Decreasing order of reactivity in Williamson synthesis of the following :
I. $M e_{3} C C H_{2} B r$ II. $C H_{3} C H_{2} C H_{2} B r$
III. $C H_{2} = C H C H_{2} C l$ IV. $C H_{3} C H_{2} C H_{2} C l$

I I I > I I > I V > I

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I > I I > I V > I I I

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I I > I I I > I V > I

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I > I I I > I I > I V

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Solution

The correct option is D $I I > I I I > I V > I$
C-Br bond is weaker than C-Cl bond, therefore, alkyl bromide (II) reacts faster than alkyl chlorides, (III) and (IV) . Since $C H_{2} = C H -$ is electron withdrawing therefore, $C H_{2}$ has more positive charge on III than on IV. In other words, nucleophilic attack occurs faster on III than on IV. Further, since Williamson synthesis occurs by $S_{N}^{2}$ mechanism, therefore, due to steric hindrance alkyl bromide (I) is the least reactive. Thus, the decreasing order of reactivity is $I I > I I I > I V > I$ .
III. $C H_{2} = C H \leftarrow δ δ + C H_{2} - C l$
IV. $C H_{3} - C H_{2} \to δ + C H_{2} - C l$

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Order of decreasing acidity of the following is

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(II) ${C H}_{3} C O O H$

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R - C H = C H_{2},

M e

E t

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M e_{3} C

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