The reagent used in the Stephen reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → SnCl2 + HCl
The Stephen reaction converts a nitrile into an aldehyde with the same number of carbon atoms. Stannous chloride in hydrochloric acid reduces the nitrile to an imine (as its salt), and hydrolysis of the imine then gives the aldehyde. For example, ethanenitrile gives ethanal:
\( \displaystyle \ce{CH3CN} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{SnCl2/HCl}} \ce{CH3CHO} \)
Why the other options are incorrect
Option (1) Pd-BaSO4 + H2 — Incorrect. This is the reagent of the Rosenmund reduction, in which an acyl chloride (not a nitrile) is hydrogenated over palladium on barium sulphate to an aldehyde:
\( \displaystyle \ce{RCOCl + H2} \xrightarrow{\ce{Pd-BaSO4}} \ce{RCHO + HCl} \)
Option (3) AlH (i-Bu)2 + H2O — Incorrect. AlH(i-Bu)2 is diisobutylaluminium hydride (DIBAL-H). It also reduces a nitrile to an imine, which water hydrolyses to an aldehyde, but this is an alternative to the Stephen reaction, not the Stephen reaction itself:
\( \displaystyle \ce{CH3CN} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{AlH(i-Bu)2}} \ce{CH3CHO} \)
Option (4) DIBAL-H + H2O — Incorrect. This is the same reagent as Option (3), written by its short name. Besides nitriles, DIBAL-H also reduces esters to aldehydes.
NCERT (2026-27 edition):
From nitriles and esters – “Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde.
\( \displaystyle \ce{RCN + SnCl2 + HCl -> RCH=NH} \xrightarrow{\ce{H3O+}} \ce{RCHO} \)
This reaction is called Stephen reaction.
Alternatively, nitriles are selectively reduced by diisobutylaluminium hydride, (DIBAL-H) to imines followed by hydrolysis to aldehydes:
\( \displaystyle \ce{RCN} \xrightarrow[\text{2. }\ce{H2O}]{\text{1. }\ce{AlH(i-Bu)2}} \ce{R-CHO} \)
Similarly, esters are also reduced to aldehydes with DIBAL-H.”