Consider the following bromides: The correct order of $S_{N} 1$ reactivity is:

B > C > A

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B > A > C

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C > B > A

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A > B > C

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Solution

The correct option is D $B > C > A$
In B, the carbocation formed is resonance stabiised. So this is the most reactive compound.
In C, the carbocation formed is $2^{o}$ . In A, the carbocation formed is $1^{o}$ . Therefore, C is more stable than A. So, the order is $B > C > A$ .

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