Question

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 \(\mathrm{CH}_{3} \mathrm{NHCO}_{2} \mathrm{CH}_{3}\)

Short answer

In summary: (a) Hydrolysis: Tertiary amine shows no change, isocyanide turns litmus paper red. (b) Sodium nitroprusside: Nitroalkane produces deep-red coloration, amide shows no change. (c) Tollens' reagent: Nitrile has no precipitate, alkyne forms a silver mirror. (d) AgNO3 reaction: Heat first, ammonium chloride sublimes, amine hydrochloride does not. (e) Base hydrolysis: Urea derivative releases ammonia gas (pungent odor), amide shows no change.

Step-by-step solution

Pair (a) Test

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) Test

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) Test

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) Test

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) Test

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.

Key concepts

Hydrolysis Reaction

Hydrolysis is a chemical reaction where a molecule of water is used to break down the bonds of a particular substance. In the context of distinguishing between compounds, hydrolysis is used to observe changes that occur when water interacts with different chemical groups.
For example, in pair (a) from the original exercise, a tertiary amine and an isocyanide are tested.

• Tertiary amines, like \((\mathrm{CH}_{3})_{3} \mathrm{CNH}_{2}\), do not undergo hydrolysis and remain unchanged in the presence of water.
• Isocyanides, such as \((\mathrm{CH}_{3})_{2} \mathrm{NC}_{2} \mathrm{H}_{5}\), do undergo hydrolysis, forming an amine and a carboxylic acid. This acid alters the pH of the solution, signaling hydrolysis has occurred.

Hydrolysis reactions can thus be helpful for differentiating between compounds based on observable changes like a shift in pH.

Sodium Nitroprusside Test

The Sodium Nitroprusside test is a qualitative test used to identify functional groups capable of forming colored complexes with nitroprusside ions. This test is especially useful for detecting nitro compounds, as demonstrated in pair (b) of the original exercise.

• Nitroalkanes, such as \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NO}_2\), react with sodium nitroprusside, resulting in a noticeable deep-red color change. This indicates the presence of nitro groups.
• In contrast, amides like \(\mathrm{CH}_3 \mathrm{CONH}_2\) do not undergo a similar reaction, hence no color change is observed.

This test serves as a convenient and visual method for differentiating between nitroalkanes and other compounds like amides.

Tollens' Reagent Test

Tollens' reagent test is a classic chemical test used to determine the presence of aldehydes and some alkynes. This works by using the Tollens' reagent, a colorless solution that contains diamminesilver(I) ions, as an oxidizing agent that can deposit metallic silver when reduced.
In pair (c) of the original exercise, this test is applied to a nitrile and an alkyne:

• Nitriles, like \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}\), do not react with Tollens' reagent, so no silver precipitate forms.
• However, terminal alkynes such as \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}_{2}\), react to produce a silver mirror or precipitate due to their ability to reduce the diamminesilver(I) complex.

The test's specificity for certain functional groups makes it a powerful tool for distinguishing between structurally similar compounds.

Silver Nitrate Reaction

The Silver Nitrate Reaction is a common laboratory technique used to identify halide ions based on the formation of precipitates with silver ions. In certain conditions, this reaction can be used to distinguish between different salts.
For pair (d) in the original exercise, this reaction distinguishes between an amine hydrochloride salt and an ammonium chloride salt.

• Both salts will react with silver nitrate to form white silver chloride precipitates. The difference lies in their behavior when heated before the test.
• Upon heating, ammonium chloride sublimes completely, leaving no residue to react. In contrast, amine hydrochlorides do not sublime, making the presence of precipitate after heating a distinguishing factor.

This method provides a practical approach to tell apart salts with similar ionic components.

Base Hydrolysis

Base hydrolysis involves breaking chemical bonds using a strong base, often resulting in the liberation of a distinct product like ammonia. This reaction is particularly useful in distinguishing between compounds with different acyl groups.
In pair (e) from the original exercise, base hydrolysis helps differentiate between a urea derivative and an amide:

• The urea derivative \(\mathrm{CH}_{3} \mathrm{NHCOCH}_{3}\) reacts with a strong base such as KOH to produce ammonia gas. This gas can be detected by its characteristic pungent smell.
• The amide \(\mathrm{CH}_{3} \mathrm{NHCO}_2 \mathrm{CH}_{3}\) does not hydrolyze under similar conditions, making it inactive in the reaction.

This differentiation allows for the selective identification of compounds that otherwise might appear quite similar.