Question

What mechanism would be followed for the following reaction?
$\left(Ph\right)(C{H}_3{)}_{2}C-Br+NaOMe\to$

• A. $S_{N}2$
• B. $S_{N}1$ and $E_1$
• C. $E_2$
• D. $S_{N}1$ and $E_2$

Answer 1

The correct option is C $E_2$

$NaOMe$ is a strong base. It act as base than a nucleophile in presence of acidic hydrogens.
Substrate is a tertiary alkyl halide which is highly steric for the $S_{N}2$ reaction to occur. Also, only weak nucleophile favour $S_{N}1$ reaction.
Hence, due to $NaOMe$ strong basic nature it will favour $E2$ mechanism over $E1$.