Arrange the following alcohols in the increasing order of their react…

The correct answer is Option (3) → B, D, A, C

Benzyl alcohols react with HBr by the S_N1 path. The –OH group is protonated and leaves as water, and the benzylic carbocation formed is then attacked by \( \ce{Br^{-}} \):

    \( \displaystyle \ce{ArCH2OH + HBr -> ArCH2Br + H2O} \)

The more stable the carbocation, the faster the reaction. A group at the para position changes this stability:

• Electron donating group (–CH₃, +I and hyperconjugation): pushes electrons towards the positive carbon, stabilises the carbocation and increases the rate.

• Weakly withdrawing group (–Cl, −I stronger than +R): slightly destabilises the carbocation and lowers the rate.

• Strongly withdrawing group (–NO₂, −I and −R): strongly destabilises the carbocation and makes the reaction slowest.

Combined order of reactivity:

    \( \displaystyle p\text{-}\ce{NO2} < p\text{-}\ce{Cl} < \ce{H} < p\text{-}\ce{CH3} \)

Matching the given alcohols:

• (B) p-nitrobenzyl alcohol: strong electron withdrawing group

• (D) p-chlorobenzyl alcohol: weak electron withdrawing group

• (A) benzyl alcohol: no substituent

• (C) p-methylbenzyl alcohol: electron donating group

Shortcut: The nitro compound (B) is slowest and the methyl compound (C) is fastest, which leaves Options (2) and (3). Chlorine withdraws electrons, so D is slower than A. That gives Option (3).