CBSE Class 10 Science — Carbon and Its Compounds

Chapter Overview & Weightage

Carbon and Its Compounds is one of the highest-scoring chapters in CBSE Class 10 Science. It carries approximately 8–10 marks in the board exam and is a favourite for both objective and long-answer questions.

This chapter is the foundation of organic chemistry. Board exams test: naming of carbon compounds, properties and reactions of functional groups, and distinguishing between saturated/unsaturated compounds. One 5-mark long-answer question on this chapter is almost guaranteed.

Topic	Marks weightage
Covalent bonding in carbon	1–2 marks
Naming and classification	2–3 marks
Properties/reactions of functional groups	3–4 marks
Soaps, detergents, and their action	1–2 marks

Key Concepts You Must Know

Covalent bonding: Carbon forms 4 covalent bonds (tetravalency). It can bond with C, H, O, N, S, halogens.
Catenation: Carbon’s unique ability to form chains with other carbon atoms — single, double, or triple bonds.
Homologous series: A series of compounds with the same functional group and general formula, differing by –CH₂– (14 mass units).
Functional groups: –OH (alcohol), –CHO (aldehyde), –COOH (carboxylic acid), C=C (alkene), C≡C (alkyne), halogens.
IUPAC naming: Prefix (substituent) + Root (number of carbons) + Suffix (functional group).
Saturated vs unsaturated: Saturated = only single bonds (alkanes). Unsaturated = one or more double/triple bonds (alkenes, alkynes).
Addition reaction: Unsaturated compounds react with $H_2$ , $Cl_2$ , $HCl$ across double/triple bonds.
Substitution reaction: Saturated compounds (alkanes) replace one H with halogen in presence of sunlight.
Combustion: Complete combustion gives CO₂ + H₂O (clean blue flame). Incomplete gives CO + soot (yellow flame).
Oxidation: Alcohols → aldehydes → carboxylic acids using oxidising agents like $K_2Cr_2O_7/H_2SO_4$ .
Esterification: Alcohol + Carboxylic acid $\rightleftharpoons$ Ester + Water (acid catalyst, reversible).
Saponification: Hydrolysis of ester (fat) with NaOH to give soap (sodium salt of fatty acid) + glycerol.
Micelle formation: Soaps work because one end is hydrophobic (fatty acid chain) and the other is hydrophilic (–COO⁻Na⁺). They form micelles to trap oil in water.

Important Formulas

Series	General Formula	Example
Alkane	CₙH₂ₙ₊₂	CH₄, C₂H₆
Alkene	CₙH₂ₙ	C₂H₄, C₃H₆
Alkyne	CₙH₂ₙ₋₂	C₂H₂, C₃H₄
Alcohol	CₙH₂ₙ₊₁OH	CH₃OH, C₂H₅OH
Carboxylic acid	CₙH₂ₙO₂	HCOOH, CH₃COOH

\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}

Ethanoic acid + Ethanol → Ethyl ethanoate (fruity smell) + Water

Solved Previous Year Questions

PYQ 1 — Functional groups and naming (3 marks)

Q: Name the functional groups present in: (a) $\text{CH}_3\text{CHO}$ (b) $\text{CH}_3\text{COOH}$ (c) $\text{C}_2\text{H}_5\text{OH}$ .

Answer: (a) $\text{CH}_3\text{CHO}$ — Aldehyde group (–CHO). Name: Ethanal. (b) $\text{CH}_3\text{COOH}$ — Carboxyl group (–COOH). Name: Ethanoic acid (acetic acid). (c) $\text{C}_2\text{H}_5\text{OH}$ — Hydroxyl group (–OH). Name: Ethanol.

PYQ 2 — Reaction type identification (2 marks)

Q: What type of reaction does ethene undergo with bromine water? Write the equation.

Answer: Addition reaction (unsaturated compound, C=C present).

$\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}\text{--}\text{CH}_2\text{Br}$

Bromine water decolourises — this is a test for unsaturation.

PYQ 3 — Soaps and detergents (5 marks)

Q: Explain the cleansing action of soap. Why don’t soaps work in hard water?

Answer: Soap molecules have a long non-polar (hydrophobic) hydrocarbon tail and a polar (hydrophilic) ionic head (–COO⁻Na⁺). When soap is mixed with oily water, the hydrophobic tails cluster around the oil droplet while the hydrophilic heads face the water. This forms a cluster called a micelle — the oil is trapped inside and can be washed away.

Hard water contains $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ ions. These react with soap to form insoluble calcium/magnesium salts (scum) that are useless for cleaning. Detergents don’t have this problem because their ionic end doesn’t react with these ions.

Difficulty Distribution

Level	Marks	Example
Easy (1 mark)	MCQ, fill in blank	Identify functional group
Medium (2–3 marks)	Short answer	Write reaction, name compound
Hard (5 marks)	Long answer	Explain micelle, full reaction sequence

Expert Strategy

The chapter has two parts: naming/classification (easy to memorise, guaranteed marks) and reactions (needs understanding). Focus first on getting the naming right — prefixes meth-, eth-, prop-, but-, pent- mapped to carbon counts 1–5.

Then learn reactions grouped by functional group. Don’t try to memorise all reactions at once; learn one functional group at a time.

For a 5-mark board answer on carbon compounds, always use: (1) a definition, (2) a balanced equation, (3) conditions above the arrow, (4) a real-world application. This structure earns full marks.

Common Traps

Trap 1: Writing combustion as incomplete by default. For organic compounds burning in excess oxygen, always write the complete combustion product (CO₂ + H₂O), not CO.

Trap 2: Confusing saponification and esterification. Saponification = fat + NaOH → soap (irreversible). Esterification = alcohol + acid → ester + water (reversible, acid catalyst).

Trap 3: Naming the carbon chain from the wrong end. IUPAC rules say: number the chain from the end nearer to the functional group (or branch if no functional group).

Trap 4: Soap vs. detergent. Soaps are sodium salts of long-chain fatty acids (natural). Detergents are synthetic and work in hard water. The question may ask you to distinguish — know both the structural and practical differences.