Which of the following reaction will not produce ketones?

The correct answer is Option (4) → Stephen's reaction

Ketones are prepared by several routes: ozonolysis of suitably substituted alkenes, hydration of alkynes other than ethyne, and the reaction of nitriles with Grignard reagents. The Stephen reaction belongs to the methods for aldehydes.

• Option (1) Hydration of alkynes — Produces a ketone. With dilute \( \ce{H2SO4} \) and \( \ce{HgSO4} \), ethyne gives acetaldehyde, but all other alkynes give ketones.

      \( \ce{CH3-C#CH + H2O} \xrightarrow[\text{dil. }\ce{H2SO4}]{\ce{HgSO4}} \ce{CH3COCH3} \)

• Option (2) Ozonolysis of substituted alkenes — Produces a ketone. Ozonolysis followed by Zn/water puts a C=O on each carbon of the double bond. A carbon that carries two alkyl groups becomes a ketone.

      \( \ce{(CH3)2C=CH2} \xrightarrow[\text{(ii) Zn/}\ce{H2O}]{\text{(i) }\ce{O3}} \ce{CH3COCH3 + HCHO} \)

• Option (3) Treating nitrile with a Grignard reagent — Produces a ketone. The Grignard reagent adds to the C≡N group, and hydrolysis of the imine salt gives a ketone.

      \( \ce{CH3CN + C6H5MgBr} \xrightarrow{\text{ether}} \ce{C6H5C(CH3)=NMgBr} \xrightarrow{\ce{H3O+}} \ce{C6H5COCH3} \)

• Option (4) Stephen's reaction — Does not produce a ketone. In the Stephen reaction a nitrile is reduced by stannous chloride and hydrochloric acid to an imine, and hydrolysis of the imine gives an aldehyde. The carbon of –CHO keeps one hydrogen, so no ketone can form.

      \( \ce{RCN + SnCl2 + HCl -> RCH=NH} \xrightarrow{\ce{H3O+}} \ce{RCHO} \)

NCERT (2026-27 edition):
From Hydrocarbons

From Nitriles and Esters

From Nitriles

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