CUET UG Chemistry Booster Test - 2 Properties and Reactions of Amines -

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CUET UG Chemistry Booster Test - 2 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 - 2 Properties and Reactions of Amines

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QUESTION 1 OF 20

Which of the following specific amines is most likely to exist as a gas with a fishy odour at room temperature?

Answer Review

1 Which of the following specific amines is most likely to exist as a gas with a fishy odour at room temperature?

A Methanamine

B Hexane-1,6-diamine

C Benzenamine

D Propan-1-amine

💡 Short Explanation

�� Methanamine is a lower aliphatic amine. �� Lower aliphatic amines are gases. �� They have fishy odour.

✅ Correct Answer Explanation

Methanamine is the simplest lower aliphatic amine. Lower aliphatic amines exist as gases at room temperature and possess a fishy odour. Hence, methanamine best matches the given condition.

❓ Why Other Options Are Incorrect

�� Option B → Hexane-1,6-diamine is a higher amine, not a gaseous lower amine.
�� Option C → Benzenamine is aromatic and not the best example of lower aliphatic gaseous amine.
�� Option D → Propan-1-amine is generally liquid at room temperature.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate higher, aromatic, and liquid amines.

Final Logic:

�� Lower aliphatic amine = gas with fishy odour.

🧠 Memory Trick

Methanamine = Fishy Gas

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

2 As the carbon chain grows, primary amines transition from liquids to higher solids. This physical state change is primarily correlated with:

A The loss of an unshared pair of electrons.

B An increase in the size of the hydrophobic alkyl part, which increases molecular mass.

C A decrease in hydrogen bonding capacity.

D The conversion of the amine to an aromatic system.

💡 Short Explanation

�� Longer alkyl chains increase molecular mass. �� Intermolecular forces become stronger. �� Higher amines may become solids.

✅ Correct Answer Explanation

As the alkyl chain becomes larger, the hydrophobic part and molecular mass increase. This increases intermolecular attractions, causing higher amines to shift from liquid to solid state.

❓ Why Other Options Are Incorrect

�� Option A → Amines do not lose their lone pair.
�� Option C → The main reason is increased molecular size, not loss of H-bonding capacity.
�� Option D → Chain growth does not convert amines into aromatic compounds.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect physical state trend with molecular size.

Final Logic:

�� Larger alkyl chain increases molecular mass and changes physical state.

🧠 Memory Trick

Long Chain = Heavy = Solid

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Physical State and Molecular Mass

3 Identify the accepted IUPAC name of the simplest aromatic amine, which is colourless but becomes coloured on storage due to oxidation.

A N-Methylethanamine

B Benzenamine

C Prop-2-en-1-amine

D N,N-Dimethylmethanamine

💡 Short Explanation

�� Simplest aromatic amine is aniline. �� IUPAC name of aniline is benzenamine. �� It becomes coloured due to atmospheric oxidation.

✅ Correct Answer Explanation

The simplest aromatic amine is C₆H₅NH₂, commonly called aniline. Its accepted IUPAC name is benzenamine. Arylamines like aniline are colourless but become coloured on storage due to atmospheric oxidation.

❓ Why Other Options Are Incorrect

�� Option A → N-Methylethanamine is an aliphatic secondary amine.
�� Option C → Prop-2-en-1-amine is an unsaturated aliphatic amine.
�� Option D → N,N-Dimethylmethanamine is a tertiary aliphatic amine.

🎯 Strategy

Used

Elimination

Application:

�� Identify the only aromatic amine among the options.

Final Logic:

�� Aniline = Benzenamine.

🧠 Memory Trick

Benzene + NH₂ = Benzenamine

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Arylamines and Aniline

4 The transformation of colourless aniline into coloured products upon storage in air is chemically classified as:

A Acid-base neutralization

B Atmospheric oxidation

C Electrophilic substitution

D Nucleophilic substitution

💡 Short Explanation

�� Aniline is colourless when pure. �� It becomes coloured on exposure to air. �� This is due to oxidation.

✅ Correct Answer Explanation

Aniline and other arylamines become coloured on storage because atmospheric oxygen oxidizes them slowly. The coloured products formed are responsible for the observed colour change.

❓ Why Other Options Are Incorrect

�� Option A → No acid-base salt formation is involved.
�� Option C → Electrophilic substitution is not the storage effect.
�� Option D → Nucleophilic substitution is unrelated here.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Link colour formation on storage with air oxidation.

Final Logic:

�� Air exposure causes atmospheric oxidation.

🧠 Memory Trick

Aniline + Air = Oxidation Colour

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Atmospheric Oxidation of Arylamines

5 Consider the solubility of butan-1-ol and butan-1-amine in water. Which statement correctly explains their relative solubility?
(I) Butan-1-ol is more soluble because oxygen is more electronegative than nitrogen (3.5 vs 3.0), forming stronger hydrogen bonds.
(II) Butan-1-amine is more soluble because it has more hydrogen atoms available for bonding.

A Only (I)

B Only (II)

C Both (I) and (II)

D Neither (I) nor (II)

💡 Short Explanation

�� Alcohols form stronger hydrogen bonds than amines. �� Oxygen is more electronegative than nitrogen. �� Therefore, butan-1-ol is more water soluble.

✅ Correct Answer Explanation

Butan-1-ol forms stronger hydrogen bonds with water because oxygen is more electronegative than nitrogen. Amines can also form hydrogen bonds, but they are generally weaker than those formed by alcohols.

❓ Why Other Options Are Incorrect

�� Option B → Statement II is incorrect; more hydrogen atoms alone does not make butan-1-amine more soluble.
�� Option C → Statement II is incorrect.
�� Option D → Statement I is correct.

🎯 Strategy

Used

Option Grouping

Application:

�� Compare hydrogen bond strength of O–H and N–H groups.

Final Logic:

�� Stronger O–H hydrogen bonding makes alcohol more soluble.

🧠 Memory Trick

O bonds stronger than N

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

6 Arrange the following compounds in decreasing order of their solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂

A C₆H₅NH₂ > (C₂H₅)₂NH > C₂H₅NH₂

B C₂H₅NH₂ > (C₂H₅)₂NH > C₆H₅NH₂

C (C₂H₅)₂NH > C₂H₅NH₂ > C₆H₅NH₂

D C₆H₅NH₂ > C₂H₅NH₂ > (C₂H₅)₂NH

💡 Short Explanation

�� Smaller aliphatic amines are more soluble. �� Larger hydrophobic groups reduce solubility. �� Aromatic amines are least soluble here.

✅ Correct Answer Explanation

Ethylamine is a lower aliphatic primary amine and forms hydrogen bonds with water effectively. Diethylamine has a larger hydrophobic part, so it is less soluble. Aniline has an aromatic ring, making it least soluble among the given compounds.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly places aniline as most soluble.
�� Option C → Incorrectly places diethylamine above ethylamine.
�� Option D → Again wrongly places aniline highest.

🎯 Strategy

Used

Ordering

Application:

�� Compare hydrophobic alkyl/aryl size and hydrogen bonding.

Final Logic:

�� Smaller aliphatic amine > larger aliphatic amine > aromatic amine.

🧠 Memory Trick

Small aliphatic dissolves best

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

7 Which of the following amines will lack intermolecular hydrogen bonding completely?

A Propan-1-amine

B N-Methylethanamine

C N,N-Diethylbutan-1-amine

D Ethanamine

💡 Short Explanation

�� Tertiary amines lack N–H bonds. �� Intermolecular hydrogen bonding requires N–H hydrogen. �� N,N-Diethylbutan-1-amine is tertiary.

✅ Correct Answer Explanation

N,N-Diethylbutan-1-amine has three carbon groups attached to nitrogen and no hydrogen directly bonded to nitrogen. Therefore, it cannot form intermolecular hydrogen bonds among its own molecules.

❓ Why Other Options Are Incorrect

�� Option A → Propan-1-amine is primary and has N–H bonds.
�� Option B → N-Methylethanamine is secondary and has one N–H bond.
�� Option D → Ethanamine is primary and has two N–H bonds.

🎯 Strategy

Used

Elimination

Application:

�� Identify the tertiary amine with no N–H bond.

Final Logic:

�� No N–H bond means no intermolecular hydrogen bonding.

🧠 Memory Trick

Tertiary = 0 N–H

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

8 Why does a secondary amine possess a lower degree of intermolecular association compared to a primary amine?
(I) It has only one hydrogen atom available for hydrogen bond formation.
(II) It completely lacks an unshared pair of electrons.

A Only (I)

B Only (II)

C Both (I) and (II)

D Neither (I) nor (II)

💡 Short Explanation

�� Secondary amines have one N–H hydrogen. �� Primary amines have two N–H hydrogens. �� Amines still possess a lone pair.

✅ Correct Answer Explanation

Secondary amines have the formula R₂NH, so only one hydrogen atom is available for hydrogen bond formation. Primary amines have two such hydrogens, so they show stronger intermolecular association. → Statement II is incorrect because secondary amines still contain an unshared pair on nitrogen.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application:

�� Compare primary and secondary amine structures.

Final Logic:

�� Secondary amine has fewer N–H hydrogens than primary amine.

🧠 Memory Trick

Secondary = 1 H only

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Intermolecular Association

9 Arrange the following compounds of similar molecular masses in decreasing order of their boiling points:
n-C₄H₉OH, n-C₄H₉NH₂, (C₂H₅)₂NH, C₂H₅N(CH₃)₂

A n-C₄H₉OH > n-C₄H₉NH₂ > (C₂H₅)₂NH > C₂H₅N(CH₃)₂

B C₂H₅N(CH₃)₂ > (C₂H₅)₂NH > n-C₄H₉NH₂ > n-C₄H₉OH

C n-C₄H₉OH > (C₂H₅)₂NH > n-C₄H₉NH₂ > C₂H₅N(CH₃)₂

D n-C₄H₉NH₂ > n-C₄H₉OH > (C₂H₅)₂NH > C₂H₅N(CH₃)₂

💡 Short Explanation

�� Alcohols form stronger hydrogen bonds than amines. �� Primary amines form stronger H-bonds than secondary amines. �� Tertiary amines lack N–H hydrogen.

✅ Correct Answer Explanation

n-C₄H₉OH has the highest boiling point because alcohols form stronger hydrogen bonds due to more electronegative oxygen. → Among amines, boiling point order is: Primary amine > Secondary amine > Tertiary amine Therefore: n-C₄H₉OH > n-C₄H₉NH₂ > (C₂H₅)₂NH > C₂H₅N(CH₃)₂

❓ Why Other Options Are Incorrect

�� Option B → Gives nearly reverse order.
�� Option C → Incorrectly places secondary amine above primary amine.
�� Option D → Incorrectly places primary amine above alcohol.

🎯 Strategy

Used

Ordering

Application:

�� Rank based on hydrogen-bonding strength.

Final Logic:

�� Alcohol > Primary amine > Secondary amine > Tertiary amine.

🧠 Memory Trick

O–H strongest, then 1°, 2°, 3° amines

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

10 The boiling point of N,N-Dimethylethanamine is 310.5 K, which is significantly lower than that of its isomer n-Butan-1-amine (350.8 K). This is fundamentally because N,N-Dimethylethanamine:

A Possesses a larger molar mass.

B Is highly soluble in water.

C Lacks a hydrogen atom available for intermolecular hydrogen bond formation.

D Has stronger intermolecular Van der Waals forces.

💡 Short Explanation

�� N,N-Dimethylethanamine is tertiary. �� Tertiary amines lack N–H bonds. �� Hence intermolecular hydrogen bonding is absent.

✅ Correct Answer Explanation

N,N-Dimethylethanamine is a tertiary amine. It has no hydrogen atom attached directly to nitrogen, so it cannot form intermolecular hydrogen bonds with its own molecules. → n-Butan-1-amine is a primary amine and forms intermolecular hydrogen bonds, giving it a higher boiling point.

❓ Why Other Options Are Incorrect

�� Option A → Both are isomeric and have similar molecular formula/molar mass.
�� Option B → Water solubility is not the main boiling point reason.
�� Option D → Tertiary amine does not have stronger intermolecular attraction than primary amine.

🎯 Strategy

Used

Elimination

Application:

�� Compare hydrogen-bonding ability of isomeric amines.

Final Logic:

�� No N–H bond means lower boiling point.

🧠 Memory Trick

3° Amine = No H-bond Donor

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

List 1 (Basicity Term)	List 2 (Mathematical Expression / Meaning)
1. Base dissociation constant (Kb)	a. [RNH₃⁺][OH⁻] / [RNH₂]
2. pKb	b. –log Kb
3. Stronger base	c. Larger Kb or smaller pKb
4. Weaker base	d. Smaller Kb or larger pKb

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

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

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

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

💡 Short Explanation

�� Kb measures basic strength. �� pKb is the negative logarithm of Kb. �� Stronger bases have larger Kb values.

✅ Correct Answer Explanation

The base dissociation constant (Kb) is given by: Kb = [RNH₃⁺][OH⁻] / [RNH₂] → pKb is defined as: pKb = –log Kb → A stronger base has a larger Kb and therefore a smaller pKb. → A weaker base has a smaller Kb and a larger pKb. Hence: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Interchanges Kb and pKb definitions.
�� Option C → Incorrectly assigns stronger base relation.
�� Option D → Gives incorrect expression for Kb.

🎯 Strategy

Used

Option Grouping

Application:

�� Match each basicity term with its mathematical definition.

Final Logic:

�� Larger Kb means stronger basic character.

🧠 Memory Trick

High Kb = High Base

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

12 When an amine salt is treated with an aqueous solution of sodium hydroxide (NaOH), the parent amine is regenerated. This specific transformation represents:

A Electrophilic addition

B Neutralization of an acidic salt by a strong base

C Alkylation

D Diazotisation

💡 Short Explanation

�� Amine salts are acidic in nature. �� NaOH removes the proton. �� Free amine is regenerated.

✅ Correct Answer Explanation

Amines react with acids to form ammonium salts. Example: RNH₂ + HCl → RNH₃⁺Cl⁻ → Treatment with NaOH regenerates the amine: RNH₃⁺Cl⁻ + NaOH → RNH₂ + NaCl + H₂O → This is an acid-base neutralization/displacement reaction.

❓ Why Other Options Are Incorrect

�� Option A → No addition reaction occurs.
�� Option C → No alkyl group is introduced.
�� Option D → Nitrous acid is required for diazotisation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the role of NaOH in the reaction.

Final Logic:

�� Base regenerates amine from its salt.

🧠 Memory Trick

Salt + Base → Free Amine

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Basic Nature and Salt Formation

📄 Passage

pKb value of aniline is quite high. Why is it so? It is because in aniline or other arylamines, the –NH₂ group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation. If you write different resonating structures of aniline, you will find that aniline is a resonance hybrid of the following five structures. On the other hand, anilinium ion obtained by accepting a proton can have only two resonating structures.

Based strictly on the passage, why are alkylamines (like methanamine) inherently stronger bases than aniline?

A Alkylamines have an unshared electron pair that is in conjugation with a ring.

B Alkylamines completely lack unshared electron pairs.

C In alkylamines, the unshared electron pair is not involved in resonance conjugation and remains fully available for protonation.

D Alkylamines have five resonating structures.

💡 Short Explanation

�� Alkylamines do not undergo resonance with a benzene ring. �� Lone pair remains available. �� Protonation occurs more readily.

✅ Correct Answer Explanation

In aniline, the nitrogen lone pair participates in resonance with the benzene ring. → Because of this delocalization, the lone pair becomes less available for protonation. → In alkylamines, such resonance does not occur. Therefore, the lone pair remains localized and can readily accept a proton. Hence alkylamines are stronger bases than aniline.

❓ Why Other Options Are Incorrect

�� Option A → This is true for aniline, not alkylamines.
�� Option B → Alkylamines possess a lone pair.
�� Option D → Five resonance structures belong to aniline.

🎯 Strategy

Used

Elimination

Application:

�� Compare lone-pair availability in aniline and alkylamines.

Final Logic:

�� Greater lone-pair availability means stronger basicity.

🧠 Memory Trick

No Resonance = Stronger Base

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

📄 Passage

According to the passage, the basic nature of aniline is less than that of ammonia because:

A The anilinium ion has five resonating structures, making it highly stable.

B Aniline has five resonating structures while the anilinium ion has only two, making the unprotonated aniline more stable relative to its cation.

C Ammonia is a resonance hybrid of five structures.

D Aniline forms an unstable carbocation.

💡 Short Explanation

�� Aniline is resonance stabilized. �� Lone pair is delocalized. �� Protonation becomes less favorable.

✅ Correct Answer Explanation

The lone pair on nitrogen in aniline participates in resonance with the benzene ring. → Aniline is stabilized by several resonance structures. → Upon protonation, the anilinium ion loses much of this stabilization and has fewer resonance forms. → Therefore, aniline is less willing to accept a proton and behaves as a weaker base than ammonia.

❓ Why Other Options Are Incorrect

�� Option A → Five resonance structures belong to aniline, not anilinium ion.
�� Option C → Ammonia does not show this resonance.
�� Option D → Carbocation formation is irrelevant.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Focus on resonance stabilization described in the passage.

Final Logic:

�� Greater stabilization of aniline reduces proton acceptance.

🧠 Memory Trick

More Resonance → Less Basic

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Resonance and Basicity of Aniline

15 Arrange the following methyl-substituted amines in decreasing order of basic strength in aqueous solution:
(A) (CH₃)₂NH
(B) CH₃NH₂
(C) (CH₃)₃N
(D) NH₃

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

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

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

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

💡 Short Explanation

�� Aqueous basicity depends on +I effect and solvation. �� Secondary amines are best balanced. �� Tertiary amines suffer steric hindrance.

✅ Correct Answer Explanation

In aqueous solution, basicity depends on: +I effect of alkyl groups Solvation of the conjugate acid Steric hindrance → Secondary amines receive sufficient electron donation and are effectively solvated. → Tertiary amines have stronger +I effect but poorer solvation due to steric hindrance. Thus: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly places tertiary amine first.
�� Option C → Places primary amine above secondary amine.
�� Option D → Gives reverse trend.

🎯 Strategy

Used

Ordering

Application:

�� Apply aqueous basicity order from NCERT.

Final Logic:

�� Solvation makes secondary amines strongest in water.

🧠 Memory Trick

2° > 1° > 3° > NH₃

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

16 Based on the dimensionless pKb values provided in the text (Methanamine = 3.38, N,N-Dimethylmethanamine = 4.22, Benzenamine = 9.38), which of the following represents the weakest base?

A Methanamine

B N,N-Dimethylmethanamine

C Benzenamine

D Ammonia (pKb = 4.75)

💡 Short Explanation

�� Higher pKb means weaker base. �� Benzenamine has the highest pKb value. �� Therefore, it is the weakest base.

✅ Correct Answer Explanation

Basic strength is inversely related to pKb value. A smaller pKb indicates a stronger base, while a larger pKb indicates a weaker base. Given values: Methanamine = 3.38 N,N-Dimethylmethanamine = 4.22 Ammonia = 4.75 Benzenamine = 9.38 Since Benzenamine has the highest pKb value, it is the weakest base among the given options.

❓ Why Other Options Are Incorrect

�� Option A → Methanamine has a lower pKb, so it is stronger than benzenamine.
�� Option B → N,N-Dimethylmethanamine has a lower pKb than benzenamine.
�� Option D → Ammonia is weaker than many aliphatic amines but stronger than benzenamine here.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Compare the numerical pKb values.

Final Logic:

�� Highest pKb = weakest base.

🧠 Memory Trick

High pKb = Poor base

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

17 Arrange the following alkyl halides in decreasing order of their reactivity with amines during an alkylation reaction:

A RI, RBr, RCl

B RCl, RBr, RI

C RBr, RI, RCl

D RI, RCl, RBr

💡 Short Explanation

�� Reactivity depends on leaving group ability. �� Iodide is the best leaving group. �� Chloride is the least reactive.

✅ Correct Answer Explanation

In alkylation of amines, alkyl halides react through nucleophilic substitution. The reactivity depends mainly on the ease of C–X bond cleavage. Leaving group ability: I⁻ > Br⁻ > Cl⁻ Therefore, the decreasing order is: RI > RBr > RCl

❓ Why Other Options Are Incorrect

�� Option B → Gives reverse order.
�� Option C → Places RBr above RI incorrectly.
�� Option D → Places RCl above RBr incorrectly.

🎯 Strategy

Used

Ordering

Application:

�� Arrange alkyl halides based on leaving group ability.

Final Logic:

�� Better leaving group gives faster reaction.

🧠 Memory Trick

I > Br > Cl

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

18 What is the primary disadvantage of preparing primary amines via the ammonolysis of alkyl halides?
(I) It requires an extremely high temperature that destroys the amine.
(II) It yields a mixture of primary, secondary, and tertiary amines, and also a quaternary ammonium salt.

A Only (I)

B Only (II)

C Both (I) and (II)

D Neither (I) nor (II)

💡 Short Explanation

�� Primary amine formed can react further. �� Further alkylation gives higher amines. �� Final product may include quaternary ammonium salt.

✅ Correct Answer Explanation

During ammonolysis, the primary amine formed still has a lone pair on nitrogen and can react further with alkyl halide. This leads to formation of secondary amines, tertiary amines, and quaternary ammonium salts. Hence, the major disadvantage is product mixture formation. Statement I is incorrect because destruction by extremely high temperature is not the main disadvantage.

❓ Why Other Options Are Incorrect

�� Option A → Statement I is incorrect.
�� Option C → Statement I is incorrect.
�� Option D → Statement II is correct.

🎯 Strategy

Used

Option Grouping

Application:

�� Identify the actual limitation of ammonolysis.

Final Logic:

�� Further alkylation causes mixture formation.

🧠 Memory Trick

Ammonolysis = Mixture problem

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Ammonolysis of Alkyl Halides

19 During the acylation of a primary amine with an acid chloride, a stronger base like pyridine is typically added. What is the chemical purpose of pyridine in this mixture?

A It acts as an oxidizing agent.

B It performs electrophilic addition on the benzene ring.

C It removes the HCl formed and shifts the equilibrium to the right hand side.

D It replaces the acyl group through substitution.

💡 Short Explanation

�� Acylation produces HCl. �� Pyridine removes HCl. �� This drives product formation.

✅ Correct Answer Explanation

In acylation of amines with acid chlorides, HCl is produced as a by-product. Pyridine, being a stronger base, removes this HCl and prevents the amine from being converted into its salt. This shifts the equilibrium towards amide formation.

❓ Why Other Options Are Incorrect

�� Option A → Pyridine is not used as an oxidizing agent.
�� Option B → Electrophilic addition is not involved.
�� Option D → Pyridine does not replace the acyl group.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the role of base in acylation.

Final Logic:

�� Removing HCl favours amide formation.

🧠 Memory Trick

Pyridine pulls HCl

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

20 When methanamine reacts with benzoyl chloride, the hydrogen atom of the amine is replaced by a benzoyl group. What is the IUPAC name of the product?

A N-Methylbenzamide

B N-Phenylacetamide

C Benzenamine

D Benzamide

💡 Short Explanation

�� Methanamine contains a methyl group on nitrogen. �� Benzoyl chloride introduces benzoyl group. �� Product is N-Methylbenzamide.

✅ Correct Answer Explanation

Methanamine, CH₃NH₂, reacts with benzoyl chloride, C₆H₅COCl, by acylation. One hydrogen attached to nitrogen is replaced by the benzoyl group. Product: C₆H₅CONHCH₃ This compound is named N-Methylbenzamide because the methyl group is attached to the nitrogen atom of benzamide.

❓ Why Other Options Are Incorrect

�� Option B → N-Phenylacetamide has a phenyl group on nitrogen and an acetyl group, not benzoyl.
�� Option C → Benzenamine is aniline, not the acylation product.
�� Option D → Benzamide lacks the N-methyl group.

🎯 Strategy

Used

Substitution

Application:

�� Add benzoyl group to methanamine and name the N-substituted amide.

Final Logic:

�� Benzoyl chloride + methanamine = N-Methylbenzamide.

🧠 Memory Trick

Methanamine + Benzoyl = N-Methylbenzamide

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

CUET UG Chemistry Booster Test – 2 Properties and Reactions of Amines

CUET UG Chemistry Booster Test - 2 Properties and Reactions of Amines

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