Propionitrile when treated with phenyl magnesium bromide followed by acidic hydrolysis gives (P) which further undergoes Clemmensen reduction to give
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → n-propylbenzene.
Two standard reactions are joined here.
Nitrile + Grignard reagent, then hydrolysis, gives a ketone. The R group of RMgX adds to the carbon of C≡N to form an imine salt, which is hydrolysed to C=O. The ketone carries the alkyl group of the nitrile and the group of the Grignard reagent.
Clemmensen reduction (Zn–Hg and concentrated HCl) turns the C=O group of an aldehyde or ketone into \( \ce{CH2} \).
Propionitrile (\( \ce{CH3CH2CN} \)) with phenylmagnesium bromide gives propiophenone, P:
\( \displaystyle \ce{CH3CH2CN} \xrightarrow[\text{(ii) }\ce{H3O^+}]{\text{(i) }\ce{C6H5MgBr}} \ce{CH3CH2COC6H5} \)
Clemmensen reduction of P then gives the hydrocarbon:
\( \displaystyle \ce{CH3CH2COC6H5} \xrightarrow{\ce{Zn-Hg}\text{, conc. }\ce{HCl}} \ce{CH3CH2CH2C6H5} \)
So the final product is n-propylbenzene.
Why the other options are incorrect
Option (1) Propiophenone — Incorrect. This is P, the ketone formed before the Clemmensen reduction, not the final product.
Option (2) Benzophenone — Incorrect. \( \ce{C6H5COC6H5} \) would need benzonitrile, not propionitrile, and it would still be reduced further.
Option (4) Diphenylethane — Incorrect. It has two phenyl groups, but only one phenyl group enters from the Grignard reagent.
NCERT (2026-27 edition):
From Nitriles
Reduction – “… (ii) Reduction to hydrocarbons: The carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with zinc- amalgam and concentrated hydrochloric acid [Clemmensen reduction] or with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol (Wolff-Kishner reduction).”