An alkyl chloride 'X' reacts with magnesium metal in dry ether to...

The reaction of an alkyl chloride $X$ ($R-Cl$) with magnesium in dry ether forms a Grignard reagent $RMgCl$.

$R-Cl + Mg \xrightarrow{\text{dry ether}} RMgCl$

The Grignard reagent reacts with solid $CO_2$ followed by acid hydrolysis to form a carboxylic acid $RCOOH$.

$RMgCl + CO_2 \xrightarrow{H_3O^+} RCOOH$

The product is given as $2$-methylpropanoic acid, which has the structure $(CH_3)_2CH-COOH$.

Comparing $RCOOH$ with $(CH_3)_2CH-COOH$, the alkyl group $R$ is the isopropyl group, $(CH_3)_2CH-$.

Therefore, the alkyl chloride $X$ is isopropyl chloride, which is $2$-chloropropane, $(CH_3)_2CH-Cl$.

Answer: $2$-Chloropropane