Question
The best reagent for converting 2-phenylpropanamide into 2-phenylpropanamine is ________.
A.
excess $${H_2}$$
B.
$$B{r_2}$$ in aqueous $$NaOH$$
C.
iodine in the presence of red phosphorus
D.
$$LiAl{H_4}$$ in ether
Answer :
$$LiAl{H_4}$$ in ether
Solution :
\[\underset{\text{2-Phenylpropanamide}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix}
\,\,\,{{C}_{6}}{{H}_{5}} \\
|\,\,\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,CON{{H}_{2}}}}\,\xrightarrow{LiAl{{H}_{4}}/\operatorname{ether}}\] \[\underset{\text{2-Phenylpropanamine}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix}
\,\,\,{{C}_{6}}{{H}_{5}} \\
|\,\,\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}N{{H}_{2}}}}\,\]
No. of $$C$$ atoms remains same from amide to amine.