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 sp³ as 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.

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.

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).