Question

Alkyl bromides react with sodium iodine (Nal) to yield alkyl iodides by a reaction that is a biomolecular nucleophillic substitution. Which compound will react faster: tert-butyl bromide or methyl bromide?

Answer 1

Ans- methyl bromide

Explanation-

A bimolecular nucleophillic substitution i.e SN2 reaction is a substitution reacation of nucleophile which is concerted in nature.

That means the elimination of one gropu from the reactant and the substitution of the incoming nucleophile takes place  in one step.

So to have such a mechanism, primary C-chains (i.e the C-atom attached to the Br or anny leaving group shold be bonded with only 1 C-atom or no C-atom i.e Methyl) is the most suitable one.

Where as if we takke a tertiary C-chains (i.e the C-atom attached to the Br or any leaving group bonded with 3 more C-atoms ) is the most unsuitable one.

Because in the later case, the bulky size of the tertiary group make the incoming nucleophile difficult to reach the target annd simultaneously removel of the leaving group in one step.

So for the SN2 reaction of Alkyl bromides with sodium iodine, methyl bromide (primary) is faster than tert-butyl bromide (tertiary)

However if we tak about SN1 raection, then tert-butyl bromide will have better reaction than methyl bromide