Amines — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Amines" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Amines" — 7 important questions with detailed answers for CBSE board exam preparatio…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Chemistry — Amines. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Amines — All 7 Questions Solved
Q1. Define amines. What is the difference between primary, secondary, and tertiary amines? Give one example of each.
Amines: Organic compounds derived from ammonia (NH3) by replacement of one, two, or three hydrogen atoms with alkyl or aryl groups.
General formula: R-NH2, R2N-H, or R3N
Classification:
Primary amines (1° amine):
- One hydrogen of ammonia is replaced by an alkyl or aryl group
- General formula: R-NH2
- Contains one N-H bond and one N-C bond
- Example: CH3CH2NH2 (ethylamine or ethanamine)
- Structure: C-NH2
Secondary amines (2° amine):
- Two hydrogens of ammonia are replaced by alkyl or aryl groups
- General formula: R2N-H or R-NH-R'
- Contains one N-H bond and two N-C bonds
- Example: CH3-NH-CH3 (N-methylmethanamine or dimethylamine)
- Structure: C-NH-C
Tertiary amines (3° amine):
- All three hydrogens of ammonia are replaced by alkyl or aryl groups
- General formula: R3N or R-N(R')-R''
- Contains no N-H bond and three N-C bonds
- Example: (CH3)3N or CH3-N(CH3)-CH3 (trimethylamine or N,N-dimethylmethanamine)
- Structure: C-N(-C)-C
Key differences:
Primary Secondary Tertiary
General formula R-NH2 R2N-H R3N
N-H bonds 1 1 0
N-C bonds 1 2 3
Boiling point Intermediate High Low (no H-bonding)
H-bonding Yes (donor & acceptor) Yes (acceptor only) No
Basicity Strong Moderate Strong (but different reason)
Solubility High (water) Moderate Low (no H-bonding with water)
Examples:
1. Primary: Aniline (C6H5NH2), methylamine (CH3NH2)
2. Secondary: N-methylaniline (C6H5NHCH3), diethylamine (C2H5-NH-C2H5)
3. Tertiary: Trimethylamine ((CH3)3N), triethylamine ((C2H5)3N)
Q2. Explain the basicity of amines. Why are amines basic in character? Compare the basicity of primary, secondary, and tertiary amines.
Basicity of amines:
Amines are basic because they contain a lone pair of electrons on the nitrogen atom. This lone pair can accept a proton (H+) from an acid.
General reaction:
R-NH2 + HX → R-NH3+ + X- (or R-NH3+X-)
The amine acts as a proton acceptor (Lewis base):
R-NH2 + H+ → R-NH3+
Basic strength comparison:
Primary (RNH2) > Tertiary (R3N) > Secondary (R2NH)
Detailed explanation:
1. Basicity order and reasons:
- Primary amines: Strongest (in aliphatic series)
- Tertiary amines: Comparable or slightly weaker (due to steric hindrance in protonation)
- Secondary amines: Generally weakest
2. Factors affecting basicity:
a) Inductive effect (electron-donating alkyl groups):
- More alkyl groups → more electron density on N → stronger base
- Tertiary > secondary > primary (alkyl effect)
b) Solvation effect:
- Primary (RNH3+) can form hydrogen bonds (as both donor and acceptor) → stabilized
- Secondary (R2NH2+) forms fewer hydrogen bonds → less stabilized
- Tertiary (R3NH+) cannot form hydrogen bonds → not stabilized
- This makes primary most basic in aqueous solution
c) Steric hindrance:
- In gas phase: Less steric hindrance in tertiary amines
- In aqueous solution: Tertiary amines face difficulty in solvation of the conjugate acid
- Bulky groups around N hinder approach of water molecules
3. Quantitative comparison (pKb values in water):
- Primary aliphatic amines: pKb ≈ 3-4 (strong bases)
- Tertiary aliphatic amines: pKb ≈ 3-4 (strong bases)
- Secondary aliphatic amines: pKb ≈ 3-4 (intermediate)
- Aniline (C6H5NH2): pKb ≈ 9 (weak base, due to resonance delocalization)
4. Effect of aryl substitution:
- Aliphatic amines are stronger bases
- Aromatic amines (anilines) are weaker bases
- Reason: In anilines, the lone pair on N is delocalized into the benzene ring through resonance
- This reduces the electron density available for proton acceptance
- Structure: C6H5-N with lone pair resonating into the ring
5. Comparison of specific examples:
- CH3NH2 (methylamine): Very strong base, pKb ≈ 3.4
- (CH3)2NH (dimethylamine): Strong base, pKb ≈ 3.3
- (CH3)3N (trimethylamine): Strong base, pKb ≈ 4.2
- C6H5NH2 (aniline): Weak base, pKb ≈ 9.4
- C6H5-NH-CH3 (N-methylaniline): Intermediate, pKb ≈ 9.0
Summary of basicity order in aqueous solution:
Aliphatic primary ≈ tertiary > secondary > aromatic amines
But in gas phase (no solvation):
Tertiary > secondary > primary (due to inductive effect dominating)
Q3. Write the IUPAC name of (CH3)2N-CH2-CH3. Describe its structure and classify this amine.
Structure of (CH3)2N-CH2-CH3:
CH3
|
CH3-N-CH2-CH3
|
CH3
Alternative representation: CH3-N(CH3)-CH2-CH3
IUPAC name: N,N-dimethylethanamine
Naming explanation:
1. Main chain: ethane (2 carbons attached to N in the longest chain)
2. Functional group: amine (-amine suffix)
3. The nitrogen is attached to:
- Two methyl groups (methyl substituents on N)
- One ethyl group (main chain)
4. Numbering: The carbon attached to N is position 1
5. Substituents on N: N,N-dimethyl (two methyl groups on the same nitrogen)
6. Full name: N,N-dimethylethanamine
Common name: N,N-dimethylethylamine or dimethylethylamine
Classification:
- Tertiary amine (3° amine or tertiary aliphatic amine)
- The nitrogen has three carbon groups attached (no N-H bond)
- All three bonds to N are C-N bonds
Structural features:
- One N-C bond connecting to the ethyl chain (CH2-CH3)
- Two N-C bonds connecting to methyl groups (CH3)
- No N-H bond (characteristic of tertiary amine)
- The nitrogen is sp3 hybridized
- The nitrogen has a lone pair
Basicity:
- Tertiary amines are strong bases
- This compound can accept a proton on the lone pair
- Protonation: (CH3)2N-CH2-CH3 + H+ → [(CH3)2N+-CH2-CH3]
Physical properties:
- Lower boiling point than primary or secondary amines of similar molar mass
- Cannot form hydrogen bonds with itself (no N-H bond)
- Can act as hydrogen bond acceptor with water or other polar molecules
- Liquid at room temperature
- Relatively insoluble in water compared to primary amines
Reactivity:
- Can undergo quaternization with alkyl halides:
(CH3)2N-CH2-CH3 + CH3I → [(CH3)3N-CH2-CH3]+ I-
- Acts as nucleophile in SN2 reactions
- Can form complexes with Lewis acids
- Undergoes acylation with acid anhydrides and acid chlorides
Q4. Describe the reaction of amines with acids. Write the equation for the reaction of aniline with hydrochloric acid.
Reaction of amines with acids:
Amines react with acids to form salts (ammonium salts). The nitrogen lone pair acts as a nucleophile and accepts a proton from the acid.
General reaction:
R3N + HX → R3N-H+ X-
For primary amines:
R-NH2 + HX → R-NH3+ X-
For secondary amines:
R2N-H + HX → R2N-H2+ X-
Specific example with aniline and HCl:
Equation: C6H5NH2 + HCl → C6H5NH3+ Cl-
Or written as: C6H5NH2·HCl (aniline hydrochloride)
Alternatively: C6H5NH2 + HCl → [C6H5NH3]+Cl-
The aniline hydrochloride salt can also be written as:
Phenylammonium chloride
Mechanism:
1. The lone pair on nitrogen attacks the H+ from HCl
2. Formation of N-H bond
3. The chloride ion balances the positive charge on nitrogen
4. Product is an ionic compound (salt)
Characteristics of the reaction:
- This is an acid-base reaction
- The amine acts as a base
- The acid acts as an acid donor
- The reaction is typically fast and essentially complete
- The product is an ionic salt, not a covalent compound
Properties of aniline hydrochloride:
- White crystalline solid
- Soluble in water (ionic)
- Can be separated from organic solutions by extraction with dilute acid
- The salt can be converted back to free aniline by adding a strong base
C6H5NH3+ Cl- + NaOH → C6H5NH2 + NaCl + H2O
Other examples with different acids:
1. With H2SO4:
2 C6H5NH2 + H2SO4 → (C6H5NH3+)2 SO4^2-
(Aniline sulfate)
2. With HNO3:
C6H5NH2 + HNO3 → C6H5NH3+ NO3-
(Aniline nitrate)
3. With acetic acid (weak acid):
C6H5NH2 + CH3COOH → C6H5NH3+ CH3COO-
(Aniline acetate)
Applications:
1. Purification of amines (salt formation and recrystallization)
2. Identification of amines (formation of characteristic salts)
3. Preparation of pharmaceutical amine salts
4. Separation of amines from mixtures (using differential acid solubility)
Reverse reaction (salt dissociation):
When a strong base is added to an amine salt, the free amine is regenerated:
R-NH3+ Cl- + NaOH → R-NH2 + NaCl + H2O
This is used in the extraction of amines from aqueous acidic solutions to organic solvents.
Q5. Explain the Hofmann elimination reaction. Write the equation for the Hofmann elimination of 1-bromopropane.
Hofmann elimination reaction: Elimination reaction of quaternary ammonium salts (produced by exhaustive methylation of primary amines) to form alkenes and a tertiary amine as byproducts.
Note: The reaction is actually with 1-bromopropane (primary alkyl halide, not amine). The actual Hofmann reaction requires a quaternary ammonium salt. Let me address the correct interpretation:
For a primary amine (to demonstrate Hofmann elimination):
Example: CH3CH2CH2NH2 (1-propylamine or propan-1-amine)
Hofmann elimination steps:
Step 1: Convert primary amine to quaternary ammonium salt (exhaustive methylation)
CH3CH2CH2NH2 + CH3I → [CH3CH2CH2N(CH3)3]+ I-
(with excess CH3I)
Step 2: Hofmann elimination with strong base (aqueous NaOH or KOH, heating)
[CH3CH2CH2N(CH3)3]+ I- + NaOH → CH3CH=CH2 + N(CH3)3 + NaI + H2O
(heat)
Product: Alkene (prop-1-ene or propene) + trimethylamine
Alternative reagent (Hofmann's original):
Using Ag2O in water:
[CH3CH2CH2N(CH3)3]+ OH- + Ag2O → CH3CH=CH2 + N(CH3)3 + AgI + H2O
Mechanism (E2 elimination):
1. The quaternary ammonium cation [R-N(CH3)3]+ is the leaving group
2. Strong base (OH-) abstracts a β-hydrogen
3. Simultaneous formation of C=C double bond
4. The trimethylammonium becomes the leaving group
5. Products: alkene + trimethylamine + water
Zaitsev vs Hofmann selectivity:
Zaitsev rule (normal elimination):
- Forms the more substituted (more stable) alkene
- Preference for anti-periplanar geometry
Hofmann elimination:
- Forms the LESS substituted alkene (terminal alkene preferred)
- Opposite of Zaitsev rule
- Due to steric hindrance of the large N(CH3)3+ leaving group
- The base attacks the less hindered β-hydrogen
Example comparison:
For CH3CH2CH(CH3)NH2:
Normal elimination (with HBr/heat):
CH3CH2CH(CH3)OH + HBr → CH3CH=CHCH3 (major, Zaitsev product, more substituted)
Hofmann elimination:
[CH3CH2CH(CH3)N(CH3)3]+ + NaOH → CH2=CHCH2CH3 (major, Hofmann product, less substituted)
Why Hofmann rule operates:
1. The bulky quaternary ammonium group is a poor leaving group
2. Its large size creates steric hindrance
3. The base preferentially abstracts a hydrogen from the less hindered position
4. This is usually at the terminal carbon (least substituted position)
5. Forms the less substituted (terminal) alkene
Applications:
1. Synthesis of terminal alkenes (difficult to make by other methods)
2. Structure determination of amines
3. Converting primary amines to alkenes
4. Degradation studies
Note on the original question:
If the question meant Hofmann degradation of 1-bromopropane (as a haloalkane, not amine):
CH3CH2CH2Br + NaOH (alc.) → CH3CH=CH2 + NaBr + H2O (E2 elimination, not Hofmann)
But true Hofmann elimination specifically refers to the reaction of quaternary ammonium salts derived from amines.
Q6. What happens when aniline is treated with nitrous acid at 0°C? Write the equation and explain the product.
Reaction of aniline with nitrous acid (HNO2) at 0°C:
Equation:
C6H5NH2 + HNO2 + HCl (at 0°C) → [C6H5N2]+Cl- + 2 H2O
Or more clearly:
C6H5NH2 + NaNO2 + HCl (0°C) → [C6H5N2]+Cl- (diazonium salt) + NaCl + 2 H2O
Product name: Benzenediazonium chloride (or phenyldiazonium chloride)
Product structure: C6H5-N≡N+ Cl-
Step-by-step mechanism:
Step 1: Formation of nitrous acid in situ
NaNO2 + HCl → HNO2 + NaCl
HNO2 is the active reagent
Step 2: Protonation of aniline
C6H5NH2 + H+ → C6H5NH3+
Step 3: Formation of N-nitroso intermediate
C6H5NH3+ + HNO2 → C6H5N=N-OH + H2O
The nitroso compound forms (N bonded to N with O)
Step 4: Rearrangement and dehydration to form diazonium salt
C6H5N=N-OH → C6H5-N≡N+ + OH-
The diazonium cation forms with a triple bond between the two nitrogen atoms
Characteristics of diazonium salts:
1. Structure: R-N≡N+ with a triple bond between two nitrogens
2. Very unstable and reactive (especially at room temperature)
3. Must be kept at 0°C to prevent decomposition
4. Can undergo various substitution reactions
5. Highly polar due to positive charge on nitrogen
6. The N2 group is an excellent leaving group
Importance of low temperature (0°C):
- Diazonium salts are unstable and decompose easily
- At 0°C, the salt is stable for a short time (minutes to hours)
- At room temperature, rapid decomposition occurs
- Spontaneous decomposition: C6H5N2+ → C6H5+ + N2↑
- The resulting phenyl cation can lead to various products (phenol, dimers, etc.)
Uses of benzenediazonium salt:
1. Sandmeyer reaction (substitution with Cu(I) salt):
C6H5N2+Cl- + CuCl → C6H5Cl + N2 + Cu+
(Formation of chlorobenzene)
2. Gattermann reaction (with Cu powder):
Similar to Sandmeyer but with copper powder
3. Schiemann reaction (with HPF6):
C6H5N2+BF4- → C6H5F + N2 + BF3
(Formation of fluorobenzene)
4. Formation of aryl iodides:
C6H5N2+ + I- → C6H5I + N2
5. Coupling reactions (with phenols or amines):
C6H5N2+ + C6H5OH → C6H5-N=N-C6H4-OH (azo dye)
6. Reduction to form hydrazines:
C6H5N2+ + reducing agent → C6H5-NH-NH2 (phenylhydrazine)
Comparison with secondary amines:
Secondary amines (R2NH) form N-nitroso compounds that are stable:
R2NH + HNO2 → R2N-NO + H2O (N-nitroso compound, yellow, unstable)
No diazonium salt forms
Comparison with tertiary amines:
Tertiary amines (R3N) do not react with nitrous acid under normal conditions (no N-H group)
Q7. Describe the preparation of primary amines from alkyl halides using the Gabriel phthalimide synthesis. Write the equation and explain the mechanism.
Gabriel phthalimide synthesis: A method for the preparation of primary amines from alkyl halides using phthalimide as a protecting group.
Overall reaction:
R-X + KOH/phthalimide → R-NH2 (primary amine)
Detailed steps with equation:
Step 1: Nucleophilic substitution on phthalimide (with KOH)
Phthalimide + KOH → Potassium salt of phthalimide (K+N-phthalimide-)
Structure of phthalimide:
O
||
/--C
| \
| N-H
\--C
||
O
(The N-H is weakly acidic)
Potassium phthalimide:
O
||
/--C
| \
| N-K+
\--C
||
O
Step 2: SN2 reaction with alkyl halide
Potassium phthalimide + R-X → R-N-phthalimide + KX
Example (using bromoethane):
K+ N(phthalimide)- + CH3CH2Br → CH3CH2-N-phthalimide + KBr
The alkyl halide undergoes nucleophilic substitution with the N- acting as nucleophile
Step 3: Hydrolysis of phthalimide (with aqueous HCl or KOH)
R-N-phthalimide + H2O + HCl → R-NH2·HCl (amine salt) + phthalic acid
With aqueous HCl:
CH3CH2-N-phthalimide + H2O/HCl → CH3CH2-NH3+Cl- + phthalic acid
Or with aqueous KOH:
CH3CH2-N-phthalimide + H2O/KOH → CH3CH2-NH2 + phthalate salt + H2O
Complete reaction sequence:
C6H4(CO)2NH + KOH → [C6H4(CO)2N]- K+ + H2O
(Potassium phthalimide)
[C6H4(CO)2N]- K+ + R-X → C6H4(CO)2N-R + KX
(N-alkylphthalimide)
C6H4(CO)2N-R + H2O/HCl → R-NH3+Cl- + phthalic acid
Or: C6H4(CO)2N-R + H2O/KOH → R-NH2 + potassium phthalate
Mechanism:
Step 1: Deprotonation of phthalimide
- KOH removes the N-H proton from phthalimide
- Forms potassium phthalimide with N- as a nucleophile
- Resonance stabilization of the negative charge on N
Step 2: Nucleophilic substitution (SN2)
- N- attacks the carbon bearing the halogen
- Displacement of halide as leaving group
- Inversion of configuration at the carbon (if stereospecific)
- Formation of N-R bond
Step 3: Hydrolysis
- In acidic aqueous solution: Protonation of N-, hydrolysis of C-N bond in phthalimide, and breaking of N-CO bonds
- The phthalimide ring opens, releasing the primary amine
- Conversion to amine salt (in acidic conditions)
Advantages of Gabriel synthesis:
1. Selectivity: Only primary amines are produced (no secondary or tertiary amine byproducts)
2. No elimination: Unlike with NaOH/aqueous solution, no competing E2 elimination
3. No over-alkylation: The phthalimide intermediate cannot be alkylated further (no free N-H)
4. High yields: Typically good yields with primary and secondary alkyl halides
5. Phthalimide can be recovered (useful in some variations)
Limitations:
1. Not suitable for aryl halides (C6H5-X) - too unreactive for SN2
2. Tertiary alkyl halides may undergo elimination instead of substitution
3. Does not work well with very bulky primary alkyl halides (steric hindrance)
4. Hydrolysis step uses harsh conditions (concentrated HCl)
5. Phthalic acid byproduct must be separated
Example synthesis:
Prepare ethylamine from bromoethane:
C6H4(CO)2NH + KOH → [C6H4(CO)2N]-K+ + H2O
[C6H4(CO)2N]-K+ + CH3CH2Br → C6H4(CO)2N-CH2CH3 + KBr
(N-ethylphthalimide)
C6H4(CO)2N-CH2CH3 + H2O/HCl → CH3CH2-NH3+Cl- + phthalic acid
Final product: Ethylammonium chloride (CH3CH2NH3+Cl-) which can be converted to free ethylamine by adding NaOH
Amines — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the IUPAC name for (CH₃)₃N?
- A. Trimethylamine
- B. N,N-Dimethylmethanamine
- C. Methylmethanamine
- D. N-Trimethylamine
Answer: B. N,N-Dimethylmethanamine — In IUPAC nomenclature for tertiary amines, the longest carbon chain attached to nitrogen is chosen as the parent amine, and the other alkyl groups are designated as N-substituents.
MCQ 2. Which of the following is the strongest base in aqueous solution?
- A. Aniline
- B. Methylamine
- C. Dimethylamine
- D. Trimethylamine
Answer: C. Dimethylamine — Aliphatic amines are stronger bases than aromatic amines. Among methyl-substituted amines in aqueous solution, the basicity order is generally secondary > primary > tertiary > NH₃ due to a combined effect of +I effect, solvation, and steric hindrance.
MCQ 3. Identify the product 'X' in the following reaction: C₆H₅NH₂ + CHCl₃ + 3KOH (alc) → X + 3KCl + 3H₂O
- A. C₆H₅CN
- B. C₆H₅NC
- C. C₆H₅N₂Cl
- D. C₆H₅NO₂
Answer: B. C₆H₅NC — This is the carbylamine reaction (or isocyanide test), characteristic of primary amines. Aniline reacts with chloroform and alcoholic KOH to form phenyl isocyanide (phenyl carbylamine), which has a foul smell.
MCQ 4. Which of the following reagents can be used to convert R-CN into R-CH₂NH₂?
- A. LiAlH₄
- B. H₂O/H⁺
- C. NaOH
- D. Cl₂/UV light
Answer: A. LiAlH₄ — Nitriles can be reduced to primary amines using strong reducing agents like lithium aluminium hydride (LiAlH₄) or by catalytic hydrogenation (H₂/Ni, Pt, or Pd).
MCQ 5. Amines have lower boiling points than alcohols of comparable molar mass. This is because:
- A. Amines are less polar than alcohols.
- B. Hydrogen bonding in amines is weaker than in alcohols.
- C. Amines are more volatile.
- D. Amines have higher vapor pressure.
Answer: B. Hydrogen bonding in amines is weaker than in alcohols. — The N-H bond is less polar than the O-H bond, and nitrogen is less electronegative than oxygen. Consequently, the hydrogen bonds formed between amine molecules are weaker than those formed between alcohol molecules, leading to lower boiling points.
MCQ 6. An organic compound 'A' gives a pungent smelling gas when heated with CHCl₃ and alcoholic KOH. 'A' reacts with benzenesulphonyl chloride to give a product that is soluble in aqueous NaOH. 'A' is:
- A. (CH₃)₂NH
- B. CH₃NH₂
- C. (CH₃)₃N
- D. C₆H₅-NH-CH₃
Answer: B. CH₃NH₂ — The carbylamine test indicates 'A' is a primary amine. Primary amines react with benzenesulphonyl chloride (Hinsberg's reagent) to form N-alkylbenzenesulphonamide, which is acidic due to the hydrogen attached to nitrogen and thus soluble in aqueous NaOH.
MCQ 7. Benzene diazonium chloride on reaction with CuCl/HCl gives:
- A. Chlorobenzene
- B. Aniline
- C. Phenol
- D. Benzene
Answer: A. Chlorobenzene — This is the Sandmeyer reaction, where the diazonium group is replaced by a chlorine atom using cuprous chloride (CuCl) in the presence of HCl, yielding chlorobenzene.
MCQ 8. Why is aniline a weaker base than methylamine?
- A. Aniline has a larger molecular mass.
- B. The lone pair of electrons on nitrogen in aniline is delocalized due to resonance.
- C. Aniline is less soluble in water.
- D. Aniline is a primary amine.
Answer: B. The lone pair of electrons on nitrogen in aniline is delocalized due to resonance. — In aniline, the lone pair of electrons on the nitrogen atom participates in resonance with the benzene ring, making it less available for protonation. This delocalization reduces its basicity compared to methylamine, where the lone pair is localized and inductively enhanced.
MCQ 9. The Hoffmann bromamide degradation reaction is used for the preparation of:
- A. Alcohols
- B. Aldehydes
- C. Primary amines
- D. Carboxylic acids
Answer: C. Primary amines — Hoffmann bromamide degradation converts a primary amide into a primary amine with one carbon atom less than the original amide. It involves reaction with bromine and a strong base.
MCQ 10. Which of the following compounds will give a yellow oily liquid when treated with nitrous acid (NaNO₂/HCl) at 0-5°C?
- A. Aniline
- B. N-Methylaniline
- C. N,N-Dimethylaniline
- D. Ethylamine
Answer: C. N,N-Dimethylaniline — Tertiary aromatic amines like N,N-dimethylaniline react with nitrous acid to form p-nitroso-N,N-dimethylaniline, which is a yellow oily liquid.
MCQ 11. The functional group in amines is:
- A. -COOH
- B. -CHO
- C. -NH₂
- D. -OH
Answer: C. -NH₂ — Amines are organic compounds characterized by the presence of an amino group (-NH₂), which contains a nitrogen atom bonded to hydrogen atoms and/or alkyl/aryl groups.
MCQ 12. Gabriel phthalimide synthesis is used for the preparation of:
- A. Primary aromatic amines
- B. Secondary amines
- C. Primary aliphatic amines
- D. Tertiary amines
Answer: C. Primary aliphatic amines — Gabriel phthalimide synthesis is a method primarily used for the synthesis of pure primary aliphatic amines. Aromatic primary amines cannot be prepared by this method.
MCQ 13. Aniline undergoes bromination with bromine water at room temperature to give a white precipitate of:
- A. o-Bromoaniline
- B. m-Bromoaniline
- C. p-Bromoaniline
- D. 2,4,6-Tribromoaniline
Answer: D. 2,4,6-Tribromoaniline — The amino group (-NH₂) is a powerful activating group and o,p-director. In the presence of bromine water, it activates the benzene ring so strongly that bromination occurs at all available ortho and para positions, leading to 2,4,6-Tribromoaniline.
MCQ 14. Arrange the following in increasing order of their basic strength in aqueous solution: CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, NH₃
- A. NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
- B. NH₃ < CH₃NH₂ < (CH₃)₂NH < (CH₃)₃N
- C. (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH < NH₃
- D. (CH₃)₂NH < CH₃NH₂ < (CH₃)₃N < NH₃
Answer: A. NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH — The basicity of amines in aqueous solution is influenced by inductive effect, solvation effect, and steric hindrance. For methyl amines, the observed order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.
MCQ 15. Which of the following reactions is used to test for primary amines?
- A. Reimer-Tiemann reaction
- B. Carbylamine reaction
- C. Fittig reaction
- D. Wurtz reaction
Answer: B. Carbylamine reaction — The carbylamine reaction (isocyanide test) is a specific test for primary amines (both aliphatic and aromatic). It produces foul-smelling isocyanides when a primary amine is heated with chloroform and alcoholic KOH.
MCQ 16. Which of the following compounds gives a red-orange dye with β-naphthol in alkaline medium?
- A. Aniline
- B. Phenol
- C. Benzene diazonium chloride
- D. N,N-Dimethylaniline
Answer: C. Benzene diazonium chloride — This is a coupling reaction, characteristic of diazonium salts. Benzene diazonium chloride reacts with β-naphthol in an alkaline medium to form an azo dye, which is typically red-orange in color.
MCQ 17. How can nitrobenzene be converted to N-ethylaniline?
- A. Reduction followed by acetylation and then reduction with LiAlH₄.
- B. Reduction followed by reaction with ethyl iodide and then reduction.
- C. Reduction with Fe/HCl, then reaction with CH₃CHO and finally reduction.
- D. Reduction with Fe/HCl, then reaction with CH₃Cl and then reduction.
Answer: C. Reduction with Fe/HCl, then reaction with CH₃CHO and finally reduction. — Nitrobenzene is first reduced to aniline (e.g., with Fe/HCl). Aniline then undergoes reductive amination with acetaldehyde (CH₃CHO) and a reducing agent (e.g., H₂/Ni or NaBH₃CN) to yield N-ethylaniline.
MCQ 18. What is the hybridization of the nitrogen atom in amines?
- A. sp
- B. sp²
- C. sp³
- D. sp³d
Answer: C. sp³ — The nitrogen atom in amines is sp³ hybridized, with three sigma bonds to alkyl/aryl groups or hydrogen and one lone pair of electrons. This gives it a pyramidal geometry.
MCQ 19. Which reagent is used to distinguish between primary, secondary, and tertiary amines?
- A. Fehling's solution
- B. Tollen's reagent
- C. Hinsberg's reagent
- D. Lucas reagent
Answer: C. Hinsberg's reagent — Hinsberg's reagent (benzenesulphonyl chloride) reacts differently with primary, secondary, and tertiary amines, forming products with distinct solubilities in alkali, allowing their differentiation.
MCQ 20. What is the major product when aniline reacts with acetic anhydride?
- A. Acetophenone
- B. N-Acetylaniline
- C. Benzoic acid
- D. Anilinium acetate
Answer: B. N-Acetylaniline — Aniline, a primary amine, undergoes acylation with acetic anhydride to form N-phenylacetamide (acetanilide or N-Acetylaniline), an N-substituted amide.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.