In the following sequence of reaction identify the compound [C]:

The correct answer is Option (4) → 2-Methylpropan-2-ol

• First: the starting compound is propan-2-amine, a primary aliphatic amine. With nitrous acid it gives an unstable diazonium salt, which loses \( \ce{N2} \) and gives the alcohol:
      \( \ce{(CH3)2CH-NH2 + HNO2 -> (CH3)2CH-OH + N2 + H2O} \) → [A] = propan-2-ol

• Second: oxidation of the secondary alcohol gives a ketone → [B] = propanone:
      \( \ce{(CH3)2CHOH} \xrightarrow{\ce{K2Cr2O7/H+}} \ce{(CH3)2C=O} \)

• Third: \( \ce{CH3MgBr} \) adds to the C=O group, and hydrolysis gives a tertiary alcohol:
      \( \ce{(CH3)2C=O} \xrightarrow[\text{(ii) }\ce{H+/H2O}]{\text{(i) }\ce{CH3MgBr}} \ce{(CH3)3C-OH} \) → [C] = 2-methylpropan-2-ol

Why the other options are incorrect

• Option (1) Butan-1-ol — Incorrect. A primary alcohol; a Grignard reagent gives a primary alcohol only with methanal, not with a ketone.

• Option (2) Butan-2-ol — Incorrect. A secondary alcohol, which would need \( \ce{CH3MgBr} \) with propanal (an aldehyde), not propanone.

• Option (3) 2-Methylpropan-1-ol — Incorrect. A primary alcohol, \( \ce{(CH3)2CHCH2OH} \); the OH would be on a \( \ce{CH2} \), but \( \ce{CH3MgBr} \) adds to the carbonyl carbon of propanone, which becomes the C–OH carbon bearing three \( \ce{CH3} \) groups.

NCERT (2026-27 edition):
Reaction with Nitrous Acid – “… (a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols.”

From Grignard Reagents – “… You will notice that the reaction produces a primary alcohol with methanal, a secondary alcohol with other aldehydes and tertiary alcohol with ketones.”