Question

Identify the reagents X, Y and Z.
${C_2}{H_5}Cl\xrightarrow{X}{C_2}{H_5}CN\xrightarrow{Y}{C_2}{H_5}C{H_2}N{H_2}\xrightarrow{Z}{ C_2}{H_5}C{H_2}NHCOC{H_3}$

Answer 1

The reagents X, Y and Z are alcoholic $KCN,{\text{ }}LiAl{H_4},{\text{ }}C{H_3}COCl$ respectively. Here $LiAl{H_4}$ is a reducing agent and used here for the reduction of
cyanide group to amine group.
Complete step by step answer:
Here the reaction involved is shown below:
Step 1:${C_2}{H_5}Cl + \mathop {KCN}\limits_{Alc.} \to {C_2}{H_5}CN + HCl$
The ethyl chloride reacts with alcoholic $KCN$ to give ethyl cyanide and $HCl$ as a side product.
This reaction is a substitution reaction in which the cyanide ion substitutes the chloride ion. The
chlorine atom acts as a good leaving group and immediately leaves through a process of $S{N^2}$
mechanism without any intermediate formation.
Step 2: Then ethyl cyanide reacts with reducing agent $LiAl{H_4}$ to give propyl amine as given
below:
${C_2}{H_5}CN\xrightarrow{{LiAl{H_4}}}{C_2}{H_5}C{H_2}N{H_2}$
$LiAl{H_4}$ has the tendency to reduce cyanide group to give amine group with one more number
of carbon as given in reactant. Lithium aluminium hydride is a very strong reducing agent.
Step 3: In this step, the propyl amine is acetylated with acetyl chloride to give acetamide as the
product. The reaction involved as
${C_2}{H_5}C{H_2}N{H_2} + C{H_3}COCl \to {C_2}{H_5}C{H_2}NHCOC{H_3}$
The process of introducing an acyl group into any molecule is called acylation.
Hence, the correct answer is the reagents used are $alc.{\text{ }}KCN,{\text{ }}LiAl{H_4}$ and acetyl
chloride.

Note:
Instead of acetyl chloride we can also use acid anhydride. The reaction carried out by acid
anhydride is also known as acylation. Both acyl chloride and acid anhydride are used as acylating
agent. In case of aromatic amine acylation is carried out in the presence of catalyst like pyridine
which removes $HCl$ formed during the reaction and shifts the equilibrium in forward reaction.