Distinguish between the following pairs:
`{:((I),,.(II)),(a.(PhNH_(3))_(2)^(+)SO_(4)^(2-)`,and,,(b.`Me_(4)N^(o+)CI^(Theta)`,and,`[Me_(3)overset(o+)NH]CI^(Theta)),(c.PhNMe_(2)`,and,PHNHMe),(d.`Me_(2)N-CH_(2)OH`,and,`Me_(4)overset(o+)Noverset(Theta)OH),(e.Phoverset(o+)(NH_(3))CI^(Theta)`,and,(f.`PhNH_(2)`,and,(g.PhNHCOMe,and,`PhNH_(2)):}`

Explanation

(a) Add `BaCl_(2)` solution (I) is a sulphate salt and will give a white precipitate of `BaSO_(4)`. `(II)` is sulphonate (sulphonic acid salt) and will not give any precipitate.
(b) Add concentrated `NaOH` and heat the mixture. `(II)` will change to a volatile free base `Me_(3)N`, which has a typical `NH_(3)` odour. compound `(I)` will not react.
(c) With `HNO_(2) (NaNO_(2)+HCl),(II)` (`2^(@)` amine) will give yellow oil of nitrosamine, which is insoluble in acid. `(I)` will react with `HNO_(2)` to give `p`-nitro compound which is still basic and is soluble in acid.

`(II)` is ionic and thus gives white precipitate of `Ba(OH)_(2)` on addition of `BaCl_(2)` solution. it is a strong base and its solution turns red litmus blue.
(I) is a covalent amino alcohol and does not respond to the above tests.
(e) (I) is ionic and gives white precipate of `AgCl` with `AgNO_(3)` solution whereas (II) does not.
(f) (I) is `1^(@)` aromatic amine and gives dye test, whereas (II) does not.
(g) (II) is basic and (I) (amide) is amphoteric, so (II) dissolves in aqueous `HCl`, whereas (I) Does not.