Question
The best reagent for converting, 2-phenylpropanamide into 1-phenylethanamine is _______.
A.
$${\text{excess}}\,\,{H_2}/Pt$$
B.
$$NaOH/B{r_2}$$
C.
$$NaB{H_4}/{\text{methanol}}$$
D.
$$LiAl{H_4}/{\text{ether}}$$
Answer :
$$NaOH/B{r_2}$$
Solution :
By Hoffmann bromamide degradation reaction, the amine formed contains one carbon less than that present in the amide.
\[\underset{2\text{-Phenylpropanamide}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix}
\,\,\,{{C}_{6}}{{H}_{5}} \\
|\,\,\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,CON{{H}_{2}}}}\,+B{{r}_{2}}+4NaOH\to \] \[\underset{\text{1-Phenylethanamine}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix}
\,\,\,{{C}_{6}}{{H}_{5}} \\
|\,\,\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,N{{H}_{2}}}}\,+N{{a}_{2}}C{{O}_{3}}+\] \[2NaBr+2{{H}_{2}}O\]