Question

$C_6{H}_5{NH}_2\underset{{NaNO}_2+HCl}{\overset{{0-5}^{0}C}{\to }}\left(A\right)\underset{{Cu}_2{\left(CN\right)}_2}{\overset{KCN}{\to }}\left(B\right)\stackrel{H^{+}/H_{2}O}{\to }\left(C\right)$

The product (C) is :

• A. $C_6{H}_5{CH}_2{NH}_2$
• B. $C_6{H}_{5}COOH$
• C. $C_6{H}_{5}COC{H}_3$
• D. All of the above

Answer 1

The correct option is B $C_6{H}_{5}COOH$

$C_6{H}_{5}N{H}_2+NaN{O}_2+HCl\to Ph{N}_2^{+}C{l}^{-}$

Now,
$Ph{N}_2^{+}C{l}^{-}+KCN\left(C{u}_2\right(CN{)}_2)\to PhCN$

This reaction is used to replace the amino group with the cyano group.
Now, the above-formed cyanide will hydrolyze to give carboxylic acid.

$PhCN+H_3{O}^{+}\to PhCOOH$