CUET UG Chemistry Booster Test - 1 Properties and Reactions of Amines -

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CUET UG Chemistry Booster Test - 1 Properties and Reactions of Amines

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 1 Properties and Reactions of Amines

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QUESTION 1 OF 20

Even though lower amines and corresponding alcohols both form hydrogen bonds, lower amines remain gases at room temperature while corresponding alcohols are liquids. This implies that:

Answer Review

1 Even though lower amines and corresponding alcohols both form hydrogen bonds, lower amines remain gases at room temperature while corresponding alcohols are liquids. This implies that:

A Amines have a much higher molar mass than alcohols.

B Nitrogen is more electronegative than Oxygen.

C Alcohols are more polar than amines and form stronger intermolecular hydrogen bonds.

D Amines do not form hydrogen bonds in the pure state.

💡 Short Explanation

�� Both alcohols and amines exhibit hydrogen bonding. �� Oxygen is more electronegative than nitrogen. �� Alcohols therefore form stronger intermolecular hydrogen bonds.

✅ Correct Answer Explanation

Lower amines can form intermolecular hydrogen bonds because they contain N–H bonds and a lone pair on nitrogen. → However, oxygen is more electronegative than nitrogen, making O–H bonds more polar than N–H bonds. → Consequently, alcohols exhibit stronger intermolecular hydrogen bonding than corresponding amines. → Stronger intermolecular forces result in higher boiling points, causing lower alcohols to be liquids while lower amines remain gases.

❓ Why Other Options Are Incorrect

�� Option A → Amines generally have comparable or lower molar masses.
�� Option B → Oxygen is more electronegative than nitrogen.
�� Option D → Amines do form hydrogen bonds.

🎯 Strategy

Used

Conceptual Elimination

Application:

�� Compare intermolecular forces in alcohols and amines.

Final Logic:

�� Stronger hydrogen bonding in alcohols causes higher boiling points.

🧠 Memory Trick

Oxygen Bonds Stronger than Nitrogen

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

2 Regarding the phase state of aliphatic amines:
(I) An increase in the size of the hydrophobic alkyl part increases Van der Waals forces, pushing the phase from gas to liquid to solid.
(II) All aromatic amines are inherently highly volatile gases at room temperature.

A Only (I)

B Only (II)

C Both (I) and (II)

D Neither (I) nor (II)

💡 Short Explanation

�� Larger alkyl groups increase intermolecular attraction. �� Physical state changes from gas → liquid → solid. �� Aromatic amines are generally not gases.

✅ Correct Answer Explanation

As carbon chain length increases, molecular size and Van der Waals forces increase. → Lower aliphatic amines are gases, medium-chain amines are liquids, and higher amines are solids. → Aromatic amines such as aniline are liquids or solids and are not highly volatile gases.

❓ Why Other Options Are Incorrect

�� Option B → Statement II is incorrect.
�� Option C → Statement II is false.
�� Option D → Statement I is true.

🎯 Strategy

Used

Elimination

Application:

�� Check each statement independently.

Final Logic:

�� Only Statement I is correct.

🧠 Memory Trick

Bigger Chain → Stronger Forces

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

3 Methanamine is recognized by a characteristic fishy odour. What is its exact molar mass?

A 17 g mol⁻¹

B 31 g mol⁻¹

C 45 g mol⁻¹

D 59 g mol⁻¹

💡 Short Explanation

�� Formula of methanamine = CH₃NH₂. �� Calculate atomic masses. �� Total equals 31 g mol⁻¹.

✅ Correct Answer Explanation

Methanamine formula: CH₃NH₂ Molar mass: C = 12 H₅ = 5 N = 14 Total = 12 + 5 + 14 = 31 g mol⁻¹

❓ Why Other Options Are Incorrect

�� Option A → Underestimated value.
�� Option C → Does not match formula.
�� Option D → Too large for methanamine.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Calculate molecular mass directly.

Final Logic:

�� CH₃NH₂ = 31 g mol⁻¹.

🧠 Memory Trick

12 + 5 + 14 = 31

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Physical Properties of Lower Amines

4 Aniline is observed to change from colourless to a coloured state upon prolonged storage. This transformation, an atmospheric oxidation, is facilitated because the amino group is a powerful:

A Deactivating group that removes electron density from the ring.

B Activating group that increases electron density on the ring, making it susceptible to oxidation.

C Steric blocking group.

D Lewis acid.

💡 Short Explanation

�� –NH₂ donates electron density to benzene ring. �� Ring becomes highly reactive. �� Oxidation occurs on exposure to air.

✅ Correct Answer Explanation

The amino group exhibits a strong +M effect. → It increases electron density in the aromatic ring, especially at ortho and para positions. → The electron-rich ring becomes more susceptible to atmospheric oxidation, causing aniline to develop colour during storage.

❓ Why Other Options Are Incorrect

�� Option A → –NH₂ is activating, not deactivating.
�� Option C → Steric effects are not responsible.
�� Option D → Aniline behaves as a base, not a Lewis acid.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect oxidation tendency with electron density.

Final Logic:

�� Electron-rich aromatic rings oxidize more easily.

🧠 Memory Trick

NH₂ Activates Ring

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Aromatic Amines and Oxidation

5 Why does the solubility of aliphatic amines in water decrease sharply as the molar mass increases?
(I) The hydrophobic alkyl part increases in size, which resists solvation.
(II) Higher molar mass amines undergo rapid hydrolysis.

A Only (I)

B Only (II)

C Both (I) and (II)

D Neither (I) nor (II)

💡 Short Explanation

�� Hydrogen bonding favors solubility. �� Larger alkyl chains are hydrophobic. �� Solubility decreases as chain length increases.

✅ Correct Answer Explanation

Lower amines dissolve in water through hydrogen bonding. → As alkyl chain size increases, the nonpolar hydrophobic portion becomes dominant. → This reduces interaction with water molecules and lowers solubility. → Hydrolysis is unrelated to the solubility trend.

❓ Why Other Options Are Incorrect

�� Option B → Hydrolysis is not the reason.
�� Option C → Statement II is incorrect.
�� Option D → Statement I is correct.

🎯 Strategy

Used

Elimination

Application:

�� Identify the factor affecting water solubility.

Final Logic:

�� Hydrophobic alkyl chains reduce solubility.

🧠 Memory Trick

Long Chain = Less Solubility

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

6 List 1 (Physical Property / Solubility Characteristic) | List 2 (Amine Class)

List 1	List 2
1. Highly soluble in water, fishy odour	a. Lower aliphatic amines
2. Insoluble in water, highly soluble in benzene	b. Higher aliphatic amines
3. No intermolecular hydrogen bonding in pure liquid	c. Tertiary amines

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

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

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

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

💡 Short Explanation

�� Lower amines are water soluble. �� Higher amines dissolve in organic solvents. �� Tertiary amines lack N–H bonds.

✅ Correct Answer Explanation

Lower aliphatic amines are water soluble and possess fishy odour. → Higher amines become insoluble in water and dissolve readily in organic solvents. → Tertiary amines lack N–H bonds and therefore cannot form intermolecular hydrogen bonds among themselves. Thus: 1 → a 2 → b 3 → c

❓ Why Other Options Are Incorrect

�� Options B, C, D incorrectly assign amine classes.

🎯 Strategy

Used

Option Grouping

Application:

�� Match physical properties with amine types.

Final Logic:

�� Solubility and H-bonding determine the classification.

🧠 Memory Trick

Lower = Water, Higher = Benzene, Tertiary = No N–H

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

7 The presence of an unshared pair of electrons and covalent bonds dictates the geometry around the nitrogen atom in amines. What is the geometry of trimethylamine?

A Tetrahedral

B Pyramidal

C Trigonal planar

D Linear

💡 Short Explanation

�� Nitrogen is sp³ hybridized. �� One lone pair is present. �� Shape becomes pyramidal.

✅ Correct Answer Explanation

Nitrogen in amines possesses three bond pairs and one lone pair. → The electron pair geometry is tetrahedral. → Due to the lone pair, the molecular geometry becomes trigonal pyramidal.

❓ Why Other Options Are Incorrect

�� Option A → Electron pair geometry, not molecular geometry.
�� Option C → Requires sp² hybridization.
�� Option D → Requires sp hybridization.

🎯 Strategy

Used

Conceptual Elimination

Application:

�� Recall geometry of amines.

Final Logic:

�� Lone pair converts tetrahedral arrangement into pyramidal shape.

🧠 Memory Trick

3 Bonds + 1 Lone Pair = Pyramid

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

8 Arrange the following substituted ammonium ions in decreasing order of their stabilization via hydrogen bonding with water (solvation effect):

A RNH₃⁺ > R₂NH₂⁺ > R₃NH⁺

B R₃NH⁺ > R₂NH₂⁺ > RNH₃⁺

C R₂NH₂⁺ > RNH₃⁺ > R₃NH⁺

D RNH₃⁺ > R₃NH⁺ > R₂NH₂⁺

💡 Short Explanation

�� Smaller ions are better solvated. �� More N–H bonds allow stronger hydration. �� Solvation decreases with increasing alkyl substitution.

✅ Correct Answer Explanation

Solvation depends on hydrogen bonding with water. → Primary ammonium ions possess the greatest capacity for hydrogen bonding. → Increasing alkyl substitution introduces steric hindrance. Therefore: RNH₃⁺ > R₂NH₂⁺ > R₃NH⁺

❓ Why Other Options Are Incorrect

�� Options B, C, D contradict solvation trends.

🎯 Strategy

Used

Ordering

Application:

�� Compare extent of hydrogen bonding.

Final Logic:

�� Greater hydrogen bonding gives greater stabilization.

🧠 Memory Trick

1° > 2° > 3° (Solvation)

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Solvation Effect on Basicity

9 Evaluate the reasoning for boiling point differences:
(I) n-Butan-1-amine (350.8 K) has a higher boiling point than N,N-Dimethylethanamine (310.5 K) strictly because primary amines form extensive intermolecular hydrogen bonds, while tertiary amines cannot.
(II) The boiling points of isomeric amines follow the order: Tertiary > Secondary > Primary.

A Only (I)

B Only (II)

C Both (I) and (II)

D Neither (I) nor (II)

💡 Short Explanation

�� Primary amines form strong H-bonds. �� Tertiary amines cannot self-associate through N–H bonds. �� Boiling point order is Primary > Secondary > Tertiary.

✅ Correct Answer Explanation

Primary amines possess N–H bonds that participate extensively in intermolecular hydrogen bonding. → Tertiary amines lack N–H bonds and cannot form intermolecular hydrogen bonds among themselves. → Hence primary amines have higher boiling points. → Statement II reverses the actual order.

❓ Why Other Options Are Incorrect

�� Option B → Statement I is correct.
�� Option C → Statement II is false.
�� Option D → Statement I is true.

🎯 Strategy

Used

Elimination

Application:

�� Compare hydrogen-bonding capability.

Final Logic:

�� Greater H-bonding results in higher boiling point.

🧠 Memory Trick

1° > 2° > 3° Boiling Point

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

10 A specific amine of formula C₃H₉N exists as a liquid with the lowest boiling point among its isomers. What is its IUPAC name?

A Propan-1-amine

B N-Methylethanamine

C N,N-Dimethylmethanamine

D Propan-2-amine

💡 Short Explanation

�� Tertiary amines show the lowest boiling points. �� They cannot form intermolecular hydrogen bonds. �� N,N-Dimethylmethanamine is a tertiary amine.

✅ Correct Answer Explanation

Among C₃H₉N isomers: • Propan-1-amine → Primary • Propan-2-amine → Primary • N-Methylethanamine → Secondary • N,N-Dimethylmethanamine → Tertiary → Tertiary amines lack N–H bonds and therefore have the weakest intermolecular association. → Hence they possess the lowest boiling point.

❓ Why Other Options Are Incorrect

�� Options A and D → Primary amines.
�� Option B → Secondary amine.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the tertiary amine among isomers.

Final Logic:

�� Tertiary amines have the lowest boiling points.

🧠 Memory Trick

Tertiary = Lowest BP

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Physical Properties and Boiling Points

11 The base dissociation constant, Kb, and pKb determine basic strength. The pKb of ammonia is 4.75. If a new amine has a pKb of 3.00, it is:

A A weaker base than ammonia

B A stronger base than ammonia

C A strong acid

D Completely insoluble in water

💡 Short Explanation

�� Lower pKb means stronger base. �� 3.00 is less than 4.75. �� Therefore, the new amine is stronger than ammonia.

✅ Correct Answer Explanation

Basic strength is inversely related to pKb value. → Smaller pKb indicates greater basic strength. Given: New amine pKb = 3.00 Ammonia pKb = 4.75 Since 3.00 < 4.75, the new amine is a stronger base than ammonia.

❓ Why Other Options Are Incorrect

�� Option A → Higher pKb would indicate weaker base, but the new amine has lower pKb.
�� Option C → pKb relates to basicity, not acidity.
�� Option D → Solubility cannot be concluded from pKb alone.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Compare numerical pKb values.

Final Logic:

�� Smaller pKb = stronger base.

🧠 Memory Trick

Low pKb = Strong Base

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Basic Strength and pKb Values

12 When aniline reacts with highly concentrated sulphuric acid, it initially forms anilinium hydrogensulphate. This indicates aniline is acting as a base. In electrophilic substitution, the resulting anilinium ion acts as a:

A Strong activating group

B Meta-directing group

C Ortho/para-directing group

D Nucleophile

💡 Short Explanation

�� Aniline gets protonated in strong acid. �� Anilinium ion is electron-withdrawing. �� It directs electrophilic substitution to meta position.

✅ Correct Answer Explanation

In highly concentrated sulphuric acid, aniline accepts a proton and forms the anilinium ion. → The –NH₃⁺ group withdraws electron density from the benzene ring. → Because of this deactivating effect, the anilinium ion behaves as a meta-directing group in electrophilic substitution reactions.

❓ Why Other Options Are Incorrect

�� Option A → Anilinium ion is deactivating, not strongly activating.
�� Option C → Neutral –NH₂ is ortho/para-directing, but –NH₃⁺ is meta-directing.
�� Option D → The anilinium ion does not act as a nucleophile in electrophilic substitution.

🎯 Strategy

Used

Elimination

Application:

�� Distinguish neutral aniline from protonated anilinium ion.

Final Logic:

�� Protonated –NH₃⁺ group is meta-directing.

🧠 Memory Trick

NH₂ = o,p; NH₃⁺ = meta

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Electrophilic Substitution in Aniline

13 Arrange the following amines in decreasing order of their basic strength in the aqueous phase:
(C₂H₅)₂NH, C₂H₅NH₂, C₆H₅NH₂

A C₆H₅NH₂ > C₂H₅NH₂ > (C₂H₅)₂NH

B (C₂H₅)₂NH > C₂H₅NH₂ > C₆H₅NH₂

C C₂H₅NH₂ > (C₂H₅)₂NH > C₆H₅NH₂

D (C₂H₅)₂NH > C₆H₅NH₂ > C₂H₅NH₂

💡 Short Explanation

�� Aliphatic amines are stronger bases than arylamines. �� Secondary ethylamine is stronger than primary ethylamine in water. �� Aniline is weakest due to resonance.

✅ Correct Answer Explanation

In aqueous solution, ethyl-substituted aliphatic amines are stronger bases than aniline because alkyl groups increase electron density on nitrogen through the +I effect. → Diethylamine is stronger than ethylamine because it has greater electron donation and sufficient solvation. → Aniline is weakest because the lone pair of nitrogen is delocalized into the benzene ring, making it less available for protonation. Therefore: (C₂H₅)₂NH > C₂H₅NH₂ > C₆H₅NH₂

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly places aniline as strongest.
�� Option C → Incorrectly places ethylamine above diethylamine.
�� Option D → Incorrectly places aniline above ethylamine.

🎯 Strategy

Used

Ordering

Application:

�� Compare +I effect, solvation, and resonance.

Final Logic:

�� Aliphatic secondary > aliphatic primary > aromatic amine.

🧠 Memory Trick

Diethyl > Ethyl > Aniline

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

14 In substituted anilines, electron-releasing groups increase basic strength, while electron-withdrawing groups decrease it. Arrange p-nitroaniline, aniline, and p-toluidine in increasing order of basic strength:

A p-nitroaniline < aniline < p-toluidine

B p-toluidine < aniline < p-nitroaniline

C aniline < p-nitroaniline < p-toluidine

D p-nitroaniline < p-toluidine < aniline

💡 Short Explanation

�� –NO₂ decreases basicity. �� –CH₃ increases basicity. �� Therefore, p-toluidine is strongest.

✅ Correct Answer Explanation

p-Nitroaniline contains the –NO₂ group, which is strongly electron-withdrawing. It reduces electron density on nitrogen and decreases basic strength. → Aniline has no extra electron-donating or withdrawing substituent. → p-Toluidine contains the –CH₃ group, which is electron-releasing and increases electron density on nitrogen. Increasing basic strength: p-nitroaniline < aniline < p-toluidine

❓ Why Other Options Are Incorrect

�� Option B → Gives reverse logic.
�� Option C → Places aniline below p-nitroaniline incorrectly.
�� Option D → Places p-toluidine below aniline incorrectly.

🎯 Strategy

Used

Ordering

Application:

�� Rank substituents by electron-withdrawing/electron-releasing effects.

Final Logic:

�� EWG lowers basicity; ERG raises basicity.

🧠 Memory Trick

NO₂ Down, CH₃ Up

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Effect of Substituents on Basicity of Aniline

📄 Passage

In the aqueous phase, the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group (+I) but also by solvation with water molecules. The greater the size of the ion, lesser will be the solvation and the less stabilised is the ion. Secondly, when the alkyl group is small, like –CH₃ group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than CH₃ group, there will be steric hinderance to H-bonding.

According to the passage, if solvation and steric effects are entirely absent (as in the gas phase), the basicity of ethyl amines would be dictated solely by the inductive effect (+I). What is the theoretical basicity order in the gaseous phase?

A Primary > Secondary > Tertiary > Ammonia

B Tertiary > Secondary > Primary > Ammonia

C Ammonia > Primary > Secondary > Tertiary

D Secondary > Tertiary > Primary > Ammonia

💡 Short Explanation

�� Gas phase has no solvation effect. �� Basicity depends mainly on +I effect. �� More alkyl groups increase electron density on nitrogen.

✅ Correct Answer Explanation

In the gas phase, there is no solvation of ammonium ions and steric hindrance to hydration does not apply. → Therefore, the basic strength depends mainly on the electron-releasing +I effect of alkyl groups. → Tertiary amines have three alkyl groups, secondary amines have two, primary amines have one, and ammonia has none. Thus: Tertiary > Secondary > Primary > Ammonia

❓ Why Other Options Are Incorrect

�� Option A → Gives opposite order of +I effect.
�� Option C → Places ammonia incorrectly as strongest.
�� Option D → Places secondary above tertiary without solvation logic.

🎯 Strategy

Used

Ordering

Application:

�� Count the number of electron-releasing alkyl groups.

Final Logic:

�� More alkyl groups = stronger +I effect = stronger base.

🧠 Memory Trick

3R > 2R > 1R > NH₃

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Gas Phase Basicity of Amines

16 List 1 (Factor Affecting Basicity) | List 2 (Specific Role in Aqueous Solution)

List 1	List 2
1. Inductive (+I) Effect	a. Pushes electron density toward nitrogen
2. Solvation with Water	b. Stabilises substituted ammonium ion through H-bonding
3. Steric Hindrance	c. Restricts effective solvation of larger alkyl-substituted ions
4. Resonance in Aniline	d. Delocalises lone pair, decreasing basicity

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

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

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

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

💡 Short Explanation

�� +I effect increases electron density on nitrogen. �� Solvation stabilizes ammonium ions. �� Steric hindrance reduces hydration. �� Resonance decreases availability of lone pair.

✅ Correct Answer Explanation

Alkyl groups exhibit a +I effect and push electron density toward nitrogen, increasing basicity. → Solvation stabilizes the conjugate acid (ammonium ion) through hydrogen bonding with water molecules. → Larger alkyl groups hinder effective hydrogen bonding and decrease solvation. → In aniline, the lone pair is delocalized into the benzene ring by resonance, reducing basic strength. Thus: 1 → a 2 → b 3 → c 4 → d

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly swaps inductive effect and solvation roles.
�� Option C → Steric hindrance does not stabilize ammonium ions.
�� Option D → Resonance does not increase electron density on nitrogen.

🎯 Strategy

Used

Option Grouping

Application:

�� Match each factor with its direct effect on basicity.

Final Logic:

�� Correct understanding of +I effect, solvation, steric effect, and resonance gives the required matching.

🧠 Memory Trick

I–S–S–R = Increase, Solvate, Stop, Resonance

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Factors Affecting Basic Strength of Amines

17 Amines undergo a specific reaction with alkyl halides to form a carbon-nitrogen bond. What is this reaction known as?

A Acylation

B Alkylation

C Diazotisation

D Bromination

💡 Short Explanation

�� Alkyl halides transfer alkyl groups to amines. �� A new C–N bond is formed. �� This process is called alkylation.

✅ Correct Answer Explanation

Amines possess a lone pair on nitrogen and behave as nucleophiles. → They attack alkyl halides and replace the halogen atom. → As a result, an alkyl group becomes attached to nitrogen. → The reaction is therefore known as alkylation.

❓ Why Other Options Are Incorrect

�� Option A → Acylation introduces an acyl group (RCO–), not an alkyl group.
�� Option C → Diazotisation involves aromatic primary amines and nitrous acid.
�� Option D → Bromination introduces bromine into a molecule.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the reaction specifically involving alkyl group transfer.

Final Logic:

�� Alkyl halide + amine = Alkylation.

🧠 Memory Trick

Alkyl Halide → Alkylation

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

18 When a primary amine continuously reacts with alkyl halides, it forms secondary and tertiary amines, and finally yields which class of compounds?

A Primary amides

B Quaternary ammonium salts

C Diazonium salts

D Sulphonamides

💡 Short Explanation

�� Amines continue reacting with alkyl halides. �� Successive alkylation occurs. �� Final product is a quaternary ammonium salt.

✅ Correct Answer Explanation

A primary amine formed initially still contains a lone pair and remains nucleophilic. → It reacts further with alkyl halides to form secondary and tertiary amines. → Continued alkylation finally produces a positively charged nitrogen species having four alkyl groups. → Such compounds are called quaternary ammonium salts.

❓ Why Other Options Are Incorrect

�� Option A → Amides are formed by acylation.
�� Option C → Diazonium salts arise from diazotisation.
�� Option D → Sulphonamides result from reaction with sulphonyl chlorides.

🎯 Strategy

Used

Conceptual Elimination

Application:

�� Follow the sequence of repeated alkylation.

Final Logic:

�� Exhaustive alkylation ends at quaternary ammonium salts.

🧠 Memory Trick

1° → 2° → 3° → Quaternary

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

19 Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides, and esters. What is the name of the product class formed by this acylation reaction?

A Nitriles

B Amides

C Imines

D Isocyanides

💡 Short Explanation

�� Acylation introduces an acyl group. �� Amines react with acid derivatives. �� Products formed are amides.

✅ Correct Answer Explanation

Primary and secondary amines react with acid chlorides, acid anhydrides, and esters. → The acyl group (RCO–) becomes attached to nitrogen. → The resulting compounds contain the –CONH– linkage characteristic of amides.

❓ Why Other Options Are Incorrect

�� Option A → Nitriles contain –C≡N groups.
�� Option C → Imines contain C=N bonds.
�� Option D → Isocyanides contain –NC groups.

🎯 Strategy

Used

Elimination

Application:

�� Identify the product of acyl group transfer.

Final Logic:

�� Acylation of amines gives amides.

🧠 Memory Trick

Acylation → Amide

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

20 The specific reaction of amines with benzoyl chloride (C₆H₅COCl) is known as:

A Acetylation

B Benzoylation

C Sandmeyer reaction

D Carbylamine reaction

💡 Short Explanation

�� Benzoyl chloride provides a benzoyl group. �� Amines undergo acylation with benzoyl chloride. �� This special acylation is called benzoylation.

✅ Correct Answer Explanation

Benzoyl chloride (C₆H₅COCl) reacts with primary and secondary amines. → The benzoyl group (C₆H₅CO–) becomes attached to nitrogen. → This reaction is specifically known as benzoylation and is commonly used in the Hinsberg test.

❓ Why Other Options Are Incorrect

�� Option A → Acetylation involves acetyl chloride or acetic anhydride.
�� Option C → Sandmeyer reaction involves diazonium salts.
�� Option D → Carbylamine reaction produces isocyanides.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Match reagent name with reaction name.

Final Logic:

�� Benzoyl chloride gives benzoylation.

🧠 Memory Trick

Benzoyl Chloride → Benzoylation

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Benzoylation of Amines (Hinsberg Reagent)

CUET UG Chemistry Booster Test – 1 Properties and Reactions of Amines

CUET UG Chemistry Booster Test - 1 Properties and Reactions of Amines

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