Question

A single alkyl bromide reactant theoretically yields either of the given products, depending on the reaction conditions. Draw the structure of the alkyl bromide compound. alkyl bromide H3C (one compound only) CH OH CH3 CH3 H3C CH3 CH, O DMSO CH3

A single alkyl bromide reactant theoretically yields either of the given products, depending on the reaction conditions. Draw the structure of the alkyl bromide compound.

Answer 1

Concepts and reason

Two reactions with their corresponding products are given whose common reactant needs to be determined based on the type of reagent given. Generally, alkyl halides in the presence on base undergo either substitution reaction or elimination reaction depending on the reaction condition.

The given products indicate the substitution reaction which can be formed by ${{\rm{S}}_{\rm{N}}}1$ and ${{\rm{S}}_{\rm{N}}}2$ reaction. Thus, draw the correct structure of the alkyl bromide.

Fundamentals

• ${{\rm{S}}_{\rm{N}}}1$ reaction is unimolecular nucleophilic substitution reaction which takes place between a haloalkane and a nucleophile. The reaction takes place in two steps through a carbocation intermediate and the reaction product are formed with both retention as well as inversion of configuration. Consider the following scheme of a ${{\rm{S}}_{\rm{N}}}1$ reaction:

In the reaction, the nucleophile can attack from the back side as well as from the front side of the reactant.

• ${{\rm{S}}_{\rm{N}}}2$ reaction is bimolecular nucleophilic substitution reaction which takes place between a haloalkane and a nucleophile. The reaction takes place in one step through a transition state and the reaction is stereospecific in nature. Consider the following scheme of a ${{\rm{S}}_{\rm{N}}}2$ reaction:

In the reaction, the nucleophile attacks from the back side of the reaction as the front side is hindered by the leaving group.

The reaction product shows the substitution product, therefore the possible structures of the alkyl halide involved the reaction is as follows:

${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}$ is a weak nucleophile and the reaction in the presence of methanol generally follows ${{\rm{S}}_{\rm{N}}}1$ mechanism of substitution while methoxide ion is a strong nucleophile and the substitution reaction is followed by ${{\rm{S}}_{\rm{N}}}2$ mechanism. Hence, the correct structure of the common reactant is as follows:

Ans:

Hence, the correct structure of the common alkyl bromide of the given reaction is as follows: