Question

Addition of one equivalent of ammonia to 1-bromoheptane gives a mixture of heptan-1-amine, some dialkylamine, some trialkylamine, and even some tetraalkylammonium bromide.

1. Give a mechanism to show how this reaction takes place, as far as the dialkylamine.
2. How would you modify the procedure to get an acceptable yield of heptan-1-amine?

Short answer

(a)

(b)Using a large excess of ammonia to avoid multiple alkylations of each nitrogen, we can acceptable yield of heptan-1-amine.

Step-by-step solution

Step-1. Explanation of part (a):

reaction occurs when ammonia molecule when reacts with 1-bromoheptane and ammonia molecule acts as a nucleophile and attacks at the carbon to which bromine is attached as leaving group. Then, further ammonia acts as a base and abstracts hydrogen so that charge neutralization can occur and heptan-1-amine forms. Then, product obtained reacts with another molecule of given reactant which leads to formation of dialkylamine which on further charge neutralization forms a good nucleophile and alkylation process can continue.

Mechanism of formation of various products

Step-2. Getting acceptable yield of heptan-1-amine:

When reactive nucleophiles like amines are reacted with electrophiles like alkyl halide, over-alkylation is problematic. This tends to be most acute when the desired products are primary or secondary amines. Using a large excess of ammonia can avoid multiple alkylations of each nitrogen and this can lead to obtaining an acceptable yield of heptan-1-amine.