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Home/Practice questions/Chemistry
ChemistryOrganic: Aldehydes, Ketones and Carboxylic Acids

What is the correct order of reactivity towards nucleophilic addition?

What is the correct order of reactivity towards nucleophilic addition?

Answer & explanation+

Correct answer: option 4

The correct answer is Option (4) → Benzaldehyde > Acetophenone > Benzophenone

In nucleophilic addition the nucleophile attacks the carbonyl carbon. The reaction is faster when this carbon is more electrophilic and less crowded. Large groups hinder the approach of the nucleophile, and a phenyl group attached to the carbonyl carbon lowers its electrophilicity by resonance.

  • Benzaldehyde, \( \ce{C6H5CHO} \) — first. As an aldehyde it has only one large group (phenyl) and an H on the carbonyl carbon, so it is the least hindered and the most electrophilic of the three.

  • Acetophenone, \( \ce{C6H5COCH3} \) — second. As a ketone it carries two groups (phenyl and methyl); they crowd the carbonyl carbon more and reduce its electrophilicity more than in benzaldehyde.

  • Benzophenone, \( \ce{C6H5COC6H5} \) — third. It carries two phenyl groups: the largest steric hindrance, and resonance with two rings lowers the electrophilicity of the carbonyl carbon the most.

NCERT (2026-27 edition):
8.4 Chemical Reactions – “… 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 than in aldehydes having only one such substituent. Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former.”

“The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance … and hence it is less reactive than propanal.”

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