Chapter 9: Amines Short Questions | chemistry Class 12 CBSE

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Chapter 9: Amines Short Questions | chemistry Class 12 CBSE | ByteNotes.in

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Question 1

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What is the hybridisation and geometry of nitrogen in amines? How does the lone pair affect bond angles?

The nitrogen atom in amines is sp3 hybridised, with three bonding orbitals and one orbital containing the lone pair.
The geometry of amines is pyramidal, similar to ammonia.
The lone pair causes the C–N–C or C–N–H bond angle to be slightly less than 109.5°; for example, trimethylamine has a C–N–C angle of 108°.

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Question 2

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Classify amines into primary, secondary, and tertiary with one example each.

Primary amines (1°): one hydrogen of NH3 is replaced by an alkyl or aryl group; example: CH3NH2 (methylamine).
Secondary amines (2°): two hydrogens of NH3 are replaced; example: (CH3)2NH (dimethylamine).
Tertiary amines (3°): all three hydrogens of NH3 are replaced; example: (CH3)3N (trimethylamine).

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Question 3

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Write the IUPAC names of (i) CH3NHCH2CH3 and (ii) (CH3CH2)3N.

CH3NHCH2CH3: the principal chain is ethanamine; the methyl group is on nitrogen; IUPAC name is N-methylethanamine.
(CH3CH2)3N: principal chain is ethanamine; two ethyl groups are on nitrogen; IUPAC name is N,N-diethylethanamine.
In IUPAC nomenclature of secondary and tertiary amines, the locant N is used to designate substituents attached to the nitrogen atom.

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Question 4

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Why is reduction with iron scrap and hydrochloric acid preferred over other reductants in the reduction of nitro compounds to amines?

FeCl2 formed during the reaction gets hydrolysed to release hydrochloric acid back into the reaction mixture.
Because HCl is regenerated in situ, only a small initial amount of hydrochloric acid is needed to initiate the reaction.
This makes the process economical and avoids the use of expensive reagents like Sn or catalytic hydrogen.

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Question 5

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What is ammonolysis? Why does it yield a mixture of products? How is a primary amine obtained as the major product?

Ammonolysis is the cleavage of the C–X bond in an alkyl or benzyl halide by reaction with ethanolic ammonia via nucleophilic substitution, forming an amine.
The primary amine formed is itself a nucleophile and reacts further with the alkyl halide to form secondary, tertiary amines, and finally a quaternary ammonium salt, giving a mixture.
A primary amine is obtained as the major product by using a large excess of ammonia, which suppresses further alkylation.

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Question 6

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How is reduction of nitriles useful in the preparation of amines? What is the significance of this method?

Nitriles on reduction with LiAlH4 or by catalytic hydrogenation (H2/Ni) yield primary amines.
This reaction is used for the ascent of amine series, i.e., for preparation of a primary amine containing one more carbon atom than the starting amine.
For example, R–C≡N is reduced to R–CH2–NH2, giving a primary amine with one additional carbon.

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Question 7

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Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Gabriel phthalimide synthesis involves nucleophilic substitution of the alkyl halide by the potassium salt of phthalimide.
Aryl halides do not undergo nucleophilic substitution reactions under normal conditions because the benzene ring is electron-rich and resists attack by the nucleophilic phthalimide anion.
Hence, aromatic primary amines such as aniline cannot be prepared by this method.

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Question 8

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State the Hoffmann bromamide degradation reaction. What is the structural relationship between the amide and the amine formed?

In Hoffmann bromamide degradation, an amide is treated with Br2 in aqueous or ethanolic NaOH to yield a primary amine.
During this reaction, an alkyl or aryl group migrates from the carbonyl carbon of the amide to the nitrogen atom.
The primary amine formed contains one carbon atom less than the starting amide, making this a carbon-descending reaction.

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Question 9

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Why do lower aliphatic amines dissolve in water, but higher amines do not?

Lower aliphatic amines can form hydrogen bonds with water molecules due to the electronegative nitrogen atom, making them soluble.
As the molar mass and size of the alkyl group increase, the hydrophobic part of the molecule grows larger.
The increase in the hydrophobic alkyl portion outweighs hydrogen bonding capability, causing higher amines to be essentially insoluble in water.

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Question 10

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Explain why the boiling point order of isomeric amines is primary > secondary > tertiary.

Primary amines have two N–H hydrogens available for intermolecular hydrogen bonding, resulting in strong intermolecular association and the highest boiling point.
Secondary amines have one N–H hydrogen, so they form fewer hydrogen bonds than primary amines and have intermediate boiling points.
Tertiary amines have no N–H hydrogen atoms and hence cannot form intermolecular hydrogen bonds, giving them the lowest boiling points among isomers.

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Question 11

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How do amines behave as Lewis bases? Give the reaction of methylamine with HCl.

Amines possess an unshared pair of electrons on the nitrogen atom, which they can donate to a proton or Lewis acid, making them Lewis bases.
Methylamine reacts with HCl to form methylammonium chloride: CH3NH2 + HCl → CH3NH3+Cl–.
The amine salt formed is soluble in water; treatment with NaOH regenerates the free amine.

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Question 12

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Why are aliphatic amines stronger bases than ammonia? Explain with reference to the +I effect.

Alkyl groups are electron-releasing (+I effect), which pushes electron density towards the nitrogen atom, increasing availability of the lone pair for protonation.
The substituted ammonium cation formed after protonation is also stabilised by dispersal of positive charge through the +I effect of alkyl groups.
Both factors make aliphatic amines stronger proton acceptors (lower pKb) than ammonia, whose pKb is 4.75 compared to 3–4.22 for alkylamines.

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Question 13

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Why is aniline a weaker base than ammonia? Use resonance to explain.

In aniline, the lone pair on nitrogen is in conjugation with the benzene ring pi system, as shown by five resonating structures of aniline.
This delocalisation makes the lone pair less available for donation to a proton.
The anilinium ion (formed by protonation) has only two Kekulé resonance structures, making it less stable than aniline; therefore aniline is less prone to accept a proton, making it a weaker base (pKb = 9.38) than ammonia (pKb = 4.75).

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Chapter 9: Amines | Short Questions

What is the hybridisation and geometry of nitrogen in amines? How does the lone pair affect bond angles?

Classify amines into primary, secondary, and tertiary with one example each.

Write the IUPAC names of (i) CH3NHCH2CH3 and (ii) (CH3CH2)3N.

Why is reduction with iron scrap and hydrochloric acid preferred over other reductants in the reduction of nitro compounds to amines?

What is ammonolysis? Why does it yield a mixture of products? How is a primary amine obtained as the major product?

How is reduction of nitriles useful in the preparation of amines? What is the significance of this method?

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

State the Hoffmann bromamide degradation reaction. What is the structural relationship between the amide and the amine formed?

Why do lower aliphatic amines dissolve in water, but higher amines do not?

Explain why the boiling point order of isomeric amines is primary > secondary > tertiary.

How do amines behave as Lewis bases? Give the reaction of methylamine with HCl.

Why are aliphatic amines stronger bases than ammonia? Explain with reference to the +I effect.

Why is aniline a weaker base than ammonia? Use resonance to explain.

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