Class 11 : Chemistry (In English) – Chapter 8: Organic Chemistry-some basic principles and techniques

EXPLANATION & SUMMARY

🔵 Introduction
Organic Chemistry 🧪 is the chemistry of carbon compounds (excluding simple oxides like CO, CO₂, carbonates, etc.). Carbon, due to its tetravalency (4 bonds) and catenation (ability to form chains), creates an enormous variety of compounds 🌍 — fuels, plastics, medicines, DNA, proteins. To study this vast field systematically, we need fundamental rules, nomenclature, classification, electronic effects, and techniques.
This chapter lays down the basic language of Organic Chemistry 📝.

🟢 General Characteristics of Organic Compounds
Carbon is tetravalent (forms 4 bonds).
Shows catenation: ability to form long chains, rings, branches.
Bonds are strong covalent bonds.
Variety in hybridization → sp³ (alkanes), sp² (alkenes), sp (alkynes).
Exhibits isomerism (same formula, different structure).

🔵 Classification of Organic Compounds
1️⃣ Open Chain Compounds (Acyclic)
Straight or branched chains.
Example: CH₃–CH₂–CH₃ (propane).
2️⃣ Closed Chain Compounds (Cyclic)
Ring structures:
Alicyclic (cycloalkanes, e.g., cyclohexane).
Aromatic (benzene, C₆H₆).
3️⃣ Based on Functional Group
Alcohols (–OH), Aldehydes (–CHO), Ketones (–CO–), Carboxylic acids (–COOH), Amines (–NH₂), etc.
4️⃣ Based on Saturation
Saturated: single bonds only (alkanes).
Unsaturated: double/triple bonds (alkenes/alkynes).

🟡 Nomenclature of Organic Compounds
IUPAC rules provide systematic names 🌍.
✔ Steps for IUPAC Naming:
Select longest carbon chain → parent name.
Number the chain → lowest locants for substituents.
Identify and name substituents.
Place substituents alphabetically with positions.
Add suffix for functional group.
Examples:
CH₃–CH₂–CH₃ → Propane.
CH₂=CH–CH₃ → Prop-1-ene.
CH₃–CH(OH)–CH₃ → Propan-2-ol.
💡 Functional Group Priority Order: Carboxylic acid > Aldehyde > Ketone > Alcohol > Alkene > Alkyne > Alkane.

🔴 Isomerism
Compounds with same molecular formula but different arrangements.
Structural Isomerism
Chain (butane vs isobutane).
Position (but-1-ene vs but-2-ene).
Functional group (alcohol vs ether).
Metamerism (ethers, ketones with different alkyl groups).
Tautomerism (keto–enol equilibrium).
Stereoisomerism
Geometrical (cis–trans in alkenes).
Optical (chiral carbon with 4 different groups → enantiomers).

🟢 Electronic Displacements in Covalent Bonds
These effects explain reactivity of organic molecules ⚡.
1️⃣ Inductive Effect (I-effect)
Electron displacement along sigma bond due to electronegativity differences.
–I groups (–NO₂, –Cl) withdraw electrons.
+I groups (–CH₃, –C₂H₅) release electrons.
Influences acidity/basicity.
2️⃣ Electromeric Effect (E-effect)
Temporary electron shift in double bond when attacking reagent approaches.
3️⃣ Resonance Effect
Delocalization of electrons represented by resonance structures.
Example: Benzene (delocalized π electrons → stability).
4️⃣ Hyperconjugation
Delocalization of σ-electrons of C–H bond adjacent to double bond or carbocation.
“No bond resonance.”
Explains stability of carbocations, alkenes.

🔵 Types of Organic Reactions
1️⃣ Substitution → one atom/group replaced.
Example: CH₄ + Cl₂ → CH₃Cl + HCl.
2️⃣ Addition → molecules add across double/triple bonds.
Example: CH₂=CH₂ + H₂ → CH₃–CH₃.
3️⃣ Elimination → atoms/groups removed forming double/triple bonds.
Example: CH₃–CH₂–Br → CH₂=CH₂ + HBr.
4️⃣ Rearrangement → atoms/groups rearrange within molecule.
Example: Pinacol–Pinacolone rearrangement.

🟡 Reaction Intermediates
Organic reactions often proceed through unstable intermediates:
Carbocations (C⁺): positively charged carbon.
Carbanions (C⁻): negatively charged carbon.
Free radicals (C•): carbon with unpaired electron.
Carbenes: neutral, divalent carbon species.
Stability order:
Carbocation: 3° > 2° > 1° > CH₃⁺ (due to +I and hyperconjugation).
Carbanion: CH₃⁻ > 1° > 2° > 3° (due to destabilizing +I).
Free radical: 3° > 2° > 1° > CH₃•.

🔴 Purification of Organic Compounds
Since organic compounds are often obtained with impurities, purification techniques are essential.
1️⃣ Crystallization → purification based on solubility differences.
2️⃣ Sublimation → separation of sublimable solids (camphor, iodine).
3️⃣ Distillation
Simple distillation (different b.p.).
Fractional distillation (close b.p.).
Steam distillation (volatile compounds).
Distillation under reduced pressure (high b.p. compounds).
4️⃣ Chromatography → separation based on adsorption/partition.
5️⃣ Differential Extraction → organic vs aqueous layer separation.

🟢 Detection of Elements in Organic Compounds
Lassaigne’s Test: Fusion with sodium → extract tested for elements.
Nitrogen: NaCN formed → Prussian blue with Fe²⁺/Fe³⁺.
Sulphur: Na₂S → black ppt with Pb²⁺.
Halogens: NaX → white ppt with AgNO₃.

🔵 Quantitative Analysis
Carbon & Hydrogen → CO₂, H₂O produced, measured.
Nitrogen → Kjeldahl’s method.
Halogens → Carius method (AgX formed).
Sulphur & Phosphorus → converted to sulphate/phosphate, weighed.

🟡 Empirical and Molecular Formula
Empirical formula → simplest whole-number ratio.
Molecular formula = (Empirical formula) × n
n = Molar mass / Empirical formula mass.
Example: Acetic acid → Empirical = CH₂O, Molecular = C₂H₄O₂.

🔴 Significance of the Chapter
Provides foundation for naming and classifying compounds.
Explains stability and reactivity (inductive, resonance, hyperconjugation).
Introduces types of reactions and intermediates.
Essential purification and detection techniques.
Lays base for advanced organic topics (hydrocarbons, functional groups, biomolecules).

✨ Conclusion
Organic chemistry is vast but logical 🌍. This chapter equips us with the grammar of organic language — how to name, classify, analyze, and understand carbon compounds. It provides the stepping stone to the whole world of medicines, fuels, plastics, and biomolecules.

📌 Organic Chemistry – Some Basic Principles and Techniques
🔵 General Features
Carbon shows tetravalency and catenation.
Variety of compounds due to hybridization, functional groups.
🟢 Classification
Open chain, closed chain (alicyclic, aromatic).
Based on saturation: alkanes, alkenes, alkynes.
Based on functional groups: alcohols, aldehydes, acids, amines.
🔴 Nomenclature (IUPAC)
Longest chain, numbering, substituents, functional group priority.
Examples: propan-1-ol, but-2-ene.
🟡 Isomerism
Structural: chain, position, functional.
Stereoisomerism: geometrical, optical.
💡 Electronic Effects
Inductive effect (–I, +I).
Electromeric effect.
Resonance (delocalization).
Hyperconjugation (no-bond resonance).
📚 Reactions
Substitution, addition, elimination, rearrangement.
Intermediates: carbocations, carbanions, free radicals, carbenes.
🌟 Purification Techniques
Crystallization, sublimation, distillation, chromatography, extraction.
🧪 Elemental Analysis
Lassaigne’s Test for N, S, X.
Quantitative: C, H (CO₂/H₂O measurement), N (Kjeldahl), halogens (Carius).
📍 Formulas
Empirical vs molecular formula.
Relation: n = Molar mass / Empirical mass.
✨ Importance
Basis for studying all organic reactions.
Explains structure, reactivity, and analysis.
Provides lab techniques essential for chemistry.

📝 Quick Recap:
✔ Organic chemistry studies carbon compounds (except CO, CO₂, etc.).
✔ Compounds classified into open/closed chain, saturated/unsaturated, functional group-based.
✔ IUPAC rules → systematic naming of molecules.
✔ Isomerism (structural, stereoisomerism) gives diversity.
✔ Electronic effects (inductive, resonance, hyperconjugation) explain reactivity.
✔ Organic reactions: substitution, addition, elimination, rearrangement.
✔ Intermediates: carbocations, carbanions, radicals, carbenes.
✔ Purification techniques: crystallization, distillation, chromatography.
✔ Elemental detection (Lassaigne’s test) and quantitative analysis provide formulas.
✔ Foundation chapter for all organic chemistry.

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QUESTIONS FROM TEXTBOOK

🔵 Q8.1
What are hybridisation states of each carbon atom in the following compounds?
CH₂=C=O, CH₃CH=CH₂, CH₃COCH₃, CH₂=CHCN, C₆H₆
🟢 Answer
Let’s examine each carbon atom:
(i) CH₂=C=O
First C (CH₂): sp² (forms 3 σ bonds: 2 with H, 1 with next C)
Second C (=O): sp (forms 2 σ bonds: 1 with first C, 1 with O)
O: sp (not asked)
(ii) CH₃–CH=CH₂
C₁ (CH₃): sp³
C₂ (CH=): sp²
C₃ (=CH₂): sp²
(iii) CH₃–CO–CH₃
Terminal CH₃ carbons: sp³
Central carbonyl C: sp²
(iv) CH₂=CH–CN
C₁ (CH₂): sp²
C₂ (CH=): sp²
C₃ (–CN): sp (triple bond)
(v) C₆H₆ (Benzene)
Each carbon: sp² hybridised
Planar hexagonal structure with delocalised π electrons
✔ Final:
Compound Hybridisation of Carbons
CH₂=C=O sp², sp
CH₃CH=CH₂ sp³, sp², sp²
CH₃COCH₃ sp³, sp², sp³
CH₂=CHCN sp², sp², sp
C₆H₆ all sp²

🔵 Q8.2
Indicate the σ and π bonds in the following molecules:
C₂H₆, C₃H₈, CH₂=CH₂, CH≡CH, CH₂=CHCN, CH₃NO₂, HCONH₂
🟢 Answer
👉 Rule:
Single bond = 1 σ
Double bond = 1 σ + 1 π
Triple bond = 1 σ + 2 π
Now count bonds:
(i) C₂H₆
All single → 7 σ, 0 π
(ii) C₃H₈
All single → 10 σ, 0 π
(iii) CH₂=CH₂
1 C=C double = 1 σ + 1 π
4 C–H = 4 σ
→ Total: 5 σ + 1 π
(iv) CH≡CH
1 C≡C triple = 1 σ + 2 π
2 C–H = 2 σ
→ 3 σ + 2 π
(v) CH₂=CH–CN
C=C double = 1 σ + 1 π
C–C single = 1 σ
C≡N triple = 1 σ + 2 π
C–H = 3 σ
→ Total = 6 σ + 3 π
(vi) CH₃NO₂
3 C–H = 3 σ
1 C–N = 1 σ
2 N–O = 1 σ + 1 σ
1 π (delocalised between N=O bonds)
→ 6 σ + 1 π
(vii) HCONH₂
C=O → 1 σ + 1 π
C–N, C–H, N–H all single
→ 7 σ + 1 π

🔵 Q8.3
Write bond line formulas for the following:
Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one
🟢 Answer
💠 (i) Isopropyl alcohol (C₃H₈O):
Bond-line:
OH
|
CH3–CH–CH3
💠 (ii) 2,3-Dimethylbutanal (C₆H₁₂O):
CH3
|
CH3–CH–CH–CH=O
|
CH3
💠 (iii) Heptan-4-one (C₇H₁₄O):
CH3–CH2–CO–CH2–CH2–CH2–CH3

🔵 Q8.4
Give IUPAC names of the following compounds:
(a)
CH3
|
CH3–CH–CH2–CH3
(b)
CH3–CH2–CH2–CN
(c)
CH3–CH(Br)–CH2–CH3
(d)
CH3–CH(Cl)–CH2–CH(Br)–CH3
(e)
CH3–CH=CH–CH2–CH3
(f)
ClCH2–CH(OH)–CH3
🟢 Answer
(a) 2-Methylbutane
(b) Butanenitrile
(c) 2-Bromobutane
(d) 2-Chloro-4-bromopentane
(e) Pent-2-ene
(f) 3-Chloro-2-propanol

🔵 Q8.5
Which of the following represents correct IUPAC name?
(a) 2,2-Dimethylpentane or 2-Dimethylpentane
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne
🟢 Answer
✅ Correct IUPAC:
(a) 2,2-Dimethylpentane (lowest locants)
(b) 2,4,7-Trimethyloctane (sum of locants lowest)
(c) 2-Chloro-4-methylpentane (alphabetical order + low locants)
(d) But-3-yn-1-ol (OH gets priority, lowest number)

🔵 Q8.6
Draw formulas for first five members of homologous series:
(a) Carboxylic acids (R–COOH)
(b) Ketones (R–CO–R’)
(c) Alkenes (CₙH₂ₙ)
🟢 Answer
(a) Carboxylic acids:
HCOOH (Methanoic)
CH₃COOH (Ethanoic)
C₂H₅COOH (Propanoic)
C₃H₇COOH (Butanoic)
C₄H₉COOH (Pentanoic)
(b) Ketones:
(Not exist for 1C, 2C)
3C: CH₃COCH₃ (propanone)
4C: CH₃COC₂H₅ (butan-2-one)
5C: CH₃COC₃H₇ (pentan-2-one)
6C: C₂H₅COC₂H₅ (hexan-3-one)
(c) Alkenes:
C₂H₄ (Ethene)
C₃H₆ (Propene)
C₄H₈ (Butene)
C₅H₁₀ (Pentene)
C₆H₁₂ (Hexene)

🔵 Q8.7
Give condensed and bond line structural formulas and identify functional groups (if any) for:
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propane tricarboxylic acid
(c) Hexanedial
🟢 Answer
(a) 2,2,4-Trimethylpentane (Iso-octane)
Condensed: CH₃–C(CH₃)₂–CH(CH₃)–CH₂–CH₃
Functional group: Alkane
(b) 2-Hydroxy-1,2,3-propane tricarboxylic acid (Citric acid)
Condensed: HOOC–CH₂–C(OH)(COOH)–CH₂–COOH
Functional groups: –COOH and –OH
(c) Hexanedial
Condensed: OHC–(CH₂)₄–CHO
Functional group: –CHO (aldehyde ×2)


🔵 Question 8.8:
Identify the functional groups in the following compounds:
(a) 
(b) ₂)
(c)
🟢 Answer:
✔ (a) Functional groups:
–CHO (Aldehyde)
–OH (Hydroxyl / Alcohol)
–OCH₃ (Methoxy / Ether)
✔ (b) Functional groups:
–NH₂ (Amino group)
–OCH₂CH₂N(C₂H₅)₂ (Ether and tertiary amine group)
✔ (c) Functional groups:
–CH=CH– (Alkene)
–NO₂ (Nitro group)
🧠 Hence, each compound contains multiple functional groups responsible for its reactivity.

🔵 Question 8.9:
Which of the two is expected to be more stable and why?
(i) O⁻CH₂CH₂O⁻
(ii) CH₃CH₂O⁻
🟢 Answer:
✔ Compound (i) O⁻CH₂CH₂O⁻ is less stable because of negative–negative charge repulsion between two adjacent oxyanions.
✔ Compound (ii) CH₃CH₂O⁻ is more stable as it has only one negative charge, and the –CH₃ group donates electrons slightly by +I effect, dispersing the charge.
💡 Conclusion: CH₃CH₂O⁻ is more stable due to less repulsion and charge delocalisation.

🔵 Question 8.10:
Explain why alkyl groups act as electron donors when attached to a π system.
🟢 Answer:
✔ Alkyl groups show +I effect (inductive effect), pushing electron density through σ-bonds.
✔ When attached to a π-system (like a benzene ring), this extra electron density is delocalised into the ring, stabilising positive charge.
✔ Hence, alkyl groups act as electron donors by inductive effect and sometimes hyperconjugation.
💡 Example: In toluene, –CH₃ donates electrons into benzene ring, activating it toward electrophilic substitution.

🔵 Question 8.11:
Draw the resonance structures for the following compounds. Show electron shift using curved-arrow notation.
(a) C₆H₅OH
(b) C₆H₅NO₂
(c) CH₂=CH–CHO
(d) C₆H₅–CHO
(e) C₆H₅–CH=CH₂
🟢 Answer:
✔ (a) Phenol (C₆H₅OH):
– Lone pair on oxygen delocalises into ring → ortho/para resonance forms with negative charge on ring atoms.
✔ (b) Nitrobenzene (C₆H₅NO₂):
– Electrons of ring delocalise toward –NO₂ group (electron-withdrawing) → decreased electron density on ring (especially ortho/para).
✔ (c) Acrolein (CH₂=CH–CHO):
– π-electrons shift toward oxygen → resonance form CH₂–CH=CH–O⁻ ↔ CH₂=CH–CH=O⁺.
✔ (d) Benzaldehyde (C₆H₅–CHO):
– π-bond of –CHO interacts with aromatic ring; resonance reduces charge on carbon.
✔ (e) Styrene (C₆H₅–CH=CH₂):
– π-electrons conjugate with aromatic ring → resonance stabilisation across ring and double bond.
💡 Resonance = delocalisation of π-electrons → stabilisation of molecule.

🔵 Question 8.12:
What are electrophiles and nucleophiles? Explain with examples.
🟢 Answer:
✔ Electrophiles:
→ Electron-deficient species, seek electrons.
→ Act as Lewis acids.
🧪 Examples: H⁺, NO₂⁺, BF₃, AlCl₃.
✔ Nucleophiles:
→ Electron-rich species, donate electrons.
→ Act as Lewis bases.
🧪 Examples: OH⁻, CN⁻, NH₃, H₂O.
💡 In a reaction, nucleophile attacks electrophilic centre.

🔵 Question 8.13:
Identify reagents in bold as nucleophiles or electrophiles:
(a) CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O
(b) CH₃CH=CH₂ + H⁺ → CH₃CH⁺–CH₃
(c) C₆H₆ + CH₃⁺ → C₆H₆CH₃
(d) CH₃Cl + :OH⁻ → CH₃OH + Cl⁻
🟢 Answer:
✔ (a) HO⁻ → Nucleophile (donates electron pair).
✔ (b) H⁺ → Electrophile (accepts electrons).
✔ (c) CH₃⁺ → Electrophile (positively charged carbocation).
✔ (d) :OH⁻ → Nucleophile (attacks carbon).

🔵 Question 8.14:
Classify the following reactions as per type:
(a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻
(b) (CH₃)₂C=CH₂ + HCl → (CH₃)₂CCl–CH₃
(c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻
(d) (CH₃)₃C–CH₂OH + HBr → (CH₃)₃CBrCH₂CH₃ + H₂O
🟢 Answer:
✔ (a) Nucleophilic substitution (Br replaced by SH).
✔ (b) Electrophilic addition (H⁺, Cl⁻ added).
✔ (c) Elimination reaction (dehydrohalogenation).
✔ (d) Nucleophilic substitution (–OH replaced by –Br).

🔵 Question 8.15:
What is the relationship between the following pairs of structures?
(a) CH₃–CH₂–CH=CH₂ and CH₂=CH–CH₂–CH₃
(b) CH₃CH=CH₂ and CH₂=CHCH₃
(c) CH₃CH₂OH and CH₃CH₂OH
🟢 Answer:
✔ (a) Geometrical isomers (cis–trans).
✔ (b) Structural isomers (different structure but same formula).
✔ (c) Same compound (identical).
💡 Relationship classification depends on connectivity and spatial arrangement.

🔵 Question 8.16:
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediate produced as free radical, carbocation, or carbanion.
(a) CH₃–O–CH₃ → CH₃O• + •OCH₃
(b) R–O–H → R–O⁻ + H⁺
(c) R–Br → R⁺ + Br⁻
(d) C₆H₆–E → C₆H₆⁺ + E⁻
🟢 Answer:
✔ (a) Equal sharing of electrons → Homolytic cleavage → Forms free radicals (CH₃O• and •OCH₃).
✔ (b) Unequal sharing → Heterolytic cleavage → Forms carbanion (R–O⁻) and cation (H⁺).
✔ (c) Heterolytic cleavage → Gives carbocation (R⁺) and anion (Br⁻).
✔ (d) Heterolytic cleavage → Gives carbocation (C₆H₆⁺) and anion (E⁻).
💡 Homolysis → equal electron division → free radicals.
💡 Heterolysis → unequal electron division → ions (R⁺/R⁻).

🔵 Question 8.17:
Explain the terms Inductive effect and Electromeric effect. Which effect explains the following order of acidity?
(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
🟢 Answer:
✔ Inductive effect (+I/–I): Permanent displacement of σ-electrons along a chain.
–I effect: Electron-withdrawing groups (Cl) increase acidity.
+I effect: Electron-donating alkyl groups decrease acidity.
✔ Electromeric effect: Temporary electron shift in π-bonds when an attacking reagent approaches.
✔ (a) Cl atoms show –I effect, pulling electrons from –COOH → stabilises conjugate base → more acidic.
➡ Order: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
✔ (b) Alkyl groups show +I effect, donate electrons → destabilise conjugate base → less acidic.
➡ Order: CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
💡 Thus, inductive effect explains both orders.

🔵 Question 8.18:
Give a brief description of the principles of the following techniques, taking an example in each case:
(a) Crystallisation
(b) Distillation
(c) Chromatography
🟢 Answer:
✔ (a) Crystallisation:
Used for purification of solids.
Principle: Substance dissolves in a solvent at high temperature and crystallises upon cooling.
🧪 Example: Purification of benzoic acid.
✔ (b) Distillation:
Used for liquid purification.
Principle: Based on difference in boiling points.
🧪 Example: Separation of acetone and water.
✔ (c) Chromatography:
Used for mixture separation.
Principle: Components distribute differently between stationary and mobile phase.
🧪 Example: Separation of pigments from ink.

🔵 Question 8.19:
Describe the method used to separate two compounds with different solubilities in a solvent S.
🟢 Answer:
✔ Fractional Crystallisation:
Based on difference in solubility.
Dissolve both solids in hot solvent S.
Cool slowly → less soluble compound crystallises first.
Filter → crystallise second compound from filtrate.
💡 Example: Separation of KNO₃ and NaCl.

🔵 Question 8.20:
What is the difference between distillation, distillation under reduced pressure, and steam distillation?
🟢 Answer:
✔ Distillation:
→ Used for liquids with different boiling points.
→ Example: Water & alcohol.
✔ Reduced pressure distillation:
→ For high boiling liquids that decompose at normal bp.
→ Low pressure lowers bp.
→ Example: Glycerol.
✔ Steam distillation:
→ For temperature-sensitive liquids immiscible with water.
→ Boil below 100°C.
→ Example: Essential oils.

🔵 Question 8.21:
Discuss the chemistry of Lassaigne’s test.
🟢 Answer:
✔ Used to detect N, S, and halogens in organic compounds.
✔ Organic compound fused with Na metal → converts covalent elements into ionic forms:
NaCN, Na₂S, NaX.
✔ Aqueous extract = Lassaigne’s extract.
✔ Test each ion separately:
NaCN → Prussian blue test
Na₂S → Black ppt with Pb²⁺
NaX → AgX ppt with AgNO₃.

🔵 Question 8.22:
Differentiate between principles of nitrogen estimation by Dumas method and Kjeldahl’s method.
🟢 Answer:
✔ Dumas Method:
Organic sample burnt in excess O₂.
N converted to N₂ gas → measured.
Suitable for all compounds.
✔ Kjeldahl Method:
Compound digested with conc. H₂SO₄ → NH₃ formed.
NH₃ absorbed in acid → titrated.
Not suitable for N in ring (NO₂, N=N).
💡 Dumas → gaseous estimation; Kjeldahl → titrimetric.

🔵 Question 8.23:
Discuss the principle of estimation of halogens, sulphur, and phosphorus present in an organic compound.
🟢 Answer:
✔ Principle:
Fuse compound with Na metal → convert into ionic forms:
NaX (halogen), Na₂S (sulphur), Na₃PO₄ (phosphorus).
✔ Extract ions → test quantitatively.
➡ Halogens: Precipitated with AgNO₃ → weighed as AgX.
➡ Sulphur: Precipitated as BaSO₄ → weighed.
➡ Phosphorus: Precipitated as MgNH₄PO₄ → ignited to Mg₂P₂O₇ → weighed.
💡 Method = Gravimetric analysis.

🔵 Question 8.24:
Explain the principle of paper chromatography.
🟢 Answer:
✔ Paper chromatography separates components of a mixture based on their different partitioning between two phases:
Stationary phase: Thin film of water molecules adsorbed on the cellulose of filter paper.
Mobile phase: A solvent (or solvent mixture) moving through the paper by capillary action.
✔ Each component has a distinct partition coefficient (K), so it travels a different distance.
✔ The retention factor (Rf) identifies compounds:
Rf = (distance moved by substance) ÷ (distance moved by solvent front)
💡 Example: Separation of amino acids using a mixture of n-butanol : acetic acid : water.

🔵 Question 8.25:
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
🟢 Answer:
✔ To acidify the extract and destroy interfering ions like CN⁻ and S²⁻ that could form misleading precipitates (AgCN or Ag₂S).
✔ HNO₃ oxidises these ions to volatile products (HCN, H₂S), ensuring only halide ions remain.
✔ Then AgNO₃ gives characteristic precipitates:
AgCl → white  AgBr → pale-cream  AgI → yellow.

🔵 Question 8.26:
Explain the reason for fusion with metallic sodium when testing nitrogen, sulphur, and halogens.
🟢 Answer:
✔ In organic compounds, N, S, and halogens are covalently bound and not directly testable.
✔ Fusion with sodium metal converts them into ionic, water-soluble salts:
NaCN, Na₂S, NaX (X = Cl, Br, I).
✔ These dissolve in water to form Lassaigne’s extract, enabling specific ionic tests.

🔵 Question 8.27:
Name a suitable technique to separate a mixture of calcium sulphate and camphor.
🟢 Answer:
✔ Sublimation
Camphor sublimes (solid → vapour → solid).
Calcium sulphate is non-volatile.
✔ Gentle heating → camphor vapour condenses on a cool surface → separated from residue.

🔵 Question 8.28:
Explain why an organic liquid vaporises below its boiling point during steam distillation.
🟢 Answer:
✔ For immiscible liquids (organic compound + water), total vapour pressure = P(organic) + P(water).
✔ Boiling occurs when total pressure = atmospheric pressure → at a lower temperature than each component’s individual boiling point.
✔ Thus, steam distillation allows distillation without decomposition of heat-sensitive substances (e.g. essential oils).

🔵 Question 8.29:
Will CCl₄ give a white precipitate of AgCl on heating with AgNO₃? Give reason.
🟢 Answer:
❌ No.
✔ Chlorine atoms in CCl₄ are covalently bonded, not ionised.
✔ AgNO₃ reacts only with free Cl⁻ ions; hence no AgCl precipitate forms.
✔ Detection requires conversion to ionic chloride via sodium fusion.

🔵 Question 8.30:
Why is a KOH solution used to absorb CO₂ evolved during carbon estimation?
🟢 Answer:
✔ In the combustion method, carbon forms CO₂.
✔ Potassium hydroxide absorbs CO₂ completely, producing potassium carbonate:
2 KOH + CO₂ → K₂CO₃ + H₂O
✔ This ensures accurate quantitative measurement of carbon.

🔵 Question 8.31:
What do you understand by the term hybridisation? Explain sp³, sp², and sp hybridisation with examples.
🟢 Answer:
✔ Hybridisation = Process of mixing atomic orbitals of nearly equal energy on the same atom to form new equivalent orbitals called hybrid orbitals.
(i) sp³ Hybridisation:
1 s + 3 p orbitals → 4 sp³ orbitals.
Geometry: Tetrahedral; angle = 109.5°.
Example: CH₄ (each C–H bond = sp³–s σ bond).
(ii) sp² Hybridisation:
1 s + 2 p → 3 sp² orbitals.
Geometry: Trigonal planar; angle = 120°.
Example: C₂H₄ (ethylene).
(iii) sp Hybridisation:
1 s + 1 p → 2 sp orbitals.
Geometry: Linear; angle = 180°.
Example: C₂H₂ (acetylene).
💡 Hybridisation explains shape and bond angles.

🔵 Question 8.32:
What are resonance structures? Draw resonance structures for benzene.
🟢 Answer:
✔ When a molecule can be represented by two or more Lewis structures differing only in position of electrons, not atoms → such structures are resonance forms.
✔ The actual molecule is a hybrid of all forms and more stable.
Example: Benzene (C₆H₆):
Two Kekulé structures:
Alternating double bonds starting at C₁–C₂
Alternating double bonds starting at C₂–C₃
💡 Actual benzene: all C–C bonds equal (140 pm), electron delocalised over ring.

🔵 Question 8.33:
Define electromeric effect and differentiate it from inductive effect.
🟢 Answer:
✔ Electromeric effect (E-effect):
Temporary shift of π electrons toward one atom under the attack of a reagent.
Reversible, only in presence of attacking reagent.
Example:
 C=O + H⁺ → C⁺–O⁻ (π electrons shift to O).
✔ Inductive effect (I-effect):
Permanent displacement of σ-electrons due to electronegativity difference; transmitted along chain.
Feature Electromeric Inductive
Nature Temporary Permanent
Electrons shifted π σ
Condition Only when reagent attacks Always present
Example C=O + H⁺ –NO₂ shows –I

🔵 Question 8.34:
What is hyperconjugation? Explain with an example.
🟢 Answer:
✔ Definition: Delocalisation of σ-electrons of C–H bond adjacent to a double bond or positive centre into π-system or vacant p-orbital.
✔ Known as no-bond resonance.
Example: Propene (CH₃–CH=CH₂):
C–H σ bond of CH₃ overlaps with π bond → stabilisation.
✔ More hyperconjugative structures = greater stability.
Order (carbocation): 3° > 2° > 1° > CH₃⁺.
💡 Explains stability of alkenes and carbocations.

🔵 Question 8.35:
What is carbocation? Classify and discuss stability order.
🟢 Answer:
✔ Carbocation = species with positive charge on carbon (C⁺), sp² hybridised, planar.
Types:
Primary (1°): charge on C attached to 1 C
Secondary (2°): charge on C attached to 2 C
Tertiary (3°): charge on C attached to 3 C
Stability order:
3° > 2° > 1° > CH₃⁺
(because of +I effect and hyperconjugation)
Example: (CH₃)₃C⁺ most stable.

🔵 Question 8.36:
Define free radical. Discuss its structure and stability.
🟢 Answer:
✔ Free radical = species with odd electron (unpaired), electrically neutral.
✔ Carbon is sp² hybridised; geometry = planar; unpaired electron in p orbital.
Stability order:
3° > 2° > 1° > CH₃•
Due to hyperconjugation and inductive effect.
Example: CH₃•, C₂H₅•, (CH₃)₂CH•, (CH₃)₃C•.

🔵 Question 8.37:
Define carbanion. Discuss its structure and stability.
🟢 Answer:
✔ Carbanion = species with negative charge on carbon (C⁻), with lone pair.
✔ Hybridisation: sp³; geometry: pyramidal.
Stability order:
CH₃⁻ > 1° > 2° > 3°
Because alkyl groups donate electrons (+I) → destabilise negative charge.
💡 Electron-withdrawing groups increase stability.

🔵 Question 8.38:
Explain electrophilic substitution reaction with example.
🟢 Answer:
✔ Reaction where an electrophile replaces a hydrogen atom in an aromatic compound.
Example:
Nitration of benzene:
C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (in presence of H₂SO₄)
Electrophile = NO₂⁺.
Mechanism:
Generation of electrophile
Attack on ring
Loss of proton → substitution product.

🔵 Question 8.39:
Explain nucleophilic substitution reaction with example.
🟢 Answer:
✔ Reaction in which a nucleophile replaces another atom/group (usually halide).
Example:
CH₃Br + OH⁻ → CH₃OH + Br⁻
Nucleophile: OH⁻; leaving group: Br⁻.
Types: SN1 (2-step) and SN2 (1-step).

🔵 Question 8.40:
Explain elimination reaction with example.
🟢 Answer:
✔ In elimination, two atoms/groups are removed from adjacent carbons forming multiple bond.
Example:
CH₃CH₂Br + KOH (alc.) → CH₂=CH₂ + HBr + K⁺Br⁻
Called dehydrohalogenation.
💡 Competes with substitution; favoured by strong base and heat.

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OTHER IMPORTANT QUESTIONS FOR EXAMS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

Section A (Q1–Q16): MCQs (1 mark each)
Question 1. Which of the following represents the correct IUPAC name?
2-methylpropane
2-propane methyl
methyl-2-propane
propane-2-methyl
Answer: 1
Question 2. The bond angle in methane (CH₄) is:
90°
109.5°
120°
104.5°
Answer: 2
Question 3. The number of σ and π bonds in benzene is:
6 σ and 6 π
12 σ and 3 π
12 σ and 6 π
6 σ and 3 π
Answer: 3
Question 4. Which effect explains the polarity of chlorobenzene?
Inductive effect
Resonance effect
Hyperconjugation
Electromeric effect
Answer: 2
Question 5. The functional group –CHO belongs to:
Ketones
Aldehydes
Carboxylic acids
Alcohols
Answer: 2
Question 6. The test used to detect unsaturation in an organic compound is:
Fehling’s test
Bromine water test
Tollen’s test
Lassaigne’s test
Answer: 2
Question 7. Which of the following is the strongest acid?
Ethanol
Phenol
Acetic acid
Water
Answer: 3
Question 8. The hybridisation of carbon in ethyne (C₂H₂) is:
sp³
sp²
sp
dsp²
Answer: 3
Question 9. Tetrahedral carbon is a feature of:
Methane
Ethene
Ethyne
Benzene
Answer: 1
Question 10. Which of the following is used as a nucleophile?
NH₃
BF₃
H⁺
AlCl₃
Answer: 1
Assertion–Reason Type (insert key before first A/R)
Options:
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A
A is true, but R is false
A is false, but R is true
Question 11. Assertion (A): Electrophiles are electron-loving species.
Reason (R): They are electron-deficient and accept electrons.
Answer: 1
Question 12. Assertion (A): Resonance hybrid is more stable than any of its canonical structures.
Reason (R): Delocalisation of electrons lowers the energy of the system.
Answer: 1
Question 13. Assertion (A): Hyperconjugation is also called no-bond resonance.
Reason (R): It involves delocalisation of σ electrons of C–H bond with adjacent π bond or vacant p orbital.
Answer: 1
Question 14. Assertion (A): Lassaigne’s test is used for halogens.
Reason (R): Sodium fusion converts covalent halogens into ionic halides.
Answer: 1
Question 15. Which of the following has highest boiling point?
Butane
Butanol
Butanal
Butanone
Answer: 2
Question 16. The correct order of stability of carbocations is:
1° < 2° < 3°
3° < 2° < 1°
2° < 1° < 3°
1° < 3° < 2°
Answer: 1

Section B (Q17–Q21): Very Short Answer (2 marks each)
Q17. Define inductive effect. Give one example.
🟦 Inductive effect is the permanent displacement of σ electrons along a chain due to electronegativity difference.
🟩 Example: –I effect of –NO₂ group in nitroalkanes reduces electron density.

Q18. Give two differences between nucleophiles and electrophiles.
🔷 Nucleophiles — electron-rich species, donate electrons (e.g., OH⁻).
🔶 Electrophiles — electron-deficient species, accept electrons (e.g., H⁺).

Q19. Write the formula of one compound each containing: (i) alcoholic group (ii) carboxylic group.
🟪 Alcoholic group: C₂H₅OH (ethanol)
🟩 Carboxylic group: CH₃COOH (acetic acid)

Q20. A compound contains 40% C, 6.7% H, and 53.3% O by mass. Calculate its empirical formula.
➤ Assume 100 g → C = 40 g, H = 6.7 g, O = 53.3 g
➤ Moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33
➤ Ratio = C : H : O = 1 : 2 : 1
✅ Empirical formula = CH₂O

Q21. State two differences between heterolytic and homolytic fission.
🟦 Homolytic fission: bond breaks evenly → free radicals.
🟨 Heterolytic fission: bond breaks unevenly → cation and anion.

Section C (Q22–Q28): Short Answer (3 marks each)
Q22. Explain the classification of organic compounds based on functional groups with examples.
🔷 Hydroxyl group → Alcohols (C₂H₅OH)
🔶 Carbonyl group → Aldehydes (CH₃CHO), Ketones (CH₃COCH₃)
🧪 Carboxyl group → Carboxylic acids (CH₃COOH)

Q23. Write three points of difference between inductive effect and resonance effect.
🟦 Inductive effect — operates through σ bonds, decreases with distance.
🟨 Resonance effect — involves delocalisation of π electrons, distance-independent.
🎯 Inductive effect is weaker; resonance stabilises molecules more effectively.

Q24. Calculate the empirical formula of a compound containing 85.7% C and 14.3% H by mass.
➤ Assume 100 g → C = 85.7 g, H = 14.3 g
➤ Moles: C = 85.7/12 = 7.14, H = 14.3/1 = 14.3
➤ Ratio = C : H = 1 : 2
✅ Empirical formula = CH₂

Q25. Explain Markovnikov’s rule with example.
🧪 Markovnikov’s rule: In the addition of HX to unsymmetrical alkenes, the negative part attaches to the carbon with fewer hydrogens.
⚗ Example: CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃

Q26. Write the steps involved in the mechanism of nitration of benzene.
🧪 Step 1 — Generation of nitronium ion: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
🧪 Step 2 — Electrophilic attack on benzene ring → σ complex
🧪 Step 3 — Deprotonation regenerates aromaticity → Nitrobenzene
✅ Product: C₆H₅NO₂

Q27. How is Lassaigne’s test performed for nitrogen?
➜ Sodium fusion → NaCN formed.
➜ Extract treated with FeSO₄ and NaOH, then acidified with H₂SO₄.
🔬 Formation of Prussian blue colour confirms nitrogen.

Q28. Distinguish between SN1 and SN2 mechanisms.
🟦 SN1 — two steps, carbocation intermediate, rate ∝ [substrate], favoured by polar protic solvents.
🟩 SN2 — one step, backside attack, rate ∝ [substrate][nucleophile], inversion of configuration.

Section D (Q29–Q30: Case-Based, 4 marks each)
Q29. Read the passage and answer the questions below:
During elemental analysis of organic compounds, the Lassaigne’s test is used. Sodium fusion converts covalently bonded elements into ionic form, which can then be detected easily.
a) What is the product formed when sodium fusion extract containing nitrogen is boiled with FeSO₄ and NaOH, then acidified with H₂SO₄? (1 mark)
b) State the colour observed. (1 mark)
c) Write the reaction involved. (2 marks)
🟦 a) Product formed: Sodium ferrocyanide → converted to Prussian blue complex.
🟨 b) Colour observed: Prussian blue.
🧪 c) Reaction:
NaCN + FeSO₄ → Na₄[Fe(CN)₆]
Na₄[Fe(CN)₆] + Fe³⁺ → Fe₄[Fe(CN)₆]₃ (Prussian blue).

Q30. Read the passage and answer the questions below:
Markovnikov’s rule governs the orientation of addition in unsymmetrical alkenes. However, in presence of peroxides, the rule is not followed due to free radical mechanism.
a) State Markovnikov’s rule. (1 mark)
b) Name the exception to Markovnikov’s rule. (1 mark)
c) Write the mechanism for addition of HBr to propene in presence of peroxides. (2 marks)
🟦 a) Markovnikov’s rule: In the addition of HX to unsymmetrical alkenes, the halide attaches to the more substituted carbon.
🟨 b) Exception: Anti-Markovnikov addition (Kharasch effect) in presence of peroxides.
🧪 c) Mechanism (Free radical):
Initiation: ROOR → 2RO·
RO· + HBr → ROH + Br·
Propagation:
Br· + CH₃–CH=CH₂ → CH₃–CH·–CH₂Br
CH₃–CH·–CH₂Br + HBr → CH₃–CH₂–CH₂Br + Br·
✅ Product: 1-bromopropane.

Section E (Q31–Q33: Long Answer, 5 marks each with OR)
Q31. Explain the steps involved in the determination of empirical formula of an organic compound. Illustrate with an example.
🟦 Step 1 — % composition of elements is determined by analysis.
🟨 Step 2 — Convert % into grams (assume 100 g sample).
🧮 Step 3 — Calculate moles = mass / atomic mass.
🟪 Step 4 — Divide all values by the smallest number of moles.
🟩 Step 5 — Multiply to get whole number ratio → empirical formula.
Example: A compound contains 40% C, 6.7% H, and 53.3% O.
➤ Moles: C = 3.33, H = 6.7, O = 3.33.
➤ Ratio = 1 : 2 : 1.
✅ Empirical formula = CH₂O.
OR
Describe the Kjeldahl’s method for determination of nitrogen in an organic compound.
🔷 Organic compound is digested with conc. H₂SO₄ → (NH₄)₂SO₄ formed.
🔶 Ammonia liberated by NaOH → absorbed in standard H₂SO₄/HCl.
🧮 Back titration with alkali → % N calculated:
% N = (1.4 × volume of acid neutralised × N) / weight of sample.

Q32. Explain the following terms with examples: (a) Inductive effect, (b) Resonance effect, (c) Hyperconjugation.
🟦 (a) Inductive effect — permanent displacement of σ electrons due to electronegativity difference. Example: –I effect of NO₂ in nitroalkanes.
🟨 (b) Resonance effect — delocalisation of π electrons stabilises molecule. Example: Benzene resonance structures.
🟩 (c) Hyperconjugation — delocalisation of C–H σ electrons with adjacent π bond. Example: Stability of 2° carbocation > 1° due to hyperconjugation.
OR
Discuss the differences between electrophiles and nucleophiles with suitable examples.
🔷 Electrophiles: electron-deficient species; accept electron pair (H⁺, BF₃).
🔶 Nucleophiles: electron-rich species; donate electron pair (OH⁻, NH₃).
🎯 Difference lies in electron density, reaction role, and examples.

Q33. Write the mechanism of nitration of benzene. Why is it called an electrophilic substitution reaction?
🧪 Step 1 — Generation of electrophile:
HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
🧪 Step 2 — Electrophilic attack: NO₂⁺ attacks benzene → σ complex.
🧪 Step 3 — Deprotonation: H⁺ removed, aromaticity restored.
✅ Product: Nitrobenzene.
⚗ It is called electrophilic substitution because an electrophile replaces a hydrogen atom on the aromatic ring.
OR
Explain the difference between SN1 and SN2 reactions with suitable examples.
🟦 SN1: Two-step mechanism, carbocation intermediate, rate depends on substrate (e.g., hydrolysis of tert-butyl bromide).
🟩 SN2: One-step mechanism, nucleophile attacks from backside, inversion occurs (e.g., hydrolysis of methyl bromide).
🎯 Key difference — SN1 gives racemisation, SN2 gives inversion.

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