The increasing order of nucleophilicity of the following nucleophiles is : (a) $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$ (b) $\mathrm{H}_{2} \mathrm{O}$ (c) $\mathrm{CH}_{3} \mathrm{SO}_{3}^{-}$ (d) $\overline{\mathrm{O}} \mathrm{H}$

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

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

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

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

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · General Organic Chemistry

The increasing order of nucleophilicity of the following nucleophiles is :
(a)

CH_{3} CO_{2}^{-}

(b)

H_{2} O

(c)

CH_{3} SO_{3}^{-}

(d)

\overline{O} H

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

Answer: (A)

Solution

Nucleophilicity here tracks how freely the lone pair is offered. Water is neutral and least willing, so it ranks lowest. The methanesulfonate ion spreads its negative charge over three oxygens, which ties the lone pair up and leaves it only a little better than water. Acetate spreads its charge over just two oxygens, so it is a stronger donor than either. Hydroxide keeps its charge concentrated on a single oxygen with no delocalisation, making it the strongest of the set. The increasing order is

H X_{2} O < CH X_{3} SO X_{3} X^{-} < CH X_{3} CO X_{2} X^{-} < OH X^{-}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · General Organic Chemistry

Solution

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