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Chemistry — Class 12 — L24: Hydrocarbons

NIOS Code 313 · Module 7 · Chemistry of Organic Compounds

Notes extracted from NIOS Chemistry Course (313), Lesson 24 — Hydrocarbons (313_Chemistry_Eng_Lesson24.pdf). Content covers sections 24.1–24.4.
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Overview — Hydrocarbons

Hydrocarbons contain only carbon and hydrogen. They are fuels, lubricants, solvents and starting materials for almost all organic chemistry. Sources: petroleum (fractional distillation) and coal (destructive distillation). This lesson covers alkanes, alkenes, alkynes and aromatic hydrocarbons (benzene) — preparation, physical and chemical properties, tests, aromaticity and directive effects.

Section 1: Alkanes (24.1)

Saturated hydrocarbons CnH2n+2 — little reactivity (“paraffins”).

Preparation: (1) Haloalkanes — Zn/HCl, HI/red P, H₂/Pt; Grignard RMgX + H₂O/ROH → RH; Wurtz 2RX + 2Na → R–R. (2) Hydrogenation of alkenes/alkynes (Ni/Pt/Pd). (3) Alcohols/aldehydes/ketones + HI/red P. (4) Carboxylic acids: soda lime decarboxylation (RCOONa + NaOH/CaO → RH); reduction with HI; Kolbe electrolysis of RCOO⁻ → R–R.

2RX + 2Na → R–R  |  RMgX + H₂O → RH  |  RCOONa + NaOH → RH
Wurtz · Grignard · soda-lime decarboxylation — classic alkane routes
Ethane Conformations Staggeredmore stable Eclipsedhigher energy Rotation about C–C σ bond · staggered preferred (~12.5 kJ mol⁻¹ lower)
Conformations of ethane — staggered is the energy minimum.

Physical: C₁–₄ gases, C₅–₁₇ liquids, higher solids. b.p. rises with mass; branching lowers b.p. (smaller surface). Density < water. Melting points irregular (packing of odd/even chains).

Chemical: Free-radical halogenation (Cl₂, light) — initiation, propagation, termination; F₂ > Cl₂ > Br₂ > I₂. Combustion (complete → CO₂; limited O₂ → CO). Cracking at high T. Isomerisation with AlCl₃/HCl. Uses: fuels, LPG (propane/butane), solvents, methane for synthesis.

Cl₂ + hν → 2Cl·  →  CH₄ → CH₃Cl → … → CCl₄
Free-radical chain halogenation of methane

Section 2: Alkenes (24.2)

Unsaturated CnH2n with C=C (“olefins”). Prep: dehydrohalogenation of RX with alc. KOH (Saytzeff — more substituted alkene major); dehydration of alcohols (H₂SO₄ or Al₂O₃).

Markovnikov Addition of HBr propene H⁺ 1° C⁺ minor 2° C⁺ major 2-Br H to C with more H · Br to more substituted C · peroxide reverses HBr
Markovnikov’s rule explained by preferential formation of the more stable carbocation.

Addition: H₂/Ni; X₂ (Br₂ colour test); HX (Markovnikov; HBr + peroxide = anti-Markovnikov); H₂O/H⁺; H₂SO₄; polymerisation → polyethene.

Oxidation: cold alk. KMnO₄ (Baeyer) → diol, purple discharged; hot KMnO₄ cleaves C=C; O₂/Ag → ethylene oxide; ozonolysis → carbonyls; combustion.

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃  |  + HBr/ROOR → CH₃CH₂CH₂Br
Markovnikov vs anti-Markovnikov (peroxide effect) for HBr only

Uses: plastics (polyethene), ethanol, ethanal, antifreeze (ethanediol).

Section 3: Alkynes (24.3)

CnH2n−2 with C≡C. Ethyne: CaC₂ + 2H₂O → HC≡CH; dihaloalkanes + alc. KOH; RC≡CNa + R′X for higher alkynes.

Physical: gases then liquids/solids with size; slightly polar (higher b.p. than alkanes); ethyne garlic odour; slightly soluble in water, soluble in acetone.

Chemical: sequential addition of H₂, X₂, HX; H₂O/Hg²⁺/H₂SO₄ → ethanal (via enol); oxidation with KMnO₄; combustion (oxyacetylene flame ~2800°C); ozonolysis to dicarbonyls without chain break.

Acidic nature: sp carbon 50% s-character → more electronegative → terminal ≡C–H acidic; Na or NaNH₂ → acetylides; ammoniacal AgNO₃ (white) or Cu₂Cl₂ (red) ppt for terminal alkynes only.

% s-Character & Acidity sp³ alkane25% s sp² alkene33% s sp alkyne50% s · acidic H Higher s-character → more acidic terminal hydrogen
Hybridisation explains why ethyne forms metal acetylides but ethane does not.
CaC₂ + 2H₂O → HC≡CH  |  HC≡CH + 2Na → NaC≡CNa + H₂
Industrial/lab ethyne · acidic hydrogen confirmation

Distinction table: Br₂ and Baeyer — positive for alkenes and alkynes, not alkanes. Ammoniacal AgNO₃/Cu₂Cl₂ — only terminal alkynes.

Uses: welding, fruit ripening, synthesis of ethanal, polymers, Orlon.

Section 4: Aromatic Hydrocarbons — Benzene (24.4)

From coal tar (destructive distillation). Formula C₆H₆ — unsaturated by addition of H₂/Cl₂, but prefers electrophilic substitution; does not decolourise Br₂ water or Baeyer’s reagent under mild conditions.

Structure: Kekulé alternate double bonds inadequate (all C–C 139 pm; one ortho product). Resonance hybrid of two forms; circle notation; resonance energy ~150 kJ mol⁻¹ (heat of hydrogenation lower than expected for three double bonds). MO: six sp² carbons, delocalised π cloud above and below ring.

Benzene — Resonance & Aromaticity Hückel: 4n+2 n=1 → 6 e⁻ planar · conjugated C–C all 139 pm Resonance energy ~150 kJ mol⁻¹ · substitution preferred
Benzene is aromatic by Hückel’s rule and resonance-stabilised.
Hückel: 4n + 2 π electrons  |  benzene n = 1
Planar monocyclic conjugated system · aromatic if rule satisfied

Electrophilic aromatic substitution (EAS): halogenation (FeX₃); nitration (NO₂⁺ from HNO₃/H₂SO₄); sulphonation (oleum); Friedel–Crafts alkylation/acylation (AlCl₃).

Directive influence: o/p directors (–OH, –CH₃, –NH₂, –OR); m directors (–NO₂, –COOH, –SO₃H, –CN).

Physical: colourless liquids with characteristic odour; immiscible with water; dissolve fats. b.p. rises with size (benzene 353 K, toluene 383 K).

Carcinogenicity: some polycyclic aromatics (coal tar fractions) are toxic/carcinogenic. Uses: solvents, dyes, drugs, plastics precursors.

C₆H₆ + HNO₃/H₂SO₄ → C₆H₅NO₂  |  + CH₃Cl/AlCl₃ → C₆H₅CH₃
Nitration (NO₂⁺) · Friedel–Crafts alkylation — classic EAS
EAS Reactions of Benzene Halogenation Nitration Sulphonation Friedel–Crafts o/p: OH, CH₃, NH₂ · m: NO₂, COOH, SO₃H
Four major electrophilic substitution reactions and directing effects.

Exam Connections and Chapter Summary

High-yield: alkane prep (Wurtz, Kolbe, soda lime); free-radical chlorination mechanism; ethane conformations; alkene prep and Saytzeff; Markovnikov + mechanism + peroxide exception; Baeyer and Br₂ tests; CaC₂ ethyne; acidic alkyne + acetylide tests; distinction table; benzene structure/resonance/Hückel; EAS reagents; o/p vs m directors.

Builds on L23 (naming, mechanisms, isomerism). Next lessons add functional groups (haloalkanes, alcohols, carbonyls). Practice writing mechanisms for HX addition and free-radical halogenation — they are exam staples.

Intext drills: Grignard and active hydrogen; isomers of pentane and b.p. order; chlorination steps; Saytzeff major product; conditions for hydrogenation; CaC₂ reaction; % s-character ethane/ethene/ethyne; distinction tests; Hückel check for benzene; nitration electrophile; o/p vs m directors. Coal tar fractions: light oil (benzene), middle oil (phenol), heavy oil (naphthalene), green oil (anthracene), pitch.

MCQ Quiz — L24 Hydrocarbons

0 / 10 correct

Flashcards — L24

1 / 18

Golden Rules — L24 Hydrocarbons

Most exam-important points from this chapter:

Alkanes

Prep: Wurtz, Grignard, Kolbe, H₂/Ni, soda lime. Free-radical X₂. Staggered ethane preferred. Fuels & LPG.

Alkenes

Prep by elimination (Saytzeff). Additions: Markovnikov; HBr/peroxide anti-M. Baeyer & Br₂ tests. Polymers.

Alkynes

CaC₂ → ethyne. Acidic ≡CH (sp). Ag⁺/Cu⁺ ppt for terminal. Distinction table vs alkanes/alkenes.

Benzene

Resonance hybrid; Hückel 6π. Substitution not addition. EAS: X₂, NO₂, SO₃H, R, COR.

Directing effects

o/p: activating groups (OH, CH₃). m: deactivating (NO₂, COOH, SO₃H).

Wurtz · Grignard · Kolbe
Staggered > eclipsed
Free-radical halogenation
Saytzeff · Markovnikov
Anti-Markovnikov (HBr/ROOR)
Baeyer’s · Br₂ test
CaC₂ → C₂H₂ · acidic ≡CH
Hückel 4n+2 · benzene
EAS: X₂ · NO₂ · SO₃H · R

Section 1: Alkanes

NIOS Chemistry 313, Module 7 — Hydrocarbons (sections 24.1–24.4).

Preparation of Alkanes

Haloalkane + Zn/HCl or HI/red P or H₂/Pt → RH

RMgX + H₂O/ROH → RH · Wurtz: 2RX + 2Na → R–R

Alkene/alkyne + H₂/Ni → alkane · RCOONa + NaOH/CaO → RH (soda lime)

Kolbe: 2RCOO⁻ electrolysis → R–R at anode

Properties & Conformations

C₁–₄ gases · C₅–₁₇ liquids · higher solids · b.p. ↑ with mass · branching ↓ b.p.

Ethane: staggered more stable than eclipsed (~12.5 kJ mol⁻¹)

Halogenation free radical: Cl₂/hν · initiation · propagation · termination

Combustion → CO₂ + H₂O · cracking · isomerisation (AlCl₃/HCl)

Section 2: Alkenes & Alkynes

Alkenes

Prep: dehydrohalogenation (alc. KOH) · dehydration of alcohols

Saytzeff: more substituted alkene major

Markovnikov: H of HX to C with more H · mechanism via more stable carbocation

Anti-Markovnikov: HBr + peroxide → terminal Br

Tests: Br₂/CCl₄ decolourises · cold alk. KMnO₄ (Baeyer) decolourises · ozonolysis · polymerisation

Alkynes

CaC₂ + 2H₂O → HC≡CH · dihaloalkane + alc. KOH · RC≡CNa + R′X

Addition of H₂, X₂, HX, H₂O/Hg²⁺/H⁺ → ethanal from ethyne

Acidic: sp 50%s · Na or NaNH₂ → acetylide · Ag⁺/Cu⁺ ammoniacal ppt (terminal only)

Distinction: Br₂ & KMnO₄ for unsaturation · AgNO₃/Cu₂Cl₂ only alkynes (terminal)

Section 3: Aromatic Hydrocarbons

Benzene & Aromaticity

C₆H₆ · resonance hybrid · all C–C 139 pm · resonance energy ~150 kJ mol⁻¹

Hückel: planar cyclic conjugated 4n+2 π e⁻ · benzene n=1 (6π)

EAS: halogenation (FeX₃) · nitration (HNO₃/H₂SO₄, NO₂⁺) · sulphonation · Friedel–Crafts R/acyl + AlCl₃

o/p directors: –OH, –CH₃, –NH₂ · m directors: –NO₂, –COOH, –SO₃H

Section 2: Definitions

Paraffin: Alkane — little affinity (low reactivity).

Markovnikov’s rule: In HX addition to unsymmetrical alkene, H goes to C with more hydrogens.

Saytzeff’s rule: Elimination gives more substituted alkene preferentially.

Hückel’s rule: Aromatic if planar, cyclic, conjugated, 4n+2 π electrons.

Baeyer’s reagent: Cold alkaline KMnO₄ — tests unsaturation (purple discharged).

Section 3: Visual Map

L24 Map — Hydrocarbons Alkanes Alkenes Alkynes Benzene Wurtz · Markovnikov · Saytzeff · CaC₂ · Hückel · EAS Tests: Br₂ · Baeyer · Ag⁺/Cu⁺ for terminal alkyne

Section 5: Q&A (12 Questions)

Q1: Wurtz reaction equation?

2CH₃Br + 2Na (dry ether) → CH₃–CH₃ + 2NaBr

Q2: Why staggered ethane more stable?

Less torsional strain; H atoms farther apart than in eclipsed form.

Q3: Free-radical chlorination steps?

Initiation (Cl₂ → 2Cl·); propagation (Cl· + CH₄ → ·CH₃ + HCl; ·CH₃ + Cl₂ → CH₃Cl + Cl·); termination.

Q4: Markovnikov product of propene + HBr?

2-Bromopropane (via secondary carbocation).

Q5: Anti-Markovnikov condition?

HBr in presence of organic peroxide (free-radical path).

Q6: Baeyer’s test observation?

Purple KMnO₄ discharged by alkenes/alkynes (cold alkaline).

Q7: Ethyne from calcium carbide?

CaC₂ + 2H₂O → HC≡CH + Ca(OH)₂

Q8: Why is ethyne acidic?

sp carbon 50% s-character → more electronegative → H more acidic; forms acetylides with Na/NaNH₂.

Q9: Distinguish ethene from ethyne?

Ammoniacal AgNO₃ or Cu₂Cl₂: only terminal alkyne gives ppt; both decolourise Br₂/KMnO₄.

Q10: Hückel rule for benzene?

6 π electrons → 4n+2 with n=1; planar conjugated monocycle.

Q11: Electrophile in nitration of benzene?

NO₂⁺ (nitronium ion) from HNO₃ + H₂SO₄.

Q12: –OH vs –NO₂ directing effect?

–OH ortho/para director; –NO₂ meta director.

Section 6: Tips & Exam Hacks

Memory Aids

  • Markovnikov: "H to rich in H"
  • Saytzeff: "More substituted alkene wins"
  • Peroxide: "HBr only anti-M"
  • s-Character: "sp 50% most acidic H"
  • Hückel: "4n+2 aromatic"

Exam Tips

  • Write mechanism for Markovnikov via 2° carbocation
  • Free radical halogenation needs light/heat
  • Benzene prefers substitution not addition
  • Resonance energy explains benzene stability
  • LPG = propane + butane; ethyne for welding flame

Section 8: Quick Reference

• Alkanes: Wurtz, Grignard, Kolbe, H₂/Ni · free-radical X₂ · combustion

• Alkenes: elimination · Markovnikov / anti-M · Baeyer · Br₂ · ozonolysis

• Alkynes: CaC₂ · acidic ≡CH · acetylides · distinction table

• Benzene: resonance · Hückel · EAS (X, NO₂, SO₃H, R, COR) · o/p vs m directors

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L24 — Hydrocarbons only. Use Model Answer for marking points; Explanation for concept clarity.

L24 — Hydrocarbons

13 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. Alkanes undergo pyrolysis — (A) at a very high pressure and in the presence of air   (B) at a very low temperature and in the absence of air   (C) at a very high temperature and in the absence of air   (D) at a very high temperature and in the presence of air Eoë

1 mark · Q13 · 313/MAY/205A

Model Answer

Answer: (B) at a very low temperature and in the absence of air

Explanation

Pyrolysis is thermal decomposition of alkanes at high temperature.

Map to L24 syllabus. Paper 313/MAY/205A, Q13 (1 mark).

PYQ2. The correct order of stability of 1°, 2° and 3° carbocations is — (A) 1°, 2° Am¡a 3°

1 mark · Q14 · 313/MAY/205A

Model Answer

Answer using key concepts from L24 (definitions, equations, and one example where useful). Stay within the suggested word range for a 1-mark NIOS question.

Explanation

Cross-check with L24 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 1 mark — concise correct choice/fact. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q14 · 1 mark(s) · L24.

PYQ3. Write True (T) for correct statement and False (F) for incorrect statement : Acetylene has a bond order of 3. The lone pair-bond pair repulsion is intermediate between lone pair- lone pair and bond pair-bond pair repulsion. ghr

2 marks · Q19 · 313/MAY/205A

Model Answer

Answer using key concepts from L24 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L24 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q19 · 2 mark(s) · L24.

PYQ4. Complete and balance the following reactions : CaC2 + 2H2O 40% H SO 1% HgSO

2 marks · Q25 · 313/MAY/205A

Model Answer

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂ (ethyne). Used industrially for acetylene generation.

Explanation

Ionic carbide reacts with water; product is the simplest alkyne (L24).

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q25 · 2 mark(s) · L24.

PYQ5. Alkanes undergo pyrolysis — (A) at a very high pressure and in the presence of air   (B) at a very low temperature and in the absence of air   (C) at a very high temperature and in the absence of air   (D) at a very high temperature and in the presence of air Eoë

1 mark · Q5 · 313/MAY/205B

Model Answer

Answer: (B) at a very low temperature and in the absence of air

Explanation

Pyrolysis is thermal decomposition of alkanes at high temperature.

Map to L24 syllabus. Paper 313/MAY/205B, Q5 (1 mark).

PYQ6. The correct order of stability of 1°, 2° and 3° carbocations is — (A) 1°, 2° Am¡a 3°   (B) Note : Question Nos. 17 to 28 are objective type questions of 2 marks each. {ZX}e : à0 g§0 17 go 2

1 mark · Q16 · 313/MAY/205B

Model Answer

Answer: (A) 1°, 2° Am¡a 3°

Explanation

Carbocation stability order: 3° > 2° > 1°.

Map to L24 syllabus. Paper 313/MAY/205B, Q16 (1 mark).

PYQ7. Write True (T) for correct statement and False (F) for incorrect statement : Acetylene has a bond order of 3. The lone pair-bond pair repulsion is intermediate between lone pair- lone pair and bond pair-bond pair repulsion. ghr

2 marks · Q22 · 313/MAY/205B

Model Answer

Answer using key concepts from L24 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L24 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q22 · 2 mark(s) · L24.

PYQ8. Complete and balance the following reactions : CaC2 + 2H2O 40% H SO 1% HgSO

2 marks · Q26 · 313/MAY/205B

Model Answer

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂ (ethyne). Used industrially for acetylene generation.

Explanation

Ionic carbide reacts with water; product is the simplest alkyne (L24).

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q26 · 2 mark(s) · L24.

PYQ9. The correct order of stability of 1°, 2° and 3° carbocations is — (A) 1°, 2° Am¡a 3°

1 mark · Q5 · 313/MAY/205C

Model Answer

Answer using key concepts from L24 (definitions, equations, and one example where useful). Stay within the suggested word range for a 1-mark NIOS question.

Explanation

Cross-check with L24 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 1 mark — concise correct choice/fact. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q5 · 1 mark(s) · L24.

PYQ10. Alkanes undergo pyrolysis — (A) at a very high pressure and in the presence of air   (B) at a very low temperature and in the absence of air   (C) at a very high temperature and in the absence of air   (D) at a very high temperature and in the presence of air Eoë

1 mark · Q15 · 313/MAY/205C

Model Answer

Answer: (B) at a very low temperature and in the absence of air

Explanation

Pyrolysis is thermal decomposition of alkanes at high temperature.

Map to L24 syllabus. Paper 313/MAY/205C, Q15 (1 mark).

PYQ11. Complete and balance the following reactions : CaC2 + 2H2O 40% H SO 1% HgSO

2 marks · Q22 · 313/MAY/205C

Model Answer

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂ (ethyne). Used industrially for acetylene generation.

Explanation

Ionic carbide reacts with water; product is the simplest alkyne (L24).

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q22 · 2 mark(s) · L24.

PYQ12. Write True (T) for correct statement and False (F) for incorrect statement : Acetylene has a bond order of 3. The lone pair-bond pair repulsion is intermediate between lone pair- lone pair and bond pair-bond pair repulsion. ghr

2 marks · Q28 · 313/MAY/205C

Model Answer

Answer using key concepts from L24 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L24 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q28 · 2 mark(s) · L24.

PYQ13. Read the passage given below and answer the following questions (out of four attempt any two) : Nitro compounds are those derivatives of hydrocarbons in which a hydrogen atom is replaced by a nitro group. These may be aliphatic or aromatic. Nitroalkanes are divided into primary, secondary or tertiary depending upon the attachment of nitro group to primary, secondary or tertiary carbon atom respectively. Name the following compound according to the IUPAC nomenclature : CH —CH—CH—CH —CH What happens when propane reacts with nitric acid at 680 K? Give the chemical equation. Write the reduction of nitrobenzene in alkaline medium. Compare the boiling points nitro compounds with the corresponding alkanes. Justify your answer.

2 marks · Q26 · 313/TUS/105A

Model Answer

Present a clear comparison in 2–3 points (definition / structure / property / example). Use a table style in prose: A vs B for each criterion.

Explanation

Comparison answers need parallel points. Award marks for each distinct difference.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q26 · 2 mark(s) · L24.

Problem Solving — L24 Hydrocarbons

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6Types

Draw/label alkane, alkene, alkyne bonding. Give one example of each.

Pencil sketch (labelled)

Hydrocarbon types alkane C–C (single) alkene C=C (double) alkyne C≡C (triple) aromatic: benzene ring
Pencil sketch: bond types in hydrocarbons

Solution — step by step with formulas

  1. Alkane C–C (ethane); alkene C=C (ethene); alkyne C≡C (ethyne).

Final answer: Ethane / ethene / ethyne

Textbook formal language

Hydrocarbons contain only C and H; classified by bonds.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Single, double, triple carbon–carbon bonds define the family.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Hydrocarbon classes

Aromatic hydrocarbons contain benzene-type rings.

Linked to chapter notes (L24). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Markownikov

State Markownikov’s rule for HBr addition to propene.

Solution — step by step with formulas

  1. H adds to carbon with more hydrogens; Br to more substituted carbon → 2-bromopropane major.

Final answer: 2-bromopropane major

Textbook formal language

Electrophilic addition via more stable carbocation.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

H goes where there are already more H’s; Br goes to the busier carbon.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Addition to alkenes

Peroxide effect gives anti-Markownikov with HBr.

Linked to chapter notes (L24). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Benzene

State Hückel’s rule briefly.

Solution — step by step with formulas

  1. Planar cyclic conjugated system with (4n+2) π electrons is aromatic.

Final answer: (4n+2) π electrons

Textbook formal language

Explains special stability of benzene.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Magic electron counts 6,10,14… for aromatic rings.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Aromaticity

Benzene undergoes substitution more than addition.

Linked to chapter notes (L24). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Combustion

Write balanced combustion of CH₄.

Solution — step by step with formulas

  1. CH₄ + 2O₂ → CO₂ + 2H₂O.

Final answer: CH₄ + 2O₂ → CO₂ + 2H₂O

Textbook formal language

Complete combustion yields CO₂ and H₂O.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Burn methane fully: carbon dioxide and water.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Combustion

Incomplete combustion can give CO/soot.

Linked to chapter notes (L24). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Isomer

Give chain isomers of C₄H₁₀.

Solution — step by step with formulas

  1. Butane and 2-methylpropane.

Final answer: n-butane & isobutane

Textbook formal language

Same formula, different carbon skeleton.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Straight chain vs branched.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Chain isomerism

Affects boiling points.

Linked to chapter notes (L24). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Preparation

How is ethene prepared from ethanol in lab (outline)?

Solution — step by step with formulas

  1. Acid-catalysed dehydration (conc. H₂SO₄, heat).

Final answer: Dehydration of ethanol

Textbook formal language

Elimination of water forms the double bond.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Heat ethanol with strong acid—lose water, get ethene gas.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Ethene from ethanol

Industrial: cracking of alkanes.

Linked to chapter notes (L24). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.