jmc

First, we need to multiply the units digit: $5_6\times 4_6=20_{10}=32_6$. Hence, we write down a $2$ and carry-over the $3$. Evaluating the next digit, we need to multiply $1_6\times 4_6+3_6=7_{10}=11_6$. Thus, the next digit is a $1$ and $1$ is carried over. Finally, the leftmost digits are given by $3_6\times 4_6+1_6=13_{10}=21_6$. Writing this out: \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & & \stackrel{1}{3} & \stackrel{3}{1} & \stackrel{}{5}_6 \\ & & & \times & & & 4_6 \\ \cline{4-7} & & & 2 & 1 & 1 & 2_6 \\ \end{array}Thus, the answer is $\boxed{2112_6}$.