Learnosha CBSE 12
Chemistry Amines 4 Marks
Appeared in: 2025 (3 sets)
Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes. Answer the following questions : (a) How can you convert the following ? (i) Ethanoic acid to methanamine (ii) Propanenitrile to 1-aminopropane (b) Why is $\text{pK}_\text{b}$ value of aniline more than that of methylamine ? (c) (i) Arrange the following in increasing order of their basic strength in aqueous solution : $\text{CH}_3-\text{NH}_2$, $(\text{CH}_3)_2\text{NH}$, $(\text{CH}_3)_3\text{N}$ OR (c) (ii) Give the structures of A and B in the following reaction : $\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{A} \xrightarrow[273\text{ K}]{\text{HNO}_2} \text{B}$
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CBSE Class 12 Chemistry: Marking Scheme Solution

(a) Conversions

  • (i) Ethanoic acid to methanamine: [1 Mark]

    Step 1: Convert Ethanoic acid to Ethanamide using $\text{NH}_3$ and heat. Step 2: Perform Hofmann Bromamide degradation.

    $$\text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{NH}_2$$
  • (ii) Propanenitrile to 1-aminopropane: [1 Mark]

    Reduction of nitrile using strong reducing agents like $\text{LiAlH}_4$ or $\text{H}_2/\text{Ni}$.

    $$\text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$

(b) $\text{pK}_\text{b}$ value of aniline vs methylamine: [1 Mark]

Step 1 (Concept): $\text{pK}_\text{b}$ is inversely proportional to basic strength. Aniline is a weaker base than methylamine.

Step 2 (Reasoning): In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring due to resonance, making it less available for protonation. In methylamine, the electron-releasing inductive effect ($+\text{I}$) of the methyl group increases electron density on nitrogen, making it more available for protonation.

Step 3 (Conclusion): Since aniline is a weaker base, its $\text{K}_\text{b}$ is lower, and therefore its $\text{pK}_\text{b}$ ($-\log \text{K}_\text{b}$) is higher.

(c) (i) Increasing order of basic strength: [1 Mark]

Step 1 (Governing Factors): Basic strength in aqueous solution depends on the combined effect of inductive effect ($+\text{I}$), solvation effect (H-bonding), and steric hindrance.

Step 2 (Ordering): For methyl-substituted amines, the order is $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N}$.

Step 3 (Final Answer): The increasing order is: $(\text{CH}_3)_3\text{N} < \text{CH}_3\text{NH}_2 < (\text{CH}_3)_2\text{NH}$

OR (c) (ii) Structures of A and B: [1 Mark]

Step 1 (Reaction A): Reduction of nitrobenzene with $\text{Fe/HCl}$ yields Aniline.

Step 2 (Reaction B): Diazotization of aniline with $\text{HNO}_2$ at $273\text{--}278\text{ K}$ yields Benzenediazonium chloride.

Step 3 (Final Answer):

  • A: $\text{C}_6\text{H}_5\text{NH}_2$ (Aniline)
  • B: $\text{C}_6\text{H}_
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