Which is the correct decreasing order of boiling point A. 1-Bromoetha…

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

For compounds with the same halogen, the boiling point is decided by the strength of the van der Waals forces, which grow as the molecule becomes bigger and has more electrons (higher molar mass):

• C. Bromobenzene, \( \ce{C6H5Br} \) (first, highest) — 157 g mol⁻¹, b.p. 429 K; the flat, polarisable ring gives the strongest attractions.

• B. 1-Bromopentane, \( \ce{C5H11Br} \) (second) — 151 g mol⁻¹, b.p. 403 K.

• D. 1-Bromopropane, \( \ce{C3H7Br} \) (third) — 123 g mol⁻¹, b.p. 344 K.

• A. Bromoethane, \( \ce{C2H5Br} \) (last, lowest) — 109 g mol⁻¹, b.p. 311 K; the smallest molecule.

Shortcut: bromoethane (A), the smallest molecule, boils lowest and must come last. Options (2) and (3) end with A; of these, 1-bromopentane (B) must come before 1-bromopropane (D), which leaves Option (2).

NCERT (2026-27 edition):
Melting and Boiling Points – “… The attractions get stronger as the molecules get bigger in size and have more electrons. … For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.”