India's #1 AI Tutorformula-sheet · Chemistry · Chapter 9

Class 12 Chemistry Chapter 9 Amines — Formulas & Key Points

Q: What is the difference between primary, secondary, and tertiary amines?

Primary (1°) amines have one alkyl/aryl group on nitrogen (RNH₂), secondary (2°) have two (R₂NH), and tertiary (3°) have three (R₃N). The classification affects reactivity, basicity, and behavior in tests like Hinsberg and carbylamine.

Q: Why is the basicity order different in gas phase and aqueous solution?

In gas phase, only the +I effect of alkyl groups matters, so 3° > 2° > 1° > NH₃. In aqueous solution, solvation stabilizes smaller cations better, reversing the order to 2° > 1° > 3° > NH₃ for aliphatic amines because the bulky tertiary ammonium ion is less solvated.

Chapter 9 Amines in NCERT Class 12 Chemistry covers nitrogen-containing organic compounds that act as bases. This formula sheet organizes every reaction—preparation via Gabriel synthesis, reduction, Hoffmann degradation, diazonium salt chemistry, and distinction tests—into tables for rapid revision. Each formula is paired with the exact conditions and common pitfalls. Whether you are solving CBSE board numericals or memorizing reactions the night before your exam, this page gives you one-stop access to all key points.

Your child's private AI tutor — trained on NCERT.

3-day free trial · ₹1 to start · Cancel anytime.

Start 3-day free trial →

Key takeaways

✓Amines are classified as 1°, 2°, or 3° based on the number of alkyl/aryl groups attached to nitrogen, affecting their reactivity and basicity order.
✓Gabriel synthesis converts phthalimide to primary aliphatic amines; Hoffmann bromamide degradation shortens the carbon chain by one atom.
✓Basicity order in gaseous phase differs from aqueous phase due to solvation: aliphatic amines are stronger bases than aromatic amines in both.
✓Diazonium salts (Ar–N₂⁺X⁻) are formed only from primary aromatic amines at 0–5 °C and undergo coupling, Sandmeyer, and replacement reactions.
✓Carbylamine test (foul smell) is specific for primary amines; Hinsberg reagent (benzenesulfonyl chloride) distinguishes 1°, 2°, and 3° amines.
✓Aromatic amines are weaker bases than ammonia because the lone pair on nitrogen delocalizes into the benzene ring (resonance).
✓All amines show basic character due to the lone pair on nitrogen; they form salts with acids and can be acylated or alkylated.

Classification and Nomenclature of Amines

Amines are derivatives of ammonia (NH₃) where one or more hydrogen atoms are replaced by alkyl or aryl groups. Classification depends on how many carbon groups are bonded to the nitrogen atom. Primary amines (1°) have one carbon group, secondary (2°) have two, and tertiary (3°) have three. The same nitrogen can appear in different positions within a molecule, so functional isomerism is common. IUPAC nomenclature uses the suffix '-amine' for simple amines; for complex molecules, the amino group is treated as a substituent ('-amino'). Common names use 'alkylamine' format. Aromatic amines like aniline (C₆H₅NH₂) are named with the parent aromatic ring. Quaternary ammonium salts carry four carbon groups on nitrogen and bear a permanent positive charge. The table below summarizes classification with examples, essential for CBSE board exams and NCERT Class 12 Chemistry solutions.

Primary (1°): One carbon group on N, e.g. CH₃NH₂ (methylamine), C₆H₅NH₂ (aniline).
Secondary (2°): Two carbon groups on N, e.g. (CH₃)₂NH (dimethylamine), C₆H₅NHCH₃ (N-methylaniline).
Tertiary (3°): Three carbon groups on N, e.g. (CH₃)₃N (trimethylamine), C₆H₅N(CH₃)₂ (N,N-dimethylaniline).
Quaternary ammonium salts: Four carbon groups, e.g. [(CH₃)₄N]⁺Cl⁻ (tetramethylammonium chloride).
IUPAC: Longest chain + suffix '-amine' or prefix 'amino-' when amine is substituent.

Structure and Physical Properties

Nitrogen in amines is sp³ hybridized with a pyramidal geometry around it; the lone pair occupies the fourth sp³ orbital. This lone pair is responsible for basicity and hydrogen bonding. Primary and secondary amines can form intermolecular hydrogen bonds (N–H⋯N), leading to higher boiling points than hydrocarbons of similar molecular mass but lower than alcohols. Tertiary amines lack N–H bonds and cannot hydrogen-bond among themselves, so they have lower boiling points than 1° and 2° amines of comparable mass. Lower aliphatic amines (up to C₃) are gases with a fishy smell; C₄–C₁₂ are liquids, and higher amines are solids. Aromatic amines like aniline are liquids at room temperature but turn brown on standing due to air oxidation. Solubility in water decreases with increasing molar mass because the hydrophobic alkyl part dominates. The bond angle at nitrogen is approximately 108° (close to tetrahedral 109.5°), slightly compressed by the lone pair's repulsion. These points appear frequently in CBSE Class 12 Chemistry Chapter 9 theory questions.

Hybridization: sp³; geometry: pyramidal; bond angle ≈108°.
Hydrogen bonding: 1° and 2° amines form N–H⋯N bonds; 3° amines do not.
Boiling point order: 1° > 2° > 3° (for same carbon number) and amines < alcohols.
Lower amines (C₁–C₃) are gases; C₄–C₁₂ liquids; higher amines solids.
Solubility in water decreases with molecular mass; all amines dissolve in dilute acids.

All Preparation Methods — Formulas and Conditions

NCERT Class 12 Chemistry prescribes six major routes to prepare amines, each suited to different starting materials and target amine classes. The table below lists every method with exact reagents, conditions, and product types. Reduction of nitro compounds (using Sn/HCl, Fe/HCl, or catalytic H₂/Ni) is the most common industrial route for aromatic amines. Reduction of nitriles (using LiAlH₄ or H₂/Ni) yields primary amines with one extra carbon than the nitrile. Reduction of amides (LiAlH₄) also gives primary amines but retains the same carbon count as the amide. Gabriel phthalimide synthesis is specific for 1° aliphatic amines and avoids over-alkylation. Hoffmann bromamide degradation shortens the amide by one carbon, producing a 1° amine. Reductive amination of aldehydes or ketones with NH₃ and a reducing agent (H₂/Ni or NaBH₃CN) gives 1°, 2°, or 3° amines depending on stoichiometry. Memorize reagents and conditions—CBSE board exams test these in 3-mark numerical problems and 5-mark reaction-sequence questions.

Basicity Order and pKₐ Values

Amines are basic because the lone pair on nitrogen can accept a proton, forming the ammonium ion RNH₃⁺. The strength of this basicity depends on electron availability at nitrogen and solvation effects. In the gas phase (no solvent), the basicity order is 3° > 2° > 1° > NH₃ because alkyl groups are electron-donating (+I effect), stabilizing the positive charge on the conjugate acid. In aqueous solution, however, the order changes to 2° > 1° > 3° for aliphatic amines due to solvation: smaller ions are better solvated, and tertiary ammonium ions are crowded, making them less stable. For aromatic amines, the lone pair delocalizes into the benzene ring (resonance), reducing electron density on nitrogen and making them much weaker bases than aliphatic amines and even ammonia. Electron-donating groups on the ring (like –CH₃, –OCH₃) increase basicity; electron-withdrawing groups (–NO₂, –Cl, –CN) decrease it. Aniline has pKₐ (of conjugate acid) ≈4.6, whereas methylamine has pKₐ ≈10.6. This 6-unit difference is crucial for CBSE Class 12 Chemistry Chapter 9 distinction questions and reaction predictions.

Gas phase: 3° > 2° > 1° > NH₃ (only +I effect matters).
Aqueous phase (aliphatic): 2° > 1° > 3° > NH₃ (solvation + inductive effects).
Aromatic amines: weaker than NH₃ due to resonance delocalization of lone pair.
Electron-donating substituents on benzene ring increase basicity; electron-withdrawing decrease it.
pKₐ of conjugate acid: higher pKₐ = stronger base. Methylamine (10.6) > Aniline (4.6).

Chemical Reactions of Amines — Comprehensive Table

Amines undergo four main reaction types: alkylation and acylation (both nucleophilic substitution on carbon), reactions with nitrous acid (differ by amine class), and electrophilic substitution on the aromatic ring (for aromatic amines). Alkylation with alkyl halides (R–X) can produce 1°, 2°, 3° amines and even quaternary salts; it is difficult to control and gives mixtures. Acylation with acid chlorides (RCOCl) or anhydrides yields N-substituted amides and is cleaner. Primary amines react with nitrous acid (NaNO₂/HCl) to form alcohols (aliphatic 1°) or diazonium salts (aromatic 1°, stable only at 0–5 °C). Secondary amines give yellow oily N-nitroso compounds; tertiary aliphatic amines form soluble salts, while tertiary aromatic amines undergo electrophilic para-nitrosation. Aromatic amines activate the benzene ring (ortho/para-directing) but react vigorously; protecting the –NH₂ as acetamide moderates reactivity. The table below consolidates every NCERT-prescribed reaction with reagents and products. This is the single most important section for Class 12 Chemistry solutions and board exam reaction-writing questions.

Diazonium Salts — Formation and Reactions

Diazonium salts have the general formula Ar–N≡N⁺X⁻ (X = Cl, Br, HSO₄) and are formed exclusively from primary aromatic amines by treatment with sodium nitrite and dilute HCl at 0–5 °C (diazotization). Aliphatic diazonium salts are too unstable even at low temperature and decompose immediately to alcohols and N₂. Aromatic diazonium salts are stable in cold aqueous solution but decompose on warming. They are powerful intermediates in synthesis because the diazo group (–N₂⁺) is an excellent leaving group. Reactions fall into two categories: replacement reactions (where –N₂⁺ is replaced by another group) and coupling reactions (where the diazo group stays and couples with activated aromatic compounds). Sandmeyer reactions (CuCl, CuBr, CuCN) replace –N₂⁺ with –Cl, –Br, –CN. Gattermann reactions use Cu/HX for –Cl or –Br. Replacement with –F uses HBF₄ (Balz-Schiemann), with –I uses KI, with –OH uses warm water, and with –H uses H₃PO₂. Coupling with phenols or aromatic amines in alkaline solution produces brightly colored azo dyes (–N=N– linkage). This chemistry is central to CBSE Class 12 Chemistry Chapter 9 and appears in nearly every board paper.

Formation: Ar–NH₂ + NaNO₂ + 2 HCl (0–5 °C) → Ar–N₂⁺Cl⁻ + NaCl + 2 H₂O.
Sandmeyer (CuX): –N₂⁺ → –Cl, –Br, –CN.
Gattermann (Cu/HX): –N₂⁺ → –Cl, –Br.
Balz-Schiemann (HBF₄, heat): –N₂⁺ → –F.
With KI: –N₂⁺ → –I.
With H₂O (warm): –N₂⁺ → –OH.
With H₃PO₂: –N₂⁺ → –H (reduction).
Coupling: Ar–N₂⁺ + Ar'–OH (or Ar'–NHR) in alkaline medium → Ar–N=N–Ar' (azo dye, colored).

Distinction Tests for Amines

CBSE board exams frequently ask you to distinguish between primary, secondary, and tertiary amines or between aliphatic and aromatic amines using specific reagents. The carbylamine test is positive only for primary amines: heating with chloroform (CHCl₃) and alcoholic KOH produces isocyanide (R–N≡C) with an unbearable foul smell. Secondary and tertiary amines do not respond. The Hinsberg test uses benzenesulfonyl chloride (C₆H₅SO₂Cl) in aqueous KOH. Primary amines form a sulfonamide that is soluble in alkali (due to acidic N–H); secondary amines form an insoluble sulfonamide (no acidic H); tertiary amines do not react and remain as a separate layer. Aromatic amines can be distinguished from aliphatic ones by the azo-dye test: only aromatic primary amines form stable diazonium salts at 0–5 °C that couple with β-naphthol to give orange-red dye. Nitrous acid reactions also differ: aliphatic 1° → alcohol + N₂ (brisk effervescence); aromatic 1° → diazonium salt (clear solution at 0–5 °C); 2° → yellow oily nitrosoamine; 3° aliphatic → soluble salt (no visible reaction), 3° aromatic → green para-nitroso compound. Master these tests for 2-mark and 3-mark short-answer questions in Class 12 Chemistry solutions.

Carbylamine test (CHCl₃ + alc. KOH, heat): foul smell only for 1° amines.
Hinsberg reagent (C₆H₅SO₂Cl, aq. KOH): 1° → soluble; 2° → insoluble; 3° → no reaction.
Nitrous acid (NaNO₂/HCl): 1° aliph. → alcohol + N₂; 1° arom. → diazonium salt; 2° → yellow oil; 3° aliph. → salt; 3° arom. → green p-nitroso derivative.
Azo-dye test (β-naphthol coupling): orange-red color only for aromatic 1° amines.

Common Mistakes, Sign Conventions, and Unit Reminders

Students often confuse the basicity order in gas phase versus aqueous solution—remember that solvation reverses the simple inductive trend for aliphatic amines. Another frequent error is writing aromatic diazonium salt formation at room temperature; the reaction must be kept at 0–5 °C or the salt decomposes. When writing Hoffmann bromamide degradation, do not forget that the product amine has one carbon fewer than the starting amide. In Gabriel synthesis, the final hydrolysis step requires strong base or acid; weak base will not cleave the phthalimide ring. For nomenclature, the prefix 'N-' indicates that a substituent is on nitrogen, not on the carbon chain (e.g. N-methylaniline vs. o-toluidine). Aromatic amines are often written as C₆H₅NH₂, but make sure to show resonance structures when explaining reduced basicity. In coupling reactions, the phenol or amine partner must be activated (electron-rich); coupling with nitrobenzene does not occur. pKₐ values refer to the conjugate acid (RNH₃⁺), so higher pKₐ means stronger base—students sometimes invert this. Units: basicity constants (Kₐ) are dimensionless in logarithmic form; molecular mass in g/mol. Bond angles in amines are around 108°, not 120° (that is sp² trigonal). These pitfalls cost marks in CBSE board exams, so review them before your Chemistry paper.

Diazonium salt formation: always 0–5 °C, never room temperature.
Hoffmann degradation: product has (n–1) carbons if amide has n carbons.
Gabriel synthesis: final step needs strong base (KOH) or acid (HCl), not weak base.
Nomenclature: 'N-' prefix for nitrogen substituent, e.g. N,N-dimethylaniline.
pKₐ refers to RNH₃⁺; higher pKₐ = stronger base (more stable conjugate acid).
Bond angle at N: ≈108° (sp³ pyramidal), not 120° (sp² planar).
Coupling reactions: partner must be electron-rich (phenol, aniline), not deactivated ring.

Memory Tricks and Mnemonics for Amines

Mnemonics help you recall reaction sequences and reagent lists under exam pressure. For basicity order in aqueous solution, remember 'STOAT': Secondary > Tertiary > One (primary) > Ammonia—though the real order is 2° > 1° > 3° > NH₃, the mnemonic reminds you that secondary is strongest. For Sandmeyer reactions, 'CBC' stands for CuCl (chloro), CuBr (bromo), CuCN (cyano). To remember that Hoffmann degradation loses one carbon, think 'Hoffmann Hacks off one carbon.' Gabriel synthesis is 'Phthalimide Plus Potassium gives Primary amine.' For the carbylamine test, 'Chloroform Creates a Choking smell for 1° only.' Resonance in aniline: 'Aniline's Alone pair Abandoned into the ring, so Basicity Drops.' For diazonium coupling, 'Cold Couples, Warm Walks away' (diazonium salts couple at low temperature but decompose on warming). The order of activating power for electrophilic substitution: 'NOAD'—NH₂ > OH > Alkyl > Deactivators. Use these tricks during quick weekend revision with NCERT Class 12 Chemistry notes, and you will retain formulas faster. Share them with classmates or write them on flashcards for last-minute recap before the CBSE board exam.

Aqueous basicity: 'Second Seat Tops' → 2° > 1° > 3° > NH₃.
Sandmeyer: 'CBC' → CuCl, CuBr, CuCN.
Hoffmann: 'Hacks off one carbon' → product has (n–1) carbons.
Gabriel: 'Phthalimide + Potassium → Primary amine.'
Carbylamine: 'Chloroform Choking smell for 1° only.'
Aniline resonance: 'Alone pair Abandoned → Basicity Drops.'
Diazonium: 'Cold Couples, Warm Walks away.'
Activating power: 'NOAD' → NH₂ > OH > Alkyl > Deactivators.

Three Solved Mini-Examples Applying the Formulas

Example 1: Identify A, B, C in the sequence: CH₃CONH₂ →[Br₂/KOH] A →[NaNO₂/HCl, 0–5°C] B →[CuCN/HCl] C. Solution: Step 1 is Hoffmann bromamide degradation, so A = CH₃NH₂ (methylamine, one carbon less). Step 2 is diazotization, but aliphatic primary amines do not form stable diazonium salts; instead they decompose immediately to alcohols. So actually B = CH₃OH + N₂. Wait—Sandmeyer needs aromatic diazonium. The sequence as written is faulty for aliphatic. Correct interpretation: if the starting material were benzamide (C₆H₅CONH₂), then A = C₆H₅NH₂ (aniline), B = C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride), C = C₆H₅CN (benzonitrile). This example teaches you to check whether the substrate is aromatic or aliphatic before applying diazonium chemistry. Example 2: Arrange in decreasing basicity: aniline, p-nitroaniline, p-methoxyaniline, methylamine. Solution: Methylamine is aliphatic (strongest). Among aromatics, p-OCH₃ is electron-donating (+R), increasing basicity; p-NO₂ is electron-withdrawing (–R), decreasing it. Order: CH₃NH₂ > p-methoxyaniline > aniline > p-nitroaniline. Example 3: A compound C₇H₉N gives a yellow oily product with nitrous acid and is insoluble in aqueous KOH after treatment with benzenesulfonyl chloride. Identify the class of amine. Solution: Yellow oily product with HNO₂ indicates secondary amine (N-nitrosoamine). Insoluble sulfonamide with Hinsberg reagent also confirms 2°. So the compound is a secondary amine, e.g. N-methylaniline (C₆H₅NHCH₃). These examples mirror CBSE board exam 3-mark and 5-mark problems and are ideal practice for Class 12 Chemistry solutions.

One-Glance Last-Minute Revision Box

Use this box 24 hours before your CBSE board exam. Classification: 1° (RNH₂), 2° (R₂NH), 3° (R₃N), 4° (R₄N⁺X⁻). Basicity aqueous: 2° > 1° > 3° > NH₃ (aliphatic); aromatic < NH₃. Preparation: Gabriel (1° aliphatic), Hoffmann (−1 carbon, 1°), reduction of nitro/nitrile/amide. Carbylamine test: 1° only (foul smell). Hinsberg: 1° soluble, 2° insoluble, 3° no reaction. Diazonium formation: Ar–NH₂ + NaNO₂/HCl at 0–5 °C → Ar–N₂⁺Cl⁻. Sandmeyer: CuCl (Cl), CuBr (Br), CuCN (CN). Replacement: HBF₄/heat (F), KI (I), H₂O warm (OH), H₃PO₂ (H). Coupling: Ar–N₂⁺ + phenol/aniline in alkaline → azo dye (colored). Aromatic substitution: –NH₂ is ortho/para-directing, highly activating; protect as –NHCOCH₃ to moderate. Key pKₐ: aniline conjugate acid ≈4.6, methylamine ≈10.6. Common mistakes: writing diazonium at room temp, forgetting −1 carbon in Hoffmann, confusing gas vs. aqueous basicity. Equations to memorize: all in the tables above. Practice writing mechanisms for Gabriel, Hoffmann, and Sandmeyer at least twice before the exam. CBSETUTOR.ai offers a 24×7 AI tutor where you can snap a photo of any amine reaction problem and get instant step-by-step solutions, mechanism explanations, and basicity comparisons—all for ₹999/month, one price for Class 6 to 12, with a 3-day free trial to test before your board exams. Revise this box, solve five past-year CBSE questions, and you are ready for Chapter 9.

Frequently asked questions

What is the difference between primary, secondary, and tertiary amines?+

Primary (1°) amines have one alkyl/aryl group on nitrogen (RNH₂), secondary (2°) have two (R₂NH), and tertiary (3°) have three (R₃N). The classification affects reactivity, basicity, and behavior in tests like Hinsberg and carbylamine.

Why is the basicity order different in gas phase and aqueous solution?+

In gas phase, only the +I effect of alkyl groups matters, so 3° > 2° > 1° > NH₃. In aqueous solution, solvation stabilizes smaller cations better, reversing the order to 2° > 1° > 3° > NH₃ for aliphatic amines because the bulky tertiary ammonium ion is less solvated.

How does Gabriel synthesis ensure only primary amines are formed?+

Phthalimide has one acidic N–H, so it can be deprotonated only once by KOH. The resulting anion undergoes a single SN² alkylation, preventing over-alkylation. Hydrolysis releases a 1° amine, avoiding the mixtures typical of direct alkylation with RX.

Why are aromatic amines weaker bases than aliphatic amines?+

The lone pair on nitrogen in aniline delocalizes into the benzene π-system through resonance, reducing its availability to accept a proton. Aliphatic amines lack this delocalization, so their lone pairs are more available, making them stronger bases.

What happens if diazonium salt formation is done at room temperature?+

Aromatic diazonium salts decompose rapidly above 5 °C, releasing nitrogen gas and forming phenols or other by-products. The reaction must be kept at 0–5 °C to maintain the diazonium ion long enough for further synthetic steps like Sandmeyer or coupling.

Which test distinguishes primary, secondary, and tertiary amines in one step?+

The Hinsberg test using benzenesulfonyl chloride (C₆H₅SO₂Cl) in aqueous KOH. Primary amines give a clear alkaline solution, secondary amines form an insoluble precipitate, and tertiary amines remain unreacted as a separate oily layer.

Why does Hoffmann bromamide degradation produce an amine with one less carbon?+

The mechanism involves rearrangement where the alkyl or aryl group migrates from carbonyl carbon to nitrogen as the bromine and hydroxide ion attack. The carbonyl carbon is lost as CO₃²⁻, so the product amine has (n−1) carbons if the amide had n carbons.

How do electron-withdrawing groups affect the basicity of aniline?+

Groups like –NO₂, –Cl, –CN withdraw electrons from the benzene ring through –I or –R effects, further reducing the already low electron density on nitrogen. This makes the lone pair even less available, decreasing basicity below that of unsubstituted aniline.

What is the Sandmeyer reaction and when is it used?+

Sandmeyer reactions replace the diazonium group (–N₂⁺) with –Cl, –Br, or –CN using copper(I) salts (CuCl, CuBr, CuCN) in HCl. They are used to introduce halogens or cyano groups onto an aromatic ring in positions that are hard to achieve by direct electrophilic substitution.

Can CBSETUTOR.ai help me solve amine reaction mechanisms step-by-step?+

Yes. CBSETUTOR.ai provides a 24×7 AI tutor where you upload a photo of any amine problem—Gabriel synthesis, Hoffmann, Sandmeyer, or coupling—and receive instant worked solutions with mechanisms, electron shifts, and explanations. One flat fee of ₹999/month covers all subjects for Class 6–12, with a 3-day free trial to try before board exams.

Related resources

Important Questions: CBSE Class 12 Chemistry Chapter 9 Amines CBSE Class 12 Chemistry Chapter 9 Amines Worksheet with Answers CBSE Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids Worksheet with Answers Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids — Formulas & Key Points AI Tutor for Class 12: The Smart Alternative to Tuition AI Tutor for Class 12 Accountancy: Learn Faster with Instant Help Online Class 9 Chemistry Tutor in Mumbai — NCERT-Aligned Board Exam Coaching Class 9 Mathematics Chapter 2 Polynomials — Formulas & Key Points

Keep learning — related guides

Class 12Chemistry

Class 12 Chemistry Tutor in Rushikonda, Visakhapatnam

Class 12Chemistry

Class 12 Chemistry Chapter 10 Biomolecules — Formulas & Key Points

Class 12Chemistry

CBSE Class 12 Chemistry Chapter 10 Biomolecules Worksheet with Answers

Class 12Chemistry

Important Questions: CBSE Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids

Class 12Chemistry

Important Questions: CBSE Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers

Class 12Chemistry

Important Questions: CBSE Class 12 Chemistry Chapter 10 Biomolecules

Ready to give your Class 12 child the tutor that never sleeps?

CBSETUTOR.ai covers every chapter in the Class 12 NCERT syllabus — Maths, Science, Social Science, English, Hindi and more. 24×7. Patient. Unlimited. 3-day free trial.

Start your child's 3-day free trial →