1
Question
Give one chemical test to distinguish between the following pairs of compounds.
A Methylamine and dimethylamine
B Secondary and tertiary amines
C Ethylamine and aniline
D Aniline and benzylamine
E Aniline and N-methylaniline.
A Methylamine and dimethylamine
B Secondary and tertiary amines
C Ethylamine and aniline
D Aniline and benzylamine
E Aniline and N-methylaniline.
Open in App
Solution
A. Dimethylamine & Methylamine can be differentiated by the carbylamine test. When Aliphatic & aromatic primary amines are heated with CHCl3 & alc. KOH, foul-smelling isocyanides or carbylamine are formed.
As methylamine is an aliphatic primary amine it gives a positive carbylamine test, but dimethylamine does not give this test as that is secondary amine.
CH3−NH2+CHCl3+3KOH↓Δ CH3−NC+3KCl+3H2O( Foul smell)( CH3)2NH+CHCl3+3KOH↓Δ No reaction
B. Secondary amines & tertiary amines can be differentiated by Hinsberg’s reagent test (benzene sulphonyl chloride, C6H5SO2Cl). When Secondary amines react with Hinsberg’s reagent, a product which is insoluble in alkali is formed.
For example, when N, N - diethyl amine (secondary amine) reacts with Hinsberg’s reagent, it forms N, N-diethyl benzene sulphonamide, which is insoluble in an alkali. While tertiary amines, however, do not react with Hinsberg’s reagent.
C. Ethylamine and aniline can be differentiated by azo-dye test.
When aromatic amines are made to react with HNO2(NaNO2+dil.HCl) at 0−5∘C, followed by alkaline solution of 2 –naphthol, a dye is obtained.
The dye is observed to be in the following colours: yellow, red and orange. A brisk effervescence is given out by aliphatic amines due to the evolution of N2 gas under analogous conditions.
Conclusion
Aniline being an aromatic amine gives orange colour dye.
While ethylamine being an aliphatic amine gives a brisk effervescence due to the evolution of N2 gas.
C2H5NH2+HONO+HCl↓273−278k C2H5OH+CH2=CH2+C2H5Cl+N2
D. Benzylamine & Aniline can be differentiated by reacting them with nitrous acid. When benzylamine reacts with nitrous acid, it forms unstable diazonium salt which gives a by – product as alcohol along with the evolution of N2 gas.
C6H5CH2−NH2+HNO2↓ NaNO2+HCl[ C6H5CH2−N+2Cl−]( Unstable)
[ C6H5CH2−N+2Cl−]( Unstable)↓ H2O N2↑+C6H5CH2−OH+HCl
When aniline reacts with HNO2 at a very low temperature, it forms a stable diazonium salt.
Hence, the evolution of nitrogen gas does not happen.
C6H5NH2 273−278K↓ NaNO2+HCl C6H5N+2Cl−+NaCl+2H2O
E. N-methylaniline & aniline can be differentiated by using the Carbylamine test.
When aliphatic & aromatic primary amines are heated with CHCl3 & alc. KOH, foul-smelling isocyanide or carbylamine is formed.
As aniline is an aromatic primary amine, it gives a positive carbylamine test but N- methyl aniline does not give this test as that is secondary amine.
C6H5−NH2+CHCl3+3KOH↓Δ C6H5−NC+3KCl+3H2O( Foul smell) C6H5NHCH3+CHCl3+3KOH↓Δ No reaction
As methylamine is an aliphatic primary amine it gives a positive carbylamine test, but dimethylamine does not give this test as that is secondary amine.
CH3−NH2+CHCl3+3KOH↓Δ CH3−NC+3KCl+3H2O( Foul smell)( CH3)2NH+CHCl3+3KOH↓Δ No reaction
B. Secondary amines & tertiary amines can be differentiated by Hinsberg’s reagent test (benzene sulphonyl chloride, C6H5SO2Cl). When Secondary amines react with Hinsberg’s reagent, a product which is insoluble in alkali is formed.
For example, when N, N - diethyl amine (secondary amine) reacts with Hinsberg’s reagent, it forms N, N-diethyl benzene sulphonamide, which is insoluble in an alkali. While tertiary amines, however, do not react with Hinsberg’s reagent.
C. Ethylamine and aniline can be differentiated by azo-dye test.
When aromatic amines are made to react with HNO2(NaNO2+dil.HCl) at 0−5∘C, followed by alkaline solution of 2 –naphthol, a dye is obtained.
The dye is observed to be in the following colours: yellow, red and orange. A brisk effervescence is given out by aliphatic amines due to the evolution of N2 gas under analogous conditions.
Conclusion
Aniline being an aromatic amine gives orange colour dye.
While ethylamine being an aliphatic amine gives a brisk effervescence due to the evolution of N2 gas.
C2H5NH2+HONO+HCl↓273−278k C2H5OH+CH2=CH2+C2H5Cl+N2
D. Benzylamine & Aniline can be differentiated by reacting them with nitrous acid. When benzylamine reacts with nitrous acid, it forms unstable diazonium salt which gives a by – product as alcohol along with the evolution of N2 gas.
C6H5CH2−NH2+HNO2↓ NaNO2+HCl[ C6H5CH2−N+2Cl−]( Unstable)
[ C6H5CH2−N+2Cl−]( Unstable)↓ H2O N2↑+C6H5CH2−OH+HCl
When aniline reacts with HNO2 at a very low temperature, it forms a stable diazonium salt.
Hence, the evolution of nitrogen gas does not happen.
C6H5NH2 273−278K↓ NaNO2+HCl C6H5N+2Cl−+NaCl+2H2O
E. N-methylaniline & aniline can be differentiated by using the Carbylamine test.
When aliphatic & aromatic primary amines are heated with CHCl3 & alc. KOH, foul-smelling isocyanide or carbylamine is formed.
As aniline is an aromatic primary amine, it gives a positive carbylamine test but N- methyl aniline does not give this test as that is secondary amine.
C6H5−NH2+CHCl3+3KOH↓Δ C6H5−NC+3KCl+3H2O( Foul smell) C6H5NHCH3+CHCl3+3KOH↓Δ No reaction
Suggest Corrections
1
Similar questions