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.
