The correct order of leaving group ability in a nucleophilic substitution reaction is:

B r^{-} > C l^{-} > C H_{3} C O_{2}^{-} > O H^{-} > H^{-}

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H^{-} > O H^{-} > C H_{3} C O_{2}^{-} > C l^{-} > B r^{-}

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B r^{-} > C H_{3} C O_{2}^{-} > C l^{-} > O H^{-} > H^{-}

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C H_{3} C O_{2}^{-} > B r^{-} > C l^{-} > O H^{-} > H^{-}

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Solution

The correct option is D $B r^{-} > C l^{-} > C H_{3} C O_{2}^{-} > O H^{-} > H^{-}$
1. If we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases.

2. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them.

Thus, $B r^{-} > C l^{-} > C H_{3} C O_{2}^{-} > O H^{-} > H^{-}$

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Similar questions

Which of the following is not correct?

$O r d e r o f b a s i c i t y O r d e r o f n u c l e o p h i l i c i t y (a) F^{-} > C l^{-} > B r^{-} B r_{-} > C l^{-} > F^{-} (b) H - O - O^{-} > O H^{-} H - O - O^{-} > O H^{-} (c) C H_{3} > N H_{2}^{-} > O H - C H_{3} > N H_{2}^{-} > O H^{-} (d) H_{2} O > H_{2} S H_{2} O < H_{2} S$

Q. The IUPAC name of the compound is

C H_{3} - C H_{2} - B r | C | O H - C l | C H - C H_{2} - C l

C H_{3} - C | H C l - C | H B r - C | H O H - C H_{3}

The IUPAC name of the compound is

Q. Few pairs of molecule are given below. Which bond of the molecule of the pair is more polar?
(i)

H_{3} C - H, H_{3} C - B r

(ii)

H_{3} C - N H_{2}, H_{3} C - O H

(iii)

H_{3} C - O H, H_{3} C - S H

(iv)

H_{3} C - C l, H_{3} C - B r

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