Amines

Let’s analyze each statement about amines:

of aniline is more than that of methylamine: True. Aniline is a weaker base than methylamine due to resonance delocalization of the lone pair on nitrogen into the benzene ring. Hence, it has a higher .

Ethylamine is soluble in water whereas aniline is not: True. Ethylamine, being a small and polar molecule, forms hydrogen bonds with water and is soluble. Aniline is less polar and has limited solubility because of the bulky aromatic ring.

Ethanamide on reaction with and NaOH gives ethylamine: False. This is incorrect. The reaction of ethanamide with bromine and sodium hydroxide (Hofmann bromamide reaction) gives methylamine, not ethylamine.

Ethylamine reacts with nitrous acid to give ethanol: True. Primary aliphatic amines like ethylamine react with nitrous acid to form ethanol through diazonium salt decomposition.

Aniline does not undergo Friedel–Crafts reaction: True. The lone pair on nitrogen in aniline interacts with the Lewis acid catalyst (such as AlCl₃), which deactivates the aromatic ring and prevents the Friedel–Crafts reaction.

Correct Answer: (C)

The Kjeldahl's method cannot be applied to pyridine.

Explanation:

Kjeldahl's method is a widely used technique for the estimation of nitrogen in organic compounds. However, it cannot be applied to pyridine because pyridine is a basic compound and can react with the strong acids used in the Kjeldahl method, interfering with the process. The method works by converting the nitrogen into ammonium salt, but pyridine forms pyridinium salts, which prevent the complete conversion of nitrogen. Therefore, the presence of pyridine would lead to inaccurate results.

The reaction of benzene diazonium fluoroborate with aqueous sodium nitrite in the presence of copper leads to the formation of nitrobenzene. This reaction is a typical example of the Sandmeyer reaction, where a diazonium salt undergoes a substitution reaction in the presence of copper and sodium nitrite, forming a nitro group on the benzene ring.

The correct option is (E) : nitrobenzene

In aqueous solution, the basicity of amines is affected by the electron-donating or electron-withdrawing effects of substituents on the nitrogen atom.

• Methanamine (CH₃NH₂) is a stronger base compared to benzenamine (C₆H₅NH₂) because the methyl group in methanamine is an electron-donating group, which increases the electron density on nitrogen, enhancing its ability to accept a proton.
• Benzenamine (Aniline) has a weaker basicity because the phenyl group (C₆H₅) is electron-withdrawing through resonance, reducing the electron density on nitrogen and making it less available for protonation.
• Ethanamine (C₂H₅NH₂) is a stronger base than methanamine due to the ethyl group being a stronger electron-donating group than the methyl group.
• N, N-dimethylaniline has reduced basicity compared to ethanamine because the methyl groups on nitrogen withdraw electron density, making it less basic.

Thus, the weakest base is benzenamine and the strongest base is ethanamine.

The correct option is (D) : benzenamine and ethanamine

In aqueous solution, basicity depends on the availability of the lone pair on the nitrogen atom for protonation. Dimethylamine is the strongest base, followed by methylamine, trimethylamine, and ammonia.
Thus, the correct order of basicity is (A).

i) Non-reducing disaccharide: Sucrose is a non-reducing disaccharide because it does not reduce Fehling's solution. It is made up of glucose and fructose, but their reducing ends are involved in the glycosidic bond, making it non-reducing.

ii) Reducing disaccharide: Maltose is a reducing disaccharide because it has a free reducing end that can reduce Fehling's solution. It is made up of two glucose units.

iii) Aldohexose: Glucose is an aldohexose, which means it is a six-carbon sugar with an aldehyde group (-CHO) at one end.

iv) Ketohexose: Fructose is a ketohexose because it is a six-carbon sugar with a ketone group (-C=O) at the second carbon.

v) Polysaccharide: Lactose is a disaccharide, but polysaccharides like starch or cellulose would be examples of longer chains of sugars.

Thus, the correct matches are as follows:
- Non-reducing disaccharide → Sucrose
- Reducing disaccharide → Maltose
- Aldohexose → Glucose
- Ketohexose → Fructose
- Polysaccharide → Lactose