Arrange the following bromides in increasing order of their reactivit…

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

In an S_N2 reaction the nucleophile must reach the carbon that carries the halogen, from the side opposite to it. Bulky groups on that carbon block the approach, so the reactivity falls as the carbon carries more alkyl groups: methyl > primary > secondary > tertiary.

(A) 2-Bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃ — first (least reactive). It is a tertiary bromide: the carbon bearing Br carries three alkyl groups, which hinder the nucleophile most.

(D) 2-Bromo-3-methylbutane, (CH₃)₂CHCH(Br)CH₃ — second. It is a secondary bromide, with two alkyl groups on the carbon bearing Br.

(B) 1-Bromo-3-methylbutane, (CH₃)₂CHCH₂CH₂Br — third. It is a primary bromide; its branch is far from the carbon bearing Br, so the approach is easy.

(C) Bromomethane, CH₃Br — fourth (most reactive). The carbon carries only three small hydrogen atoms, so nothing hinders the nucleophile.

NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “Since this reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon atom have a dramatic inhibiting effect. Of the simple alkyl halides, methyl halides react most rapidly in S_N2 reactions because there are only three small hydrogen atoms. Tertiary halides are the least reactive because bulky groups hinder the approaching nucleophiles. Thus the order of reactivity followed is: Primary halide > Secondary halide > Tertiary halide.”