CUET UG Chemistry Booster Test - 3 Fundamentals of Amines -

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CUET UG Chemistry Booster Test - 3 Fundamentals of Amines

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 3 Fundamentals of Amines

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QUESTION 1 OF 20

Arrange the following compounds in decreasing order of the number of hydrogen atoms directly attached to the nitrogen atom (reflecting the replacement sequence):
(A) Ammonia
(B) Ethanamine
(C) N-Methylethanamine
(D) N,N-Diethylethanamine

Answer Review

1 Arrange the following compounds in decreasing order of the number of hydrogen atoms directly attached to the nitrogen atom (reflecting the replacement sequence):
(A) Ammonia
(B) Ethanamine
(C) N-Methylethanamine
(D) N,N-Diethylethanamine

A (A), (B), (C), (D)

B (D), (C), (B), (A)

C (B), (A), (D), (C)

D (A), (C), (B), (D)

💡 Short Explanation

�� Ammonia has three N–H bonds. �� Ethanamine has two N–H bonds. �� Secondary and tertiary amines have one and zero N–H bonds respectively.

✅ Correct Answer Explanation

Number of hydrogen atoms directly attached to nitrogen: → Ammonia = NH₃ = 3 → Ethanamine = C₂H₅NH₂ = 2 → N-Methylethanamine = secondary amine = 1 → N,N-Diethylethanamine = tertiary amine = 0 So, the decreasing order is: (A), (B), (C), (D)

❓ Why Other Options Are Incorrect

�� Option B → Gives increasing order instead of decreasing order.
�� Option C → Ethanamine is placed before ammonia incorrectly.
�� Option D → N-Methylethanamine and Ethanamine are interchanged.

🎯 Strategy

Used

Substitution

Application:

�� Write the general formula and count N–H bonds.

Final Logic:

�� 3 > 2 > 1 > 0 gives Ammonia → Ethanamine → N-Methylethanamine → N,N-Diethylethanamine.

🧠 Memory Trick

NH₃ has 3H, 1° has 2H, 2° has 1H, 3° has 0H

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Classification of Amines

2 Analyze the conceptual definition of amines:
(A) Replacement of one H yields a 1° amine.
(B) Quaternary ammonium salts are strictly classified as secondary amines.
(C) Alkyl or aryl groups are the sole substituents in the core replacement definition of amines.
(D) Replacement of all three H atoms yields a tertiary amine.

A A, B, C

B A, C, D

C B, C, D

D A, D

💡 Short Explanation

�� One H replacement gives primary amine. �� Alkyl/aryl groups replace ammonia hydrogens. �� Three H replacements give tertiary amine.

✅ Correct Answer Explanation

Amines are derivatives of ammonia formed by replacing one or more hydrogen atoms with alkyl or aryl groups. ✓ Statement A is correct: one hydrogen replacement gives a primary amine. ✓ Statement C is correct: alkyl or aryl groups replace hydrogen atoms in the basic definition. ✓ Statement D is correct: replacement of all three hydrogens gives a tertiary amine. ✗ Statement B is incorrect because quaternary ammonium salts have four groups attached to nitrogen and are not secondary amines.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement B, which is incorrect.
�� Option C → Includes Statement B, which is incorrect.
�� Option D → Omits Statement C, which is correct.

🎯 Strategy

Used

Option Grouping

Application:

�� Check each statement with the basic definition of amines.

Final Logic:

�� A, C and D correctly explain the replacement concept of amines.

🧠 Memory Trick

1H → 1°, 3H → 3°

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Definition and Classification of Amines

3 Match the following

List 1 (Biological/Synthetic Compound)	List 2 (Amine characteristic/Function)
1. Novocain	a. Used as surfactants
2. Benadryl	b. Contains a secondary amino group, increases blood pressure
3. Quaternary ammonium salts	c. Synthetic amino compound used as a dentistry anaesthetic
4. Adrenaline	d. Antihistaminic drug containing a tertiary amino group

A 1-c, 2-d, 3-a, 4-b

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

Novocain is a dental local anaesthetic. Benadryl is an antihistaminic containing a tertiary amino group. Quaternary ammonium salts are widely used as surfactants, while adrenaline contains a secondary amino group.

✅ Correct Answer Explanation

According to NCERT, amines and their derivatives have numerous biological and industrial applications. Novocain is a synthetic amino compound used as a local anaesthetic in dentistry. Benadryl is an antihistaminic drug containing a tertiary amino group, used to relieve allergic reactions. Quaternary ammonium salts possess surface-active properties and are widely used as surfactants and disinfectants. Adrenaline (epinephrine) contains a secondary amino group and plays an important physiological role by increasing blood pressure during emergency conditions. Therefore, the correct matching is: 1 → c 2 → d 3 → a 4 → b Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because Novocain is not a surfactant, Benadryl does not increase blood pressure, and quaternary ammonium salts are not dentistry anaesthetics.
Option C → Incorrect because Novocain is not an antihistaminic, Benadryl is not a dentistry anaesthetic, and quaternary ammonium salts do not contain a secondary amino group.
Option D → Incorrect because multiple compound–function pairs are incorrectly matched.

🎯 Strategy

Used: Option Grouping

Application: Match the well-known drug–function pairs first, then identify the remaining industrial and biological applications.

Final Logic:

Novocain → Dental anaesthetic
Benadryl → Antihistaminic
Quaternary ammonium salts → Surfactants
Adrenaline → Secondary amine; increases blood pressure

🧠 Memory Trick

"Novo Numbs, Benadryl Battles Allergy, Quats Clean, Adrenaline Raises BP."

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Importance and Uses of Amines

4 According to the text, diazonium salts, which are related to amines, serve primarily as intermediates in the synthesis of what specific class of aromatic compounds?

A Aliphatic polymers

B Quaternary ammonium surfactants

C A variety of aromatic compounds including dyes

D Naturally occurring hormones

💡 Short Explanation

�� Diazonium salts are important synthetic intermediates. �� They are used in preparing aromatic compounds. �� Azo dyes are important products.

✅ Correct Answer Explanation

Diazonium salts are prepared from aromatic primary amines and are highly useful intermediates in organic synthesis. They are especially important in preparing a variety of aromatic compounds, including azo dyes.

❓ Why Other Options Are Incorrect

�� Option A → Diazonium salts are mainly linked with aromatic synthesis, not aliphatic polymers.
�� Option B → Quaternary ammonium surfactants are different compounds.
�� Option D → Naturally occurring hormones are not the primary synthetic use of diazonium salts.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect diazonium salts with their standard NCERT application.

Final Logic:

�� Diazonium salts are used for synthesis of aromatic compounds including dyes.

🧠 Memory Trick

Diazonium → Dyes

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Diazonium Salts and Their Importance

📄 Passage

Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp³ hybridised and the geometry of amines is pyramidal. Each of the three sp³ hybridised orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines. The fourth orbital of nitrogen in all amines contains an unshared pair of electrons. Due to the presence of unshared pair of electrons, the angle C–N–E, (where E is C or H) is less than 109.5°; for instance, it is 108° in case of trimethylamine.

If an amine molecule were to undergo a theoretical change where its unshared pair of electrons is donated to fully form a dative bond with a proton (H⁺), what would be the resulting structural geometry of the central nitrogen atom, based on its initial sp³ hybridisation?

A It would become trigonal planar.

B It would become perfectly tetrahedral (109.5°).

C It would remain pyramidal with a compressed angle.

D It would become linear.

💡 Short Explanation

�� Lone pair becomes a bond pair after protonation. �� Nitrogen then has four bond pairs. �� Four bond pairs give tetrahedral geometry.

✅ Correct Answer Explanation

In amines, nitrogen is sp³ hybridised with three bond pairs and one lone pair, giving pyramidal geometry. When the lone pair is donated to H⁺, it forms a coordinate bond. Now nitrogen has four bond pairs and no lone pair. Four bond pairs around sp³ nitrogen produce tetrahedral geometry with an angle close to 109.5°, as in ammonium-type ions.

❓ Why Other Options Are Incorrect

�� Option A → Trigonal planar geometry corresponds to sp² arrangement, not protonated sp³ nitrogen.
�� Option C → Pyramidal shape occurs when a lone pair remains on nitrogen.
�� Option D → Linear geometry is not possible with four electron domains.

🎯 Strategy

Used

Substitution

Application:

�� Replace the lone pair with a bond pair and predict geometry.

Final Logic:

�� Four bond pairs around sp³ nitrogen produce tetrahedral geometry.

🧠 Memory Trick

Lone pair + H⁺ = 4 bonds

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Structure, Hybridisation and Basicity of Amines

📄 Passage

The pyramidal geometry of amines is an outcome of sp³ hybridisation combined with lone pair-bond pair repulsion. Which of the following best contrasts with this by representing a perfect tetrahedral geometry (109.5°)?

A Trimethylamine

B Ammonia

C The ammonium ion (NH₄⁺)

D Methanamine

💡 Short Explanation

�� NH₄⁺ has four bond pairs. �� It has no lone pair on nitrogen. �� Therefore, it shows tetrahedral geometry.

✅ Correct Answer Explanation

Amines and ammonia are pyramidal because nitrogen has three bond pairs and one lone pair. In ammonium ion, NH₄⁺, nitrogen forms four N–H bonds and has no lone pair. Hence, the bond pairs are arranged symmetrically in tetrahedral geometry with bond angle 109.5°.

❓ Why Other Options Are Incorrect

�� Option A → Trimethylamine has a lone pair and a compressed bond angle.
�� Option B → Ammonia has one lone pair and pyramidal geometry.
�� Option D → Methanamine also has one lone pair and pyramidal geometry.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the only species without a lone pair on nitrogen.

Final Logic:

�� NH₄⁺ has four bond pairs and no lone pair, so it is tetrahedral.

🧠 Memory Trick

NH₄⁺ = No Lone Pair = Tetrahedral

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Geometry of Amines and Ammonium Ions

7 Evaluate the statements regarding the bonding in primary amines:
(A) Three sp³ hybridised orbitals of nitrogen are available for overlap.
(B) Exactly two sp³ orbitals overlap with hydrogen orbitals.
(C) Exactly one sp³ orbital overlaps with a carbon orbital.
(D) The fourth sp³ orbital overlaps with an oxygen atom.

A A, B, C

B A, C, D

C B, C, D

D A, B

💡 Short Explanation

�� Primary amine has formula RNH₂. �� Nitrogen forms two N–H bonds and one N–C bond. �� The fourth orbital contains a lone pair.

✅ Correct Answer Explanation

In a primary amine, RNH₂, nitrogen is sp³ hybridised. Three sp³ orbitals form sigma bonds: two overlap with hydrogen orbitals and one overlaps with a carbon orbital of the alkyl/aryl group. The fourth sp³ orbital contains the unshared electron pair and does not overlap with oxygen.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement D, which is incorrect.
�� Option C → Includes Statement D and omits Statement A.
�� Option D → Omits Statement C, which is correct.

🎯 Strategy

Used

Option Grouping

Application:

�� Use RNH₂ structure and count the overlaps.

Final Logic:

�� A primary amine has two N–H overlaps, one N–C overlap, and one lone pair.

🧠 Memory Trick

RNH₂ = 1 C + 2 H + 1 Lone Pair

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Bonding in Primary Amines

8 Identify the classification of the atomic domain responsible for amines acting as Lewis bases, as described in their bonding structure:

A The sp² carbon-nitrogen double bond

B The unshared (lone) pair of electrons in the fourth sp³ orbital

C The delocalised pi electrons in the aryl ring

D The covalent overlap between nitrogen and hydrogen

💡 Short Explanation

�� Lewis bases donate electron pairs. �� Amines contain a lone pair on nitrogen. �� This lone pair is present in the fourth sp³ orbital.

✅ Correct Answer Explanation

Amines behave as Lewis bases because the nitrogen atom has an unshared pair of electrons. This lone pair can be donated to electron-deficient species such as H⁺. Hence, the lone pair in the fourth sp³ orbital is responsible for the Lewis basic nature of amines.

❓ Why Other Options Are Incorrect

�� Option A → Amines generally do not contain an sp² C=N double bond.
�� Option C → Aryl π electrons are not the main basic site in amines.
�� Option D → A covalent N–H bond is not donated as an electron pair.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Link Lewis base behaviour with electron pair donation.

Final Logic:

�� The nitrogen lone pair is donated; therefore it causes Lewis basicity.

🧠 Memory Trick

Base = Lone Pair Donor

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Basic Character of Amines

9 Arrange the following in decreasing order of their typical expected bond angles based on the structural descriptions provided:
(A) Standard tetrahedral angle (no lone pairs)
(B) Trimethylamine C–N–C angle
(C) Compressed angle due to severe steric hindrance in extremely bulky tertiary amines (theoretical relative to ammonia) -> Wait, let's use exact text data. Correct formulation: Arrange in decreasing order of magnitude: (A) Standard tetrahedral angle (B) The C–N–C bond angle in trimethylamine

A (A) > (B)

B (B) > (A)

C (A) = (B)

D Cannot be determined

💡 Short Explanation

�� Standard tetrahedral angle is 109.5°. �� Trimethylamine bond angle is 108°. �� Therefore, standard tetrahedral angle is larger.

✅ Correct Answer Explanation

The standard tetrahedral bond angle is 109.5°. In trimethylamine, due to the lone pair on nitrogen, lone pair-bond pair repulsion compresses the C–N–C bond angle to about 108°. Since 109.5° is greater than 108°, the correct decreasing order is: (A) > (B)

❓ Why Other Options Are Incorrect

�� Option B → Trimethylamine angle is smaller, not larger.
�� Option C → The two angles are not equal.
�� Option D → The values are given conceptually and can be determined.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Compare the numerical angle values.

Final Logic:

�� 109.5° > 108°, so standard tetrahedral angle is greater.

🧠 Memory Trick

Tetra 109.5, TMA 108

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Bond Angle in Amines

10 The value of the compression of the bond angle in degrees for trimethylamine, relative to the standard uncompressed tetrahedral angle (109.5°), is exactly:

A 0.5

B 1.0

C 1.5

D 2.0

💡 Short Explanation

�� Standard tetrahedral angle is 109.5°. �� Trimethylamine angle is 108°. �� Difference = 109.5 − 108 = 1.5°.

✅ Correct Answer Explanation

The bond angle in trimethylamine is 108°. The standard tetrahedral angle is 109.5°. Therefore, compression in bond angle is: 109.5° − 108° = 1.5° So, the compression is 1.5°.

❓ Why Other Options Are Incorrect

�� Option A → 0.5° is too small.
�� Option B → 1.0° is not the correct difference.
�� Option D → 2.0° is greater than the actual difference.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Subtract the observed angle from the standard tetrahedral angle.

Final Logic:

�� 109.5° − 108° = 1.5°.

🧠 Memory Trick

109.5 − 108 = 1.5

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Bond Angle and Geometry of Amines

11 Identify the structural classification of the amine compound known as N,N-Dimethylaniline:

A Primary aromatic amine

B Secondary mixed amine

C Tertiary mixed aromatic amine

D Simple tertiary amine

💡 Short Explanation

�� N,N-Dimethylaniline has nitrogen attached to one phenyl group and two methyl groups. �� It has three carbon-containing groups attached to nitrogen. �� Hence, it is a tertiary mixed aromatic amine.

✅ Correct Answer Explanation

N,N-Dimethylaniline has the structure C₆H₅N(CH₃)₂. The nitrogen atom is attached to one aryl group, phenyl, and two alkyl groups, methyl. Since nitrogen is bonded to three carbon-containing groups, it is a tertiary amine. Since the groups are not all identical, it is a mixed amine. Since one group is aromatic, it is classified as a tertiary mixed aromatic amine.

❓ Why Other Options Are Incorrect

�� Option A → It is not primary because nitrogen has no N–H bond and is attached to three carbon groups.
�� Option B → It is not secondary because nitrogen is attached to three carbon-containing groups.
�� Option D → It is not simple because all the groups attached to nitrogen are not identical.

🎯 Strategy

Used

Substitution

Application:

�� Write the structure C₆H₅N(CH₃)₂ and count the groups attached to nitrogen.

Final Logic:

�� Three groups on nitrogen with phenyl and methyl groups make it a tertiary mixed aromatic amine.

🧠 Memory Trick

N,N-Dimethyl + Aniline = Tertiary Mixed

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Classification and Nomenclature of Aromatic Amines

12 If an amine has two identical alkyl groups and one distinct aryl group replacing the three hydrogen atoms of ammonia, what is its complete structural classification?

A Simple secondary amine

B Mixed tertiary amine

C Simple tertiary amine

D Mixed secondary amine

💡 Short Explanation

�� Three hydrogen atoms of ammonia are replaced. �� Hence, it is a tertiary amine. �� Since the groups are not all identical, it is mixed.

✅ Correct Answer Explanation

When all three hydrogen atoms of ammonia are replaced, the compound becomes a tertiary amine. Here, nitrogen is attached to two identical alkyl groups and one distinct aryl group. Since the three groups are not all the same, the amine is classified as mixed. Therefore, the complete classification is mixed tertiary amine.

❓ Why Other Options Are Incorrect

�� Option A → It is not secondary because three groups are attached to nitrogen.
�� Option C → It is not simple because all three groups are not identical.
�� Option D → It is not secondary because all three hydrogen atoms are replaced.

🎯 Strategy

Used

Option Grouping

Application:

�� First decide primary/secondary/tertiary by number of groups, then simple/mixed by similarity of groups.

Final Logic:

�� Three groups = tertiary; different groups = mixed.

🧠 Memory Trick

3 Groups + Not Same = Mixed Tertiary

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Simple and Mixed Amines

List 1 (Amine)	List 2 (Classification based on one substitution)
1. Propan-1-amine	a. Primary aromatic amine
2. Benzenamine	b. Primary unsaturated aliphatic amine
3. Allylamine (Prop-2-en-1-amine)	c. Primary aliphatic amine
4. 2-Methylaniline	d. Primary aromatic amine with ortho substitution

A 1-c, 2-a, 3-b, 4-d

B 1-a, 2-b, 3-c, 4-d

C 1-b, 2-c, 3-d, 4-a

D 1-d, 2-a, 3-b, 4-c

💡 Short Explanation

�� Propan-1-amine is a primary aliphatic amine. �� Benzenamine is a primary aromatic amine. �� Allylamine is a primary unsaturated aliphatic amine.

✅ Correct Answer Explanation

Propan-1-amine has an –NH₂ group attached to an aliphatic propyl chain, so it is a primary aliphatic amine. Benzenamine has –NH₂ directly attached to a benzene ring, so it is a primary aromatic amine. Allylamine or prop-2-en-1-amine contains an alkene group and –NH₂, so it is a primary unsaturated aliphatic amine. 2-Methylaniline has –NH₂ on benzene with a methyl group at the ortho position, so it is a primary aromatic amine with ortho substitution. Correct matching: 1-c, 2-a, 3-b, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Propan-1-amine and Benzenamine classifications are interchanged.
�� Option C → Multiple amine classifications are mismatched.
�� Option D → Propan-1-amine is not an ortho-substituted aromatic amine.

🎯 Strategy

Used

Option Grouping

Application:

�� Identify aliphatic/aromatic nature and then check primary classification.

Final Logic:

�� Each compound contains –NH₂ but differs by carbon framework: aliphatic, aromatic, unsaturated, or ortho-substituted aromatic.

🧠 Memory Trick

Propyl = Aliphatic, Benzene = Aromatic, Allyl = Unsaturated

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Classification of Primary Amines

14 Analyze the structural traits of an RNH₂ type amine:
(A) It contains two hydrogen atoms available for hydrogen bonding.
(B) It exhibits stronger intermolecular association than secondary amines.
(C) It forms a pyramidal geometry with a lone pair.
(D) It is classified as a secondary amine.

A A, C

B B, C, D

C A, B, C

D C, D

💡 Short Explanation

�� RNH₂ is a primary amine. �� It has two N–H bonds. �� It has pyramidal geometry due to lone pair on nitrogen.

✅ Correct Answer Explanation

RNH₂ represents a primary amine. It contains two hydrogen atoms directly attached to nitrogen, so it can participate in intermolecular hydrogen bonding. Primary amines generally show stronger intermolecular association than secondary amines because they have more N–H bonds. Nitrogen is sp³ hybridised and contains one lone pair, giving pyramidal geometry. ✓ A, B and C are correct. ✗ D is incorrect because RNH₂ is a primary amine, not a secondary amine.

❓ Why Other Options Are Incorrect

�� Option A → Omits Statement B, which is correct.
�� Option B → Includes Statement D, which is incorrect.
�� Option D → Includes Statement D and omits correct statements A and B.

🎯 Strategy

Used

Option Grouping

Application:

�� Check each statement using the structure RNH₂.

Final Logic:

�� RNH₂ is primary, has two N–H bonds, and pyramidal geometry with a lone pair.

🧠 Memory Trick

RNH₂ = Primary + 2H + Lone Pair

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Primary Amines and Hydrogen Bonding

15 Which of the following compounds fundamentally represents a secondary amine where the two substitutions consist of an ethyl and a methyl group?

A Propan-2-amine

B N-Methylethanamine

C Ethanamine

D N,N-Dimethylmethanamine

💡 Short Explanation

�� Secondary amines have two carbon groups attached to nitrogen. �� N-Methylethanamine has methyl and ethyl groups. �� Therefore, it is a mixed secondary amine.

✅ Correct Answer Explanation

N-Methylethanamine has the structure CH₃CH₂NHCH₃. Nitrogen is attached to one ethyl group and one methyl group, with one hydrogen remaining. Since two carbon-containing groups are attached to nitrogen, it is a secondary amine. Since the two groups are different, it is a mixed secondary amine.

❓ Why Other Options Are Incorrect

�� Option A → Propan-2-amine is a primary amine because nitrogen is attached to one alkyl group and two hydrogens.
�� Option C → Ethanamine is a primary amine.
�� Option D → N,N-Dimethylmethanamine is a tertiary amine.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate primary and tertiary amines, then identify the compound with ethyl and methyl groups.

Final Logic:

�� N-Methylethanamine contains ethyl + methyl groups on nitrogen, making it secondary.

🧠 Memory Trick

N-Methyl + Ethanamine = Methyl + Ethyl on N

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Secondary Amines

16 The accepted IUPAC name for the secondary amine where one hydrogen is replaced by a phenyl group and the other by a methyl group is:

A N-Methylaniline

B N-Methylbenzenamine

C Benzylmethylamine

D Aminomethylbenzene

💡 Short Explanation

�� The compound has phenyl and methyl groups attached to nitrogen. �� It is a secondary aromatic amine. �� IUPAC name is N-Methylbenzenamine.

✅ Correct Answer Explanation

The compound contains a benzene ring directly attached to nitrogen and a methyl group also attached to nitrogen. The parent structure is benzenamine, and the methyl group attached to nitrogen is indicated using the prefix N-methyl. Hence, the accepted IUPAC name is N-Methylbenzenamine.

❓ Why Other Options Are Incorrect

�� Option A → N-Methylaniline is a common/retained style name, not the accepted IUPAC form asked here.
�� Option C → Benzylmethylamine has a benzyl group, C₆H₅CH₂–, not a phenyl group directly attached to nitrogen.
�� Option D → Aminomethylbenzene represents a methyl group bearing –NH₂ attached to benzene, not phenyl and methyl directly attached to nitrogen.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate names with incorrect parent structure or common naming style.

Final Logic:

�� Benzenamine parent + N-methyl substituent gives N-Methylbenzenamine.

🧠 Memory Trick

Phenyl–N–Methyl = N-Methylbenzenamine

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Nomenclature of Secondary Aromatic Amines

17 Assess the characteristics of tertiary (R₃N type) amines:
(A) They have three alkyl/aryl groups attached to the nitrogen.
(B) They possess hydrogen atoms directly attached to the nitrogen.
(C) They do not exhibit intermolecular hydrogen bonding among themselves.
(D) They act as Lewis bases due to the unshared electron pair.

A A, B, C

B A, C, D

C B, C, D

D A, D

💡 Short Explanation

�� Tertiary amines have three carbon-containing groups attached to nitrogen. �� They have no N–H bond. �� They act as Lewis bases due to the lone pair.

✅ Correct Answer Explanation

Tertiary amines have the general formula R₃N, where nitrogen is bonded to three alkyl or aryl groups. Since all hydrogen atoms of ammonia are replaced, no hydrogen remains directly attached to nitrogen. Therefore, tertiary amines cannot form intermolecular hydrogen bonding among themselves. However, they still contain a lone pair of electrons on nitrogen, so they can act as Lewis bases. ✓ Statements A, C and D are correct. ✗ Statement B is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement B, which is incorrect.
�� Option C → Includes Statement B, which is incorrect.
�� Option D → Omits Statement C, which is correct.

🎯 Strategy

Used

Option Grouping

Application:

�� Check each statement using the formula R₃N.

Final Logic:

�� R₃N has three groups, no N–H bond, no self H-bonding, and one lone pair.

🧠 Memory Trick

Tertiary = 3R, 0H, Lone Pair

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Tertiary Amines and Hydrogen Bonding

18 Identify the classification type of the amine formed by the exhaustive substitution of ammonia with three ethyl groups:

A Primary simple amine

B Secondary mixed amine

C Tertiary simple amine

D Tertiary mixed amine

💡 Short Explanation

�� Three hydrogen atoms of ammonia are replaced. �� All three replacing groups are ethyl groups. �� Hence, it is tertiary and simple.

✅ Correct Answer Explanation

Exhaustive substitution of ammonia with three ethyl groups gives triethylamine, (C₂H₅)₃N. Since nitrogen is attached to three alkyl groups, the compound is a tertiary amine. Since all three groups are identical ethyl groups, it is a simple tertiary amine.

❓ Why Other Options Are Incorrect

�� Option A → It is not primary because three groups are attached to nitrogen.
�� Option B → It is not secondary and not mixed.
�� Option D → It is not mixed because all groups are ethyl groups.

🎯 Strategy

Used

Substitution

Application:

�� Count the number of substituted hydrogens and check whether the substituent groups are identical.

Final Logic:

�� Three identical ethyl groups on nitrogen = tertiary simple amine.

🧠 Memory Trick

3 Same Groups = Simple Tertiary

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Simple and Mixed Tertiary Amines

19 Arrange the following amines in increasing order of the number of distinct (different) alkyl/aryl groups they contain (to distinguish simple vs mixed complexity):
(A) N,N-Dimethylmethanamine
(B) N-Ethyl-N-methylpropan-1-amine
(C) N-Methylethanamine

A (A), (C), (B)

B (B), (C), (A)

C (C), (A), (B)

D (A), (B), (C)

💡 Short Explanation

�� N,N-Dimethylmethanamine has only one type of group: methyl. �� N-Methylethanamine has two different groups: methyl and ethyl. �� N-Ethyl-N-methylpropan-1-amine has three different groups.

✅ Correct Answer Explanation

Count the number of distinct alkyl groups attached to nitrogen: → N,N-Dimethylmethanamine = methyl, methyl, methyl = 1 distinct group → N-Methylethanamine = methyl and ethyl = 2 distinct groups → N-Ethyl-N-methylpropan-1-amine = ethyl, methyl and propyl = 3 distinct groups Therefore, the increasing order is: (A), (C), (B)

❓ Why Other Options Are Incorrect

�� Option B → Starts with the compound having three distinct groups.
�� Option C → Places compound with two distinct groups before compound with one distinct group.
�� Option D → Places compound with three distinct groups before compound with two distinct groups.

🎯 Strategy

Used

Option Grouping

Application:

�� Group attached substituents as same or different and count distinct types.

Final Logic:

�� 1 distinct group < 2 distinct groups < 3 distinct groups.

🧠 Memory Trick

Same → Simple; More Different → More Mixed

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Simple and Mixed Amines

20 A researcher synthesizes an amine and notes that the nitrogen is bonded to a methyl group, an ethyl group, and a propyl group. Based on the classification rules, this compound is strictly a:

A Simple tertiary amine

B Mixed tertiary amine

C Mixed secondary amine

D Complex primary amine

💡 Short Explanation

�� Nitrogen is attached to three alkyl groups. �� Therefore, it is tertiary. �� Since the groups are different, it is mixed.

✅ Correct Answer Explanation

The nitrogen atom is bonded to methyl, ethyl and propyl groups. Since three alkyl groups are attached to nitrogen, the amine is tertiary. Because all three groups are different, it is classified as a mixed tertiary amine.

❓ Why Other Options Are Incorrect

�� Option A → It is not simple because the attached groups are not identical.
�� Option C → It is not secondary because nitrogen is attached to three alkyl groups.
�� Option D → It is not primary because nitrogen has more than one alkyl group attached.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate primary and secondary classifications first, then decide simple or mixed.

Final Logic:

�� Three different alkyl groups attached to nitrogen = mixed tertiary amine.

🧠 Memory Trick

Methyl + Ethyl + Propyl = Mixed Tertiary

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Classification of Mixed Tertiary Amines

CUET UG Chemistry Booster Test – 3 Fundamentals of Amines

CUET UG Chemistry Booster Test - 3 Fundamentals of Amines

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