Amines - 1 - JEE MCQ

Equation for the Reaction:

CH₃NH₂+HCl→CH₃NH₃+Cl⁻

Explanation:

• Methylamine (CH₃NH₂) is a weak base.
• Hydrochloric acid (HCl) is a strong acid.
• When methylamine reacts with HCl, the lone pair on nitrogen accepts a proton (H⁺) from HCl, forming methylammonium ion (CH₃NH₃⁺).
• This results in the formation of methylammonium chloride (CH₃NH₃⁺ Cl⁻), which is a salt.

Final Answer:

Option A: Methylammonium chloride.

A quaternary ammonium salt is formed by multiple methods:

1. Serial nucleophilic substitution (Option A): A primary amine undergoes successive alkylation, forming a quaternary ammonium salt (R₄N⁺X⁻).
2. Reaction with alkyl or benzyl halide (Option B): Ammonia or amines react with alkyl/benzyl halides, leading to quaternary ammonium salt formation.
3. Nucleophilic substitution (Option C): Amines act as nucleophiles, replacing the halide group in alkyl halides, forming quaternary ammonium salts.

Since all these processes can lead to quaternary ammonium salt formation, the correct answer is: Option D: All of these

• The methoxy group (-OCH₃) at the para position significantly increases basicity compared to the -CH₃ group due to its strong +M (mesomeric) effect, which donates electron density to the benzene ring.

• The presence of the nitro group (-NO₂) at the para position decreases basicity due to its strong -M (mesomeric) and -I (inductive) effects, which withdraw electron density from the benzene ring, reducing the availability of the lone pair on nitrogen.

• The -OCH₃ group at the para position increases basicity more than the -CH₃ group at the para position, while the presence of the -NO₂ group at the para position significantly decreases basicity.

Hence, the correct answer is Option B.

(a): Reduction of hydrolysed product of ester by LiAlH₄ produces two alcohols. ​

Aromatic diazonium salts are more stable due to dispersal of the positive charge in the benzene ring.

In the carbylamine reaction, primary amines on heating with chloroform in the presence of alcoholic KOH form isocyanides (or carbylamines). It is used to distinguish 1° amines from 2° and 3° amines.

(d): 2° aliphatic and aromatic amines react with nitrous acid to form N-nitrosoamine.

(c): FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH

Acidity decreases as the -I effect of the group decreases. Fluorine (F) is the most electronegative atom and hence has the highest -I effect among the halogens. ​

When PCl₅ reacts with a carboxylic acid, an acyl halide is formed. Here's how it works:

• PCl₅ is a chemical reagent used to replace the -OH group of a carboxylic acid with a chlorine atom.
• This results in the formation of an acyl chloride, a type of acyl halide.
• Acyl halides are reactive and commonly used in organic synthesis.

(c): Electrolytic reduction of nitrobenzene in weakly acidic medium gives aniline, but in strongly acidic medium, it gives p-aminophenol through the acid-catalyzed rearrangement of the initially formed phenylhydroxylamine.

(b): ‘C’ must be an isocyanide and obtained from a 1° amine by the Carbylamine reaction (CHCl₃ + KOH). Further, the 1° amine must be obtained by reduction of nitrohydrocarbon. So, ‘A’ is nitrobenzene.

(NaOH + I₂) / NaOI is the best suitable reagent for the above reaction.

(b): Basic strength decreases as:
cyclohexylamine > aniline > benzamide.

Lesser basicity in aniline and benzamide is due to the participation of the lone pair of electrons of the –NH₂ group in resonance.

The reaction involves a racemic acid (CH₃CHClCOOH) reacting with (S)-2-methylbutan-1-ol. The mixture is then distilled, resulting in:

• Formation of two different compounds: (R, S) and (S, S).
• These compounds are diastereomers.
• Diastereomers have different physical properties (such as boiling points) and can be separated by distillation.
• Each fraction obtained is optically active.
• The racemic mixture reacts separately with (S)-2-methylbutan-1-ol, forming (R, S) and (S, S) esters, which are diastereomers.

Final Answer: Option A: Two fractions, each optically active.

(b): In benzylamine, the electron pair present on the nitrogen is not delocalized with the benzene ring.

since –CH₂–NH₂ is more basic.
The resulting amide will fail to react further. Had it been possible, imide formation would have occurred at both the sites.