Question
By which mechanism does the above reaction proceed?

Answer

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 $C−Br$ bond.
This indicates $E_2$​ mechanism for elimination.

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