Arrange the steps followed during nucleophilic addition on a carbonyl…

The correct answer is Option (2) → B → A → C → D

The carbonyl carbon is electrophilic because the C=O bond is polarised (C^δ+–O^δ−). In nucleophilic addition the new bond to carbon must form before anything else can change at that carbon:

• B (first) — the nucleophile attacks the electrophilic carbon; this slow step starts the reaction.

• A (second) — the change from sp² to sp³ happens as the new C–Nu bond forms and the π electrons move onto oxygen; it cannot come before the attack.

• C (third) — the result is the tetrahedral alkoxide ion, with the negative charge on oxygen.

• D (last) — the alkoxide oxygen takes a proton in the fast second step, giving the neutral product.

    \( \displaystyle \ce{R2C=O + Nu^-} \xrightarrow{\text{slow}} \ce{R2C(Nu)-O^-} \xrightarrow[\text{fast}]{\ce{H+}} \ce{R2C(Nu)-OH} \)

Shortcut: the attack of the nucleophile must come first and the protonation last. Only Option (2) starts with B and ends with D.

NCERT (2026-27 edition):
Mechanism of Nucleophilic Addition Reactions – “A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp² hybridised orbitals of carbonyl carbon (Fig. 8.2). The hybridisation of carbon changes from sp² to sp³ in this process, and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product.”