Match LIst-I with List-II. Match the test used for distinguishing the organic compounds

List-I	List-II
(A) \((CH_3)_3N, CH_3CH_2CH_2NH_2\)	(I) Tollen's test
(B) \(HCOOH, CH_3COOH\)	(II) Lucas test
	(III) Carbylamine test
	(IV) Iodoform test

\(A-III, B-I, C-IV, D-II\)

The correct answer is (C) \(A-III, B-I, C-IV, D-II\)

(A) \((CH_3)_3N, CH_3CH_2CH_2NH_2\):

The carbylamine reaction is also known as Hofmann isocyanide synthesis. It is the reaction of a primary amine, chloroform and a base to synthesize isocyanides. The dichlorocarbene intermediate is very important for this conversion. The carbylamine reaction cannot be used to synthesize isocyanides from secondary or tertiary amines.

Trimethyl amine \((CH_3)_3N\) is a tertiary amine and does not give give carbylamine test whereas propanamine gives a carbylamine test.

(B) \(HCOOH, CH_3COOH\):

One way to distinguish between formic acid and acetic acid is Tollen’s test. This test is also known as the silver mirror test. Formic acid gives Tollens test whereas acetic acid does not give this test. Tollen’s reagent is an ammoniacal silver nitrate solution. When formic acid is heated with Tollen’s reagent, a silver mirror is formed on the inner sides of the test tube. Acetic acid does not give this test.

\(\underset{\text{Formic acid}}{HCOOH} + \underset{\text{Tollen's reagent}}{2[Ag(NH_3)_2]^{2+}} + 2OH^- \longrightarrow \underset{\text{Silver mirror}}{2Ag↓} + CO_2 + 2H_2O + 4NH_3\)

\(\underset{\text{Acetic acid}}{CH_3COOH} + \underset{\text{Tollen's reagent}}{2[Ag(NH_3)_2]^{2+}} + 2OH^- \longrightarrow\text{ Silver mirror is not formed}\)

Pentan-2-one is a methyl ketone and hence responds to the iodoform test, but Pentan-3-one is not. As a result, it does not respond to the iodoform test and no reaction occurs.

We have been given a pair of alcohols as n-butanol and tert. Butyl alcohol. In n-butanol, the OH group is attached with one carbon atom, so it is primary \(1^∘\) alcohol. In tert – butyl alcohol, the OH group is attached with three carbon atoms, so it is tertiary \(3^∘\) alcohol.

These \(1^o\text{ and }3^o\) alcohols are distinguished by Lucas reagent test. Lucas reagent is anhydrous zinc chloride and hydrochloric acid in the ratio 1:1 \(ZnCl_2 + HCl\). By reacting with Lucas reagent, alcohols form halides (alkyl chlorides) which have variable solubility. So, on reaction, n-butanol is a primary alcohol, so it will have no reaction with Lucas reagent as

tert – butyl alcohol will show immediate turbidity in the reaction as it is a tertiary halide: