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For the following reactions CH3CH2CH2Br + Z⊖ → CH3CH2CH2Z + Br⊖ (substitution) CH3CH2CH2Br + Z⊖ → CH3CH=CH2 + HZ + Br⊖ (elimination) where, Z⊖ = CH3CH2O⊖ (A) or H3C–C(O⊖)(CH3)2 (B) ks and ke are, respectively, the rate constants for substitution and elimination, and μ = ks / ke, the correct option is ________

1. μB > μA and ke(A) < ke(B)
2. μB > μA and ke(B) > ke(A)
3. μA > μB and ke(B) < ke(A)
4. μA > μB and ke(A) > ke(B)

Correct answer: μB > μA and ke(A) < ke(B)

Solution

The correct option indicates that the substitution reaction with Z⊖ (A) is slower than with Z⊖ (B), leading to a lower rate constant for substitution (ke(A) < ke(B)). Additionally, the higher ratio of rate constants (μB > μA) suggests that the elimination pathway is favored for Z⊖ (B), which is more sterically hindered, thus promoting elimination over substitution.

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