Give reasons for the following: (i) Aniline does not undergo Friedel-Crafts reaction. (ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution. (iii) Primary amines have higher boiling points than tertiary amines.

(i) Aniline does not undergo Friedel-craft's reactions because the reagent AlCl3 (the Lewis acid which is used as a catalyst in Friedel-Crafts reaction), being electron deficient acts as a Lewis base and attacks on the lone pair of nitrogen present in aniline to form an insoluble complex which precipitates out and the reaction does not proceed. (ii) With an increase in alkyl groups, the +I effect will increase which will increase the ease of donation of a lone pair of electrons. But in water, one other factor is controlling the strength of basicity. Amine will accept a proton and form a cation. This cation will be stabilised in water by solvation. The better the solvation by hydrogen bonding, the higher will be the basic strength (refer image). Thus with an increase in methyl groups, hydrogen bonding and stabilisation by solvation decreases. The net effect is when we move from secondary to tertiary amine, basic strength decreases. (iii) In a tertiary amine, there aren't any hydrogen atoms attached directly to the nitrogen. That means that hydrogen bonding between tertiary amine molecules is impossible. That's why the boiling point is much lower. The small amines of all types are very soluble in water.