Question

Which of the following products are not correctly matched in the given reactions?
(A) ${C_2}{H_5}OC{H_3} + HBr\xrightarrow{{373K}}{C_2}{H_5}OH + C{H_3}Br$
(B) ${C_2}{H_5}O{C_2}{H_5} + 2HI\xrightarrow{{}}{C_2}{H_5}I + {C_2}{H_5}OH$
(C) $[{C_2}{H_5}O{C_2}{H_5} + excess HCl\xrightarrow{{\operatorname{co} ld}}{({C_2}{H_5})_2}{O^ + }HC{l^ - }]$
(D) ${(C{H_3})_3}CO{C_2}{H_5}\xrightarrow{{HI}}{(C{H_3})_3}CI + {C_2}{H_5}OH$

Answer 1

All the reactions shown above are reactions taking place between an unsymmetrical or mixed ether and halogen halide that means $HX(X = Br,I)$. All the reactions between ether and halogen halides are $S{N^2}$ reactions. The $S{N^2}$ reactions are bimolecular nucleophilic substitution reactions and they are one step process. This means that the attack of nucleophiles and the removal of leaving groups takes place simultaneously. As you can see, during the reaction between a hydrogen halide and an ether, the cleavage of alkyl ethers takes place.

Complete step-by-step answer:
Ethers are relatively stable towards bases, oxidising agents and reducing agents. The ether linkage in ethers is broken by action of acids by concentrated acids and at high temperature. When an ether reacts with hydrogen halides, $HX(X = Br,I)$ at 373 K, alcohols and alkyl halides will form through the $S{N^2}$ mechanism. Given below is the general chemical equation for the same.
$R - O - R' + HX\xrightarrow{{373K}}R' - OH + R - X$.

A dialkyl ether will produce an alcohol and alkyl halide on acidic cleavage by hydrogen halide at first and in the next step, the alcohol may react further to form a second mole of an alkyl halide.
The order of reactivity of hydrogen halides with ether is $HI > HBr > HCl$

(A) The reaction in option (A) is taking place between methoxyethane or ethyl methyl ether with hydrogen bromide at 373K. The reaction proceeds through $S{N^2}$ mechanism to produce ethyl alcohol and methyl bromide.
$$$$${C_2}{H_5}OC{H_3} + HBr\xrightarrow{{373K}}{C_2}{H_5}OH + C{H_3}Br$

(B) When diethyl ether reacts with excess of hydrogen iodide, it will produce ethyl iodide and water. Let us see how this happens.
In the first step, ether reacts with concentrated $HI$ and ether gets protonated forming an oxonium ion that means a hydrogen atom from $HI$ is added to ether as shown below.
[{C_2}{H_5} - O - {C_2}{H_5} + HI \rightleftarrows [{C_2}{H_5} - {\underset{\raise0.3em\hbox{\smash{\scriptscriptstyle\cdot\cdot}}}{\ddot O} ^ + } - {C_2}{H_5}]H + {I^ - }]

In the second step, iodine being a good nucleophile will attack one of the carbon atoms of oxonium ion formed in step 1. The nucleophile will help by separating alcohol molecules from the oxonium ion through the $S{N^2}$ mechanism. This will result in formation of ethyl alcohol and ethyl iodide. If excess of $HI$ is present during the reaction, the excess of $HI$ will react with alcohol at high temperature formed to produce ethyl iodide along with water.
$[{C_2}{H_5} - OH + HI \rightleftarrows [{C_2}{H_5} - {O^ + }{H_2}] + {I^ - }]$
$[{I^ - } + [{C_2}{H_5} - {O^ + }{H_2}] \to {C_2}{H_5} - I + {H_2}O]$
Therefore, the reaction shown in option (B) is incorrect. The correct reaction is shown below.
${C_2}{H_5}O{C_2}{H_5} + 2HI \to {C_2}{H_5}I + {H_2}O$

(C) Diethyl Ether when treated with excess of $HCl$does not show any reaction and hence, option (C) is correct.
$[{C_2}{H_5}O{C_2}{H_5} + \mathop {HCl}\limits^{excess} \xrightarrow{{\operatorname{co} ld}}\left( {{{({C_2}{H_5})}_2}\mathop O\limits^ + H} \right)C{l^ - }]$

(D) When a mixed ether with hydrogen iodide and produces an alkyl halide and an alcohol through $S{N^2}$ mechanism.
${(C{H_3})_3}CO{C_2}{H_5}\xrightarrow{{HI}}(C{H_3})CI + {C_2}{H_5}OH$

Therefore, the correct answer is option (B).

Note: Student should keep in mind that the ether cleavage will take place by reaction between an ether and both $HI$ and $HBr$, but not by $HCl$. This is because $HI$ and $HBr$ are stronger nucleophiles as compared to that of $HCl$. Also, $HCl$ is weak acid and cleavage of bonds in $HCl$ is quite difficult. Hence, it does not react with ethers through the $S{N^2}$ mechanism.