4
Question
(a) Illustrate the following reactions giving suitable example in each case :
(i) Hoffmann bromaimide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Distinguish between the following pairs of compounds :
(i) Aniline and N-methylaniline
(ii) (CH3)2NH and (CH3)3N
(i) Hoffmann bromaimide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Distinguish between the following pairs of compounds :
(i) Aniline and N-methylaniline
(ii) (CH3)2NH and (CH3)3N
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Solution
(a) (i) Hoffmann bromamide degradation reaction: It involves the converstion of a primary amide to a primary amine containing one carbon atom less than original amide with a mixture of Br2 in presence of NaOH or KOH.
e.g.,CH3CONH2Acetamide+Br2+4NaOHΔ−→ CH3NH2Methyl amine+2NaBr+Na2CO3+2H2O
C6H5CONH2Benzamide+Br2+4KOHΔ−→ C6H5NH2Aniline+2KBr+K2CO3+2H2O
(ii) Diazotisation: It involves the formation of arenediazonium salt when a cold solution of a primary aromatic amine in a dilute mineral acid (HCl or H2SO4) is treated with a cold solution of nitrous acid (generated in situ by the action of dil.HCl or dil.H2SO4 on NaNO2 at 273 -278 K).
e.g., NaNO2+HCl→HONONitrousacid+NaCl
(iii) Gabriel phthalimide synthesis: It involves the conversion of phthalimide to its potassium salt by treating it with alcoholic potassium hydroxide. Potassium phthalimide is then treated with an alkyl halide to yield an N-alkylphthalimide which is hydrolysed to phthalic acid and a primary amine by heating with HCl or KOH solution.
(b) (i) Distinction between Aniline and N-methylaniline
Azo dye test: Aniline on treatment on treatment with HNO2(NaNO2+dil.HCl) at 273 -278 K gives benzene diazonium chloride which on subsequent treatment with an alkaline solution of β-Naphthol gives a brilliant red azo dye.
On the other hand, N-methyl aniline under these conditions gives yellow coloured oily N-methyl-N-nitroaniline.
C6H5−NH|CH3N-methylaniline+HO−N=O→C6H5−N|CH3N-Methyl-N-nitroaniline(Yellow oil)−N=O+H2O
N-Methyl-N-nitroaniline on warming with a crystal of phenol and conc. H2SO4 gives a green solution which when made alkaline with aq. NaOH turns deep blue and then red on dilution.
(ii) Distinction between (CH3)2NH and (CH3)N
(CH3)2NH reacts with nitrous acid to give N-nitrosoamine which being insoluble indilute mineral and separates out as a neutral yellow oily compound.
NaNO2+HCl→HONONitrous acid+NaCl(CH3)2NHDimethylamine+HONO→(CH3)2N−N=ON-Nitroso dimethylamine(Yellow oil)+H2O
N-Nitrosoamine on heating with a crystal of phenol and a few drops of conc. H2SO4 form a green solution which when made alkaline with aqueous NaOH, turns deep blue and then red on dilution.
(CH3)3NTrimethylamine+HNO2→[(CH3)3NH+]NO−2Trimethylammoniumnitrite (soluble)
e.g.,CH3CONH2Acetamide+Br2+4NaOHΔ−→ CH3NH2Methyl amine+2NaBr+Na2CO3+2H2O
C6H5CONH2Benzamide+Br2+4KOHΔ−→ C6H5NH2Aniline+2KBr+K2CO3+2H2O
(ii) Diazotisation: It involves the formation of arenediazonium salt when a cold solution of a primary aromatic amine in a dilute mineral acid (HCl or H2SO4) is treated with a cold solution of nitrous acid (generated in situ by the action of dil.HCl or dil.H2SO4 on NaNO2 at 273 -278 K).
e.g., NaNO2+HCl→HONONitrousacid+NaCl
(iii) Gabriel phthalimide synthesis: It involves the conversion of phthalimide to its potassium salt by treating it with alcoholic potassium hydroxide. Potassium phthalimide is then treated with an alkyl halide to yield an N-alkylphthalimide which is hydrolysed to phthalic acid and a primary amine by heating with HCl or KOH solution.
(b) (i) Distinction between Aniline and N-methylaniline
Azo dye test: Aniline on treatment on treatment with HNO2(NaNO2+dil.HCl) at 273 -278 K gives benzene diazonium chloride which on subsequent treatment with an alkaline solution of β-Naphthol gives a brilliant red azo dye.
On the other hand, N-methyl aniline under these conditions gives yellow coloured oily N-methyl-N-nitroaniline.
C6H5−NH|CH3N-methylaniline+HO−N=O→C6H5−N|CH3N-Methyl-N-nitroaniline(Yellow oil)−N=O+H2O
N-Methyl-N-nitroaniline on warming with a crystal of phenol and conc. H2SO4 gives a green solution which when made alkaline with aq. NaOH turns deep blue and then red on dilution.
(ii) Distinction between (CH3)2NH and (CH3)N
(CH3)2NH reacts with nitrous acid to give N-nitrosoamine which being insoluble indilute mineral and separates out as a neutral yellow oily compound.
NaNO2+HCl→HONONitrous acid+NaCl(CH3)2NHDimethylamine+HONO→(CH3)2N−N=ON-Nitroso dimethylamine(Yellow oil)+H2O
N-Nitrosoamine on heating with a crystal of phenol and a few drops of conc. H2SO4 form a green solution which when made alkaline with aqueous NaOH, turns deep blue and then red on dilution.
(CH3)3NTrimethylamine+HNO2→[(CH3)3NH+]NO−2Trimethylammoniumnitrite (soluble)
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