Increasing order of reactivity of the following compounds for $\mathrm{S}_{\mathrm{N}} 1$ substitution is:

(C) (b) $<$ (a) $<$ (d) $<$ (c)

(A) (b) $<$ (c) $<$ (d) $<$ (a)

(B) (a) $<$ (b) $<$ (d) $<$ (c)

(D) (b) $<$ (c) $<$ (a) $<$ (d)

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Haloalkanes and Haloarenes

Increasing order of reactivity of the following compounds for

S_{N} 1

substitution is:

(A)(b) $<$ (c) $<$ (d) $<$ (a)
(B)(a) $<$ (b) $<$ (d) $<$ (c)
(C)(b) $<$ (a) $<$ (d) $<$ (c)✓ Correct
(D)(b) $<$ (c) $<$ (a) $<$ (d)

Answer: (C)

Solution

Reactivity toward

S_{N} 1

substitution follows the stability of the carbocation each substrate can form. The primary halide (b) gives the least stable cation and so reacts slowest. The secondary halide (a) comes next. The benzyl halide (d) forms a resonance-stabilised benzylic cation, which is more stable still, and the p-methoxybenzyl halide (c) is best of all because the electron-donating

- OCH3

feeds extra electron density into the benzylic cation by resonance. The increasing order is therefore (b) < (a) < (d) < (c).

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Haloalkanes and Haloarenes

Solution

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