Compound 'A' undergoes following sequence of reactions to give compound 'B'. The correct structure and chirality of compound 'B' is [where Et is $$-C_2H_5$$]

The starting compound is $$\mathrm{2\text{-}Bromo\text{-}3\text{-}methylbutane}$$.

It reacts with magnesium in dry ether $$\mathrm{(Mg/Et_2O)}$$.

Magnesium inserts into the $$\mathrm{C-Br}$$ bond to form a Grignard reagent.

$$\mathrm{CH_3-CH(Br)-CH(CH_3)-CH_3 \xrightarrow{Mg/Et_2O} CH_3-CH(MgBr)-CH(CH_3)-CH_3}$$

The Grignard reagent reacts with heavy water $$\mathrm{(D_2O)}$$ and abstracts $$\mathrm{D^+}$$.

$$\mathrm{CH_3-CH(MgBr)-CH(CH_3)-CH_3 + D_2O \longrightarrow CH_3-CH(D)-CH(CH_3)-CH_3}$$

The final product is $$\mathrm{2\text{-}Deuterio\text{-}3\text{-}methylbutane}$$.

The carbon containing deuterium is attached to $$\mathrm{-H,\ -D,\ -CH_3,\ -CH(CH_3)_2}$$.

Since all four groups are different, the carbon atom becomes a $$\mathrm{Chiral\ Centre}$$.

Therefore, compound $$\mathrm{B}$$ is $$\mathrm{Chiral}$$.

Correct option: $$\mathrm{C}$$