Match the following

$$\text { (A) Match with (Q) }$$
The formation of $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ can be explained on the basis of the fact that $\mathrm{C}-\mathrm{D}$ bond is much stronger than $\mathrm{C}-\mathrm{H}$ bond. (B) match with (Q) Reactivity of $\mathrm{PhCHBrCH}_3$ is greater than $\mathrm{Ph} \mathrm{CHBrCD}_3$ because $\mathrm{C}-\mathrm{D}$ bond is more stronger than $\mathrm{C}-\mathrm{H}$ bond. $$\text { (C) match with (R) and (S) }$$

In the step (II), a slow unimolecular elimination occurs in the conjugate base of the reactant and hence this mechanism is called $E_1 C B$ or carbanion mechanism. Since step (I) must be reversible, if ethanol containing $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ is used as solvent, it would be expected that the original bromide would incorporate deuterium (D). $$\text { (D) Match with (P) and (S) }$$ Step I. $\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Slow }}{\longrightarrow} \mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-}$ Step II. $\mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{Ph}-$\mathrm{CH}=\mathrm{CH}$_2+\mathrm{H}^{-}$ Rate $\propto\left[\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}\right]$ Similarly Step II. $\mathrm{PhCD}_2 \stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow}$\mathrm{PhCD}=\mathrm{CHD}$+\mathrm{H}^{+}$ Rate $\propto\left[\mathrm{PhCD}_2-\mathrm{CH}_2 \mathrm{Br}\right]$ Hence, $E_1$ reaction and first order kinetics.