitron which is used to create durable gaskets.
• 31.4: Step-Growth Polymers
Often, the reactions used to link these monomers include multiple nucleophilic acyl substitutions. Step-growth polymerizations usually use two different monomers, neither of which would undergo polymerization on its own. The two monomers are multifuntional and complementary to each other, such that each provides the other with a reactive partner. In this section, we will be focusing on monomers which are difunctional, meaning they contain two of the same reactive functional group.
• 31.5: Olefin Metathesis Polymerization
Alkene metathesis reactions are gaining wide popularity in synthesizing unsaturated olefinic compounds. Central to this catalysis is a metal carbene intermediate that reacts with olefins to give different olefinic compounds. When two different olefin substrates are used, the reaction is called the “cross metathesis” owing to the fact that the olefinic ends are exchanged. In a process called, olefin metathesis polymerization, unsaturated olefinic polymers can be created by a metathesis reaction.
• 31.6: Polymer Structure and Physical Properties
To account for the physical differences between the different types of polymers, the nature of the aggregate macromolecular structure, or morphology, of each substance must be considered. Because polymer molecules are so large, they generally pack together in a non-uniform fashion, with ordered or crystalline-like regions, called crystallites, mixed together with disordered or amorphous domains. In some cases the entire solid may be amorphous, composed entirely of tangled macromolecules.
Chapter 31: Synthetic Polymers
Polymers resulting from additions to alkenes monomers are chain-growth polymers. In these processes each addition step results in a longer chain which ends in a reactive site. The mechanism of each addition step is the same, and each addition step adds another monomer to extend the chain by one repeating unit. 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.
Each step in this polymer formation is an addition to an alkene. The mechanism is in most cases a free radical addition. In free radical reactions the pi pair of electrons separates. One of these electrons pairs with an electron from the attacking reagent to form a sigma bond with one of the alkene carbons. and the other electron remains attached to the other alkene carbon. (Curved arrows with only one "barb" on a point are used to follow the path of a single electron in the same way that "double-headed" arrows follow the path of an electron pair.) Intermediates with an unpaired electron are called free radicals, so this step can be described as adding a free radical to an alkene to lengthen the chain by two carbons and generate a new free radical. In its turn this new free radical can add to another molecule of monomer and continue the process.
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.
Cationic Chain-Growth Polymerization
Cationic polymerizations are typically acid-catalyzed. Electrophilic addition of H+ to a double bond forms a carbocation which is propagated by repeated reactions with addition alkene monomers. Alkene monomers bearing cation stabilizing groups, such as alkyl, phenyl or vinyl can be polymerized by cationic processes.
Polymerization of isobutylene (2-methylpropene) by traces of strong acids is an example of cationic polymerization. The initiation reagent cationic polymerization is commonly the Lewis acid, boron tribluoride (BF3), along with traces of water to form the acidic BF3OH-H+ catalyst. The polyisobutylene product is a soft rubbery solid, 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).
Anionic Chain-Growth Polymerization
Only monomers having anion stabilizing electron-withdrawing groups (EWG) substituents, such as phenyl, cyano or carbonyl can undergo anionic polymerization.
Species that have been used to initiate anionic polymerization include alkali metals, alkali amides, alkyl lithiums and various electron sources. The fundamental reaction for anion polymerization is a conjugate nucleophilic addition (Section 18-13) 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.
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. Because the monomer has two electron withdrawing groups the polymerization particularly rapid.
Exercise \(1\)
1) Anionic polymerization of p-substituted styrene proceeds very well when the substituent is an electron-withdrawing group such as nitrile. Explain the reason for the success of this approach.
2) In each group, select the alkene monomer most suitable for cationic polymerization.
3) Provide a mechanism for the formation of a protic initiator from the interaction of boron trifluoride with water.
4) Anethole in a naturally-occurring compound that has been used in cationic polymerizations. Show why anethole should be a good monomer for this method.
5) Show why radicals formed from the following monomers are relatively stable:
a) acrylonitrile, CH2=CHCN
b) methyl acrylate, CH2=CHCO2Me
Answer
1) The additional resonance stabilization of the anion by the nitrile group makes anionic polymerization proceed smoothly.
2)
3)
4)
5) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_31%3A_Synthetic_Polymers/31.1%3A_Chain-Growth_Polymers.txt |
Ziegler-Natta Catalytic Polymerization
When propene is enchained into a polymer, a new chiral center is formed at every position where a methyl group branches from the backbone. Rather than trying to assign each of these chiral centers with stereochemical configurations (R) or (S), we instead describe the relative stereochemical relationships along the backbone. The term used to describe these relationships is "tacticity". If there is no apparent relationship between the projection of the methyl groups along the backbone, the polymer is termed "atactic". If the methyl groups alternate, pointing first one direction, then the other, all the way along the chain, then the polymer is termed "syndiotactic". If the methyl groups all project the same direction, the polymer is described as "isotactic".
An efficient and stereospecific catalytic polymerization procedure was developed by Karl Ziegler (Germany) and Giulio Natta (Italy) in the 1950's. 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 is commonly used. Ziegler-Natta catalysts allowed for the first time, the stereochemically controlled synthesis of polymers with virtually no branching. By changing the catalyst, pure isotactic, syndiotactic, or atactic polymers could be created. For this important discovery, Ziegler and Natta received the 1963 Nobel Prize in chemistry. For example, the polymerization of ethylene, using a Ziegler-Natta catalyst produces a stronger (more crystalline) and more heat resistant product, called high-density polyethylene (HDPE), than typical radical polymerizations which produces low-density polyethlene (LDPE). HDPE is normally produced with molecular weights in the range of 200,000 to 500,000, but it can be made even higher. Polyethylene with molecular weights of three to six million is referred to as ultra-high molecular weight polyethylene, or UHMWPE. UHMWPE can be used to make fibers which are so strong they replaced Kevlar for use in bullet proof vests. Large sheets of it can be used instead of ice for skating rinks.
The following diagram presents one mechanism of the Ziegler-Natta polymerization. Formation of the Ziegler-Natta catalyst, adds an alkyl group ligand to create an organo transition metal compound with a vacant coordination site. An alkene ligand is then coordinated to the transition metal which is followed by a 1,2-insertion of the alkene into the metal-carbon bond. The insertion creates a vacant coordination site which can react with another alkene. These same elementary steps continue to occur to provide the polymerization.
Exercise \(1\)
When the monomer vinylidene fluoride, H2C=CF2, is polymerized it does not create isotactic, syndiotactic, and atactic forms. Please explain
Answer
When vinylidene fluoride is polymerized it does not create a chiral center due to the symmertry created by the two fluorides. Chirality is necessary to create isotactic, syndiotactic, and atactic polymer forms.
31.3: Copolymers
Homopolymers are made with a single monomer and are made up of identical repeating units. Copolymers is made when two or more different monomers are polymerized together to create a polymer with variable repeating units. For example the monomers hexafluoropropene and vinylidene fluoride can be polymerized together to create the copolymer vitron which is used to create durable gaskets.
The synthesis of macromolecules composed of more than one repeating unit has been explored as a means of controlling the properties of the resulting material. In this respect, it is useful to distinguish several ways in which different monomeric units might be incorporated in a polymer. The following examples refer to a two component system, in which one monomer is designated A and the other B.
Statistical Copolymers: Also called random copolymers. Here the monomeric units are distributed randomly, and sometimes unevenly, in the polymer chain:
~ABBAAABAABBBABAABA~
Most direct copolymerizations of equimolar mixtures of different monomers give statistical copolymers. If you take a mixture of alkenes that are capable of forming polymers and you polymerize them together, you may well get them randomly enchained into a growing polymer. For example, polymerizing propene and vinyl chloride together creates a polymer with random monmer unit.
Alternating Copolymers: Here the monomeric units are distributed in a regular alternating fashion, with nearly equimolar amounts of each in the chain:
~ABABABABABABABAB~
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.
In some of the examples alternating copolymers, the chain is actually composed of two different monomers. This is the case in polyamides such as nylon-6,6, which is a chain is composed of difunctional amines alternating with difunctional carboxyloids (such as carboxylic acids or acid chlorides). Because of their complementary reactivity, the monomers have to alternate: an amine and then a carboxyloid, to form an amide, and so on. We can think of these polymers as "alternating co-polymers" because the two different monomers alternate with each other along the chain.
Block Copolymers: Instead of a mixed distribution of monomeric units, a long sequence or block of one monomer is joined to a block of the second monomer:
~AAAAA-BBBBBBB~AAAAAAA~BBB~
Several different techniques for preparing block copolymers have been developed, with a common example being anionic polymerization. In the anionic polymerization of styrene, a reactive site remains at the end of the chain until it is quenched. The unquenched polymer has been termed a living polymer because the polymerization can continue as long as monomer is present. If different suitable monomer, methyl methacrylate, is added the chain will continue grow by adding methyl methacrylate units and a block polymer will form. This is illustrated for in the following diagram.
Graft Copolymers: As the name suggests, side chains of a given monomer are "grafted" to the main chain of a different monomer:
~AAAAAAA(BBBBBBB~)AAAAAAA(BBBB~)AAA~.
Graft polymers can be made in great profusion by attaching chains of one kind of polymer to the middle of another. A particularly simple but uncontrollable way of doing this is to knock groups off a polymer chain with x-ray or alpha radiation in the presence of a monomer. The polymer radicals so produced then can grow side chains made of the new monomer. A more elegant procedure is to use a photochemical reaction to dissociate groups from the polymer chains and form radicals capable of polymerization with an added monomer.
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
Exercise \(1\)
Draw the structure of an alternating segment of Saran, a copolymer of vinyl chloride (chloroethene) and vinylidene chloride (1,1-dichloroethene).
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_31%3A_Synthetic_Polymers/31.2%3A_Stereochemistry_of_Polymerization%3A_Ziegler-Natta_Catalysts.txt |
The process of step-growth polymerizations are fundamentally different than in chain-growth. In step-growth polymerizations, monomers are generally linked by a carbon-heteroatom bond (C-O & C-N) formed in non-sequential steps. Often, the reactions used to link these monomers include multiple nucleophilic acyl substitutions. Step-growth polymerizations usually use two different monomers, neither of which would undergo polymerization on its own. The two monomers are multifuntional and complementary to each other, such that each provides the other with a reactive partner. In this section, we will be focusing on monomers which are difunctional, meaning they contain two of the same reactive functional group. A step-growth polymerization starts with two complementary functional groups on different monomers reacting to form a dimer. Because both monomers were difunctional, each retains a reactive group and can react with additional complementary monomers.
In fact the difuctionality of the monomers, allows step-growth polymers to grow in two directions at once. First, two complementary monomers react with each other to form a dimer. Assuming that monomers react at roughly similar rates, when one end of the dimer reacts again it will likely find another dimer and form a tetramer. Then when the tetramer goes to react again it will most likely find another tetramer and form an octamer. This process is repeated allowing the polymer to grow in two directions at the same time.
Virtually all fibers are made from some form of polymer. In particular, silk and wool are composed of a naturally occurring protein polymer. The monomers of proteins are called amino acid residues. These residues are connected by amide linkages which are also called peptide bonds. Many of the early efforts of polymer chemistry were to artificially create fibers which mimicked the properties of silk and wool.
Polyamides
The first fully synthetic polymer fiber, nylon-6,6, was produced in 1938 by the company DuPont. The lead chemist of DuPont's work was Wallace H. Carothers, who reasoned that the properties of silk could be mimicked by constructing a polymer chain created with repeating amide bonds, just like the proteins in silk. Nylon-6,6 was created by first reacting 1,6-hexanedioic acid (adipic acid) and 1,6-hexanediamine to give a salt which was then heated creating multiple amide bonds through nucleophilic acyl substitution. The product of this particular reaction is a polyamide called nylon-6,6. The numbers of the name indicate how many carbons are contained in each monomer. The first “6” stands for the number of carbons in the diamine monomer while the second number indicated the number of carbons in the dicarboxylic acid. Simply by varying the number of carbons in each monomer, a wide variety of nylon polymers can be made.
Nylons are among the most widely used synthetic fibers—for example, they are used in ropes, sails, carpets, clothing, tires, brushes, and parachutes. Known for their high strength and abrasion resistance, nylons can be molded into blocks for use in electrical equipment, gears, bearings, and valves. The strength of nylon fibers comes, in part, from their ability to form strong hydrogen bonding intermolecular forces with each other in much the same fashion as proteins.
Polyesters
Esterfication, via nucleophilic acyl substitutions, can also be used to form the primary linkages in step-growth polymers. A polyester is typically produced when a dicarboxylic acid and a diol are reacted together. After the initial reaction, the ester product contains a free (unreacted) carboxyl group at one end and a free alcohol group at the other. Further esterification using a step-growth polymerization, produces a polyester. The most important polyester, polyethylene terephthalate (PET), is made from the reaction of 1,4-benzenedicarboxylic (terephthalic acid) and 1,2-ethanediol (ethylene glycol) monomers.
Polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes.
Polycarbonates
Beyond carboxylic acid derivatives, virtually any reaction which involves reactive species on two different molecules can be used to perform a step-growth polymerization. A variation involves using a monomer containing a carbonate functional group (−O−(C=O)−O−). A carbonate acts like a double ester and can undergo a type of double transesterification reaction with two alcohols to form a new carbonate containing compound.
In effect, a carbonate is difunctional and can be reacted with a diol to form polymers containing repeated carbonate groups in their structure called polycarbonates. An example of a polycarbonate is the polymer, Lexan, which is created when diphenyl carbonate and bisphenol A (a diol) are reacted together.
Bisphenol (BPA), primarily used to make polycarbonate, is one of the highest-volume chemicals produced in the world, with over six billion pounds made each year. Because polycarbonate is used to make plastic bottles, the lining for food cans, and the lining for beverage cans there has been much concern about trace amounts of BPA leaching from the containers and being ingested. A study conducted in 2003 and 2004 by the Center for Disease Control and Prevention found trace amounts BPA in the tissues of 93% of people in the United States. Consequently, this has led to many beverage companies switching to non-polycarbonate polymers.
Polyurethane
A urethane is a functional group similar to a carbonate and a urea. A urethane has an -OR and a -NR2 group attached to the carbonyl carbon.
Polyurethane (PUR and PU) is a polymer composed of organic units joined by carbamate (urethane) links. Polyurethanes are produced by reacting an isocyanate containing two or more isocyanate groups per molecule (R−(N=C=O)n) with a polyol containing on average two or more hydroxyl groups per molecule (R′−(OH)n) in the presence of a catalyst or by activation with ultraviolet light.
The very widely used polyurethane foams can be considered to be either block polymers or copolymers. The essential ingredients are a diisocyanate and a diol. The diisocyanate most used is 2,4-diisocyano-1-methylbenzene, and the diol can be a polyether or a polyester with hydroxyl end groups. The isocyano groups react with the hydroxyl end groups to form initially an addition polymer, which has polycarbamate (polyurethane) links, and isocyano end groups:
A foam is formed by addition of the proper amount of water. The water reacts with the isocyanate end groups to form carbamic acids which decarboxylate to give amine groups:
The carbon dioxide evolved is the foaming agent, and the amino groups formed at the same time extend the polymer chains by reacting with the residual isocyano end groups to form urea linkages:
$\ce{R'N=C=O} + \ce{RNH_2} \rightarrow \ce{R'NHCONHR} \nonumber$
RN=C=O+RNH2RNHCONPolyurethanes are used in the manufacture of high-resilience foam seating, rigid foam insulation panels, microcellular foam seals and gaskets, durable elastomeric wheels and tires (such as roller coaster, escalator, shopping cart, elevator, and skateboard wheels), automotive suspension bushings, electrical potting compounds, high performance adhesives, surface coatings and surface sealants, synthetic fibers (e.g., Spandex), carpet underlay, hard-plastic parts (e.g., for electronic instruments), condoms,[1] and hoses.
Common Step Growth Polymers
Exercise $1$
1) The following difunctional monomers would undergo polymerization together. Show the product that results.
2) Provide structures of the polyesters made from the monomers as indicated.
Answer
1)
2) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_31%3A_Synthetic_Polymers/31.4%3A_Step-Growth_Polymers.txt |
Olefin Metathesis
The application of organometallic chemistry in homogenous catalysis is progressively increasing with the fast pace of discovery of new catalysts in the area. Alkene metathesis reactions are gaining wide popularity in synthesizing unsaturated olefinic compounds. Central to this catalysis is a metal carbene intermediate that reacts with olefins to give different olefinic compounds. When two different olefin substrates are used, the reaction is called the “cross metathesis” owing to the fact that the olefinic ends are exchanged. In a process called, olefin metathesis polymerization, unsaturated olefinic polymers can be created by a metathesis reaction.
Olefin metathesis catalysts contain a metal carbon double bond with two of the most notable being are the Grubb’s Ru catalyst and the Schrock’s Mo catalyst.
The Grubb’s and the Schrock’s Catalyst
Mechanism Cross Metathesis
An olefin metathesis catalyst is a transition metal compound that is capable of splitting the double bond of an alkene in half and putting the two pieces together with other alkenes. The key part of an olefin metathesis catalyst is a metal-carbon double bond. That is the group that is capable of switching the ends of alkenes around with different partners. The catalyst react reversibly with an alkene to produce a 4−membered metalacyclobutane intermediate called a metallacycle. The ring promptly opens to produce a different catalyst and alkene. The reaction of this new catalyst with the second alkene produces a second metallacycle intermediate. This ring opens to produce the metathesis product and yet another form of the catalyst. The ring opening and closing continues as the reaction moves forward.
Olefin Metathesis Polymerization The variants of metathesis often used in producing polymers are, (i) the Acyclic Diene Metathesis (ADMET) and (ii) the Ring Opening Metathesis Polymerization (ROMP), both of which produce long chain polymers
One method, called ring-opening metathesis polymerization, or ROMP, involves the use of a moderately strained cycloalkene, such as cyclopentene. The strain of the ring favors ring-opening, thereby driving formation of the open-chain product. The polymer that results has double bonds spaced regularly along the chain, allowing for either hydrogenation or further functionalization if desired.
Ring-opening metathesis polymerization (ROMP) takes a cyclic alkene, splits open its double bond, and knitting it together with other cyclic alkenes to produce a long polymer chain with regularly spaced double bonds. The ring-strain of the cycloalkenes favors the open-chain product and drives the polymerization reaction forward.
During a ROMP, a cyclopentene ring is opened up at the double bond and reaches out to join with other cyclopentene rings on either side of it.
The product polymer of a ROMP using cyclopentene rings would look something like this:
Which can draw it in an usual zig-zag conformation.
Mechanism
The mechanism starts with a cycloalkene and a metathesis catalyst reacting to form a four-member metallacycle. Opening of the metallacycle forms a new alkene and catalyst both of which are part of the same intermediate molecule. The catalyst end of the intermediate molecule can then react with a different cycloalkene to form a new metallacycle. This process is repeated causing the polymer chain to from.
The second method is called acyclic diene metathesis (ADMET) which uses long open-chain substrates, which contain a double bond at both ends of the main chain, such as 1,6-heptadiene. A byproduct of the reaction, gaseous ethylene (H2C=CH2), escapes which drives the equilibrium forward.
Exercise $1$
The ring openning metathesis polymerization of norborene produces a commerical polymer called Norsore which is used a sealing material. Please draw the structure of Norsore.
Answer | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_31%3A_Synthetic_Polymers/31.5%3A_Olefin_Metathesis_Polymerization.txt |
Polymer Crystallinity
To account for the physical differences between the different types of polymers, the nature of the aggregate macromolecular structure, or morphology, of each substance must be considered. Because polymer molecules are so large, they generally pack together in a non-uniform fashion, with ordered or crystalline-like regions, called crystallites, mixed together with disordered or amorphous domains. In some cases the entire solid may be amorphous, composed entirely of coiled and tangled macromolecular chains. Crystallinity occurs when linear polymer chains are structurally oriented in a uniform three-dimensional matrix. In the diagram on the right, crystalline domains are colored blue. Increased crystallinity is associated with an increase in rigidity, tensile strength and opacity (due to light scattering). Amorphous polymers are usually less rigid, weaker and more easily deformed. They are often transparent.
Schematic diagram of crystallites (in blue) in a largely crystalline polymer.
Three factors that influence the degree of crystallinity are:
1. Chain length: Longer polymer chains tend to have greater van der Waals forces and increased crystallinity than do shorter chains (Section 2-12)
2. Chain branching: Polymer chains without branching can pack closely together. Lack of branching increases the crystallinity of polymers.
3. Interchain interactions: (Intermolecular forces and Crosslinking). Crystallinity increases as the strength of the intermolecular forces between polymer chains increases. Croslining between polymer chains tends to increase crystallinity.
The importance of the first two factors is nicely illustrated by the differences between LDPE and HDPE. As noted earlier (Section 31-2), HDPE is composed of very long unbranched hydrocarbon chains. These pack together easily in crystalline domains that alternate with amorphous segments, and the resulting material, while relatively strong and stiff, retains a degree of flexibility. In contrast, LDPE is composed of smaller and more highly branched chains which do not easily adopt crystalline structures. This material is therefore softer, weaker, less dense and more easily deformed than HDPE.
The forces between the chains in the crystallites of polyethene are the so-called van der Waals or dispersion forces, which are the same forces acting between smaller hydrocarbon molecules. Although these forces are relatively weak, they increase with the size of the molecule. Polymer chains are large enough that the van der Waals forces create a strong and stiff material.
Representation of attractive interactions between the hydrogens in a crystallite of polyethene
In other kinds of polymers, even stronger intermolecular forces can be produced by hydrogen bonding. This is especially important in the polyamides, such as the nylons, of which nylon 66 is most widely used. The increase strength of the interactions between polymer chains makes polyamides among some of the strongest polymer materials known.
Possible hydrogen-bonded structure for crystallites of nylon 66, an amide-type polymer of hexanedioic acid and 1,6-hexanediamine.
Natural rubber is a completely amorphous polymer. Unfortunately, the potentially useful properties of raw latex rubber are limited by temperature dependence; however, these properties can be modified by chemical change. If the chains of rubber molecules are slightly cross-linked by sulfur atoms, a process called vulcanization which was discovered by Charles Goodyear in 1839, the desirable elastomeric properties of rubber are substantially improved. At 2 to 3% crosslinking a useful soft rubber, that no longer suffers stickiness and brittleness problems on heating and cooling, is obtained. At 25 to 35% crosslinking a rigid hard rubber product is formed. The following illustration shows a cross-linked section of amorphous rubber.
Glass Transition and Melt Transition Temperatures
The effect of temperature on the physical properties of polymers is very important to their practical uses. At low temperatures, polymers become hard and glasslike because the motions of the segments of the polymer chains with relation to each other are slow. The approximate temperature below which glasslike behavior is apparent is called the glass transition temperature and is symbolized by Tg. When a polymer containing crystallites is heated, the crystallites ultimately melt, and this temperature is usually called the melt transition temperature and is symbolized as Tm. As the amount of crystallinity in a polymer increases, so does Tm.
Tg often depends on the history of the sample, particularly previous heat treatment, mechanical manipulation and annealing. It is sometimes interpreted as the temperature above which significant portions of polymer chains are able to slide past each other in response to an applied force. The introduction of relatively large and stiff substituents (such as benzene rings) will interfere with this chain movement, thus increasing Tg (note polystyrene below). The introduction of small molecular compounds called plasticizers (discussed later in this section) into the polymer matrix increases the interchain spacing, allowing chain movement at lower temperatures. with a resulting decrease in Tg.
Tm and Tg values for some common addition polymers are listed below.
Polymer LDPE HDPE PP PVC PS PAN PTFE PMMA Rubber
Tm (ºC) 110 130 175 180 175 >200 330 180 30
Tg (ºC) _110 _100 _10 80 90 95 _110 105 _70
Polymer Categories
Thermoplastics
Most of the polymers described in this chapter are classified as thermoplastics. Plastics that soften when heated and become firm again when cooled. This is the more popular type of plastic because the heating and cooling may be repeated and the thermoplastic may be reformed. These polymers tend to have high Tg so they are hard solids at room temperature, however, above Tg they they become malleable and may be shaped, pressed into molds, spun, or cast from melts.
Polyethylene (poly(ethylene terephthalate) or PET is the most common thermoplastic . In 2017, PET made up 34% of the total plastics market with over 100 million tones of polyethylene resins being produced. PET is partially crystalline and is used to create clear plastic bottles such as 2-liter beverage bottles, milk jugs, detergent bottles, and water bottles.
Polystyrene is also a common thermoplastic. The polymer is making it a solid but rather rather brittle at room temperature. Polystyrene is used to make hard clear plastic cups, foam cups, eating utensils, deli food containers, toy model kits, some packing popcorn.
Plasticizers
Plasticizers or dispersants are additives that increase the plasticity or decrease the viscosity of a material. These substances are compounded into certain types of plastics to render them more flexible by lowering the glass transition temperature. They accomplish this by taking up space between the polymer chains and acting as lubricants to enable the chains to more readily slip over each other. Many (but not all) are small enough to be diffusible and a potential source of health problems. Substantial concerns have been expressed over the safety of some plasticizers, especially because some low molecular weight ortho-phthalates have been classified as potential endocrine disruptors with some developmental toxicity reported.
Polyvinyl chloride (PVC) polymers are one of the most widely-plasticized types. PVC is usually is not very crystalline and is relatively brittle and glassy. The properties of polyvinyl chloride can be improved by blending it with substances of low volatility which tend to break down its glasslike structure. Common plasticizers used with PVC are tris-(2-methylphenyl) phosphate (tricresyl phosphate) and dibutyl benzene-1,2-dicarboxylate (dibutyl phthalate). Plasticized polyvinyl chloride is reasonably flexible and is widely used for flexible vinyl materials such as garden hoses, waterbeds, cheap shower curtains, raincoats and upholstery. The pungent oder associated with these products are a testament to the ability of plasticizers to migrate into the environment.
Fibers
Fibers are drawn into thin threads by forcing the melted or dissolved polymer through a spinneret to generate filaments that can be woven into fabrics. Part of this process, called cold-drawing, the polymer material is subjected to strong stress in one direction causing the material to elongate and the crystallites to be drawn together and oriented along the direction of the applied stress. Having the crystallites in a polymer oriented with respect to one another gives the polymer a much higher tensile strength than an unoriented polymer. Polymers such as nylon, which has strong intermolecular forces, has the crystallinity required to be drawn in oriented fibers.
Schematic representation of an oriented crystalline polymer produced by drawing the polymer in the horizontal direction. The crystalline regions are enclosed with dashed lines.
Elastomers
Elastomers usually are amorphous polymers which have the ability to be stretched. The key to this elastic behavior is polymer chains with weak forces between the chains and a sufficiently irregular structure to be unstable in the crystalline state. A useful elastomer needs to have some kind of cross-linking. The important difference between an elastomer and a crystalline polymer is the size of the amorphous regions. When tension is applied and the material elongates, the chains in the amorphous regions straighten out and become more nearly parallel. The forces between the chains are too weak to maintain the crystalline state in the absence of tension. Thus when tension is released, contraction occurs and the original, amorphous polymer is produced. The entropy of the chains is more favorable in the relaxed state than in the stretched state.
Schematic representation of an elastomer in relaxed and stretched configurations.
A good elastomer should not undergo plastic flow in either the stretched or relaxed state, and when stretched should have a "memory" of its relaxed state. These conditions are best achieved with natural rubber (cis-poly-2-methyl-1,3-butadiene, cis-polyisoprene) by curing (vulcanizing) with sulfur. Natural rubber (Section 14-6) is tacky and undergoes plastic flow rather readily, but when it is heated with elemental sulfur, sulfur cross-links are introduced between the chains. These cross-links reduce plastic flow and provide a reference framework for the stretched polymer to return to when it is allowed to relax. Also the double bonds in rubber all have a Z-configuration, which causes this macromolecule to adopt a kinked or coiled conformation.
However, the gutta-percha (structure above) E-isomer of rubber adopts a uniform zig-zag conformation, which produces greater crystallinity making it not an elastomer.
Thermosets
Thermosets are plastics that soften when heated and can be molded, but harden permanently. In a thermoset, crosslinks connect the different chains in the material, forming bridges that span from chain to chain to chain, essentially uniting the material into one big molecule. If it is one big molecule, the chains can never move completely independently of each other, and the material cannot form a new shape.
Schematic representation of the conversion of an uncross-linked thermosetting polymer to a highly cross-linked polymer. The cross-links are shown in a two-dimensional network, but in practice three-dimensional networks are formed.
One of the oldest known thermosetting synthetic polymers, Bakelite, is made by condensation of phenol with formaldehyde. During the thermoset process water is lost and many crosslinks are formed. produce (4-hydroxyphenyl)methanol. Bakelite was patented on December 7, 1909 and was revolutionary for its electrical nonconductivity and heat-resistant properties used in electrical insulators, radio and telephone casings and such diverse products as kitchenware, jewelry, pipe stems, and children's toys.
Exercise $1$
1) Propose a mechanism for the base-catalyzed polymerization of phenol and formaldehyde to form Bakelite.
2) Would you expect the catalytic hydrogenation of gutta-percha to produce a product that is syndiotactic, atactic, or isotactic?
Answer
1) One of the oldest known thermosetting synthetic polymers is made by condensation of phenols with aldehydes using basic catalysts. The resins that are formed are known as Bakelites. The initial stage is the base-induced reaction of benzenol and methanal to give a (4-hydroxyphenyl)methanol, and this reaction closely resembles an aldol addition and can take place at either the 2- or the 4-position of the benzene ring:
The next step in the condensation is formation of a bis(hydroxyphenyl)methane derivative, which for convenience is here taken to be the 4,4'-isomer:
This reaction is probably a Michael type of addition to a base-induced dehydration product of the (4-hydroxyphenyl)methanol:
Continuation of these reactions at the 2-, 4-, and 6-positions of the benzenol leads to the cross-linked three-dimensional Bakelite resin:
2) Atactic. The methyl groups in gutta-percha lack stereochemistry because they are attached to double bond. Catalytic hydrogenation does not provide stereochemical control so H2 could attack from either side of the double bond to produce a chiral center involving the methyl group which is R or S. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/Chapter_31%3A_Synthetic_Polymers/31.6%3A_Polymer_Structure_and_Physical_Properties.txt |
Chapter Objectives
This chapter provides a review of material covered in a standard freshman general-chemistry course through a discussion of the following topics:
• the differences between organic and inorganic chemistry.
• the shapes and significance of atomic orbitals.
• electron configurations.
• ionic and covalent bonding.
• molecular orbital theory.
• hybridization.
• the structure and geometry of the compounds methane, ethane, ethylene and acetylene.
• 1.1: Why This Chapter?
Organic compounds contain carbon atoms bonded hydrogen and other carbon atoms. Organic chemistry studies the properties and reactions of organic compounds.
• 1.2: Atomic Structure - The Nucleus
Atoms are comprised of protons, neutrons and electrons. Protons and neutrons are found in the nucleus of the atom, while electrons are found in the electron cloud around the nucleus. The relative electrical charge of a proton is +1, a neutron has no charge, and an electron’s relative charge is -1. The number of protons in an atom’s nucleus is called the atomic number, Z. The mass number, A, is the sum of the number of protons and the number of neutrons in a nucleus.
• 1.3: Atomic Structure - Orbitals
An atomic orbital is the probability description of where an electron can be found. The four basic types of orbitals are designated as s, p, d, and f.
• 1.4: Atomic Structure - Electron Configurations
The order in which electrons are placed in atomic orbitals is called the electron configuration and is governed by the aufbau principle. Electrons in the outermost shell of an atom are called valence electrons. The number of valence electrons in any atom is related to its position in the periodic table. Elements in the same periodic group have the same number of valence electrons.
• 1.5: Development of Chemical Bonding Theory
Lewis Dot Symbols are a way of indicating the number of valence electrons in an atom. They are useful for predicting the number and types of covalent bonds within organic molecules. The molecular shape of molecules is predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. The shapes of common organic molecules are based on tetrahedral, trigonal planar or linear arrangements of electron groups.
• 1.6: Describing Chemical Bonds - Valence Bond Theory
Covalent bonds form as valence electrons are shared between two atoms. Lewis Structures and structural formulas are common ways of showing the covalent bonding in organic molecules. Formal charge describes the changes in the number of valence electrons as an atom becomes bonded into a molecule. If the atom has a net loss of valence electrons it will have a positive formal charge. If the atom has a net gain of valence electrons it will have a negative formal charge.
• 1.7: sp³ Hybrid Orbitals and the Structure of Methane
The four identical C-H single bonds in methane form as the result of sigma bond overlap between the sp3 hybrid orbitals of carbon and the s orbital of each hydrogen.
• 1.8: sp³ Hybrid Orbitals and the Structure of Ethane
The C-C bond in ethane forms as the result of sigma bond overlap between a sp³ hybrid orbital on each carbon. and the s orbital of each hydrogen. The six identical C-H single bonds in form as the result of sigma bond overlap between the sp³ hybrid orbitals of carbon and the s orbital of each hydrogen.
• 1.9: sp² Hybrid Orbitals and the Structure of Ethylene
The C=C bond in ethylene forms as the result of both a sigma bond overlap between a sp2 hybrid orbital on each carbon and a pi bond overlap of a p orbital on each carbon
• 1.10: sp Hybrid Orbitals and the Structure of Acetylene
The carbon-carbon triple bond in acetylene forms as the result of one sigma bond overlap between a sp hybrid orbital on each carbon and two pi bond overlaps of p orbitals on each carbon.
• 1.11: Hybridization of Nitrogen, Oxygen, Phosphorus and Sulfur
The atomic orbitals of nitrogen, oxygen, phosphorus and sulfur can hybridize in the same way as those of carbon.
• 1.12: Describing Chemical Bonds - Molecular Orbital Theory
Molecular Orbital theory (MO) is a more advanced bonding model than Valence Bond Theory, in which two atomic orbitals overlap to form two molecular orbitals – a bonding MO and an anti-bonding MO.
• 1.13: Drawing Chemical Structures
Kekulé Formulas or structural formulas display the atoms of the molecule in the order they are bonded. Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space. Skeleton formulas or Shorthand formulas or line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. Isomers have the same molecular formula, but different structural formulas
• 1.14: Chemistry Matters—Organic Foods- Risk versus Benefit
• 1.15: Key Terms
• 1.16: Summary
• 1.17: Additional Problems
01: Structure and Bonding
Chapter Contents
1.1 Atomic Structure: The Nucleus
1.2 Atomic Structure: Orbitals 1.3 Atomic Structure: Electron Configurations 1.4 Development of Chemical Bonding Theory 1.5 Describing Chemical Bonds: Valence Bond Theory 1.6 sp3 Hybrid Orbitals and the Structure of Methane 1.7 sp3 Hybrid Orbitals and the Structure of Ethane 1.8 sp2 Hybrid Orbitals and the Structure of Ethylene 1.9 sp Hybrid Orbitals and the Structure of Acetylene 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur 1.11 Describing Chemical Bonds: Molecular Orbital Theory 1.12 Drawing Chemical Structures
We’ll ease into the study of organic chemistry by first reviewing some ideas about atoms, bonds, and molecular geometry that you may recall from your general chemistry course. Much of the material in this chapter and the next is likely to be familiar to you, but it’s nevertheless a good idea to make sure you understand it before moving on.
What is organic chemistry, and why should you study it? The answers to these questions are all around you. Every living organism is made of organic chemicals. The proteins that make up your hair, skin, and muscles; the DNA that controls your genetic heritage; the foods that nourish you; and the medicines that heal you are all organic chemicals. Anyone with a curiosity about life and living things, and anyone who wants to be a part of the remarkable advances taking place in medicine and the biological sciences, must first understand organic chemistry. Look at the following drawings for instance, which show the chemical structures of some molecules whose names might be familiar to you. Although the drawings may appear unintelligible at this point, don’t worry. They’ll make perfectly good sense before long, and you’ll soon be drawing similar structures for any substance you’re interested in.
Historically, the term organic chemistry dates to the mid-1700s, when it was used to mean the chemistry of substances found in living organisms. Little was known about chemistry at that time, and the behavior of the “organic” substances isolated from plants and animals seemed different from that of the “inorganic” substances found in minerals. Organic compounds were generally low-melting solids and were usually more difficult to isolate, purify, and work with than high-melting inorganic compounds.
By the mid-1800s, however, it was clear that there was no fundamental difference between organic and inorganic compounds. The only distinguishing characteristic of organic compounds is that all contain the element carbon.
Organic chemistry, then, is the study of carbon compounds. But why is carbon special? Why, of the more than 197 million presently known chemical compounds, do almost all of them contain carbon? The answers to these questions come from carbon’s electronic structure and its consequent position in the periodic table (Figure 1.2). As a group 4A element, carbon can share four valence electrons and form four strong covalent bonds. Furthermore, carbon atoms can bond to one another, forming long chains and rings. Carbon, alone of all elements, is able to form an immense diversity of compounds, from the simple methane, with one carbon atom, to the staggeringly complex DNA, which can have more than 100 million carbons.
Not all carbon compounds are derived from living organisms, however. Modern chemists have developed a remarkably sophisticated ability to design and synthesize new organic compounds in the laboratory—medicines, dyes, polymers, and a host of other substances. Organic chemistry touches the lives of everyone. Its study can be a fascinating undertaking. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.01%3A_Why_This_Chapter.txt |
As you might remember from your general chemistry course, an atom consists of a dense, positively charged nucleus surrounded at a relatively large distance by negatively charged electrons (Figure 1.3). The nucleus consists of subatomic particles called neutrons, which are electrically neutral, and protons, which are positively charged. Because an atom is neutral overall, the number of positive protons in the nucleus and the number of negative electrons surrounding the nucleus are the same.
than at the gray mesh surface.
Although extremely small—about 10–14 to 10–15 meter (m) in diameter—the nucleus nevertheless contains essentially all the mass of the atom. Electrons have negligible mass and circulate around the nucleus at a distance of approximately 10–10 m. Thus, the diameter of a typical atom is about 2 × 10–10 m, or 200 picometers (pm), where 1 pm = 10–12 m. To give you an idea of how small this is, a thin pencil line is about 3 million carbon atoms wide. Although most chemists throughout the world use the International System (SI) of units and describe small distances in picometers, many organic chemists and biochemists in the United States still use the unit angstrom (Å) to express atomic distances, where 1 Å = 100 pm = 10–10 m. As you probably did in your general chemistry course, however, we’ll stay with SI units in this book.
A specific atom is described by its atomic number (Z), which gives the number of protons (or electrons) it contains, and its mass number (A), which gives the total number of protons and neutrons in its nucleus. All the atoms of a given element have the same atomic number: 1 for hydrogen, 6 for carbon, 15 for phosphorus, and so on; but they can have different mass numbers depending on how many neutrons they contain. Atoms with the same atomic number but different mass numbers are called isotopes. The element carbon, for instance, has three isotopes that occur naturally, with mass numbers of 12, 13, and 14. Carbon-12 has a natural abundance of 98.89%, carbon-13 has a natural abundance of 1.11%, and carbon-14 has only a negligible natural abundance.
The weighted-average of an element’s naturally occurring isotopes is called atomic weight and is given in unified atomic mass units (u) or daltons (Da) where 1 u or 1 Da is defined as one twelfth the mass of one atom of carbon-12. Thus, the atomic weight is 1.008 u for hydrogen, 12.011 u for carbon, 30.974 u for phosphorus, and so on. Atomic weights of all elements are given in the periodic table in Appendix D.
1.03: Atomic Structure - Orbitals
How are the electrons distributed in an atom? You might recall from your general chemistry course that, according to the quantum mechanical model, the behavior of a specific electron in an atom can be described by a mathematical expression called a wave equation—the same type of expression used to describe the motion of waves in a fluid. The solution to a wave equation is called a wave function, or orbital, and is denoted by the lowercase Greek letter psi (ψ).
When the square of the wave function, ψ2, is plotted in three-dimensional space, an orbital describes the volume of space around a nucleus that an electron is most likely to occupy. You might therefore think of an orbital as looking like a photograph of the electron taken at a slow shutter speed. In such a photo, the orbital would appear as a blurry cloud, indicating the region of space where the electron has been. This electron cloud doesn’t have a sharp boundary, but for practical purposes we can set the limits by saying that an orbital represents the space where an electron spends 90% to 95% of its time.
What do orbitals look like? There are four different kinds of orbitals, denoted s, p, d, and f, each with a different shape. Of the four, we’ll be concerned primarily with s and p orbitals because these are the most common in organic and biological chemistry. An s orbital has a spherical shape, with the nucleus at its center; a p orbital has a dumbbell shape with two parts, or lobes; and four of the five d orbitals have a cloverleaf shape with four lobes, as shown in Figure 1.4. The fifth d orbital is shaped like an elongated dumbbell with a doughnut around its middle.
The orbitals in an atom are organized into different layers around the nucleus called electron shells, which are centered around the nucleus and have successively larger size and energy. Different shells contain different numbers and kinds of orbitals, and each orbital within a shell can be occupied by two electrons. The first shell contains only a single s orbital, denoted 1s, and thus holds only 2 electrons. The second shell contains one 2s orbital and three 2p orbitals and thus holds a total of 8 electrons. The third shell contains a 3s orbital, three 3p orbitals, and five 3d orbitals, for a total capacity of 18 electrons. These orbital groupings and their energy levels are shown in Figure 1.5.
orbital; the second shell holds a maximum of 8 electrons in one 2s and three 2p orbitals; the third shell holds a maximum of 18 electrons in one 3s, three 3p, and 3d orbitals; and so on. The two electrons in each orbital are represented by five up and down arrows, $⇅.⇅.$ Although not shown, the energy level of the 4s orbital falls between 3p and 3d.
The three different p orbitals within a given shell are oriented in space along mutually perpendicular directions, denoted px, py, and pz. As shown in Figure 1.6, the two lobes of each p orbital are separated by a region of zero electron density called a node. Furthermore, the two orbital regions separated by the node have different algebraic signs, + and −, in the wave function, as represented by the different colors in Figure 1.4 and Figure 1.6. As we’ll see in Section 1.11, these algebraic signs for different orbital lobes have important consequences with respect to chemical bonding and chemical reactivity.
1.04: Atomic Structure - Electron Configurations
The lowest-energy arrangement, or ground-state electron configuration, of an atom is a list of the orbitals occupied by its electrons. We can predict this arrangement by following three rules.
RULE 1
The lowest-energy orbitals fill up first, $1s→2s→2p→3s→3p→4s→3d1s→2s→2p→3s→3p→4s→3d$, according to the following graphic, a statement called the Aufbau principle. Note that the 4s orbital lies between the 3p and 3d orbitals in energy.
RULE 2
Electrons act in some ways as if they were spinning around an axis, somewhat as the earth spins. This spin can have two orientations, denoted as up () and down (). Only two electrons can occupy an orbital, and they must have opposite spins, a statement called the Pauli exclusion principle.
RULE 3
If two or more empty orbitals of equal energy are available, one electron occupies each with spins parallel until all orbitals are half-full, a statement called Hund’s rule.
Some examples of how these rules apply are shown in Table 1.1. Hydrogen, for instance, has only one electron, which must occupy the lowest-energy orbital. Thus, hydrogen has a 1s ground-state configuration. Carbon has six electrons and the ground-state configuration 1s22s22px12py1, and so forth. Note that a superscript is used to represent the number of electrons in a particular orbital.
Table 1.1 Ground-State Electron Configurations of Some Elements
Element Atomic number Configuration
Hydrogen 1
Carbon 6
Phosphorus 15
Problem 1-1 What is the ground-state electron configuration of each of the following elements: (a)
Oxygen
(b) Nitrogen (c) Sulfur
Problem 1-2 How many electrons does each of the following biological trace elements have in its outermost electron shell? (a)
Magnesium
(b) Cobalt (c) Selenium | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.02%3A_Atomic_Structure_-_The_Nucleus.txt |
By the mid-1800s, the new science of chemistry was developing rapidly, especially in Europe, and chemists had begun to probe the forces holding compounds together. In 1858, the German chemist August Kekulé and the Scottish chemist Archibald Couper independently proposed that, in all organic compounds, carbon is tetravalent—it always forms four bonds when it joins other elements to form stable compounds. Furthermore, said Kekulé, carbon atoms can bond to one another to form extended chains of linked atoms. In 1865, Kekulé provided another major advance when he suggested that carbon chains can double back on themselves to form rings of atoms.
Although Kekulé and Couper were correct in describing the tetravalent nature of carbon, chemistry was still viewed in a two-dimensional way until 1874. In that year, the Dutch chemist Jacobus van’t Hoff and French chemist Joseph Le Bel added a third dimension to our ideas about organic compounds when they proposed that the four bonds of carbon are not oriented randomly but have specific spatial directions. Van’t Hoff went even further and suggested that the four atoms to which carbon is bonded sit at the corners of a regular tetrahedron, with carbon in the center.
A representation of a tetrahedral carbon atom is shown in Figure 1.7. Note the conventions used to show three-dimensionality: solid lines represent bonds in the plane of the page, the heavy wedged line represents a bond coming out of the page toward the viewer, and the dashed line represents a bond receding back behind the page, away from the viewer. Get used to them; these representations will be used throughout the text.
Why, though, do atoms bond together, and how can chemical bonds be described electronically? The why question is relatively easy to answer: atoms bond together because the compound that results is more stable and lower in energy than the separate atoms. Energy—usually as heat—is always released and flows out of the chemical system when a bond forms. Conversely, energy is added to the chemical system when a bond breaks. Making bonds always releases energy, and breaking bonds always absorbs energy. The how question is more difficult. To answer it, we need to know more about the electronic properties of atoms.
We know through observation that eight electrons (an electron octet) in an atom’s outermost shell, or valence shell, impart special stability to the noble-gas elements in group 8A of the periodic table: Ne (2 + 8); Ar (2 + 8 + 8); Kr (2 + 8 + 18 + 8). We also know that the chemistry of the main-group elements on the left and right sides of the periodic table is governed by their tendency to take on the electron configuration of the nearest noble gas. The alkali metals such as sodium in group 1A, for example, achieve a noble-gas configuration by losing the single s electron from their valence shell to form a cation, while the halogens such as chlorine in group 7A achieve a noble-gas configuration by gaining a p electron to fill their valence shell and form an anion. The resultant ions are held together in compounds like Na+ Cl by the electrical attraction of unlike charges that we call an ionic bond.
But how do elements closer to the middle of the periodic table form bonds? Look at methane, CH4, the main constituent of natural gas, for example. The bonding in methane is not ionic because it would take too much energy for carbon (1s2 2s2 2p2) either to gain or lose four electrons to achieve a noble-gas configuration. Instead, carbon bonds to other atoms, not by gaining or losing electrons, but by sharing them. Such a shared-electron bond, first proposed in 1916 by the American chemist G. N. Lewis, is called a covalent bond. The neutral collection of atoms held together by covalent bonds is called a molecule. Ionic compounds such as sodium chloride, however, are not called molecules.
A simple way of indicating the covalent bonds in molecules is to use what are called Lewis structures, or electron-dot structures, in which the valence-shell electrons of an atom are represented as dots. Thus, hydrogen has one dot representing its 1s electron, carbon has four dots (2s2 2p2), oxygen has six dots (2s2 2p4), and so on. A stable molecule results whenever a noble-gas configuration of eight dots (an octet) is achieved for all main-group atoms or two dots for hydrogen. Even simpler than Lewis structures is the use of Kekulé structures, or line-bond structures, in which the two-electron covalent bonds are indicated as lines drawn between atoms.
The number of covalent bonds an atom forms depends on how many additional valence electrons it needs to reach a noble-gas configuration. Hydrogen has one valence electron (1s) and needs only one more to reach the helium configuration (1s2), so it forms one bond. Carbon has four valence electrons (2s2 2p2) and needs four more to reach the neon configuration (2s2 2p6), so it forms four bonds. Nitrogen has five valence electrons (2s2 2p3), needs three more, and forms three bonds; oxygen has six valence electrons (2s2 2p4), needs two more, and forms two bonds; and the halogens have seven valence electrons, need one more, and form one bond.
Valence electrons that are not used for bonding remain as dots in structures and are called lone-pair electrons, or nonbonding electrons. The nitrogen atom in ammonia, NH3, for instance, shares six valence electrons in three covalent bonds and has its remaining two valence electrons as two dots in a nonbonding lone pair. As a time-saving shorthand, nonbonding electrons are often omitted when drawing line-bond structures, but you still have to keep them in mind since they’re often crucial in chemical reactions.
Worked Example 1.1
Predicting the Number of Bonds Formed by Atoms in Molecules
How many hydrogen atoms does phosphorus bond to in forming phosphine, PH??
Strategy
Identify the periodic group of phosphorus, and find from that how many electrons (bonds) are needed to make an octet.
Solution
Phosphorus is in group 5A of the periodic table and has five valence electrons. It thus needs to share three more electrons to make an octet and therefore bonds to three hydrogen atoms, giving PH3.
Worked Example 1.2
Drawing Electron-Dot and Line-Bond Structures
Draw both electron-dot and line-bond structures for chloromethane, CH3Cl.
Strategy
Remember that a covalent bond—that is, a pair of shared electrons—is represented as a line between atoms.
Solution
Hydrogen has one valence electron, carbon has four valence electrons, and chlorine has seven valence electrons. Thus, chloromethane is represented as
Problem 1-3
Draw a molecule of chloroform, CHCl3, using solid, wedged, and dashed lines to show its tetrahedral geometry.
Problem 1-4
Convert the following representation of ethane, C2H6, into a conventional drawing that uses solid, wedged, and dashed lines to indicate tetrahedral geometry around each carbon (black = C, gray = H).
Problem 1-5 What are likely formulas for the following substances? (a)
CCl?
(b) AlH? (c) CH?Cl2 (d) SiF (e) CH3NH?
Problem 1-6 Write line-bond structures for the following substances, showing all nonbonding electrons: (a)
CHCl3, chloroform
(b) H2S, hydrogen sulfide (c) CH3NH2, methylamine (d)
CH3Li, methyllithium
Problem 1-7
Why can’t an organic molecule have the formula C2H7? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.05%3A_Development_of_Chemical_Bonding_Theory.txt |
How does electron sharing lead to bonding between atoms? Two models have been developed to describe covalent bonding: valence bond theory and molecular orbital theory. Each model has its strengths and weaknesses, and chemists tend to use them interchangeably depending on the circumstances. Valence bond theory is the more easily visualized of the two, so most of the descriptions we’ll use in this book derive from that approach.
According to valence bond (VB) theory, a covalent bond forms when two atoms approach each other closely and a singly occupied orbital on one atom overlaps a singly occupied orbital on the other atom. The electrons are now paired in the overlapping orbitals and are attracted to the nuclei of both atoms, thus bonding the atoms together. In the H2 molecule, for instance, the H–H bond results from the overlap of two singly occupied hydrogen 1s orbitals.
The overlapping orbitals in the H2 molecule have the elongated egg shape we might get by pressing two spheres together. If a plane were to pass through the middle of the bond, the intersection of the plane and the overlapping orbitals would be a circle. In other words, the H–H bond is cylindrically symmetrical, as shown in Figure 1.8. Such bonds, which are formed by the head-on overlap of two atomic orbitals along a line drawn between the nuclei, are called sigma (σ) bonds.
During the bond-forming reaction 2H· ⟶ H2, 436 kJ/mol (104 kcal/mol) of energy is released. Because the product H2 molecule has 436 kJ/mol less energy than the starting 2 H· atoms, the product is more stable than the reactant and we say that the H–H bond has a bond strength of 436 kJ/mol. In other words, we would have to put 436 kJ/mol of energy into the H–H bond to break the H2 molecule apart into two H atoms (Figure 1.9). For convenience, we’ll generally give energies in both kilocalories (kcal) and the SI unit kilojoules (kJ): 1 kJ = 0.2390 kcal; 1 kcal = 4.184 kJ.
. Conversely, 436 kJ/mol is absorbed when the H–H bond breaks.
How close are the two nuclei in the H2 molecule? If they are too close, they will repel each other because both are positively charged. Yet if they’re too far apart, they won’t be able to share the bonding electrons. Thus, there is an optimum distance between nuclei that leads to maximum stability (Figure 1.10). Called the bond length, this distance is 74 pm in the H–H molecule. Every covalent bond has both a characteristic bond strength and bond length.
.
1.07: sp Hybrid Orbitals and the Structure of Methane
The bonding in the hydrogen molecule is fairly straightforward, but the situation is more complicated in organic molecules with tetravalent carbon atoms. Take methane, CH4, for instance. As we’ve seen, carbon has four valence electrons (2s2 2p2) and forms four bonds. Because carbon uses two kinds of orbitals for bonding, 2s and 2p, we might expect methane to have two kinds of C–H bonds. In fact, though, all four C–H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron, as shown previously in Figure 1.7. How can we explain this?
An answer was provided in 1931 by Linus Pauling, who showed mathematically how an s orbital and three p orbitals on an atom can combine, or hybridize, to form four equivalent atomic orbitals with tetrahedral orientation. Shown in Figure 1.11, these tetrahedrally oriented orbitals are called sp3 hybrid orbitals. Note that the superscript 3 in the name sp3 tells how many of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it.
, oriented toward the corners of a regular tetrahedron, are formed by the combination of an s orbital and three p orbitals (red/blue). The sp3 hybrids have two lobes and are unsymmetrical about the nucleus, giving them a directionality and allowing them to form strong bonds to other atoms.
The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer to why. When an s orbital hybridizes with three p orbitals, the resultant sp3 hybrid orbitals are unsymmetrical about the nucleus. One of the two lobes is larger than the other and can therefore overlap more effectively with an orbital from another atom to form a bond. As a result, sp3 hybrid orbitals form stronger bonds than do unhybridized s or p orbitals.
The asymmetry of sp3 orbitals arises because, as noted previously, the two lobes of a p orbital have different algebraic signs, + and –, in the wave function. Thus, when a p orbital hybridizes with an s orbital, the positive p lobe adds to the s orbital but the negative p lobe subtracts from the s orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction.
When each of the four identical sp3 hybrid orbitals of a carbon atom overlaps with the 1s orbital of a hydrogen atom, four identical C–H bonds are formed and methane results. Each C–H bond in methane has a strength of 439 kJ/mol (105 kcal/mol) and a length of 109 pm. Because the four bonds have a specific geometry, we also can define a property called the bond angle. The angle formed by each H–C–H is 109.5°, the so-called tetrahedral angle. Methane thus has the structure shown in Figure 1.12. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.06%3A_Describing_Chemical_Bonds_-_Valence_Bond_Theory.txt |
The same kind of orbital hybridization that accounts for the methane structure also accounts for the bonding together of carbon atoms into chains and rings to make possible many millions of organic compounds. Ethane, C2H6, is the simplest molecule containing a carbon–carbon bond.
We can picture the ethane molecule by imagining that the two carbon atoms bond to each other by head-on sigma (σ) overlap of an sp3 hybrid orbital from each (Figure 1.13). The remaining three sp3 hybrid orbitals on each carbon overlap with the 1s orbitals of three hydrogens to form the six C–H bonds. The C–H bonds in ethane are similar to those in methane, although a bit weaker: 421 kJ/mol (101 kcal/mol) for ethane versus 439 kJ/mol for methane. The C–C bond is 153 pm in length and has a strength of 377 kJ/mol (90 kcal/mol). All the bond angles of ethane are near, although not exactly at, the tetrahedral value of 109.5°.
. For clarity, the smaller lobes of the sp3 hybrid orbitals are not shown.
Problem 1-8
Draw a line-bond structure for propane, CH3CH2CH3. Predict the value of each bond angle, and indicate the overall shape of the molecule.
Problem 1-9
Convert the following molecular model of hexane, a component of gasoline, into a line-bond structure (black = C, gray = H).
1.09: sp Hybrid Orbitals and the Structure of Ethylene
The bonds we’ve seen in methane and ethane are called single bonds because they result from the sharing of one electron pair between bonded atoms. It was recognized nearly 150 years ago, however, that carbon atoms can also form double bonds by sharing two electron pairs between atoms or triple bonds by sharing three electron pairs. Ethylene, for instance, has the structure $H2C=CH2H2C=CH2$ and contains a carbon–carbon double bond, while acetylene has the structure $HC≡CHHC≡CH$ and contains a carbon–carbon triple bond.
How are multiple bonds described by valence bond theory? When we discussed sp3 hybrid orbitals in Section 1.6, we said that the four valence-shell atomic orbitals of carbon combine to form four equivalent sp3 hybrids. Imagine instead that the 2s orbital combines with only two of the three available 2p orbitals. Three sp2 hybrid orbitals result, and one 2p orbital remains unchanged. Like sp3 hybrids, sp2 hybrid orbitals are unsymmetrical about the nucleus and are strongly oriented in a specific direction so they can form strong bonds. The three sp2 orbitals lie in a plane at angles of 120° to one another, with the remaining p orbital perpendicular to the sp2 plane, as shown in Figure 1.14.
lie in a plane at angles of 120° to one another, and a single unhybridized p orbital (red/blue) is perpendicular to the sp2 plane.
When two carbons with sp2 hybridization approach each other, they form a strong σ bond by sp2sp2 head-on overlap. At the same time, the unhybridized p orbitals interact by sideways overlap to form what is called a pi (π) bond. The combination of an sp2sp2 σ bond and a 2p–2p π bond results in the sharing of four electrons and the formation of a carbon–carbon double bond (Figure 1.15). Note that the electrons in the σ bond occupy the region centered between nuclei, while the electrons in the π bond occupy regions above and below a line drawn between nuclei.
To complete the structure of ethylene, four hydrogen atoms form σ bonds with the remaining four sp2 orbitals. Ethylene thus has a planar structure, with H–C–H and H–C–C bond angles of approximately 120°. (The actual values are 117.4° for the H–C–H bond angle and 121.3° for the H–C–C bond angle.) Each C–H bond has a length of 108.7 pm and a strength of 464 kJ/mol (111 kcal/mol).
, and the other part results from π (sideways) overlap of unhybridized p orbitals (red/blue). The π bond has regions of electron density above and below a line drawn between nuclei.
As you might expect, the carbon–carbon double bond in ethylene is both shorter and stronger than the single bond in ethane because it has four electrons bonding the nuclei together rather than two. Ethylene has a $C=CC=C$ bond length of 134 pm and a strength of 728 kJ/mol (174 kcal/mol) versus a C–C length of 153 pm and a strength of 377 kJ/mol for ethane. The carbon–carbon double bond is less than twice as strong as a single bond because the sideways overlap in the π part of the double bond is not as great as the head-on overlap in the σ part.
Worked Example 1.3
Drawing Electron-Dot and Line-Bond Structures
Commonly used in biology as a tissue preservative, formaldehyde, CH2O, contains a carbon–oxygen double bond. Draw electron-dot and line-bond structures of formaldehyde, and indicate the hybridization of the carbon orbitals.
Strategy
We know that hydrogen forms one covalent bond, carbon forms four, and oxygen forms two. Trial and error, combined with intuition, is needed to fit the atoms together.
Solution
There is only one way that two hydrogens, one carbon, and one oxygen can combine:
Like the carbon atoms in ethylene, the carbon atom in formaldehyde is in a double bond and its orbitals are therefore sp2-hybridized.
Problem 1-10
Draw a line-bond structure for propene, CH3CH$\text{=}$CH2. Indicate the hybridization of the orbitals on each carbon, and predict the value of each bond angle.
Problem 1-11
Draw a line-bond structure for 1,3-butadiene, H2C$\text{=}$CH–CH$\text{=}$CH2. Indicate the hybridization of the orbitals on each carbon, and predict the value of each bond angle.
Problem 1-12
A molecular model of aspirin (acetylsalicylic acid) is shown. Identify the hybridization of the orbitals on each carbon atom in aspirin, and tell which atoms have lone pairs of electrons (black = C, red = O, gray = H). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.08%3A_sp_Hybrid_Orbitals_and_the_Structure_of_Ethane.txt |
In addition to forming single and double bonds by sharing two and four electrons, respectively, carbon can also form a triple bond by sharing six electrons. To account for the triple bond in a molecule such as acetylene, $H−C≡C−HFigure 1.16.$
are oriented 180° away from each other, perpendicular to the two remaining p orbitals (red/blue).
When two sp-hybridized carbon atoms approach each other, sp hybrid orbitals on each carbon overlap head-on to form a strong spsp σ bond. At the same time, the pz orbitals from each carbon form a pzpz π bond by sideways overlap, and the py orbitals overlap similarly to form a pypy π bond. The net effect is the sharing of six electrons and formation of a carbon–carbon triple bond. Each of the two remaining sp hybrid orbitals forms a σ bond with hydrogen to complete the acetylene molecule (Figure 1.17).
As suggested by sp hybridization, acetylene is a linear molecule with H–C–C bond angles of 180°. The C–H bonds have a length of 106 pm and a strength of 558 kJ/mol (133 kcal/mol). The C–C bond length in acetylene is 120 pm, and its strength is about 965 kJ/mol (231 kcal/mol), making it the shortest and strongest of any carbon–carbon bond. A comparison of sp, sp2, and sp3 hybridization is given in Table 1.2.
Table 1.2 Comparison of C−C and C−H Bonds in Methane, Ethane, Ethylene, and Acetylene
Molecule Bond Bond strength Bond length (pm)
(kJ/mol) (kcal/mol)
Methane, CH4 (sp3) C−H 439 105 109
Ethane, CH3CH3 (sp3) C−C (sp3) 377 90 153
(sp3) C−H 421 101 109
Ethylene, H2C=CH2 (sp2) $HC≡CHHC≡CH$ (sp) $\text{≡}$CH. Indicate the hybridization of the orbitals on each carbon, and predict a value for each bond angle.
1.11: Hybridization of Nitrogen Oxygen Phosphorus and Sulfur
1.10 • Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur
The valence-bond concept of orbital hybridization described in the previous four sections is not limited to carbon. Covalent bonds formed by other elements can also be described using hybrid orbitals. Look, for instance, at the nitrogen atom in methylamine (CH3NH2), an organic derivative of ammonia (NH3) and the substance responsible for the odor of rotting fish.
The experimentally measured H–N–H bond angle in methylamine is 107.1°, and the C–N–H bond angle is 110.3°, both of which are close to the 109.5° tetrahedral angle found in methane. We therefore assume that nitrogen forms four sp3-hybridized orbitals, just as carbon does. One of the four sp3 orbitals is occupied by two nonbonding electrons (a lone pair), and the other three hybrid orbitals have one electron each. Overlap of these three half-filled nitrogen orbitals with half-filled orbitals from other atoms (C or H) gives methylamine. Note that the unshared lone pair of electrons in the fourth sp3 hybrid orbital of nitrogen occupies as much space as an N–H bond does and is very important to the chemistry of methylamine and other nitrogen-containing organic molecules.
Like the carbon atom in methane and the nitrogen atom in methylamine, the oxygen atom in methanol (methyl alcohol) and many other organic molecules can be described as sp3-hybridized. The C–O–H bond angle in methanol is 108.5°, very close to the 109.5° tetrahedral angle. Two of the four sp3 hybrid orbitals on oxygen are occupied by nonbonding electron lone pairs, and two are used to form bonds.
In the periodic table, phosphorus and sulfur are the third-row analogs of nitrogen and oxygen, and the bonding in both can be described using hybrid orbitals. Because of their positions in the third row, however, both phosphorus and sulfur can expand their outer-shell octets and form more than the typical number of covalent bonds. Phosphorus, for instance, often forms five covalent bonds, and sulfur often forms four.
Phosphorus is most commonly encountered in biological molecules in compounds called organophosphates, which contain a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon. Methyl phosphate, CH3OPO32, is the simplest example. The O–P–O bond angle in such compounds is typically in the range 110° to 112°, implying sp3 hybridization for phosphorus orbitals.
Sulfur is most commonly encountered in biological molecules either in compounds called thiols, which have a sulfur atom bonded to one hydrogen and one carbon, C–S–H or in sulfides, which have a sulfur atom bonded to two carbons, C–S–C. Produced by some bacteria, methanethiol (CH3SH) is the simplest example of a thiol, and dimethyl sulfide, H3C–S–CH3, is the simplest example of a sulfide. Both can be described by approximate sp3 hybridization around sulfur, although both have significant deviation from the 109.5° tetrahedral angle.
Problem 1-14 Identify all nonbonding lone pairs of electrons in the following molecules, and tell what geometry you expect for each of the indicated atoms. (a)
The oxygen atom in dimethyl ether, CH3–O–CH3
(b)
The nitrogen atom in trimethylamine,
(c) The phosphorus atom in phosphine, PH3 (d)
The sulfur atom in the amino acid methionine, | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.10%3A_sp_Hybrid_Orbitals_and_the_Structure_of_Acetylene.txt |
We said in Section 1.5 that chemists use two models for describing covalent bonds: valence bond theory and molecular orbital theory. Having now seen the valence bond approach, which uses hybrid atomic orbitals to account for geometry and assumes the overlap of atomic orbitals to account for electron sharing, let’s look briefly at the molecular orbital approach to bonding. We’ll return to this topic in Chapters 14, 15, and 30 for a more in-depth discussion.
Molecular orbital (MO) theory describes covalent bond formation as arising from a mathematical combination of atomic orbitals (wave functions) on different atoms to form molecular orbitals, so called because they belong to the entire molecule rather than to an individual atom. Just as an atomic orbital, whether unhybridized or hybridized, describes a region of space around an atom where an electron is likely to be found, so a molecular orbital describes a region of space in a molecule where electrons are most likely to be found.
Like an atomic orbital, a molecular orbital has a specific size, shape, and energy. In the H2 molecule, for example, two singly occupied 1s atomic orbitals combine to form two molecular orbitals. There are two ways for the orbital combination to occur—an additive way and a subtractive way. The additive combination leads to formation of a molecular orbital that is lower in energy and roughly egg-shaped, while the subtractive combination leads to a molecular orbital that is higher in energy and has a node between nuclei (Figure 1.18). Note that the additive combination is a single, egg-shaped, molecular orbital; it is not the same as the two overlapping 1s atomic orbitals of the valence bond description. Similarly, the subtractive combination is a single molecular orbital with the shape of an elongated dumbbell.
is filled, and the higher-energy, antibonding MO is unfilled.
The additive combination is lower in energy than the two hydrogen 1s atomic orbitals and is called a bonding MO because electrons in this MO spend most of their time in the region between the two nuclei, thereby bonding the atoms together. The subtractive combination is higher in energy than the two hydrogen 1s orbitals and is called an antibonding MO because any electrons it contains can’t occupy the central region between the nuclei, where there is a node, and thus can’t contribute to bonding. The two nuclei therefore repel each other.
Just as bonding and antibonding σ molecular orbitals result from the head-on combination of two s atomic orbitals in H2, so bonding and antibonding π molecular orbitals result from the sideways combination of two p atomic orbitals in ethylene. As shown in Figure 1.19, the lower-energy, π bonding MO has no node between nuclei and results from the combination of p orbital lobes with the same algebraic sign. The higher-energy, π antibonding MO has a node between nuclei and results from the combination of lobes with opposite algebraic signs. Only the bonding MO is occupied; the higher-energy, antibonding MO is vacant. We’ll see in Chapters 14, 15, and 30 that molecular orbital theory is particularly useful for describing π bonds in compounds that have more than one double bond.
bonding MO results from an additive combination of p orbital lobes with the same algebraic sign and is filled. The higher-energy, π antibonding MO results from a subtractive combination of p orbital lobes with opposite algebraic signs and is unfilled. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.12%3A_Describing_Chemical_Bonds_-_Molecular_Orbital_Theory.txt |
Let’s cover just one more point before ending this introductory chapter. In the structures we’ve been drawing until now, a line between atoms has represented the two electrons in a covalent bond. Drawing every bond and every atom is tedious, however, so chemists have devised several shorthand ways for writing structures. In condensed structures, carbon–hydrogen and carbon–carbon single bonds aren’t shown; instead, they’re understood. If a carbon has three hydrogens bonded to it, we write CH3; if a carbon has two hydrogens bonded to it, we write CH2; and so on. The compound called 2-methylbutane, for example, is written as follows:
Note that the horizontal bonds between carbons aren’t shown in condensed structures—the CH3, CH2, and CH units are simply placed next to each other—but vertical carbon–carbon bonds like that of the first of the condensed structures drawn above is shown for clarity. Notice also in the second of the condensed structures that the two CH3 units attached to the CH carbon are grouped together as (CH3)2.
Even simpler than condensed structures are skeletal structures such as those shown in Table 1.3. The rules for drawing skeletal structures are straightforward.
RULE 1
Carbon atoms aren’t usually shown. Instead, a carbon atom is assumed to be at each intersection of two lines (bonds) and at the end of each line. Occasionally, a carbon atom might be indicated for emphasis or clarity.
RULE 2
Hydrogen atoms bonded to carbon aren’t shown. Because carbon always has a valence of 4, we mentally supply the correct number of hydrogen atoms for each carbon.
RULE 3
Atoms other than carbon and hydrogen are shown.
One further comment: Although such groupings as –CH3, –OH, and –NH2 are usually written with the C, O, or N atom first and the H atom second, the order of writing is sometimes inverted to H3C–, HO–, and H2N– if needed to make the bonding connections clearer. Larger units such as –CH2CH3 are not inverted, though; we don’t write H3CH2C– because it would be confusing. There are, however, no well-defined rules that cover all cases; it’s largely a matter of preference.
Table 1.3 Line-bond and Skeletal Structures for Some Compounds
Compound Line-bond structure Skeletal structure
Isoprene, C5H8
Methylcyclohexane, C7H14
Phenol, C6H6O
Worked Example 1.4
Interpreting a Line-Bond Structure
Carvone, a substance responsible for the odor of spearmint, has the following structure. Tell how many hydrogens are bonded to each carbon, and give the molecular formula of carvone.
Strategy
The end of a line represents a carbon atom with 3 hydrogens, CH3; a two-way intersection is a carbon atom with 2 hydrogens, CH2; a three-way intersection is a carbon atom with 1 hydrogen, CH; and a four-way intersection is a carbon atom with no attached hydrogens.
Solution
(b)
Problem 1-16 Propose skeletal structures for compounds that satisfy the following molecular formulas: There is more than one possibility in each case. (a) C5H12 (b)
C2H7N
(c) C3H6O (d) C4H9Cl
Problem 1-17
The following molecular model is a representation of para-aminobenzoic acid (PABA), the active ingredient in many sunscreens. Indicate the positions of the multiple bonds, and draw a skeletal structure (black = C, red = O, blue = N, gray = H). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.13%3A_Drawing_Chemical_Structures.txt |
1 • Chemistry Matters 1 • Chemistry Matters
Contrary to what you may hear in supermarkets or on television, all foods are organic—that is, complex mixtures of organic molecules. Even so, when applied to food, the word organic has come to mean an absence of synthetic chemicals, typically pesticides, antibiotics, and preservatives. How concerned should we be about traces of pesticides in the food we eat? Or toxins in the water we drink? Or pollutants in the air we breathe?
Life is not risk-free—we all take many risks each day without even thinking about it. We decide to ride a bike rather than drive, even though there is a ten times greater likelihood per mile of dying in a bicycling accident than in a car. We decide to walk down stairs rather than take an elevator, even though 32,000 people die from falls each year in the United States. Some of us decide to smoke cigarettes, even though it increases our chance of getting cancer by 50%. But what about risks from chemicals like pesticides?
One thing is certain: without pesticides, whether they target weeds (herbicides), insects (insecticides), or molds and fungi (fungicides), crop production would drop significantly, food prices would increase, and famines would occur in less developed parts of the world. Take the herbicide atrazine, for instance. In the United States alone, approximately 100 million pounds of atrazine are used each year to kill weeds in corn, sorghum, and sugarcane fields, greatly improving the yields of these crops. Nevertheless, the use of atrazine continues to be a concern because traces persist in the environment. Indeed, heavy atrazine exposure can pose health risks to humans and some animals. Because of these risks, the United States Environmental Protection Agency (EPA) has decided not to ban its use because doing so would result in lower crop yields and increased food costs, and because there is no suitable alternative herbicide available.
How can the potential hazards from a chemical like atrazine be determined? Risk evaluation of chemicals is carried out by exposing test animals, usually mice or rats, to the chemical and then monitoring the animals for signs of harm. To limit the expense and time needed, the amounts administered are typically hundreds or thousands of times greater than those a person might normally encounter. The results obtained in animal tests are then distilled into a single number called an LD50, the amount of substance per kilogram of body weight that is a lethal dose for 50% of the test animals. For atrazine, the LD50 value is between 1 and 4 g/kg depending on the animal species. Aspirin, for comparison, has an LD50 of 1.1 g/kg, and ethanol (ethyl alcohol) has an LD50 of 10.6 g/kg.
Table 1.4 lists the LD50 for some other familiar substances. The lower the value, the more toxic the substance. Note, though, that LD50 values only pertain to the effects of heavy exposure for a relatively short time. They say nothing about the risks of long-term exposure, such as whether the substance can cause cancer or interfere with development in the unborn.
Table 1.4 Some LD50 Values
Substance LD50 (g/kg) Substance LD50 (g/kg)
Strychnine 0.005 Chloroform 1.2
Arsenic trioxide 0.015 Iron(II) sulfate 1.5
DDT 0.115 Ethyl alcohol 10.6
Aspirin 1.1 Sodium cyclamate 17
So, should we still use atrazine? All decisions involve tradeoffs, and the answer is rarely obvious. Does the benefit of increased food production outweigh possible health risks of a pesticide? Do the beneficial effects of a new drug outweigh a potentially dangerous side effect in a small number of users? Different people will have different opinions, but an honest evaluation of facts is surely the best way to start. As of June 2022, atrazine was still approved for continued use in the United States because the EPA believes that the benefits of increased food production outweigh possible health risks. At the same time, atrazine is little used, though not banned, in the European Union. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.14%3A_Chemistry_MattersOrganic_Foods-_Risk_versus_Benefit.txt |
1 • Key Terms 1 • Key Terms
• antibonding MO
• atomic number (Z)
• Aufbau principle
• bond angle
• bond length
• bond strength
• bonding MO
• condensed structure
• covalent bond
• electron shell
• electron-dot structure
• ground-state electron configuration
• Hund’s rule
• ionic bond
• isotope
• Kekulé structure
• Lewis structure
• line-bond structure
• lone-pair electrons
• mass number (A)
• molecular orbital (MO) theory
• molecule
• node
• nonbonding electron
• orbital
• organic chemistry
• Pauli exclusion principle
• pi (π) bond
• sigma (σ) bond
• skeletal structure
• sp hybrid orbital
• sp2 hybrid orbital
• sp3 hybrid orbital
• valence bond (VB) theory
• valence shell
1.16: Summary
1 • Summary 1 • Summary
The purpose of this chapter has been to get you up to speed—to review some ideas about atoms, bonds, and molecular geometry. As we’ve seen, organic chemistry is the study of carbon compounds. Although a division into organic and inorganic chemistry occurred historically, there is no scientific reason for the division.
An atom consists of a positively charged nucleus surrounded by one or more negatively charged electrons. The electronic structure of an atom can be described by a quantum mechanical wave equation, in which electrons are considered to occupy orbitals around the nucleus. Different orbitals have different energy levels and different shapes. For example, s orbitals are spherical and p orbitals are dumbbell-shaped. The ground-state electron configuration of an atom can be found by assigning electrons to the proper orbitals, beginning with the lowest-energy ones.
A covalent bond is formed when an electron pair is shared between atoms. According to valence bond (VB) theory, electron sharing occurs by the overlap of two atomic orbitals. According to molecular orbital (MO) theory, bonds result from the mathematical combination of atomic orbitals to give molecular orbitals, which belong to the entire molecule. Bonds that have a circular cross-section and are formed by head-on interaction are called sigma (σ) bonds; bonds formed by sideways interaction of p orbitals are called pi (π) bonds.
In the valence bond description, carbon uses hybrid orbitals to form bonds in organic molecules. When forming only single bonds with tetrahedral geometry, carbon uses four equivalent sp3 hybrid orbitals. When forming a double bond with planar geometry, carbon uses three equivalent sp2 hybrid orbitals and one unhybridized p orbital. When forming a triple bond with linear geometry, carbon uses two equivalent sp hybrid orbitals and two unhybridized p orbitals. Other atoms such as nitrogen, phosphorus, oxygen, and sulfur also use hybrid orbitals to form strong, oriented bonds.
Organic molecules are usually drawn using either condensed structures or skeletal structures. In condensed structures, carbon–carbon and carbon–hydrogen bonds aren’t shown. In skeletal structures, only the bonds and not the atoms are shown. A carbon atom is assumed to be at the ends and at the junctions of lines (bonds), and the correct number of hydrogens is supplied mentally.
Why You Should Work Problems
There’s no surer way to learn organic chemistry than by working problems. Although careful reading and rereading of this text are important, reading alone isn’t enough. You must also be able to use the information you’ve read and be able to apply your knowledge in new situations. Working problems gives you practice at doing this.
Each chapter in this book provides many problems of different sorts. The in-chapter problems are placed for immediate reinforcement of ideas just learned, while end-of-chapter problems provide additional practice and come in several forms. They often begin with a short section called “Visualizing Chemistry,” which helps you see the microscopic world of molecules and provides practice for working in three dimensions. After the visualizations are many further problems, which are organized by topic. Early problems are primarily of the drill type, providing an opportunity for you to practice your command of the fundamentals. Later problems tend to be more thought-provoking, and some are real challenges.
As you study organic chemistry, take the time to work the problems. Do the ones you can, and ask for help on the ones you can’t. If you’re stumped by a particular problem, check the accompanying Study Guide and Student Solutions Manual for an explanation that should help clarify the difficulty. Working problems takes effort, but the payoff in knowledge and understanding is immense. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.15%3A_Key_Terms.txt |
1 • Additional Problems 1 • Additional Problems
Visualizing Chemistry
Problem 1-18
Convert each of the following molecular models into a skeletal structure, and give the formula of each. Only the connections between atoms are shown; multiple bonds are not indicated (black = C, red = O, blue = N, gray = H).
(a)
(b)
Problem 1-19 The following model is a representation of citric acid, the key substance in the so-called citric acid cycle, by which food molecules are metabolized in the body. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure by indicating the positions of multiple bonds and lone-pair electrons (black = C, red = O, gray = H). Problem 1-20 The following model is a representation of acetaminophen, a pain reliever sold in drugstores under a variety of names, including Tylenol. Identify the hybridization of each carbon atom in acetaminophen, and tell which atoms have lone pairs of electrons (black = C, red = O, blue = N, gray = H). Problem 1-21 The following model is a representation of aspartame, C14H18N2O5, known commercially under many names, including NutraSweet. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure for aspartame, and indicate the positions of multiple bonds (black = C, red = O, blue = N, gray = H). Electron Configurations Problem 1-22 How many valence electrons does each of the following dietary trace elements have? (a) Zinc (b)
Iodine
(c) Silicon (d) Iron
Problem 1-23 Give the ground-state electron configuration for each of the following elements: (a)
Potassium
(b) Arsenic (c) Aluminum (d) Germanium
Electron-Dot and Line-Bond Structures
Problem 1-24 What are likely formulas for the following molecules? (a)
NH?OH
(b) AlCl? (c) CF2Cl? (d) CH?O
Problem 1-25 Why can’t molecules with the following formulas exist? (a)
CH5
(b) C2H6N (c) C3H5Br2
Problem 1-26
Draw an electron-dot structure for acetonitrile, C2H3N, which contains a carbon–nitrogen triple bond. How many electrons does the nitrogen atom have in its outer shell? How many are bonding, and how many are nonbonding?
Problem 1-27
Draw a line-bond structure for vinyl chloride, C2H3Cl, the starting material from which PVC poly(vinyl chloride) plastic is made.
Problem 1-28 Fill in any nonbonding valence electrons that are missing from the following structures: (a) (b) (c) Problem 1-29 Convert the following line-bond structures into molecular formulas: (a) (b) (c) (d) Problem 1-30 Convert the following molecular formulas into line-bond structures that are consistent with valence rules: (a) C3H8 (b)
CH5N
(c) C2H6O (2 possibilities) (d) C3H7Br (2 possibilities) (e) C2H4O (3 possibilities) (f) C3H9N (4 possibilities)
Problem 1-31
Draw a three-dimensional representation of the oxygen-bearing carbon atom in ethanol, CH3CH2OH, using the standard convention of solid, wedged, and dashed lines.
Problem 1-32
Oxaloacetic acid, an important intermediate in food metabolism, has the formula C4H4O5 and contains three C$\text{=}$O bonds and two O–H bonds. Propose two possible structures.
Problem 1-33 Draw structures for the following molecules, showing lone pairs: (a)
Acrylonitrile, C3H3N, which contains a carbon–carbon double bond and a carbon–nitrogen triple bond
(b) Ethyl methyl ether, C3H8O, which contains an oxygen atom bonded to two carbons (c) Butane, C4H10, which contains a chain of four carbon atoms (d) Cyclohexene, C6H10, which contains a ring of six carbon atoms and one carbon–carbon double bond
Hybridization
Problem 1-34
What is the hybridization of each carbon atom in acetonitrile (Problem 1-26)?
Problem 1-35 What kind of hybridization do you expect for each carbon atom in the following molecules? (a) (b) (c) (d) Problem 1-36 What is the shape of benzene, and what hybridization do you expect for each carbon? Problem 1-37 What bond angle do you expect for each of the indicated atoms, and what kind of hybridization do you expect for the central atom in each molecule? (a) (b) (c) Problem 1-38 Propose structures for molecules that meet the following descriptions: (a) Contains two sp2-hybridized carbons and two sp3-hybridized carbons (b)
Contains only four carbons, all of which are sp2-hybridized
(c) Contains two sp-hybridized carbons and two sp2-hybridized carbons
Problem 1-39 What kind of hybridization do you expect for each carbon atom in the following molecules: (a) (b) Problem 1-40 Pyridoxal phosphate, a close relative of vitamin B6, is involved in a large number of metabolic reactions. What is the hybridization and the bond angle for each nonterminal atom? Skeletal Structures Problem 1-41 Convert the following structures into skeletal drawings: (a) (b) (c) (d) Problem 1-42 How many hydrogens are bonded to each carbon atom in the following substances, and what is the molecular formula of each? (a) (b) (c) Problem 1-43 Quetiapine, marketed as Seroquel, is a heavily prescribed antipsychotic drug used in the treatment of schizophrenia and bipolar disorder. Convert the following representation into a skeletal structure, and give the molecular formula of quetiapine. Problem 1-44 How many hydrogens are bonded to each carbon atom in (a) the antiinfluenza agent oseltamivir, marketed as Tamiflu, and (b) the platelet aggregation inhibitor clopidogrel, marketed as Plavix? Give the molecular formula of each. (a) (b) General Problems Problem 1-45 Why do you suppose no one has ever been able to make cyclopentyne as a stable molecule? Problem 1-46 Allene, H2C$\text{=}$C$\text{=}$CH2, has two adjacent double bonds. Draw a picture showing the orbitals involved in the σ and π bonds of allene. Is the central carbon atom sp2- or sp-hybridized? What about the hybridization of the terminal carbons? What shape do you predict for allene? Problem 1-47 Allene (see Problem 1-46) is structurally related to carbon dioxide, CO2. Draw a picture showing the orbitals involved in the σ and π bonds of CO2, and identify the likely hybridization of carbon. Problem 1-48 Complete the electron-dot structure of caffeine, showing all lone-pair electrons, and identify the hybridization of the indicated atoms. Problem 1-49 Most stable organic species have tetravalent carbon atoms, but species with trivalent carbon atoms also exist. Carbocations are one such class of compounds. (a) How many valence electrons does the positively charged carbon atom have? (b)
What hybridization do you expect this carbon atom to have?
(c) What geometry is the carbocation likely to have?
Problem 1-50 A carbanion is a species that contains a negatively charged, trivalent carbon. (a)
What is the electronic relationship between a carbanion and a trivalent nitrogen compound such as NH3?
(b) How many valence electrons does the negatively charged carbon atom have? (c) What hybridization do you expect this carbon atom to have? (d) What geometry is the carbanion likely to have?
Problem 1-51
Divalent carbon species called carbenes are capable of fleeting existence. For example, methylene, :CH2, is the simplest carbene. The two unshared electrons in methylene can be either paired in a single orbital or unpaired in different orbitals. Predict the type of hybridization you expect carbon to adopt in singlet (spin-paired) methylene and triplet (spin-unpaired) methylene. Draw a picture of each, and identify the valence orbitals on carbon.
Problem 1-52
Two different substances have the formula C4H10. Draw both, and tell how they differ.
Problem 1-53
Two different substances have the formula C3H6. Draw both, and tell how they differ.
Problem 1-54
Two different substances have the formula C2H6O. Draw both, and tell how they differ.
Problem 1-55
Three different substances contain a carbon–carbon double bond and have the formula C4H8. Draw them, and tell how they differ.
Problem 1-56 Among the most common over-the-counter drugs you might find in a medicine cabinet are mild pain relievers such ibuprofen (Advil, Motrin), naproxen (Aleve), and acetaminophen (Tylenol). (a)
How many sp3-hybridized carbons does each molecule have?
(b) How many sp2-hybridized carbons does each molecule have? (c) What similarities can you see in their structures? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding/1.17%3A_Additional_Problems.txt |
Chapter Objectives
This chapter provides a review of the more advanced material covered in a standard introductory chemistry course through a discussion of the following topics:
• the use of electronegativity to determine bond polarity, and the application of this knowledge to determine whether a given molecule possesses a dipole moment.
• the drawing and interpretation of organic chemical structures.
• the concept and determination of formal charge.
• resonance and drawing of resonance forms
• the Brønsted-Lowry and Lewis definitions of acids and bases, acidity constants and acid-base reactions.
• intermolecular forces
• 2.1: Why This Chapter?
• 2.2: Polar Covalent Bonds - Electronegativity
Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity, defined as the relative ability of an atom to attract electrons to itself in a chemical compound.
• 2.3: Polar Covalent Bonds - Dipole Moments
Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment.
• 2.4: Formal Charges
A formal charge is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.
• 2.5: Resonance
Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structure are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds.
• 2.6: Rules for Resonance Forms
The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite formal charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on one terminal oxygen can be delocalized by resonance through the other terminal oxygen.
• 2.7: Drawing Resonance Forms
Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer.
• 2.8: Acids and Bases - The Brønsted-Lowry Definition
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.
• 2.9: Acid and Base Strength
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.
• 2.10: Predicting Acid-Base Reactions from pKa Values
pKa values can be used to predict the equilibrium of an acid-base reaction. The equilibrium will favor the side with the weaker acid.
• 2.11: Organic Acids and Organic Bases
In the absence of pKa values, the relative strength of an organic acid can be predicted based on the stability of the conjugate base that it forms. The acid that forms the more stable conjugate base will be the stronger acid. The common factors that affect the conjugate base's stability are 1) the size and electronegativity of the the atom that has lost the proton, 2) resonance effects, 3) inductive effects, and 4) solvation effects.
• 2.12: Acids and Bases - The Lewis Definition
A broader definition is provided by the Lewis theory of acids and bases, in which a Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. This definition covers Brønsted-Lowry proton transfer reactions, but also includes reactions in which no proton transfer is involved.
• 2.13: Noncovalent Interactions Between Molecules
In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. The most common intermolecular forces in organic chemistry are from strongest to weakest are hydrogen bonds, dipole-dipole interactions, and London Dispersion (van der Waals) forces.
• 2.14: Chemistry Matters—Alkaloids- From Cocaine to Dental Anesthetics
• 2.15: Key Terms
• 2.16: Summary
• 2.17: Additional Problems
02: Polar Covalent Bonds Acids and Bases
Chapter Contents
2.1 Polar Covalent Bonds and Electronegativity
2.2 Polar Covalent Bonds and Dipole Moments 2.3 Formal Charges 2.4 Resonance 2.5 Rules for Resonance Forms 2.6 Drawing Resonance Forms 2.7 Acids and Bases: The Brønsted–Lowry Definition 2.8 Acid and Base Strength 2.9 Predicting Acid–Base Reactions from pKa Values 2.10 Organic Acids and Organic Bases 2.11 Acids and Bases: The Lewis Definition 2.12 Noncovalent Interactions between Molecules
Understanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we’ll look at some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation to understand the specific reactions discussed in subsequent chapters. Topics such as bond polarity, the acid–base behavior of molecules, and hydrogen-bonding are a particularly important part of that foundation.
We saw in the previous chapter how covalent bonds between atoms are described, and we looked at the valence bond model, which uses hybrid orbitals to account for the observed shapes of organic molecules. Before going on to a systematic study of organic chemistry, however, we still need to review a few fundamental topics. In particular, we need to look more closely at how electrons are distributed in covalent bonds and at some of the consequences that arise when the electrons in a bond are not shared equally between atoms. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.01%3A_Why_This_Chapter.txt |
Up to this point, we’ve treated chemical bonds as either ionic or covalent. The bond in sodium chloride, for instance, is ionic. Sodium transfers an electron to chlorine to produce Na+ and Cl ions, which are held together in the solid by electrostatic attractions between unlike charges. The C–C bond in ethane, however, is covalent. The two bonding electrons are shared equally by the two equivalent carbon atoms, resulting in a symmetrical electron distribution in the bond. Most bonds, however, are neither fully ionic nor fully covalent but are somewhere between the two extremes. Such bonds are called polar covalent bonds, meaning that the bonding electrons are attracted more strongly by one atom than the other so that the electron distribution between atoms is not symmetrical (Figure 2.2).
The symbol δ (lowercase Greek letter delta) means partial charge, either partial positive (δ+) for the electron-poor atom or partial negative (δ–) for the electron-rich atom.
Bond polarity is due to differences in electronegativity (EN), the intrinsic ability of an atom to attract the shared electrons in a covalent bond. As shown in Figure 2.3, electronegativities are based on an arbitrary scale, with fluorine the most electronegative (EN = 4.0) and cesium the least (EN = 0.7). Metals on the left side of the periodic table attract electrons weakly and have lower electronegativities, while oxygen, nitrogen, and halogens on the right side of the periodic table attract electrons strongly and have higher electronegativities. Carbon, the most important element in organic compounds, has an intermediate electronegativity value of 2.5.
Electronegativity generally increases from left to right across the periodic table and decreases from top to bottom. The values are on an arbitrary scale, with F = 4.0 and Cs = 0.7. Elements in red are the most electronegative, those in yellow are medium, and those in green are the least electronegative.
As a rough guide, bonds between atoms whose electronegativities differ by less than 0.5 are nonpolar covalent, bonds between atoms whose electronegativities differ by 0.5 to 2 are polar covalent, and bonds between atoms whose electronegativities differ by more than 2 are largely ionic. Carbon–hydrogen bonds, for example, are relatively nonpolar because carbon (EN = 2.5) and hydrogen (EN = 2.1) have similar electronegativities. Bonds between carbon and more electronegative elements such as oxygen (EN = 3.5) and nitrogen (EN = 3.0), by contrast, are polarized so that the bonding electrons are drawn away from carbon toward the electronegative atom. This leaves carbon with a partial positive charge, δ–, and the electronegative atom with a partial negative charge, δ–. An example is the C–O bond in methanol, CH3OH (Figure 2.4a). Bonds between carbon and less electronegative elements are polarized so that carbon bears a partial negative charge and the other atom bears a partial positive charge. An example is the C–Li bond in methyllithium, CH3Li (Figure 2.4b).
: (a) Methanol, CH3OH, has a polar covalent C–O bond, and (b) methyllithium, CH3Li, has a polar covalent C–Li bond. The computer-generated representations, called electrostatic potential maps, use color to show calculated charge distributions, ranging from red (electron-rich; δ−) to blue (electron-poor; δ+).
Note in the representations of methanol and methyllithium in Figure 2.4 that a crossed arrow is used to indicate the direction of bond polarity. By convention, electrons are displaced in the direction of the arrow. The tail of the arrow (which looks like a plus sign) is electron-poor (δ+), and the head of the arrow is electron-rich (δ–).
Note also in Figure 2.4 that calculated charge distributions in molecules can be displayed visually with what are called electrostatic potential maps, which use color to indicate electron-rich (red; δ–) and electron-poor (blue; δ+) regions. In methanol, oxygen carries a partial negative charge and is colored red, while the carbon and hydrogen atoms carry partial positive charges and are colored blue-green. In methyllithium, lithium carries a partial positive charge (blue), while carbon and the hydrogen atoms carry partial negative charges (red). Electrostatic potential maps are useful because they show at a glance the electron-rich and electron-poor atoms in molecules. We’ll make frequent use of these maps throughout the text and will see many examples of how electronic structure correlates with chemical reactivity.
When speaking of an atom’s ability to polarize a bond, we often use the term inductive effect. An inductive effect is simply the shifting of electrons in a σ bond in response to the electronegativity of nearby atoms. Metals, such as lithium and magnesium, inductively donate electrons, whereas reactive nonmetals, such as oxygen and nitrogen, inductively withdraw electrons. Inductive effects play a major role in understanding chemical reactivity, and we’ll use them many times throughout this text to explain a variety of chemical observations.
Problem 2-1 Which element in each of the following pairs is more electronegative? (a)
Li or H
(b) B or Br (c) Cl or I (d) C or H
Problem 2-2 Use the δ+/δ– convention to indicate the direction of expected polarity for each of the bonds indicated. (a)
H3CCl
(b) H3CNH2 (c) H2NH (d) H3CSH (e) H3CMgBr (f) H3CF
Problem 2-3
Use the electronegativity values shown in Figure 2.3 to rank the following bonds from least polar to most polar: H3CLi, H3CK, H3CF, H3CMgBr, H3COH
Problem 2-4
Look at the following electrostatic potential map of methylamine, a substance responsible for the odor of rotting fish, and tell the direction of polarization of the C–N bond: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.02%3A_Polar_Covalent_Bonds_-_Electronegativity.txt |
Just as individual bonds are often polar, molecules as a whole are often polar as well. Molecular polarity results from the vector summation of all individual bond polarities and lone-pair contributions in the molecule. As a practical matter, strongly polar substances are often soluble in polar solvents like water, whereas less polar substances are insoluble in water.
Net polarity is measured by a quantity called the dipole moment and can be thought of in the following way: assume that there is a center of mass of all positive charges (nuclei) in a molecule and a center of mass of all negative charges (electrons). If these two centers don’t coincide, then the molecule has a net polarity.
The dipole moment, μ (lowercase Greek letter mu), is defined as the magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges, μ = Q × r. Dipole moments are expressed in debyes (D), where 1 D = 3.336 × 10–30 coulomb meters (C · m) in SI units. For example, the unit charge on an electron is 1.60 × 10–19 C. Thus, if one positive charge and one negative charge are separated by 100 pm (a bit less than the length of a typical covalent bond), the dipole moment is 1.60 × 10–29 C · m, or 4.80 D.
$μ=Q×r μ=(1.60× 10 −19 C)(100× 10 −12 m) 1 D 3.336 × 10 −30 C ⋅ m =4.80 D μ=Q×r μ=(1.60× 10 −19 C)(100× 10 −12 m) 1 D 3.336 × 10 −30 C ⋅ m =4.80 D$
Dipole moments for some common substances are given in Table 2.1. Of the compounds shown in the table, sodium chloride has the largest dipole moment (9.00 D) because it is ionic. Even small molecules like water (μ = 1.85 D), methanol (CH3OH; μ = 1.70 D), and ammonia (μ = 1.47 D), have substantial dipole moments, however, both because they contain strongly electronegative atoms (oxygen and nitrogen) and because all three molecules have lone-pair electrons. The lone-pair electrons on oxygen and nitrogen stick out into space away from the positively charged nuclei, giving rise to a considerable charge separation and making a large contribution to the dipole moment.
Table 2.1 Dipole Moments of Some Compounds
Compound Dipole moment (D) Compound Dipole moment (D)
NaCl 9.00 NH3 1.47
CH2O 2.33 CH3NH2 1.31
CH3Cl 1.87 CO2 0
H2O 1.85 CH4 0
CH3OH 1.70 CH3CH3 0
CH3CO2H 1.70 0
CH3SH 1.52
In contrast with water, methanol, and ammonia, molecules such as carbon dioxide, methane, ethane, and benzene have zero dipole moments. Because of the symmetrical structures of these molecules, the individual bond polarities and lone-pair contributions exactly cancel.
Worked Example 2.1
Predicting the Direction of a Dipole Moment
Make a three-dimensional drawing of methylamine, CH3NH2, and show the direction of its dipole moment (μ = 1.31).
Strategy
Look for any lone-pair electrons, and identify any atom with an electronegativity substantially different from that of carbon. (Usually, this means O, N, F, Cl, or Br.) Electron density will be displaced in the general direction of the electronegative atoms and the lone pairs.
Solution
Methylamine contains an electronegative nitrogen atom with a lone pair of electrons. The dipole moment thus points generally from –CH3 toward the lone pair.
Problem 2-5
Ethylene glycol, HOCH2CH2OH, may look nonpolar when drawn, but an internal hydrogen bond between the two –OH groups results in a dipole moment. Explain.
Problem 2-6 Make three-dimensional drawings of the following molecules, and predict whether each has a dipole moment. If you expect a dipole moment, show its direction. (a)
H2C$\text{=}$CH2
(b) CHCl3 (c) CH2Cl2 (d) H2C$\text{=}$CCl2 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.03%3A_Polar_Covalent_Bonds_-_Dipole_Moments.txt |
Closely related to the ideas of bond polarity and dipole moment is the assignment of formal charges to specific atoms within a molecule, particularly atoms that have an apparently “abnormal” number of bonds. Look at dimethyl sulfoxide (CH3SOCH3), for instance, a solvent commonly used for preserving biological cell lines at low temperature. The sulfur atom in dimethyl sulfoxide has three bonds rather than the usual two and has a formal positive charge. The oxygen atom, by contrast, has one bond rather than the usual two and has a formal negative charge. Note that an electrostatic potential map of dimethyl sulfoxide shows the oxygen as negative (red) and the sulfur as relatively positive (blue), in accordance with the formal charges.
Formal charges, as the name suggests, are a formalism and don’t imply the presence of actual ionic charges in a molecule. Instead, they’re a device for electron “bookkeeping” and can be thought of in the following way: A typical covalent bond is formed when each atom donates one electron. Although the bonding electrons are shared by both atoms, each atom can still be considered to “own” one electron for bookkeeping purposes. In methane, for instance, the carbon atom owns one electron in each of the four C–H bonds. Because a neutral, isolated carbon atom has four valence electrons, and because the carbon atom in methane still owns four, the methane carbon atom is neutral and has no formal charge.
The same is true for the nitrogen atom in ammonia, which has three covalent N–H bonds and two nonbonding electrons (a lone pair). Atomic nitrogen has five valence electrons, and the ammonia nitrogen also has five—one in each of three shared N–H bonds plus two in the lone pair. Thus, the nitrogen atom in ammonia has no formal charge.
The situation is different in dimethyl sulfoxide. Atomic sulfur has six valence electrons, but the dimethyl sulfoxide sulfur owns only five—one in each of the two S–C single bonds, one in the S–O single bond, and two in a lone pair. Thus, the sulfur atom has formally lost an electron and therefore has a positive formal charge. A similar calculation for the oxygen atom shows that it has formally gained an electron and has a negative charge. Atomic oxygen has six valence electrons, but the oxygen in dimethyl sulfoxide has seven—one in the O–S bond and two in each of three lone pairs. Thus, the oxygen has formally gained an electron and has a negative formal charge.
To express the calculations in a general way, the formal charge on an atom is equal to the number of valence electrons in a neutral, isolated atom minus the number of electrons owned by that bonded atom in a molecule. The number of electrons in the bonded atom, in turn, is equal to half the number of bonding electrons plus the nonbonding, lone-pair electrons.
A summary of commonly encountered formal charges and the bonding situations in which they occur is given in Table 2.2. Although only a bookkeeping device, formal charges often give clues about chemical reactivity, so it’s helpful to be able to identify and calculate them correctly.
Table 2.2 A Summary of Common Formal Charges
Atom C N O S P
Structure
Valence electrons 4 4 4 5 5 6 6 6 6 5
Number of bonds 3 3 3 4 2 3 1 3 1 4
Number of nonbonding electrons 1 0 2 0 4 2 6 2 6 0
Formal charge 0 +1 –1 +1 –1 +1 –1 +1 –1 +1
Problem 2-7 Calculate formal charges for the nonhydrogen atoms in the following molecules: (a)
(b)
(c)
Problem 2-8
Organic phosphate groups occur commonly in biological molecules. Calculate formal charges on the four O atoms in the methyl phosphate dianion. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.04%3A_Formal_Charges.txt |
Most substances can be represented unambiguously by the Kekulé line-bond structures we’ve been using up to this point, but an interesting problem sometimes arises. Look at the acetate ion, for instance. When we draw a line-bond structure for acetate, we need to show a double bond to one oxygen and a single bond to the other. But which oxygen is which? Should we draw a double bond to the “top” oxygen and a single bond to the “bottom” oxygen, or vice versa?
Although the two oxygen atoms in the acetate ion appear different in line-bond structures, experiments show that they are equivalent. Both carbon–oxygen bonds, for instance, are 127 pm in length, midway between the length of a typical C–O single bond (135 pm) and a typical $C=OC=O$ double bond (120 pm). In other words, neither of the two structures for acetate is correct by itself. The true structure is intermediate between the two, and an electrostatic potential map shows that both oxygen atoms share the negative charge and have equal electron densities (red).
The two individual line-bond structures for acetate ion are called resonance forms, and their special resonance relationship is indicated by the double-headed arrow between them. The only difference between the two resonance forms is the placement of the π and nonbonding valence electrons. The atoms themselves occupy exactly the same place in both resonance forms, the connections between atoms are the same, and the three-dimensional shapes of the resonance forms are the same.
A good way to think about resonance forms is to realize that a substance like the acetate ion is the same as any other. Acetate doesn’t jump back and forth between two resonance forms, spending part of the time looking like one and part of the time looking like the other. Rather, acetate has a single unchanging structure that we say is a resonance hybrid of the two individual forms and has characteristics of both. The only “problem” with acetate is that we can’t draw it accurately using a familiar line-bond structure—line-bond structures just don’t work for resonance hybrids. The difficulty, however, is with the representation of acetate on paper, not with acetate itself.
Resonance is a very useful concept that we’ll return to on numerous occasions throughout the rest of this book. We’ll see in Chapter 15, for instance, that the six carbon–carbon bonds in aromatic compounds, such as benzene, are equivalent and that benzene is best represented as a hybrid of two resonance forms. Although each individual resonance form seems to imply that benzene has alternating single and double bonds, neither form is correct by itself. The true benzene structure is a hybrid of the two individual forms, and all six carbon–carbon bonds are equivalent. This symmetrical distribution of electrons around the molecule is evident in an electrostatic potential map.
2.06: Rules for Resonance Forms
When first dealing with resonance forms, it’s useful to have a set of guidelines that describe how to draw and interpret them. The following rules should be helpful:
RULE 1
Individual resonance forms are imaginary, not real. The real structure is a composite, or resonance hybrid, of the different forms. Species such as the acetate ion and benzene are no different from any other. They have single, unchanging structures, and they don't switch back and forth between resonance forms. The only difference between these and other substances is in the way they are represented in drawings.
RULE 2
Resonance forms differ only in the placement of their π or nonbonding electrons. Neither the position nor the hybridization of any atom changes from one resonance form to another. In the acetate ion, for instance, the carbon atom is sp2-hybridized and the oxygen atoms remain in exactly the same place in both resonance forms. Only the positions of the π electrons in the $C=OC=O$ bond and the lone-pair electrons on oxygen differ from one form to another. This movement of electrons from one resonance structure to another can be indicated with curved arrows. A curved arrow always indicates the movement of electrons, not the movement of atoms. An arrow shows that a pair of electrons moves from the atom or bond at the tail of the arrow to the atom or bond at the head of the arrow.
The situation with benzene is similar to that with acetate. The π electrons in the double bonds move, as shown with curved arrows, but the carbon and hydrogen atoms remain in place.
RULE 3
Different resonance forms of a substance don’t have to be equivalent. As an example, we’ll see in Chapter 22 that a compound such as acetone, which contains a $C=OC=O$ bond, can be converted into its anion by reaction with a strong base. The resultant anion has two resonance forms. One form contains a carbon–oxygen double bond and has a negative charge on carbon; the other contains a carbon–carbon double bond and has a negative charge on oxygen. Even though the two resonance forms aren’t equivalent, both contribute to the overall resonance hybrid.
When two resonance forms are nonequivalent, the actual structure of the resonance hybrid resembles the more stable form. Thus, we might expect the true structure of the acetone anion to be more like that of the form that places the negative charge on the electronegative oxygen atom rather than on the carbon.
RULE 4
Resonance forms obey normal rules of valency. A resonance form is like any other structure: the octet rule still applies to second-row, main-group atoms. For example, one of the following structures for the acetate ion is not a valid resonance form because the carbon atom has five bonds and ten valence electrons:
RULE 5
The resonance hybrid is more stable than any individual resonance form. In other words, resonance leads to stability. Generally speaking, the larger the number of resonance forms a substance has, the more stable the substance is, because its electrons are spread out over a larger part of the molecule and are closer to more nuclei. We’ll see in Chapter 15, for instance, that a benzene ring is more stable because of resonance than might otherwise be expected. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.05%3A_Resonance.txt |
Look back at the resonance forms of the acetate ion and the acetone anion shown in the previous section. The pattern seen there is a common one that leads to a useful technique for drawing resonance forms. In general, any three-atom grouping with a p orbital on each atom has two resonance forms:
The atoms X, Y, and Z in the general structure might be C, N, O, P, S, or others, and the asterisk (*) might mean that the p orbital on atom Z is vacant, that it contains a single electron, or that it contains a lone pair of electrons. The two resonance forms differ simply by an exchange in position of the multiple bond and the asterisk from one end of the three-atom grouping to the other.
By learning to recognize such three-atom groupings within larger structures, resonance forms can be systematically generated. Look, for instance, at the anion produced when H+ is removed from 2,4-pentanedione by reaction with a base. How many resonance structures does the resultant anion have?
The 2,4-pentanedione anion has a lone pair of electrons and a formal negative charge on the central carbon atom, next to a $C=OC=O$ bond on the left. The $O=C–C:−O=C–C:−$ grouping is a typical one for which two resonance structures can be drawn.
Just as there is a $C=OC=O$ bond to the left of the lone pair, there is a second $C=OC=O$ bond to the right. Thus, we can draw a total of three resonance structures for the 2,4-pentanedione anion.
Worked Example 2.2
Drawing Resonance Forms for an Anion
Draw three resonance structures for the carbonate ion, CO32.
Strategy
Look for three-atom groupings that contain a multiple bond next to an atom with a p orbital. Then exchange the positions of the multiple bond and the electrons in the p orbital. In the carbonate ion, each singly bonded oxygen atom with three lone pairs and a negative charge is adjacent to the $C=OC=O$ double bond, giving the grouping .
Solution
Exchanging the position of the double bond and an electron lone pair in each grouping generates three resonance structures.
Worked Example 2.3
Drawing Resonance Forms for a Radical
Draw three resonance forms for the pentadienyl radical, where a radical is a substance that contains a single, unpaired electron in one of its orbitals, denoted by a dot (·).
Strategy
Find the three-atom groupings that contain a multiple bond next to an atom with a p orbital.
Solution
The unpaired electron is on a carbon atom next to a $C=CC=C$ bond, giving a typical three-atom grouping that has two resonance forms.
In the second resonance form, the unpaired electron is next to another double bond, giving another three-atom grouping and leading to another resonance form.
Thus, the three resonance forms for the pentadienyl radical are:
Problem 2-9
Which of the following pairs of structures represent resonance forms, and which do not? Explain.
(a) (b)
Problem 2-10 Draw the indicated number of resonance forms for each of the following substances: (a)
The methyl phosphate anion, CH3OPO32– (3)
(b) The nitrate anion, NO3 (3) (c) The allyl cation, H2C$\text{=}$CH–CH2+ (2) (d)
The benzoate anion (2)
2.08: Acids and Bases - The Brnsted-Lowry Definition
Perhaps the most important of all concepts related to electronegativity and polarity is that of acidity and basicity. We’ll soon see, in fact, that the acid–base behavior of organic molecules explains much of their chemistry. You may recall from a course in general chemistry that two definitions of acidity are frequently used: the Brønsted–Lowry definition and the Lewis definition. We’ll look at the Brønsted–Lowry definition in this and the following three sections and then discuss the Lewis definition in Section 2.11.
A Brønsted–Lowry acid is a substance that donates a hydrogen ion, H+, and a Brønsted–Lowry base is a substance that accepts a hydrogen ion. (The name proton is often used as a synonym for H+ because loss of the valence electron from a neutral hydrogen atom leaves only the hydrogen nucleus—a proton.) When gaseous hydrogen chloride dissolves in water, for example, a polar HCl molecule acts as an acid and donates a proton, while a water molecule acts as a base and accepts the proton, yielding chloride anion (Cl) and hydronium cation (H3O+). This and other acid–base reactions are reversible, so we’ll write them with double, forward-and-backward arrows.
Chloride ion, the product that results when the acid HCl loses a proton, is called the conjugate base of the acid, and hydronium ion, the product that results when the base H2O gains a proton, is called the conjugate acid of the base. Other common mineral acids such as H2SO4 and HNO3 behave similarly, as do organic acids such as acetic acid, CH3CO2H.
In a general sense,
For example:
Notice that water can act either as an acid or as a base, depending on the circumstances. In its reaction with HCl, water is a base that accepts a proton to give the hydronium ion, H3O+. In its reaction with ammonia (NH3), however, water is an acid that donates a proton to give ammonium ion (NH4+) and hydroxide ion, HO.
Problem 2-11
Nitric acid (HNO3) reacts with ammonia (NH3) to yield ammonium nitrate. Write the reaction, and identify the acid, the base, the conjugate acid product, and the conjugate base product. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.07%3A_Drawing_Resonance_Forms.txt |
Different acids differ in their ability to donate H+. Stronger acids, such as HCl, react almost completely with water, whereas weaker acids, such as acetic acid (CH3CO2H), react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant (Ka) for the acid-dissociation equilibrium.
$\ce{HA(aq) + H2O(l) <=> H3O^{+}(aq) + A^{-}(aq)} \nonumber$
the acid ionization constant is written
$K_{ a }=\frac{[\ce{H3O^{+}}] [\ce{A^{-}}]}{[\ce{HA}]} \nonumber$
Recall from general chemistry that the solvent concentration does not appear in the equilibrium expression, and that brackets [ ] around a substance refer to the concentration of the enclosed species in moles per liter (molarity).
Note
Equilibrium constant expressions are actually ratios of activities, and the value of K is determined at the limit of infinite dilution of the solutes. In these very dilute solutions, the activity of the solvent has a value of unity (1) and the activity of each solute can be approximated by its molar concentration.
Stronger acids have their equilibria toward the right and thus have larger acidity constants, whereas weaker acids have their equilibria toward the left and have smaller acidity constants. The range of Ka values for different acids is enormous, running from about 1015 for the strongest acids to about 10–60 for the weakest. Common inorganic acids such as H2SO4, HNO3, and HCl have Ka’s in the range of 101 to 109, while organic acids generally have Ka’s in the range of 10–5 to 10–15. As you gain experience, you’ll develop a rough feeling for which acids are “strong” and which are “weak” (always remembering that the terms are relative, not absolute).
Acid strengths are normally expressed using pKa values rather than Ka values, where the pKa is the negative common logarithm of the Ka:
pKa=log Ka
A stronger acid (larger Ka) has a smaller pKa, and a weaker acid (smaller Ka) has a larger pKa. Table 2.3 lists the pKa’s of some common acids in order of their strength, and a more comprehensive table is given in Appendix B.
Notice that the pKa value shown in Table 2.3 for water is 14.00, which results from the following calculation.
$\ce{H2O(l) + H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)} \label{autoionization}$
with
$K_w=K_a =[\ce{H3O^{+}}][\ce{OH^{-}}] \label{Kw}$
As explained above, because the water is the solvent and has an activity of unity (1), water is not shown explicitly in the equilibrium constant expression for Kw
Notice also in Table 2.3 that there is an inverse relationship between the acid strength of an acid and the base strength of its conjugate base. A strong acid has a weak conjugate base, and a weak acid has a strong conjugate base. To understand this inverse relationship, think about what is happening to the acidic hydrogen in an acid–base reaction. A strong acid is one that loses H+ easily, meaning that its conjugate base holds the H+ weakly and is therefore a weak base. A weak acid is one that loses H+ with difficulty, meaning that its conjugate base holds the proton tightly and is therefore a strong base. The fact that HCl is a strong acid, for example, means that Cl does not hold H+ tightly and is thus a weak base. Water, on the other hand, is a weak acid, meaning that OH holds H+ tightly and is a strong base.
Table 2.3 Relative Strengths of Some Common Acids and Their Conjugate Bases
Acid Name pKa Conjugate base Name
CH3CH2OH Ethanol 16.00 CH3CH2O Ethoxide ion
H2O Water 14.00 HO Hydroxide ion
HCN Hydrocyanic acid 9.31 CN Cyanide ion
H2PO4 Dihydrogen phosphate ion 7.21 HPO42 Hydrogen phosphate ion
CH3CO2H Acetic acid 4.76 CH3CO2 Acetate ion
H3PO4 Phosphoric acid 2.16 H2PO4 Dihydrogen phosphate ion
H3O+ Hydronium ion 0.0 H2O Water
HNO3 Nitric acid –1.3 NO3 Nitrate ion
HCl Hydrochloric acid –7.0 Cl Chloride ion
Problem 2-12
The amino acid phenylalanine has pKa = 1.83, and tryptophan has pKa = 2.83. Which is the stronger acid?
Problem 2-13
Amide ion, H2N, is a much stronger base than hydroxide ion, HO. Which is the stronger acid, NH3 or H2O? Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.09%3A_Acid_and_Base_Strength.txt |
Compilations of pKa values like those in Table 2.3 and Appendix B are useful for predicting whether a given acid–base reaction will take place, because H+ will always go from the stronger acid to the stronger base. That is, an acid will donate a proton to the conjugate base of a weaker acid, and the conjugate base of a weaker acid will remove a proton from a stronger acid. Since water (pKa = 14.00) is a weaker acid than acetic acid (pKa = 4.76), for example, hydroxide ion holds a proton more tightly than acetate ion does. Hydroxide ion will therefore react to a large extent with acetic acid, CH3CO2H, to yield acetate ion and H2O.
Another way to predict acid–base reactivity is to remember that the product conjugate acid in an acid–base reaction must be weaker and less reactive than the starting acid and the product conjugate base must be weaker and less reactive than the starting base. In the reaction of acetic acid with hydroxide ion, for example, the product conjugate acid (H2O) is weaker than the starting acid (CH3CO2H), and the product conjugate base (CH3CO2) is weaker than the starting base (OH).
Worked Example 2.4
Predicting Acid Strengths from pKa Values
Water has pKa = 14.00, and acetylene has pKa = 25. Which is the stronger acid? Does hydroxide ion react to a significant extent with acetylene?
Strategy
In comparing two acids, the one with the lower pKa is stronger. Thus, water is a stronger acid than acetylene and gives up H+ more easily.
Solution
Because water is a stronger acid and gives up H+ more easily than acetylene, the HO ion must have less affinity for H+ than the $HC≡C:−HC≡C:−$ ion. In other words, the anion of acetylene is a stronger base than hydroxide ion, and the reaction will not proceed significantly as written.
Worked Example 2.5
Calculating Ka from pKa
According to the data in Table 2.3, acetic acid has pKa = 4.76. What is its Ka?
Strategy
Since pKa is the negative logarithm of Ka, it’s necessary to use a calculator with an ANTILOG or INV LOG function. Enter the value of the pKa (4.76), change the sign (–4.76), and then find the antilog (1.74 × 10–5).
Solution
Ka = 1.74 × 10–5.
Problem 2-14
Will either of the following reactions take place to a significant extent as written, according to the data in Table 2.3?
(a)
(b)
Problem 2-15
Ammonia, NH3, has pKa ≈ 36, and acetone has pKa ≈ 19. Will the following reaction take place to a significant extent?
Problem 2-16
What is the Ka of HCN if its pKa = 9.31?
2.11: Organic Acids and Organic Bases
Many of the reactions we’ll be seeing in future chapters, including practically all biological reactions, involve organic acids and organic bases. Although it’s too early to go into the details of these processes now, you might keep the following generalities in mind:
Organic Acids
Organic acids are characterized by the presence of a positively polarized hydrogen atom (blue in electrostatic potential maps) and are of two main kinds: acids such as methanol and acetic acid that contain a hydrogen atom bonded to an electronegative oxygen atom (O–H) and those such as acetone (Section 2.5) that contain a hydrogen atom bonded to a carbon atom next to a $C=OC=O$ bond ($O=C–C–HO=C–C–H$).
Methanol contains an O–H bond and is a weak acid, while acetic acid also contains an O–H bond and is a somewhat stronger acid. In both cases, acidity is due to the fact that the conjugate base resulting from loss of H+ is stabilized by having its negative charge on a strongly electronegative oxygen atom. In addition, the conjugate base of acetic acid is stabilized by resonance (Section 2.4 and Section 2.5).
The acidity of acetone and other compounds with $C=OC=O$ bonds is due to the fact that the conjugate base resulting from loss of H+ is stabilized by resonance. In addition, one of the resonance forms stabilizes the negative charge by placing it on an electronegative oxygen atom.
Electrostatic potential maps of the conjugate bases from methanol, acetic acid, and acetone are shown in Figure 2.5. As you might expect, all three show a substantial amount of negative charge (red) on oxygen.
The electronegative oxygen atoms stabilize the negative charge in all three.
Compounds called carboxylic acids, which contain the –CO2H grouping, occur abundantly in all living organisms and are involved in almost all metabolic pathways. Acetic acid, pyruvic acid, and citric acid are examples. You might note that at the typical pH of 7.3 found within cells, carboxylic acids are usually dissociated and exist as their carboxylate anions, –CO2.
Organic Bases
Organic bases are characterized by the presence of an atom (reddish in electrostatic potential maps) with a lone pair of electrons that can bond to H+. Nitrogen-containing compounds such as methylamine are the most common organic bases and are involved in almost all metabolic pathways, but oxygen-containing compounds can also act as bases when reacting with a sufficiently strong acid. Note that some oxygen-containing compounds can act both as acids and as bases depending on the circumstances, just as water can. Methanol and acetone, for instance, act as acids when they donate a proton but as bases when their oxygen atom accepts a proton.
We’ll see in Chapter 26 that substances called amino acids, so-named because they are both amines (–NH2) and carboxylic acids (–CO2H), are the building blocks from which the proteins in all living organisms are made. Twenty different amino acids go into making up proteins—alanine is an example. Interestingly, alanine and other amino acids exist primarily in a doubly charged form called a zwitterion rather than in the uncharged form. The zwitterion form arises because amino acids have both acidic and basic sites within the same molecule and therefore undergo an internal acid–base reaction. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.10%3A_Predicting_Acid-Base_Reactions_from_pKa_Values.txt |
The Lewis definition of acids and bases is more encompassing than the Brønsted–Lowry definition because it’s not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond.
Lewis Acids and the Curved Arrow Formalism
The fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate H+ (which has an empty 1s orbital). Thus, the Lewis definition of acidity includes many species in addition to H+. For example, various metal cations, such as Mg2+, are Lewis acids because they accept a pair of electrons when they form a bond to a base. We’ll also see in later chapters that certain metabolic reactions begin with an acid–base reaction between Mg2+ as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base.
In the same way, compounds of group 3A elements, such as BF3 and AlCl3, are Lewis acids because they have unfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in Figure 2.6. Similarly, many transition-metal compounds, such as TiCl4, FeCl3, ZnCl2, and SnCl4, are Lewis acids.
The Lewis acid accepts a pair of electrons, and the Lewis base donates a pair of nonbonding electrons. Note how the movement of electrons from the Lewis base to the Lewis acid is indicated by a curved arrow. Note also how, in electrostatic potential maps, the boron becomes more negative (red) after reaction because it has gained electrons and the oxygen atom becomes more positive (blue) because it has donated electrons.
Look closely at the acid–base reaction in Figure 2.6, and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in BF3, a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in Section 2.5. We’ll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it.
Some further examples of Lewis acids follow:
Lewis Bases
The Lewis definition of a base—a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid—is similar to the Brønsted–Lowry definition. Thus, H2O, with its two pairs of nonbonding electrons on oxygen, acts as a Lewis base by donating an electron pair to an H+ in forming the hydronium ion, H3O+.
In a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an H+ but as bases when their oxygen atom accepts an H+.
Note also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. Reaction normally occurs only once in such instances, and the more stable of the two possible protonation products is formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms.
Worked Example 2.6
Using Curved Arrows to Show Electron Flow
Using curved arrows, show how acetaldehyde, CH3CHO, can act as a Lewis base.
Strategy
A Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid.
Solution
Problem 2-17 Using curved arrows, show how the species in part (a) can act as Lewis bases in their reactions with HCl, and show how the species in part (b) can act as Lewis acids in their reaction with OH–. (a)
CH3CH2OH, HN(CH3)2, P(CH3)3
(b) H3C+, B(CH3)3, MgBr2
Problem 2-18 Imidazole, which forms part of amino acid histidine, can act as both an acid and a base. (a)
Look at the electrostatic potential map of imidazole, and identify the most acidic hydrogen atom and the most basic nitrogen atom.
(b) Draw structures for the resonance forms of the products that result when imidazole is protonated by an acid and deprotonated by a base. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.12%3A_Acids_and_Bases_-_The_Lewis_Definition.txt |
When thinking about chemical reactivity, chemists usually focus their attention on bonds, the covalent interactions between atoms within molecules. Also important, however, particularly in large biomolecules like proteins and nucleic acids, are a variety of interactions between molecules that strongly affect molecular properties. Collectively called either intermolecular forces, van der Waals forces, or noncovalent interactions, they are of several different types: dipole–dipole forces, dispersion forces, and hydrogen bonds.
Dipole–dipole forces occur between polar molecules as a result of electrostatic interactions among dipoles. The forces can be either attractive or repulsive depending on the orientation of the molecules—attractive when unlike charges are together and repulsive when like charges are together. The attractive geometry is lower in energy and therefore predominates (Figure 2.7).
Dispersion forces occur between all neighboring molecules and arise because the electron distribution within molecules is constantly changing. Although uniform on a time-averaged basis, the electron distribution even in nonpolar molecules is likely to be nonuniform at any given instant. One side of a molecule may, by chance, have a slight excess of electrons relative to the opposite side, giving the molecule a temporary dipole. This temporary dipole in one molecule causes a nearby molecule to adopt a temporarily opposite dipole, resulting in a tiny attraction between the two (Figure 2.8). Temporary molecular dipoles have only a fleeting existence and are constantly changing, but their cumulative effect is often strong enough to hold molecules close together so that a substance is a liquid or solid rather than a gas.
Perhaps the most important noncovalent interaction in biological molecules is the hydrogen bond, an attractive interaction between a hydrogen atom bonded to an electronegative O or N atom and an unshared electron pair on another O or N atom. In essence, a hydrogen bond is a very strong dipole–dipole interaction involving polarized O–H or N–H bonds. Electrostatic potential maps of water and ammonia clearly show the positively polarized hydrogens (blue) and the negatively polarized oxygens and nitrogens (red).
Hydrogen bonding has enormous consequences for living organisms. Hydrogen bonds cause water to be a liquid rather than a gas at ordinary temperatures, they hold enzymes in the shapes necessary for catalyzing biological reactions, and they cause strands of deoxyribonucleic acid (DNA) to pair up and coil into the double helix that stores genetic information.
One further point before leaving the subject of noncovalent interactions: biochemists frequently use the term hydrophilic, meaning “water-loving,” to describe a substance that is attracted to water and the term hydrophobic, meaning “water-fearing,” to describe a substance that is not strongly attracted to water. Hydrophilic substances, such as table sugar, often have a number of –OH groups in their structure so they can form hydrogen bonds and dissolve in water, whereas hydrophobic substances, such as vegetable oil, do not have groups that form hydrogen bonds and do not dissolve in water.
Problem 2-19
Of the two vitamins A and C, one is hydrophilic and water-soluble while the other is hydrophobic and fat-soluble. Which is which?
2.14: Chemistry MattersAlkaloids- From Cocaine to Dental Anesthetics
2 • Chemistry Matters 2 • Chemistry Matters
Just as ammonia (NH3) is a weak base, there are a large number of nitrogen-containing organic compounds called amines that are also weak bases. In the early days of organic chemistry, basic amines derived from natural sources were known as vegetable alkali, but they are now called alkaloids. More than 20,000 alkaloids are known. Their study provided much of the impetus for the growth of organic chemistry in the nineteenth century and remains today an active and fascinating area of research.
Many alkaloids have pronounced biological properties, and approximately 50% of pharmaceutical agents used today are derived from naturally occurring amines. As just three examples, morphine, an analgesic agent (painkiller), is obtained from the opium poppy Papaver somniferum. Ephedrine, a bronchodilator, decongestant, and appetite suppressant, is obtained from Ephedra sinica, an evergreen shrub native to Mongolia and northeastern China. Cocaine, both an anesthetic and a stimulant, is obtained from the coca bush Erythroxylon coca, endemic to the upland rain forest areas of central South America. (And yes, there really was a small amount of cocaine in the original Coca-Cola recipe, although it was removed in 1906.)
Cocaine itself is rarely used medically because it is too addictive, but its anesthetic properties provoked a long search for related but nonaddictive compounds. This search ultimately resulted in the synthesis of the “caine” anesthetics that are commonly used today in dental and surgical anesthesia. Procaine, the first such compound, was synthesized in 1898 and marketed under the name Novocain. It was rapidly adopted and remains in use today as a topical anesthetic. Other related compounds with different activity profiles followed: Lidocaine, marketed as Xylocaine, was introduced in 1943, and mepivacaine (Carbocaine) in the early 1960s. More recently, bupivacaine (Marcaine) and prilocaine (Citanest) have gained popularity. Both are quick-acting, but the effects of bupivacaine last for 3 to 6 hours while those of prilocaine fade after 45 minutes. Notice some structural similarity of all the caines to cocaine itself.
An estimate from the U.S. National Academy of Sciences is that less than 1% of all living species have been characterized. Thus, alkaloid chemistry remains an active area of research, and innumerable substances with potentially useful properties have yet to be discovered. Undoubtedly even the caine anesthetics will become obsolete at some point, perhaps supplanted by newly discovered alkaloids. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.13%3A_Noncovalent_Interactions_Between_Molecules.txt |
2 • Key Terms 2 • Key Terms
• acidity constant (Ka)
• alkaloid
• Brønsted–Lowry acid
• Brønsted–Lowry base
• conjugate acid
• conjugate base
• dipole moment ($μμ$)
• dispersion force
• electronegativity (EN)
• electrostatic potential map
• formal charge
• hydrogen bond
• hydrophilic
• hydrophobic
• inductive effect
• intermolecular force
• Lewis acid
• Lewis base
• noncovalent interaction
• pKa
• polar covalent bond
• resonance form
• resonance hybrid
• van der Waals force
2.16: Summary
2 • Summary 2 • Summary
Understanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we’ve reviewed some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation for understanding the specific reactions that will be discussed in subsequent chapters.
Organic molecules often have polar covalent bonds as a result of unsymmetrical electron sharing caused by differences in the electronegativity of atoms. A carbon–oxygen bond is polar, for example, because oxygen attracts the shared electrons more strongly than carbon does. Carbon–hydrogen bonds are relatively nonpolar. Many molecules as a whole are also polar, owing to the presence of individual polar bonds and electron lone pairs. The polarity of a molecule is measured by its dipole moment, μ.
Plus (+) and minus (–) signs are often used to indicate the presence of formal charges on atoms in molecules. Assigning formal charges to specific atoms is a bookkeeping technique that makes it possible to keep track of the valence electrons around an atom and offers some clues about chemical reactivity.
Some substances, such as acetate ion and benzene, can’t be represented by a single line-bond structure and must be considered as a resonance hybrid of two or more structures, none of which would be correct by themselves. The only difference between two resonance forms is in the location of their π and nonbonding electrons. The nuclei remain in the same places in both structures, and the hybridization of the atoms remains the same.
Acidity and basicity are closely related to the ideas of polarity and electronegativity. A Brønsted–Lowry acid is a compound that can donate a proton (hydrogen ion, H+), and a Brønsted–Lowry base is a compound that can accept a proton. The strength of a Brønsted–Lowry acid or base is expressed by its acidity constant, Ka, or by the negative logarithm of the acidity constant, pKa. The larger the pKa, the weaker the acid. More useful is the Lewis definition of acids and bases. A Lewis acid is a compound that has a low-energy empty orbital that can accept an electron pair; Mg2+, BF3, AlCl3, and H+ are examples. A Lewis base is a compound that can donate an unshared electron pair; NH3 and H2O are examples. Most organic molecules that contain oxygen and nitrogen can act as Lewis bases toward sufficiently strong acids.
A variety of noncovalent interactions have a significant effect on the properties of large biomolecules. Hydrogen bonding—the attractive interaction between a positively polarized hydrogen atom bonded to an oxygen or nitrogen atom with an unshared electron pair on another O or N atom, is particularly important in giving proteins and nucleic acids their shapes. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.15%3A_Key_Terms.txt |
2 • Additional Problems 2 • Additional Problems
Visualizing Chemistry
Problem 2-20
Fill in the multiple bonds in the following model of naphthalene, C10H8 (black = C, gray = H). How many resonance structures does naphthalene have? Draw them.
Problem 2-21
The following model is a representation of ibuprofen, a common over-the-counter pain reliever. Indicate the positions of the multiple bonds, and draw a skeletal structure (black = C, red = O, gray = H).
Problem 2-22
cis-1,2-Dichloroethylene and trans-1,2-dichloroethylene are isomers, compounds with the same formula but different chemical structures. Look at the following electrostatic potential maps, and tell whether either compound has a dipole moment.
Problem 2-23 The following molecular models are representations of (a) adenine and (b) cytosine, constituents of DNA (deoxyribonucleic acid). Indicate the positions of multiple bonds and lone pairs for both, and draw skeletal structures (black = C, red = O, blue = N, gray = H). (a) (b) Mechanism Problems Problem 2-24 Predict the product(s) of the following acid/base reactions. Draw curved arrows to show the formation and breaking of bonds. (a) (b) (c) Problem 2-25 Use curved arrows to draw the protonated form of the following Lewis bases. (a) (b) (c) (d) Problem 2-26 Use the curved-arrow formalism to show how the electrons flow in the resonance form on the left to give the one on the right. (a) (b) (c) Problem 2-27 Double bonds can also act like Lewis bases, sharing their electrons with Lewis acids. Use curved arrows to show how each of the following double bonds will react with HCl and draw the resulting carbocation. (a) (b) (c) Electronegativity and Dipole Moments Problem 2-28 Identify the most electronegative element in each of the following molecules: (a) CH2FCl (b)
FCH2CH2CH2Br
(c) HOCH2CH2NH2 (d) CH3OCH2Li
Problem 2-29 Use the electronegativity table given in Figure 2.3 to predict which bond in each of the following pairs is more polar, and indicate the direction of bond polarity for each compound. (a)
H3CCl or ClCl
(b) H3CH or HCl (c) HOCH3 or (CH3)3SiCH3 (d) H3CLi or LiOH
Problem 2-30 Which of the following molecules has a dipole moment? Indicate the expected direction of each. (a) (b) (c) (d) Problem 2-31 (a) The H–Cl bond length is 136 pm. What would the dipole moment of HCl be if the molecule were 100% ionic, H+ Cl−? (b)
The actual dipole moment of HCl is 1.08 D. What is the percent ionic character of the H–Cl bond?
Problem 2-32
Phosgene, Cl2C$\text{=}$O, has a smaller dipole moment than formaldehyde, H2C$\text{=}$O, even though it contains electronegative chlorine atoms in place of hydrogen. Explain.
Problem 2-33
Fluoromethane (CH3F, μ = 1.81 D) has a smaller dipole moment than chloromethane (CH3Cl, μ = 1.87 D) even though fluorine is more electronegative than chlorine. Explain.
Problem 2-34
Methanethiol, CH3SH, has a substantial dipole moment (μ = 1.52) even though carbon and sulfur have identical electronegativities. Explain.
Formal Charges
Problem 2-35 Calculate the formal charges on the atoms shown in red. (a) (b) (c) (d) (e) (f) Problem 2-36 Assign formal charges to the atoms in each of the following molecules: (a) (b) (c) Resonance Problem 2-37 Which of the following pairs of structures represent resonance forms? (a) (b) (c) (d) Problem 2-38 Draw as many resonance structures as you can for the following species: (a) (b) (c) (d) (e) Problem 2-39 1,3-Cyclobutadiene is a rectangular molecule with two shorter double bonds and two longer single bonds. Why do the following structures not represent resonance forms? Acids and Bases Problem 2-40 Alcohols can act either as weak acids or as weak bases, just as water can. Show the reaction of methanol, CH3OH, with a strong acid such as HCl and with a strong base such as Na+ –NH2 Problem 2-41 The O–H hydrogen in acetic acid is more acidic than any of the C–H hydrogens. Explain this result using resonance structures. Problem 2-42 Draw electron-dot structures for the following molecules, indicating any unshared electron pairs. Which of the compounds are likely to act as Lewis acids and which as Lewis bases? (a) AlBr3 (b)
CH3CH2NH2
(c) BH3 (d) HF (e) CH3SCH3 (f) TiCl4
Problem 2-43 Write the products of the following acid–base reactions: (a)
CH3OH + H2SO4 ⇄ ?
(b) CH3OH + NaNH2 ⇄ ? (c) CH3NH3+ Cl + NaOH ⇄ ?
Problem 2-44
Rank the following substances in order of increasing acidity:
Problem 2-45
Which, if any, of the substances in Problem 2-44 is a strong enough acid to react almost completely with NaOH? (The pKa of H2O is 15.74.)
Problem 2-46
The ammonium ion (NH4+, pKa = 9.25) has a lower pKa than the methylammonium ion (CH3NH3+, pKa = 10.66). Which is the stronger base, ammonia (NH3) or methylamine (CH3NH2)? Explain.
Problem 2-47
Is tert-butoxide anion a strong enough base to react significantly with water? In other words, can a solution of potassium tert-butoxide be prepared in water? The pKa of tert-butyl alcohol is approximately 18.
Problem 2-48
Predict the structure of the product formed in the reaction of the organic base pyridine with the organic acid acetic acid, and use curved arrows to indicate the direction of electron flow.
Problem 2-49 Calculate Ka values from the following pKa’s: (a)
Acetone, pKa = 19.3
(b) Formic acid, pKa = 3.75
Problem 2-50 Calculate pKa values from the following Ka’s: (a)
Nitromethane, Ka = 5.0 × 10–11
(b) Acrylic acid, Ka = 5.6 × 10–5
Problem 2-51
What is the pH of a 0.050 M solution of formic acid, pKa = 3.75?
Problem 2-52
Sodium bicarbonate, NaHCO3, is the sodium salt of carbonic acid (H2CO3), pKa = 6.37. Which of the substances shown in Problem 2-44 will react significantly with sodium bicarbonate?
General Problems
Problem 2-53
Maleic acid has a dipole moment, but the closely related fumaric acid, a substance involved in the citric acid cycle by which food molecules are metabolized, does not. Explain.
Problem 2-54
Assume that you have two unlabeled bottles, one of which contains phenol (pKa = 9.9) and one of which contains acetic acid (pKa = 4.76). In light of your answer to Problem 2-52, suggest a simple way to determine what is in each bottle.
Problem 2-55 Identify the acids and bases in the following reactions: (a) (b) (c) (d) Problem 2-56 Which of the following pairs represent resonance structures? (a) (b) (c) (d) Problem 2-57 Draw as many resonance structures as you can for the following species, adding appropriate formal charges to each: (a) (b) (c) Problem 2-58 Carbocations, which contain a trivalent, positively charged carbon atom, react with water to give alcohols: How can you account for the fact that the following carbocation gives a mixture of two alcohols on reaction with water? Problem 2-59 We’ll see in the next chapter that organic molecules can be classified according to the functional groups they contain, where a functional group is a collection of atoms with a characteristic chemical reactivity. Use the electronegativity values given in Figure 2.3 to predict the direction of polarization of the following functional groups. (a) (b) (c) (d) Problem 2-60 The azide functional group, which occurs in azidobenzene, contains three adjacent nitrogen atoms. One resonance structure for azidobenzene is shown. Draw three additional resonance structures, and assign appropriate formal charges to the atoms in all four. Problem 2-61 Phenol, C6H5OH, is a stronger acid than methanol, CH3OH, even though both contain an O–H bond. Draw the structures of the anions resulting from loss of H+ from phenol and methanol, and use resonance structures to explain the difference in acidity. Problem 2-62 Thiamin diphosphate (TPP), a derivative of vitamin B1 required for glucose metabolism, is a weak acid that can be deprotonated by a base. Assign formal charges to the appropriate atoms in both TPP and its deprotonation product. Problem 2-63 Which of the following compounds or ions have a dipole moment? (a) Carbonate ion (CO32–) (b) (c) Problem 2-64 Use the pKa table in Appendix B to determine in which direction the equilibrium is favored. (a) (b) (c) Problem 2-65 Which intermolecular force is predominantly responsible for each observation below? (a) CH3(CH2)29CH3, a component found in paraffin wax, is a solid at room temperature while CH3(CH2)6CH3 is a liquid. (b)
CH3CH2CH2OH has a higher boiling point than CH4.
(c) CH3CO2H, which is found in vinegar, will dissolve in water but not in oil. Assume that oil is CH3(CH2)4CH3. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.17%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 3, you should be able to
1. fulfill the detailed objectives listed under each section.
2. identify some of the commonest functional groups.
3. write the structures and names of the first ten straight-chain alkanes.
4. recognize and name the simple alkyl substituents, and give the systematic names for branched-chain alkanes.
5. briefly describe some of the processes used during the refining of petroleum.
6. briefly describe the physical properties of alkanes.
7. draw a number of possible conformations of some simple alkanes and alkane-like compounds, and represent the energies of such conformations on energy versus rotation diagrams.
8. define, and use in context, the key terms introduced in this chapter.
This chapter begins with an introduction to the concept of the functional group, a concept that facilitates the systematic study of organic chemistry. Next, we introduce the fundamentals of organic nomenclature (i.e., the naming of organic chemicals) through examination of the alkane family of compounds. We then discuss, briefly, the occurrence and properties of alkanes, and end with a description of cis-trans isomerism in cycloalkanes.
• 3.1: Why This Chapter?
• 3.2: Functional Groups
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups
• 3.3: Alkanes and Alkane Isomers
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 CnH2n+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.
• 3.4: Alkyl Groups
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.
• 3.5: Naming Alkanes
There are too many organic molecules to memorize a name for each one. The IUPAC nomenclature system provides an unique name for each different molecule based on functional groups, the longest carbon chain and other attached substituents.
• 3.6: Properties of Alkanes
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.
• 3.7: Conformations of Ethane
Conformational isomerism involves rotation about sigma bonds, and does not involve any differences in the connectivity 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 terms of the angle about one or more sigma bonds.
• 3.8: Conformations of Other Alkanes
Ethane has only two conformers of note - staggered and eclipsed. Alkanes that are more complex than ethane, like propane and butane have a greater variety in possible conformers and their relative energies.
• 3.9: Chemistry Matters—Gasoline
The petroleum that is pumped out of the ground at locations around the world 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 abundances of the components vary depending on the source.
• 3.10: Key Terms
• 3.11: Summary
• 3.12: Additional Problems
03: Organic Compounds- Alkanes and Their Stereochemistry
The group of organic compounds called alkanes are simple and relatively unreactive, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we’ll use alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules, a topic of particular importance in understanding biological organic chemistry.
According to Chemical Abstracts, the publication that abstracts and indexes the chemical literature, there are more than 195 million known organic compounds. Each of these compounds has its own physical properties, such as melting point and boiling point, and each has its own chemical reactivity.
Chemists have learned through years of experience that organic compounds can be classified into families according to their structural features and that the members of a given family have similar chemical behavior. Instead of 195 million compounds with random reactivity, there are a few dozen families of organic compounds whose chemistry is reasonably predictable. We’ll study the chemistry of specific families throughout much of this book, beginning in this chapter with a look at the simplest family, the alkanes. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.01%3A_Why_This_Chapter.txt |
The structural features that make it possible to classify compounds into families are called functional groups. A functional group is a group of atoms within a molecule that has a characteristic chemical behavior. Chemically, a given functional group behaves in nearly the same way in every molecule it’s a part of. For example, compare ethylene, a plant hormone that causes fruit to ripen, with menthene, a much more complicated molecule found in peppermint oil. Both substances contain a carbon–carbon double-bond functional group, and both therefore react with Br2 in the same way to give a product in which a Br atom has added to each of the double-bond carbons (Figure 3.2). This example is typical: the chemistry of every organic molecule, regardless of size and complexity, is determined by the functional groups it contains.
Look at Table 3.1, which lists many of the common functional groups and gives simple examples of their occurrence. Some functional groups have only carbon–carbon double or triple bonds; others have halogen atoms; and still others contain oxygen, nitrogen, or sulfur. Much of the chemistry you’ll be studying is the chemistry of these functional groups.
Functional Groups with Carbon–Carbon Multiple Bonds
Alkenes, alkynes, and arenes (aromatic compounds) all contain carbon–carbon multiple bonds. Alkenes have a double bond, alkynes have a triple bond, and arenes have alternating double and single bonds in a six-membered ring of carbon atoms. They look different, but because of their structural similarities, they also have chemical similarities.
Table 3.1 Structures of Some Common Functional Groups
Name Structure* Name ending Example
Alkene (double bond) -ene $H2C═CH2H2C═CH2$
Ethene
Alkyne (triple bond) $–C≡C––C≡C–$ -yne $HC≡CHHC≡CH$
Ethyne
Arene (aromatic ring) None
Halide
(X=F, Cl, Br, I)
None CH3Cl
Chloromethane
Alcohol -ol CH3OH
Methanol
Ether ether CH3OCH3
Dimethyl ether
Monophosphate phosphate CH3OPO32
Methyl phosphate
Diphosphate diphosphate CH3OP2O63
Methyl diphosphate
Amine -amine CH3NH2
Methylamine
Imine (Schiff base) None
Nitrile $–C≡N–C≡N$ -nitrile $CH3C≡NCH3C≡N$
Ethanenitrile
Thiol -thiol CH3SH
Methanethiol
Sulfide sulfide CH3SCH3
Dimethyl sulfide
Disulfide disulfide CH3SSCH3
Dimethyl disulfide
Sulfoxide sulfoxide
Aldehyde -al
Ketone -one
Carboxylic acid -oic acid
Ester -oate
Thioester -thioate
Amide -amide
Acid chloride -oyl chloride
Carboxylic acid anhydride -oic anhydride
*The bonds whose connections aren’t specified are assumed to be attached to carbon or hydrogen atoms in the rest of the molecule.
Functional Groups with Carbon Singly Bonded to an Electronegative Atom
Alkyl halides (haloalkanes), alcohols, ethers, alkyl phosphates, amines, thiols, sulfides, and disulfides all have a carbon atom singly bonded to an electronegative atom—halogen, oxygen, nitrogen, or sulfur. Alkyl halides have a carbon atom bonded to halogen (–X), alcohols have a carbon atom bonded to the oxygen of a hydroxyl group (–OH), ethers have two carbon atoms bonded to the same oxygen, organophosphates have a carbon atom bonded to the oxygen of a phosphate group (–OPO32), amines have a carbon atom bonded to a nitrogen, thiols have a carbon atom bonded to the sulfur of an –SH group, sulfides have two carbon atoms bonded to the same sulfur, and disulfides have carbon atoms bonded to two sulfurs that are joined together. In all cases, the bonds are polar, with the carbon atom bearing a partial positive charge (δ+) and the electronegative atom bearing a partial negative charge (δ–).
Functional Groups with a Carbon–Oxygen Double Bond (Carbonyl Groups)
The carbonyl group, $C=OTable 3.1. Carbonyl groups are present in a majority of organic compounds and in practically all biological molecules. These compounds therefore behave similarly in many respects but differ depending on the identity of the other atoms bonded to the carbonyl-group carbon.$Aldehydes have at least one hydrogen bonded to the $C=OC=O$, ketones have two carbons bonded to the $C=OC=O$, carboxylic acids have an –OH group bonded to the $C=OC=O$, esters have an ether-like oxygen bonded to the $C=OC=O$, thioesters have a sulfide-like sulfur bonded to the $C=OC=O$, amides have an amine-like nitrogen bonded to the $C=OC=O$, acid chlorides have a chlorine bonded to the $C=OC=O$, and so on. In all these functional groups, the carbonyl carbon atom bears a partial positive charge (δ+), and the oxygen bears a partial negative charge (δ–).
Problem 3-1
Use Table 3.1 to identify the functional groups in each of the following molecules: (a)
b)
(c)
Problem 3-2
Propose structures for simple molecules that contain the following functional groups:
1. Alcohol
2. Aromatic ring
3. Carboxylic acid
4. Amine
5. Both ketone and amine
6. Two double bonds
Problem 3-3
Identify the functional groups in the following model of arecoline, a veterinary drug used to control worms in animals. Convert the drawing into a line-bond structure and a molecular formula (red = O, blue = N, black = C, gray = H). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.02%3A_Functional_Groups.txt |
Before beginning a systematic study of the different functional groups, let’s look first at the simplest family of molecules to develop some general ideas that apply to all families. We saw in Section 1.7 that the carbon–carbon single bond in ethane results from σ (head-on) overlap of carbon sp3 hybrid orbitals. If we imagine joining three, four, five, or even more carbon atoms by C–C single bonds, we can generate the large family of molecules called alkanes.
Alkanes are often described as saturated hydrocarbons: hydrocarbons because they contain only carbon and hydrogen; saturated because they have only C–C and C–H single bonds and thus contain the maximum possible number of hydrogens per carbon. They have the general formula CnH2n+2, where n is an integer. Alkanes are also occasionally called aliphatic compounds, a name derived from the Greek aleiphas, meaning “fat.” We’ll see in Section 27.1 that many animal fats contain long carbon chains similar to alkanes.
Think about the ways that carbon and hydrogen might combine to make alkanes. With one carbon and four hydrogens, only one structure is possible: methane, CH4. Similarly, there is only one combination of two carbons with six hydrogens (ethane, CH3CH3) and only one combination of three carbons with eight hydrogens (propane, CH3CH2CH3). When larger numbers of carbons and hydrogens combine, however, more than one structure is possible. For example, there are two substances with the formula C4H10: the four carbons can all be in a row (butane), or they can branch (isobutane). Similarly, there are three C5H12 molecules, and so on for larger alkanes.
Compounds like butane and pentane, whose carbons are all connected in a row, are called straight-chain alkanes, or normal alkanes. Compounds like 2-methylpropane (isobutane), 2-methylbutane, and 2,2-dimethylpropane, whose carbon chains branch, are called branched-chain alkanes.
Compounds like the two C4H10 molecules and the three C5H12 molecules, which have the same formula but different structures, are called Isomers, from the Greek isos + meros, meaning “made of the same parts.” Isomers have the same numbers and kinds of atoms but differ in the way the atoms are arranged. Compounds like butane and isobutane, whose atoms are connected differently, are called constitutional isomers. We’ll see shortly that other kinds of isomers are also possible, even among compounds whose atoms are connected in the same order. As Table 3.2 shows, the number of possible alkane isomers increases dramatically with the number of carbon atoms.
Table 3.2 Number of Alkane Isomers
Formula Number of isomers Formula Number of isomers
C6H14 5 C10H22 75
C7H16 9 C15H32 4347
C8H18 18 C20H42 366,319
C9H20 35 C30H62 4,111,846,763
Constitutional isomerism is not limited to alkanes—it occurs widely throughout organic chemistry. Constitutional isomers may have different carbon skeletons (as in isobutane and butane), different functional groups (as in ethanol and dimethyl ether), or different locations of a functional group along the chain (as in isopropylamine and propylamine). Regardless of the reason for the isomerism, constitutional isomers are always different compounds with different properties but with the same formula.
A given alkane can be drawn in many ways. For example, the straight-chain, four-carbon alkane called butane can be represented by any of the structures shown in Figure 3.3. These structures don’t imply any particular three-dimensional geometry for butane; they indicate only the connections among atoms. In practice, as noted in Section 1.12, chemists rarely draw all the bonds in a molecule and usually refer to butane by the condensed structure, CH3CH2CH2CH3 or CH3(CH2)2CH3. Still more simply, butane can be represented as n-C4H10, where n denotes normal (straight-chain) butane.
Straight-chain alkanes are named according to the number of carbon atoms they contain, as shown in Table 3.3. With the exception of the first four compounds—methane, ethane, propane, and butane—whose names have historical roots, the alkanes are named based on Greek numbers. The suffix -ane is added to the end of each name to indicate that the molecule identified is an alkane. Thus, pentane is the five-carbon alkane, hexane is the six-carbon alkane, and so on. We’ll soon see that these alkane names form the basis for naming all other organic compounds, so at least the first ten should be memorized.
Table 3.3 Names of Straight-Chain Alkanes
Number of carbons (n) Name Formula (CnH2n+2) Number of carbons (n) Name Formula (CnH2n+2)
1 Methane CH4 9 Nonane C9H20
2 Ethane C2H6 10 Decane C10H22
3 Propane C3H8 11 Undecane C11H24
4 Butane C4H10 12 Dodecane C12H26
5 Pentane C5H12 13 Tridecane C13H28
6 Hexane C6H14 20 Icosane C20H42
7 Heptane C7H16 30 Triacontane C30H62
8 Octane C8H18
Worked Example 3.1: Drawing the Structures of Isomers
Propose structures for two isomers with the formula C2H7N.
Strategy
We know that carbon forms four bonds, nitrogen forms three, and hydrogen forms one. Write down the carbon atoms first, and then use trial and error plus intuition to put the pieces together.
Solution
There are two isomeric structures. One has the connection C–C–N, and the other has the connection C–N–C.
Problem 3-4
Draw structures of the five isomers of C6H14.
Problem 3-5
Propose structures that meet the following descriptions:
1. Two isomeric esters with the formula C5H10O2:
2. Two isomeric nitriles with the formula C4H7N
3. Two isomeric disulfides with the formula C4H10S2
Problem 3-6
How many isomers are there with the following descriptions?
1. Alcohols with the formula C3H8O
2. Bromoalkanes with the formula C4H9Br
3. Thioesters with the formula C4H8OS
3.04: Alkyl Groups
If you imagine removing a hydrogen atom from an alkane, the partial structure that remains is called an alkyl group. Alkyl groups are not stable compounds themselves, they are simply parts of larger compounds and are named by replacing the -ane ending of the parent alkane with an -yl ending. For example, removal of a hydrogen from methane, CH4, generates a methyl group, –CH3, and removal of a hydrogen from ethane, CH3CH3, generates an ethyl group, –CH2CH3. Similarly, removal of a hydrogen atom from the end carbon of any straight-chain alkane gives the series of straight-chain alkyl groups shown in Table 3.4. Combining an alkyl group with any of the functional groups listed earlier makes it possible to generate and name many thousands of compounds. For example:
Table 3.4 Some Straight-Chain Alkyl Groups
Alkane Name Alkyl group Name (abbreviation)
CH4 Methane –CH3 Methyl (Me)
CH3CH3 Ethane –CH2CH3 Ethyl (Et)
CH3CH2CH3 Propane –CH2CH2CH3 Propyl (Pr)
CH3CH2CH2CH3 Butane –CH2CH2CH2CH3 Butyl (Bu)
CH3CH2CH2CH2CH3 Pentane –CH2CH2CH2CH2CH3 Pentyl, or amyl
Just as straight-chain alkyl groups are generated by removing a hydrogen from an end carbon, branched alkyl groups are generated by removing a hydrogen atom from an internal carbon. Two 3-carbon alkyl groups and four 4-carbon alkyl groups are possible (Figure 3.4).
One further comment about naming alkyl groups: the prefixes sec- (for secondary) and tert- (for tertiary) used for the C4 alkyl groups in Figure 3.4 refer to the number of other carbon atoms attached to the branching carbon atom. There are four possibilities: primary (1°), secondary (2°), tertiary (3°), and quaternary (4°).
The symbol R is used here and throughout organic chemistry to represent a generalized organic group. The R group can be methyl, ethyl, propyl, or any of a multitude of others. You might think of R as representing the Rest of the molecule, which isn’t specified.
The terms primary, secondary, tertiary, and quaternary are routinely used in organic chemistry, and their meanings need to become second nature. For example, if we were to say, “Citric acid is a tertiary alcohol,” we would mean that it has an alcohol functional group (–OH) bonded to a carbon atom that is itself bonded to three other carbons.
In addition to speaking of carbon atoms as being primary, secondary, or tertiary, we speak of hydrogens in the same way. Primary hydrogen atoms are attached to primary carbons (RCH3), secondary hydrogens are attached to secondary carbons (R2CH2), and tertiary hydrogens are attached to tertiary carbons (R3CH). There is, however, no such thing as a quaternary hydrogen. (Why not?)
Problem 3-7
Draw the eight 5-carbon alkyl groups (pentyl isomers).
Problem 3-8
Identify the carbon atoms in the following molecules as primary, secondary, tertiary, or quaternary:
(a)
(b)
(c)
Problem 3-9
Identify the hydrogen atoms on the compounds shown in Problem 3-8 as primary, secondary, or tertiary.
Problem 3-10
Draw structures of alkanes that meet the following descriptions:
1. An alkane with two tertiary carbons
2. An alkane that contains an isopropyl group
3. An alkane that has one quaternary and one secondary carbon | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.03%3A_Alkanes_and_Alkane_Isomers.txt |
In earlier times, when relatively few pure organic chemicals were known, new compounds were named at the whim of their discoverer. Thus, urea (CH4N2O) is a crystalline substance isolated from urine; morphine (C17H19NO3) is an analgesic (painkiller) named after Morpheus, the Greek god of dreams; and acetic acid, the primary organic constituent of vinegar, is named from the Latin word for vinegar, acetum.
As the science of organic chemistry slowly grew in the 19th century, so too did the number of known compounds and the need for a systematic method of naming them. The system of naming (nomenclature) we’ll use in this book is that devised by the International Union of Pure and Applied Chemistry (IUPAC, usually spoken as eye-you-pac).
A chemical name typically has four parts in the IUPAC system: parent, prefix, locant, and suffix. The parent name identifies the main part of the molecule and tells how many carbon atoms are in that part. Prefixes identify the various substituent groups attached to the parent. Locants give the positions of the attached substituents. And the suffix identifies the primary functional group attached to the parent.
As we cover new functional groups in later chapters, the applicable IUPAC rules of nomenclature will be given. In addition, Appendix A at the back of this book gives an overall view of organic nomenclature and shows how compounds that contain more than one functional group are named. (If preferred, you can study that appendix now.) For the present, let’s see how to name branched-chain alkanes and learn some general rules that are applicable to all compounds.
All but the most complex branched-chain alkanes can be named by following four steps. For a very few compounds, a fifth step is needed.
STEP 1
Identify the parent hydrocarbon.
(a) Find the longest continuous chain of carbon atoms in the molecule, and use the name of that chain as the parent name. The longest chain may not always be apparent from the manner of writing; you may have to “turn corners.”
(b) If two different chains of equal length are present, choose the one with the larger number of branch points as the parent.
STEP 2
Number the atoms in the longest chain.
(a) Beginning at the end nearer the first branch point, number each carbon atom in the parent chain.
The first branch occurs at C3 in the proper system of numbering, not at C4.
(b) If there is branching an equal distance away from both ends of the parent chain, begin numbering at the end nearer the second branch point.
STEP 3
Identify and number the substituents.
(a) Assign a number to each substituent to locate its point of attachment to the parent chain.
(b) If there are two substituents on the same carbon, give both the same number. There must be as many numbers in the name as there are substituents.
STEP 4
Write the name as a single word.
Use hyphens to separate the different prefixes, and use commas to separate numbers. If two or more different substituents are present, cite them in alphabetical order. If two or more identical substituents are present on the parent chain, use one of the multiplier prefixes di-, tri-, tetra-, and so forth, but don’t use these prefixes for alphabetizing. Full names for some of the examples we have been using are as follows:
STEP 5
Name a branched substituent as though it were itself a compound.
In some particularly complex cases, a fifth step is necessary. It occasionally happens that a substituent on the main chain is itself branched. In the following case, for instance, the substituent at C6 is a three-carbon chain with a methyl group. To name the compound fully, the branched substituent must first be named.
Number the branched substituent beginning at the point of its attachment to the main chain, and identify it—in this case, a 2-methylpropyl group. The substituent is treated as a whole and is alphabetized according to the first letter of its complete name, including any numerical prefix. It is set off in parentheses when naming the entire molecule.
As a further example:
For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic, common names, as noted earlier.
The common names of these simple alkyl groups are so well entrenched in the chemical literature that IUPAC rules make allowance for them. Thus, the following compound is properly named either 4-(1-methylethyl)heptane or 4-isopropylheptane. There’s no choice but to memorize these common names; fortunately, there are only a few of them.
When writing an alkane name, the nonhyphenated prefix iso- is considered part of the alkyl-group name for alphabetizing purposes, but the hyphenated and italicized prefixes sec- and tert- are not. Thus, isopropyl and isobutyl are listed alphabetically under i, but sec-butyl and tert-butyl are listed under b.
Worked Example 3.2: Naming Alkanes
What is the IUPAC name for the following alkane?
Strategy
Find the longest continuous carbon chain in the molecule, and use that as the parent name. This molecule has a chain of eight carbons—octane—with two methyl substituents. (You have to turn corners to see it.) Numbering from the end nearer the first methyl substituent indicates that the methyls are at C2 and C6.
Worked Example 3.3: Converting a Chemical Name into a Structure
Draw the structure of 3-isopropyl-2-methylhexane.
Strategy
This is the reverse of Worked Example 3.2 and uses a reverse strategy. Look at the parent name (hexane), and draw its carbon structure.
C–C–C–C–C–C Hexane
Next, find the substituents (3-isopropyl and 2-methyl), and place them on the proper carbons.
Finally, add hydrogens to complete the structure.
Solution
Problem 3-11
Give IUPAC names for the following compounds: (a)
(b)
(c)
(d)
Problem 3-12
Draw structures corresponding to the following IUPAC names: (a) 3,4-Dimethylnonane (b) 3-Ethyl-4,4-dimethylheptane
(c) 2,2-Dimethyl-4-propyloctane (d) 2,2,4-Trimethylpentane
Problem 3-13
Name the eight 5-carbon alkyl groups you drew in Problem 3-7.
Problem 3-14
Give the IUPAC name for the following hydrocarbon, and convert the drawing into a skeletal structure. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.05%3A_Naming_Alkanes.txt |
Alkanes are sometimes referred to as paraffins, a word derived from the Latin parum affinis, meaning “little affinity.” This term aptly describes their behavior, for alkanes show little chemical affinity for other substances and are chemically inert to most laboratory reagents. They are also relatively inert biologically and are not often involved in the chemistry of living organisms. Alkanes do, however, react with oxygen, halogens, and a few other substances under appropriate conditions.
Reaction with oxygen occurs during combustion in an engine or furnace when an alkane is used as a fuel. Carbon dioxide and water are formed as products, and a large amount of heat is released. For example, methane (natural gas) reacts with oxygen according to the equation
$CH4+2O2⟶CO2+2H2O+890kJ/mol(213kcal/mol)CH4+2O2⟶CO2+2H2O+890kJ/mol(213kcal/mol)$
The reaction of an alkane with Cl2 occurs when a mixture of the two is irradiated with ultraviolet light (denoted hυ, where υ is the Greek letter nu). Depending on the time allowed and the relative amounts of the two reactants, a sequential substitution of the alkane hydrogen atoms by chlorine occurs, leading to a mixture of chlorinated products. Methane, for instance, reacts with Cl2 to yield a mixture of CH3Cl, CH2Cl2, CHCl3, and CCl4. We’ll look at this reaction in more detail in Section 6.6.
Alkanes show regular increases in both boiling point and melting point as molecular weight increases (Figure 3.5), an effect due to the presence of weak dispersion forces between molecules (Section 2.12). Only when sufficient energy is applied to overcome these forces does the solid melt or liquid boil. As you might expect, dispersion forces increase as molecular size increases, accounting for the higher melting and boiling points of larger alkanes.
Another effect seen in alkanes is that increased branching lowers an alkane’s boiling point. Thus, pentane has no branches and boils at 36.1 °C, isopentane (2-methylbutane) has one branch and boils at 27.85 °C, and neopentane (2,2-dimethylpropane) has two branches and boils at 9.5 °C. Similarly, octane boils at 125.7 °C, whereas isooctane (2,2,4-trimethylpentane) boils at 99.3 °C. Branched-chain alkanes are lower-boiling because they are more nearly spherical than straight-chain alkanes, have smaller surface areas, and consequently have smaller dispersion forces.
3.07: Conformations of Ethane
Up until now, we’ve viewed molecules primarily in a two-dimensional way and have given little thought to any consequences that might arise from the spatial arrangement of atoms in molecules. Now it’s time to add a third dimension to our study. Stereochemistry is the branch of chemistry concerned with the three-dimensional aspects of molecules. We’ll see on many occasions in future chapters that the exact three-dimensional structure of a molecule is often crucial to determining its properties and biological behavior.
We know from Section 1.5 that σ bonds are cylindrically symmetrical. In other words, the intersection of a plane cutting through a carbon–carbon single-bond orbital looks like a circle. Because of this cylindrical symmetry, rotation is possible around carbon–carbon bonds in open-chain molecules. In ethane, for instance, rotation around the C–C bond occurs freely, constantly changing the spatial relationships between the hydrogens on one carbon and those on the other (Figure 3.6).
The different arrangements of atoms that result from bond rotation are called conformations, and molecules that have different arrangements are called conformational isomers, or conformers. Unlike constitutional isomers, however, different conformers often can’t be isolated because they interconvert too rapidly.
Conformational isomers are represented in two ways, as shown in Figure 3.7. A sawhorse representation views the carbon–carbon bond from an oblique angle and indicates spatial orientation by showing all C–H bonds. A Newman projection views the carbon–carbon bond directly end-on and represents the two carbon atoms by a circle. Bonds attached to the front carbon are represented by lines to the center of the circle, and bonds attached to the rear carbon are represented by lines to the edge of the circle.
Despite what we’ve just said, we actually don’t observe perfectly free rotation in ethane. Experiments show that there is a small (12 kJ/mol; 2.9 kcal/mol) barrier to rotation and that some conformations are more stable than others. The lowest-energy, most stable conformation is the one in which all six C–H bonds are as far away from one another as possible—staggered when viewed end-on in a Newman projection. The highest-energy, least stable conformation is the one in which the six C–H bonds are as close as possible—eclipsed in a Newman projection. At any given instant, about 99% of ethane molecules have an approximately staggered conformation and only about 1% are near the eclipsed conformation.
The extra 12 kJ/mol of energy present in the eclipsed conformation of ethane is called torsional strain. Its cause has been the subject of controversy, but the major factor is an interaction between C–H bonding orbitals on one carbon and antibonding orbitals on the adjacent carbon, which stabilizes the staggered conformation relative to the eclipsed one. Because a total strain of 12 kJ/mol arises from three equal hydrogen–hydrogen eclipsing interactions, we can assign a value of approximately 4.0 kJ/mol (1.0 kcal/mol) to each single interaction. The barrier to rotation that results can be represented on a graph of potential energy versus degree of rotation, in which the angle between C–H bonds on the front and back carbons as viewed end-on (the dihedral angle) goes full circle from 0 to 360°. Energy minima occur at staggered conformations, and energy maxima occur at eclipsed conformations, as shown in Figure 3.8. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.06%3A_Properties_of_Alkanes.txt |
Propane, the next-higher member in the alkane series, also has a torsional barrier that results in hindered rotation around the carbon–carbon bonds. The barrier is slightly higher in propane than in ethane—a total of 14 kJ/mol (3.4 kcal/mol) versus 12 kJ/mol.
The eclipsed conformation of propane has three interactions—two ethane-type hydrogen–hydrogen interactions and one additional hydrogen–methyl interaction. Since each eclipsing H ⟷ H interaction is the same as that in ethane and thus has an energy “cost” of 4.0 kJ/mol, we can assign a value of 14 – (2 × 4.0) = 6.0 kJ/mol (1.4 kcal/mol) to the eclipsing H ⟷ CH3 interaction (Figure 3.9).
The conformational situation becomes more complex for larger alkanes because not all staggered conformations have the same energy and not all eclipsed conformations have the same energy. In butane, for instance, the lowest-energy arrangement, called the anti conformation, is the one in which the two methyl groups are as far apart as possible—180° away from each other. As rotation around the C2–C3 bond occurs, an eclipsed conformation is reached where there are two CH3 ⟷ H interactions and one H ⟷ H interaction. Using the energy values derived previously from ethane and propane, this eclipsed conformation is more strained than the anti conformation by 2 × 6.0 kJ/mol + 4.0 kJ/mol (two CH3 ⟷ H interactions plus one H ⟷ H interaction), for a total of 16 kJ/mol (3.8 kcal/mol).
As bond rotation continues, an energy minimum is reached at the staggered conformation where the methyl groups are 60° apart. Called the gauche conformation, it lies 3.8 kJ/mol (0.9 kcal/mol) higher in energy than the anti conformation even though it has no eclipsing interactions. This energy difference occurs because the hydrogen atoms of the methyl groups are near one another in the gauche conformation, resulting in what is called steric strain. Steric strain is the repulsive interaction that occurs when atoms are forced closer together than their atomic radii allow. It’s the result of trying to force two atoms to occupy the same space.
As the dihedral angle between the methyl groups approaches zero, an energy maximum is reached at a second eclipsed conformation. Because the methyl groups are forced even closer together than in the gauche conformation, both torsional strain and steric strain are present. A total strain energy of 19 kJ/mol (4.5 kcal/mol) has been estimated for this conformation, making it possible to calculate a value of 11 kJ/mol (2.6 kcal/mol) for the CH3 ⟷ CH3 eclipsing interaction: total strain of 19 kJ/mol minus the strain of two H ⟷ H eclipsing interactions (2 × 4.0 kcal/mol) equals 11 kJ/mol.
After 0°, the rotation becomes a mirror image of what we’ve already seen: another gauche conformation is reached, another eclipsed conformation, and finally a return to the anti conformation. A plot of potential energy versus rotation about the C2–C3 bond is shown in Figure 3.10.
The notion of assigning definite energy values to specific interactions within a molecule is very useful, and we’ll return to it in the next chapter. A summary of what we’ve seen thus far is given in Table 3.5.
The same principles just developed for butane apply to pentane, hexane, and all higher alkanes. The most favorable conformation for any alkane has the carbon–carbon bonds in staggered arrangements, with large substituents arranged anti to one another. A generalized alkane structure is shown in Figure 3.11.
Table 3.5 Energy Costs for Interactions in Alkane Conformers
Interaction Cause Energy cost
(kJ/mol) (kcal/mol)
H ⟷ H eclipsed Torsional strain 4.0 1.0
H ⟷ CH3 eclipsed Mostly torsional strain 6.0 1.4
CH3 ⟷ CH3 eclipsed Torsional and steric strain 11.0 2.6
CH3 ⟷ CH3 gauche Steric strain 3.8 0.9
One final point: saying that one particular conformer is “more stable” than another doesn’t mean the molecule adopts and maintains only the more stable conformation. At room temperature, rotations around σ bonds occur so rapidly that all conformers are in equilibrium. At any given instant, however, a larger percentage of molecules will be found in a more stable conformation than in a less stable one.
Worked Example 3.4: Newman Projections
Sight along the C1–C2 bond of 1-chloropropane, and draw Newman projections of the most stable and least stable conformations.
Strategy
The most stable conformation of a substituted alkane is generally a staggered one in which large groups have an anti relationship. The least stable conformation is generally an eclipsed one in which large groups are as close as possible.
Problem 3-15
Make a graph of potential energy versus angle of bond rotation for propane, and assign values to the energy maxima.
Problem 3-16 Sight along the C2–C1 bond of 2-methylpropane (isobutane).
1. Draw a Newman projection of the most stable conformation.
2. Draw a Newman projection of the least stable conformation.
3. Make a graph of energy versus angle of rotation around the C2–C1 bond.
4. Assign relative values to the maxima and minima in your graph, given that an H ⟷ H eclipsing interaction costs 4.0 kJ/mol and an H ⟷ CH3 eclipsing interaction costs 6.0 kJ/mol.
Problem 3-17
Sight along the C2–C3 bond of 2,3-dimethylbutane, and draw a Newman projection of the most stable conformation.
Problem 3-18
Draw a Newman projection along the C2–C3 bond of the following conformation of 2,3-dimethylbutane, and calculate a total strain energy: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.08%3A_Conformations_of_Other_Alkanes.txt |
British Foreign Minister Ernest Bevin once said that “The Kingdom of Heaven runs on righteousness, but the Kingdom of Earth runs on alkanes.” (Actually, he said “runs on oil” not “runs on alkanes,” but they’re essentially the same.) By far, the major sources of alkanes are the world’s natural gas and petroleum deposits. Laid down eons ago, these deposits are thought to be derived primarily from the decomposition of tiny single-celled marine organisms called foraminifera. Natural gas consists chiefly of methane but also contains ethane, propane, and butane. Petroleum is a complex mixture of hydrocarbons that must be separated into fractions and then further refined before it can be used.
The petroleum era began in August 1859, when the world’s first oil well was drilled by Edwin Drake near Titusville, Pennsylvania. The petroleum was distilled into fractions according to boiling point, but it was high-boiling kerosene, or lamp oil, rather than gasoline that was primarily sought. Literacy was becoming widespread at the time, and people wanted better light for reading than was available from candles. Gasoline was too volatile for use in lamps and was initially considered a waste by-product. The world has changed greatly since those early days, however, and it is now gasoline rather than lamp oil that is prized.
Petroleum refining begins by fractional distillation of crude oil into three principal cuts according to boiling point (bp): straight-run gasoline (bp 30–200 °C), kerosene (bp 175–300 °C), and heating oil, or diesel fuel (bp 275–400 °C). Further distillation under reduced pressure then yields lubricating oils and waxes and leaves a tarry residue of asphalt. The distillation of crude oil is only the first step in gasoline production, however. Straight-run gasoline turns out to be a poor fuel in automobiles because of engine knock, an uncontrolled combustion that can occur in a hot engine causing potentially serious damage.
The octane number of a fuel is the measure by which its antiknock properties are judged. It was recognized long ago that straight-chain hydrocarbons are far more prone to inducing engine knock than highly branched compounds. Heptane, a particularly bad fuel, is assigned a base value of 0 octane number, and 2,2,4-trimethylpentane, commonly known as isooctane, has a rating of 100.
Because straight-run gasoline burns so poorly in engines, petroleum chemists have devised numerous methods for producing higher-quality fuels. One of these methods, catalytic cracking, involves taking the high-boiling kerosene cut (C11–C14) and “cracking” it into smaller branched molecules suitable for use in gasoline. Another process, called reforming, is used to convert C6–C8 alkanes to aromatic compounds such as benzene and toluene, which have substantially higher octane numbers than alkanes. The final product that goes in your tank has an approximate composition of 15% C4–C8 straight-chain alkanes, 25% to 40% C4–C10 branched-chain alkanes, 10% cyclic alkanes, 10% straight-chain and cyclic alkenes, and 25% arenes (aromatics).
3.10: Key Terms
• alcohol
• aldehyde
• aliphatic
• alkane
• alkene
• alkyl group
• alkyl halide
• alkyne
• amide
• amine
• anti conformation
• arene
• branched-chain alkane
• carbonyl group
• carboxylic acid
• conformation
• conformational isomer
• conformer
• constitutional isomer
• eclipsed conformation
• ester
• ether
• functional group
• gauche conformation
• hydrocarbon
• isomer
• ketone
• Newman projection
• nitrile
• R group
• saturated
• sawhorse representation
• staggered conformation
• stereochemistry
• steric strain
• straight-chain alkane
• substituent
• sulfide
• thiol
• torsional strain
3.11: Summary
Alkanes are relatively unreactive and rarely involved in chemical reactions, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we’ve used alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules.
A functional group is a group of atoms within a larger molecule that has a characteristic chemical reactivity. Because functional groups behave in approximately the same way in all molecules where they occur, the chemical reactions of an organic molecule are largely determined by its functional groups.
Alkanes are a class of saturated hydrocarbons with the general formula CnH2n+2. They contain no functional groups, are relatively inert, and can be either straight-chain (normal) or branched. Alkanes are named by a series of IUPAC rules of nomenclature. Compounds that have the same chemical formula but different structures are called isomers. More specifically, compounds such as butane and isobutane, which differ in their connections between atoms, are called constitutional isomers.
Carbon–carbon single bonds in alkanes are formed by σ overlap of carbon sp3 hybrid orbitals. Rotation is possible around σ bonds because of their cylindrical symmetry, and alkanes therefore exist in a large number of rapidly interconverting conformations. Newman projections make it possible to visualize the spatial consequences of bond rotation by sighting directly along a carbon–carbon bond axis. Not all alkane conformations are equally stable. The staggered conformation of ethane is 12 kJ/mol (2.9 kcal/mol) more stable than the eclipsed conformation because of torsional strain. In general, any alkane is most stable when all its bonds are staggered. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.09%3A_Chemistry_MattersGasoline.txt |
Visualizing Chemistry
Problem 3-19
Identify the functional groups in the following substances, and convert each drawing into a molecular formula (red = O, blue = N).
(a)
(b)
Problem 3-20 Give IUPAC names for the following alkanes, and convert each drawing into a skeletal structure.
(a)
(b)
(c)
(d)
Problem 3-21
Draw a Newman projection along the C2–C3 bond of the following conformation of 2-butanol.
Functional Groups
Problem 3-22 Locate and identify the functional groups in the following molecules.
(a)
(b)
(c)
(d)
(e)
(f)
Problem 3-23 Propose structures that meet the following descriptions: (a) A ketone with five carbons
(b) A four-carbon amide
(c) A five-carbon ester (d) An aromatic aldehyde (e) A keto ester (f) An amino alcohol
Problem 3-24 Propose structures for the following:
(a) A ketone, C4H8O
(b) A nitrile, C5H9N (c) A dialdehyde, C4H6O2 (d) A bromoalkene, C6H11Br (e) An alkane, C6H14 (f) cyclic saturated hydrocarbon, C6H12 (g) A diene (dialkene), C5H8 (h) A keto alkene, C5H8O
Problem 3-25 Predict the hybridization of the carbon atom in each of the following functional groups:
(a) Ketone
(b) Nitrile (c) Carboxylic acid
Problem 3-26 Draw the structures of the following molecules:
(a) Biacetyl, C4H6O2, a substance with the aroma of butter; it contains no rings or carbon–carbon multiple bonds.
(b) Ethylenimine, C2H5N, a substance used in the synthesis of melamine polymers; it contains no multiple bonds. (c) Glycerol, C3H8O3, a substance isolated from fat and used in cosmetics; it has an –OH group on each carbon.
Isomers
Problem 3-27 Draw structures that meet the following descriptions (there are many possibilities):
(a) Three isomers with the formula C8H18
(b) Two isomers with the formula C4H8O2
Problem 3-28
Draw structures of the nine isomers of C7H16.
(a)
(b)
(c)
Problem 3-30
Seven constitutional isomers have the formula C4H10O. Draw as many as you can.
Problem 3-31 Draw as many compounds as you can that fit the following descriptions:
(a) Alcohols with formula C4H10O
(b) Amines with formula C5H13N
(c) Ketones with formula C5H10O (d) Aldehydes with formula C5H10O (e) Esters with formula C4H8O2 (f) Ethers with formula C4H10O
Problem 3-32 Draw compounds that contain the following:
(a) A primary alcohol
(b) A tertiary nitrile (c) A secondary thiol (d) Both primary and secondary alcohols (e) An isopropyl group (f) A quaternary carbon
Naming Compounds
Problem 3-33
Draw and name all monobromo derivatives of pentane, C5H11Br.
Problem 3-34
Draw and name all monochloro derivatives of 2,5-dimethylhexane, C8H17Cl.
Problem 3-35 Draw structures for the following:
(a) 2-Methylheptane
(b) 4-Ethyl-2,2-dimethylhexane (c) 4-Ethyl-3,4-dimethyloctane (d) 2,4,4-Trimethylheptane (e) 3,3-Diethyl-2,5-dimethylnonane (f) 4-Isopropyl-3-methylheptane
Problem 3-36 Draw a compound that:
(a) Has only primary and tertiary carbons
(b) Has no secondary or tertiary carbons (c) Has no secondary or tertiary carbons
Problem 3-37 Draw a compound that:
(a) Has nine primary hydrogens
(b) Has only primary hydrogens
(a)
(b)
(c)
(d)
(e)
(f)
Problem 3-39
Name the five isomers of C6H14.
Problem 3-40 Explain why each of the following names is incorrect:
(a) 2,2-Dimethyl-6-ethylheptane
(b) 4-Ethyl-5,5-dimethylpentane
(c) 3-Ethyl-4,4-dimethylhexane (d) 5,5,6-Trimethyloctane (e) 2-Isopropyl-4-methylheptane
Problem 3-41 Propose structures and give IUPAC names for the following:
(a) A diethyldimethylhexane
(b) A (3-methylbutyl)-substituted alkane
Conformations
Problem 3-42 Consider 2-methylbutane (isopentane). Sighting along the C2–C3 bond:
(a) Draw a Newman projection of the most stable conformation.
(b) Draw a Newman projection of the least stable conformation. (c) If a CH3 ⟷ CH3 eclipsing interaction costs 11 kJ/mol (2.5 kcal/mol) and a CH3 ⟷ CH3 gauche interaction costs 3.8 kJ/mol (0.9 kcal/mol), make a quantitative plot of energy versus rotation about the C2–C3 bond.
Problem 3-43
What are the relative energies of the three possible staggered conformations around the C2–C3 bond in 2,3-dimethylbutane? (See Problem 3-42.)
Problem 3-44
Construct a qualitative potential-energy diagram for rotation about the C–C bond of 1,2-dibromoethane. Which conformation would you expect to be most stable? Label the anti and gauche conformations of 1,2-dibromoethane.
Problem 3-45
Which conformation of 1,2-dibromoethane (Problem 3-44) would you expect to have the largest dipole moment? The observed dipole moment of 1,2-dibromoethane is μ = 1.0 D. What does this tell you about the actual conformation of the molecule?
Problem 3-46
Draw the most stable conformation of pentane, using wedges and dashes to represent bonds coming out of the paper and going behind the paper, respectively.
Problem 3-47
Draw the most stable conformation of 1,4-dichlorobutane, using wedges and dashes to represent bonds coming out of the paper and going behind the paper, respectively.
General Problems
(a)
(b)
(c)
(d)
(e)
(f)
Problem 3-49 Malic acid, C4H6O5, has been isolated from apples. Because this compound reacts with 2 molar equivalents of base, it is a dicarboxylic acid.
(a) Draw at least five possible structures.
(b) If malic acid is a secondary alcohol, what is its structure?
Problem 3-50
Formaldehyde, H2C$\text{=}$O, is known to all biologists because of its usefulness as a tissue preservative. When pure, formaldehyde trimerizes to give trioxane, C3H6O3, which, surprisingly enough, has no carbonyl groups. Only one monobromo derivative (C3H5BrO3) of trioxane is possible. Propose a structure for trioxane.
Problem 3-51 The barrier to rotation about the C–C bond in bromoethane is 15 kJ/mol (3.6 kcal/mol).
(a) What energy value can you assign to an H ↔ Br eclipsing interaction?
(b) Construct a quantitative diagram of potential energy versus bond rotation for bromoethane.
Problem 3-52 Increased substitution around a bond leads to increased strain. Take the four substituted butanes listed below, for example. For each compound, sight along the C2–C3 bond and draw Newman projections of the most stable and least stable conformations. Use the data in Table 3.5 to assign strain-energy values to each conformation. Which of the eight conformations is most strained? Which is least strained?
(a) 2-Methylbutane
(b) 2,2-Dimethylbutane (c) 2,3-Dimethylbutane (d) 2,2,3-Trimethylbutane
Problem 3-53
The cholesterol-lowering agents called statins, such as simvastatin (Zocor) and pravastatin (Pravachol), are among the most widely prescribed drugs in the world, with annual sales estimated at approximately \$25 billion. Identify the functional groups in both, and tell how the two substances differ.
Problem 3-54
In the next chapter we’ll look at cycloalkanes—saturated cyclic hydrocarbons—and we’ll see that the molecules generally adopt puckered, nonplanar conformations. Cyclohexane, for instance, has a puckered shape like a lounge chair rather than a flat shape. Why?
Problem 3-55
We’ll see in the next chapter that there are two isomeric substances, both named 1,2-dimethylcyclohexane. Explain. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.12%3A_Additional_Problems.txt |
Learning Objectives
After you have completed Chapter 4, you should be able to
1. fulfill all of the detailed objectives listed under each individual section.
2. draw the cis-trans isomers of some simple disubstituted cycloalkanes, and write the IUPAC names of such compounds.
3. define, and use in context, the key terms introduced in this chapter.
This chapter deals with the concept of stereochemistry and conformational analysis in cyclic compounds. The causes of various ring strains and their effects on the overall energy level of a cycloalkane are discussed. We shall stress the stereochemistry of alicyclic compounds.
• 4.1: Why This Chapter?
We’ll see numerous instances in future chapters where the chemistry of a given functional group is affected by being in a ring rather than an open chain. Because cyclic molecules are encountered in most pharmaceuticals and in all classes of biomolecules, including proteins, lipids, carbohydrates, and nucleic acids, it’s important to understand the behavior of cyclic structures.
• 4.2: Naming Cycloalkanes
Cycloalkanes have one or more rings of carbon atoms, and contain only carbon-hydrogen and carbon-carbon single bonds. The naming of cycloalkanes follows a set of rules similar to that used for naming alkanes.
• 4.3: Cis-Trans Isomerism in Cycloalkanes
Stereoisomers are molecules that have the same molecular formula, the same atom connectivity, but they differ in the relative spatial orientation of the atoms. Di-substituted cycloalkanes are one class of molecules that exhibit stereoisomerism. Di-substituted cycloalkane stereoisomers are designated by the nomenclature prefixes cis (Latin, meaning on this side) and trans (Latin, meaning across).
• 4.4: Stability of Cycloalkanes - Ring Strain
Small cycloalkanes, like cyclopropane, are dramatically less stable than larger cycloalkanes due to ring strain. Ring strain is caused by increased torsional strain, steric strain, and angle strain, in the small, nearly planar ring of cyclopropane. Larger rings like cyclohexane, have much lower ring straing because they adopt non-planar conformations.
• 4.5: Conformations of Cycloalkanes
Overall ring strain decreases in cycloalkane rings that are large enough to allow the carbon-carbon bonds to rotate away from planar structures. For this reason, cyclopentane is significantly more stable, than cyclopropane and cyclobutane.
• 4.6: Conformations of Cyclohexane
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.
• 4.7: Axial and Equatorial Bonds in Cyclohexane
The hydrogens of cyclohexane exist in two distinct locations - axial and equatorial. The two chair conformations of cyclohexane rapidly interconverts through a process called ring flip.
• 4.8: Conformations of Monosubstituted Cyclohexanes
Mono-substituted cyclohexane prefers the ring flip conformer in which the substituent is equatorial. 1,3-diaxial interactions occur when the substituent is axial, instead of equatorial. The larger the substituent, the more pronounced the preference.
• 4.9: Conformations of Disubstituted Cyclohexanes
The most stable configurational isomer of a disubstituted cyclohexane will be the isomer that has the most stable conformational isomer.
• 4.10: Conformations of Polycyclic Molecules
Polycyclic molecules are common and important in nature. Biologically important polycyclic molecules are found in cholesterol, sex hormones, birth control pills, cortisone, and anabolic steroids
• 4.11: Chemistry Matters—Molecular Mechanics
• 4.12: Key Terms
• 4.13: Summary
• 4.14: Additional Problems
Thumbnail: Ball-and-stick model of cyclobutane. (Public Domain; Ben Mills via Wikipedia)
04: Organic Compounds - Cycloalkanes and their Stereochemistry
We’ll see numerous instances in future chapters where the chemistry of a given functional group is affected by being in a ring rather than an open chain. Because cyclic molecules are encountered in most pharmaceuticals and in all classes of biomolecules, including proteins, lipids, carbohydrates, and nucleic acids, it’s important to understand the behavior of cyclic structures.
Although we’ve only discussed open-chain compounds up to now, most organic compounds contain rings of carbon atoms. Chrysanthemic acid, for instance, whose esters occur naturally as the active insecticidal constituents of chrysanthemum flowers, contains a three-membered (cyclopropane) ring.
Prostaglandins, potent hormones that control an extraordinary variety of physiological functions in humans, contain a five-membered (cyclopentane) ring.
Steroids, such as cortisone, contain four rings joined together—three six-membered (cyclohexane) and one five-membered. We’ll discuss steroids and their properties in more detail in Sections 27.6 and 27.7. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.01%3A_Why_This_Chapter.txt |
Saturated cyclic hydrocarbons are called cycloalkanes, or alicyclic compounds (aliphatic cyclic). Because cycloalkanes consist of rings of −CH2−units, they have the general formula (CH2)n, or CnH2n, and can be represented by polygons in skeletal drawings.
Substituted cycloalkanes are named by rules similar to those we saw in (Section 3.4) for open-chain alkanes. For most compounds, there are only two steps.
STEP 1
Find the parent.
Count the number of carbon atoms in the ring and the number in the largest substituent. If the number of carbon atoms in the ring is equal to or greater than the number in the substituent, the compound is named as an alkyl-substituted cycloalkane. If the number of carbon atoms in the largest substituent is greater than the number in the ring, the compound is named as a cycloalkyl-substituted alkane. For example:
STEP 2
Number the substituents, and write the name.
For an alkyl- or halo-substituted cycloalkane, choose a point of attachment as carbon 1 and number the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found.
(a) When two or more different alkyl groups are present that could potentially take the same numbers, number them by alphabetical priority, ignoring numerical prefixes such as di- and tri-.
(b) If halogens are present, treat them just like alkyl groups.
Some additional examples follow:
Problem 4-1
Give IUPAC names for the following cycloalkanes:
(a)
(b)
(c)
(d)
(e)
(f)
Problem 4-2 Draw structures corresponding to the following IUPAC names:
1. 1,1-Dimethylcyclooctane
2. 3-Cyclobutylhexane
3. 1,2-Dichlorocyclopentane
4. 1,3-Dibromo-5-methylcyclohexane
Problem 4-3
Name the following cycloalkane:
4.03: Cis-Trans Isomerism in Cycloalkanes
In many respects, the chemistry of cycloalkanes is like that of open-chain alkanes: both are nonpolar and fairly inert. There are, however, some important differences. One difference is that cycloalkanes are less flexible than open-chain alkanes. In contrast with the relatively free rotation around single bonds in open-chain alkanes (Section 3.6 and Section 3.7), there is much less freedom in cycloalkanes. Cyclopropane, for example, must be a rigid, planar molecule because three points (the carbon atoms) define a plane. No bond rotation can take place around a cyclopropane carbon–carbon bond without breaking open the ring (Figure 4.2).
Larger cycloalkanes have increasing rotational freedom, and very large rings (C25 and up) are so floppy that they are nearly indistinguishable from open-chain alkanes. The common ring sizes (C3–C7), however, are severely restricted in their molecular motions.
Because of their cyclic structures, cycloalkanes have two faces when viewed edge-on, a “top” face and a “bottom” face. As a result, isomerism is possible in substituted cycloalkanes. For example, there are two different 1,2-dimethylcyclopropane isomers, one with the two methyl groups on the same face of the ring and one with the methyl groups on opposite faces (Figure 4.3). Both isomers are stable compounds, and neither can be converted into the other without breaking and reforming chemical bonds.
Unlike the constitutional isomers butane and isobutane, which have their atoms connected in a different order (Section 3.2), the two 1,2-dimethylcyclopropanes have the same order of connections but differ in the spatial orientation of the atoms. Such compounds, with atoms connected in the same order but differing in three-dimensional orientation, are called stereochemical isomers, or stereoisomers. As we saw in Section 3.6, the term stereochemistry is used generally to refer to the three-dimensional aspects of structure and reactivity.
The 1,2-dimethylcyclopropanes are members of a subclass of stereoisomers called cis–trans isomers. The prefixes cis- (Latin “on the same side”) and trans- (Latin “across”) are used to distinguish between them. Cis–trans isomerism is a common occurrence in substituted cycloalkanes and in many cyclic biological molecules.
Worked Example 4.1: Naming Cycloalkanes
Name the following substances, including the cis- or trans- prefix:
Strategy
In these views, the ring is roughly in the plane of the page, a wedged bond protrudes out of the page, and a dashed bond recedes into the page. Two substituents are cis if they are both out of or both into the page, and they are trans if one is out of and one is into the page.
Solution
(a) trans-1,3-Dimethylcyclopentane
(b) cis-1,2-Dichlorocyclohexane
(a)
(b)
Problem 4-5 Draw the structures of the following molecules:
1. trans-1-Bromo-3-methylcyclohexane
2. cis-1,2-Dimethylcyclobutane
3. trans-1-tert-Butyl-2-ethylcyclohexane
Problem 4-6
Prostaglandin F2α, a hormone that causes uterine contraction during childbirth, has the following structure. Are the two hydroxyl groups (–OH) on the cyclopentane ring cis or trans to each other? What about the two carbon chains attached to the ring?
Problem 4-7 Name the following substances, including the cis- or trans- prefix (red-brown = Br): (a)
(b) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.02%3A_Naming_Cycloalkanes.txt |
Chemists in the late 1800s knew that cyclic molecules existed, but the limitations on ring size were unclear. Although numerous compounds containing five-membered and six-membered rings were known, smaller and larger ring sizes had not been prepared, despite many attempts.
A theoretical interpretation of this observation was proposed in 1885 by Adolf von Baeyer, who suggested that small and large rings might be unstable due to angle strain—the strain induced in a molecule when bond angles are forced to deviate from the ideal 109° tetrahedral value. Baeyer based his suggestion on the simple geometric notion that a three-membered ring (cyclopropane) should be an equilateral triangle with bond angles of 60° rather than 109°, a four-membered ring (cyclobutane) should be a square with bond angles of 90°, a five-membered ring should be a regular pentagon with bond angles of 108°, and so on. Continuing this argument, large rings should be strained by having bond angles that are much greater than 109°.
What are the facts? To measure the amount of strain in a compound, we have to measure the total energy of the compound and then subtract the energy of a strain-free reference compound. The difference between the two values should represent the amount of extra energy in the molecule due to strain. The simplest experimental way to do this for a cycloalkane is to measure its heat of combustion, the amount of heat released when the compound burns completely with oxygen. The more energy (strain) the compound contains, the more energy (heat) is released by combustion.
$\ce{(CH2)_{n} + {3 n}/{2} O2 -> n CO2 + n H2O + Heat (add)} \nonumber$
Because the heat of combustion of a cycloalkane depends on size, we need to look at heats of combustion per CH2 unit. Subtracting a reference value derived from a strain-free acyclic alkane and then multiplying by the number of CH2 units in the ring gives the overall strain energy. Figure 4.4 shows the results.
The data in Figure 4.4 show that Baeyer’s theory is only partially correct. Cyclopropane and cyclobutane are indeed strained, just as predicted, but cyclopentane is more strained than predicted, and cyclohexane is strain-free. Cycloalkanes of intermediate size have only modest strain, and rings of 14 carbons or more are strain-free. Why is Baeyer’s theory wrong?
Baeyer’s theory is wrong for the simple reason that he assumed all cycloalkanes to be flat. In fact, as we’ll see in the next section, most cycloalkanes are not flat; instead, they adopt puckered three-dimensional conformations that allow bond angles to be nearly tetrahedral. As a result, angle strain occurs only in three- and four-membered rings, which have little flexibility. For most ring sizes, particularly the medium-ring (C7–C11) cycloalkanes, torsional strain caused by H ⟷ H eclipsing interactions at adjacent carbons (Section 3.6) and steric strain caused by the repulsion between nonbonded atoms that approach too closely (Section 3.7) are the most important factors. Thus, three kinds of strain contribute to the overall energy of a cycloalkane.
• Angle strain—the strain due to expansion or compression of bond angles
• Torsional strain—the strain due to eclipsing of bonds between neighboring atoms
• Steric strain—the strain due to repulsive interactions when atoms approach each other too closely
Problem 4-8
Each H ⟷ H eclipsing interaction in ethane costs about 4.0 kJ/mol. How many such interactions are present in cyclopropane? What fraction of the overall 115 kJ/mol (27.5 kcal/mol) strain energy of cyclopropane is due to torsional strain?
Problem 4-9
cis-1,2-Dimethylcyclopropane has more strain than trans-1,2-dimethylcyclopropane. How can you account for this difference? Which of the two compounds is more stable? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.04%3A_Stability_of_Cycloalkanes_-_Ring_Strain.txt |
Cyclopropane
Cyclopropane is the most strained of all rings, primarily due to the angle strain caused by its 60° C−C−C bond angles. In addition, cyclopropane has considerable torsional strain because the C−H bonds on neighboring carbon atoms are eclipsed (Figure 4.5).
How can the hybrid-orbital model of bonding account for the large distortion of bond angles from the normal 109° tetrahedral value to 60° in cyclopropane? The answer is that cyclopropane has bent bonds. In an unstrained alkane, maximum bonding is achieved when two atoms have their overlapping orbitals pointing directly toward each other. In cyclopropane, though, the orbitals can’t point directly toward each other; instead, they overlap at a slight angle. The result is that cyclopropane bonds are weaker and more reactive than typical alkane bonds—255 kJ/mol (61 kcal/mol) for a C−C bond in cyclopropane versus 370 kJ/mol (88 kcal/mol) for a C−C bond in open-chain propane.
Cyclobutane
Cyclobutane has less angle strain than cyclopropane but has more torsional strain because of its larger number of ring hydrogens. As a result, the total strain for the two compounds is nearly the same—110 kJ/mol (26.4 kcal/mol) for cyclobutane versus 115 kJ/mol (27.5 kcal/mol) for cyclopropane. Cyclobutane is not quite flat but is slightly bent so that one carbon atom lies about 25° above the plane of the other three (Figure 4.6). The effect of this slight bend is to increase angle strain but to decrease torsional strain, until a minimum-energy balance between the two opposing effects is achieved.
Cyclopentane
Cyclopentane was predicted by Baeyer to be nearly strain-free, but it actually has a total strain energy of 26 kJ/mol (6.2 kcal/mol). Although planar cyclopentane has practically no angle strain, it has a large torsional strain. Cyclopentane therefore twists to adopt a puckered, nonplanar conformation that strikes a balance between increased angle strain and decreased torsional strain. Four of the cyclopentane carbon atoms are in approximately the same plane, with the fifth carbon atom bent out of the plane. Most of the hydrogens are nearly staggered with respect to their neighbors (Figure 4.7).
Problem 4-10
How many H ⟷ H eclipsing interactions would be present if cyclopentane were planar? Assuming an energy cost of 4.0 kJ/mol for each eclipsing interaction, how much torsional strain would planar cyclopentane have? Since the measured total strain of cyclopentane is 26 kJ/mol, how much of the torsional strain is relieved by puckering?
Problem 4-11
Two conformations of cis-1,3-dimethylcyclobutane are shown. What is the difference between them, and which do you think is likely to be more stable?
(a) (b)
4.06: Conformations of Cyclohexane
Substituted cyclohexanes are the most common cycloalkanes and occur widely in nature. A large number of compounds, including steroids and many pharmaceutical agents, have cyclohexane rings. The flavoring agent menthol, for instance, has three substituents on a six-membered ring.
Cyclohexane adopts a strain-free, three-dimensional shape that is called a chair conformation because of its similarity to a lounge chair, with a back, seat, and footrest (Figure 4.8). Chair cyclohexane has neither angle strain nor torsional strain—all C−C−C bond angles are near the 109° tetrahedral value, and all neighboring C−H bonds are staggered.
The easiest way to visualize chair cyclohexane is to build a molecular model if you have access to a model kit, or alternatively to explore with one of the many computer-based modeling programs you may have access to.
The chair conformation of cyclohexane can be drawn in three steps.
STEP 1
Draw two parallel lines, slanted downward and slightly offset from each other. This means that four of the cyclohexane carbons lie in a plane.
STEP 2
Place the topmost carbon atom above and to the right of the plane of the other four, and connect the bonds.
STEP 3
Place the bottommost carbon atom below and to the left of the plane of the middle four, and connect the bonds. Note that the bonds to the bottommost carbon atom are parallel to the bonds to the topmost carbon.
When viewing cyclohexane, it’s helpful to remember that the lower bond is in front and the upper bond is in back. If this convention isn’t defined, it can appear that the reverse is true. For clarity, all cyclohexane rings drawn in this book will have the front (lower) bond heavily shaded to indicate nearness to the viewer.
In addition to the chair conformation of cyclohexane, there is an alternative conformation of cyclohexane that bears a slight resemblance to a twisted boat. Called the twist-boat conformation, it is nearly free of angle strain. It does, however, have both steric strain and torsional strain and is about 23 kJ/mol (5.5 kcal/mol) higher in energy than the chair conformation. As a result, molecules adopt the twist-boat geometry only rarely. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_(OpenStax)/04%3A_Organic_Compounds_-_Cycloalkanes_and_their_Stereochemistry/4.05%3A_Conformations_of_Cycloalkanes.txt |
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