Question

The major product obtained in the following reaction is

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

The substrate is 2-methyl-6-oxoheptanal, OHC-CH(CH₃)-CH₂CH₂CH₂-CO-CH₃. Intramolecular aldol condensation that closes a stable five-membered ring is the preferred path.

Answer 1

The Correct Option is C

The reactant is OHC-CH(CH₃)-CH₂-CH₂-CH₂-C(=O)-CH₃ (2-methyl-6-oxoheptanal).
With NaOH and heat it undergoes intramolecular aldol condensation; only 5- or 6-membered rings form readily.
Enolate from the terminal CH₃ of the ketone attacking the CHO would give a 7-membered ring, so it is not preferred.
Enolate from the ring-side alpha-CH₂ of the ketone attacking the (more electrophilic) aldehyde carbon closes a 5-membered ring: the ring contains the alpha-carbon (bearing COCH₃), two CH₂, CH(CH₃) and the former CHO carbon (now CHOH).
Heating eliminates water to give the conjugated enone: C(COCH₃)=CH in the ring, with the CH(CH₃) carbon next to the =CH.
Product: 1-acetyl-3-methylcyclopent-1-ene, which is option (3). Option (4) wrongly places the methyl on the double bond and the acetyl on an sp³ carbon; options (1) and (2) cannot form from this substrate.
Answer: (3) 1-acetyl-3-methylcyclopent-1-ene (structure shown as option 3)