Alkyl Halides - Result Question 44
####44. Match the following :
(2006, 3M)
| 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. | $Ph-CH _2-CH _2 Br$ on treatment with $C _2 H _5 OD / C _2 H _5 O^{-}$gives $Ph-CD=CH _2$ as the major product. |
r. | E1CB reaction |
| D. | PhCH $CH _2 Br$ and $PhCD _2 CH _2 Br$ react with same rate. |
s. | First order reaction |
Show Answer
Answer:
Correct Answer: 44.
$A \longrightarrow q,$ $B \longrightarrow q,$ $C \longrightarrow r,$ $D \longrightarrow p,s$
Solution:
- A. $CH _3-CHBr-CD _3 \xrightarrow[E 2]{\text { Alc. } KOH} CH _2=CH-CD _3$
E2 reaction is a single-step reaction in which both deprotonation from $\beta-C$ and loss of leaving group from $\alpha-C$ occur simultaneously in the rate-determining step.
$C-D$ bond is stronger than $C-H$ bond, $C-H$ is preferably broken in elimination.
B. $Ph-CHBr-CH _3$ reacts faster than $Ph-CHBr-CD _3$ in $E 2$ reaction because in latter case, stronger $C-D$ bond is to be broken in the rate determining step.
C. $Ph-CH _2-CH _2 Br \xrightarrow[C _2 H _5 O^{-}]{C _2 H _5 OD} Ph-CD=CH _2$
Deuterium incorporation in the product indicates E1CB mechanism
$$ \begin{aligned} & Ph-CH _2-CH _2 Br \stackrel{C _2 H _5 O^{-}}{\rightleftharpoons} Ph-\stackrel{\overline{C} H-CH _2 Br}{\text { carbanion }} \\ & \stackrel{C _2 H _5 OD}{\rightleftharpoons} Ph-\underset{I}{CHD}-CH _2 Br \\ & I \xrightarrow{C _2 H _5 O^{-}} Ph-\stackrel{\left.\right| _{-} ^{D}}{-} CH _2 \xrightarrow{Cr} \longrightarrow Ph-\underset{D}{C}=CH _2 \end{aligned} $$
D. Both $PhCH _2 CH _2 Br$ and $PhCD _2 CH _2 Br$ will react at same rate in $E 1$ reaction because $C-H$ bond is broken in fast non rate determining step. Also E1 reaction follow first order kinetics.
