Statement 1: Alkaline Hydrolysis of methyl bromide gives product with inverted configuration

Statement 2: Alkaline Hydrolysis of methyl bromide follows ${\mathrm{S}}_{\mathrm{N}}2$ mechanism.

Methyl bromide on alkaline hydrolysis gives ${\mathrm{CH}}_3\mathrm{OH}$.

Mechanism is ${\mathrm{S}}_{\mathrm{N}}2$ because steric hindrance in ${\mathrm{CH}}_3\mathrm{Br}$ is minimum.

It is to be noted that the product has no chiral centre.

So the term “inversion in configuration” cannot be used.