Question

Convert Ethyl chloride to propanamine.

Answer 1

The reactant ethyl chloride contains a halogen atom and the product has a primary amine functional group. One can think about the possible reagents which can form this product. This is a multistep reaction. The reactions of ${S_N}1$ and reduction can be a good method to prepare this product.

Complete step by step answer:

1. First of all we will analyze the functional groups that are given in reactant and product in order to know which reactants can be used in the given conversion. The ethyl chloride has a halogen atom in it and the propanamine has a primary amine functional group.
2. Now let's discuss how we can do this conversion. The reactant has halogen atom chlorine and we know those halogen atoms are a good leaving group which means we can use ${S_N}1$ reaction and add the group of cyanide which contains nitrogen and one carbon atom which will also increase the initial chain of carbon. The reaction will go as below,
   ${\text{C}}{{\text{H}}_3} - C{H_2} - Cl\xrightarrow[{Ethnol({\text{Protic solvent}})}]{{NaCN}}C{H_3} - C{H_2} - CN$
3. Now as we added a ${\text{ - CN}}$ group which increased the carbon chain and a nitrogen atom. The next step is about the conversion of the cyanide group into the primary amine group which we can do by reducing the cyanide group that will break the triple bond between carbon and nitrogen and forms a primary amine. The reaction will go as below,
   $C{H_3} - C{H_2} - CN\xrightarrow[{(reduction)}]{{LiAl{H_4}}}C{H_3} - C{H_2} - C{H_2} - N{H_2}$
4. In the above reaction the ethyl chloride is converted to propanamine.

Note:
The reagent $LiAl{H_4}$ is a strong reducing agent that reduces the carbon-nitrogen triple bond to the amine group. This reagent is a good source of hydride i.e. ${H^ - }$ which reduces double and triple bonds. Aldehydes, ketones, and esters are converted to alcohol by using this reagent.