Aldol condensation is a popular and classical method to prepare α, β-unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?

Solution

Chemistry ma’am · 3 steps

In an intramolecular aldol the new C=C closes the ring, so an exocyclic =CH₂ gives the game away.

1. Work backwards. In aldol condensation the α-carbon of one carbonyl bonds to the carbonyl carbon of the other, and that bond becomes the α,β C=C. Break that C=C and put C=O back on the β-carbon. This gives you the starting material.
2. Intramolecular test. If both carbonyl groups sit in one molecule, the new C–C bond closes a ring, so the C=C is part of that ring. If breaking the C=C splits the product into two separate molecules, the reaction was intermolecular.

   Option	Break the C=C, add back C=O	Ring formed?
   (a) bicyclic enone, C=C at the ring fusion	cyclodecane-1,6-dione, one molecule	yes, intramolecular (transannular)
   (b) 2-methylene-1-tetralone	1-tetralone + HCHO, two molecules	no, the C=C is exocyclic
   (c) cyclohex-2-enone	CH₃CO(CH₂)₃CHO (5-oxohexanal)	yes, 6-membered ring
   (d) inden-1-one	2-acetylbenzaldehyde	yes, 5-membered ring

3. So. Option (b) comes from a crossed aldol between 1-tetralone and formaldehyde, which is an intermolecular reaction. It is not a product of intramolecular aldol condensation, so the answer is (b).

The idea to keep. To find an aldol precursor, cut the C=C of the enone and put C=O on the β-carbon. If one molecule remains, the aldol was intramolecular. If two pieces fall out, it was intermolecular.

Answer from the NTA final answer key. Solution by PrepWiser.