Question

From the list of structures on the right, select the major product formed when the following alkyl bromide:

1) is treated with sodium methoxide in DMSO.

2) is treated with sodium t-butoxide in DMSO.

Answer 1

Concepts and reason

The concept used to solve the given problem is the knowledge of nucleophilic substitution reaction and elimination reaction and its reactivity variations due to the effect of substrate and reagents.

Fundamentals

Nucleophilic substitution reaction involves the primarily attack of nucleophile to the carbon of alkyl halide, which results in removal of the halogen.

The elimination reaction proceeds with the removal of the leaving group and the adjacent ${\rm{\beta }}$ - H which results in the formation of respective alkene.

In the case of bulkier reagent or the bulkier substrate, the elimination reaction is preferred over the substitution reaction.

(1)

The product of the reaction cannot be molecule A which is shown as follows:

The product of the reaction cannot be molecule B which is shown as follows:

The product of the reaction cannot be molecule D which is shown as follows:

The product of the reaction cannot be molecule E which is shown as follows:

The product of the reaction cannot be molecule F which is shown as follows:

Formation of the stable tertiary carbocation from the tertiary alkyl bromide takes place.

The formation of tertiary carbocation via ${{\rm{E}}_{\rm{1}}}$ mechanism is shown below:

The removal of ${\rm{\beta }}$ - H on the attack of sodium methoxide in presence of polar aprotic solvent DMSO leads to the formation of more substituted alkene.

The reaction of the removal of ${\rm{\beta }}$ - H by the methoxide ion which acts as a base is as follows:

(2)

The product of the reaction cannot be molecule A which is shown as follows:

The product of the reaction cannot be molecule B which is shown as follows:

The product of the reaction cannot be molecule C which is shown as follows:

The product of the reaction cannot be molecule D which is shown as follows:

The product of the reaction cannot be molecule F which is shown as follows:

Formation of the stable tertiary carbocation from the tertiary alkyl bromide.

The formation of tertiary carbocation via ${{\rm{E}}_{\rm{1}}}$ mechanism is shown below:

The removal of ${\rm{\beta }}$ - H on the attack of sodium t-butoxide in presence of polar aprotic solvent DMSO leads to the formation of less substituted alkene.

The reaction of the removal of ${\rm{\beta }}$ - H by the t-butoxide ion which acts as a base is as follows:

Ans: Part 1

The major product formed when the alkyl bromide is treated with sodium methoxide in DMSO is C which is as shown as follows: