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:

  1. 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.



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