Question

$R-C\equiv N\stackrel{reagent}{\to }R-CHO$, reagent can be:

• A. $SnC{l}_2/HCl/H_{2}O$
• B. $DIBAL-H,H^{+}/H_{2}O$
• C. $Pd-BaS{O}_4/$ boiling xylene
• D. Both 1 and 2

Answer 1

The correct option is D Both 1 and 2

a) This is Stephen's reaction, which involves the conversion of an alkyl cyanide to an aldehyde using $SnC{l}_2/HCl$ as a reducing agent.
b) $DIBAL-H$, also reduces the nitrile group to an aldehyde.
c) $Pd-BaS{O}_4$ will not reduce alkyl cyanide into aldehyde.

$R-C\equiv N\stackrel{SnC{l}_2/HCl}{\to }R-CHO$
$R-C\equiv N\stackrel{DIBAL-H,H^{+}/H_{2}O}{\to }R-CHO$