Butan-1-amine (I), N-ethylethanamine (II) and N,N-dimethylethanamine (III) have the same molar mass. The increasing order of their boiling points is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → III < II < I
The three amines are isomers (C4H11N), so their boiling points depend on intermolecular hydrogen bonding, which needs an N–H hydrogen. The more N–H hydrogens, the more association and the higher the boiling point.
(III) N,N-Dimethylethanamine, \( \ce{(CH3)2NCH2CH3} \) — lowest. A tertiary amine has no N–H hydrogen, so there is no intermolecular hydrogen bonding.
(II) N-Ethylethanamine, \( \ce{(CH3CH2)2NH} \) — middle. A secondary amine has one N–H hydrogen for hydrogen bonding.
(I) Butan-1-amine, \( \ce{CH3CH2CH2CH2NH2} \) — highest. A primary amine has two N–H hydrogens, so its molecules are associated the most.
NCERT (2026-27 edition):
9.5 Physical Properties – “Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows: Primary > Secondary > Tertiary”