The correct answer is Option (2) → (D), (B), (C), (A)
Factors Influencing Reactivity
The reactivity of a carbonyl compound ($C=O$) towards a nucleophile is governed by two main factors:
- Steric Hindrance: Nucleophilic attack involves a nucleophile approaching the carbonyl carbon. Large alkyl groups attached to this carbon block the nucleophile's path. Therefore, more alkyl groups = less reactive.
- Electronic (+I) Effect: Alkyl groups are electron-donating. They push electron density toward the carbonyl carbon, reducing its partial positive charge ($\delta^+$). Since a nucleophile is attracted to a positive charge, more alkyl groups = less electrophilic = less reactive.
Analyzing the Compounds
Based on these factors, aldehydes are always more reactive than ketones.
- (A) Ethanal ($CH_3CHO$): An aldehyde with one small methyl group. It has the least steric hindrance and the highest electrophilicity among the choices. (Most Reactive)
- (C) Propanal ($CH_3CH_2CHO$): An aldehyde with one ethyl group. The ethyl group is bulkier and more electron-donating than a methyl group, making it slightly less reactive than ethanal.
- (B) Propanone ($CH_3COCH_3$): A ketone with two methyl groups. The presence of two electron-donating groups significantly reduces reactivity compared to aldehydes.
- (D) Butanone ($CH_3COCH_2CH_3$): A ketone with one methyl and one ethyl group. It has the most steric hindrance and the greatest +I effect, making it the least reactive. (Least Reactive)
Increasing Order of Reactivity
Starting from the least reactive to the most reactive:
Butanone < Propanone < Propanal < Ethanal
(D) < (B) < (C) < (A) |
