enolates

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.

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.

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.

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.

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!

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.