91Ó°ÊÓ

N-Methylpyrrolidine has a boiling point of${\mathbf{81}}^{\mathbf{ο}}\mathbf{C}$, and piperidine has a boiling point of${106}^{\mathrm{ο}}\mathrm{C}$, this is because piperidine has an $N-H$bond so it can hydrogen bond with other molecules of itself. N-methylpyrrolidine has no $N-H$bond so it cannot form hydrogen bond and will require less energy to separate one molecule from another, hence will have less boiling point. Intermolecular hydrogen bonding increases boiling point of the molecule.

Tetrahydropyran has a boiling point of${88}^{\mathrm{ο}}\mathrm{C}$, and cyclopentanol has a boiling point of${141}^{\mathrm{ο}}C$. These two isomers have a boiling point difference of ${53}^{\mathrm{ο}}C$.This is because, the OH group present in cyclopentanol is involved in intermolecular hydrogen bonding with other molecules of cyclopentanol whereas tetrahydropyran does not and that’s the reason why boiling point of cyclopentanol is more than that of tetrahydropyran. Due to hydrogen bonding, the force of attraction between the molecules increases, and hence, more energy will be required to separate the two molecules, thereby increasing the boiling point of the molecule.

N,N-Dimethylformamide has a boiling point of${150}^{\mathrm{ο}}\mathrm{C}$, and N-methylacetamide has a boiling point of${206}^{\mathrm{ο}}C$, for a difference of${56}^{\mathrm{ο}}\mathrm{C}$. This is due to the resonance structures of these two compounds.The first structure, that is of N,N-dimethylformamide has no hydrogen bonding because it has no $\mathrm{O}-\mathrm{H}$or $\mathrm{N}-\mathrm{H}$ bond but it is a highly polar structure and has large dipole moment. In the second structure, that is of N-methylacetamide, hydrogen bonding exists in addition to the strong dipole-dipole interaction reflected in its higher boiling point.

Both amides boil higher than the four other compounds because of the dipole-dipole interactions that exist due to the resonance forms of the amide due to which boiling point increases.