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Give one chemical tet to distinguish between the following pairs of compounds: (i) Methylamine and dimethylamine. (ii) Secondary and tertiary aniline. (iii) Ethylamine and aniline. (iv) Aniline and benzylamine. (v) Aniline and N-methylaniline. |
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Answer» Solution :(i) These can be distinguished by the carbylamine test. Methylamine is a PRIMARY amine, THEREFORE, it gives carbylamine test, i.e., when heated with an alcoholic solution of KOH and `CHCl_(3)`, it gives offensive smell of methyl isocyanide. In constrast, dimethylamine is a secondary amine and hence does not give this test. `underset("Methylamine "(1^(@)" Amine"))(CH_(3)NH_(2))+CHCl_(2)+ underset((alc))(3KOH)overset(Delta)to underset("Methyl isocyanide (offensive smell)")(CH_(3)N overset(to)(=)C+3KCl +3H_(2)O` `underset("Dimethylamine "(2^(@)" Amine"))((CH_(3))_(2)NH) underset(Delta)overset(CHCl_(3)//KOH(alc))to ` No REACTION. (II) These can be distinguished by Liebermann nitrosoaine test since `2^(@)` amines give Liebermann nitrosoaine test while `3^(@)` amines do not. `2^(@)` amines on treatment with `HNO_(2)` (generated in situ by the action of dol. HCl on `NaNO_(2)`) gives yellow coloured oily N-nitrosoamines. for example, `underset("Diethylamine")((CH_(3)CH_(2))_(2)NH)+HO-N=O to underset("N-Nitrosodiethylamine (Yellow colour)")((CH_(3)CH_(2))_(2)N-N=O)+H_(2)O` N-Nitrosodiethylamine on warming with a crystal of phenol and conc. `H_(2)SO_(4)`, gives a gree solution which when made alkaline with aqueous, NaOH turns deep blue and then red on dilution. Tertiary amines, on th other hand, do not give this test. |
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