Question

Pure $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ and pure $ {\text{CH}}{{\text{I}}_{\text{3}}} $ can be distinguished by:
(A) Treating with litmus paper
(B) Treating with aq. $ {\text{KOH}} $
(C) Treating with $ {\text{HCl}} $
(D) Treating with $ {\text{AgN}}{{\text{O}}_{\text{3}}} $

Answer 1

Hint: To answer this question, you must recall the reactions occurring when alkenes and alkynes are reacted with the given reagents in the compounds. The reaction in which only one of the hydrocarbons reacts selectively or compounds with different properties are formed or changes like change in colour, precipitation etc. are observed can be used to distinguish them. The test used is known as iodoform test.

Complete step by step solution
Treating with $ {\text{AgN}}{{\text{O}}_{\text{3}}} $ : When reacted with silver nitrate, silver halide is formed. We know that silver chloride is readily soluble in aqueous solution while silver iodide on the other hand is insoluble in water and forms a precipitate. Thus, we can use this method to distinguish between pure $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ and pure $ {\text{CH}}{{\text{I}}_{\text{3}}} $ .
The correct answer is D.

Note
Litmus paper is used as an indicator for testing whether a given substance or solution is acidic or basic or neutral. Both chloroform and iodoform are neutral substances. They are neither acidic nor basic. There will be no change in the colour of litmus paper with either of the compounds. Thus, we cannot use this method for distinguishing pure $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ and pure $ {\text{CH}}{{\text{I}}_{\text{3}}} $ .
On treatment with aq. $ {\text{KOH}} $ , the halide atoms are replaced by the hydroxyl groups one by one. All the three halogen atoms are replaced in both chloroform as well as iodoform. Three hydroxyl groups cannot co- exist on a single carbon atom. Thus the molecule undergoes loss of a water molecule to form formic acid. Aqueous potassium hydroxide is a strong base and thus reacts with the formic acid to give potassium formate. Thus, we cannot use this method for distinguishing pure $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ and pure $ {\text{CH}}{{\text{I}}_{\text{3}}} $ .
Treating with $ {\text{HCl}} $ also doesn’t give any such visible changes that can help us distinguish.