The figure shows the mechanism of the nucleophilic addition reaction of an aldehyde or a ketone.
What is the hybridisation of the carbonyl carbon in the starting compound and in the tetrahedral intermediate, respectively?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → sp2 and sp3
The figure labels the starting carbonyl compound planar and the species formed in step 1 a tetrahedral intermediate. A planar carbon with three groups is sp2; a tetrahedral carbon with four groups is sp3.
Starting compound — sp2. The carbonyl carbon forms three σ bonds and one π bond with oxygen. It and the three atoms attached to it lie in one plane, with bond angles of about 120°.
Tetrahedral intermediate — sp3. In step 1 the nucleophile bonds to the carbonyl carbon, and the π bond breaks. The carbon now has four groups around it, so it becomes sp3 and tetrahedral.
Why the other options are incorrect
Option (1) sp2 and sp — Incorrect. An sp carbon is linear, with two groups; the carbon in the intermediate has four groups.
Option (3) sp3 and sp2 — Incorrect. This reverses the change. The planar starting compound is sp2, and the intermediate is sp3.
Option (4) sp and sp3 — Incorrect. The carbonyl carbon of the starting compound has three groups in one plane, so it is sp2, not sp.
NCERT (2026-27 edition):
8.4 Chemical 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.”
8.1.2 Structure of the Carbonyl Group – “The carbonyl carbon atom is sp2-hybridised and forms three sigma (σ) bonds. … Thus, the carbonyl carbon and the three atoms attached to it lie in the same plane …”
Fig. 8.2: Nucleophilic attack on carbonyl carbon