$\mathrm{CH}_{3} \mathrm{Br}+\mathrm{Nu}^{-} \longrightarrow \mathrm{CH}_{3}-\mathrm{Nu}+\mathrm{Br}^{-}$ The decreasing order of the rate of the above reaction with nucleophiles ( $\mathrm{Nu}^{-}$) A to D is $\left[\mathrm{Nu}^{-}=(\mathrm{A}) \mathrm{PhO}^{-}\right.$, (B) $\mathrm{AcO}^{-}$, (C) $\mathrm{HO}^{-}$, (D) $\left.\mathrm{CH}_{3} \mathrm{O}^{-}\right]$

(A) $\mathrm{D}>\mathrm{C}>\mathrm{A}>\mathrm{B}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Haloalkanes and Haloarenes

CH_{3} Br + Nu^{-} ⟶ CH_{3} - Nu + Br^{-}

The decreasing order of the rate of the above reaction with nucleophiles (

Nu^{-}

) A to D is

[Nu^{-} = (A) PhO^{-}

, (B)

AcO^{-}

, (C)

HO^{-}

, (D)

CH_{3} O^{-}]

(A) $D > C > A > B$ ✓ Correct
(B)D $>$ C $>$ B $>$ A
(C)A $>$ B $>$ C $>$ D
(D)B $>$ D $>$ C $>$ A

Answer: (A)

Solution

With a simple primary substrate like

C H_{3} Br

the reaction runs by the

S_{N} 2

path, so the rate tracks the nucleophilicity of the attacking anion. Among these oxygen nucleophiles methoxide is the most reactive because its negative charge is localised and least stabilised; hydroxide follows; phenoxide is weaker because resonance into the ring delocalises its charge; and acetate is weakest of all, its charge spread over two oxygens. The rate therefore decreases in the order

C H_{3} O^{-} > H O^{-} > Ph O^{-} > Ac O^{-}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Haloalkanes and Haloarenes

Solution

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