The compound will react most readily with ${\rm{NaOH}}$ to form methanol is

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_4}{{\rm{N}}^ + }{{\rm{I}}^ - }$

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

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{{\rm{S}}^ + }{{\rm{I}}^ - }$

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

NEET UG CHEMISTRY MCQ: AMINES

1 Hard

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Analyzing +4 | -1

The compound will react most readily with

\mathrm{NaOH}

to form methanol is

View answer and explanation

Correct option: A

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_4}{\mathrm{N}^ + }{\mathrm{I}^ - }

Option A (

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_4}{\mathrm{N}^ + }{\mathrm{I}^ - }

) is correct because it's a quaternary ammonium salt that undergoes Hoffmann elimination upon reaction with NaOH to form methanol and trimethylamine.