Question

Consider Anisole is treated with $ HI $ in two conditions; determine the nature from A to D, it will be:
 $ C + D\mathop \leftarrow \limits^{HI(g)} C_{6}H_{5}OCH_{3}\mathop \to \limits^{conc.HI} A + B $
(A) A, B $ \to $ $ CH_{3}I $ and $ C_{6}H_{5}OH $ ; C , D $ \to $ $ CH_{3}OH $ and $ C_{6}H_{5}I $
(B) A, B $ \to $ $ CH_{3}OH $ and $ C_{6}H_{5}I $ , while C, D $ \to $ $ CH_{3}I $ and $ C_{6}H_{5}OH $
(C) All A, B, C and D are $ CH_{3}I $ and $ C_{6}H_{5}OH $
(D) A, B $ \to $ $ CH_{3}I $ and $ C_{6}H_{5}OH $ and no reaction in C,D

Answer 1

Hint : Anisole is a benzene ring with a substituted methoxy ( $ - O - CH_{3} $ ) group. We need to find out, on reaction of anisole with regular and concentrated Hydrogen iodide, it will yield only two products or four products. Since anisole has aromatic and aliphatic carbon-oxygen bonds, in comparison, the aliphatic bond is much easier to break. Remember the procedure of SN1 and SN2 nucleophilic reactions.

Complete Step By Step Answer:
Let us consider Hydrogen Iodide;
It is an acid, and an acid can dissociate into protons and electrons, here the $ {I^ - } $ ions are the electrons.
If we consider the reaction between Hydrogen Iodide and Anisole; Anisole will react with the proton $ {H^ + } $ given by the acid and so it is able to form another ion called the methyl(phenyl) oxonium ion. So now there is a $ - CH_{3} $ available for reaction. The iodide ion is able to react with the previously formed ion, but it needs to break either the aromatic carbon-oxygen bond or aliphatic carbon-oxygen bond. Since it is easier to break the aliphatic bond, the iodide ion will react with the methyl cation to form iodomethane, and in turn allows the previous product to be phenol.
But we need to know that both $ HI $ and $ conc.HI $ give react with different nucleophilic substitution reactions: SN1 and SN2.
With $ HI(g) $ the reaction follows an SN1 substitution because of the carbocation stability.
 $ C_{6}H_{5} - O - CH_{3}\mathop \to \limits_{HI(g)}^{SN2} \;CH_{3}I + C_{6}H_{5}OH $
With $ conc.HI $ , due to steric hindrance, we proceed with SN2 substitution.We understand that while forming the cation, the phenyl cation ( $ C6H{5^ + } $ ) is less stable than $ CH{3^ + } $ , that is why instead of phenyl cation, $ CH{3^ + } $ was formed after SN1 reaction.
 $ C_{6}H_{5} - O - CH_{3}\mathop \to \limits_{HI(conc.)}^{SN1} CH_{3}I + C_{6}H_{5}OH $
So now we can see that in both cases of regular and concentrated Hydrogen iodide, the products yielded are $ CH_{3}I $ and $ C_{6}H_{5}OH $ .
The correct answer is: option (C) All A, B, C and D are $ CH_{3}I $ and $ C_{6}H_{5}OH $ .

Note :
Benzene is a highly flammable substance. It can be formed by either natural processes or also by human activity. It is used for making detergents, pesticides and drugs. If exposed to, it is extremely hazardous to one’s health. Tobacco smoke and other emissions from industries and motor vehicles contain low levels of benzene, while paint and detergents contain higher levels of benzene.