Textbook: Brown 6th Ed. (2011)

Chapter 19: Enolate Anions and Enamines

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Individual Problems

Individual Problems

Problem # 738

Carbonyls are in equilibrium with their enol forms. This process is called keto-enol tautomerization.

This equilibrium happens in both acid and base.

Let's go through this equilibrium under acidic conditions. Draw a mechanism using curved arrows for each reaction below.

Remember that under acidic conditions, most species are either neutral or positively charged, and rarely negatively charged. So your structures will contain either ROH or ROH₂⁺, but not RO^-.

a) Carbonyl to Enol (acidic)

b) Enol to Carbonyl (acidic)

Solution

Enol and enolate mechanisms always involve the alpha position of the carbonyl (one away from the carbonyl carbon). This the alpha hydrogen is acidic, and so can be deprotonated.

To form an enol (or enolate) from a carbonyl, the alpha position of the carbonyl must be deprotonated, and the double bond (pi electrons) travels "goes up" to the oxygen to become a lone pair.

To form a carbonyl from an enol (or enolate), the alpha position of the carbonyl must be protonated, and a lone pair on the oxygen "comes down" to reform the carbonyl.

This "up and down" mechanism is similar to the Nucleophilic Acyl Addition/Substitution mechanisms discussed in problems 705 and 706, with the difference being that the carbon carbonyl is not where bond formation takes place. In enol/enolate mechanisms, new bond formation occurs at the alpha position.

a) Carbonyl to Enol (acidic) "UP"

Because this mechanism takes place under acidic conditions, the carbonyl oxygen must be protonated before its double bond "goes up" to form a lone pair.

b) Enol to Carbonyl (acidic) "DOWN"

Most people remember that a lone pair from the enol oxygen "comes down." But many forget that you have to add something back to the alpha position! In this case, it's just a hydrogen- you reprotonate the alpha position to form the carbonyl.

Problem Details

MendelSet practice problem # 738 submitted by Matt on July 26, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enols, enolates, keto-enol tautomerization

Problem # 739

Carbonyls are in equilibrium with their enol forms. An enolate is the deprotonated form of an enol.

Enolates are formed from carbonyls under basic conditions.

Let's go through this equilibrium under basic conditions. Draw a mechanism using curved arrows for each reaction below.

Remember that under basic conditions, most species are either neutral or negatively charged, and rarely positively charged. So your structures will contain either ROH or RO^-, but not ROH₂⁺.

a) Carbonyl to Enolate (basic)

b) Enolate to Carbonyl (basic)

Solution

These mechanisms are similar to those in problem 738, only under basic conditions.

a) Carbonyl to Enolate (basic)

In the mechanism below the carbonyl was deprotonated at the alpha position, and the enolate was then drawn as a resonance form. You can also go straight from carbonyl to enolate in one reaction arrow. (rip off the proton, form the new double bond, and send the carbonyl "up" to form a negative charged oxygen).

b) Enolate to Carbonyl (basic) "DOWN"

Like a), this reaction can be done in one reaction arrow (negative charge comes "down" and double bond attacks a proton)

It's important that you become familiar with the mechanisms for enol and enolate formation in both acid (Q738) and in base (this problem), as well as carbonyl hydrate formation in both acid (Q706) and base (Q705).

If you can draw these four mechanisms you will be able to figure most of the reactions in second semester organic chemistry!

Problem Details

MendelSet practice problem # 739 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enols, enolates, keto-enol tautomerization

Problem # 740

Enolates are formed from carbonyls by adding a strong base, such as lithium diisopropyl amide (LDA), to deprotonate the alpha position. The enolate can then act as a nucleophile and attack an electrophile (such as an alkyl halide), to form a new bond at the alpha position. This is called a carbonyl alpha substitution reaction.

Let's go through the mechanism of how enolates are formed and how they react with electrophiles.

Draw in the curved arrows to show the formation of the enolate (middle compound), and draw the structure of the carbonyl product (right compound)

Solution

Unlike in problem 739 in which enolates were formed in two steps, enolate formation is usually drawn in one step, as shown below.

Enolates then can attack electrophiles (alkyl halides, carbonyls, etc.) to make new bonds at the alpha position.

Notice that after an alpha substitution reaction the carbonyl is reformed, and so can react again.

Problem Details

MendelSet practice problem # 740 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enolates, carbonyl alpha substitution

Problem # 741

Like enolates (Q740), enols can also act as nucleophiles and make new bonds at the alpha position.

Let's go through the mechanism for how this happens, using alpha bromination as our example.

In the reaction below, the carbonyl form is in equilibrium with its enol form (Q738), which can then attack molecular bromine.

Complete the structure of the protonated carbonyl, and show the curved arrows that transform it into the final carbonyl product.

Solution

Enols are just like enolates in that they attack electrophiles and add to the alpha position.

Problem Details

MendelSet practice problem # 741 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enols, carbonyl alpha substitution, alpha halogenation

Problem # 742

Enamines are similar to enols and enolates in that they also undergo alpha substitution reactions.

The process of performing a carbonyl alpha substitution reaction via an enamine intermediate is called the Stork enamine synthesis. Let's work through this reaction.

Draw in the structures for the enamine and show it attacking the alkyl halide to form the "3º imine." Also draw the structure of the final and carbonyl product.

Solution

Enamines attack alkyl halides just as enolates do.

The trick to the Stock enamine synthesis is to remember how enamines (and imines) are related to carbonyls:

Enamines are formed from carbonyls and 2º amines.
Enamines (and imines) are converted back to carbonyls with H₃O⁺ (acidic hydrolysis).

Problem Details

MendelSet practice problem # 742 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): carbonyl alpha substitution, Stork enamine synthesis

Problem # 306

Rank each group of acids in order of decreasing acidity. (1 = most acidic)

Explain your reasoning. You will have to use more than one rule in your explanation (resonance, electronegativity, atomic radius, etc.).

Solution

Compounds 1 and 2 are more acidic than compound 3 because their conjugate bases have more resonance forms than that of compound 3.

Compound 1 is more acidic than compound 2 because the resonance form of its conjugate base has two oxygen atoms and is more stable than that of compound 2, which has one oxygen and one nitrogen; a negative oxygen atom is more stable than a negative nitrogen atom.

Problem Details

MendelSet practice problem # 306 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, acid strength

Problem # 743

Enolates are nucleophiles and react with a variety of electrophiles.

Carbonyls are electrophiles. But aldehydes/ketones and esters/acid chlorides often form different products.

Use curved arrows to draw a mechanism for each reaction below. How do the two products differ?

Solution

When enolates attack, the arrow is drawn from the double bond to the electrophile, and NOT from the negatively charged oxygen! The enolate's carbonyl reforms, and a new bond is created at the alpha position.

But Aldehydes and ketones react differently than ester and acid chlorides.

a) Aldehydes and ketones are oxidation state II carbonyls (the carbonyl has carbon-heteroatom bonds- the double bonded oxygen counts twice).

They do not have a built-in leaving group, and so undergo nucleophilic acyl addition reactions. The carbonyl on the electrophile becomes an alcohol.

So the product from this reaction is a beta hydroxyl carbonyl. If this β-hydroxyl carbonyl is heated up, the hydroxyl will eliminate to from an alpha, beta unsaturated carbonyl.

The mechanism for a) is very similiar to that of an aldol condensation reaction.

b) Esters (and acid chlorides) are oxidation state III carbonyls (the carbonyl carbon has three carbon-heteroatom bonds). They have a built-in leaving group (^-OR in the case of an ester, and Cl^- for an acid chloride), and so undergo nucleophilic acyl substitution reactions. ("up, down, kick"). The carbonyl on the electrophile is reformed.

So the product from this reaction is a beta keto carbonyl.

The mechanism for b) is very similiar to that of a Claisen condensation reaction.

Problem Details

MendelSet practice problem # 743 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork), Carbonyl Condensation Reactions (Aldol, Claisen, Dieckmann, Robinson)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition, enolates, carbonyl alpha substitution, aldol condensation, Claisen condensation

Problem # 744

After a sample of optically pure (S)-2-ethyl-cyclohexanone is dissolved in an aqueous solution for several hours, a significant loss of optical activity is observed. Explain.

Solution

Carbonyls are in equilibrium with their enol forms. This process is called keto-enol tautomerization. (See problems 738 and 739). So when this compound is placed in water a small amount of it will constantly be converted to its enol form, and then back again to the carbonyl form.

But the enol form is achiral! (the alpha carbon is sp² instead of sp³as in the carbonyl form).

So when the achiral enol goes back to its carbonyl form, half will become (S) and the other half will become (R). The stereocenter becomes "scrambled." The process of an optically pure compound becoming a racemic mixture is called racemization.

Problem Details

MendelSet practice problem # 744 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Stereochemistry, Chirality, and Optical Activity, Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Concept (explain why this is so..)

Keyword(s): keto-enol tautomerization, racemization

Problem # 748

Alpha bromination is usually carried out under acidic conditions via the enol intermediate.

Alpha bromination is uncontrollable under basic conditions, which goes through the enolate intermediate. Let's explore why.

a) Rank each carbonyl below in order of decreasing alpha-proton acidity (1= most acidic). Explain.

b) based on a), why does the reaction below lead to polyhalogenation?

Solution

a) Bromine is an electron withdrawing group (EWG), which makes nearby protons more acidic. So the carbonyl with two bromines is the most acidic.

b) To perform alpha bromination, the enolate (or enol) must first be formed, which then attacks Br₂.

The problem with doing this under basic conditions is that each successive bromination leads to a carbonyl that is more acidic, and so forms an enolate even more easily. So once this reaction starts, it's difficult to control.

Enols are less nucleophilic than enolates, so this react can be controlled under acidic conditions. (Br₂/H3O⁺ instead of Br₂/NaOH).

Problem Details

MendelSet practice problem # 748 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength, EDG and EWG, carbonyl alpha substitution, alpha halogenation

Problem # 749

Show how to prepare each compound below from propanal. I've marked the "cuts" for you.

Solution

a) Adding to alpha position, so this must have come from the combination of an enolate and an electrophile.

b) Adding to the beta position, so this must have come from a Michael addition (also called a conjugate addition).

To carry out a Michael addition, the first step is to convert propanal to an alpha, beta unsaturated carbonyl.

To do this, perform an alpha bromination with (Br₂/H₃O⁺) and then do an E2 elimination (tBuOK/heat).

Then add a Gilman reagent to the alpha,beta unsaturated aldehyde to add a carbon-carbon bond at the beta position.

Problem Details

MendelSet practice problem # 749 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork), Carbonyl Conjugate Additions (Michael Addition, Soft Nucleophiles)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): carbonyl alpha substitution, Michael addition

Problem # 745

The molecule below has five different types of hydrogens (A through E). Rank each in order of decreasing acidity.

(1 = most acidic). Explain your reasoning.

Solution

Hydrogens D and E are alpha to two carbonyls and so will be the more acidic than C, which is only alpha to one carbonyl. This is because the enolates that arise from deprotonate at D and E have more resonance forms than the enolate that arrises at C.

Because ketones are more electron withdrawing than esters, D will be more acidic than E. So far we have:

D > E > C > (other)

The carbanions that would arrise from deprotonation of carbons A and B both do not have any resonance forms, so we don't expect either to be acidic. But because A is next to a fluorine atom, which is an electron withdrawing group, A will be more acidic than B. So the overall order is:

(most acidic) D > E > C > A > B (least acidic)

Problem Details

MendelSet practice problem # 745 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength

Problem # 746

Show what combination of aldehyde, ketone, and/or ester can prepare each compound below. Every compound is a Claisen or aldol product.

Solution

When doing synthesis problems involving enolates (carbonyl alpha substitutions, aldol and Claisen condensations), there are some things to keep in mind:

The alpha carbon must have been the enolate (nucleophile). So it's importnant to identify the alpha carbon!
The bond is always made between the alpha and beta carbon.
Beta hydroxyl products come from aldol condensations (aldehydes/ketones), beta carbonyl products come from Claisen condensations (esters/acid chlorides). (See problem 743 for these mechanisms.)

Also, α,β-unsaturated products come from β-hydroxy products, which come from aldol condensations.

After finding the alpha carbon, it's a good idea to mark the "cut" or "disconnection" where the new bond was formed. This will always be between the alph and beta carbons. I mark this with a dotted line.

a) This product is an α,β-unsaturated carbonyl, which must have come from a β-hydroxyl carbonyl. So this is an aldol condensation.

The α carbon must have come form the enolate. Because the β carbon is an alcohol, the electrophile must have been an aldehyde or ketone. So the two carbons on the right were the enolate, and the two carbons (and phenyl) on the left were an aldehyde.

b) The product is β-keto so this must have been a Claisen condensation.

The β carbon is an ketone, so the electrophile must have been an ester.

There are actually two different combinations of esters that would result in this product. I arbitrarily chose the alpha carbon to belong to the left carbonyl.

c) β-keto product so this must have been a Claisen style condensation. It's easiest if you have the alpha carbon belong to the left carbonyl, so an intramolecule reaction isn't necessary.

The β carbon is already an ester, so the electrophile had to have been more than an ester, as regular esters become ketones after a Claisen condensation.

d) This is a Robinson annulation product, which comes from an intramolecular condensation.

The α,β-unsaturated carbonyl must have come from a β-hydroxyl carbonyl. (aldol condensation)

Because the β carbon became an alcohol, the electrophile must have been an aldehyde or ketone.

Problem Details

MendelSet practice problem # 746 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Condensation Reactions (Aldol, Claisen, Dieckmann, Robinson)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): enolates, carbonyl alpha substitution, aldol condensation, Claisen condensation

Problem # 747

Show a combination of enolate (nucleophile) and electrophile that can produce each compound below.

Remember that all enolates come from carbonyls.

Solution

This problem is similar to problem 746, except each synthesis is a little harder.

The general strategy for carbonyl alpha substitution synthesis problems is the same: locatte the alpha carbon, and remember that the bond is always formed between the alpha and beta carbons.

a) The carbonyl on the left contains the alpha carbon, so that must have been the enolate.

So the beta carbon on the right must have been the electrophile (and been attacked by the enolate).

So how can the beta carbon keep its carbonyl AND keep an -OEt group? Doesn't the -OEt group act as a leaving group in a Claisen condensation? The starting ester must have had two -OEt groups!

b) This one is intimidating because it's intramolecular, but the same logic applies. The left carbonyl has the alpha carbon and so must have been the enolate.

The beta carbon is now an alcohol, so before it was attacked it must have been a ketone (aldol condensation).

c) This is one of the hardest synthesis problems you will see relating to enolates and carbonyl alpha substitution reactions. Why is it hard? Because you have to use a special electrophile.

Usually you add a carbon chain to the alpha position of a carbonyl by reacting an enolate with an alkyl halide, suhc as propyl bromide.

So how do you add a carbon chain that wraps and meets again at the alpha carbon? By using 1,3-dibromo propane and doing two consecutive alkyl additions.