Dehydrohalogenation in presence of OH⁻ is correctly represented by
(A)An incomplete arrow-pushing scheme omitting the departing bromide's electron flow
(B)An arrow-pushing scheme with the wrong bond breaking to form the new π-bond
(C)A scheme combining the attacking base and the leaving group into one ambiguous step
(D)The full three-arrow E2 scheme: OH⁻ removes the β-H, that C–H pair forms the new C=C π-bond, and the C–Br σ-bond breaks to release Br⁻, all concertedlyCorrect
Answer: (D)
Solution
A properly drawn E2 elimination mechanism requires three curved arrows in one concerted step: the base's lone pair attacking the β-hydrogen, the
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