${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{CONa}}$ on reaction with ${\rm{C}}{{\rm{H}}_3}{\rm{Br}}$ will give:

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{COC}}{{\rm{H}}_3}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{COC}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}$

${\rm{C}}{{\rm{H}}_3}{\rm{OC}}{{\rm{H}}_3}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OC}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}$

NEET UG CHEMISTRY MCQ: ALCOHOLS & PHENOLS & ETHERS

1 Medium

1 of 1

Application +4 | -1

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{CONa}

on reaction with

\mathrm{C}{\mathrm{H}_3}\mathrm{Br}

will give:

View answer and explanation

Correct option: D

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{COC}{\mathrm{H}_3}

The correct answer is D (

{{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_3}\mathrm{COC}{\mathrm{H}_3}}

) because tert-butoxide reacts with methyl bromide via an SN2 mechanism to form tert-butyl methyl ether.