In a reaction, $\left(\mathrm{CH}_3\right)_2 \mathrm{CHMgBr}+\mathrm{CO}_2 \xrightarrow[\text { ether }]{\text { dry }} \mathrm{A} \xrightarrow[\text { dil HCl }]{\mathrm{H} \cdot \mathrm{OH}} \mathrm{~B} .$ Find the product ' $B$ ' of above reaction.

The Grignard reagent isopropyl magnesium bromide, $(\mathrm{CH}_3)_2\mathrm{CHMgBr}$, undergoes carboxylation with $\mathrm{CO}_2$ in dry ether to form a magnesium carboxylate intermediate.

$\left(\mathrm{CH}_3\right)_2\mathrm{CHMgBr} + \mathrm{CO}_2 \to \left(\mathrm{CH}_3\right)_2\mathrm{CHCOO}^- \mathrm{MgBr}^+$

Hydrolysis with dilute $\mathrm{HCl}$ yields the carboxylic acid product.

$\left(\mathrm{CH}_3\right)_2\mathrm{CHCOO}^- \mathrm{MgBr}^+ + \mathrm{H}_3\mathrm{O}^+ \to \left(\mathrm{CH}_3\right)_2\mathrm{CHCOOH} + \mathrm{Mg}^{2+} + \mathrm{Br}^- + \mathrm{H}_2\mathrm{O}$

The $\mathrm{IUPAC}$ name of $(\mathrm{CH}_3)_2\mathrm{CHCOOH}$ is 2-methylpropanoic acid, corresponding to option B.