Chemistry - ALCOHOL & ETHER Question with Solution | TestHub

Presence of excess of HI favours $S_{N}1$ mechanism. So, formation of products is controlled by the stability of the carbocation resulting in the cleavage of $\mathrm{C}-\mathrm{O}$ bond in protonated ether. Thus the product for given equation are ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{I},{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I},\mathrm{HOCH}{\mathrm{H}}_2-{\mathrm{CH}}_2\mathrm{OH}$.