For the elimination of $PhSO X_{2} CH X_{2} CH X_{2} X$ in $NaOD / D X_{2} O$ the rate of beta-H/D exchange is 40 times the rate of alkene formation. The most reasonable conclusion is

AE1cB with slow irreversible proton removal, loss of $X X^{-}$ fast
BE2, a concerted single step
CE1cB with fast reversible carbanion formation and slow loss of $X X^{-}$
DE1, with rate-determining ionisation of C-X

Answer: C

Exchange means the carbanion forms and is reprotonated many times (about 40 per elimination event). So carbanion formation is fast and reversible and loss of

X X^{-}

from the carbanion is the slow step. If proton removal were slow and irreversible, exchange could not exceed elimination.

Practice Organic Reaction Mechanisms IV: Elimination More on E1cB mechanism

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For the elimination of PhSOX2​CHX2​CHX2​X in NaOD/DX2​O the rate of beta-H/D exchange is 40 times the rate of alkene formation. The most reasonable conclusion is

Answer: C

More questions on E1cB mechanism

For the elimination of $PhSO X_{2} CH X_{2} CH X_{2} X$ in $NaOD / D X_{2} O$ the rate of beta-H/D exchange is 40 times the rate of alkene formation. The most reasonable conclusion is