Predict the product (A) for the following rea | vedclass

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(A) The reaction is an elimination reaction (dehydrohalogenation) using a strong base,potassium tert-butoxide $(K^+ \, ^-OC(CH_3)_3)$.
This reaction

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Ph-$C^* \equiv C-C_6H_4-Br$

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(A) The reaction is an elimination reaction (dehydrohalogenation) using a strong base,potassium tert-butoxide $(K^+ \, ^-OC(CH_3)_3)$.
This reaction proceeds via an $E2$ mechanism.
In the $E2$ elimination of vinyl halides,the base removes the proton $(H^+)$ and the leaving group $(Br^-)$ must be in an anti-periplanar orientation.
The starting material is $Ph-C^*(=CHBr)-C_6H_4-Br$.
The proton $(H)$ on the terminal carbon and the bromine $(Br)$ on the internal carbon are eliminated to form an alkyne.
Since the $H$ is on the terminal carbon and the $Br$ is on the carbon attached to the phenyl ring,the triple bond forms between the two carbons.
The carbon marked with the asterisk $(C^*)$ is the one attached to the phenyl group $(Ph)$.
Following the anti-elimination,the product is $Ph-C^* \equiv C-C_6H_4-Br$.

Predict the product $(A)$ for the following reaction:
$Ph-C^(=CHBr)-C_6H_4-Br \xrightarrow{K^+ \, ^-OC(CH_3)_3, \text{ Heat}} (A)$(Note: $C^ = ^{14}C$)

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Predict the product $(A)$ for the following reaction: $Ph-C^*(=CHBr)-C_6H_4-Br \xrightarrow{K^+ \, ^-OC(CH_3)_3, \text{ Heat}} (A)$(Note: $C^* = ^{14}C$)