(N/A) $(i)$ Amines undergo deprotonation to give amide ions (amide anions).
$R-NH_2 \to R-NH^- + H^+$
Similarly,alcohol loses a proton to give an alkoxide ion.
$R-OH \to R-O^- + H^+$
In an amide ion,the negative charge is on the $N$ atom,whereas in an alkoxide ion,the negative charge is on the $O$ atom. Since $O$ is more electronegative than $N$,$O$ can accommodate the negative charge more easily than $N$. As a result,the amide ion is less stable than the alkoxide ion. Hence,amines are less acidic than alcohols of comparable molecular masses.
$(ii)$ In a molecule of a tertiary amine,there are no $H$ atoms attached to the nitrogen atom,whereas in primary amines,two hydrogen atoms are present. Due to the presence of $N-H$ bonds,primary amines undergo extensive intermolecular hydrogen bonding.
As a result,extra energy is required to separate the molecules of primary amines. Hence,primary amines have higher boiling points than tertiary amines.
$(iii)$ Due to the $-R$ (resonance) effect of the benzene ring,the lone pair electrons on the $N$ atom are delocalized into the ring in aromatic amines. Therefore,the electrons on the $N$ atom in aromatic amines are less available for donation. This explains why aliphatic amines are stronger bases than aromatic amines.