CBSE Class 12 Chemistry Chapter 26: Amines NCERT Solutions

The IUPAC names and classifications of the given compounds are as follows:

1. (i) $(\backslash text\{CH\}\_3)\_2\; \backslash text\{CHNH\}\_2$

   IUPAC Name: Propan-2-amine

   Classification: Primary amine (1°)

   Explanation: The longest carbon chain has three carbons (propane). The amino group ($-\backslash text\{NH\}\_2$) is attached to the second carbon. The nitrogen atom is bonded to only one carbon atom.

2. (ii) $\backslash text\{CH\}\_3(\backslash text\{CH\}\_2)\_2\backslash text\{NH\}\_2$

   IUPAC Name: Propan-1-amine

   Classification: Primary amine (1°)

   Explanation: The longest carbon chain has three carbons (propane). The amino group ($-\backslash text\{NH\}\_2$) is attached to the first carbon. The nitrogen atom is bonded to only one carbon atom.

3. (iii) $\backslash text\{CH\}\_3\backslash text\{NHCH\}(\backslash text\{CH\}\_3)\_2$

   IUPAC Name: N-Methylpropan-2-amine

   Classification: Secondary amine (2°)

   Explanation: The longest carbon chain is propane. The nitrogen atom is bonded to a methyl group and a propan-2-yl group. Since the nitrogen is bonded to two carbon atoms (one from methyl and one from the isopropyl group), it is a secondary amine. The name indicates a methyl group attached to the nitrogen of propan-2-amine.

4. (iv) $(\backslash text\{CH\}\_3)\_3\backslash text\{CNH\}\_2$

   IUPAC Name: 2-Methylpropan-2-amine

   Classification: Primary amine (1°)

   Explanation: The longest carbon chain has three carbons (propane), with a methyl group on the second carbon. The amino group ($-\backslash text\{NH\}\_2$) is attached to the tertiary carbon atom (the central carbon bonded to three other carbons). However, the nitrogen atom itself is bonded to only one carbon atom (the tertiary carbon), making it a primary amine.

5. (v) $\backslash text\{C\}\_6\backslash text\{H\}\_5\backslash text\{NHCH\}\_3$

   IUPAC Name: N-Methylbenzenamine

   Classification: Secondary amine (2°)

   Explanation: This compound is N-methylaniline. The nitrogen atom is bonded to a phenyl group ($\backslash text\{C\}\_6\backslash text\{H\}\_5$) and a methyl group ($-\backslash text\{CH\}\_3$). Since the nitrogen is bonded to two carbon atoms (one in the phenyl ring and one in the methyl group), it is a secondary amine.

6. (vi) $(\backslash text\{CH\}\_3\backslash text\{CH\}\_2)\_2\backslash text\{NCH\}\_3$

   IUPAC Name: N-Ethyl-N-methylethanamine

   Classification: Tertiary amine (3°)

   Explanation: The nitrogen atom is bonded to two ethyl groups and one methyl group. Since the nitrogen is bonded to three carbon atoms, it is a tertiary amine.

7. (vii) $m-\backslash text\{BrC\}\_6\backslash text\{H\}\_4\backslash text\{NH\}\_2$

   IUPAC Name: 3-Bromobenzenamine

   Classification: Primary amine (1°)

   Explanation: This is an aromatic amine where the amino group ($-\backslash text\{NH\}\_2$) is attached to a benzene ring substituted with a bromine atom at the meta position. The nitrogen atom is bonded to only one carbon atom (in the benzene ring), making it a primary amine.

1. Distinguishing Methylamine and Dimethylamine:

   The carbylamine test can be used. Methylamine is a primary aliphatic amine, while dimethylamine is a secondary amine.

   Procedure: Heat a small amount of the compound with chloroform ( \text{CHCl}_3 ) and an alcoholic solution of potassium hydroxide (KOH).

   Observation:

   Methylamine (primary amine) will produce a foul-smelling isocyanide (carbylamine), indicating a positive test.

   Dimethylamine (secondary amine) will not produce any such foul smell, indicating a negative test.

   Reactions:

   \text{CH}_3-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{CH}_3-\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}

   (Methylamine) (Methyl isocyanide - foul smell)

   (\text{CH}_3)_2\text{NH} + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{No reaction}

   (Dimethylamine)

2. Distinguishing Secondary and Tertiary Amines:

   The Hinsberg's test can be used. This test utilizes benzenesulphonyl chloride ( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} ), also known as Hinsberg's reagent.

   Procedure: React the amine with benzenesulphonyl chloride in the presence of aqueous KOH or NaOH.

   Observation:

   Secondary amines react to form a sulfonamide derivative which is insoluble in the alkali. For example, N,N-diethylamine reacts to form N,N-diethylbenzenesulphonamide, which precipitates out.

   Tertiary amines do not react with benzenesulphonyl chloride under these conditions.

   Reactions:

   For a secondary amine (e.g., diethylamine):

   (\text{CH}_3\text{CH}_2)_2\text{NH} + \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \xrightarrow{\text{KOH}} (\text{CH}_3\text{CH}_2)_2\text{NSO}_2\text{C}_6\text{H}_5 + \text{KCl} + \text{H}_2\text{O}

   (Diethylamine) (N,N-Diethylbenzenesulphonamide - insoluble in alkali)

   For a tertiary amine (e.g., trimethylamine):

   (\text{CH}_3)_3\text{N} + \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \xrightarrow{\text{KOH}} \text{No reaction}

   (Trimethylamine)

3. Distinguishing Ethylamine and Aniline:

   The azo-dye test can be used. Ethylamine is an aliphatic primary amine, while aniline is an aromatic primary amine.

   Procedure: React the amine with nitrous acid ( \text{HONO} , prepared in situ from \text{NaNO}_2 and dilute \text{HCl} ) at 0-5°C, followed by coupling with alkaline 2-naphthol.

   Observation:

   Aniline (aromatic amine) will form a diazonium salt which couples with 2-naphthol to produce a brightly coloured azo dye (usually orange or red).

   Ethylamine (aliphatic amine) reacts with nitrous acid to produce nitrogen gas with brisk effervescence, and no dye is formed.

   Reactions:

   For Aniline:

   \text{C}_6\text{H}_5\text{NH}_2 + \text{HONO} \xrightarrow{273-278 \text{ K}} [\text{C}_6\text{H}_5\text{N}_2]^+\text{Cl}^- + 2\text{H}_2\text{O}

   (Aniline) (Benzenediazonium chloride)

   [\text{C}_6\text{H}_5\text{N}_2]^+\text{Cl}^- + \text{2-Naphthol} \xrightarrow{\text{NaOH, pH 9-10}} \text{Orange dye}

   For Ethylamine:

   \text{CH}_3\text{CH}_2-\text{NH}_2 + \text{HONO} \xrightarrow{0-5^{\circ} \text{C}} \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 \uparrow + \text{H}_2\text{O}

   (Ethylamine) (Ethanol) (Nitrogen gas)

4. Distinguishing Aniline and Benzylamine:

   The reaction with nitrous acid at 0-5°C can distinguish them. Aniline is an aromatic primary amine, and benzylamine is an aliphatic primary amine.

   Procedure: Treat both amines with nitrous acid ( \text{NaNO}_2 + \text{dil. HCl} ) at 0-5°C.

   Observation:

   Aniline forms a stable diazonium salt, which can be detected by coupling with alkaline 2-naphthol to form an azo dye.

   Benzylamine reacts to form benzyl alcohol and nitrogen gas, as it is an aliphatic amine. It does not form a stable diazonium salt under these conditions and does not give the azo dye test.

   Reactions:

   For Aniline (as described in point 3): Forms a diazonium salt, leading to azo dye formation.

   For Benzylamine:

   \text{C}_6\text{H}_5\text{CH}_2-\text{NH}_2 + \text{HONO} \xrightarrow{0-5^{\circ} \text{C}} \text{C}_6\text{H}_5\text{CH}_2-\text{OH} + \text{N}_2 \uparrow + \text{H}_2\text{O}

   (Benzylamine) (Benzyl alcohol) (Nitrogen gas)

5. Distinguishing Aniline and N-methylaniline:

   The reaction with nitrous acid can distinguish them. Aniline is a primary aromatic amine, while N-methylaniline is a secondary amine.

   Procedure: Treat both compounds with nitrous acid ( \text{NaNO}_2 + \text{dil. HCl} ) at 0-5°C.

   Observation:

   Aniline (primary aromatic amine) will react to form a diazonium salt, which can then be coupled with 2-naphthol to form an azo dye.

   N-methylaniline (secondary amine) will react with nitrous acid to form an N-nitroso compound, which is typically an oily yellow substance. It does not form a diazonium salt and hence does not give the azo dye test.

   Reactions:

   For Aniline:

   \text{C}_6\text{H}_5\text{NH}_2 + \text{HONO} \xrightarrow{273-278 \text{ K}} [\text{C}_6\text{H}_5\text{N}_2]^+\text{Cl}^- \xrightarrow{\text{2-Naphthol}} \text{Azo dye}

   For N-methylaniline:

   \text{C}_6\text{H}_5\text{NHCH}_3 + \text{HONO} \xrightarrow{0-5^{\circ} \text{C}} \text{C}_6\text{H}_5\text{N}(\text{CH}_3)\text{NO} + \text{H}_2\text{O}

   (N-methylaniline) (N-methyl-N-nitrosoaniline - yellow oil)