Compound '$\mathrm{A}$' undergoes following sequence of reactions to give compound '$\mathrm{B}$'. The correct structure and chirality of compound '$\mathrm{B}$' is [where $\mathrm{Et}$ is $-{\mathrm{C}}_2{\mathrm{H}}_5$]

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Compound '$\mathrm{A}$'

In the first step the given alkyl halide that is 2-bromo-3-methyl butane reacts with Mg in presence of dry ether to form a grignard reagent,This grignard reagent reacts with ${\mathrm{D}}_2\mathrm{O}$ to form a hydrocarbon (B)which has a  chiral carbon.

In the reaction with ${\mathrm{D}}_2\mathrm{O}$ grignard reagent behaves as a strong base.