A chemist has 18.5 g of 1-chlorobutane (92.5 g/mol) and wants an amine…

A chemist has $18.5\ \text{g}$ of 1-chlorobutane ($92.5\ \text{g/mol}$) and wants an amine. Chemists compare routes by the amount of product as well as by the number of steps. Assume that the potassium cyanide and the reducing agent are used in excess. Route 1 treats the halide with potassium cyanide to give pentanenitrile in a yield of $85\ \%$, and then reduces the nitrile with lithium aluminium hydride to pentan-1-amine in a yield of $70\ \%$. Route 2 is the Gabriel synthesis of butan-1-amine, with an overall yield of $60\ \%$. The route 1 amine has one carbon more than the halide. Use pentan-1-amine $87\ \text{g/mol}$ and butan-1-amine $73\ \text{g/mol}$. Find the amount of nitrile formed in route 1, the amount and the mass of pentan-1-amine, and the mass of butan-1-amine from route 2. Then say, in the solution, why a student who wanted butan-1-amine could not use route 1.