Give reasons for the following –
(A) Aniline does not participate in the Friedel–Crafts reaction.
(B) In an aqueous solution, dimethylamine ((CH₃)₂NH) is more basic than trimethylamine ((CH₃)₃N).
(C) Primary amines have a boiling point that is higher than that of tertiary amines.
Hint:
(A) Aniline acts as a potent Lewis base, while the catalyst \((\text{AlCl}_3) \) is a powerful Lewis acid.
(B) Take into account the inductive, solvation, and steric effects. The molecule benefiting from multiple effects will be the preferred answer.
(C) Examine the variation in the degree of hydrogen bonding between the two compounds.
Complete-step- by- step answer:
(A) The Friedel–Crafts reaction is an organic transformation where electrophilic aromatic substitution takes place, leading to the addition of substituent groups onto an aromatic ring.
Catalysts such as \( \text{AlCl}_3 \) or \( \text{FeCl}_3 \) are used in this process. These catalysts function by removing chlorine atoms from the reactant, aiding in the chlorination of aromatic rings and facilitating the release of HCl from the aromatic compound.
As an example, in Friedel–Crafts alkylation, the reaction mechanism proceeds as illustrated below.
\( \text{R-Cl} + \text{FeCl}_3 \rightarrow \text{R}^+ + \text{FeCl}_4^- \)
As shown, the catalyst is crucial to the reaction, as it creates the alkyl carbocation that subsequently reacts with benzene to form alkylbenzene.
However, for aniline, the Friedel–Crafts reaction does not proceed. The nitrogen in aniline has a lone pair of electrons, making aniline a strong Lewis base. Additionally, the catalyst \((\text{AlCl}_3) \) is a very strong Lewis acid. Consequently, an acid-base reaction occurs between aniline and \( \text{AlCl}_3 \), resulting in the formation of a salt. This salt acts as a deactivating group, preventing the electrophilic substitution from taking place.
Therefore, aniline does not participate in the Friedel–Crafts reaction.
(B) Dimethylamine ((CH₃)₂NH) is a secondary amine, while trimethylamine ((CH₃)₃N) is a tertiary amine. To understand why the secondary amine is more basic than the tertiary, we need to look at the accessibility of the nitrogen’s lone pair, which determines amine basicity.
Firstly, consider the inductive effect. With more alkyl groups, the tertiary amine would typically be more stabilized according to this effect.
However, in an aqueous solution, the solvation effect must also be considered. In water, the stability of the resulting ion depends on hydrogen bonding. Since secondary amines have more hydrogen atoms available for bonding, they are more stabilized in solution. Thus, in terms of the solvation effect alone, the secondary amine is more stable.
When we account for steric hindrance, the secondary amine also has an advantage, as it experiences less crowding around the nitrogen.
Since two of these factors support the secondary amine, we can conclude that (CH₃)₂NH is more basic than (CH₃)₃N in aqueous solution.
C) To understand why primary amines have a higher boiling point than tertiary amines, we need to consider the number of hydrogen atoms available for bonding. Primary amines have more hydrogen atoms than tertiary amines, which allows for a stronger degree of intermolecular hydrogen bonding.
The stronger the intermolecular bonding, the more energy required to break these bonds. Thus, breaking the bonds in primary amines demands more energy than in tertiary amines.
In conclusion, primary amines have a higher boiling point compared to tertiary amines.
Note: For part (B), make sure to take all three effects into account. Relying on just one effect could lead to an incorrect answer.
