As compound $A$ does not react with sodium metal or potassium permanganate, it cannot be an alcohol.
As compound $A$ dissolves in conc. $H_2SO_4$, it may be an ether.
As compound $A$ on heating with excess of $HI$ gives a single alkyl halide, therefore, compound $A$ must be a symmetrical ether.
The only symmetrical ether having molecular formula $C_4H_{10}O$ is diethyl ether. Thus compound $'A\ '$ is diethylether, $\ce{CH_3-CH_2-O-CH_2-CH_3.}$

$\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3\xrightarrow{\text{conc.}\text{H}_2\text{SO}_4}\big[\text{CH}_3-\text{CH}_2-\stackrel{{+}}{\hbox{O}}-\text{CH}_2-\stackrel{{. \ .}}{\hbox{O}}-\text{CH}_3\big]$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}$
$\ \ \ \ \ \ \ \ \ \ \text{Diethyl ether} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Soluble oxonium salt}$
$\text{CH}_3\text{ CH}_2\text{ OCH}_2\text{ CH}_3\xrightarrow{ \ \ \ \ \ \ \text{m} \ \ \ \ \ \ \ }2\text{CH}_3\text{ CH}_2-\text{I}$
$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta \ \ \ \ \ \ \ \ \ \text{Ethyl iodide}$
$\ \ \ \ \ \ \ \ \ \ \ \ \text{Diethyl ether}$

Alcohols, Phenols, and Ethers — Chemistry STD 12 Science — Question

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