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Give reasons for the following:
(i)Aniline does not undergo Friedel- Crafts reaction.
(ii)${(C{H_3})_2}NH$ is more basic than ${(C{H_3})_3}N$ in an aqueous solution.
(iii)Primary amines have higher boiling point than tertiary amines.

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Last updated date: 20th Jun 2024
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Answer
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Hint: (i) $AlC{l_3}$ is used as a catalyst and is acidic in nature i.e. Lewis acid whereas aniline is a strong Lewis base. Thus, aniline reacts with $AlC{l_3}$ to form a salt.
(ii)The +I effect will increase in the alkyl group that results in increasing the case of donation of lone pair electrons. Amine accepts a proton and forms cation which will be stabilized in water by solvation. Higher the solvation by hydrogen bonding, higher will be the basic strength.
(iii)In primary amines, two H-atoms are present. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding. As a result, extra energy is required to separate the molecules of primary amine.

Complete Step by step answer: (i) The Lewis acid used in Friedel-Crafts reactions as a catalyst is $AlC{l_3}$ , and we also know that aniline is electron deficient and due to which it acts as a Lewis base and which attacks on the lone pair of Nitrogen present in aniline to form an insoluble complex which precipitates out and the reaction does not proceeds further. And hence, Aniline does not undergo Friedel-Crafts Reaction.
(ii) As the alkyl group increases, the +I effect also increases which in turn increases the ease of donation of lone pairs of electrons. Besides this, there is one more thing which we should keep in mind while concluding the strength of basicity. And that is, here Amine will accept a proton and form cation which will be stabilized in water by salvation. And Better the salvation by H-bonding, the higher will be the basic strength.
And so, with increase in methyl group, hydrogen bonding and stabilization by salvation decreases. Therefore we can conclude that ${(C{H_3})_2}NH$ is more basic than ${(C{H_3})_3}N$ in an aqueous solution.
(iii)For comparison of boiling points in amines, we consider the extent of Hydrogen bonding. The more the H-bonding is present, the higher the boiling point. And in a tertiary amine there aren't any hydrogen atoms attached directly to the Nitrogen-atom. Therefore, hydrogen bonding between tertiary amine molecules is impossible. That's why the boiling point is much lower.

Note: friedel-Crafts reaction is carried out in the presence of $AlC{l_3}$ . But $AlC{l_3}$ used as a catalyst and is acidic in nature.
With the increase in methyl group, hydrogen bonding, and stabilization by solvation decreases. This net effect results in a decrease of basic strength from secondary to the tertiary amine. The more the H-bonds present in Amines, the higher the boiling point.