Arrange the following:

In increasing order of boiling point:

C2H5OH, (CH3)2NH, C2H5NH2

As we know that boiling point of compounds depend upon the formation of H-bonding. Amines have higher boiling points than hydrocarbons of simple molecular masses. This is due to the reason that amines being polar, form intermolecular H-bonding(except tertiary amine which do not have hydrogen atom linked to N-atom, i.e., R3N)

Further, since the electronegativity (tendency to attract a shared pair of electrons) of nitrogen in amine is lower (3.0) than that of oxygen (3.5) in alcohol, therefore, amines form weaker H-bonds than electronegative oxygen atom.

Hence, C2H5OH has higher boiling point than (CH3)2NH and C2H5NH2. Because in C2H5OH, the electronegativity of O-atom is higher than H-atom which makes strong intermolecular H-bonding whereas in ( CH3)2NH and C2H5NH2, the electronegativity of N-atom is higher than H-atom but lower than O-atom in C2H5OH which makes weak intermolecular H-bonding.

Further, since, the extent of H-bonding depends upon the number of H-atoms on the N-atom. More the no. of H-atoms linked to nitrogen, the higher the boiling point. Since in(CH3)2NH have one hydrogen atom and in C2H5NH2 have two H-atoms linked to nitrogen, therefore,

C2H5NH2 has higher boiling point than (CH3)2NH.

Combining all these facts, the boiling point of the given three compounds increases in the order:

(CH3)2NH < C2H5NH2 < C2H5OH