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Give the product of the bimolecular elimination from each of the following isomeric halogenated compounds. KOtBu +HOtBu +Br Br H CH3 One of these compounds undergoes elimination 50x faster than the other. Which one and why? (scroll down A because the conformation needed for elimination places the phenyl groups anti to each other.

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In presence of strong bulky base (KOtBu). alkyl halides undergo E2 elimination reaction In E2 elimination reaction, both eliminating groups are anti to each other In compound-A, Br (hashed), and H (hashed) are not anti-periplanar to each otheir CH3 KOtBu HOtBu Ph Ph Br Cis product less stable Compound-A In compound-B, Br (hashed), and H (wedge) are not anti to each other Ph Br KOtBu + HOtBu CH Br Ph Trans product More stable Compound-B So, compound-B undergoes elimination 50 times faster than compound-A. B because the conformation needed for elimination places the phenyl groups anti to each other

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