Chemistry Class 11 Part II…
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Fundamentals Purification

🎓 Class 11 Chemistry CBSE Theory Ch 8 – Organic Chemistry: Some Basic Principles and Techniques ⏱ ~14 min
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Electronic Effects, Bond Cleavage and Purification Methods

Introduction: How Bonds Break and How to Purify a Flask

Every organic reaction is essentially a re-arrangement of electrons — bonds break, electrons shift, new bonds form. To read mechanisms we must learn how a covalent bond can split and what the resulting fragments look like. Once a product has been synthesised, the next practical problem is separating it from side products and solvents. That is where the classical purification techniques — sublimation, crystallisation, distillation, chromatography — come in.

8.7 Fundamental Concepts in Organic Reaction Mechanisms

(a) Bond Fission: Homolytic vs Heterolytic L2 Understand

Homolytic fission: each atom keeps one of the two bonding electrons. Two neutral free radicals (species with an unpaired electron) are produced. Curly half-arrows (fish-hooks) show single-electron movement.
A–B → A• + •B

Heterolytic fission: one atom takes both electrons; the other leaves empty-handed. A pair of ions is produced. A full curly arrow shows the movement of the pair.
A–B → A+ + B− (or A− + B+, depending on which is more electronegative)
Homolytic A : B → A • + • B two free radicals Heterolytic A : B → A + + B − : cation + anion
Fig 8.9 — Two ways a bond can break.

(b) Reactive Carbon Intermediates

SpeciesStructure on CChargeHybridisationShape
Carbocation3 bonds, 0 lone pair, empty p-orbital+1sp²Trigonal planar
Carbanion3 bonds + 1 lone pair−1sp³Pyramidal
Free radical3 bonds + 1 unpaired electron0sp² (approx.)Nearly planar
Stability order
Carbocations: 3° > 2° > 1° > methyl (alkyl groups donate electrons by +I and hyperconjugation).
Carbanions: methyl > 1° > 2° > 3° (reverse order; alkyl groups destabilise by donating electrons to an already electron-rich carbon).
Free radicals: 3° > 2° > 1° > methyl (similar to cations).

(c) Electrophiles and Nucleophiles L2 Understand

Electrophile (E+): an electron-seeking species. Includes cations (H+, NO2+, R+, Cl+) and neutral electron-deficient molecules (BF3, AlCl3, SO3, carbonyl carbon).
Nucleophile (Nu−/Nu:): a nucleus-seeker, electron-rich. Includes anions (OH−, CN−, RO−, halides) and neutral molecules with lone pairs (NH3, H2O, ROH, R2S).
E⁺ electrophile δ⁻ site electron-rich Nu⁻ nucleophile attacks δ+ site
Fig 8.10 — Electrophiles go where electrons are plentiful; nucleophiles go where they are scarce.

(d) Electronic Effects

Inductive effect (−I, +I)

Transmitted through sigma bonds. An atom or group more electronegative than carbon withdraws electrons (−I); less electronegative groups release electrons (+I). The effect weakens with distance — typically negligible beyond the third carbon.

Order of −I groups: NO2 > CN > F > Cl > Br > I > OH > OR > COOH > C6H5
Order of +I groups: (CH3)3C– > (CH3)2CH– > CH3CH2– > CH3–

Resonance (mesomeric) effect

Delocalisation of π electrons or lone pairs through a conjugated system. Multiple "resonance structures" differ only in the location of π electrons; the real molecule is a hybrid (average) and is more stable than any single contributor.

Carboxylate ion: R–CO2− ⟷ R–(CO2)− (negative charge shared equally over two O atoms) Benzene: three alternating single/double arrangements ⟷ equivalent structures

Electromeric effect

A temporary shift of a π pair that occurs only in the presence of an attacking reagent. Denoted +E or −E depending on the direction.

Hyperconjugation (no-bond resonance)

Delocalisation of the σ electrons of a C–H bond adjacent to an sp² centre (a π bond or an empty p-orbital). Each α-hydrogen adds one hyperconjugating structure. This is the principal reason 3° carbocations (9 α-H's) are more stable than 2° (6) or 1° (3).

Worked Example 1 — Identify fission type
Problem: Classify each step: (a) Cl2 hν→ 2 Cl• (b) CH3COCl + AlCl3 → CH3CO+ + AlCl4− (c) (CH3)3C–Br → (CH3)3C+ + Br−
(a) Two neutral radicals — homolytic.
(b) Cation + complex anion — heterolytic.
(c) 3° carbocation + Br− — heterolytic (the classic SN1 first step).
Worked Example 2 — Electrophile or nucleophile?
Problem: Classify: NH3, NO2+, BF3, CN−, H2O, SO3.
Nucleophiles (have lone pair / negative charge): NH3, CN−, H2O.
Electrophiles (electron-poor): NO2+, BF3 (incomplete octet on B), SO3 (S is δ+).
Worked Example 3 — Acidity via inductive effect
Problem: Arrange in order of acid strength: CH3COOH, ClCH2COOH, Cl2CHCOOH, Cl3CCOOH.
Reasoning: Each Cl exerts a −I effect, pulling electron density away from the –COO− anion and stabilising it. More chlorines → more stabilisation → stronger acid.
Order of acidity: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH.
Worked Example 4 — Carbocation stability
Problem: Rank the stability of CH3+, CH3CH2+, (CH3)2CH+, (CH3)3C+.
Alkyl groups donate electrons by +I and by hyperconjugation (each α-C–H bond contributes). Counting α-H's: methyl 0, ethyl 3, isopropyl 6, t-butyl 9.
Order: (CH3)3C+ > (CH3)2CH+ > CH3CH2+ > CH3+.

8.8 Methods of Purification of Organic Compounds

A synthesised crude contains unreacted starting material, side products and solvent. The method of purification chosen depends on the physical state and the properties (volatility, polarity, solubility, thermal stability) of the target compound.

(a) Sublimation

Used for solids that pass directly from solid to vapour on heating and vice versa on cooling. Impurities that do not sublime remain behind. Examples: camphor, naphthalene, anthracene, benzoic acid, iodine.

(b) Crystallisation

Based on the difference in solubility of the compound and its impurities in a suitable solvent. A hot, saturated solution of the crude is prepared; on cooling, the pure compound crystallises out while impurities stay in solution. Multiple recrystallisations give successively purer material.

(c) Distillation

Exploits the difference in boiling points to separate liquids. The liquid is vaporised, the vapour is condensed elsewhere, and the condensate collected.

(i) Simple distillation

Works when the boiling-point gap between components exceeds ~25 °C. Typical setup: round-bottom flask + thermometer + condenser + receiver.

(ii) Fractional distillation

For mixtures whose components have close boiling points. A fractionating column packed with glass beads or plates provides many successive evaporation–condensation cycles, enriching the lighter component with each pass. This is how crude oil is split into LPG, petrol, kerosene, diesel, lubricating oils and bitumen.

(iii) Distillation under reduced pressure

For compounds that decompose before reaching their normal boiling point. Lowering the pressure lowers the boiling point. Used for glycerol (b.p. 290 °C at 1 atm; decomposes) and sugar syrups.

(iv) Steam distillation

For compounds that are steam-volatile and insoluble in water. Steam is passed through the hot crude; the compound co-distils at a temperature below 100 °C, escaping thermal decomposition. Used to purify aniline and to extract essential oils (lemongrass, rose, eucalyptus).

Simple distillation condenser distillate heat Fractional distillation fractionating column distillate heat
Fig 8.11 — Simple distillation (left) vs fractional distillation with a packed column (right).

(d) Differential Extraction

Two immiscible liquids (usually an organic solvent and water) are shaken in a separating funnel. The compound distributes itself between them according to its solubility. Repeated extractions transfer the compound quantitatively into the organic layer, which is then drained off. Iodine can be extracted from water into CCl4 in this way.

(e) Chromatography

The powerhouse technique. A mixture is carried by a mobile phase over a stationary phase; components that interact more strongly with the stationary phase lag behind, separating the band.

  • Column chromatography (adsorption): packed column of silica gel or alumina; the mobile phase is a solvent. Components elute in order of increasing affinity for the adsorbent.
  • Thin-layer chromatography (TLC): a glass / plastic plate coated with silica or alumina. Ideal for checking reaction progress.
  • Paper chromatography (partition): partition of the solute between the water held on cellulose fibres (stationary) and an organic solvent (mobile).
Retention factor: \[ R_f \;=\; \dfrac{\text{distance travelled by solute}}{\text{distance travelled by solvent front}} \] Rf values are between 0 and 1 and serve as fingerprints for compounds under specified conditions.
baseline solvent front A (low R_f) B (mid) C (high)
Fig 8.12 — TLC plate after development; components separated by their Rf values.
Worked Example 5 — Calculate Rf
Problem: On a TLC plate the solvent front advances 8.0 cm while component X travels 3.2 cm from the baseline. Calculate Rf.
\( R_f = 3.2 / 8.0 = 0.40 \). Any compound under identical conditions sharing this value is a candidate for X.

8.9 Qualitative Analysis of Organic Compounds

(a) Detection of Carbon and Hydrogen

The compound is heated with dry cupric oxide (CuO). Any carbon is oxidised to CO2, which turns lime water milky. Hydrogen is oxidised to H2O, which turns anhydrous (white) CuSO4 blue.

C (in compound) + 2 CuO → CO2 + 2 Cu    [lime water → milky] 2 H (in compound) + CuO → H2O + Cu   [anhydrous CuSO4 → blue]

(b) Lassaigne's Test — Detecting N, S, Halogens

The organic compound is fused with a piece of sodium metal. Covalently bound N, S and halogens are converted to ionic Na+ salts, soluble in water. The aqueous "sodium fusion extract" (or Lassaigne's extract) is then tested:

ElementSpecies in extractTest & observation
NitrogenNaCNAdd FeSO4, warm, acidify with dil. H2SO4, add a drop of FeCl3 → Prussian blue (Fe4[Fe(CN)6]3) precipitate.
SulphurNa2S(i) Violet colour with sodium nitroprusside [Na2Fe(CN)5NO], or (ii) black PbS with lead acetate.
N + S togetherNaSCNBlood-red colour with FeCl3.
HalogensNaXAcidify with dil. HNO3, add AgNO3 → AgCl white, AgBr pale yellow, AgI bright yellow precipitate.

8.10 Quantitative Analysis — Combustion

A known mass of the organic compound is burnt in a stream of dry O2 inside a heated combustion tube. CO2 is absorbed in KOH solution; H2O is absorbed in anhydrous CaCl2. From the increases in mass of these absorbers, % C and % H are calculated:

\[ \%\,C = \dfrac{12}{44}\times\dfrac{\text{mass of CO}_2}{\text{mass of sample}}\times 100 \] \[ \%\,H = \dfrac{2}{18}\times\dfrac{\text{mass of H}_2O}{\text{mass of sample}}\times 100 \]
Worked Example 6 — Combustion analysis
Problem: 0.246 g of an organic compound on combustion gave 0.198 g of H2O and 0.486 g of CO2. Find % C and % H.
\( \% C = (12/44)\times(0.486/0.246)\times 100 \)
\( = 0.2727 \times 1.9756 \times 100 \approx 53.9\% \)
\( \% H = (2/18)\times(0.198/0.246)\times 100 = 0.1111 \times 0.8049 \times 100 \approx 8.95\% \)
Remaining 37.1% is probably oxygen (by difference).
Activity 8.3 — Paper chromatography of inkL3 Apply

Aim: Separate the pigments of a black sketch-pen using paper chromatography.

  1. Cut a strip of filter paper 2 cm × 15 cm. Draw a faint pencil line 2 cm from one end.
  2. Place a small dot of black ink on the line. Let it dry; repeat three times to concentrate the sample.
  3. Pour a little water (or water + a few drops of ethanol) into a tall jar — just below the pencil line.
  4. Hang the strip so the bottom dips into the solvent but the ink spot stays dry. Cover the jar.
  5. Watch the solvent climb. When it nears the top, remove the paper and mark the solvent front.
Predict: How many distinct bands will appear? Which colour will climb fastest?

Black ink typically separates into 3–5 coloured bands — commonly yellow, blue, red and violet — with the most water-soluble pigment travelling farthest (highest Rf). The experiment shows "black" is a blend of dyes and demonstrates partition chromatography in the simplest possible form.

Interactive: Purification Method Chooser

Pick a property of your crude compound and receive a recommended technique.

(Recommendation will appear here)

Competency-Based Questions

A student purifies a crude mixture that looks like a pale-brown oil with a faint smell of fish. She suspects it contains aniline (b.p. 184 °C, steam-volatile) plus some tar-like decomposition products. She plans her purification strategy and carries out Lassaigne's test on the purified sample.

1. The most suitable method of purifying the aniline is:

  • A. Simple distillation at 1 atm
  • B. Fractional distillation
  • C. Steam distillation
  • D. Sublimation
C. Aniline is steam-volatile and decomposes near its b.p.; steam distillation lets it distil below 100 °C.

2. Short answer: Why must the organic compound be fused with sodium (not simply dissolved) before Lassaigne's test?

Fusion converts the covalent N, S and X atoms into ionic sodium salts (NaCN, Na2S, NaX). Only ionic species give the colour / precipitate tests.

3. Fill in the blank: In the –I series NO2 > CN > F > Cl > ……, the next group is ________.

Br (bromine).

4. True/False: Heterolytic fission always gives two ions of opposite charge.

True. One atom takes both electrons (becomes anion), the other leaves without (becomes cation).

5. HOT: 0.20 g of an organic compound gave 0.147 g of AgCl on Carius estimation of chlorine. Calculate % Cl.

Mass of Cl in 0.147 g AgCl = (35.5/143.5) × 0.147 = 0.0364 g. % Cl = (0.0364/0.20)×100 = 18.2 %.

Assertion–Reason Questions

Options: A both true and R correctly explains A · B both true but R does not explain A · C A true R false · D A false R true.

A: A tertiary carbocation is more stable than a primary one.

R: Alkyl groups stabilise positive charge by donating electrons through +I and hyperconjugation.

A. Both true and R explains A.

A: Steam distillation is used to purify aniline.

R: Aniline is very soluble in water and hence steam carries it across.

C. Assertion is true, but R is false — aniline is only sparingly soluble in water. Steam distillation works because aniline is steam-volatile.

A: OH− is a nucleophile.

R: It bears a negative charge and possesses lone pairs of electrons that can be donated.

A. Both true and R is the correct explanation.

Frequently Asked Questions — Electronic Effects, Bond Cleavage and Purification Methods

What is the inductive effect?
The inductive effect is the polarisation of a covalent bond that is transmitted through subsequent sigma (σ) bonds along the carbon chain, weakening with distance (negligible after the third carbon). NCERT Class 11 Chemistry Chapter 8 distinguishes: (1) -I effect (electron-withdrawing) — groups like -NO₂, -CN, -COOH, -F, -Cl, -Br pull electrons toward themselves; (2) +I effect (electron-donating) — alkyl groups like -CH₃, -C₂H₅ push electrons away. The inductive effect explains acidity differences (Cl-CH₂-COOH stronger than CH₃-COOH due to -I effect of Cl) and stability of carbocations.
What is the resonance effect?
The resonance effect (mesomeric effect) is the transfer of pi (π) electrons through conjugated systems where alternating single and double bonds exist. NCERT Class 11 Chemistry Chapter 8 distinguishes: (1) +R or +M (electron-donating) — groups like -OH, -OR, -NH₂, -X donate lone pairs into the system; (2) -R or -M (electron-withdrawing) — groups like -NO₂, -CN, -COOH, -CHO withdraw electrons through resonance. Unlike the inductive effect (transmitted through σ bonds), resonance is transmitted through π bonds and is generally a stronger and longer-range effect. It explains stability of benzene, phenol acidity and aniline basicity.
What is hyperconjugation?
Hyperconjugation is the stabilising interaction between σ bonds (usually C-H or C-C of an alkyl group) and an adjacent empty p-orbital, π bond or anti-bonding orbital. It is sometimes called 'no-bond resonance'. NCERT Class 11 Chemistry Chapter 8 uses hyperconjugation to explain: (1) stability of carbocations (3° > 2° > 1° > methyl) — more α-H means more hyperconjugation, more stability; (2) stability of free radicals; (3) stability of alkenes (more substituted alkenes are more stable); (4) bond length variation. Each α-H provides one hyperconjugative structure. tert-Butyl carbocation has 9 hyperconjugative structures.
What is the difference between homolytic and heterolytic bond cleavage?
Homolytic cleavage is the symmetric breaking of a covalent bond in which each atom retains one of the two shared electrons, producing two neutral free radicals. Example: Cl-Cl → Cl• + Cl•. Heterolytic cleavage is the asymmetric breaking in which one atom takes both shared electrons, producing a cation and an anion. Example: H-Br → H⁺ + Br⁻. NCERT Class 11 Chemistry Chapter 8 explains that homolytic cleavage requires non-polar bonds and energy input (UV light, heat), while heterolytic cleavage occurs in polar bonds in polar solvents. Both are fundamental to understanding reaction mechanisms.
What are electrophiles and nucleophiles?
Electrophiles ('electron-loving') are electron-deficient species that accept an electron pair to form a new bond. Examples in NCERT Class 11 Chemistry Chapter 8: H⁺, NO₂⁺, SO₃, BF₃, carbocations (R⁺), Cl⁺. Nucleophiles ('nucleus-loving') are electron-rich species that donate an electron pair to form a new bond. Examples: OH⁻, CN⁻, NH₃, H₂O, RNH₂, Cl⁻, carbanions (R⁻). Electrophiles and nucleophiles drive most organic reactions: electrophilic substitution/addition, nucleophilic substitution/addition. Their behaviour parallels Lewis acid/base reactions: electrophile = Lewis acid, nucleophile = Lewis base.
What are common purification methods for organic compounds?
NCERT Class 11 Chemistry Chapter 8 describes several methods to purify organic compounds: (1) sublimation — purifies solids that sublime (camphor, naphthalene, anthracene) by heating and condensing vapour on a cool surface; (2) crystallisation — dissolves the impure solid in a hot solvent, then cools to crystallise pure compound, leaving impurities in solution; (3) distillation — separates liquids with different boiling points; simple distillation for large ΔT, fractional distillation for small ΔT, steam distillation for steam-volatile compounds, vacuum distillation for thermally unstable compounds; (4) differential extraction — uses immiscible solvents; (5) chromatography — column, TLC, paper chromatography for separating mixtures.
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