A secondary alkyl halide can undergo both and reactions since they share the same rate-determining step. How can the pathway be heavily favored?

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Question

A secondary alkyl halide can undergo both $S_N1$ and $E1$ reactions since they share the same rate-determining step. How can the $E1$ pathway be heavily favored?

Prashant Kotian · Accepted Answer

Correct answer: C. High temperatures heavily favor elimination over substitution because eliminations produce more molecules (gas/fragments), leading to a favorable increase in entropy ().

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