Chemistry - ALCOHOL & ETHER Question with Solution | TestHub

Phenols do not react further to give halides because the ${\mathrm{Sp}}^2$ - hybridized carbon of phenol cannot undergo nucleophilic substitution reaction needed for conversion to the halide.

Question Explanation:

Cleavage of an unsymmetrical ether with excess HI.

Concept:

Cleavage of aryl-alkyl ethers.

Solution:

The reactant is Benzyl phenyl ether. Cleavage by $\mathrm{HI}$ occurs at the bond that forms the most stable carbocation or is least hindered for ${\mathrm{S}}_{\mathrm{N}}2$. The $\mathrm{O}-{\mathrm{C}}_{phenyl}$ bond is strong due to resonance and does not cleave easily.

First Cleavage: The ether protonates, and ${\mathrm{I}}^{-}$ attacks the benzylic carbon (stable carbocation/less hindered), yielding Benzyl iodide (${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{I}$) and Phenol.

1. (${\mathbf{C}}_6{\mathbf{H}}_5\mathbf{OH}$).

$C_6{H}_{5}C{H}_2-O-C_6{H}_5+HI\to C_6{H}_{5}C{H}_{2}I+C_6{H}_{5}OH$

Second Cleavage (Excess HI, $\mathbf{\Delta }$): Since $\mathbf{HI}$ is in excess and

heated, the Phenol formed in step 1 reacts further with $HI$ to convert the $OH$

2. group to an iodide, despite the stronger $C-O$ bond.

$C_6{H}_{5}OH+HI\stackrel{excessHI,\Delta }{\to }{\mathbf{C}}_6{\mathbf{H}}_5\mathbf{I}+H_{2}O$

The final products are Benzyl iodide and Iodo benzene .