The reaction products of ${{\rm{C}}_6}{{\rm{H}}_5}{\rm{OC}}{{\rm{H}}_3} + {\rm{HI}}\mathop \to \limits^\Delta $ is:

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{OH}} + {\rm{C}}{{\rm{H}}_3}{\rm{I}}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{I}} + {\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_3} + {\rm{HOI}}$

${{\rm{C}}_6}{{\rm{H}}_6} + {\rm{C}}{{\rm{H}}_3}{\rm{OI}}$

JEE CHEMISTRY MCQ: ALCOHOLS & PHENOLS & ETHERS

1 Easy

1 of 1

Understanding +4 | -1

The reaction products of

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OC}{\mathrm{H}_3} + \mathrm{HI}\mathop \to \limits^\Delta

is:

View answer and explanation

Correct option: A

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH} + \mathrm{C}{\mathrm{H}_3}\mathrm{I}

Option A (

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH} + \mathrm{C}{\mathrm{H}_3}\mathrm{I}

) is correct because the reaction of anisole (

\mathrm{C}_6{\mathrm{H}_5}\mathrm{OC}{\mathrm{H}_3}

) with HI results in the cleavage of the C-O bond, forming phenol and methyl iodide.