Nucleophilic addition reaction on carbonyl carbon results in change o…

The correct answer is Option (1) → sp² to sp³

The carbonyl carbon of an aldehyde or ketone is \(sp^2\) hybridised: it forms three \(\sigma\) bonds in one plane (trigonal planar, bond angles about 120°) and a \(\pi\) bond with oxygen. In a nucleophilic addition the nucleophile attacks this carbon from above or below the plane. The \(\pi\) bond breaks, its electron pair moves onto oxygen, and the carbon now carries four \(\sigma\) bonds, so it becomes \(sp^3\) hybridised and tetrahedral. This gives the tetrahedral alkoxide intermediate, which then takes up a proton:

    \( \displaystyle \ce{R2C=O + Nu^- -> R2C(Nu)-O^-} \xrightarrow{\ce{H^+}} \ce{R2C(Nu)-OH} \)

So during the reaction the carbonyl carbon changes from \(sp^2\) to \(sp^3\).

Why the other options are incorrect

• Option (2) sp to sp² — Incorrect. An sp carbon has two \(\pi\) bonds, as in alkynes and nitriles; the carbonyl carbon is not sp hybridised to begin with.

• Option (3) sp² to sp² — Incorrect. The hybridisation does change: the carbon goes from three \(\sigma\) bonds and a \(\pi\) bond to four \(\sigma\) bonds.

• Option (4) sp³ to sp³ — Incorrect. The carbonyl carbon is not \(sp^3\) to begin with; it is \(sp^2\), with a \(\pi\) bond to oxygen.

NCERT (2026-27 edition):
Mechanism of Nucleophilic Addition Reactions

Note: The same/similar question has been asked in different years by NTA as per the details given below: