JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Haloalkanes and Haloarenes

For the following reactions:

where,

k_{s}

and

k_{e}

are respectively, the rate constant for substitution and elimination, and

μ = \frac{k _{s}}{k _{e}}

, the correct option is ____

(A) $μ_{B} > μ_{A}$ and $k_{e} (A) > k_{e} (B)$
(B) $μ_{A} > μ_{B}$ and $k_{e} (A) > k_{e} (B)$
(C) $μ_{A} > μ_{B}$ and $k_{e} (B) > k_{e} (A)$ ✓ Correct
(D) $μ_{B} > μ_{A}$ and $k_{e} (B) > k_{e} (A)$

Answer: (C)

Solution

The base here is bulky tert-butoxide, which pushes the reaction toward elimination rather than substitution. The more hindered substrate leans even harder on elimination, so its elimination rate constant is the larger,

k_{e} (B) > k_{e} (A)

. Since

μ = \frac{k _{s}}{k _{e}}

, a bigger elimination rate means a smaller

μ

, so

μ_{A} > μ_{B}

. Putting the two together,

μ_{A} > μ_{B}

while

k_{e} (B) > k_{e} (A)

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Haloalkanes and Haloarenes

Solution

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