Question

Compound ${C_2}{H_6}O$ has two isomers $X$ and $Y$. On reaction with ${\text{HI}}$, $X$ gives alkyl iodide and water while $Y$ gives alkyl iodide and alcohol. Compound $X$ and $Y$ are respective:
A. ${C_2}{H_5}O{C_2}{H_5}{\text{ and }}C{H_3}O{C_2}{H_5}$
B. $C{H_3}OC{H_3}{\text{ and }}{C_2}{H_5}OC{H_3}$
C. ${C_2}{H_5}OH{\text{ and }}C{H_3}OC{H_3}$
D. $C{H_3}OH{\text{ and }}C{H_3}OC{H_3}$

Answer 1

Primary alcohols react with ${\text{HI}}$ and undergoes nucleophilic substitution reaction through $S{N^2}$ mechanism, and not by $S{N^1}$ mechanism because carbocation formed by primary alcohol is not stable.

Complete step by step answer:
Step 1:
For drawing the isomers of ${C_2}{H_6}O$, it can be seen that ${C_2}{H_6}O$ fits the general formula ${C_n}{H_{2n + 2}}O$, which is the general formula for alcohol and ethers. To draw alcoholic isomers, there are only two possibilities of placing the $- OH$ group. Those are either placed next to first carbon or next to second carbon. In both the cases, the result will be the same as the carbon at which the $- OH$ group will be attached, will be considered the first carbon of the parent chain, and it will be named as ethanol or ethyl alcohol. Its formula will be $C{H_3} - C{H_2} - OH$.
Step 2:
The other possibility to put oxygen is in between the two carbons which gives such a structure where both the carbons are connected with $3$ hydrogen and the oxygen is placed between them like this $C{H_3} - O - C{H_3}$. It is commonly known as dimethyl ether, and in IUPAC, methoxy methane.
Step 3:
So, Let $X$ be $C{H_3}C{H_2}OH$ and $Y$ be $C{H_3}OC{H_3}$. It is given that $X$ on reaction with ${\text{HI}}$ gives alkyl iodide, this criteria is fulfilled by our assumptions as alcohols on reaction with ${\text{HI}}$ undergo nucleophilic substitution reaction and yields alkyl halide and water. The reaction will be:
$C{H_3} - C{H_2} - OH + HI \to {C_2}{H_5}I + {H_2}O$
Here, $C{H_3} - C{H_2}^ +$ carbocation will be formed where nucleophile ${I^ - }$ will attach to form ${C_2}{H_5}I$ and by product is ${H_2}O$.
Step 4:
Second condition is given that $Y$ reacts with ${\text{HI}}$ to form alkyl iodide and alcohol. This condition is also fulfilled as our assumption, $C{H_3} - O - C{H_3}$ undergoes a nucleophilic substitution reaction with ${\text{HI}}$ in the following manner:
$C{H_3} - O - C{H_3} + HI \to C{H_3}I + C{H_3}OH$
Here, ${\text{HI}}$ will break into ${H^ + }{\text{ and }}{I^ - }$ such that ${I^ - }$ will attack one of the carbocation $CH_{3}^{+}$ to form alkyl halide, $CH_{3}I$, and the left $H^{+}$ will react with
$CH_{3}O^ {-}$ to form methyl alcohol, $CH_{3}OH$.
So, both the given conditions are matched with the assumptions $X$ and $Y$. Hence, the assumption is valid and compound $X$ is $CH_{3}CH_{2}OH$, and compound $Y$ is $CH_{3}OCH_{3}$.
Option (a), (b), and (c) are eliminated because they do not match the general formula of the given compound, $C_{2}H_{6}O$ $C_{n}H_{2n + 2}O$.
Option (d) is the correct option.

Note: In reaction of alcohol with ${\text{HI}}$, ${\text{HI}}$ should be concentrated, otherwise, water in excess would be useless as $O{H^ - }$ will be added back to it without the reaction also, and in reaction of ether with ${\text{HI}}$, ${\text{HI}}$ should be cold. In case where both the groups of the ether are different, ${I^ - }$ will attack the less crowded group.