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Class 11 · CBSE

Organic Chemistry: Some Basic Principles and Techniques — NCERT Class 11 Chemistry

Complete NCERT Solutions for CBSE Class 11 Chemistry Chapter 12 — all intext and chapter-end exercise questions on hybridization, IUPAC nomenclature, isomerism, inductive effect, resonance, purification methods and qualitative analysis answered clearly.

Updated 11 June 202613 min read

How to Use This Page

Work through each NCERT question independently first. After writing your own answer, compare it with the solution below. Pay close attention to the reasoning steps and IUPAC naming conventions — these are the skills most frequently tested in CBSE exams.


Key Concepts (Quick Recap)

  • Tetravalency of carbon: Carbon forms four covalent bonds (sp³, sp², or sp hybridisation)
  • Hybridization: sp³ (tetrahedral, 109.5°), sp² (trigonal planar, 120°), sp (linear, 180°)
  • Homologous series: compounds differing by CH₂ unit, same functional group, similar properties
  • IUPAC nomenclature: identify parent chain, number for lowest locants, name substituents alphabetically, assign suffix by functional group priority
  • Structural isomerism: chain, position, functional group, metamerism
  • Stereoisomerism: geometric (cis/trans or E/Z) and optical (enantiomers, d/l, ±)
  • Inductive effect: +I (electron-donating, e.g., alkyl groups) and –I (electron-withdrawing, e.g., –NO₂, –CN, halogens)
  • Resonance: delocalisation of π-electrons over conjugated systems; resonance hybrid is more stable than any contributing structure
  • Hyperconjugation: σ–π conjugation, stabilises carbocations and alkenes (no-bond resonance)
  • Electrophiles: electron-deficient species (Lewis acids), e.g., H⁺, NO₂⁺, R⁺, SO₃, BF₃
  • Nucleophiles: electron-rich species (Lewis bases), e.g., OH⁻, CN⁻, NH₃, H₂O, RO⁻
  • Purification methods: crystallisation, distillation, chromatography (column, TLC, paper, gas)
  • Qualitative analysis: Lassaigne's test for N, S, halogens, P; Dumas and Kjeldahl methods for nitrogen
  • Quantitative analysis: estimation of C, H, N, halogens, sulphur, phosphorus

Intext Questions

Page 308–310 (Tetravalency, Hybridization, Shapes)

Q1. What are hybridisation states of each carbon atom in the compounds CH₂=C=O, CH₃CH=CH₂, (CH₃)₂C=O, CH₂=CHCN, C₆H₆?
Answer
  • CH₂=C=O: The first C (CH₂=) is sp², the second C (=C=O) is sp (cumulene, two double bonds)
  • CH₃CH=CH₂: C₁ (CH₃) is sp³, C₂ (CH=) is sp², C₃ (=CH₂) is sp²
  • (CH₃)₂C=O: Both CH₃ carbons are sp³, the central C is sp² (ketone)
  • CH₂=CHCN: C₁ (CH₂=) is sp², C₂ (=CH) is sp², C₃ (CN) is sp
  • C₆H₆ (benzene): All six carbon atoms are sp² hybridised
Q2. What is the shape of the following molecules: (a) CH₃⁺ (b) CH₃⁻ (c) CH₃·?
Answer

(a) CH₃⁺ (carbocation): The carbon is sp² hybridised with an empty p-orbital. Shape is trigonal planar, bond angle ≈ 120°. (b) CH₃⁻ (carbanion): The carbon is sp³ hybridised with a lone pair. Shape is pyramidal (like NH₃), bond angle ≈ 109.5°. (c) CH₃· (free radical): The carbon is sp² hybridised with an unpaired electron in the p-orbital. Shape is trigonal planar, bond angle ≈ 120°.

Q3. Explain why (CH₃)₃C⁺ is more stable than CH₃⁺.
Answer

(CH₃)₃C⁺ (tert-butyl cation) is stabilised by hyperconjugation. The nine α-hydrogen atoms from three methyl groups participate in hyperconjugation (σ C–H → p orbital overlap), delocalising the positive charge. In CH₃⁺ (methyl cation), there are no α-hydrogens for hyperconjugation, so it is the least stable carbocation. More hyperconjugative structures = more stability.

Q4. What is the relationship between the members of the following pairs? (a) CH₃CH₂CH₂CH₃ and CH₃CH(CH₃)CH₃ (b) CH₃CH₂OH and CH₃OCH₃ (c) CH₃CH₂CH₂CH₃ and CH₃CH₂CH₂CH₂CH₃
Answer

(a) Chain isomers — both are C₄H₁₀ (butane and 2-methylpropane) (b) Functional group isomers — both are C₂H₆O (ethanol and methoxymethane; alcohol vs ether) (c) They are NOT isomers — they have different molecular formulas (C₄H₁₀ and C₅H₁₂); they are homologues (differ by CH₂)

Page 318–320 (IUPAC Nomenclature Rules)

Q5. Write IUPAC names for: (a) CH₃CH₂CH₂CH₂CH₃ (b) CH₃CH(CH₃)CH₂CH₃ (c) (CH₃)₃CCH₂CH₃ (d) CH₃CH₂C(CH₃)₂CH₂CH₃
Answer

(a) Pentane (b) 2-Methylbutane (c) 2,2-Dimethylbutane (d) 3,3-Dimethylpentane

Q6. Write IUPAC names for: (a) CH₃CH=CHCH₃ (b) CH₃C≡CH (c) CH₂=CH–CH=CH₂
Answer

(a) But-2-ene (b) Propyne (c) Buta-1,3-diene

Q7. Write IUPAC names for: (a) CH₃CH₂OH (b) CH₃CHO (c) CH₃COOH (d) CH₃NH₂
Answer

(a) Ethanol (or Ethan-1-ol) (b) Ethanal (c) Ethanoic acid (d) Methanamine

Q8. Which of the following represents the correct IUPAC name for CH₃CH₂CH₂COOH?
Answer

The correct IUPAC name is butanoic acid. The parent chain has 4 carbons (butane), and the functional group is the carboxylic acid (–COOH), which receives the suffix -oic acid with locant 1 (implied).

Page 333–335 (Isomerism, Inductive Effect, Resonance)

Q9. What are the different types of structural isomerism?
Answer

Structural isomerism is classified into: 1. Chain isomerism: different carbon skeletons (e.g., n-butane and isobutane) 2. Position isomerism: same functional group at different positions (e.g., propan-1-ol and propan-2-ol) 3. Functional group isomerism: different functional groups (e.g., ethanol and methoxymethane) 4. Metamerism: different alkyl groups on either side of a polyvalent functional group (e.g., ethoxyethane and methoxypropane)

Q10. Explain the inductive effect with examples.
Answer

The inductive effect is the permanent polarisation of a σ-bond due to the electron-withdrawing or electron-donating nature of an atom or group, transmitted through σ-bonds.

  • +I effect (electron-donating): alkyl groups push electron density towards the bonded carbon, e.g., –CH₃, –C₂H₅. These stabilise carbocations.
  • –I effect (electron-withdrawing): electronegative atoms pull electron density away, e.g., –NO₂, –CN, –COOH, halogens. These destabilise carbocations but stabilise carbanions.

Order of +I: –C(CH₃)₃ > –CH(CH₃)₂ > –C₂H₅ > –CH₃ > –H Order of –I: –NO₂ > –CN > –COOH > –F > –Cl > –Br > –I

Q11. What is resonance? Explain with the example of benzene.
Answer

Resonance is the phenomenon in which a molecule cannot be represented by a single Lewis structure but is described as a hybrid of two or more contributing (canonical) structures. The actual molecule (resonance hybrid) is more stable than any individual contributing structure.

In benzene, the two Kekulé structures (with alternating single and double bonds) are the main contributing structures. The actual benzene ring has all C–C bonds of equal length (139 pm, between typical single and double bond lengths), and the π-electrons are delocalised over all six carbon atoms. The resonance stabilisation (delocalisation energy) of benzene is about 150 kJ mol⁻¹.

Q12. What is hyperconjugation? How does it stabilise a carbocation?
Answer

Hyperconjugation (also called no-bond resonance or σ–π conjugation) is the delocalisation of σ-electrons of C–H bonds into an adjacent empty or partially filled p-orbital or π-orbital. In a carbocation, the empty p-orbital on the positively charged carbon overlaps with the σ-orbital of an adjacent C–H bond, allowing the σ-electrons to partially delocalise. This spreads the positive charge and stabilises the carbocation. More α-hydrogens → more hyperconjugation structures → greater stability: 3° > 2° > 1° > methyl.

Page 351–353 (Purification Methods, Chromatography)

Q13. Differentiate between distillation and fractional distillation.
Answer

Simple distillation: Used when the boiling points of two liquids differ significantly (>25–30 K). The mixture is heated, the more volatile component vapourises first, condenses, and is collected separately.

Fractional distillation: Used when the boiling points of components are close (<25 K). A fractionating column packed with glass beads provides multiple vapourisation-condensation cycles, improving separation efficiency. Each cycle enriches the vapour in the more volatile component.

Q14. What is chromatography? Name the different types.
Answer

Chromatography is a technique for separating mixtures based on differences in the rate at which components move through a stationary phase under the influence of a mobile phase. Types include: 1. Column chromatography: stationary phase = alumina/silica gel column; mobile phase = liquid solvent 2. Thin-layer chromatography (TLC): stationary phase = silica gel on glass plate; mobile phase = solvent 3. Paper chromatography: stationary phase = cellulose (paper); mobile phase = solvent 4. Gas chromatography: stationary phase = liquid coated on solid support; mobile phase = inert gas (He, N₂)

Q15. What is Lassaigne's test? How is it performed?
Answer

Lassaigne's test (sodium fusion test) is used to detect nitrogen, sulphur, and halogens in organic compounds. The organic compound is heated with sodium metal, which converts the elements into ionic form:

  • For nitrogen: Na + C + N → NaCN; NaCN + FeSO₄ → Na₄[Fe(CN)₆]; with FeCl₃ → Prussian blue colour
  • For sulphur: Na + S → Na₂S; Na₂S + sodium nitroprusside → violet colour
  • For both N and S: Na + C + N + S → NaSCN; with FeCl₃ → blood red colour
  • For halogens: Na + X → NaX; NaX + AgNO₃ → AgX (precipitate); colour tells which halogen

Chapter-End Exercise

Q1. What are electrophiles and nucleophiles? Explain with examples.
Answer

Electrophiles are electron-seeking, electron-deficient species that accept an electron pair. They are Lewis acids. Examples: H⁺, CH₃⁺, NO₂⁺, BF₃, AlCl₃, SO₃, Br⁺, Cl⁺

Nucleophiles are nucleus-seeking, electron-rich species that donate an electron pair. They are Lewis bases. Examples: OH⁻, CN⁻, CH₃O⁻, NH₃, H₂O, ROH, Cl⁻, Br⁻, I⁻

Some species can act as both: H₂O (lone pairs make it nucleophilic; H can be electrophilic).

Q2. Classify the following as electrophiles or nucleophiles: H⁺, CH₃COO⁻, AlCl₃, OH⁻, C₂H₅⁺, NH₃, BF₃, CN⁻, NO₂⁺, H₃O⁺
Answer
  • Electrophiles: H⁺, AlCl₃, C₂H₅⁺, BF₃, NO₂⁺, H₃O⁺
  • Nucleophiles: CH₃COO⁻, OH⁻, NH₃, CN⁻
Q3. Write the IUPAC names for the following compounds: (a) CH₃CH₂CH₂CH₂CH₂CH₃ (b) CH₃CH(CH₃)CH(CH₃)CH₃ (c) CH₃CH₂C≡CCH₃ (d) CH₃CH=CHCHO
Answer

(a) Hexane (b) 2,3-Dimethylbutane (or 2,3-dimethylbutane) (c) Pent-2-yne (d) But-2-enal

Q4. Write structural formulas for: (a) 3-Methylbut-1-ene (b) 2-Ethylbutanoic acid (c) 2-Chloro-3-methylpentane (d) Hex-2-en-1-ol
Answer

(a) CH₂=CH–CH(CH₃)–CH₃ (b) CH₃CH₂CH(CH₂CH₃)COOH (c) CH₃–CH(Cl)–CH(CH₃)–CH₂–CH₃ (d) CH₃–CH=CH–CH₂–CH₂–CH₂OH → corrected: CH₃CH=CHCH₂OH (Wait, hex-2-en-1-ol: CH₃CH₂CH₂CH=CHCH₂OH... let me reconsider)

Hex-2-en-1-ol: CH₃–CH₂–CH₂–CH=CH–CH₂OH

Q5. Draw the resonance structures of: (a) CH₂=CH–CHO (b) CH₂=CH–CH=CH₂ (c) C₆H₅–NO₂ (d) CH₃COO⁻
Answer

(a) Propenal (CH₂=CH–CHO): Structure I: CH₂=CH–CH=O Structure II: CH₂⁻–CH=CH–O⁺ (or ⁻CH₂–CH=CH=O⁺)

(b) Buta-1,3-diene (CH₂=CH–CH=CH₂): Structure I: CH₂=CH–CH=CH₂ Structure II: CH₂⁻–CH=CH–CH₂⁺ (or CH₂=CH–CH=CH₂ with delocalised π-cloud)

(c) Nitrobenzene (C₆H₅–NO₂): The NO₂ group shows resonance with charge separation: N⁺(=O)–O⁻ ↔ N⁺(–O⁻)=O. Combined with benzene ring resonance.

(d) Acetate ion (CH₃COO⁻): Structure I: CH₃–C(=O)–O⁻ Structure II: CH₃–C(–O⁻)=O The two structures are equivalent, making both C–O bonds equal in length (partial double bond character).

Q6. Which of the following compounds show resonance: CH₃Cl, CH₃CH=CH₂, C₆H₆, CH₃CH₂OH, CH₂=CH–CH=CH₂?
Answer

Resonance is shown by compounds with conjugated π-systems or groups capable of electron delocalisation:

  • CH₃Cl: No resonance (no π-bonds or lone pairs adjacent to π-systems)
  • CH₃CH=CH₂: No resonance (isolated π-bond without conjugation)
  • C₆H₆: Yes, resonance (delocalised π-electrons over the ring)
  • CH₃CH₂OH: No resonance
  • CH₂=CH–CH=CH₂: Yes, resonance (conjugated diene)
Q7. Explain the following with examples: (a) Inductive effect (b) Resonance effect (c) Hyperconjugation
Answer

(a) Inductive effect: Permanent polarisation of σ-bonds due to electronegativity differences, transmitted through the chain. Example: In CH₃–CH₂–Cl, the Cl atom withdraws electron density (–I effect), making C-1 slightly positive and C-2 less positive.

(b) Resonance effect (mesomeric effect): Delocalisation of π-electrons in conjugated systems through π-bonds and lone pairs. +R effect: groups that donate electron density through resonance (e.g., –OH, –NH₂, –OR). –R effect: groups that withdraw electron density through resonance (e.g., –NO₂, –CN, –COOH).

(c) Hyperconjugation: Delocalisation of σ-electrons of C–H bonds into an adjacent empty or partially filled p-orbital or π-orbital. Example: In propene (CH₃–CH=CH₂), the three C–H bonds of the methyl group can overlap with the adjacent π-bond, stabilising the alkene. This is why more substituted alkenes are more stable.

Q8. What are the different methods of purification of organic compounds?
Answer

1. Crystallisation: Based on difference in solubilities; impure compound dissolved in hot solvent, filtered, cooled to crystallise. 2. Distillation: For volatile liquids — simple (large BP difference), fractional (close BP), vacuum (high BP or thermally sensitive). 3. Sublimation: For substances that sublime (convert directly from solid to vapour). 4. Chromatography: Separation based on differential adsorption/partition between stationary and mobile phases. 5. Differential extraction: Using two immiscible solvents to separate organic and inorganic components.

Q9. Describe the principle of chromatography.
Answer

Chromatography is based on the differential migration of components of a mixture between a stationary phase and a mobile phase. Components with greater affinity (adsorption or partition) for the stationary phase move slowly, while those with less affinity move faster with the mobile phase. This differential movement leads to separation. The component that is more strongly adsorbed remains near the point of application, while the less adsorbed component moves farther.

Q10. Explain Lassaigne's test for nitrogen.
Answer

Organic compound + Na → NaCN (sodium cyanide) The sodium fusion extract is boiled with FeSO₄ (ferrous sulphate) to form sodium hexacyanoferrate(II): 6NaCN + FeSO₄ → Na₄[Fe(CN)₆] + Na₂SO₄ On adding FeCl₃ and HCl: Na₄[Fe(CN)₆] + 4FeCl₃ → Fe₄[Fe(CN)₆]₃ (Prussian blue) + 12NaCl A Prussian blue colour confirms nitrogen.

Q11. Describe the Dumas method for estimation of nitrogen.
Answer

In the Dumas method, a known mass of organic compound is heated with copper oxide in a CO₂ atmosphere. Nitrogen is converted to N₂ gas: Compound + CuO → N₂ + CO₂ + H₂O The gaseous mixture passes over heated copper gauze (removes excess O₂), then through a KOH solution (absorbs CO₂). The volume of N₂ gas is measured, and the percentage of nitrogen is calculated: %N = (28 × Volume of N₂ at STP × 100) / (22400 × mass of compound)

Q12. How is sulphur detected by Lassaigne's test?
Answer

Organic compound + Na → Na₂S (sodium sulphide) Two tests: (a) Sodium fusion extract + sodium nitroprusside → violet colour (confirms S) (b) Sodium fusion extract + acetic acid + lead acetate → black precipitate of PbS (confirms S)

Q13. Explain the Kjeldahl method for estimation of nitrogen.
Answer

1. Organic compound + conc. H₂SO₄ + K₂SO₄ (boiling point elevator) + CuSO₄ (catalyst) → (NH₄)₂SO₄ 2. (NH₄)₂SO₄ + NaOH (excess) → NH₃ gas 3. NH₃ is absorbed in known volume of standard HCl 4. Unreacted HCl is titrated with standard NaOH 5. %N = (1.4 × M × (V₁–V₂)) / W, where M = molarity of HCl, V₁ = volume of HCl initially, V₂ = volume of HCl remaining, W = mass of compound

Q14. Give the hybridisation and shape of the following: (a) CH₄ (b) C₂H₄ (c) C₂H₂ (d) CH₃Cl
Answer

(a) CH₄: sp³ hybridised, tetrahedral (109.5°) (b) C₂H₄: sp² hybridised at each C, planar (120°) (c) C₂H₂: sp hybridised at each C, linear (180°) (d) CH₃Cl: sp³ hybridised at C, tetrahedral (approx. 109.5°)

Q15. Explain the term homologous series with an example.
Answer

A homologous series is a series of organic compounds in which each successive member differs by a –CH₂– unit. Members have the same general formula, similar chemical properties, and a gradual change in physical properties. Example: Alkanes — CH₄ (methane), C₂H₆ (ethane), C₃H₈ (propane), C₄H₁₀ (butane)... General formula: CₙH₂ₙ₊₂. Boiling points increase regularly; chemical reactions follow the same pattern.

Q16. Identify the functional groups in: (a) CH₃COOH (b) CH₃CHO (c) CH₃CH₂OH (d) CH₃COCH₃
Answer

(a) CH₃COOH — Carboxylic acid (–COOH) (b) CH₃CHO — Aldehyde (–CHO) (c) CH₃CH₂OH — Alcohol (–OH) (d) CH₃COCH₃ — Ketone (>C=O)

Q17. What is the difference between distillation, fractional distillation, and distillation under reduced pressure?
Answer
  • Simple distillation: For liquids with a large boiling point difference (>25–30 K). The mixture is heated and the vapour of the more volatile component is condensed.
  • Fractional distillation: For liquids with close boiling points (<25 K). Uses a fractionating column for repeated vapourisation-condensation cycles.
  • Distillation under reduced pressure (vacuum distillation): For liquids that decompose at or below their boiling points at atmospheric pressure. Reducing the pressure lowers the boiling point, allowing distillation at a lower temperature.
Q18. Explain the mechanism of electrophilic substitution in benzene.
Answer

Electrophilic substitution in benzene occurs in two steps: Step 1 — Attack of electrophile (slow, rate-determining): The electrophile (E⁺) attacks the π-electron cloud of benzene, forming a resonance-stabilised arenium ion (sigma complex/Wheland intermediate) in which the carbon bearing the electrophile becomes sp³ hybridised. Step 2 — Loss of proton (fast): A base removes the proton from the sp³ carbon, restoring aromaticity. The overall reaction is substitution (one H replaced by E).

Example — Nitration: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O; NO₂⁺ attacks benzene → arenium ion → loss of H⁺ → nitrobenzene.


Internal Links

  • [Chapter 12 Guide — Organic Chemistry Basics](/school/cbse/class/11/chemistry/cbse-organic-chemistry-basics)
  • [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
  • [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)

Frequently Asked Questions

IUPAC naming follows these steps: (1) Identify the longest carbon chain as the parent chain, (2) Number the chain to give the lowest locants to the substituents/functional group (lowest sum rule), (3) Name substituents as prefixes in alphabetical order, (4) Assign the appropriate suffix based on the principal functional group. For multiple identical substituents, use di-, tri-, tetra- prefixes.
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