Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Benzyl amine may be prepared by (1) \[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}\xrightarrow{LiAl{{H}_{4}}/ether}\] (2) \[{{C}_{6}}{{H}_{5}}CN\xrightarrow{LiAl{{H}_{4}}/ether}\] (3) \[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}\xrightarrow{NaOH/B{{r}_{2}}}\] (4) \[{{C}_{6}}{{H}_{5}}N{{O}_{2}}\xrightarrow{Sn/HCl}\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: B
Solution :
(1) \[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}\xrightarrow[{}]{LiAAl{{H}_{4}}/ether}\underset{benzyl\,a\min e}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}}}\,\] (2) \[{{C}_{6}}{{H}_{5}}CN\xrightarrow[{}]{LiAAl{{H}_{4}}/ether}\underset{benzyl\,a\min e}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}}}\,\] (3) \[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}\xrightarrow[{}]{NaOH+B{{r}_{2}}}\underset{aniline}{\mathop{{{C}_{6}}{{H}_{5}}N{{H}_{2}}}}\,\] (4) \[{{C}_{6}}{{H}_{5}}N{{O}_{2}}\xrightarrow[{}]{Sn/HCl}\underset{aniline}{\mathop{{{C}_{6}}{{H}_{5}}N{{H}_{2}}}}\,\] Hence, reaction (1) and (2) are used to prepare benzyl amine.
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