Question

Find the product

Answer 1

The reaction proceeds as follows-

$\underset{\left(\text{Alkyl halide}\right)}{C{H}_3-C{H}_2-Br}\stackrel{{Et}_{3}N}{\to }\underset{\left(4°\text{Ammonium salt}\right)}{{{Et}_{4}N}^{+}{Br}^{-}}$

Hence the product is triethylammonium bromide.