When the concentration of alkyl halide is triple and concentration of $O H^{-}$ is reduced to half, the rate of $S_{N} 2$ reaction is increased by :
(correct answer +2, wrong answer -0.5)

3 t i m e s

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1.5 t i m e s

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2 t i m e s

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6 t i m e s

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Solution

The correct option is B $1.5 t i m e s$
As reaction is going via $S_{N} 2$ pathway, means that the reaction rate is dependent on concentration of both substrate and $O H^{-}$ . Hence, reaction rate can be written as :
$R_{1} = k [S u b s t r a t e] [O H^{-}]$ ...(i)
If the concentration of alkyl halide i.e., substrate is tripled and that of $[O H^{-}]$ is halved. The reaction rate changes as :
$R_{2} = k [3 \times s u b s t r a t e] [\frac{O H^{-}}{2}]$ ...(ii)
On dividing (i) and (ii) , we get
$\frac{R_{2}}{R_{1}} = \frac{3}{2} = 1.5 t i m e s$

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When the concentration of alkyl halide is triple and concentration of OH− is reduced to half, the rate of SN2 reaction is increased by : (correct answer +2, wrong answer -0.5)

When the concentration of alkyl halide is triple and concentration of $O H^{-}$ is reduced to half, the rate of $S_{N} 2$ reaction is increased by :
(correct answer +2, wrong answer -0.5)