Give for each of the following pairs of compounds a chemical test, preferably a test-tube reaction which will distinguish +A between the two compounds. (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CNH}_{2}\) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NC}_{2} \mathrm{H}_{5}\) \(+A L\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}\) and \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) \(7 \mathrm{MO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}\) and \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCl}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3} \mathrm{Cl}\) F ALh (e) \(\mathrm{CH}_{3} \mathrm{NHCOCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{NHCO}_{2} \mathrm{CH}_{3}\)

Pair (a): \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CNH}_{2}\) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NC}_{2} \mathrm{H}_{5}\) To distinguish between these two compounds, we can use a hydrolysis reaction. The first compound is a tertiary amine, while the second one is an isocyanide. When we react both compounds with water separately: 1. Tertiary amine \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CNH}_{2}\) gets protonated and does not undergo hydrolysis. There won't be any observable change. 2. Isocyanide \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NC}_{2} \mathrm{H}_{5}\) will undergo hydrolysis, forming an amine and a carboxylic acid. The carboxylic acid will make the solution acidic, which can be detected through a pH change or by using a pH indicator, like litmus paper that will turn red due to acidic pH.

Pair (b): \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}\) and \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) To distinguish between these two compounds, we can use sodium nitroprusside test. 1. Nitroalkane \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}\) gives a deep-red coloration when reacted with sodium nitroprusside solution. 2. Amide \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) won't react with sodium nitroprusside solution and won't show any observable color change.

Pair (c): \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}\) and \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}_{2}\) To distinguish these two compounds, we can use the Tollens' reagent test: 1. Nitrile \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}\) won't react with the Tollens' reagent and will not form any observable precipitate. 2. Alkyne \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}_{2}\) will react with Tollens' reagent and form a silver mirror due to the presence of terminal alkyne group.

Pair (d): \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCl}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3} \mathrm{Cl}\) To test between these two compounds, we can use silver nitrate (AgNO3) solution: 1. Amine hydrochloride salt \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCl}\) will form a white precipitate of silver chloride when reacted with silver nitrate solution. 2. Ammonium chloride salt \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3} \mathrm{Cl}\) will also form a white precipitate of silver chloride. However, we can distinguish between the two compounds by heating them before the reaction with silver nitrate. The ammonium chloride salt will completely sublime upon heating while amine hydrochloride will not.

Pair (e): \(\mathrm{CH}_{3} \mathrm{NHCOCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{NHCO}_{2} \mathrm{CH}_{3}\) To distinguish between these two compounds, we can use a base hydrolysis reaction: 1. Urea derivative \(\mathrm{CH}_{3} \mathrm{NHCOCH}_{3}\) will undergo hydrolysis in the presence of a strong base like KOH, producing ammonia gas, which will create a strong pungent odor. 2. Amide \(\mathrm{CH}_{3} \mathrm{NHCO}_{2} \mathrm{CH}_{3}\) won't undergo hydrolysis under the same conditions and won't have any noticeable change.