Predict the order of decreasing reactivity of the following compounds…

The correct answer is Option (4) → D, B, C, A

An S_N1 reaction takes place in two steps. The slow, rate-determining step is the loss of the halide ion to form a carbocation, so the more stable the carbocation, the faster the reaction. Two factors stabilise the carbocation formed from these benzylic halides:

• Resonance with a phenyl group: the positive charge is spread over the benzene ring. Each extra phenyl group on the positive carbon spreads the charge further, so this is the stronger effect.

• Alkyl groups (+I effect and hyperconjugation): a methyl group pushes electron density towards the positive carbon, so 3° > 2° > 1°.

Combining the two, the stability of the carbocations decreases as follows:

    3° (two phenyls, one methyl) > 2° (two phenyls) > 2° (one phenyl, one methyl) > 1° benzylic
    \( \ce{(C6H5)2C^{+}(CH3)} \) > \( \ce{(C6H5)2CH^{+}} \) > \( \ce{C6H5CH^{+}(CH3)} \) > \( \ce{C6H5CH2^{+}} \)

Matching the given compounds:

• D. \( \ce{C6H5C(CH3)(C6H5)Br} \): 1-bromo-1,1-diphenylethane (3°, two phenyls)

• B. \( \ce{C6H5CH(C6H5)Br} \): bromodiphenylmethane (2°, two phenyls)

• C. \( \ce{C6H5CH(CH3)Br} \): 1-bromo-1-phenylethane (2°, one phenyl)

• A. \( \ce{C6H5CH2Br} \): benzyl bromide (1°, one phenyl)

Shortcut: D, with two phenyl groups and a methyl group, is the most reactive, and only Option (4) starts with D.

NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (S_N1)

Note: The same/similar question has been asked in different years by NTA as per the details given below: