What is the correct order of boiling points of isomeric primary (1°), secondary (2°) and tertiary (3°) alcohols?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1° > 2° > 3°
Isomeric alcohols of the same molar mass all form hydrogen bonds, so the difference between them comes from van der Waals forces. These depend on the surface area of the molecule: the more branched the carbon chain, the more compact the molecule, the smaller its surface area and the lower its boiling point. Taking the isomeric butanols (\( \ce{C4H9OH} \)):
Primary (1°), butan-1-ol, \( \ce{CH3CH2CH2CH2OH} \) — first (highest boiling point). It has a straight chain, the largest surface area and the strongest van der Waals forces.
Secondary (2°), butan-2-ol, \( \ce{CH3CH(OH)CH2CH3} \) — second. The –OH sits inside the chain, which makes the molecule a little more compact than the primary alcohol.
Tertiary (3°), 2-methylpropan-2-ol, \( \ce{(CH3)3COH} \) — third (lowest boiling point). It is the most branched, almost spherical, with the smallest surface area and the weakest van der Waals forces.
NCERT (2026-27 edition):
7.4.3 Physical Properties – “Boiling Points
… In alcohols, the boiling points decrease with increase of branching in carbon chain (because of decrease in van der Waals forces with decrease in surface area).”