aldol condensation

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.