Question

The best reagent for converting 2-phenylpropanamide into 2-phenylpropanolamine 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 1

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

$C{H}_3-\stackrel{C_6{H}_5}{\stackrel{|}{C}H}-CON{H}_2\stackrel{LiA{l}_4/ehter}{\to }C{H}_3-\stackrel{C_6{H}_5}{\stackrel{|}{C}H}-C{H}_{2}N{H}_2$
2-phenylpropanamide 2-phenylpropanamine