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Consider the following bromides :
The correct order of $\mathrm{S}_{\mathrm{N}} 1$ reactivity is
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The correct order of $\mathrm{S}_{\mathrm{N}} 1$ reactivity is
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The correct answer is:
$B>C>A$
$B>C>A$
$\mathrm{S}_{\mathrm{N}} 1$ proceeds via carbocation intermediate, the most stable one forming the product faster. Hence reactivity order for $\mathrm{A}, \mathrm{B}, \mathrm{C}$ depends on stability of carbocation created.
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