Carbon and its Compounds Class 10 important questions with answers

What comes in the exam

• Carbon and its Compounds is part of the Chemical Substances unit, which is worth 25 marks in all.
• This chapter brings many 5-mark questions, the biggest share, then 1-mark MCQs, and some 3-mark, Assertion–Reason and 4-mark case-study questions.
• Almost every year you get the reactions of ethanol and ethanoic acid (oxidation, esterification, saponification, dehydration), many of them as 'identify X, Y and Z' questions.
• Why carbon forms covalent bonds and so many compounds, homologous series, addition reactions and how soap cleans also come almost every year.
• Naming compounds comes often.

Top 10 Carbon and its Compounds questions from board papers

Real CBSE questions, printed as they appeared. Open each answer to see the marking and the steps.

1. Q1 Short answer · 3 marks CBSE · 2026

   Why carbon forms many compounds

   (i) State any two reasons for carbon forming a large number of compounds. Why does carbon form compounds mainly by covalent bonds? (ii) Identify the compound which contains aldehyde as its functional group: C3H7OH, C3H7Cl, C2H5CHO, CH3COCH3

   Answer and marking

   1. (i) First reason: carbon links to other carbon atoms. Catenation: carbon bonds with other carbon atoms to form long chains. 0.5 marks
   2. Second reason: its valency and size. Tetravalency and small size give strong, stable bonds. 0.5 marks
   3. Explain why carbon shares electrons instead of forming ions. Carbon has 4 valence electrons ; it cannot form C⁴⁺ or C⁴⁻ (needs too much energy, unstable) ; so it shares electrons and forms covalent bonds. 1 mark
   4. (ii) Look for the –CHO group. C₂H₅CHO 1 mark

   Answer: (i) catenation, tetravalency; cannot form C⁴⁺/C⁴⁻ so shares electrons (ii) C₂H₅CHO

   Marks: Two reasons 1 · Covalent reason 1 · Aldehyde 1

   For full marks: Say why ions are impossible: losing or gaining four electrons needs too much energy.

   Where students lose marks: Picking CH3COCH3: it has a C=O inside the chain, which is a ketone, not an aldehyde.

   CBSE marking scheme

   (i) Catenation – carbon bonds with other carbon atoms forming long chains (½); tetravalency and small size give strong, stable bonds (½). Carbon has 4 valence electrons; it cannot form C4+ or C4− ions (too much energy, unstable), so it shares electrons forming covalent bonds (1). (ii) C2H5CHO (1).

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2. Q2 Short answer · 3 marks CBSE · 2026

   Compounds A, B, C from C2H6O

   A neutral organic compound 'A' (molecular formula C2H6O) reacts with acidified K2Cr2O7 and on oxidation gives an acidic compound 'B'. Compound 'A' reacts with compound 'B' on warming, in the presence of concentrated H2SO4 to give a sweet smelling substance 'C'. What are A, B and C ? Also write their structures.

   Answer and marking

   1. A neutral C₂H₆O that oxidises to an acid is ethanol. A = ethanol, C₂H₅OH 0.5 marks
   2. Ethanol oxidises to the acid. B = ethanoic (acetic) acid, CH₃COOH 0.5 marks
   3. Acid plus alcohol with conc. H₂SO₄ gives a sweet-smelling ester. C = ethyl ethanoate (ester), CH₃COOC₂H₅ 0.5 marks
   4. Write the structural formula of ethanol, showing the –OH group. A: CH₃–CH₂–OH 0.5 marks
   5. Draw ethanoic acid with the –COOH group. B: CH₃–C(=O)–OH 0.5 marks
   6. Draw the ester. C: CH₃–C(=O)–O–CH₂–CH₃ 0.5 marks

   Answer: A ethanol, B ethanoic acid, C ethyl ethanoate

   Marks: Three names 1½ · Three structures 1½

   For full marks: Give a structure for each of A, B and C — each one is a separate ½.

   Where students lose marks: Naming C as ethanoic acid or ethene: the sweet smell shows an ester, ethyl ethanoate.

   CBSE marking scheme

   A - ethanol, C2H5OH (½); B - ethanoic (acetic) acid, CH3COOH (½); C - ethyl ethanoate (ester), CH3COOC2H5 (½). Correct structural formulae of A, B and C (½ each). (Alternatively the equations C2H5OH → CH3COOH with acidified K2Cr2O7, and CH3COOH + C2H5OH → CH3COOC2H5 + H2O with conc. H2SO4.)

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3. Q3 Case study · 4 marks CBSE · 2025

   Case: hydrocarbons and flames

   All compounds of carbon which contain only carbon and hydrogen are called hydrocarbons. Among these, saturated hydrocarbons are called 'alkanes', unsaturated compounds with one or more double bonds are called 'alkenes' and compounds containing one or more triple bonds are called 'alkynes'. (i) Identify the following compound and write its name: CH3−C≡CH (1) (ii) How do saturated and unsaturated carbon compounds differ in terms of the flame produced by them on burning? (1) (iii) Which type of hydrocarbons undergo addition reactions? Show with an example. (2) OR (iii) What are structural isomers? Draw two structural isomers of butane (C4H10). (2)

   Answer and marking

   1. (i) Three carbons and a triple bond: name it with the -yne ending. Propyne 1 mark
   2. (ii) Saturated compounds burn completely, so give their flame first. Saturated compounds burn with a clean blue flame. 0.5 marks
   3. Unsaturated compounds burn incompletely, so their flame is different. Unsaturated compounds burn with a yellow, sooty flame. 0.5 marks
   4. (iii) Only compounds with double or triple bonds can add atoms across the bond. Unsaturated hydrocarbons undergo addition reactions. 1 mark
   5. Show hydrogen adding across a double bond with a nickel or palladium catalyst. R₂C=CR₂ + H₂ →(Ni/Pd catalyst) R₂CH–CHR₂ 1 mark
   6. (iii) Define the term: same formula, different arrangement. Compounds with the same molecular formula but different structures are called structural isomers. 1 mark
   7. Draw butane as a straight chain and as a branched chain. n-butane: CH₃–CH₂–CH₂–CH₃ ; isobutane: CH₃–CH(CH₃)–CH₃ 1 mark

   Answer: (i) Propyne ; (ii) saturated: blue clean flame, unsaturated: yellow sooty flame ; (iii) (a) unsaturated hydrocarbons, R₂C=CR₂ + H₂ → R₂CH–CHR₂ / (b) same molecular formula, different structures; n-butane and isobutane

   Marks: (i) 1 · (ii) 1 · (iii) 2

   For full marks: For the isomers, both structures must show C₄H₁₀ — count four carbons and ten hydrogens in each.

   Where students lose marks: Naming saturated hydrocarbons as the ones that add hydrogen, or writing the example without the catalyst.

   CBSE marking scheme

   (i) Propyne (1) (ii) Saturated compounds: blue/clean flame (½); unsaturated compounds: yellow/sooty flame (½) (iii) Unsaturated hydrocarbons undergo addition reactions (1); example: R2C=CR2 + H2 →(Pd/Ni) R2CH–CHR2 (or any other example) (1) OR (iii) Compounds with identical molecular formula but different structures are called structural isomers (1); n-butane CH3−CH2−CH2−CH3 and isobutane CH3−CH(CH3)−CH3 (½+½)

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4. Q4 Case study · 4 marks CBSE · 2025

   Case: salts A, B and hard water

   'A' and 'B' are two salts used for washing purposes. Salt 'A' is used for bathing also. Four test tubes I, II, III and IV as mentioned below are taken. I. Rain water + solution of salt 'A'; II. Rain water + solution of salt 'B'; III. Tubewell water + solution of salt 'A'; IV. Tubewell water + solution of salt 'B'. The test tubes are shaken one by one almost identically for the same time and the lengths of foam formed in each test tube is noted. (i) In which one of the four test tubes is the foam formed the minimum? (1) (ii) Differentiate between salt A and salt B. (1) (iii) What are esters? What happens when an ester reacts with an alkali (say sodium hydroxide)? Give chemical equation for the reaction. (2) OR What is the cause of hardness of water? Sometimes it is observed that while bathing foam is formed with difficulty and an insoluble substance is formed. Name this substance and write the cause of its formation. (2)

   Answer and marking

   1. (i) Tubewell water is hard and soap lathers least in hard water. Test tube III (tubewell water + salt A, the soap) 1 mark
   2. (ii) A is soap and B is detergent; give one clear difference. A (soap) is a sodium or potassium salt of a long-chain carboxylic acid and gives no lather with hard water ; B (detergent) is a sodium salt of a sulphonic acid and lathers even in hard water 1 mark
   3. (iii) Say what an ester is and how it is made. Esters are sweet-smelling substances formed when an alcohol reacts with a carboxylic acid. 0.5 marks
   4. An alkali breaks the ester back up; name what forms and the reaction. With NaOH an ester gives back the alcohol and the sodium salt of the acid (saponification). 0.5 marks
   5. Write the equation for ethyl ethanoate with sodium hydroxide. CH₃COOC₂H₅ + NaOH → C₂H₅OH + CH₃COONa 1 mark
   6. (iii) Name the salts dissolved in water that make it hard. Hardness is caused by calcium and magnesium salts in water. 0.5 marks
   7. Name the insoluble substance seen while bathing. Scum 0.5 marks
   8. Explain how the scum forms. Soap reacts with the calcium and magnesium salts to form an insoluble substance (scum). 1 mark

   Answer: (i) Test tube III ; (ii) A soap, B detergent ; (iii) (a) esters; alcohol + sodium salt; CH₃COOC₂H₅ + NaOH → C₂H₅OH + CH₃COONa / (b) Ca and Mg salts; scum formed by soap reacting with them

   Marks: (i) 1 · (ii) 1 · (iii) 2

   For full marks: In (ii) say which salt is soap and which is detergent, then give one difference about hard water.

   Where students lose marks: Picking test tube IV — detergent lathers well even in hard water, so the least foam is with soap (III).

   CBSE marking scheme

   (i) Test tube III, tubewell (hard) water + salt A (soap) (1). (ii) A = soap, Na/K salt of long-chain carboxylic acid, no lather with hard water; B = detergent, sodium salt of sulphonic acid / ammonium salt, lathers even with hard water (any one) (1). (iii) Esters: sweet-smelling substances formed from alcohol + carboxylic acid (½); with NaOH converted back to alcohol and salt of the acid (saponification) (½); CH3COOC2H5 + NaOH → C2H5OH + CH3COONa (1). OR Ca and Mg salts in water (½); scum (½); soap reacts with Ca/Mg salts forming an insoluble substance (1).

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5. Q5 Case study · 4 marks CBSE · 2024

   Case: ethanol uses and reactions

   Ethanol (commonly known as alcohol) is an important carbon compound widely used in industries, hospitals, etc. Because it is a very good solvent, it is used in medicines such as cough syrups, tincture iodine and many tonics. As it is a cleaner fuel, ethanol is used as an additive in petrol in most of the developed countries of the world. (i) Write the structure of ethanol. (1) (ii) Name the products formed when ethanol burns in sufficient air (oxygen). (1) (iii) State what happens when 5% solution of alkaline potassium permanganate is added to warm ethanol. Draw the structure of the product formed and state the role of 5% solution of alkaline potassium permanganate in the reaction. (2) OR State what happens when ethanol is heated with excess conc. H2SO4 at 443 K. Draw the structure of the product formed and state the role of conc. H2SO4 in the reaction. (2)

   Answer and marking

   1. (i) Ethanol has two carbons with –OH on the last one. CH₃–CH₂–OH 1 mark
   2. (ii) Burning in enough oxygen gives the full oxidation products; name the first. Carbon dioxide 0.5 marks
   3. Then name the second product. Water 0.5 marks
   4. (iii) Alkaline KMnO₄ adds oxygen, so say what ethanol becomes. Ethanol is oxidised to ethanoic acid. 0.5 marks
   5. Draw the product with its –COOH group. CH₃COOH (CH₃–C(=O)–OH) 1 mark
   6. It gives oxygen to ethanol, so state its role. Alkaline KMnO₄ acts as an oxidising agent. 0.5 marks
   7. (iii) Hot conc. H₂SO₄ removes water from ethanol; name the product. Ethene, C₂H₄, is formed. 0.5 marks
   8. Draw ethene with its double bond. CH₂=CH₂ 1 mark
   9. It takes water out of ethanol, so state its role. Conc. H₂SO₄ acts as a dehydrating agent. 0.5 marks

   Answer: (i) CH₃–CH₂–OH ; (ii) CO₂ and H₂O ; (iii) (a) ethanoic acid CH₃COOH, KMnO₄ oxidising agent / (b) ethene CH₂=CH₂, conc. H₂SO₄ dehydrating agent

   Marks: (i) 1 · (ii) 1 · (iii) 2

   For full marks: Write the role in the exact words 'oxidising agent' or 'dehydrating agent' — each is a separate ½.

   Where students lose marks: Drawing ethene with a single C–C bond, or giving only one product of burning.

   CBSE marking scheme

   (i) Structural formula of ethanol, CH3–CH2–OH (1). (ii) Carbon dioxide and water (½, ½). (iii) Ethanol is oxidised to ethanoic acid (½); structure CH3COOH (1); alkaline KMnO4 acts as an oxidising agent (½). OR Ethene, C2H4, is formed (½); structure CH2=CH2 (1); conc. H2SO4 acts as a dehydrating agent (½).

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6. Q6 Case study · 4 marks CBSE · 2024

   Case: carbon's versatility

   Carbon is a versatile element that forms the basis of all living organisms and many of the things we use. A large variety of compounds is formed because of its tetravalency. Compounds of carbon are formed with oxygen, hydrogen, nitrogen, sulphur, chlorine and many other elements. (i) What are hydrocarbons? (1) (ii) List two properties by virtue of which carbon can form a large number of compounds. (1) (iii) Write the formula of the functional group present in (1) aldehydes, and (2) ketones. Write chemical equation for the reaction that occurs between ethanoic acid and ethanol in the presence of a catalyst. (2) OR (iii) What are structural isomers? Write the structures of two isomers of butane (C4H10). (2)

   Answer and marking

   1. (i) Give the definition: which two elements only. Hydrocarbons are compounds of carbon and hydrogen only. 1 mark
   2. (ii) Name the two special properties of carbon. Tetravalency and catenation 1 mark
   3. (iii) Write the aldehyde group, at the end of the chain. Aldehyde: –CHO 0.5 marks
   4. Write the ketone group, inside the chain. Ketone: >C=O 0.5 marks
   5. Acid and alcohol give an ester and water with an acid catalyst. CH₃COOH + C₂H₅OH →(acid) CH₃COOC₂H₅ + H₂O 1 mark
   6. (iii) Define the term: same formula, different structure. Compounds with identical molecular formula but different structures are structural isomers. 1 mark
   7. Draw butane as a straight chain and as a branched chain. n-butane: CH₃–CH₂–CH₂–CH₃ ; isobutane (2-methylpropane): CH₃–CH(CH₃)–CH₃ 1 mark

   Answer: (i) Compounds of C and H only ; (ii) tetravalency, catenation ; (iii) (a) –CHO, >C=O; CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O / (b) same formula, different structures; n-butane, isobutane

   Marks: (i) 1 · (ii) 1 · (iii) 2

   For full marks: Write the word 'only' in the hydrocarbon definition — carbon and hydrogen only.

   Where students lose marks: Mixing up the groups: writing >C=O for aldehydes and –CHO for ketones.

   CBSE marking scheme

   (i) Compounds of carbon and hydrogen only (1). (ii) Tetravalency and catenation (1). (iii) Aldehyde –CHO (½), ketone >C=O (½); CH3COOH + C2H5OH →(acid) CH3COOC2H5 + H2O (1). OR (iii) Compounds with identical molecular formula but different structures (1); n-butane CH3–CH2–CH2–CH3 and isobutane (2-methylpropane) CH3–CH(CH3)–CH3 (½ + ½).

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7. Q7 Long answer · 5 marks CBSE · 2026

   Benzene, C4+ ion, ethanol, ester

   (i) Write the structural formula of benzene. (ii) State the reason why carbon can neither form C4+ cation nor C4- anion. (iii) What type of bond is formed between carbon and hydrogen? (iv) How is ethanol converted to ethanoic acid? (v) Complete the following chemical equation: CH3COOC2H5 + NaOH →

   Answer and marking

   1. (i) Draw a six-carbon ring with alternate single and double bonds. Benzene C₆H₆: hexagonal ring of six C atoms, alternate single and double bonds, one H on each C 1 mark
   2. (ii) Give a reason for each ion that carbon cannot form. C⁴⁺: losing 4 electrons needs too much energy ; C⁴⁻: a nucleus with 6 protons cannot hold 10 electrons 1 mark
   3. (iii) Carbon and hydrogen share electrons, so name the bond. Covalent bond 1 mark
   4. (iv) Ethanol becomes ethanoic acid by oxidation; name the oxidising agent. By oxidation with alkaline KMnO₄ or acidified K₂Cr₂O₇ 1 mark
   5. (v) The ester splits back into alcohol and the sodium salt of the acid. CH₃COOC₂H₅ + NaOH → C₂H₅OH + CH₃COONa 1 mark

   Answer: (i) Hexagonal ring, alternate double bonds ; (ii) energy / 6 protons can't hold 10 electrons ; (iii) covalent ; (iv) oxidation by alkaline KMnO₄ or acidified K₂Cr₂O₇ ; (v) C₂H₅OH + CH₃COONa

   Marks: (i) 1 · (ii) 1 · (iii) 1 · (iv) 1 · (v) 1

   For full marks: In (iv) name the oxidising agent — 'by oxidation' without KMnO₄ or K₂Cr₂O₇ is incomplete.

   Where students lose marks: Drawing benzene with only single bonds, or forgetting the hydrogen on each carbon.

   CBSE marking scheme

   (i) Hexagonal ring of six C atoms with alternate single and double bonds, one H on each C (1). (ii) Losing 4 electrons needs too much energy; a nucleus with 6 protons cannot hold 10 electrons (1). (iii) Covalent bond (1). (iv) Oxidation with alkaline KMnO4 or acidified K2Cr2O7 (1). (v) CH3COOC2H5 + NaOH → C2H5OH + CH3COONa (1).

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8. Q8 Long answer · 5 marks CBSE · 2026

   Ethanol heated with X, sweet Y

   Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. H2SO4 and observed a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'. (i) Identify 'X', 'Y' and 'Z'. (ii) Write the role of conc. H2SO4 in the reaction. (iii) Write the chemical equations involved and name the reactions.

   Answer and marking

   1. (i) An acid plus ethanol gives a sweet ester, so X is the acid. X = CH₃COOH, ethanoic acid 0.5 marks
   2. The sweet-smelling product is the ester. Y = CH₃COOC₂H₅, ethyl ethanoate (an ester) 0.5 marks
   3. NaOH gives back ethanol and the sodium salt of the acid. Z = CH₃COONa, sodium ethanoate 0.5 marks
   4. (ii) State what the few drops of acid do. Conc. H₂SO₄ acts as a catalyst (and dehydrating agent). 0.5 marks
   5. (iii) Write the first reaction with acid over the arrow. CH₃COOH + C₂H₅OH →(acid) CH₃COOC₂H₅ + H₂O 1 mark
   6. Name it. Esterification 0.5 marks
   7. Write the reaction of the ester with sodium hydroxide. CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH 1 mark
   8. Name it. Saponification 0.5 marks

   Answer: X = CH₃COOH, Y = CH₃COOC₂H₅, Z = CH₃COONa ; H₂SO₄ catalyst ; esterification and saponification

   Marks: Identify X, Y, Z 1½ · Role ½ · Two equations 2 · Two names 1

   For full marks: Name both reactions — 'esterification' and 'saponification' each carry ½ beside their equation.

   Where students lose marks: Writing Z as ethanoic acid instead of sodium ethanoate — with NaOH you get the sodium salt.

   CBSE marking scheme

   (i) X = CH3COOH, ethanoic acid; Y = CH3COOC2H5, ethyl ethanoate (ester); Z = CH3COONa, sodium ethanoate (½ each). (ii) Catalyst / dehydrating agent (½). (iii) CH3COOH + C2H5OH → CH3COOC2H5 + H2O (acid catalyst) (1), esterification (½); CH3COOC2H5 + NaOH → CH3COONa + C2H5OH (1), saponification (½).

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9. Q9 Long answer · 5 marks CBSE · 2026

   Series; esterification; micelles

   (i) What is homologous series? Which two of the following organic compounds belong to the same homologous series? C2H4, C4H10, C3H4, C4H8 (ii) Write the product(s) in the following reactions: (I) CH3–CH2–OH + CH3COOH →(acid) (II) C2H5OH + O2 →(burning) (iii) Why does micelle formation take place when soap is added to water?

   Answer and marking

   1. (i) Define the series by its functional group and the CH₂ step. Homologous series: compounds with the same functional group replacing H in a carbon chain (same general formula) ; successive members differ by CH₂ (14 u) 1 mark
   2. Check each formula against CₙH₂ₙ; two of them fit. C₂H₄ and C₄H₈ (alkenes) 1 mark
   3. (ii)(I) Alcohol and acid with an acid catalyst give an ester and water. CH₃COOC₂H₅ + H₂O 1 mark
   4. (II) Burning ethanol gives carbon dioxide and water. C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O 1 mark
   5. (iii) Explain how each end of soap behaves. The ionic end interacts with water and the carbon chain with oil/dirt, so the molecules arrange into micelles. 1 mark

   Answer: (i) C₂H₄ and C₄H₈ ; (ii) CH₃COOC₂H₅ + H₂O ; 2CO₂ + 3H₂O ; (iii) ionic end in water, tail in oil → micelles

   Marks: Definition 1 · Same series 1 · Two products 2 · Micelle reason 1

   For full marks: Mention 'differ by CH₂' in the definition — it shows you know what links the members.

   Where students lose marks: Picking C₃H₄ with C₄H₈ — C₃H₄ fits CₙH₂ₙ₋₂ (alkyne), not CₙH₂ₙ.

   CBSE marking scheme

   (i) Series of compounds with the same functional group replacing H in a carbon chain / same general formula, successive members differ by CH2 (14 u) (1); C2H4 and C4H8 (alkenes) (1). (ii)(I) CH3COOC2H5 + H2O (1). (II) C2H5OH + 3O2 → 2CO2 + 3H2O (1) (balancing may be ignored). (iii) The ionic end of soap interacts with water and the carbon chain with oil/dirt, so molecules arrange into micelles (1).

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10. Q10 Long answer · 5 marks CBSE · 2026

    C4+ ion; series; K2Cr2O7

    (i) Carbon cannot form C4+ cation and C4– anion. Explain. (2) (ii) Define homologous series of carbon compounds. Why are the melting and boiling points of C4H10 higher than C3H8 or C2H6? (2) (iii) What happens when ethanol is heated with acidified potassium dichromate? (1)

    Answer and marking

    1. (i) Give a reason for each ion that carbon cannot form. C⁴⁺: losing four electrons needs too much energy ; C⁴⁻: a nucleus with six protons cannot hold ten electrons 2 marks
    2. (ii) Define the series by its common functional group. A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain 1 mark
    3. Melting and boiling points rise with molecular mass, so compare masses. C₄H₁₀ has a higher molecular mass, so its melting and boiling points are higher. 1 mark
    4. (iii) Acidified K₂Cr₂O₇ is an oxidising agent. Ethanol is oxidised to ethanoic acid: CH₃CH₂OH →(acidified K₂Cr₂O₇, heat) CH₃COOH 1 mark

    Answer: (i) Too much energy for C⁴⁺; 6 protons can't hold 10 electrons ; (ii) higher molecular mass ; (iii) ethanoic acid

    Marks: C⁴⁺ and C⁴⁻ reasons 2 · Definition 1 · Molecular mass 1 · Oxidation 1

    For full marks: Give a reason for BOTH ions — C⁴⁺ and C⁴⁻ each carry one mark.

    Where students lose marks: Saying only 'carbon is tetravalent' without explaining why it cannot gain or lose four electrons.

    CBSE marking scheme

    (i) Losing four electrons needs too much energy, so no C4+; a nucleus with six protons cannot hold ten electrons, so no C4– (2). (ii) Series of compounds in which the same functional group substitutes for hydrogen in a carbon chain (1); C4H10 has a higher molecular mass (1). (iii) Ethanol is oxidised to ethanoic acid: CH3CH2OH → (acidified K2Cr2O7, heat) CH3COOH (1).

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Free 10-question test: Carbon and its Compounds

MCQ and Assertion–Reason questions from CBSE papers. About 10 minutes. Your score shows straight away.

Common mistakes in Carbon and its Compounds

How to write for full marks

• For 'identify X, Y and Z', give the name and the formula of each, then the equations that link them. Each identification and each equation carries its own ½ or 1 mark.
• After each equation, name the reaction: oxidation, esterification, saponification, hydrogenation or dehydration. The name often carries its own ½ or 1 mark.
• When asked the role of a reagent, answer in a few words: conc. H₂SO₄ in esterification is a catalyst (or dehydrating agent), and at 443 K it is a dehydrating agent.
• For 'what are soaps' 5-mark questions, answer each part in order: definition, structure, cleansing action with micelles, the problem with hard water, and how it is solved. Each part carries 1 mark.
• For 'give reason' parts, write one clear sentence for each reason. Each carries 1 mark.

Not in the 2026-27 exam

Leave these out of your revision.

• Ethanol and ethanoic acid beyond their properties and uses
• Information in the NCERT boxes

Questions students ask