CBSE Class 12 Chemistry Chapter 13: Amines NCERT Solutions

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

1. (i) $(CH\_3)\_2\; CHNH\_2$

   IUPAC Name: Propan-2-amine

   Classification: Primary amine (1°)

   Explanation: The longest carbon chain has three carbons (propane). The amino group ($-NH\_2$) is attached to the second carbon. Since the nitrogen atom is bonded to only one carbon atom of the alkyl group, it is a primary amine.

2. (ii) $CH\_3(CH\_2)\_2\; NH\_2$

   IUPAC Name: Propan-1-amine

   Classification: Primary amine (1°)

   Explanation: The longest carbon chain has three carbons (propane). The amino group ($-NH\_2$) is attached to the first carbon. The nitrogen atom is bonded to only one carbon atom of the alkyl group, classifying it as a primary amine.

3. (iii) $CH\_3\; NHCH(CH\_3)\_2$

   IUPAC Name: N-Methylpropan-2-amine

   Classification: Secondary amine (2°)

   Explanation: The longest carbon chain is propane. The nitrogen atom is bonded to one methyl group and the propane chain. Since the nitrogen atom is bonded to two carbon atoms (one from the methyl group and one from the propane chain), it is a secondary amine.

4. (iv) $(CH\_3)\_3\; CNH\_2$

   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 ($-NH\_2$) is attached to the tertiary carbon atom. However, the nitrogen atom itself is bonded to only one carbon atom of the alkyl structure (the tertiary carbon), making it a primary amine.

5. (v) $C\_6H\_5\; NHCH\_3$

   IUPAC Name: N-Methylbenzenamine

   Classification: Secondary amine (2°)

   Explanation: This is an aniline derivative where one hydrogen on the nitrogen is replaced by a methyl group. The nitrogen atom is bonded to the phenyl ring (one carbon) and the methyl group (one carbon), classifying it as a secondary amine.

6. (vi) $(CH\_3CH\_2)\_2\; NCH\_3$

   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 atom is bonded to three carbon atoms (two from ethyl groups and one from the methyl group), it is a tertiary amine.

7. (vii) $m-BrC\_6H\_4NH\_2$

   IUPAC Name: 3-Bromobenzenamine

   Classification: Primary amine (1°)

   Explanation: This is a derivative of aniline (benzenamine) with a bromine atom at the meta position. The amino group ($-NH\_2$) is directly attached to the benzene ring, and the nitrogen atom is bonded to two hydrogen atoms and one carbon atom of the ring, making it a primary amine.

Here is a chemical test to distinguish between each pair of compounds:

1. Distinguishing Methylamine and Dimethylamine:

   The Carbylamine Test can be used. Methylamine is a primary amine, while dimethylamine is a secondary amine. Primary amines (both aliphatic and aromatic) react with chloroform ($CHCl\_3$) and an alcoholic solution of potassium hydroxide (KOH) upon heating to form foul-smelling isocyanides (carbylamines). Secondary amines do not give this test.

   Reaction for Methylamine (Primary):

   $$CH\_3\; -\; NH\_2\; +\; CHCl\_3\; +\; 3KOH\; \backslash xrightarrow\{\backslash Delta\}\; CH\_3\; -\; NC\; +\; 3KCl\; +\; 3H\_2O$$

   (Methylamine) (Methyl isocyanide - foul smell)

   Reaction for Dimethylamine (Secondary):

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

   Thus, the formation of a foul smell indicates the presence of methylamine.

2. Distinguishing Secondary and Tertiary Amines:

   The Hinsberg's Test is effective. This test uses benzenesulfonyl chloride ($C\_6H\_5SO\_2Cl$), also known as Hinsberg's reagent. The reaction is carried out in the presence of aqueous alkali (like NaOH or KOH).

   Reaction with Secondary Amines: Secondary amines react with benzenesulfonyl chloride to form N,N-disubstituted benzenesulfonamides. These sulfonamides are acidic due to the absence of an N-H bond and are therefore insoluble in the alkaline solution.

   Example: N,N-Diethylamine

   $$(C\_2H\_5)\_2NH\; +\; C\_6H\_5SO\_2Cl\; \backslash xrightarrow\{NaOH\}\; (C\_2H\_5)\_2N-SO\_2C\_6H\_5\; +\; NaCl$$

   (N,N-Diethylbenzenesulfonamide - insoluble in alkali)

   Reaction with Tertiary Amines: Tertiary amines do not have a hydrogen atom attached to the nitrogen that can be substituted. Therefore, they do not react with benzenesulfonyl chloride under these conditions and remain soluble in the alkaline solution.

   Thus, observing whether a precipitate forms (insoluble in alkali) or not helps distinguish between secondary and tertiary amines.

3. Distinguishing Ethylamine and Aniline:

   The Azo-Dye Test can be used. This test differentiates between aliphatic and aromatic primary amines.

   Procedure: Both amines are treated with nitrous acid ($HONO$, prepared in situ from $NaNO\_2$ and dilute $HCl$) at 0-5°C, followed by coupling with an alkaline solution of 2-naphthol.

   Reaction with Aniline (Aromatic Primary Amine): Aniline reacts with nitrous acid to form a diazonium salt (benzenediazonium chloride). This diazonium salt then couples with 2-naphthol in an alkaline medium to form a brightly colored azo dye (usually orange or red).

   $$C\_6H\_5NH\_2\; +\; HONO\; \backslash xrightarrow\{0-5^\{\backslash circ\}C\}\; [C\_6H\_5N\_2]^+Cl^-\; +\; 2H\_2O$$

   (Aniline) (Benzenediazonium chloride)

   $$[C\_6H\_5N\_2]^+Cl^-\; +\; \backslash text\{2-Naphthol\}\; \backslash xrightarrow\{NaOH,\; pH\; 9-10\}\; \backslash text\{Orange\; dye\}$$

   Reaction with Ethylamine (Aliphatic Primary Amine): Ethylamine reacts with nitrous acid at 0-5°C to form ethanol, nitrogen gas, and water. It does not form a stable diazonium salt and therefore does not undergo coupling to form an azo dye.

   $$CH\_3CH\_2NH\_2\; +\; HONO\; \backslash xrightarrow\{0-5^\{\backslash circ\}C\}\; CH\_3CH\_2OH\; +\; N\_2\; \backslash uparrow\; +\; H\_2O$$

   (Ethylamine) (Ethanol)

   The formation of a colored azo dye indicates aniline.

4. Distinguishing Aniline and Benzylamine:

   Both aniline and benzylamine are primary amines. They can be distinguished by their reaction with nitrous acid ($HONO$), prepared in situ from $NaNO\_2$ and dilute $HCl$ at 0-5°C.

   Reaction with Aniline (Aromatic Primary Amine): Aniline forms a stable diazonium salt at low temperatures, which can be isolated or used for coupling reactions.

   $$C\_6H\_5NH\_2\; +\; HONO\; \backslash xrightarrow\{0-5^\{\backslash circ\}C\}\; [C\_6H\_5N\_2]^+Cl^-\; +\; 2H\_2O$$

   Reaction with Benzylamine (Aliphatic Primary Amine): Benzylamine reacts with nitrous acid to form benzyl alcohol, nitrogen gas, and water. Unlike aromatic amines, it does not form a stable diazonium salt under these conditions.

   $$C\_6H\_5CH\_2NH\_2\; +\; HONO\; \backslash xrightarrow\{0-5^\{\backslash circ\}C\}\; C\_6H\_5CH\_2OH\; +\; N\_2\; \backslash uparrow\; +\; H\_2O$$

   (Benzylamine) (Benzyl alcohol)

   The formation of a stable diazonium salt (which can be confirmed by coupling reactions or other tests) indicates aniline, while the evolution of nitrogen gas and formation of an alcohol indicates benzylamine.

5. Distinguishing Aniline and N-methylaniline:

   Aniline is a primary amine, while N-methylaniline is a secondary amine. They can be distinguished using the Carbylamine Test or by their reaction with Nitrous Acid.

   Using Carbylamine Test:

   Aniline (primary amine) will give a positive carbylamine test (formation of foul-smelling phenyl isocyanide) when heated with chloroform and alcoholic KOH.

   $$C\_6H\_5NH\_2\; +\; CHCl\_3\; +\; 3KOH\; \backslash xrightarrow\{\backslash Delta\}\; C\_6H\_5NC\; +\; 3KCl\; +\; 3H\_2O$$

   (Phenyl isocyanide - foul smell)

   N-methylaniline (secondary amine) will not give the carbylamine test.

   Using Nitrous Acid:

   Aniline reacts with nitrous acid at 0-5°C to form a stable diazonium salt, $[C\_6H\_5N\_2]^+Cl^-$.

   N-methylaniline (a secondary amine) reacts with nitrous acid to form N-nitroso compounds, specifically N-methyl-N-nitrosoaniline, which is a yellow oily substance.

   $$C\_6H\_5NHCH\_3\; +\; HONO\; \backslash xrightarrow\{0-5^\{\backslash circ\}C\}\; C\_6H\_5N(CH\_3)NO\; +\; H\_2O$$

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

   The formation of a yellow oily N-nitroso compound indicates N-methylaniline, while the formation of a diazonium salt indicates aniline.