# Question #c5f12

Reduction of N-methylbenzamide with $L i A l H 4$ gives benzylmethylamine:
${C}_{6} {H}_{5} C O N H C {H}_{3}$ + $L i A l {H}_{4}$ $\to$ ${C}_{6} {H}_{5} C {H}_{2} N H C {H}_{3}$ + $L i A l {\left(O H\right)}_{4}$