Which of the following compound will have the maximum boiling point?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1-Bromobutane
Boiling point depends on the strength of the intermolecular forces:
A C–Br compound is polar and heavier than the parent alkane, so bromobutanes boil much higher than butane.
Among isomeric haloalkanes the boiling point decreases with branching: a straight chain has the largest surface area and the strongest van der Waals forces.
1-Bromobutane (straight chain, 375 K) > 2-bromobutane (364 K) > 2-methyl-2-bromopropane (346 K) > butane (273 K). The maximum is 1-bromobutane.
Why the other options are incorrect
Option (1) 2-Bromobutane — Incorrect. The Br on C-2 gives a slightly more branched, compact molecule; it boils lower than the 1-isomer.
Option (2) Butane — Incorrect. Non-polar and lighter; it is a gas at room temperature.
Option (4) 2-Methyl-2-bromopropane — Incorrect. The most branched (nearly spherical) isomer; it has the lowest boiling point of the three bromides.
NCERT (2026-27 edition):
Physical Properties – “… The boiling points of isomeric haloalkanes decrease with increase in branching. For example, 2-bromo-2-methylpropane has the lowest boiling point among the three isomers.”