Chemistry · Amines · NEET
It depends on how many N-H bonds the amine has for intermolecular hydrogen bonding. A primary amine (R-NH2) has two N-H bonds, a secondary amine (R2-NH) has one N-H bond, and a tertiary amine (R3-N) has NO N-H bond. More N-H bonds means more hydrogen bonds between molecules, so more energy is needed to separate them. That is why primary boils highest and tertiary lowest. Tertiary amines have no N-H, so they cannot hydrogen bond with each other at all.
Both alcohols and amines form hydrogen bonds, but the strength is different. Nitrogen is less electronegative than oxygen, so the N-H bond is less polar than the O-H bond. A weaker, less polar bond makes a weaker hydrogen bond. NCERT shows n-butylamine boils at 350.8 K while n-butanol (same molar mass ~73-74) boils at 390.3 K. So amine hydrogen bonds are weaker than alcohol hydrogen bonds, and the amine boils lower.
Alkanes are non-polar and only have weak van der Waals forces. Amines (except sometimes tertiary) can form hydrogen bonds through N-H, which are stronger than van der Waals forces. Even tertiary amines are more polar than alkanes because of the lone pair on nitrogen. So amines need more energy to boil than alkanes of similar mass. For example NCERT lists n-butylamine at 350.8 K vs the alkane C2H5CH(CH3)2 at 300.8 K.
A tertiary amine has three carbon groups on nitrogen and zero N-H bonds. With no N-H, tertiary amine molecules cannot hydrogen bond to each other. Also the three groups surround the nitrogen and block close approach. So even though all three (primary, secondary, tertiary) may have similar mass, the tertiary one has no H-bonding and boils lowest.
Lower aliphatic amines (like methylamine, ethylamine) dissolve well in water because the N-H and the nitrogen lone pair form hydrogen bonds WITH water molecules. As the carbon chain (hydrophobic part) grows, solubility drops. Aniline (C6H5NH2) has a large non-polar benzene ring, so it is only slightly soluble in water. Solubility order for a common NEET set: aniline less than diethylamine less than ethylamine.
Yes, lower tertiary amines are still fairly soluble in water. Even though a tertiary amine has no N-H to donate a hydrogen bond, its nitrogen still has a lone pair that ACCEPTS a hydrogen bond from water. So water can H-bond TO the tertiary amine. This is why solubility does not follow the same primary greater than secondary greater than tertiary pattern as boiling point does.
Arrange the following in increasing order of boiling point: C2H5OH (ethanol), (CH3)2NH (dimethylamine), C2H5NH2 (ethylamine).
Arrange the following in increasing order of solubility in water: C6H5NH2 (aniline), (C2H5)2NH (diethylamine), C2H5NH2 (ethylamine).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For isomeric amines (same molar mass) the order is primary greater than secondary greater than tertiary. Primary has two N-H bonds (most hydrogen bonding), secondary has one N-H, and tertiary has zero N-H (no hydrogen bonding), so tertiary boils lowest.
Lower. Amines and alcohols both form hydrogen bonds, but the N-H bond is weaker/less polar than the O-H bond because nitrogen is less electronegative than oxygen. So an alcohol always boils higher than an amine of the same molar mass.
Lower amines dissolve because the nitrogen lone pair and N-H bonds form hydrogen bonds with water molecules. As the carbon chain gets longer, the non-polar part increases and solubility falls. Aromatic amines like aniline are only slightly soluble.
Tertiary amines cannot form hydrogen bonds with EACH OTHER because they have no N-H bond to donate. But they can still ACCEPT a hydrogen bond from water through the nitrogen lone pair, so lower tertiary amines still dissolve in water.
Lower amines (like methylamine, trimethylamine) are volatile because of relatively weak intermolecular forces and low boiling points. They evaporate easily and have a strong fishy, ammonia-like smell. This is a physical property often mentioned in NCERT.
