Question

Identify the final product $(Z)$ in the following sequence of reactions.
$C_6H_5COOH \xrightarrow{(i) LiAlH_4, (ii) PBr_3} X \xrightarrow{KCN} Y \xrightarrow{LiAlH_4} Z$

• A. $C_{6}H_{5}CH_{2}NH_{2}$
• B. $C_{6}H_{5}CH_{2}CH_{2}NH_{2}$
• C. $C_{6}H_{5}CH_{2}CH_{2}CH_{2}NH_{2}$
• D. $C_{6}H_{5}\underset{C H_{3} \, \, }{\underset{\left|\right.}{C} H -}NH_{2}$

Answer 1

Answer: (B)

This conversion is as follows
$C_{6}H_{5}COOH\xrightarrow{L i A l H_{4}}C_{6}H_{5}CH_{2}OH\xrightarrow{P B r_{3}}\underset{\left(X\right)}{C_{6} H_{5} C H_{2} B r}\xrightarrow{K C N}\underset{\left(Y\right)}{C_{6} H_{5} C H_{2} C N}$
$\xrightarrow{L i A l H_{4}}\underset{\left(Z\right)}{C_{6} H_{5} C H_{2} C H_{2} \left(NH\right)_{2}}$