Dehydrohalogenation in the presence of OH^{-} | vedclass

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Question

(D) Dehydrohalogenation is an elimination reaction where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms to form a double bo

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(D) Dehydrohalogenation is an elimination reaction where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms to form a double bond. In the presence of a base like $OH^{-}$,the base abstracts a proton $(H^{+})$ from the $\beta$-carbon,leading to the formation of a double bond and the elimination of the halide ion $(Br^{-})$. The correct mechanism shows the $OH^{-}$ attacking the $\beta$-hydrogen,the $C-H$ bond breaking to form the $C=C$ double bond,and the $C-Br$ bond breaking to release the $Br^{-}$ ion. Option $D$ correctly depicts this concerted mechanism.

Dehydrohalogenation in the presence of $OH^{-}$ is correctly represented by:

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