When
undergoes intramolecular aldol condensation, the major product formed is :

Solution

Chemistry ma’am · 4 steps

Pick the enolate that attacks the aldehyde and closes a five-membered ring. That one wins.

1. Number the chain. The compound is 6-oxoheptanal: C1 is the CHO carbon, C6 is the ketone carbon and C7 is its CH₃. The α-carbons that can form an enolate are C2 (next to CHO), C5 and C7 (both next to the ketone).

   Enolate at	Attacks	Ring size	Product after losing water
   C7 (CH₃ of ketone)	C1 (aldehyde)	7	cyclohept-2-en-1-one, option (d)
   C5 (CH₂ next to ketone)	C1 (aldehyde)	5	1-acetylcyclopentene, option (b)
   C2 (CH₂ next to CHO)	C6 (ketone)	5	2-methylcyclopent-1-ene-1-carbaldehyde, option (c)

2. Choose the best pair. Five-membered rings form much faster and with less strain than seven-membered ones, so (d) loses. Between the two five-membered routes, the aldehyde carbon is the better electrophile: it is less hindered and has only one electron-donating alkyl group. So the ketone enolate at C5 attacks the CHO carbon.
3. Dehydrate. The aldol product has OH on C1 in the ring and the acetyl group on C5. Losing water forms the C=C between C1 and C5, conjugated with the ketone C=O.
4. So. The major product is 1-acetylcyclopentene, option (b).

The idea to keep. In intramolecular aldol, list every enolate and carbonyl pair, keep the ones that close 5 or 6 rings, and let the aldehyde act as the electrophile whenever it can.

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