MCQ
By which mechanism does the above reaction proceed? 
- A$E_1$
- ✓$E_2$
- C$\ce{E_1cb}$
- D$\gamma-$Elimination
✓
Answer
Correct option: B.
$E_2$
If the reaction proceeds by $E_1$ mechanism, then the breaking of $C−H$ and $C−D$ bonds is not involved in the slow step (or rate determining effect). Based on this both products $\ce{CH_2=CH−CD_3}$ and $\ce{CH_3−CH=CD_2}$ are expected.
However only $\ce{CH_2=CH−CD_3}$ is obtained. This indicates that breaking of $C−H$ bond $($which is faster than breaking of $C−D$ bond$)$ is involved in the mechanism along$-$with breaking of $\text{C−Br}$ bond.
This indicates $E_2$ mechanism for elimination.
However only $\ce{CH_2=CH−CD_3}$ is obtained. This indicates that breaking of $C−H$ bond $($which is faster than breaking of $C−D$ bond$)$ is involved in the mechanism along$-$with breaking of $\text{C−Br}$ bond.
This indicates $E_2$ mechanism for elimination.
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