Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

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Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

In-book Interactivity

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Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

About this textbook

Lead Authors

Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

Contributing Authors

Felix NgassaGrand Valley State University

Neil GargUCLA

Jennifer ChaytorSaginaw Valley State University

Greg DomskiAugustana College

Christian E. MaduCollin Community College

Christopher NicholsonUniversity of West Florida

Franklin OwEast Los Angeles College, UCLA

Robert S. PhillipsUniversity of Georgia

Grigoriy SeredaUniversity of South Dakota

Simon E. LopezUniversity of Florida

Brannon McCulloughNorthern Arizona University

Jason JonesKennesaw State University

José BoquinAugustana College

Stephanie BrouetSaginaw Valley State University

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Chapter 26: Condensations and Alpha Substitutions of Carbonyl Compounds

2-Heptanone, a carbonyl compound, occurs naturally in different types of cheese. [1]


Contents

Learning Objectives

  • Be able to write mechanisms of enolization under acidic and basic conditions.
  • Compare relative energies of enols and their carbonyl counterparts.
  • Be able to write mechanisms of direct aldol addition and condensation under acidic and basic conditions.
  • Be able to write mechanism of Claisen condensation.
  • Compare acidity of carbonyl compounds and β-dicarbonyl compounds and explain the difference.
  • Write the mechanism for nucleophilic addition to the C=C bond and understand when such addition is possible.
  • Understand the role of enols, enolates, and enamines in the reactions of alkylation and halogenation of carbonyl compounds.
  • Apply Michael addition to enolate nucleophiles.
  • Use reactions of enols, enolates, and enamines in planning organic synthesis.
  • Understand the role of aldol reactions, retro-aldol reactions, and enol condensations in biochemistry.
  • Know how to make ketones from β-ketoesters and carboxylic acids from malonic ester.
  • Use Michael addition to design organic syntheses.

26.1 Reactions of Enols and Enolate Ions

26.1.1 Keto-Enol Tautomerism

The carbon atom directly bonded to a carbonyl group is called an α-carbon and the hydrogens bonded to an α-carbon are referred to as α-hydrogens (alpha-hydrogens). Proximity of the α-carbon to the carbonyl group makes the alpha position an important reaction site for aldehydes and ketones. Recall that aldehydes and ketones with α-hydrogens are in equilibrium with the corresponding enol form in which there is a vinylic hydroxyl group on a carbon-carbon double bond. The interconversion between the ketoform and the enol form is called keto-enol tautomerism (Figure 26.1).

Figure 26.1. Keto-enol tautomerism.

A keto form, however, is not necessarily a ketone. The term here merely denotes a carbonyl-containing tautomer of an enol (recall that this was introduced in the chapter on Aldehydes and Ketones). 

Carbonyl groups have a significant impact on the reactivity of atoms in the α-position. In the presence of a strong base, for example NaOH, an α-hydrogen can be abstracted, forming an equilibrium with the corresponding carbanion (Figure 26.2). 

Figure 26.2. An acid-base equilibrium for the generation of carbanion.


Recall that in carboxylic acid derivatives, an electron pair on the atom next to the carbonyl group is conjugated with the C=O bond—as, for example, an oxygen atom in esters or a nitrogen atom in amides. If the electron pair belongs to a carbon atom, as in the above carbanion, resonance stabilization is also taking place. Therefore, the carbanion is resonance-stabilized, with the negative charge delocalized between the α-carbon and the carbonyl oxygen (Figure 26.3).

Figure 26.3. Resonance stabilization of enolate​.

This stabilization makes the α-hydrogen much more acidic than a hydrogen of a typical H-C bond. For example, the pKa of a α-hydrogen of an aldehyde or ketone is around 20, compared to the carbon-hydrogen bond of an alkane that has a pKa of about 50. Resonance structure B, the resonance form with the negative charge on the oxygen atom, is a better contributor than resonance structure A. This is because a more electronegative oxygen atom is able to accommodate a negative charge better than a carbon atom.The carbanion is referred to as an enolate ion because it is the anion formed upon deprotonation of an enol.


Q26.1

Which of the following are enolates of 5-hydroxy-2-pentanone?

question description
Premise
Response
1

A

A

Correct

2

B

B

Incorrect

3

C

C

Correct

4

D

D

Incorrect


If a ketone has more than one alpha-position with hydrogens, then structurally isomeric enols can be formed. In addition, cis-trans isomerism around the C=C bond is possible. For example, 2-methylcyclohexanone has two enols which are constitutional isomers. For propanal, there are two enol stereoisomers possible: Z and E. For 2-methycyclohexanone, the E-stereoisomer is locked in by the cyclic structure, so no Z-stereoisomer is possible (Figure 26.4).

Figure 26.4. Keto-enol tautomerism for 2-methylcyclohexanone and propanal​.



Q26.2

Which of the following compounds can produce three enolates?

A

Propanone

B

Propanal

C

2-Butanone

D

Butanal

E

Methanal

As a Brønsted base, the enolate can be protonated either at the oxygen or carbon, giving rise to either enol or ketoforms, respectively. Therefore, an enol and its keto form, being tautomers, are sharing a conjugate base, the enolate, which usually are rapidly interconverting in the presence of a trace amount of either base or acid. This equilibrium is an example of tautomerism, a rapid interconversion between constitutional isomers. The role of a basic catalyst in enolization is conversion of either the enol or its ketoform to the enolate (Figure 26.5). 


Figure 26.5. Keto-enol tautomerization under alkaline conditions​​.


An acid also can catalyze enolization by protonation of either enol or its ketoform to a shared conjugate acid (Figure 26.6).

Figure 26.6. Keto-enol tautomerization under acidic conditions​.

The relative energies of the enol and its ketoform determine positions of both the keto-enol equilibrium and acid-base equilibria. The acid-base equilibria favors the weaker acid, while disfavoring the stronger acid. The keto-enol equilibrium is shifted from the less stable form to the more stable form. Therefore, the less stable form is a stronger acid than the more stable form.  

Q26.3

The equilibrium constant for enolization of ethanal is 6X1076X10^{-7}. Which of the following is true?

question description
A

Ethanal is a stronger acid than its enol because the equilibrium is shifted toward the enol

B

Ethanal is a stronger acid than its enol because the equilibrium is shifted toward ethanal

C

Ethanal is a weaker acid than its enol because the equilibrium is shifted toward the enol

D

Ethanal is a weaker acid than its enolform because the equilibrium is shifted toward ethanal

E

Ethanal is a stronger acid than its enol because the hydrogen attached to the carbonyl group is highly acidic


Q26.4

The equilibrium constant for enolization of ethanal is 6X107^{-7}. Which of the following is true?

question description
A

Ethanal is a stronger base than its enol because the equilibrium is shifted toward the enol

B

Ethanal is a stronger base than its enol because the equilibrium is shifted toward ethanal

C

Ethanal is a weaker base than its enol because the equilibrium is shifted toward the enol

D

Ethanal is a weaker base than its enolform because the equilibrium is shifted toward ethanal

E

Ethanal is a stronger base than its enol because the hydrogen attached to the carbonyl group is highly acidic


Q26.5

Identify the keto-enol tautomer(s).

question description
A

1)

B

2)

C

3)

D

1 and 3

E

1 and 2


Q26.6

Identify the structure(s) representing the enolate of acetophenone.

question description
A

1)

B

2)

C

3)

D

1 and 3

E

1 and 2


Most aldehydes and ketones exist predominantly in the keto form. Propanone, for example, is 99.9997% in the keto form and only 0.0003% enol form. The main reason for the greater stability of the keto form is the stronger bond energy of the C=O and C-H bonds in the keto form compared to the weaker C=C and O-H bond energies in the enol form (Figure 26.7). 

Figure 26.7. Keto-enol equilibrium of propanone​.

If there are two carbonyl groups separated by a single CH2 group, the α-hydrogen between the two carbonyl atoms will become even more acidic because of the resonance delocalization of charge onto the two carbonyl oxygens (Figure 26.8). These types of compounds are termed β-dicarbonyl compounds, because the second carbonyl group belongs to the second carbon next to the first carbonyl group.  

Figure 26.8. Formation of two possible enols from a beta-dicarbonyl pentane-2,4-dione.​


The enol form of these β-dicarbonyl compounds can become the major tautomer. For example, 2,4-pentanedione (β-diketone) exists predominantly in the enol form (76%).Stabilization of the enol by conjugation of the C=C bond with another carbonyl group and formation of a six-membered hydrogen-bonded cycle makes concentrations of enol in the equilibrium mixture significant (Figure 26.9).

Figure 26.9: Keto-enol equilibrium of pentane-2,4-dione.


Q26.7

Which of the following compounds would favor the enol tautomer over the keto tautomer?

question description
A

1)

B

2)

C

3)

D

4)


If the C=C bond is a part of an aromatic system, the enol form, which is here called phenol, is stabilized so much that it becomes the major form. Therefore it is often the only detectable component of the equilibrium mixture (Figure 26.10). Phenols are enols you are already familiar with.

Figure 26.10. Keto-enol equilibrium of phenol​.

Replacing a polar with a non-polar solvent, like water with cyclohexane, slightly destabilizes the solvation of the more polar carbonyl form and shifts the equilibrium toward the enol. Thus, in polar water 2,4-pentanedione is 19% enol while in nonpolar cyclohexane, the percentage of the enol rises to 98% (Figure 26.11).

Figure 26.11. Solvent influence on the keto-enol equilibrium​.


Q26.8

Which enol is more stable, A or B, and why? Choose the best statement.

question description
A

Enol A is more stable because of hydrogen bonding

B

Enol B is more stable because of hydrogen bonding

C

Enol A is more stable because the double bonds are separated by a sp3sp^3 carbon

D

Enol B is more stable because the double bonds are conjugated


On the other hand, carboxylic acid derivatives enolize to an even lesser extent than aldehydes and ketones due to the additional stabilization of the carbonyl group by conjugation with the lone electron pair of the adjacent atom (Figure 26.12). 

Figure 26.12. Keto-enol equilibrium of ethyl acetate.


Figure 26.13. Comparison of the acidity of representative aldehydes, ketones, esters, and β-dicarbonyl compounds​.

The additional stabilization of the ketoform of carboxylic acid derivatives compared with aldehydes and ketones also accounts for their lower acidity. For example, the α-hydrogens in ethyl acetate are less acidic than α-hydrogens in ethanal by eight orders of magnitude. Figure 26.13 shows more examples of how the structure of a carbonyl compound affects its acidity.  

Q26.9

Which of the following compounds has the most acidic α-hydrogen?

question description
A

1)

B

2)

C

3)

D

4)


Q26.10

Which of the following compounds has the most acidic α-hydrogen?

question description
A

1)

B

2)

C

3)

D

4)


The carboxylic acid derivatives with stronger conjugation between the carbonyl and the leaving group are less acidic due to the increased stabilization of the ketoform versus the enolate. As you recall from the chemistry of carboxylic acid derivatives, more electronegative oxygen hold the electron pair better than less electronegative nitrogen, which makes conjugation less efficient for esters than for amides.


Q26.11

Which of the following statements best explains the difference in acidity between the α-hydrogen in ethyl acetate and the α-hydrogen in dimethylacetamide?

A

Dimethylacetamide is more acidic than ethyl acetate because of better stabilization of the carbonyl group in dimethylacetamide

B

Ethyl acetate is more acidic than dimethylacetamide because of better stabilization of the carbonyl group in dimethylacetamide

C

Ethyl acetate is less acidic than dimethylacetamide because of better stabilization of the carbonyl group in ethyl acetate

D

Ethyl acetate is more acidic than dimethylacetamide because of better stabilization of the carbonyl group in ethyl acetate

E

Ethyl acetate is more acidic than dimethylacetamide because of possible deprotonation of the ethyl group in ethyl acetate


Carbonyl compounds that do not contain an α-hydrogen atom cannot form enols and only exist in the keto form. Some examples are shown in Figure 26.14.

Figure 26.14. Carbonyl compounds that do not form enols.

In summary, normally the ketoform is favored over the enol form. However, this statement may not be true for β-dicarbonyl compounds. β-Dicarbonyl compounds can form stable conjugated enones that form stable six-membered rings through hydrogen bonding, which shifts the equilibrium to the enol. In the case of phenol, the stabilizing conjugation, the aromatic ring, is so significant that only the enol is present in solution (Table 26.1).

Table 26.1. Keto-enol equilibrium.


Q26.12

Match the statements with the tautomers.

question description
Premise
Response
1

1)

A

The keto tautomer is more stable becauseof the greater thermodynamic stability of the C=O C-H C-C bonds of the ketoform than the C=C. C-O O-H bonds of the enol

2

2)

B

The enol tautomer is more stable because it is stabilized by a conjugated π-system and intramolecular hydrogen bonding

3

3)

C

The enol tautomer is more stable because it is stabilized by the aromatic π-system


The molecular orbitals diagram of the enolate ion is given in Figure 26.15. The density of HOMO is mostly distributed between oxygen and the terminal carbon. Therefore, the molecular orbitals model of enolate is consistent with the resonance description, which predicts delocalization of the negative π-charge between oxygen and the terminal carbon. However, as opposed to the allyl anion, there is also some HOMO density on the central carbon, because oxygen is more electronegative than carbon and the orbitals are distorted. The overall charge distribution is made up by electrons in both occupied orbitals with the greatest electron density on the most electronegative oxygen.The HOMO has the greatest influence on the reactivity of a molecule as a nucleophile. For the enolate ion, the carbon has the greatest electron density in the HOMO, which makes the carbon more nucleophilic than the oxygen. Thus the enolate ion has more of the negative charge on the oxygen and the terminal carbon atom has the highest electron density of the HOMO.

Figure 26.15. Molecular orbital diagram of enolate.

Thus it can be expected that the oxygen will react with strong electrophiles and most other electrophiles will react at the enolate’s carbon atom, which is consistent with the largest lobe (density) of the HOMO reacting with low-lying LUMOs of electrophiles.

For example, the oxygen will react with the strongly electrophilic silicon atom of chlorotrimethylsilane and the carbon will react with less electrophilic carbon of alkyl halides (Figure 26.16). Due to this dual reactivity, the enolate is referred to as an ambident nucleophile.

Figure 26.16. Examples of ambident electrophilic reactions of acetone. a) With chlorotrimethylsilane. b) With ethylbromide. ​


Condensations and Alpha Substitution of Carbonyl

​ [2]

One notable example involving keto-enol tautomers is the molecule curcumin, which comes from the turmeric plant. The enol form of the molecule has antioxidant properties due to its radical-scavenging abilities.

Consider the two tautomeric forms of curcumin shown. Which of these is the correct structure of the enol tautomer?


The history of curcumin is thought to date back at least 5000 years! Even the famous explorer Marco Polo wrote about it in 1280 AD.

The compound has been used as a traditional medicine in India and China for centuries and, even today, curcumin is being investigated as a potential treatment for Alzheimer’s disease, rheumatoid arthritis, HIV, colorectal cancer, and several other diseases. Read more about curcumin here.

Where have you seen it before? Curcumin is commonly used in foods you may have tried, including yellow rice, curry, and mustard.

Image of Marco Polo [3] , Poha [4], ​Gourd Curry [5], mustard [6]

26.1.2 Alkylation of Enolate Ions

In organic synthesis, the introduction of an alkyl group at the carbon atom of enolates or C-alkylation is a valuable method for building the carbon skeleton of the target molecule. However, achieving practical yields of the product can be tricky due to the following reasons: 

  • The negative charge in enolates is mostly distributed between carbon and oxygen, which means there are two nucleophilic sites, carbon, or oxygen, for attack by the electrophile leading to different reaction products. Molecules that have two nucleophilic sites, such as enolates, are called ambident nucleophiles.
  • In addition to being nucleophiles, enolates are strong bases as well, and this can shift the reaction outcome from substitution to elimination.
  • The ketoform that is in equilibrium with the enolates, is an electrophile that can compete with the external electrophile for the nucleophilic enolate.
  • Non-symmetrical ketones can form two different enolates, whose C-alkylation leads to different products. 

These issues can be addressed by the following techniques of alkylation:

  • Due to the lower electronegativity of carbon versus oxygen, carbon is the more nucleophilic ‘end’ of enolate than oxygen. Therefore, an SN2 reaction, which is more sensitive to nucleophilic strength, will more likely involve the carbon end than its oxygen counterpart.
  • In order to suppress elimination, secondary, and tertiary alkylating agents, which are prone to elimination, should be avoided.
  • The enolate should be generated using a very strong base in order to keep the concentration of the ketoform negligible (Figure 26.17).
Figure 26.17. Generation of enolates by a strong base.

To put all these considerations together, for a ketone to be alkylated it should be deprotonated by at least one equivalent of a very strong base, and then treated with a methylating or a primary alkylating reagent. The most common bases for this purpose are NaH (sodium hydride) and LDA (lithium diisopropylamide). An example of alkylation of a ketone is shown in Figure 26.18.

Figure 26.18:. Ethylation of acetophenone.


Q26.13

Which of the following bases would completely convert 1,4-cyclohexanedione into an enolate?

A

LDA

B

Sodium methoxide

C

Sodium hydride

D

Both a) and c)


26.14

Acetophenone was treated by one equivalent of LDA, followed by one equivalent of tert\it{tert}-butyl bromide. What are the most likely reaction products?

question description
A

Phenyl (2,2-dimethyl-1-propyl)ketone

B

Phenyl tert\it{tert}-butyl ketone

C

4-Tert\it{Tert}-butylacetophenone

D

3-tert\it{tert}-butylacetophenone

E

Acetophenone and 2-methylpropene


  •  Sometimes selective alkylation of either α-carbon of a ketone can be achieved by utilizing different stabilities of the enolates that are formed or by utilizing the rate of formation of the enolate, which is influenced by steric factors. For instance, the deprotonation of 2-methylcyclohexanone by NaH at room temperature produces the more stable and more substituted enolate (Figure 26.19). 
Figure 26.19. Thermodynamically controlled alkylation of 2-methylcyclohexanone​.

Whereas the bulky base, LDA at -78 oC, abstracts the less sterically hindered proton, producing the faster forming and less substituted enolate (Figure 26.20). The more substituted enolate is the thermodynamically controlled product. The less substituted enolate is the kinetically controlled product.

Figure 26.20. Kinetically controlled alkylation of 2-methylcyclohexanone.


Q26.15

What is the major product in the following reaction?

question description
A

1)

B

2)

C

3)

D

4)


Q26.16

What is the major product in the following reaction?

question description
A

1)

B

2)

C

3)

D

4)


26.1.3 Formation and Alkylation of Enamines

Enamines prepared from aldehydes or ketones and secondary amines are stronger nucleophiles than enols because of lower electronegativity of nitrogen versus oxygen. However, they are weaker nucleophiles than enolates because they lack a negative charge (Figure 26.21). Fortunately, nucleophilicity of enamines is often sufficient for reactions with alkyl halides, which makes them excellent intermediates for alkylation of aldehydes and ketones. 

Utilization of enamines instead of enolates has a number of advantages. First, enamines derived from secondary amines are missing hydrogen atoms at nitrogen. Therefore, they cannot tautomerize to the ketoform, which eliminates the need for a very strong base used to generate an enolate. This is especially important for base-sensitive aldehydes and ketones. Second, a bulky secondary amine reduces the chance of N-alkylation that may compete with the desired C-alkylation. Pyrrolidine is one of the most commonly used secondary amines for this purpose. After the alkylation, the enamine group can be readily hydrolyzed back to the carbonyl compound. Another advantage is that enamines are not strong bases compared to enolates, and will not favor E2 elimination of the electrophile. Utilization of enamines for C-alkylation of aldehydes or ketones is illustrated in Figure 26.21.

Figure 26.21. Alkylation of aldehydes and ketones using enamines​.

There are other methods besides the preparation of enamines to “freeze” the enol tautomer, preventing equilibrium with the ketoform. Formation of silyl ethers is one common method (Figure 26.22).

Figure 26.22. Alkylation of aldehydes and ketones using enolsilyl ethers​.

Enol silyl ethers are weak bases that can be efficiently alkylated by tertiary halides without the unwanted competition with elimination reaction. For example, the carbocation generated from a tertiary alkyl halide by a Lewis acid such as TiCl4 will alkylate an enol silyl in SN1-like reaction (Figure 26.22).


Q26.17

In order to perform C-ethylation of acetophenone, acetophenone was first treated with one equivalent of an amine, followed by one equivalent of bromoethane, and then hydrolysis. Which of the following amines will be the best choice for this synthesis?

A

Ethylmethylamine

B

Ethylamine

C

Methylamine

D

Trimethylamine

E

2-Aminohexane


26.1.4 Alpha Halogenation of Ketones

Due to the presence of a C=C bond activated by a resonance electron donor, enols, and especially enolates are strong nucleophiles. This accounts for many reactions of carbonyl compounds with electrophilic agents. As a result of rapid equilibrium, the ketoform can completely react with an electrophile through its enol or enolate, even if the extent of enolization is very small. For instance, the enol form of a carbonyl compound rapidly reacts with molecular bromine (Br2), shifting the equilibrium from the ketoform until it is completely consumed.

Q26.18

The first step in the mechanism is the acid catalyzed generation of an enol and then electrophilic addition of bromine. Which cation is formed and why?

question description
A

1) is produced because there is less steric interaction

B

2) is produced because the positive charge is farthest away from the electronegative atoms

C

1) is produced because the electron donation through resonance is more stabilizing than the destabilizing effect of the electronegative oxygen

D

2) is produced because of the lack of the destabilizing effect of the electronegative oxygen


The carbocation in the product of addition is significantly stabilized by resonance, which makes this intermediate preferred over an alternative bromonium cation characteristic for the electrophilic bromination of alkenes. In the next step, this intermediate deprotonates to give the final α-brominated carbonyl compound (Figure 26.23).

Figure 26.23. Bromination of aldehydes and ketones​.


Mechanistically, using the enol intermediate as the nucleophile limits bromination to the α-positions with respect to the carbonyl group. This is because the C=C bond in the enol cannot extend further than the α-carbon.


Q26.19

Identify the most likely reaction product(s) in the monobromination of 2-pentanone by bromine in the presence of acid in the dark.

A

1-Bromo-2-pentanone and 3-bromo-2-pentanone

B

1-Bromo-2-pentanone, 3-bromo-2-pentanone, and 4-bromo-2-pentanone

C

1-Bromo-2-pentanone, 3-bromo-2-pentanone, 4-bromo-2-pentanone, and 5-bromo-2-pentanone

D

3-Bromo-2-pentanone

E

1-Bromo-2-pentanone


Q26.20

Which of the following is (are) the most likely reaction product(s) in the monobromination of cyclohexanone by bromine in the presence of light?

A

2-Bromocyclohexanone

B

2-Bromocyclohexanone, 3-bromocyclohexanone

C

2-Bromocyclohexanone, 3-bromocyclohexanone, 4-bromocyclohexanone

D

3-Bromocyclohexanone

E

4-Bromocyclohexanone


The second addition of bromine is substantially slower than the first addition since the intermediates formed in the second halogenation reaction are all positively charged and are destabilized by the electron withdrawing bromine atom as shown in Figure 26.23. Thus this reaction can be monitored and stopped after monohalogenation to give decent yields of the monohalogenated product. This does not mean that dibromination doesn’t occur. If two equivalents of bromine are added, dibrominated products are formed, however the second halogenation occurs on the other  carbon. For example, when acetone is reacted with two equivalents of Br2 in the presence of an acid, 1-3-dibromoacetone is produced. Notice that dibromination does not occur on the same carbon.

The reaction is autocatalytic, because the HBr byproduct, a strong acid, catalyzes the first, rate-limiting step of enolization. Therefore, to keep the overall exothermic reaction under control, one should slowly add bromine to the starting ketone under acidic conditions, for instance by using acetic acid as solvent. Chlorination proceeds similarly to bromination. 

In the presence of a base, the course of the reaction changes significantly. First, highly nucleophilic enolate rather than enol undergoes bromination. Second, due to its electronegativity, the first atom of bromine increases the acidity of the α-hydrogen and shifts the acid-base equilibrium toward the enolate. Therefore, the next atom of bromine is introduced to the molecule faster than the previous one. This means that it is difficult to isolate substantial amounts of partially brominated products until all α-hydrogens are replaced with bromines (Figure 26.24).  

Figure 26.24. Bromination of ketones under basic conditions.

Halogenation of aldehydes and ketones containing the acetyl group (CH3CO) under basic conditions proceeds even further. Substitution of all three hydrogens by bromines produces a good leaving group (-CBr3), in which the negative charge is stabilized by three electronegative halogens. Therefore, the product of tribromination can be considered as a carboxylic acid derivative. This is because hydrolysis under alkaline conditions forms a carboxylate plus the haloform (CHX3, Figure 26.25).

Figure 26.25. Haloform reaction.

You may be asking why the hydroxide ion doesn’t attack the electrophilic carbonyl carbon at the beginning of the mechanism. In fact it can. The hydroxide ion could react with the electrophilic carbon, but in the tetrahedral intermediate that is formed, the hydroxide ion is a better leaving group than a carbanion. Therefore the hydroxide is expelled, reforming the more thermodynamically stable carbonyl. The nucleophilic attack by the hydroxide ion does not have an effect until the better leaving group has been formed (-CBr3, Figure 26.26).

Figure 26.26. Ketones do not undergo nucleophilic substitution with the hydroxide ion​.

The hydrolysis mechanism of the bromination product is very similar to the hydrolysis mechanism of esters under basic conditions, where bromoform plays the role of an alcohol. Acidification of the reaction product converts the carboxylate to a carboxylic acid, so the haloform reaction can be used as a synthetic method for preparation of carboxylic acids from methylketones. 

The haloform reaction proceeds with chlorine and iodine as efficiently as with bromine. Since iodoform (CHI3) is solid, iodoform reaction (I2, NaOH, H2O) is a good quantitative test for methylketones and ethanal. Interestingly, precipitation of iodoform also slowly takes place with ethanol. First, ethanol undergoes oxidation to ethanal, which tests positive by the iodoform reaction (Figure 26.27).

Figure 26.27. Iodoform reaction of ethanol​.


Q26.21

Which of the following compounds will give a negative iodoform reaction?

A

Propanone

B

2-Butanone

C

Ethanal

D

3-Pentanone

E

2-Pentanone


Q26.22

What is the major product produced in the following reaction?

question description
A

1)

B

2)

C