Question

$X=$ A number of reactions in which the final product is an aldehyde. Find out $X$ .
$\left(a\right)Ph−C≡N\frac{\left(i\right){\left(SnCl\right)}_2+HCl}{\left(ii\right)H_3{O}^{+}}$
$\left(b\right)Ph−C≡N\stackrel{{\left(LiAIH\right)}_4}{\to }$
$\left(c\right)Ph−COCl\stackrel{H_2/Pd−{\left(BaSO\right)}_4}{\to }$
$\left(d\right)Ph−COOEt\underset{ii)H_{2}O}{\overset{i)DiBAL−H,\text{Hexane},−{\left(78\right)}^{{}^{∘}}C}{\to }}$

Answer 1

Answer: 3

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 $SnC{l}_2+HCl$ forming $Ph−CH=NH$ which further undergoes Hydrolysis in presence of water in acidic medium forming aldehyde.
$Ph−CN\stackrel{SnC{l}_2+HCl}{\to }Ph−CH=NH\stackrel{H_3{O}^{+}}{\to }Ph−CHO$
It is known as Stephen's reduction.
(b) $LiAl{H}_4$ is a reducing agent so phenyl cyanide undergoes reduction reaction in presence of it and forms $Ph−C{H}_2−N{H}_2$
(c) Benzoyl chloride undergoes reduction and forms aldehyde in presence of $H_2/Pd−BaS{O}_4$ .
$Ph−CO−Cl\stackrel{H_2/Pd−BaS{O}_4}{\to }Ph−CHO$
It is known as Rosenmund reduction.
(d) Ethyl Phenyl ester under DiBAL-H forms aldehyde group.
$Ph−COOEt\stackrel{DiBAL−H\text{Hexane}}{\to }Ph−CHO$