An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of reactions as shown below. Write the structure of A,B,C,D and E in the following reactions:
OR
(a) Write the structures of main products when aniline reacts with the following reagents:
(i) Br2 water
(ii) HCl
(iii) (CH3CO)2O/pyridine
(b) Arrange the following in the increasing order of their boiling point:
C2H5NH2 , C2H5OH, (CH3)3N
(c) Give a simple chemical test to distinguish between the following pair of compounds:
(CH3)2NH, (CH3)3N
(Hoffmann degradation)
(A= benzamide)
(B = diazonium salt)
(C= benzene)
(D = phenyl isocyanide)
(E= iodobenzene)
OR
(a) 
Bromine water reacts with aniline to give 2,4,6-tribomoaniline.
(ii) 
Aniline is a weak base. It donate the lone pair on nitrogen to proton from HCl to form phenyl ammonium chloride.
(iii) 
Aniline reacts with acetic anhydride to give acetanilide. Aniline acts as a weak base, while acetic anhydride acts as electrophile.
(b) (CH3)3N < C2H5NH2 < C2H5OH
Amines and alcohols form hydrogen bonding but the hydrogen bonding in alcohols it stronger than in amines due to greater electronegativity of oxygen with respect to nitrogen. In the above question we have a primary amine, primary alcohol and a tertiary ammine. Tertiary ammine does not have hydrogen on nitrogen so no hydrogen bonding hence has the least boiling point out of the given three.
Then the primary amine has higher boiling point due to weak hydrogen bonding, while alcohol has the highest boiling point due to strong hydrogen bonding out of the given three.
(c) (CH3)2NH is a secondary amine while (CH3)3N is a tertiary amine. Hinsberg’s test is used to differentiate between primary, secondary and tertiary amines. The amines are reacted with benzene sulfonyl chloride and depending on the type of product obtained they confirmed.
(sulphonamine insoluble in base)
