Arrange the following compounds in order of decreasing boiling points
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, D, B, C
The boiling points of haloalkanes depend on van der Waals forces, which grow with molecular mass (number of carbon atoms) and with surface area. For isomeric haloalkanes, more branching makes the molecule more compact, lowers the surface area and lowers the boiling point.
A. \( \ce{CH3CH2CH2CH2Br} \) (1-bromobutane): four carbons, unbranched — first (highest)
D. \( \ce{(CH3)3CBr} \) (2-bromo-2-methylpropane): four carbons, but highly branched, so a smaller surface area — second
B. \( \ce{CH3CH2CH2Br} \) (1-bromopropane): three carbons — third
C. \( \ce{CH3CH2Br} \) (bromoethane): two carbons — last (lowest)
Shortcut: 1-Bromobutane (A) is the highest and the two-carbon C the lowest; only Option (2) starts with A and ends with C.
NCERT (2026-27 edition):
Physical Properties – …
