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Q.
Account for the following:
(i) pkb of aniline is more than that of methylaniline.
(ii) Aniline does not undergo Friedel-Craft’s alkylation.
(iii) Primary amines have high boiling points than tertiary amines.
OR
(i) Arrange the following compounds in the increasing order of their basic strength in aqueous solution:
(ii) What is Hinsberg’s reagent?
(iii) What is the role of pyridine in the acetylation reaction of amines?
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Detailed Solution
(i) Since in the structure of methylaniline, delocalization of electrons belongs to nitrogen, which results in the resonance structures. Hence, it increases the acidity of the methylamine than aniline.
(ii) Aniline does not under Friedel-Craft reaction (alkylation and acetylation) due to the salt formation with aluminium chloride, the Lewis acid which is used as a catalyst. Due to this, the nitrogen of aniline acquires a positive charge and hence acts as a strong deactivating group for further reaction.
(iii) Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen
atom available for hydrogen bond formation. So, for this reason the boiling point is higher in primary amines.
OR
(i) Order of the basic strength is as follows, considering the steric hindrance and the +I effect of methyl groups:
(ii) Benzene sulphonyl chloride (C6H5SO2Cl) is also known as Hinsberg’s reagent.
(iii) The reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right-hand side.