Arrange the following alkyl halides in the order of increasing reacti…

The correct answer is Option (2) → A, C, B, D

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 increases as follows:

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

Matching the given compounds:

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

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

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

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

Shortcut: Benzyl bromide (A) is the least reactive and D, with two phenyl groups and a methyl group, is the most reactive. Options (2) and (4) both start with A and end with D. A second phenyl group stabilises the carbocation more than a methyl group, so B comes after C, and only Option (2) remains.

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: