Question

$X=$ A number of reactions in which the final product is an aldehyde. Find out $X$ .
$\left(a\right)Ph-C\equiv N\frac{\left(i\right) \left(SnCl\right)_{2} + HCl}{\left(ii\right) H_{3} O^{+}}$
$\left(b\right)Ph-C\equiv N\xrightarrow{\left(LiAIH\right)_{4}}$
$\left(c\right)Ph-COCl\xrightarrow{H_{2} / Pd - \left(BaSO\right)_{4}}$
$\left(d\right)Ph-COOEt \xrightarrow [ii) H_{2} O]{i) DiBAL - H , \text{Hexane} , - \left(78\right)^{^\circ } C}$

Answer 1

Of all the reactions given, reaction in option a, c and d produces aldehyde. The reactions take place as,
(a) Phenyl cyanide undergoes Reduction in presence of $SnCl_{2}+HCl$ forming $Ph-CH=NH$ which further undergoes Hydrolysis in presence of water in acidic medium forming aldehyde.
$Ph-CN \xrightarrow{SnCl_{2} + HCl}Ph-CH=NH\xrightarrow{H_{3} O^{+}}Ph-CHO$
It is known as Stephen's reduction.
(b) $LiAlH_{4}$ is a reducing agent so phenyl cyanide undergoes reduction reaction in presence of it and forms $Ph-CH_{2}-NH_{2}$
(c) Benzoyl chloride undergoes reduction and forms aldehyde in presence of $H_{2}/Pd-BaSO_{4}$ .
$Ph-CO-Cl\xrightarrow{H_{2} / Pd - BaSO_{4}} Ph-CHO$
It is known as Rosenmund reduction.
(d) Ethyl Phenyl ester under DiBAL-H forms aldehyde group.
$Ph-COOEt\xrightarrow{DiBAL - H \text{Hexane}}Ph-CHO$