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Amines Complete NCERT Resource Guide
Welcome back to examspark.in! Carbonyl compounds ke baad, organic chemistry ka ek aur crucial milestone aata hai—Amines. To help you build absolute conceptual clarity, we have crafted the ultimate guide for Amines Class 12 Solutions. This comprehensive resource features Updated NCERT Solutions alongside highly anticipated Board Exam Questions 2026 to ensure you ace your preparation effortlessly. Let's dive in!
Chapter NameAmines
SubjectChemistry
ClassClass 12
BoardCBSE & State Boards
Important TopicsBasic Strength of Amines, Hoffmann Bromamide Degradation, Gabriel Phthalimide Synthesis, Carbylamine Reaction, Diazonium Salts
Difficulty LevelModerate
Exam WeightageAround 6 Marks
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Learning Objectives
After completing this chapter, students will be able to:
- Classify amines as primary ($1^\circ$), secondary ($2^\circ$), and tertiary ($3^\circ$).
- Write correct IUPAC and common names for various aliphatic and aromatic amines.
- Explain the anomalous basic strength trends of amines in gaseous and aqueous phases.
- Master essential nitrogen-based name reactions used for amine synthesis.
- Use diazonium salts as versatile synthetic intermediates for aromatic conversions.
Key Concepts, Definitions & Formulas
- Amines: Derivatives of ammonia ($NH_3$) produced by replacing one, two, or all three hydrogen atoms with alkyl or aryl groups.
- Basic Character: Amines act as Lewis bases due to the presence of an unshared lone pair of electrons on the nitrogen atom ($R-\ddot{N}H_2$).
- Hoffmann Bromamide Degradation: A method used to prepare primary amines containing one carbon atom less than the parent amide by reacting it with $Br_2$ and aqueous $NaOH$.
- Gabriel Phthalimide Synthesis: A highly selective method used for preparing pure primary aliphatic amines. It avoids the formation of secondary and tertiary amines.
- Carbylamine Test: Primary aliphatic and aromatic amines heat up with chloroform ($CHCl_3$) and ethanolic potassium hydroxide ($KOH$) to form foul-smelling isocyanides (carbylamines).
- Hinsberg's Reagent: Benzenesulfonyl chloride ($C_6H_5SO_2Cl$), used to distinguish structurally between primary, secondary, and tertiary amines.
Full NCERT Solutions (Step-by-Step)
Here are the complete, step-by-step NCERT Solutions for Class 12 Chemistry Chapter 13 covering textbook exercise problems.
Question 13.1: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) $CH_3CH(CH_3)NH_2$
(ii) $CH_3CH_2CH_2NH_2$
(iii) $CH_3NHCH(CH_3)_2$
(iv) $(CH_3)_3CNH_2$
(v) $C_6H_5NHCH_3$
Step 1: Compound (i). IUPAC Name: Propan-2-amine. Classification: Primary ($1^\circ$) amine.
Step 2: Compound (ii). IUPAC Name: Propan-1-amine. Classification: Primary ($1^\circ$) amine.
Step 3: Compound (iii). IUPAC Name: N-Methylpropan-2-amine. Classification: Secondary ($2^\circ$) amine.
Step 4: Compound (iv). IUPAC Name: 2-Methylpropan-2-amine. Classification: Primary ($1^\circ$) amine (Note: The carbon holding the group is tertiary, but the amine itself is primary).
Step 5: Compound (v). IUPAC Name: N-Methylaniline (or N-Methylbenzenamine). Classification: Secondary ($2^\circ$) amine.
Step 2: Compound (ii). IUPAC Name: Propan-1-amine. Classification: Primary ($1^\circ$) amine.
Step 3: Compound (iii). IUPAC Name: N-Methylpropan-2-amine. Classification: Secondary ($2^\circ$) amine.
Step 4: Compound (iv). IUPAC Name: 2-Methylpropan-2-amine. Classification: Primary ($1^\circ$) amine (Note: The carbon holding the group is tertiary, but the amine itself is primary).
Step 5: Compound (v). IUPAC Name: N-Methylaniline (or N-Methylbenzenamine). Classification: Secondary ($2^\circ$) amine.
Question 13.2: Give one chemical test to distinguish between the following pairs of compounds:
(i) Methylamine and Dimethylamine
(ii) Aniline and Benzylamine
Step 1: Methylamine ($1^\circ$) and Dimethylamine ($2^\circ$).
Test: Carbylamine Test. Heat both compounds separately with chloroform ($CHCl_3$) and alcoholic $KOH$.
Observation: Methylamine ($1^\circ$) gives an extremely foul-smelling gas due to the formation of methyl isocyanide. $$CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC \uparrow \text{ (Foul Smelling)} + 3KCl + 3H_2O$$ Dimethylamine ($2^\circ$) does not give this test.
Step 2: Aniline and Benzylamine.
Test: Azo-Dye Test. Dissolve the compound in cold $HCl$ at 0-5°C, treat it with nitrous acid ($NaNO_2 + HCl$), and add an alkaline solution of $\beta$-naphthol.
Observation: Aniline forms a stable diazonium salt that undergoes a coupling reaction to yield a brilliant orange-red dye. Benzylamine does not form a stable diazonium salt under these cold conditions and releases nitrogen gas without forming an azo-dye.
Test: Carbylamine Test. Heat both compounds separately with chloroform ($CHCl_3$) and alcoholic $KOH$.
Observation: Methylamine ($1^\circ$) gives an extremely foul-smelling gas due to the formation of methyl isocyanide. $$CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC \uparrow \text{ (Foul Smelling)} + 3KCl + 3H_2O$$ Dimethylamine ($2^\circ$) does not give this test.
Step 2: Aniline and Benzylamine.
Test: Azo-Dye Test. Dissolve the compound in cold $HCl$ at 0-5°C, treat it with nitrous acid ($NaNO_2 + HCl$), and add an alkaline solution of $\beta$-naphthol.
Observation: Aniline forms a stable diazonium salt that undergoes a coupling reaction to yield a brilliant orange-red dye. Benzylamine does not form a stable diazonium salt under these cold conditions and releases nitrogen gas without forming an azo-dye.
Question 13.3: Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Primary amines have higher boiling points than tertiary amines.
Step 1: Explanation for (i). In aniline, the unshared lone pair of electrons on the nitrogen atom is delocalized into the benzene ring due to resonance. This makes the electron pair less available for protonation. In contrast, the alkyl group in methylamine exerts an electron-donating $+I$ effect, increasing electron density on nitrogen. Thus, aniline is a weaker base, resulting in a higher $pK_b$ value.
Step 2: Explanation for (ii). Ethylamine is a lower aliphatic amine that readily forms intermolecular hydrogen bonds with water molecules. Aniline contains a large, bulky hydrophobic phenyl ring ($C_6H_5-$) that interferes with hydrogen bonding, significantly cutting down its solubility in water.
Step 3: Explanation for (iii). Primary amines possess two hydrogen atoms directly bonded to nitrogen, enabling them to form extensive intermolecular hydrogen bonds. Tertiary amines do not have any hydrogen atoms attached to nitrogen, meaning they cannot form intermolecular hydrogen bonds. Consequently, primary amines require more thermal energy to boil.
Step 2: Explanation for (ii). Ethylamine is a lower aliphatic amine that readily forms intermolecular hydrogen bonds with water molecules. Aniline contains a large, bulky hydrophobic phenyl ring ($C_6H_5-$) that interferes with hydrogen bonding, significantly cutting down its solubility in water.
Step 3: Explanation for (iii). Primary amines possess two hydrogen atoms directly bonded to nitrogen, enabling them to form extensive intermolecular hydrogen bonds. Tertiary amines do not have any hydrogen atoms attached to nitrogen, meaning they cannot form intermolecular hydrogen bonds. Consequently, primary amines require more thermal energy to boil.
Question 13.4: How will you convert:
(i) Nitrobenzene to Aniline
(ii) Benzene to Aniline
Step 1: Nitrobenzene to Aniline. Reduce nitrobenzene by heating it with iron scrap ($Fe$) and concentrated hydrochloric acid ($HCl$).
$$C_6H_5NO_2 \xrightarrow{Fe / HCl} C_6H_5NH_2 + 2H_2O$$
Step 2: Benzene to Aniline. First, perform nitration on benzene using concentrated $HNO_3$ and concentrated $H_2SO_4$ to form nitrobenzene. Then, reduce it using $Sn/HCl$ or $Fe/HCl$.
$$C_6H_6 \xrightarrow{\text{conc. } HNO_3 + \text{conc. } H_2SO_4} C_6H_5NO_2 \xrightarrow{Fe / HCl} C_6H_5NH_2$$
Extra Important Questions (Board Style)
Test your knowledge with these custom practice questions created for the upcoming 2026 Board Exams.
Section A: Multiple Choice Questions (MCQs)
Q1. What is the correct increasing order of basic strength for methyl-substituted amines in an aqueous solution?
A) $NH_3 < CH_3NH_2 < (CH_3)_3N < (CH_3)_2NH$
B) $NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$
C) $(CH_3)_3N < NH_3 < CH_3NH_2 < (CH_3)_2NH$
D) $(CH_3)_2NH < CH_3NH_2 < (CH_3)_3N < NH_3$
Correct Answer Choice: B
Step 1: Metric Verification. In aqueous medium, the basic strength trend of methyl substituted amines follows the order: $2^\circ > 1^\circ > 3^\circ > NH_3$. This specific order results from a combined interplay of the $+I$ inductive effect, steric hindrance, and hydration stabilization energy.
Step 1: Metric Verification. In aqueous medium, the basic strength trend of methyl substituted amines follows the order: $2^\circ > 1^\circ > 3^\circ > NH_3$. This specific order results from a combined interplay of the $+I$ inductive effect, steric hindrance, and hydration stabilization energy.
Q2. Gabriel Phthalimide Synthesis is strictly used for the synthesis of:
A) $1^\circ$ Aromatic Amines
B) $1^\circ$ Aliphatic Amines
C) $2^\circ$ Aliphatic Amines
D) $3^\circ$ Aliphatic Amines
Correct Answer Choice: B
Step 1: Structural Verification. Aryl halides do not undergo nucleophilic substitution with potassium phthalimide easily due to partial double bond character in $C-X$. Therefore, aromatic primary amines cannot be prepared by this method.
Step 1: Structural Verification. Aryl halides do not undergo nucleophilic substitution with potassium phthalimide easily due to partial double bond character in $C-X$. Therefore, aromatic primary amines cannot be prepared by this method.
Q3. Which of the following compounds reacts with Hinsberg's reagent to produce a solid that is completely insoluble in alkali ($NaOH$)?
A) Ethylamine
B) Diethylamine
C) Triethylamine
D) Aniline
Correct Answer Choice: B
Step 1: Reaction Verification. Diethylamine is a secondary ($2^\circ$) amine. It reacts with benzenesulfonyl chloride to form N,N-diethylbenzenesulfonamide. Because this product lacks an acidic hydrogen on the nitrogen atom, it cannot dissolve in aqueous alkali.
Step 1: Reaction Verification. Diethylamine is a secondary ($2^\circ$) amine. It reacts with benzenesulfonyl chloride to form N,N-diethylbenzenesulfonamide. Because this product lacks an acidic hydrogen on the nitrogen atom, it cannot dissolve in aqueous alkali.
Section B: Short & Long Answer Questions
Q4. Write down the chemical equation for the Hoffmann Bromamide Degradation reaction of Acetamide. (2 Marks)
Step 1: Reaction Formulation. When acetamide is treated with bromine in the presence of an aqueous or ethanolic solution of sodium hydroxide, it degrades to yield methylamine, which has one less carbon atom.
$$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$$
Q5. Why does aniline undergo electrophilic substitution reactions (like bromination) at ultra-high speeds, and how can we control its reactivity? (2 Marks)
Step 1: Reactivity Mechanics. The amino group ($-NH_2$) in aniline is a highly powerful activating group because the lone pair on nitrogen readily participates in resonance.
Step 2: Control Mechanics. To avoid polysubstitution (like getting 2,4,6-tribromoaniline), its activating effect is controlled by protecting the group via acetylation with acetic anhydride, which introduces a less activating acetamido group ($-NHCOCH_3$).
Step 2: Control Mechanics. To avoid polysubstitution (like getting 2,4,6-tribromoaniline), its activating effect is controlled by protecting the group via acetylation with acetic anhydride, which introduces a less activating acetamido group ($-NHCOCH_3$).
Q6. Explain Coupling Reaction with an example equation. (2 Marks)
Step 1: Definition & Equation. Arenediazonium salts react with highly reactive aromatic compounds like phenols or amines to form brightly colored azo compounds ($Ar-N=N-Ar'$). This process is known as a coupling reaction.
$$C_6H_5-N_2^+Cl^- + C_6H_5OH \xrightarrow{\text{Mild Base } (pH 9-10)} C_6H_5-N=N-C_6H_4-OH \text{ (p-Hydroxyazobenzene, Orange dye)} + HCl$$
Q7. Identify compounds A, B, and C in the following reaction sequence: (3 Marks)
$CH_3CH_2Br \xrightarrow{KCN} A \xrightarrow{LiAlH_4} B \xrightarrow{HNO_2 \text{ at } 0^\circ\text{C}} C$
Step 1: Formation of A. Ethyl bromide reacts with potassium cyanide ($KCN$) via nucleophilic substitution to yield Propane nitrile. $A = CH_3CH_2CN \text{ (Propanenitrile)}$
Step 2: Formation of B. Reducing propanenitrile with $LiAlH_4$ converts the nitrile group into a primary amine group. $B = CH_3CH_2CH_2NH_2 \text{ (Propan-1-amine)}$
Step 3: Formation of C. Aliphatic primary amines react with nitrous acid ($HNO_2$) to form highly unstable diazonium salts, which decompose immediately to form primary alcohols. $C = CH_3CH_2CH_2OH \text{ (Propan-1-ol)}$
Step 2: Formation of B. Reducing propanenitrile with $LiAlH_4$ converts the nitrile group into a primary amine group. $B = CH_3CH_2CH_2NH_2 \text{ (Propan-1-amine)}$
Step 3: Formation of C. Aliphatic primary amines react with nitrous acid ($HNO_2$) to form highly unstable diazonium salts, which decompose immediately to form primary alcohols. $C = CH_3CH_2CH_2OH \text{ (Propan-1-ol)}$
Q8. Explain the structural mechanism of the Carbylamine reaction. (3 Marks)
Step 1: Generation of Electrophile. Chloroform ($CHCl_3$) reacts with a strong base ($OH^-$) to undergo $\alpha$-elimination, forming a highly reactive, neutral intermediate called Dichlorocarbene ($:\!CCl_2$).
Step 2: Nucleophilic Attack. The primary amine, acting as a nucleophile with its nitrogen lone pair, attacks the electron-deficient dichlorocarbene carbon atom.
Step 3: Elimination. Successive elimination of $HCl$ molecules driven by base extraction leads to the creation of the stable carbon-nitrogen triple bond found in isocyanides ($R-N\equiv C$).
Step 2: Nucleophilic Attack. The primary amine, acting as a nucleophile with its nitrogen lone pair, attacks the electron-deficient dichlorocarbene carbon atom.
Step 3: Elimination. Successive elimination of $HCl$ molecules driven by base extraction leads to the creation of the stable carbon-nitrogen triple bond found in isocyanides ($R-N\equiv C$).
Q9. Case-Based Question: Diazonium Salts (4 Marks)
Context: Diazonium salts serve as brilliant synthetic intermediates in organic chemistry. They are prepared via diazotization of primary aromatic amines at cool temperatures ($0-5^\circ C$). Once formed, the diazonium group can be cleanly replaced by an array of nucleophiles like $-Cl$, $-Br$, $-CN$, $-I$, and $-OH$, allowing chemists to bypass traditional substitution restrictions on aromatic rings.
(a) Write the chemical reagents required for the Sandmeyer reaction to convert benzenediazonium chloride to chlorobenzene. (1 Mark)
(b) How does Gattermann's reaction differ from the Sandmeyer reaction? (1 Mark)
(c) Convert Benzenediazonium chloride to Fluorobenzene. Name the reaction involved. (2 Marks)
Step 1: Part (a) Resolution. Cuprous chloride dissolved in hydrochloric acid ($Cu_2Cl_2/HCl$).
Step 2: Part (b) Resolution. In the Sandmeyer reaction, halogen ions are introduced using copper salts ($Cu_2Cl_2$ or $Cu_2Br_2$). In Gattermann's reaction, fresh copper powder mixed with the corresponding halogen acid ($Cu/HCl$ or $Cu/HBr$) is used instead.
Step 3: Part (c) Resolution. React benzenediazonium chloride with fluoroboric acid ($HBF_4$) to precipitate benzenediazonium fluoroborate. Heating this dry precipitate yields fluorobenzene. This process is called the Schiemann Reaction (or Balz-Schiemann Reaction). $$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + BF_3 + N_2 \uparrow$$
Step 2: Part (b) Resolution. In the Sandmeyer reaction, halogen ions are introduced using copper salts ($Cu_2Cl_2$ or $Cu_2Br_2$). In Gattermann's reaction, fresh copper powder mixed with the corresponding halogen acid ($Cu/HCl$ or $Cu/HBr$) is used instead.
Step 3: Part (c) Resolution. React benzenediazonium chloride with fluoroboric acid ($HBF_4$) to precipitate benzenediazonium fluoroborate. Heating this dry precipitate yields fluorobenzene. This process is called the Schiemann Reaction (or Balz-Schiemann Reaction). $$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + BF_3 + N_2 \uparrow$$
Section C: Assertion-Reason Questions (1 Mark Each)
Directions: Choose (A) if both Assertion and Reason are true and Reason is correct, (B) if both are true but Reason is incorrect, (C) if Assertion is true and Reason is false, or (D) if Assertion is false and Reason is true.
Q10. Assertion (A): Hoffmann's ammonolysis of alkyl halides usually produces a complex mixture of primary, secondary, and tertiary amines along with quaternary salts.
Reason (R): The primary amine formed initially acts as a nucleophile and continues to attack the remaining alkyl halide.
Correct Answer Choice: A
Q11. Assertion (A): In a gaseous phase, the basic strength of alkylamines follows the strict order: $3^\circ > 2^\circ > 1^\circ > NH_3$.
Reason (R): In the gas phase, the basic strength depends purely on the $+I$ inductive effect of the electron-donating alkyl groups without hydration interference.
Correct Answer Choice: A
Q12. Assertion (A): Aniline does not undergo Friedel-Crafts alkylation or acylation reactions.
Reason (R): The catalyst anhydrous $AlCl_3$ is a strong Lewis acid that coordinates with the basic nitrogen lone pair of aniline, creating a highly deactivating positive salt complex.
Correct Answer Choice: A
Q13. Assertion (A): Tertiary amines display a sharp, high-intensity absorption band in infrared spectroscopy for $N-H$ stretching frequencies.
Reason (R): Tertiary amine structures contain exactly one single hydrogen atom bound directly to nitrogen.
Correct Answer Choice: D
Step 1: Validation. Both statements are false. Tertiary amines possess zero $N-H$ bonds, meaning they do not show $N-H$ stretching bands in IR spectroscopy.
Step 1: Validation. Both statements are false. Tertiary amines possess zero $N-H$ bonds, meaning they do not show $N-H$ stretching bands in IR spectroscopy.
Q14. Assertion (A): Nitration of aniline with a concentrated acid mixture at high temperatures produces a surprisingly high yield (around 47%) of m-nitroaniline.
Reason (R): Under highly acidic conditions, aniline gets protonated into the anilinium ion, which acts as a powerful meta-directing group.
Correct Answer Choice: A
Q15. Assertion (A): Benzenediazonium chloride can be safely stored at room temperature for extended periods.
Reason (R): The diazonium group is stabilized by the extensive resonance cloud of the aromatic benzene ring.
Correct Answer Choice: D
Step 1: Validation. The Assertion is false. Aromatic diazonium salts are stable only in cold solutions ($0-5^\circ C$) and decompose rapidly at room temperature.
Step 1: Validation. The Assertion is false. Aromatic diazonium salts are stable only in cold solutions ($0-5^\circ C$) and decompose rapidly at room temperature.
Common Mistakes Students Make
- Ethyl vs Methyl Trends: Many students mix up the basic strength orders in aqueous solutions. Remember:
For Methyl groups: $2^\circ > 1^\circ > 3^\circ > NH_3$
For Ethyl groups: $2^\circ > 3^\circ > 1^\circ > NH_3$ - Gabriel Phthalimide for Aniline: Never use Gabriel Phthalimide synthesis to try to make aniline. Aryl halides will not react with the phthalimide ion.
- Anilinium Ion Direction: When asked why nitration yields a meta product, explicitly specify that it is due to the formation of the Anilinium Ion under acidic conditions.
Exam Preparation Tips
- Create an Amine Flowchart: Sketch a single-page overview connecting Nitrobenzene, Aniline, Diazonium salts, and substituted benzenes. This visual summary is perfect for last-minute review.
- Learn Nitrogen Balancing: In name reactions like Hoffmann Bromamide, balanced equations with correct coefficients ($4NaOH$) are essential for scoring full marks.
- Master Distinction Tests: Expect a structural distinction question on your exam. Master the chemical differences between $1^\circ$, $2^\circ$, and $3^\circ$ amines (using Hinsberg's reagent) and between aniline and aliphatic amines (using the Azo-dye test).
Frequently Asked Questions (FAQs)
1. Why is Gabriel Phthalimide synthesis preferred over ammonolysis for making primary amines?
Ammonolysis produces a complex mixture of $1^\circ, 2^\circ, 3^\circ$ amines and quaternary ammonium salts that are difficult to separate. Gabriel Phthalimide synthesis cleanly yields pure $1^\circ$ aliphatic amines.
2. What is Hinsberg's Reagent and how is it used?
It is Benzenesulfonyl chloride ($C_6H_5SO_2Cl$). It reacts with $1^\circ$ amines to form a product soluble in alkali, reacts with $2^\circ$ amines to form a product insoluble in alkali, and does not react with $3^\circ$ amines at all.
3. How heavy is Chapter 13 Amines in the CBSE Class 12 board examinations?
It typically carries a weightage of about 6 marks, making it a concise and highly rewarding section of organic chemistry.
4. Where can I download the fully updated NCERT Solutions Class 12 Chemistry PDF?
You can access the complete, updated text solutions right here on examspark.in in a clean, mobile-optimized format.
5. Why are aliphatic amines stronger bases than ammonia?
Aliphatic alkyl groups exert an electron-donating $+I$ inductive effect, which increases the electron density on the nitrogen atom and makes its lone pair more available for protonation compared to ammonia.
Conclusion: Amines is an accessible and rewarding chapter if you focus on basic strength trends, name reactions, and diazonium salt conversions. Practice writing out structural mechanisms regularly, verify your distinction test equations, and solve as many previous years' questions as possible. Stay focused, and let examspark.in help you secure an outstanding score in your board exams!
More Class 12 Chemistry Chapters
Ch 1: The Solid State
Ch 2: Solutions
Ch 3: Electrochemistry
Ch 4: Chemical Kinetics
Ch 5: Surface Chemistry
Ch 6: General Principles
Ch 7: The p-Block Elements
Ch 8: The d- & f-Block Elements
Ch 9: Coordination Compounds
Ch 10: Haloalkanes
Ch 11: Alcohols & Phenols
Ch 12: Aldehydes & Ketones
Ch 13: Amines
Ch 14: Biomolecules
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