Amines Study Notes

Amines Study Notes

Very Short Answer Type Questions (1 Mark)

Q1: Complete the following reaction.

  • Reaction:

    • Inputs: CoHs, Na, tce, H₂O (room temp)

    • Outputs: N₂ + ce-

    • Result: N₂ + Hu

Q2: Write the IUPAC name of the following compound: (CH₁₂N-CH₂CH₂CH₂-N-CH₂CH)

  • IUPAC Name: N,N-Dimethyl ethanamine (2017)

Q3: Out of CH₃NH₂ and CH₃OH, which has a higher boiling point?

  • Answer: The hydrogen bond in alcohol (CH₃OH) is stronger than that in amines due to the higher electronegativity of oxygen compared to nitrogen, hence CH₃OH has a higher boiling point.

Q4: Which is more basic and why: CH₃NH₂ or NH₃?

  • Answer: CH₃NH₂ (methanamine) is more basic due to the presence of an electron-donating CH₃ group, which increases electron density on the nitrogen atom (2014).

Intermediate Questions (2 Marks)

Q5: Give the structure of A and B in the following sequence:

  • Reaction: C₆H₅NO₂ + Fe/H

  • Input: NaNO₂ + HCl -5°C

  • Structure A:

    • N₂ + H₂O

  • Structure B:

    • Aniline (C₆H₅NH₂)
      (2020)

Q6: Write the IUPAC name of the given compound: Benzene diazonium chloride.

  • Answer: The compound is 2,4,6-tribromoaniline (2016).

Q7: How do you convert the following: Ethanenitrile to ethanamine?

  • Reaction:

    • Reactants: CH₃CN with LiAlH₄ or NaCH₃ followed by C₂H₅OH

    • Product: Ethanamine (C₂H₅NH₂)
      (2015)

Q8: Why does propanamine have a higher boiling point than N,N-dimethylmethanamine despite containing the same number of carbon atoms?

  • Answer: Propanamine has two hydrogen atoms on nitrogen that can form intermolecular hydrogen bonds, while no such hydrogen bonding can occur in N,N-dimethylmethanamine (2019).

Q9: Write the chemical equation involved in the following reaction: Hoffmann bromamide degradation reaction.

  • Reaction:

    • RCONH₂ + Br₂ + 4 NaOH → RNH₂ + Na₂CO₃ + 2 NaBr + 2H₂O (2019)

Q10: Write the structure of A and B in the following reactions:

  • Reaction:

    • C₆H₅N₂ + CuK →

  • Results:

    • A: Benzenediazonium chloride (C₆H₅N₂Cl)

    • B: Benzonitrile
      (2020)

Q11: Arrange the following in order of base strength in the gas phase:

  • Compounds: (C₂H₅)₃N, C₂H₅NH₂, (C₂H₅)₂NH

  • Order: C₂H₅NH₂ < (C₂H₅)₂NH < (C₂H₅)₃N (2019).

Q12: How will you convert Aniline into N-phenylethanamide?

  • Reaction:

    • Inputs: Aniline + CH₃COCl → N-Phenylethanamide (Acetanilide)
      (2014)

Short Answer Type Questions (2 Marks)

Q13: Give reasons for the following:

  1. Aniline does not undergo Friedel-Craft reaction.

    • Explanation: Aniline forms a salt with aluminium chloride, a Lewis acid catalyst, resulting in a positive charge on nitrogen, which deactivates the ring for further acylation or alkylation (2014).

  2. The pKb of methylamine is less than that of aniline.

    • Explanation: In aniline, the lone pair on the nitrogen atom is delocalized over the benzene ring, leading to decreased electron density on nitrogen. In contrast, the +I effect of the CH₃ group in methylamine increases electron density on the nitrogen atom.

Q14: Distinguish between the following pairs:

  1. CH₃CH₂NH₂ and (CH₃CH₂)₂NH

    • Answer: Ethylamine (1°) on heating with alcoholic KOH and CHCl₃ gives foul-smelling ethyl isocyanide + H₂O, whereas diethylamine (2°) does not give this test (2020).

  2. Aniline and N-methyl aniline

    • Answer: N-methyl aniline on reaction may show different solubility or chemical behavior compared to aniline.

Q15: Arrange the following compounds in increasing order of solubility in water:

  • Compounds: C₆H₅NH₂ < (CH₃)₂NH < CH₃NH₂

    • Explanation: Benzene ring decreases solubility due to hydrophobic interactions, while primary amines can hydrogen bond more effectively than secondary amines (2020).

Q16: Write the chemical reaction involved in the Diazotization process.

  • Reaction: Primary aromatic amines (R-NH₂) react with NaNO₂ in the presence of HCl to form diazonium salt R-N₂⁺ (273-278 K). Aromatic diazonium salts are more stable than aliphatic salts due to resonance stabilization.

Q17: Explain why electrophilic substitution in aromatic amines occurs more readily than in benzene.

  • Answer: The -NH₂ group of aromatic amines activates the aromatic ring through delocalization of its lone pair of electrons, which increases electron density on the ring, facilitating the electrophilic attack (2013).

Q18: Write the IUPAC name of the compound (CH₃)₂NH.

  • Answer: Dimethylamine or N,N-Dimethylamine (2023).

Q19: Arrange the following compounds in increasing order of basic strength in aqueous solution:

  • Compounds: (CH₃)₂NH < CH₃NH₂ < (CH₃)₃N (2022).

Q20: What is Hinsberg's reagent?

  • Answer: Benzene sulphonyl chloride, which is used in the Hinsberg test to distinguish between primary, secondary, and tertiary amines (2022).

Q21: Describe the role of pyridine in acylation reactions of amines.

  • Answer: Pyridine acts as a base that abstracts the proton formed during the reaction, thus shifting the equilibrium to favor the reaction product (2022).

Q22: Write the chemical reaction of methylamine with benzoyl chloride and write the IUPAC name of the product obtained.

  • Reaction: CH₃NH₂ + C₆H₅COCl → N-Methyl benzamide
    (2020).

Q23: A compound A on reduction with iron hydrochloric acid gives compound B with molecular formula C₆H₇N.

  • Answer:

    • Compound A: Nitrobenzene (C₆H₅NO₂)

    • Compound B: Aniline (C₆H₅NH₂)

    • Reaction: C₆H₅NO₂ + Fe + HCl → C₆H₅NH₂ + H₂O (2022).

Q24: Explain why aliphatic amines are stronger bases than ammonia.

  • Answer: Aliphatic amines possess alkyl groups that provide an electron-donating inductive effect, which increases the electron density on nitrogen, enhancing its basicity compared to ammonia.

Conclusion:

  • These notes provide comprehensive coverage of questions, reactions, definitions, and explanations related to amines. Use as a detailed reference for study and exam preparation.