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Question: What will happen when diethyl ether is treated with hot and concentrated HI? A.Ethyl alcohol B.E...

What will happen when diethyl ether is treated with hot and concentrated HI?
A.Ethyl alcohol
B.Ethyl iodide
C.Methyl iodide
D.Methanol

Explanation

Solution

In the given reaction diethyl ether is treated with hot and concentrated HI. HI is an acid and when it is hot and concentrated it means we are having an excess of HI. Iodide ion produced from HI in this reaction will act as a nucleophile. It is a type of nucleophilic substitution reaction.
Complete step by step answer:
Let us now look into the mechanism of the given reaction:
Firstly the HI given will get ionized:
HIH +  + I - {\text{HI}} \to {{\text{H}}^{\text{ + }}}{\text{ + }}{{\text{I}}^{\text{ - }}}
The iodide ion produced will attack one of the Carbon-oxygen bonds, gets attached to carbon, the hydrogen ion produced in the first step will get attached to a negatively charged oxygen atom:
CH3CH2O - CH2CH3 + HI{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{O - C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ + HI}} CH3CH2OH + I - CH2 - CH3 \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH + I - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{3}}}
The ethanol produced in the above step still has a carbon-oxygen bond and we are provided with an excess of HI (concentrated), so the iodide ion will now attack the carbon-oxygen bond of ethanol similarly, and the hydrogen ion will get attached to hydroxide ion produced:
CH3CH2 - OH + HICH3CH2 - I + H2O{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - OH + HI}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - I + }}{{\text{H}}_{\text{2}}}{\text{O}}
Thus, the overall reaction between diethyl ether and hot and concentrated HI will be:
CH3CH2OCH2CH3 + 2HI{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ + 2HI}} 2CH3CH2I + H2O \to 2{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{I + }}{{\text{H}}_{\text{2}}}{\text{O}}
Means, the given reaction will give two moles of ethyl iodide.
Hence, option B is the correct answer.

Note: The above reaction is an example of nucleophilic substitution, which is one of the types of substitution reactions.
During a nucleophilic substitution reaction, a nucleophile attached to a molecule is replaced by another molecule.
A nucleophile is a species which is rich in electrons, meaning having an excess of electrons.
In the above reaction ethoxide ion (a nucleophile) is replaced by iodide ion (another nucleophile)
Another type of substitution reaction is electrophilic substitution reaction, in which an electrophilic group is replaced by another electrophilic group.