Class 12 Amines MCQ
Concept 1: Introduction, Structure & Classification
A. Concept Check (Understanding)
Q1.
What are amines?
Answer: Amines are organic compounds derived from ammonia by replacing one or more hydrogen atoms with alkyl or aryl groups.
Q2.
Why are amines called derivatives of ammonia?
Answer: Because they are formed by replacing hydrogen atoms of ammonia with hydrocarbon groups.
Q3.
What type of hybridisation is present in nitrogen of amines?
Answer: sp³ hybridisation.
Q4.
What is the molecular shape of amines?
Answer: Trigonal pyramidal.
Q5.
Why is the geometry of amines pyramidal instead of tetrahedral?
Answer: Because one sp³ orbital contains a lone pair of electrons.
Q6.
Which feature makes amines basic?
Answer: Presence of a lone pair of electrons on nitrogen.
B. Definitions
Q7.
Define primary amine.
Answer: An amine in which one hydrogen atom of ammonia is replaced by an alkyl or aryl group.
Q8.
Define secondary amine.
Answer: An amine in which two hydrogen atoms of ammonia are replaced by alkyl or aryl groups.
Q9.
Define tertiary amine.
Answer: An amine in which all three hydrogen atoms of ammonia are replaced by alkyl or aryl groups.
Q10.
What are simple amines?
Answer: Amines containing identical alkyl or aryl groups.
Q11.
What are mixed amines?
Answer: Amines containing different alkyl or aryl groups.
C. Identification Questions
Q12.
Identify the class of CH₃NH₂.
Answer: Primary amine.
Q13.
Identify the class of (CH₃)₂NH.
Answer: Secondary amine.
Q14.
Identify the class of (CH₃)₃N.
Answer: Tertiary amine.
Q15.
Classify C₆H₅NH₂.
Answer: Primary aromatic amine.
Q16.
Classify C₂H₅NHCH₃.
Answer: Secondary mixed amine.
Q17.
Classify (C₂H₅)₃N.
Answer: Tertiary aliphatic amine.
D. Multiple Choice Questions
Q18.
Amines are derivatives of
A. Methane
B. Ethene
C. Ammonia
D. Benzene
Answer: C
Q19.
The nitrogen atom in amines is generally
A. sp
B. sp²
C. sp³
D. dsp²
Answer: C
Q20.
The shape of NH₃ and most amines is
A. Linear
B. Planar
C. Tetrahedral
D. Trigonal pyramidal
Answer: D
Q21.
Which contains one lone pair on nitrogen?
A. Alkene
B. Alcohol
C. Amine
D. Alkane
Answer: C
Q22.
The basic nature of amines is mainly due to
A. Hydrogen bonding
B. Lone pair on nitrogen
C. Carbon chain
D. Aromatic ring
Answer: B
Q23.
Which is a secondary amine?
A. CH₃NH₂
B. (CH₃)₂NH
C. (CH₃)₃N
D. C₆H₅NH₂
Answer: B
Q24.
Which is a tertiary amine?
A. C₂H₅NH₂
B. C₂H₅NHCH₃
C. (CH₃)₃N
D. NH₃
Answer: C
Q25.
A mixed amine contains
A. Only one alkyl group
B. Different alkyl/aryl groups
C. Only aryl groups
D. Only hydrogen atoms
Answer: B
Q26.
Which statement is incorrect?
A. Primary amines contain one alkyl/aryl group.
B. Secondary amines contain two hydrocarbon groups.
C. Tertiary amines possess three hydrocarbon groups.
D. Primary amines contain no lone pair.
Answer: D
E. Fill in the Blanks
Q27.
Amines are derivatives of __________.
Answer: ammonia
Q28.
Nitrogen in amines is __________ hybridised.
Answer: sp³
Q29.
The geometry of amines is __________.
Answer: trigonal pyramidal
Q30.
Primary amines have the general formula __________.
Answer: RNH₂
Q31.
Secondary amines have the general formula __________.
Answer: R₂NH
Q32.
Tertiary amines have the general formula __________.
Answer: R₃N
Q33.
The basic character of amines arises due to the presence of a __________ pair of electrons.
Answer: lone
F. True or False
Q34.
All amines are planar molecules.
Answer: False
Q35.
Nitrogen in amines contains one lone pair.
Answer: True
Q36.
(CH₃)₃N is a tertiary amine.
Answer: True
Q37.
Aniline is an aliphatic amine.
Answer: False
Q38.
Simple amines contain identical alkyl or aryl groups.
Answer: True
G. Assertion–Reason
Q39.
Assertion (A): Amines behave as Lewis bases.
Reason (R): Nitrogen possesses a lone pair of electrons.
A. Both A and R are true and R is the correct explanation.
B. Both are true but R is not the correct explanation.
C. A is true but R is false.
D. A is false but R is true.
Answer: A
Q40.
Assertion: Amines have trigonal pyramidal geometry.
Reason: Nitrogen is sp³ hybridised with one lone pair.
Answer: A
Q41.
Assertion: Tertiary amines contain N–H bonds.
Reason: Tertiary amines have three alkyl or aryl groups attached to nitrogen.
Answer: D (Assertion is false, Reason is true.)
H. Match the Following
| Column I | Column II |
|---|---|
| Primary amine | RNH₂ |
| Secondary amine | R₂NH |
| Tertiary amine | R₃N |
| Aniline | Aromatic amine |
I. One-Word Answer
Q42.
Which orbital on nitrogen contains the non-bonding electron pair?
Answer: sp³ orbital
Q43.
Name the simplest aromatic amine.
Answer: Aniline
Q44.
Which electron pair is responsible for the basic nature of amines?
Answer: Lone pair
Concept 2: Nomenclature of Amines
A. Concept Understanding Questions
Q1.
What is the common method of naming aliphatic primary amines?
Answer:
The alkyl group name is written before the word amine.
Example:
CH₃NH₂ → Methylamine
Q2.
How are primary amines named according to IUPAC system?
Answer:
The suffix -e of the parent alkane is replaced by -amine.
Example:
CH₃NH₂ → Methanamine
Q3.
What is the IUPAC name of CH₃CH₂NH₂?
Answer:
Ethanamine
Q4.
What is the common name of CH₃NH₂?
Answer:
Methylamine
Q5.
What suffix is used in IUPAC naming of amines?
Answer:
Amine
Q6.
How are secondary and tertiary amines named when groups attached to nitrogen are substituted?
Answer:
The substituent attached to nitrogen is indicated by the prefix N-.
B. Direct Naming Questions
Q7.
Give the IUPAC name of:
CH₃NH₂
Answer: Methanamine
Q8.
Give the IUPAC name of:
C₂H₅NH₂
Answer: Ethanamine
Q9.
Give the IUPAC name of:
CH₃CH₂CH₂NH₂
Answer: Propan-1-amine
Q10.
Give the IUPAC name of:
CH₃CH(NH₂)CH₃
Answer: Propan-2-amine
Q11.
Give the IUPAC name of:
H₂N–CH₂–CH₂–NH₂
Answer: Ethane-1,2-diamine
Q12.
Give the common name of C₆H₅NH₂.
Answer: Aniline
Q13.
Give the IUPAC name of C₆H₅NH₂.
Answer: Benzenamine
C. Secondary and Tertiary Amine Naming
Q14.
Name:
CH₃NHCH₂CH₃
Answer:
N-Methylethanamine
Q15.
Why is “N” used while naming secondary and tertiary amines?
Answer:
To show that the substituent is attached directly to nitrogen.
Q16.
Name:
(CH₃)₃N
Answer:
N,N-Dimethylmethanamine
Q17.
Name:
(C₂H₅)₃N
Answer:
N,N-Diethylethanamine
Q18.
Name:
CH₃NHCH₃
Answer:
N-Methylmethanamine
D. Aromatic Amine Naming
Q19.
What is an arylamine?
Answer:
An amine in which the amino group is directly attached to an aromatic ring.
Q20.
What is the simplest arylamine?
Answer:
Aniline
Q21.
Give the IUPAC name of aniline.
Answer:
Benzenamine
Q22.
Name:
p-BrC₆H₄NH₂
Answer:
4-Bromoaniline
(or 4-bromobenzenamine)
Q23.
Name:
o-CH₃C₆H₄NH₂
Answer:
2-Methylaniline
E. MCQs
Q24.
The IUPAC name of CH₃NH₂ is:
A. Methanamide
B. Methanamine
C. Methylamide
D. Aminomethane
Answer: B
Q25.
The common name of C₆H₅NH₂ is:
A. Benzylamine
B. Aniline
C. Phenylamide
D. Benzenamide
Answer: B
Q26.
The prefix used for a substituent attached to nitrogen is:
A. C
B. N
C. A
D. R
Answer: B
Q27.
The IUPAC name of CH₃CH₂NHCH₃ is:
A. Ethylmethylamine
B. N-Methylethanamine
C. Methylethanamide
D. N-Ethylmethanamine
Answer: B
Q28.
The correct IUPAC name of H₂N–CH₂–CH₂–NH₂ is:
A. Ethanediamine
B. Ethane-1,2-diamine
C. 1,2-aminoethane
D. Diaminoethane
Answer: B
Q29.
The suffix “amine” replaces:
A. -ol
B. -al
C. -e of alkane
D. -one
Answer: C
F. Identify Incorrect Names
Q30.
Which name is incorrect?
A. CH₃NH₂ → Methanamine
B. C₂H₅NH₂ → Ethanamine
C. C₆H₅NH₂ → Benzenamine
D. CH₃NH₂ → Ethylamine
Answer: D
Q31.
Correct the name:
CH₃NHCH₂CH₃ = Ethylmethylamine (IUPAC)
Answer:
N-Methylethanamine
G. Fill in the Blanks
Q32.
The common name of C₆H₅NH₂ is ______.
Answer: Aniline
Q33.
In IUPAC naming, alkane ending “e” is replaced by ______.
Answer: amine
Q34.
Nitrogen substituents are represented by the prefix ______.
Answer: N
Q35.
H₂N–CH₂–CH₂–NH₂ is named as ______.
Answer: Ethane-1,2-diamine
Q36.
Aromatic amines contain the –NH₂ group directly attached to a ______ ring.
Answer: benzene/aromatic
H. True or False
Q37.
Aniline is an accepted IUPAC name.
Answer: True
Q38.
Secondary amines are named using N-prefixes.
Answer: True
Q39.
CH₃NHCH₂CH₃ is named ethanamine.
Answer: False
Q40.
The amino group directly attached to benzene gives an arylamine.
Answer: True
I. Assertion–Reason
Q41.
Assertion: CH₃NHCH₂CH₃ is named N-methylethanamine.
Reason: The methyl group is attached to nitrogen.
Answer: A
(Both are true and reason correctly explains assertion.)
Q42.
Assertion: Aniline can also be called benzenamine.
Reason: In IUPAC naming, the suffix of arene is replaced by amine.
Answer: A
J. Board-Level Practice
Q43.
Write common and IUPAC names of:
(i) CH₃CH₂NH₂
(ii) C₆H₅NH₂
(iii) CH₃NHCH₃
Answer:
(i) Ethylamine — Ethanamine
(ii) Aniline — Benzenamine
(iii) Dimethylamine — N-Methylmethanamine
Q44.
Give IUPAC names:
(a) (CH₃)₂CHNH₂
(b) C₂H₅NHCH₃
(c) (C₂H₅)₃N
Answer:
(a) Propan-2-amine
(b) N-Methylethanamine
(c) N,N-Diethylethanamine
Concept 3: Preparation of Amines
A. Concept Understanding Questions
Q1.
Name the important methods used for preparation of amines.
Answer:
- Reduction of nitro compounds
- Ammonolysis of alkyl halides
- Reduction of nitriles
- Reduction of amides
- Gabriel phthalimide synthesis
- Hoffmann bromamide degradation reaction
1. Reduction of Nitro Compounds
Q2.
What happens when nitro compounds are reduced?
Answer:
Nitro compounds are converted into primary amines.
General reaction:
R–NO₂ → R–NH₂
Q3.
Name the reagents used for reduction of nitro compounds.
Answer:
- H₂/Ni, Pd or Pt
- Sn/HCl
- Fe/HCl
Q4.
Why is Fe/HCl preferred for reduction of nitro compounds?
Answer:
Because FeCl₂ formed during the reaction gets hydrolysed and regenerates HCl, so only a small amount of acid is required.
Q5.
Convert nitrobenzene into aniline.
Answer:
C₆H₅NO₂ + Fe/HCl → C₆H₅NH₂
Q6.
Nitroalkanes on reduction give:
A. Alcohols
B. Aldehydes
C. Alkanamines
D. Ketones
Answer: C
2. Ammonolysis of Alkyl Halides
Q7.
What is ammonolysis?
Answer:
The process in which alkyl halides react with ammonia to form amines is called ammonolysis.
Q8.
Write the reaction of ethyl chloride with ammonia.
Answer:
C₂H₅Cl + NH₃ → C₂H₅NH₂ + HCl
Q9.
What type of reaction occurs during ammonolysis?
Answer:
Nucleophilic substitution reaction.
Q10.
What is the role of ammonia in ammonolysis?
Answer:
Ammonia acts as a nucleophile and replaces halogen from alkyl halide.
Q11.
Why does ammonolysis produce a mixture of amines?
Answer:
Because the primary amine formed can further react with alkyl halide to form secondary, tertiary amines and quaternary ammonium salts.
Q12.
How can primary amine formation be increased in ammonolysis?
Answer:
By using a large excess of ammonia.
Q13.
Arrange the reactivity of alkyl halides in ammonolysis.
Answer:
RI > RBr > RCl
Q14.
The major product obtained by using excess ammonia in ammonolysis is:
A. Secondary amine
B. Primary amine
C. Tertiary amine
D. Quaternary salt
Answer: B
3. Reduction of Nitriles
Q15.
What is obtained when nitriles are reduced?
Answer:
Primary amines.
General reaction:
RCN → RCH₂NH₂
Q16.
Name the reducing agent used for nitrile reduction.
Answer:
Lithium aluminium hydride (LiAlH₄)
Q17.
Why is reduction of nitriles important?
Answer:
It increases the carbon chain length by one carbon atom.
Q18.
Convert ethanenitrile into propanamine.
Answer:
CH₃CN + LiAlH₄ → CH₃CH₂NH₂
Q19.
Reduction of nitriles gives:
A. Amide
B. Primary amine
C. Secondary amine
D. Alcohol
Answer: B
4. Reduction of Amides
Q20.
What happens when amides are reduced?
Answer:
Amides give amines.
Q21.
Which reagent is used for reduction of amides?
Answer:
LiAlH₄
Q22.
Convert ethanamide into ethanamine.
Answer:
CH₃CONH₂ + LiAlH₄ → CH₃CH₂NH₂
5. Gabriel Phthalimide Synthesis
Q23.
What is Gabriel phthalimide synthesis used for?
Answer:
Preparation of primary amines.
Q24.
Why is Gabriel synthesis preferred?
Answer:
Because it gives pure primary amines without formation of secondary or tertiary amines.
Q25.
Write the steps involved in Gabriel synthesis.
Answer:
Phthalimide + KOH → Potassium phthalimide
Potassium phthalimide + Alkyl halide → Alkyl phthalimide
Hydrolysis → Primary amine
Q26.
Can Gabriel synthesis prepare aromatic amines?
Answer:
No.
Q27.
Why cannot aromatic primary amines be prepared by Gabriel synthesis?
Answer:
Because aryl halides do not undergo nucleophilic substitution with phthalimide ion.
Q28.
Gabriel synthesis gives:
A. Secondary amines
B. Tertiary amines
C. Primary amines
D. Diazonium salts
Answer: C
6. Hoffmann Bromamide Degradation
Q29.
What is Hoffmann bromamide reaction?
Answer:
A reaction in which amides are converted into primary amines using bromine and sodium hydroxide.
Q30.
Write the reagents used in Hoffmann degradation.
Answer:
Br₂ + NaOH
Q31.
What is the special feature of Hoffmann bromamide reaction?
Answer:
The amine formed contains one carbon atom less than the original amide.
Q32.
Convert benzamide into aniline.
Answer:
C₆H₅CONH₂ → C₆H₅NH₂
(Br₂/NaOH)
Q33.
Convert butanamide into propanamine.
Answer:
CH₃CH₂CH₂CONH₂ → CH₃CH₂CH₂NH₂
B. MCQs
Q34.
Which method gives primary amines only?
A. Ammonolysis
B. Gabriel synthesis
C. Alkylation
D. Acylation
Answer: B
Q35.
Reduction of nitriles increases carbon chain by:
A. 0 carbon
B. 1 carbon
C. 2 carbon
D. 3 carbon
Answer: B
Q36.
Hoffmann degradation produces amines with:
A. Same number of carbons
B. One extra carbon
C. One less carbon
D. No carbon
Answer: C
Q37.
Which reagent converts nitrobenzene into aniline?
A. NaOH
B. Fe/HCl
C. Br₂/KOH
D. HNO₃
Answer: B
Q38.
Aromatic amines cannot be prepared by:
A. Reduction of nitro compounds
B. Gabriel synthesis
C. Reduction of amides
D. Diazotisation
Answer: B
C. Fill in the Blanks
Q39.
Reduction of nitro compounds gives ______ amines.
Answer: primary
Q40.
Ammonolysis uses an alcoholic solution of ______.
Answer: ammonia
Q41.
LiAlH₄ reduces nitriles into ______.
Answer: primary amines
Q42.
Gabriel synthesis involves ______.
Answer: phthalimide
Q43.
Hoffmann degradation uses bromine and ______.
Answer: sodium hydroxide
Q44.
In Hoffmann degradation, carbon atoms decrease by ______.
Answer: one
D. True/False
Q45.
Ammonolysis produces only primary amines.
Answer: False
Q46.
Gabriel synthesis is useful for preparing primary amines.
Answer: True
Q47.
Reduction of nitriles decreases carbon chain length.
Answer: False
Q48.
Hoffmann reaction produces amines with one carbon less.
Answer: True
Q49.
Aryl halides undergo easy substitution in Gabriel synthesis.
Answer: False
E. Reaction-Based Questions
Q50.
Complete:
R–NO₂ + [H] → ?
Answer:
R–NH₂
Q51.
Complete:
RCN + LiAlH₄ → ?
Answer:
RCH₂NH₂
Q52.
Complete:
RCONH₂ + Br₂ + NaOH → ?
Answer:
RNH₂
Q53.
Convert benzyl chloride into phenylethanamine.
Answer:
C₆H₅CH₂Cl
→ (KCN)
C₆H₅CH₂CN
→ (LiAlH₄)
C₆H₅CH₂CH₂NH₂
Concept 4: Physical Properties of Amines
A. Concept Understanding Questions
Q1.
What is the physical state of lower aliphatic amines?
Answer:
Lower aliphatic amines are gases with a fishy odour.
Q2.
What is the physical state of higher amines?
Answer:
Higher amines are generally liquids or solids depending on molecular mass.
Q3.
Why do lower amines have a fishy smell?
Answer:
Due to the presence of volatile amine molecules.
Q4.
What is the colour of pure aniline?
Answer:
Aniline is usually a colourless liquid.
Q5.
Why does aniline become coloured on storage?
Answer:
Due to oxidation by atmospheric oxygen.
B. Solubility of Amines
Q6.
Why are lower amines soluble in water?
Answer:
Because they form hydrogen bonds with water molecules.
Q7.
Which part of an amine decreases water solubility?
Answer:
The hydrophobic alkyl group.
Q8.
How does solubility change with increase in molecular mass of amines?
Answer:
Solubility decreases as molecular mass increases.
Q9.
Why are higher amines insoluble in water?
Answer:
Because the larger hydrocarbon part reduces interaction with water.
Q10.
Are amines soluble in organic solvents?
Answer:
Yes, amines dissolve in organic solvents like:
- Alcohol
- Ether
- Benzene
Q11.
Which is more soluble in water?
Butan-1-amine or butan-1-ol?
Answer:
Butan-1-ol.
Reason:
Alcohols form stronger hydrogen bonds because oxygen is more electronegative than nitrogen.
C. Hydrogen Bonding
Q12.
Which types of amines show intermolecular hydrogen bonding?
Answer:
Primary and secondary amines.
Q13.
Why do tertiary amines not show intermolecular hydrogen bonding?
Answer:
Because they do not contain N–H bonds.
Q14.
Which has stronger hydrogen bonding: primary or secondary amines?
Answer:
Primary amines.
Reason:
They contain two N–H bonds.
Q15.
Arrange hydrogen bonding ability:
Primary amine, secondary amine, tertiary amine
Answer:
Primary > Secondary > Tertiary
D. Boiling Point
Q16.
What is the order of boiling points among isomeric amines?
Answer:
Primary amine > Secondary amine > Tertiary amine
Q17.
Why do primary amines have higher boiling points than tertiary amines?
Answer:
Because primary amines form stronger intermolecular hydrogen bonding.
Q18.
Why do amines have lower boiling points than alcohols?
Answer:
Because N–H hydrogen bonding is weaker than O–H hydrogen bonding.
Q19.
Arrange in increasing boiling point:
Amines, alcohols, alkanes
Answer:
Alkanes < Amines < Alcohols
Q20.
Which has the highest boiling point?
A. Propane
B. Propylamine
C. Propanol
Answer:
C. Propanol
E. MCQs
Q21.
Lower aliphatic amines generally have:
A. Fruity smell
B. Fishy smell
C. No smell
D. Sweet smell
Answer: B
Q22.
Solubility of amines in water decreases with:
A. Decrease in carbon atoms
B. Increase in molecular mass
C. Decrease in alkyl group
D. Increase in polarity
Answer: B
Q23.
Which amine cannot form intermolecular hydrogen bonding?
A. CH₃NH₂
B. C₂H₅NH₂
C. (CH₃)₂NH
D. (CH₃)₃N
Answer: D
Q24.
The strongest hydrogen bonding occurs in:
A. Primary amines
B. Secondary amines
C. Tertiary amines
D. All equal
Answer: A
Q25.
The highest boiling point among similar molecular mass compounds is generally shown by:
A. Alkane
B. Amine
C. Alcohol
D. Ether
Answer: C
Q26.
Aniline becomes coloured on keeping because of:
A. Reduction
B. Polymerisation
C. Oxidation
D. Hydrolysis
Answer: C
F. Fill in the Blanks
Q27.
Lower aliphatic amines are generally ______ in nature.
Answer: gaseous
Q28.
Primary and secondary amines show ______ bonding.
Answer: hydrogen
Q29.
Tertiary amines lack ______ bonds.
Answer: N–H
Q30.
Solubility of amines decreases with increase in ______ part.
Answer: alkyl/hydrophobic
Q31.
Alcohols have higher boiling points than amines due to stronger ______ bonding.
Answer: hydrogen
Q32.
Aniline gets coloured due to atmospheric ______.
Answer: oxidation
G. True or False
Q33.
All amines are highly soluble in water.
Answer: False
Q34.
Primary amines form stronger hydrogen bonds than tertiary amines.
Answer: True
Q35.
Tertiary amines contain N–H bonds.
Answer: False
Q36.
Increasing alkyl group size decreases water solubility.
Answer: True
Q37.
Alcohols generally have higher boiling points than amines of similar mass.
Answer: True
H. Assertion–Reason
Q38.
Assertion: Primary amines have higher boiling points than tertiary amines.
Reason: Primary amines form intermolecular hydrogen bonds.
Answer: A
(Both true and reason correctly explains assertion.)
Q39.
Assertion: Higher amines are less soluble in water.
Reason: The hydrophobic alkyl group increases with molecular mass.
Answer: A
Q40.
Assertion: Tertiary amines show less hydrogen bonding.
Reason: They do not have hydrogen atoms attached to nitrogen.
Answer: A
I. Comparison Questions
Q41.
Compare solubility of methylamine and aniline.
Answer:
Methylamine is more soluble because it forms hydrogen bonds easily with water, while aniline has a hydrophobic benzene ring.
Q42.
Compare boiling points of:
CH₃NH₂ and (CH₃)₃N
Answer:
CH₃NH₂ has higher boiling point due to hydrogen bonding.
Q43.
Arrange in decreasing boiling point:
(CH₃)₃N, (CH₃)₂NH, CH₃NH₂
Answer:
CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
Q44.
Arrange in increasing solubility in water:
Aniline, diethylamine, ethylamine
Answer:
Aniline < Diethylamine < Ethylamine
J. Board-Level Questions
Q45.
Explain why amines are soluble in water.
Answer:
Amines contain nitrogen with a lone pair that forms hydrogen bonds with water molecules.
Q46.
Why is aniline less soluble in water than methylamine?
Answer:
The benzene ring in aniline increases the hydrophobic character and reduces water solubility.
Q47.
Why do primary amines have higher boiling points than tertiary amines?
Answer:
Primary amines have N–H bonds, allowing intermolecular hydrogen bonding. Tertiary amines lack these bonds.
Concept 5: Chemical Properties of Amines
Part 1 — Basic Nature of Amines
A. Concept Understanding Questions
Q1.
Why are amines basic in nature?
Answer:
Amines are basic because nitrogen contains a lone pair of electrons which can accept a proton (H⁺).
Q2.
Why are amines considered Lewis bases?
Answer:
Because they donate their lone pair of electrons to electron-deficient species.
Q3.
Write the reaction showing basic nature of methylamine.
Answer:
CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻
Q4.
What happens when an amine reacts with an acid?
Answer:
It forms an ammonium salt.
Example:
RNH₂ + HCl → RNH₃⁺Cl⁻
Q5.
Why is ammonia less basic than many aliphatic amines?
Answer:
Alkyl groups attached to nitrogen increase electron density through the +I effect, making the lone pair more available.
B. Factors Affecting Basic Strength
Q6.
What is the +I effect?
Answer:
The electron-releasing effect of alkyl groups through sigma bonds is called the +I effect.
Q7.
How does the +I effect affect basicity of amines?
Answer:
It increases electron density on nitrogen and increases basic strength.
Q8.
Why are aromatic amines less basic than aliphatic amines?
Answer:
Because the lone pair on nitrogen participates in resonance with the aromatic ring, making it less available for protonation.
Q9.
Why is aniline less basic than ammonia?
Answer:
In aniline, the nitrogen lone pair is delocalised into the benzene ring due to resonance.
Q10.
Which is more basic?
Methylamine or aniline?
Answer:
Methylamine.
C. Order of Basic Strength
Q11.
Arrange the following in decreasing basic strength:
CH₃NH₂, NH₃, C₆H₅NH₂
Answer:
CH₃NH₂ > NH₃ > C₆H₅NH₂
Q12.
Arrange:
Aniline, methylamine, ammonia
Answer:
Methylamine > Ammonia > Aniline
Q13.
Which is more basic?
(a) Ethylamine
(b) Ammonia
Answer:
Ethylamine.
Reason: Ethyl group increases electron density on nitrogen.
D. Effect of Substituents on Aniline
Q14.
How do electron-donating groups affect the basicity of aniline?
Answer:
They increase basicity by increasing electron density on nitrogen.
Q15.
How do electron-withdrawing groups affect the basicity of aniline?
Answer:
They decrease basicity.
Q16.
Which is more basic?
p-Methylaniline or aniline
Answer:
p-Methylaniline.
Reason: CH₃ group shows +I effect.
Q17.
Which is less basic?
p-Nitroaniline or aniline
Answer:
p-Nitroaniline.
Reason: NO₂ group withdraws electrons.
E. Salt Formation
Q18.
What happens when methylamine reacts with HCl?
Answer:
CH₃NH₂ + HCl → CH₃NH₃Cl
Q19.
Why are amine salts soluble in water?
Answer:
Because they are ionic compounds.
Q20.
How can an amine be separated from its salt?
Answer:
By treatment with a strong base.
F. MCQs
Q21.
Basicity of amines is due to:
A. Carbon atom
B. Nitrogen lone pair
C. Hydrogen bonding
D. Alkyl group only
Answer: B
Q22.
The strongest base among the following is:
A. Aniline
B. Ammonia
C. Methylamine
D. Nitroaniline
Answer: C
Q23.
Aniline is less basic because:
A. Nitrogen is absent
B. Lone pair participates in resonance
C. It has no electrons
D. Benzene is acidic
Answer: B
Q24.
Alkyl groups increase basicity by:
A. –I effect
B. +I effect
C. Resonance withdrawal
D. Hydrogen bonding
Answer: B
Q25.
Which group decreases basicity of aniline?
A. CH₃
B. C₂H₅
C. NO₂
D. OCH₃
Answer: C
Q26.
Amines react with acids to form:
A. Alcohols
B. Esters
C. Ammonium salts
D. Aldehydes
Answer: C
G. Fill in the Blanks
Q27.
Amines act as ______ bases.
Answer: Lewis
Q28.
The basic character of amines is due to the ______ pair of nitrogen.
Answer: lone
Q29.
Alkyl groups increase basicity by ______ effect.
Answer: +I
Q30.
Aniline is less basic due to ______ of lone pair.
Answer: resonance
Q31.
Reaction of amines with acids produces ______ salts.
Answer: ammonium
H. True or False
Q32.
All amines are stronger bases than ammonia.
Answer: False
Q33.
Alkyl groups increase electron density on nitrogen.
Answer: True
Q34.
Aniline is more basic than methylamine.
Answer: False
Q35.
Amines form salts with acids.
Answer: True
I. Assertion–Reason
Q36.
Assertion: Methylamine is more basic than ammonia.
Reason: Methyl group increases electron density on nitrogen.
Answer: A
Q37.
Assertion: Aniline is less basic than methylamine.
Reason: Lone pair on nitrogen in aniline is involved in resonance.
Answer: A
Q38.
Assertion: p-Nitroaniline is less basic than aniline.
Reason: Nitro group withdraws electrons.
Answer: A
J. Reason-Based Board Questions
Q39.
Explain why methylamine is more basic than ammonia.
Answer:
The methyl group releases electrons towards nitrogen by +I effect, increasing the availability of the lone pair.
Q40.
Explain why aniline is less basic than ammonia.
Answer:
The lone pair on nitrogen is delocalised into the benzene ring, reducing its availability for accepting H⁺.
Q41.
Why are amines converted into salts during purification?
Answer:
Salt formation increases water solubility and helps separate amines from organic impurities.
K. Reaction-Based Questions
Q42.
Complete:
RNH₂ + HCl → ?
Answer:
RNH₃⁺Cl⁻
Q43.
Complete:
C₂H₅NH₂ + H₂O ⇌ ?
Answer:
C₂H₅NH₃⁺ + OH⁻
Q44.
Write the reaction of aniline with hydrochloric acid.
Answer:
C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻
Concept 5 (Part 2): Chemical Reactions of Amines
1. Alkylation of Amines
A. Concept Questions
Q1.
What is alkylation of amines?
Answer:
The reaction in which amines react with alkyl halides to form higher amines is called alkylation.
Q2.
Which type of reaction occurs during alkylation?
Answer:
Nucleophilic substitution reaction.
Q3.
Write the reaction of methylamine with methyl chloride.
Answer:
CH₃NH₂ + CH₃Cl → (CH₃)₂NH
Q4.
Why does alkylation produce a mixture of products?
Answer:
Because the amine formed can further react with alkyl halide to give secondary, tertiary amines and quaternary ammonium salts.
Q5.
Complete the sequence:
Primary amine → Secondary amine → Tertiary amine → ?
Answer:
Quaternary ammonium salt
MCQs
Q6.
Alkylation of amines occurs with:
A. Alcohols
B. Alkyl halides
C. Aldehydes
D. Ketones
Answer: B
Q7.
Excess alkyl halide with amine produces:
A. Alcohol
B. Amide
C. Quaternary ammonium salt
D. Diazonium salt
Answer: C
Q8.
Alkylation increases:
A. Number of carbon atoms attached to nitrogen
B. Number of oxygen atoms
C. Acidity
D. Hydrogen bonding
Answer: A
2. Acylation of Amines
A. Concept Questions
Q9.
What is acylation?
Answer:
The reaction in which amines react with acid chlorides or acid anhydrides to form amides is called acylation.
Q10.
Which amines undergo acylation?
Answer:
Primary and secondary amines.
Q11.
Why do tertiary amines not undergo acylation?
Answer:
Because they do not contain an N–H bond.
Q12.
Write the reaction of methylamine with acetyl chloride.
Answer:
CH₃NH₂ + CH₃COCl → CH₃CONHCH₃ + HCl
Q13.
What is the product formed when aniline reacts with acetyl chloride?
Answer:
Acetanilide.
MCQs
Q14.
Acylation of amines gives:
A. Alcohols
B. Amides
C. Aldehydes
D. Esters
Answer: B
Q15.
Which amine cannot undergo acylation?
A. Methylamine
B. Ethylamine
C. Aniline
D. Trimethylamine
Answer: D
Q16.
The reagent used for acylation is:
A. CH₃COCl
B. NaOH
C. HCl
D. Br₂
Answer: A
3. Carbylamine Test
A. Concept Questions
Q17.
What is carbylamine reaction?
Answer:
The reaction in which primary amines react with chloroform and alcoholic KOH to form isocyanides is called carbylamine reaction.
Q18.
Which amines give carbylamine test?
Answer:
Only primary amines.
Q19.
Write the reagents used in carbylamine test.
Answer:
- Chloroform (CHCl₃)
- Alcoholic KOH
Q20.
What is the characteristic observation in carbylamine test?
Answer:
Formation of foul-smelling isocyanide.
Q21.
Why is carbylamine test used?
Answer:
To identify primary amines.
MCQs
Q22.
Carbylamine test is given by:
A. Primary amines only
B. Secondary amines only
C. Tertiary amines only
D. All amines
Answer: A
Q23.
The product formed in carbylamine reaction is:
A. Amide
B. Isocyanide
C. Alcohol
D. Nitrile
Answer: B
Q24.
Carbylamine reaction uses:
A. CHCl₃ + alcoholic KOH
B. NaNO₂ + HCl
C. Br₂ + NaOH
D. LiAlH₄
Answer: A
4. Reaction with Nitrous Acid
A. Primary Aliphatic Amines
Q25.
What happens when primary aliphatic amines react with nitrous acid?
Answer:
They form alcohols with evolution of nitrogen gas.
Q26.
Write the reaction:
RNH₂ + HNO₂ → ?
Answer:
R–OH + N₂ + H₂O
Q27.
Why is nitrogen gas evolved in this reaction?
Answer:
Because unstable diazonium intermediate decomposes.
Primary Aromatic Amines
Q28.
What happens when aniline reacts with nitrous acid at low temperature?
Answer:
It forms benzene diazonium chloride.
Q29.
What is diazotisation?
Answer:
The conversion of primary aromatic amines into diazonium salts using nitrous acid at 273–278 K is called diazotisation.
Q30.
Why is temperature maintained at 273–278 K during diazotisation?
Answer:
Because diazonium salts are stable only at low temperatures.
MCQs
Q31.
Primary aromatic amines react with nitrous acid to form:
A. Alcohol
B. Diazonium salt
C. Amide
D. Alkene
Answer: B
Q32.
Diazotisation is carried out at:
A. 373 K
B. 273–278 K
C. 500 K
D. Room temperature
Answer: B
5. Hinsberg Test
Concept Questions
Q33.
What is Hinsberg test?
Answer:
A test used to distinguish primary, secondary and tertiary amines using benzenesulphonyl chloride.
Q34.
What reagent is used in Hinsberg test?
Answer:
Benzenesulphonyl chloride.
Q35.
Which amine forms sulphonamide soluble in alkali?
Answer:
Primary amines.
Q36.
Which amines form insoluble sulphonamides?
Answer:
Secondary amines.
Q37.
Why do tertiary amines not react with Hinsberg reagent?
Answer:
Because they do not have N–H bonds.
MCQs
Q38.
Hinsberg test distinguishes:
A. Alcohols
B. Amines
C. Aldehydes
D. Ketones
Answer: B
Q39.
Primary amines form:
A. Soluble sulphonamide
B. Insoluble sulphonamide
C. Alcohol
D. Diazonium salt
Answer: A
6. Reactions of Aniline
Electrophilic Substitution
Q40.
Why is aniline highly reactive towards electrophilic substitution?
Answer:
Because the –NH₂ group increases electron density in the benzene ring.
Q41.
Which directing effect is shown by –NH₂ group?
Answer:
Ortho and para directing effect.
Q42.
What happens when aniline reacts with bromine water?
Answer:
It forms 2,4,6-tribromoaniline.
Q43.
Why does aniline undergo substitution at ortho and para positions?
Answer:
Due to electron donation by the amino group through resonance.
Q44.
Why does aniline not undergo Friedel-Crafts reaction?
Answer:
Because the amino group forms a salt with Lewis acids like AlCl₃, reducing its availability.
MCQs
Q45.
The –NH₂ group in aniline is:
A. Meta directing
B. Ortho-para directing
C. Deactivating
D. Electron withdrawing
Answer: B
Q46.
Bromination of aniline gives:
A. Bromobenzene
B. 2,4,6-tribromoaniline
C. Nitrobenzene
D. Benzene diazonium salt
Answer: B
Assertion–Reason
Q47.
Assertion: Primary amines give carbylamine test.
Reason: They contain N–H bonds.
Answer: A
Q48.
Assertion: Tertiary amines do not undergo acylation.
Reason: They do not contain N–H bonds.
Answer: A
Q49.
Assertion: Aniline is more reactive than benzene towards electrophilic substitution.
Reason: –NH₂ group increases electron density in benzene ring.
Answer: A
Q50.
Assertion: Diazonium salts are prepared at low temperature.
Reason: They are unstable at higher temperatures.
Answer: A
Concept 6: Diazonium Salts
Preparation, Reactions, Named Reactions, Coupling & Applications
A. Introduction to Diazonium Salts
Q1.
What are diazonium salts?
Answer:
Diazonium salts are compounds containing the functional group:
Ar–N₂⁺X⁻
where Ar represents an aryl group and X⁻ is an anion.
Q2.
What is the general formula of benzene diazonium chloride?
Answer:
C₆H₅N₂⁺Cl⁻
Q3.
Which type of amines form diazonium salts?
Answer:
Primary aromatic amines.
Q4.
Why are diazonium salts important in organic chemistry?
Answer:
Because they can be converted into many aromatic compounds by replacing the diazonium group.
Q5.
What is the name of the process used for preparation of diazonium salts?
Answer:
Diazotisation.
B. Diazotisation Reaction
Q6.
What is diazotisation?
Answer:
The conversion of primary aromatic amines into diazonium salts using nitrous acid at low temperature is called diazotisation.
Q7.
Which reagents are used for diazotisation?
Answer:
- Sodium nitrite (NaNO₂)
- Hydrochloric acid (HCl)
Q8.
Why is sodium nitrite and hydrochloric acid used together?
Answer:
They generate nitrous acid (HNO₂) in the reaction mixture.
Q9.
At what temperature is diazotisation carried out?
Answer:
273–278 K (0–5°C)
Q10.
Why are diazonium salts prepared at low temperature?
Answer:
Because they are stable only at low temperatures and decompose at higher temperatures.
Q11.
Write the reaction of aniline diazotisation.
Answer:
C₆H₅NH₂ + NaNO₂ + HCl
→ C₆H₅N₂⁺Cl⁻ + NaCl + H₂O
C. Preparation-Based MCQs
Q12.
Diazonium salts are prepared from:
A. Secondary amines
B. Tertiary amines
C. Primary aromatic amines
D. Alcohols
Answer: C
Q13.
The temperature required for diazotisation is:
A. 373 K
B. 273–278 K
C. 500 K
D. 200°C
Answer: B
Q14.
The reagent used for generating nitrous acid is:
A. NaOH
B. NaNO₂ + HCl
C. Br₂ + NaOH
D. LiAlH₄
Answer: B
D. Reactions of Diazonium Salts
1. Replacement by Chlorine and Bromine
Q15.
What happens when benzene diazonium chloride reacts with CuCl?
Answer:
Chlorobenzene is formed.
Q16.
Name the reaction involving replacement of diazonium group by chlorine or bromine using copper salts.
Answer:
Sandmeyer reaction.
Q17.
Write Sandmeyer reaction for chlorine.
Answer:
C₆H₅N₂⁺Cl⁻ + CuCl
→ C₆H₅Cl + N₂
Q18.
Write Sandmeyer reaction for bromine.
Answer:
C₆H₅N₂⁺Cl⁻ + CuBr
→ C₆H₅Br + N₂
2. Gattermann Reaction
Q19.
What is Gattermann reaction?
Answer:
The replacement of diazonium group by chlorine or bromine using copper powder and corresponding acid is called Gattermann reaction.
Q20.
Which metals are used in Gattermann reaction?
Answer:
Copper powder.
3. Replacement by Iodine
Q21.
How is iodobenzene prepared from diazonium salt?
Answer:
By treating benzene diazonium chloride with potassium iodide.
Q22.
Write the reaction.
Answer:
C₆H₅N₂⁺Cl⁻ + KI
→ C₆H₅I + N₂
4. Replacement by Fluorine
Q23.
Which reaction is used to prepare fluorobenzene?
Answer:
Balz–Schiemann reaction.
Q24.
What is the reagent used in Balz–Schiemann reaction?
Answer:
Tetrafluoroborate salt (HBF₄).
Q25.
Why is Balz–Schiemann reaction important?
Answer:
It is a method for introducing fluorine into aromatic compounds.
5. Replacement by Hydroxyl Group
Q26.
What happens when benzene diazonium chloride is warmed with water?
Answer:
Phenol is formed.
Q27.
Write the reaction.
Answer:
C₆H₅N₂⁺Cl⁻ + H₂O
→ C₆H₅OH + N₂ + HCl
6. Replacement by Hydrogen
Q28.
How can the diazonium group be replaced by hydrogen?
Answer:
By reduction using hypophosphorous acid (H₃PO₂).
Q29.
What is formed when benzene diazonium chloride reacts with H₃PO₂?
Answer:
Benzene.
E. Coupling Reactions
Q30.
What is coupling reaction?
Answer:
A reaction in which diazonium salts react with aromatic compounds to form azo compounds.
Q31.
What type of compounds are produced in coupling reactions?
Answer:
Azo compounds.
Q32.
Why are azo compounds coloured?
Answer:
Because they contain an extended conjugated system.
Q33.
Which compounds commonly undergo coupling with diazonium salts?
Answer:
- Phenols
- Aromatic amines
Q34.
Where does coupling occur in phenol?
Answer:
At the para position (if available).
Q35.
Where does coupling occur in aniline?
Answer:
At the para position.
F. Applications of Diazonium Salts
Q36.
Why are diazonium salts useful in organic synthesis?
Answer:
They help prepare substituted aromatic compounds.
Q37.
What are azo dyes?
Answer:
Coloured compounds containing the –N=N– azo group.
Q38.
Why are diazonium salts used in dye preparation?
Answer:
Because coupling reactions produce intensely coloured azo compounds.
G. MCQs
Q39.
The functional group present in diazonium salts is:
A. –NH₂
B. –N₂⁺
C. –NO₂
D. –CN
Answer: B
Q40.
Benzene diazonium chloride is prepared from:
A. Benzene
B. Aniline
C. Phenol
D. Nitrobenzene
Answer: B
Q41.
Sandmeyer reaction uses:
A. Copper salts
B. Sodium hydroxide
C. Lithium aluminium hydride
D. Hydrogen peroxide
Answer: A
Q42.
Balz–Schiemann reaction gives:
A. Chlorobenzene
B. Fluorobenzene
C. Phenol
D. Aniline
Answer: B
Q43.
Diazonium salts react with phenol to form:
A. Alcohol
B. Azo dye
C. Ketone
D. Amide
Answer: B
H. Fill in the Blanks
Q44.
Diazonium salts are prepared from ______ aromatic amines.
Answer: primary
Q45.
Diazotisation is carried out at ______ K.
Answer: 273–278
Q46.
The functional group of diazonium salts is ______.
Answer: –N₂⁺
Q47.
Sandmeyer reaction uses ______ salts.
Answer: copper
Q48.
Coupling reactions produce ______ compounds.
Answer: azo
Q49.
Balz–Schiemann reaction introduces ______ into aromatic compounds.
Answer: fluorine
I. True or False
Q50.
Secondary aromatic amines form diazonium salts easily.
Answer: False
Q51.
Diazonium salts are stable at high temperatures.
Answer: False
Q52.
Aniline can be converted into benzene diazonium chloride.
Answer: True
Q53.
Coupling reactions are used in preparation of dyes.
Answer: True
Q54.
Sandmeyer reaction replaces diazonium group by halogen.
Answer: True
J. Assertion–Reason
Q55.
Assertion: Diazotisation is carried out at 273–278 K.
Reason: Diazonium salts decompose at higher temperatures.
Answer: A
Q56.
Assertion: Diazonium salts are valuable intermediates.
Reason: The diazonium group can be replaced by different groups.
Answer: A
Q57.
Assertion: Azo compounds are coloured.
Reason: They contain a conjugated –N=N– group.
Answer: A
K. Conversion Questions
Q58.
Convert aniline into chlorobenzene.
Answer:
Aniline
→ Diazotisation
→ Benzene diazonium chloride
→ Sandmeyer reaction
→ Chlorobenzene
Q59.
Convert aniline into phenol.
Answer:
Aniline
→ Diazonium salt
→ Warm water
→ Phenol
Q60.
Convert aniline into fluorobenzene.
Answer:
Aniline
→ Diazonium salt
→ Balz–Schiemann reaction
→ Fluorobenzene
Final Section: Mixed Chapter Revision Question Bank
CBSE Competency-Based + HOTS + Conversions + Case-Based Questions
A. Important Concept-Based Questions
Q1.
An organic compound has molecular formula C₂H₇N. It reacts with HCl to form a salt and gives carbylamine test. Identify the compound.
Answer:
Ethylamine (C₂H₅NH₂)
Reason:
- Contains –NH₂ group → primary amine
- Primary amines give carbylamine test.
Q2.
A compound does not give carbylamine test but forms sulphonamide insoluble in alkali. Identify the class of amine.
Answer:
Secondary amine.
Reason:
Secondary amines form insoluble sulphonamides in Hinsberg test.
Q3.
An amine reacts with nitrous acid and gives alcohol with evolution of nitrogen gas. Identify the type of amine.
Answer:
Primary aliphatic amine.
Q4.
An aromatic amine is converted into a diazonium salt at 273–278 K. Name the reaction.
Answer:
Diazotisation.
Q5.
Why is aniline less basic than methylamine?
Answer:
In aniline, the lone pair on nitrogen participates in resonance with the benzene ring, making it less available.
B. Assertion–Reason Questions
Q6.
Assertion (A): Methylamine is more basic than aniline.
Reason (R): The lone pair of electrons on nitrogen in aniline is delocalised.
A. Both A and R are true and R is the correct explanation.
B. Both are true but R is not the explanation.
C. A is true but R is false.
D. Both are false.
Answer: A
Q7.
Assertion (A): Tertiary amines do not undergo acylation.
Reason (R): Tertiary amines do not have N–H bonds.
Answer: A
Q8.
Assertion (A): Gabriel synthesis gives primary amines.
Reason (R): It prevents formation of secondary and tertiary amines.
Answer: A
Q9.
Assertion (A): Diazonium salts are prepared at low temperature.
Reason (R): They are unstable at higher temperatures.
Answer: A
C. Reaction Completion Questions
Q10.
Complete:
CH₃NO₂ + [H] → ?
Answer:
CH₃NH₂
Q11.
Complete:
R–CN + LiAlH₄ → ?
Answer:
R–CH₂NH₂
Q12.
Complete:
RCONH₂ + Br₂ + NaOH → ?
Answer:
RNH₂
Q13.
Complete:
C₆H₅NH₂ + NaNO₂ + HCl (273–278 K) → ?
Answer:
C₆H₅N₂⁺Cl⁻
Q14.
Complete:
C₆H₅N₂⁺Cl⁻ + H₂O → ?
Answer:
C₆H₅OH + N₂ + HCl
Q15.
Complete:
C₆H₅N₂⁺Cl⁻ + CuCl → ?
Answer:
C₆H₅Cl + N₂
D. Conversion Questions
Q16.
Convert nitrobenzene into aniline.
Answer:
Nitrobenzene
↓ (Fe/HCl)
Aniline
Q17.
Convert aniline into chlorobenzene.
Answer:
Aniline
↓ NaNO₂/HCl (273–278 K)
Benzene diazonium chloride
↓ CuCl
Chlorobenzene
Q18.
Convert aniline into phenol.
Answer:
Aniline
→ Diazonium salt
→ Warm water
→ Phenol
Q19.
Convert aniline into fluorobenzene.
Answer:
Aniline
→ Diazonium salt
→ Balz–Schiemann reaction
→ Fluorobenzene
Q20.
Convert ethanenitrile into ethanamine.
Answer:
CH₃CN
↓ LiAlH₄
CH₃CH₂NH₂
Q21.
Convert benzamide into aniline.
Answer:
C₆H₅CONH₂
↓ Br₂/NaOH
C₆H₅NH₂
E. HOTS Questions
Q22.
Why is p-nitroaniline less basic than aniline?
Answer:
The nitro group withdraws electrons from nitrogen through its electron-withdrawing effect, reducing availability of the lone pair.
Q23.
Why does ammonolysis of alkyl halides produce a mixture of products?
Answer:
The amine formed acts as a nucleophile and reacts further with alkyl halide, producing higher amines.
Q24.
Why cannot tertiary amines give carbylamine test?
Answer:
They do not contain the N–H bond required for the reaction.
Q25.
Why are diazonium salts useful intermediates?
Answer:
The diazonium group can be replaced by several other groups to prepare different aromatic compounds.
Q26.
Why is aniline more reactive than benzene towards electrophilic substitution?
Answer:
The –NH₂ group donates electrons to the benzene ring, increasing electron density.
F. Case-Based Questions
Case Study 1
A student treats aniline with sodium nitrite and hydrochloric acid at 273–278 K. The product obtained is then reacted with CuCl.
Q27.
What is the first product formed?
Answer:
Benzene diazonium chloride.
Q28.
Name the reaction used in the first step.
Answer:
Diazotisation.
Q29.
What is the final product?
Answer:
Chlorobenzene.
Q30.
Name the reaction involving CuCl.
Answer:
Sandmeyer reaction.
Case Study 2
Three amines A, B and C are tested.
- A gives carbylamine test.
- B forms insoluble sulphonamide.
- C does not react with Hinsberg reagent.
Q31.
Identify A.
Answer:
Primary amine.
Q32.
Identify B.
Answer:
Secondary amine.
Q33.
Identify C.
Answer:
Tertiary amine.
G. Complete Chapter MCQ Revision
Q34.
Which compound gives carbylamine test?
A. Trimethylamine
B. Dimethylamine
C. Methylamine
D. Triethylamine
Answer: C
Q35.
The strongest base among the following is:
A. Aniline
B. Ammonia
C. Methylamine
D. Nitroaniline
Answer: C
Q36.
Gabriel synthesis is used for preparation of:
A. Primary amines
B. Secondary amines
C. Tertiary amines
D. Diazonium salts
Answer: A
Q37.
Hoffmann bromamide reaction decreases carbon chain by:
A. One carbon
B. Two carbons
C. Three carbons
D. No change
Answer: A
Q38.
Azo dyes contain:
A. –NH₂
B. –N=N–
C. –NO₂
D. –OH
Answer: B
Q39.
Aniline is:
A. Meta directing
B. Ortho-para directing
C. Deactivating
D. Acidic
Answer: B
Q40.
Diazonium salts are prepared from:
A. Alcohols
B. Primary aromatic amines
C. Ketones
D. Tertiary amines
Answer: B