Answer the question on the basis of passage given below:
Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon more than aldehydes having only one such substituent Electronically aldehydes are more reactive than ketones because two alkyl groups reduce the elctrophilicity of the carbonyl more effectively than in former.

4-Hydroxy-4-methylpentan-2-one and 4-methylpent-3-en-2-one are the products obtained from 'A' by aldol condensation. 'A' is:

\(CH_3COCH_3\)

The correct answer is option 1. \(CH_3COCH_3\).

Aldol condensation involves the reaction of two molecules containing carbonyl groups (aldehydes or ketones) in the presence of a base to form a β-hydroxy carbonyl compound (aldol product). This aldol product can further undergo dehydration to form an α,β-unsaturated carbonyl compound.

Structure of the Products

4-Hydroxy-4-methylpentan-2-one:

This compound has a hydroxy group \(-OH\) on the 4th carbon, a ketone group \(-CO-\) at the 2nd carbon, and a methyl group \(-CH_3\) attached to the 4th carbon.

4-Methylpent-3-en-2-one

This compound has a double bond between the 3rd and 4th carbon and a ketone group at the 2nd carbon.

Possible Aldol Reaction of Acetone

Acetone (\(CH_3COCH_3\)) can undergo self-condensation through an aldol reaction in the following manner:

The reaction described above shows that acetone (\(CH_3COCH_3\)) can undergo an aldol condensation with itself to produce both 4-Hydroxy-4-methylpentan-2-one and 4-Methylpent-3-en-2-one. Therefore, the correct starting material (compound 'A') that can produce these products through aldol condensation is acetone:option 1: \(CH_3COCH_3\).