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2-Phenylethylbromide when heated with $\mathrm{NaOEt}$, elimination takes place. No deuterium exchange takes place when the reaction is carried out in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ solvent. The mechanism will be
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The correct answer is:
E2 elimination
It is a primary bromide. So it will undergo elimination either by E2 or E1cB. Since there is no deuterium exchange in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ solvent, $\mathrm{C}-\mathrm{H}$ bond is not broken to form carbanion. Hence the actual mechanism is E2 only.
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