Match the following: Column I Column I | vedclass

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(B) . $CH_3-CHBr-CD_3$ undergoes $E2$ elimination with alc. $KOH$ to form $CH_2=CH-CD_3$ $(A-Q)$.$B$. $Ph-CHBr-CH_3$ shows a primary kinetic isotope e

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$A-Q; B-Q; C-R,S; D-P,S$

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(B) . $CH_3-CHBr-CD_3$ undergoes $E2$ elimination with alc. $KOH$ to form $CH_2=CH-CD_3$ $(A-Q)$.$B$. $Ph-CHBr-CH_3$ shows a primary kinetic isotope effect $(k_H/k_D > 1)$ in $E2$ reactions,so it reacts faster than the deuterated analog $(B-Q)$.$C$. $Ph-CH_2-CH_2Br$ undergoes $E1cb$ mechanism due to the acidic $\beta$-hydrogens stabilized by the phenyl ring,and it follows first-order kinetics $(C-R,S)$.$D$. $PhCH_2CH_2Br$ and $PhCD_2CH_2Br$ reacting at the same rate suggests the rate-determining step does not involve $C$-$H$/$C$-$D$ bond breaking,characteristic of $E1$ reactions $(D-P,S)$.

Column $I$	Column $II$
$A$. $CH_3-CHBr-CD_3$ on treatment with alc. $KOH$ gives $CH_2=CH-CD_3$ as a major product.	$P$. $E1$ reaction
$B$. $Ph-CHBr-CH_3$ reacts faster than $Ph-CHBr-CD_3$.	$Q$. $E2$ reaction
$C$. $Ph-CH_2-CH_2Br$ on treatment with $C_2H_5OD / C_2H_5O^-$ gives $Ph-CD=CH_2$ as the major product.	$R$. $E1cb$ reaction
$D$. $PhCH_2CH_2Br$ and $PhCD_2CH_2Br$ react with same rate.	$S$. First order reaction

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