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In a $S_{\mathrm{N}} 2$ substitution reaction of the type $\mathrm{R}-\mathrm{Br}+\mathrm{Cl}^{-} \stackrel{\mathrm{DMF}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}$, which one of the following has the highest relative rate?
Options:
- A $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
- B
- C
- D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$
Solution:
2222 Upvotes
Verified Answer
The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$
Key Idea: The relative reactivity of alkyl halides towards $\mathrm{S}_{\mathrm{N}} 2$ reactions is as follows:
Primary $>$ Secondary $>$ Tertiary
However, if the primary alkyl halide or the nucleophile/base is sterically hindered the nucleophile will have difficulty to getting the back side of the $\alpha$-carbon as a result of this, the elimination product will be predominant. Here $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$ is the least hindered, hence it has the highest relative rate towards $\mathrm{S}_{\mathrm{N}} 2$ reaction.
Primary $>$ Secondary $>$ Tertiary
However, if the primary alkyl halide or the nucleophile/base is sterically hindered the nucleophile will have difficulty to getting the back side of the $\alpha$-carbon as a result of this, the elimination product will be predominant. Here $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$ is the least hindered, hence it has the highest relative rate towards $\mathrm{S}_{\mathrm{N}} 2$ reaction.
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