Q3. Arrange ethanamine, hexan-1-amine and decan-1-amine in increasing order of solubility in water.
Decan-1-amine < hexan-1-amine < ethanamine is the expected trend for these straight-chain free amines under comparable conditions: the nonpolar chain grows while the amino group remains the same. Specify temperature and pH for a quantitative comparison; protonated salts behave differently.
Q4. Arrange three of the four C3H9N isomers, propan-1-amine, N-methylethanamine and N,N-dimethylmethanamine, in decreasing order of boiling point.
Propan-1-amine (1°) > N-methylethanamine (2°) > N,N-dimethylmethanamine (3°): two, one and no N–H groups for hydrogen bonding between molecules (reasoned from NCERT pages 265–266).
Q5. Propane (M 44), ethanamine (M 45) and ethanol (M 46) have almost the same molar mass. Arrange them by boiling point.
Propane < ethanamine < ethanol: propane has no hydrogen bonds, and the O–H of ethanol hydrogen bonds more strongly than the N–H of ethanamine, since oxygen (3.5) is more electronegative than nitrogen (3.0) (reasoned from NCERT pages 265–266).
Q6. Why are the higher amines essentially insoluble in water although the lower ones dissolve?
Both can hydrogen bond with water through nitrogen, but in the higher amines the hydrophobic alkyl part is so large that solubility falls away (NCERT page 265).