Question

The best reagent for converting $2-$ phenylpropanamide into $2-$phenylpropanamine is_____.

• A. Iodine in the presence of red phosphorus
• B. $B{r}_2$ in aqueous $NaOH$
• C. $LiAl{H}_4$ in ether
• D. Excess $H_2$

Answer 1

The correct option is C $LiAl{H}_4$ in ether

Analysing the options

Option $\left(A\right)$:
Only excess $H_2$ without any catalyst can’t reduce amide to amine.

Option $\left(B\right)$:
In this case Hoffmann bromamide degradation reaction will take place where primary amine with one less carbon will be obtained.


$2-$phenylpropanamide


$1-$phenylethanamine

Option $\left(C\right)$:
$2-$phenylpropanamide can not be converted into $2-$phenylpropanamine in the presence of $I_2+redP$.

Option $\left(D\right)$:
$LiAl{H}_4$ in ether reduces $2-$phenylpropanamide into $2-$phenylpropanamine.


$2-$phenylpropanamide $2-$phenylpropanamine
Hence, the correct answer is option $\left(D\right)$.