NCERT Plus One (Class 11) Chemistry • Chapter 8

Organic Chemistry – Some Basic Principles and Techniques

Carbon Tetravalence, IUPAC Nomenclature, Isomerism, Reaction Intermediates, Inductive/Resonance/Hyperconjugation, Purification & Analysis

📄 Official NCERT Textbook PDF

Topic 8.1 – 8.2: Carbon's Tetravalence, sp³/sp²/sp Hybridisation & Sigma/Pi Bond Counting

Around 1800, chemists believed living organisms possessed a mysterious "Vital Force" (proposed by Berzelius) needed to create organic molecules—like sugar, proteins, and DNA. Then in 1828, German chemist Friedrich Wöhler shattered that myth inside an ordinary glass flask! Simply by heating an inorganic salt, ammonium cyanate (NH4CNO), he produced urea (NH2CONH2). Soon Kolbe synthesised acetic acid (1845) and Berthelot made methane (1856). Why can one single element—carbon—build millions of compounds? Two superpowers: tetravalence (4 covalent bonds) and catenation (linking into endless chains and rings)!

Wöhler's Historic Synthesis (1828): NH4CNO (Ammonium cyanate, inorganic) Heat→ NH2CONH2 (Urea, organic). Carbon forms strong covalent bonds with other carbon atoms (catenation) and with H, O, N, S, P, and halogens.

1. Hybridisation, Shapes & % s-Character of Carbon

HybridisationExampleGeometry & Bond Angle% s-CharacterBond Nature & Electronegativity
sp3CH4, CH3ClTetrahedral (109.5°)25% sLongest, weakest C–H / C–C bonds; lowest EN
sp2H2C=CH2, H2C=OTrigonal planar (120°)33.3% sIntermediate bond length, enthalpy & EN
spHC≡CH, HC≡NLinear (180°)50% sShortest, strongest bonds; most electronegative C!

Why does % s-character matter? An s orbital sits closer to the positively charged nucleus than a p orbital. More s-character pulls bonding electrons tighter → shorter bond length, higher bond enthalpy, and higher electronegativity (sp > sp2 > sp3).

2. Characteristic Features of π (Pi) Bonds

  • Sideways Overlap & Planarity: Adjacent unhybridised p orbitals must stay parallel; thus in ethene (H2C=CH2), all 6 atoms lie in the same plane and both p orbitals stand perpendicular to that plane.
  • Restricted Rotation: Twisting one CH2 group breaks sideways p-overlap, so rotation around C=C is restricted.
  • High Reactivity: The π electron cloud lies above and below the molecular plane, making π electrons easily accessible to attacking reagents—so multiple bonds act as reactive centres!
NCERT Worked Problems 8.1 – 8.3:
• Problem 8.1(a) HC≡C–CH=CH–CH3: 4 C–C σ + 6 C–H σ = 10 σ bonds; 1 π(C=C) + 2 π(C≡C) = 3 π bonds.
• Problem 8.1(b) CH2=C=CH–CH3: 3 C–C σ + 6 C–H σ = 9 σ bonds; 2 π(C=C) = 2 π bonds.
• Problem 8.2 Hybridisation: CH3Cl (sp3) | (CH3)2CO (sp3, sp2) | CH3CN (sp3, sp) | HCONH2 (sp2) | CH3CH=CHCN (sp3, sp2, sp2, sp).
• Problem 8.3 Shapes: H2C=O (sp2, trigonal planar); CH3F (sp3, tetrahedral); HC≡N (sp, linear).
When counting σ and π bonds, students frequently forget the hidden C–H σ bonds! Always expand CH3 into three C–H σ bonds first. Also remember: 1 single bond = 1 σ; 1 double bond = 1 σ + 1 π; 1 triple bond = 1 σ + 2 π. Never count a π bond as σ!

Topic 8.3: Structural Representations: Complete, Condensed, Bond-Line & 3D Wedge-Dash

Writing out every single C and H for an 18-carbon fatty acid looks like a giant spiderweb that fills an entire notebook page! So chemists invented smart shortcuts—zooming out from a detailed complete dash formula to a compact condensed formula, and finally to an ultra-clean bond-line zig-zag that looks like mountain peaks while encoding every single carbon and hydrogen automatically.

1. Three 2D Ways to Write Organic Structures

  • Complete (Dash) Structural Formula: Shows every bond explicitly as a dash (– single, = double, ≡ triple). Heteroatom lone pairs (on O, N, S, X) may or may not be drawn.
  • Condensed Structural Formula: Omits most bond dashes and groups identical units with subscripts—e.g., octane becomes CH3(CH2)6CH3, and 3-methyloctane becomes CH3CH2CH(CH3)(CH2)4CH3.
  • Bond-Line Structural Formula: C and H atoms are omitted; carbon–carbon bonds are drawn in a zig-zag pattern. Every line junction (bend) and terminal end represents a carbon atom bonded to enough hydrogens to satisfy tetravalence (4 bonds), unless a heteroatom/functional group (–OH, –Cl, –CN, =O) is written!
NCERT Worked Problems 8.4 – 8.6 Practice:
• 2-Bromobutane: Condensed = CH3CHBrCH2CH3; Bond-line = 4-carbon zig-zag with Br at vertex 2.
• Cyclic Bond-Line Polygons: Triangle = Cyclopropane; Pentagon = Cyclopentane; Hexagon with –Cl = Chlorocyclohexane.
• Problem 8.5(a) HOCH2CH2CH2CH(CH3)CH(CH3)CH3 → Condensed: HO(CH2)3CH(CH3)CH(CH3)2.
• Problem 8.5(b) N≡C–CH(OH)–C≡N → Condensed: HOCH(CN)2.

2. Three-Dimensional (3D) Wedge-and-Dash & Molecular Models

To show 3D tetrahedral geometry on flat paper (Fig. 8.1 for CH4):

  • Normal line (—): Bond lying in the plane of the paper.
  • Solid wedge (▬▶): Bond projecting out of the plane towards the observer (broad end towards viewer).
  • Dashed wedge (⋯): Bond projecting away from the observer behind the paper.

Three Physical Molecular Models (Fig. 8.2): (1) Framework model (bonds only, ignores atom size), (2) Ball-and-stick model (atoms + bonds; uses springs for double bonds in ball-and-spring model), and (3) Space-filling model (no sticks; shows relative atomic sizes via van der Waals radii).

In bond-line structures like HO–CH2–... or ...–OH, the end of the line attached directly to O, Cl, N, or Br is NOT an extra CH3 methyl carbon—that bond goes straight to the heteroatom! Only a bare, unlabelled line tip counts as a –CH3 group!

Topic 8.4 – 8.5: Classification, Functional Groups, Homologous Series & IUPAC Nomenclature

Before the IUPAC system, chemists named molecules after where they discovered them—citric acid from citrus fruits and formic acid from red ants (Latin: formica)! With over ten million organic molecules today, guessing names from ants or lemons would be impossible. So the IUPAC System works like a global GPS address: Prefix (branches) + Root word (longest chain) + Primary suffix (ane/ene/yne) + Principal Functional Group suffix—telling any chemist on Earth the exact blueprint from the name alone!

1. Classification of Organic Compounds

  • I. Acyclic / Open-Chain (Aliphatic): Straight or branched chains (ethane, isobutane, acetaldehyde, acetic acid).
  • II. Cyclic / Closed-Chain:
    • (a) Alicyclic (aliphatic cyclic): Homocyclic/carbocyclic rings (cyclopropane, cyclohexane, cyclohexene) or heterocyclic non-aromatic rings (tetrahydrofuran, THF).
    • (b) Aromatic: Benzenoid (benzene, aniline, naphthalene), Non-benzenoid (tropone), and Heterocyclic aromatic (furan [O], thiophene [S], pyridine [N]).

Functional Group & Homologous Series: A functional group determines characteristic chemical reactivity. Successive members of a homologous series share the same general formula and differ by a –CH2– unit (14 u).

2. IUPAC Rules for Branched Alkanes & Substituted Benzenes

  1. Longest Continuous Chain: Choose the longest carbon chain as parent. If two chains tie in length, choose the one with more side chains.
  2. Lowest Locant Set: Number from the end giving substituents the lowest locant numbers (e.g., 2,5,6-trimethyloctane beats 3,4,7-trimethyloctane).
  3. Alphabetical Tie-Breaker: When two different substituents sit at equivalent positions, give the lower number to the one coming first alphabetically (e.g., 3-ethyl-5-methylheptane). Ignore di/tri/tetra and sec-/tert- during alphabetisation, but DO include iso- and neo-!
  4. Substituted Benzenes: Disubstituted positions 1,2 = ortho (o-), 1,3 = meta (m-), 1,4 = para (p-). For tri- or higher substitution, use lowest locants (e.g., 1-chloro-2,4-dinitrobenzene). When benzene is a substituent on a functionalised chain, call it phenyl (C6H5– / Ph).

3. Polyfunctional Priority Order (Decreasing Preference)

–COOH > –SO3H > –COOR > –COCl > –CONH2 > –CN > –CHO > >C=O > –OH > –NH2 > >C=C< > –C≡C–
Always PREFIX-only substituents: –R, –C6H5, –F, –Cl, –Br, –I, –NO2, –OR (alkoxy)
Class / GroupPrefix (Subordinate)Suffix (Principal)NCERT Example
Carboxylic acid (–COOH)carboxy--oic acid5-Oxohexanoic acid
Ester (–COOR) / Acyl halide (–COCl)alkoxycarbonyl- / halocarbonyl--oate / -oyl halideMethyl propanoate / Butanoyl chloride
Amide (–CONH2) / Nitrile (–CN)carbamoyl- / cyano--amide / -nitrileButanamide / Pentanenitrile
Aldehyde (–CHO) / Ketone (>C=O)formyl- or oxo- / oxo--al / -one6-Hydroxyheptanal / Hexane-2,4-dione
Alcohol (–OH) / Amine (–NH2)hydroxy- / amino--ol / -amine6-Methyloctan-3-ol / Butan-2-amine
Watch three classic IUPAC traps!
1. BrCH2CH=CH2 is 3-bromoprop-1-ene (NOT 1-bromoprop-2-ene)—double bonds beat halogens (which are always prefix-only)!
2. Carbon-containing principal groups like –COOH, –CHO, –CN include their own carbon atom as C-1 of the longest chain!
3. For multiple double/triple bonds (e.g., CH≡C–CH=CH–CH=CH2), retain the extra 'a' in the root: Hexa-1,3-dien-5-yne (and alkene gets lower locant than alkyne at equivalent ends)!

Topic 8.6: Isomerism: Chain, Position, Functional, Metamerism & Stereoisomerism

Give a builder the exact same box of 5 carbon bricks and 12 hydrogen connectors (C5H12). One builder snaps all 5 in a straight line (pentane, boiling at 36 °C); another builds a 4-carbon chain with a branch (isopentane, 28 °C); a third packs all 4 branches around one central carbon like a compact star (neopentane, 9.5 °C)! Same formula, completely different structures and properties—that magic is Isomerism!

Isomerism: Existence of two or more compounds having the same molecular formula but different properties. Broadly divided into Structural Isomerism (different atom connectivity) and Stereoisomerism (same connectivity, different 3D spatial arrangement: Geometrical & Optical).

Four Types of Structural Isomerism (NCERT 8.6.1)

TypeWhat Differs?Classic NCERT Example & Formula
1. Chain Isomerism Carbon skeleton (straight vs branched parent chain) C5H12: Pentane, Isopentane (2-methylbutane), Neopentane (2,2-dimethylpropane)
2. Position Isomerism Position of substituent atom, multiple bond, or functional group on same skeleton C3H8O: Propan-1-ol (CH3CH2CH2OH) & Propan-2-ol (CH3CH(OH)CH3)
3. Functional Group Isomerism Different functional groups altogether C3H6O: Propanal (CH3CH2CHO) & Propanone (CH3COCH3)
(Also C2H6O: Ethanol & Methoxymethane)
4. Metamerism Different alkyl chains on either side of a divalent/polyvalent functional group (like –O–, –S–, –NH–, –CO–) C4H10O: Methoxypropane (CH3OC3H7) & Ethoxyethane (C2H5OC2H5)

Stereoisomerism (8.6.2)

Compounds having the same constitution and sequence of covalent bonds, but differing in the relative positions of their atoms or groups in 3D space—classified into Geometrical isomerism (e.g., cis/trans across restricted C=C double bonds, Exercise 8.15b) and Optical isomerism.

Do NOT confuse Resonance Structures (like Exercise 8.15c, where only π/lone-pair electrons shift between identical nuclear positions) with Isomers! Isomers are real, isolable compounds; resonance contributors are hypothetical pictures of ONE single molecule. Also check whether carbonyl position changed (Pentan-2-one vs Pentan-3-one = Metamerism / Position) vs whether the functional group itself changed (Propanal vs Propanone = Functional isomerism)!

Topic 8.7.1 – 8.7.3: Bond Fission, Reaction Intermediates, Electrophiles & Nucleophiles

Imagine two partners breaking up a shared business. Option A (Homolytic cleavage): each partner walks away calmly with exactly half the joint account (one electron each)—leaving two restless Free Radicals. Option B (Heterolytic cleavage): the greedier partner grabs BOTH electrons and storms out negatively charged, leaving the other partner positively charged and electron-hungry! That exact split decides whether an organic reaction goes polar or free-radical.

Substrate + Attacking Reagent → [Reactive Intermediate] → Product(s) + Byproducts.

1. Homolytic vs Heterolytic Bond Fission & Curved Arrows

Cleavage TypeElectron Distribution & ArrowIntermediate FormedGeometry & Stability Order
Heterolytic
(Ionic / Polar)		 Shared e− pair goes to ONE atom.
Shown by full curved arrow (↷)		 Carbocation (C+, sextet = 6 e−)
OR Carbanion (C−, octet + lone pair)		 Carbocation: sp2, Trigonal planar (empty p orbital)
Stability: (CH3)3C+ > (CH3)2CH+ > CH3CH2+ > CH3+
Carbanion: sp3, Distorted tetrahedron
Homolytic
(Free-radical / Nonpolar)		 1 e− goes to each atom (Heat / UV light).
Shown by half-headed fish-hook arrow		 Free Radical (neutral species with unpaired e−, e.g. •CH3, •Cl) Stability increases from methyl to 3°:
(CH3)3C• > (CH3)2CH• > CH3CH2• > •CH3

2. Nucleophiles (Nu:) vs Electrophiles (E+)

  • Nucleophile (Nucleus-seeking, e− pair donor): Attacks electron-deficient sites. Includes anions with lone pairs (HO−, NC−, HS−, C2H5O−, R3C:−) and neutral molecules with lone pairs (H2O:, R3N:, R2NH).
  • Electrophile (Electron-seeking, e− pair acceptor): Attacks electron-rich sites (π bonds, lone pairs). Includes cations (CH3+, Cl+, NO2+, CH3C+=O) and neutral electron-deficient species (BF3, AlCl3) or partial-positive carbons in >C=O and R3C–X.
NCERT Problem 8.11–8.13 Highlights:
• CH3–SCH3 → CH3+ + −SCH3 (arrow to S); CH3–Cu → CH3:− + Cu+ (arrow to C because C is more EN than Cu!).
• Electrophilic carbon centres (δ+): starred carbon in CH3CH=O, H3C–C≡N, and H3C–I.
Always check arrow direction! A curved arrow starts at the electron source (bond or lone pair) and points to the electron sink (never from + charge to − charge!). Also remember: neutral molecules like H2O and NH3 are Nucleophiles (lone pair donors), while neutral BF3 (only 6 valence electrons!) is an Electrophile!

Topic 8.7.4 – 8.7.10: Electron Displacement Effects (±I, ±R, ±E, Hyperconjugation) & Reaction Types

Imagine pulling the end of a tug-of-war rope: the teammate closest to the pull feels a strong tug, the second feels less, and by the fourth person the pull has faded away—that is the Inductive Effect (±I) through σ bonds! Now imagine passing a bucket of water hand-to-hand across an alternating ring of mirrors—electrons glide effortlessly all the way around without losing steam: that is Resonance (±R) and Hyperconjugation!

1. Permanent vs Temporary Electronic Effects

EffectNature & Orbital Pathway+ Group (Donating) vs − Group (Withdrawing)Key Exam Application
Inductive Effect (±I) Permanent polarisation of σ-bonds; vanishes rapidly after 3 bonds +I: –CH3, –C2H5, alkyl groups
−I: –NO2, –CN, –COOH, –COOR, –F/Cl/Br/I, –OAr		 Acidity: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
Alkyl weakens acid: CH3CH2COOH > (CH3)2CHCOOH > (CH3)3CCOOH
Resonance / Mesomeric (±R) Permanent delocalisation of π–π or π–lone pair electrons in conjugated systems +R: –X, –OH, –OR, –OCOR, –NH2, –NHR, –NR2, –NHCOR (Aniline: high e− at o/p)
−R: –COOH, –CHO, >C=O, –CN, –NO2 (Nitrobenzene: low e− at o/p)		 Benzene C–C = 139 pm (intermediate between 154 & 134 pm). Canonical stability: more covalent bonds & complete octets > charge separation!
Hyperconjugation
(No-Bond Resonance)		 Permanent σ(C–H) → empty 2p or π* orbital overlap Stabilises carbocations, free radicals & alkenes (e.g. propene) (CH3)3C+ (9 α-H) > (CH3)2CH+ (6 α-H) > CH3CH2+ (3 α-H) > CH3+ (0 α-H)
Electromeric Effect (±E) Temporary complete transfer of π pair ONLY when reagent attacks +E: π e− shift TO atom where reagent attaches (H+ + C=C)
−E: π e− shift AWAY from attacking atom (CN− + >C=O)		 Effect disappears instantly when reagent is removed. When ±I and ±E oppose, E predominates!
NCERT Problem 8.17 & Ex 8.9:
• Resonance stability of CH2=CH–CHO: Neutral form I (all octets complete, no charge separation) > II (− on electronegative O, + on C) > III (+ on O, − on C — negligible!).
• Exercise 8.9: O2NCH2CH2O− is MORE stable than CH3CH2O− because the strong −I nitro group disperses negative charge, whereas methyl's +I effect intensifies it.

2. Four Classes of Organic Reactions (8.7.10)

(i) Substitution (CH3CH2I + KOH(aq) → CH3CH2OH + KI, nucleophilic), (ii) Addition ((CH3)2C=CH2 + HCl → (CH3)2C(Cl)CH3), (iii) Elimination (CH3CH2Br + HO− → CH2=CH2 + H2O + Br−), (iv) Rearrangement ((CH3)3CCH2OH + HBr → (CH3)2C(Br)CH2CH2CH3).

Remember TWO exam-gold facts: (1) Inductive, Resonance, and Hyperconjugation are PERMANENT; ONLY the Electromeric effect is TEMPORARY! (2) Alkyl groups act as electron donors on alkenes/arenes via Hyperconjugation (no-bond resonance, Exercise 8.10)—and in CH3+, the vacant 2p orbital is perpendicular to the C–H bonds, so hyperconjugation is ZERO!

Topic 8.8: Purification Techniques: Sublimation, Crystallisation, Distillation, Extraction & Chromatography

Suppose a perfumery in Kozhikode extracts fragrant jasmine oil, only to find it mixed with water, chlorophyll pigments, and plant wax. Boiling the mixture carelessly would burn the delicate scent! Instead, chemists match each impurity to a smart physical property—volatility, solubility at hot vs cold temperatures, vapour pressure under vacuum, or differential adsorption on silica gel—turning messy crude extracts into razor-pure crystals and liquids with sharp melting and boiling points.

Pure organic compounds exhibit sharp melting and boiling points. The five major NCERT purification techniques match specific impurity types:

MethodCore Principle & When UsedClassic NCERT Example
1. Sublimation Solid → vapour directly without melting; separates sublimable solids from non-sublimable impurities. Camphor + Calcium sulphate (Ex 8.27)
2. Crystallisation Compound is sparingly soluble at room temp, appreciably soluble in hot solvent; cool to crystallise pure solid, leaving impurities in mother liquor. Coloured impurities removed via activated charcoal. Benzoic acid / Sugar / Naphthalene in water or alcohol (mixed solvents if needed)
3a. Simple Distillation Separates volatile liquids from non-volatile impurities OR liquids with sufficient b.p. difference. Chloroform (334 K) + Aniline (457 K)
3b. Fractional Distillation Liquids with close boiling points; vapours climb a fractionating column (glass beads / bubble plates). Each condensation–vaporisation cycle = 1 theoretical plate. Crude oil fractions in petroleum industry; Acetone + Methanol
3c. Distillation Under Reduced Pressure Liquids with very high b.p. or that decompose at/below b.p. Lowering external pressure makes liquid boil at lower temperature! Glycerol from spent-lye in soap industry
3d. Steam Distillation Steam-volatile & water-immiscible liquids. Boils when patm = p1 + p2 (so organic liquid vaporises below 373 K!). Separated using separating funnel. Aniline from aniline–water mixture
4. Differential Extraction Shaking aqueous solution in a separating funnel with an immiscible organic solvent where compound is far more soluble (continuous extraction if low solubility). Benzoic acid extracted from water into ether/chloroform

5. Chromatography (Greek: Chroma = Colour)

The best and latest technique for isolation, separation, and purity testing! Components distribute between a stationary phase and a moving mobile phase (eluant).

  • Adsorption Chromatography (Adsorbent = silica gel or alumina):
    • Column Chromatography: Adsorbent packed in a glass tube with stopcock; least adsorbed component elutes fastest, most readily adsorbed stays near top.
    • Thin Layer Chromatography (TLC): ~0.2 mm adsorbent on glass plate (chromaplate). Colourless spots detected under UV light (fluorescence), in an iodine jar (brown spots), or via spray like ninhydrin (amino acids).
  • Partition Chromatography (Paper Chromatography): Special chromatography paper traps water inside it as the stationary phase; solvent rises by capillary action as the mobile phase.
Retardation Factor (Rf):   Rf = Distance moved by substance from base line (x) / Distance moved by solvent front from base line (y)   (Always 0 ≤ Rf ≤ 1)
Two high-yield distinctions in board exams:
1. TLC vs Paper Chromatography: TLC works by Adsorption (solid silica/alumina stationary phase), whereas Paper Chromatography works by Partition (water trapped inside cellulose paper acts as a liquid stationary phase)!
2. Never confuse Glycerol from spent-lye (Reduced Pressure Distillation — decomposes at normal b.p.) with Aniline from water (Steam Distillation — steam-volatile & water-immiscible, boils below 373 K)!

Topic 8.9 – 8.10: Qualitative Analysis (Lassaigne's Test) & Quantitative Estimation

Drop silver nitrate straight into liquid tetrachloromethane (CCl4) and absolutely nothing happens—no white precipitate! Why? Because chlorine in CCl4 is locked inside strong covalent bonds, not free Cl− ions! In 1843, Jean-Louis Lassaigne solved this brilliantly: fuse the organic compound with molten sodium metal to crack covalent bonds into simple ionic sodium salts (NaCN, Na2S, NaX) that light up in unmistakable test-tube colours!

1. Qualitative Detection (CuO & Lassaigne's Sodium Fusion Test)

  • C & H Detection: Heat compound with dry CuO → C oxidises to CO2 (turns limewater milky, CaCO3) and H oxidises to H2O (turns white anhydrous CuSO4 blue, CuSO4·5H2O).
  • Sodium Fusion Extract: Na + C + N → NaCN;   2Na + S → Na2S;   Na + X → NaX (X = Cl, Br, I).
ElementTest Reagents Added to Sodium ExtractObserved Result & Key Formula
Nitrogen (N)Boil with FeSO4 + acidify with conc. H2SO4Prussian Blue colour/ppt: Fe4[Fe(CN)6]3·xH2O
Sulphur (S)(a) Acetic acid + Lead acetate
(b) Sodium nitroprusside		(a) Black ppt of PbS
(b) Violet colour: [Fe(CN)5NOS]4−
Both N & SNa + C + N + S → NaSCN + Fe3+Blood-red colour: [Fe(SCN)]2+ (Excess Na breaks NaSCN → NaCN + Na2S)
Halogens (X)Boil with conc. HNO3 first, then add AgNO3Cl: White ppt (soluble in NH4OH) | Br: Pale yellow (sparingly soluble) | I: Yellow (insoluble)
Phosphorus (P)Fuse with Na2O2 → boil HNO3 + Ammonium molybdateCanary yellow ppt: (NH4)3PO4·12MoO3

2. Quantitative Analysis Formulas (NCERT 8.10)

• Liebig C & H (H2O in CaCl2, CO2 in KOH): %C = (12 × mCO₂ × 100) / (44 × m)  |  %H = (2 × mH₂O × 100) / (18 × m)
• Dumas Method (N2 over KOH): p1 = Patm − Aqueous tension; VSTP = (p1V1 × 273)/(760 × T1);   %N = (28 × VSTP × 100) / (22400 × m)
• Kjeldahl's Method (NH3 + H2SO4): %N = [1.4 × M × 2(V − V1/2)] / m  (Not valid for nitro, azo, or ring N like pyridine!)
• Carius Halogen & Sulphur: %X = (Atomic mass of X × mAgX × 100) / (Molar mass of AgX × m)  |  %S = (32 × mBaSO₄ × 100) / (233 × m)
• Phosphorus & Oxygen: %P = (31 × m1 × 100)/(1877 × m) or (62 × mMg₂P₂O₇ × 100)/(222 × m)  |  %O = 100 − Σ(others) or (32 × mCO₂ × 100)/(88 × m)
NCERT Problem Highlights (8.20 – 8.24):
• Prob 8.20 (C, H): 0.246 g gave 0.198 g CO2 & 0.1014 g H2O → %C = 21.95%, %H = 4.58%.
• Prob 8.21 (Dumas): 0.3 g gave 50 mL N2 at 300 K, 715 mm (aq. tension 15 mm → p1 = 700 mm) → VSTP = 41.9 mL → %N = 17.46%.
• Prob 8.22 (Kjeldahl): NH3 from 0.5 g neutralised 10 mL of 1 M H2SO4 (= 20 mL 1 M NH3) → %N = 56.0%.
• Prob 8.23 (Carius Br): 0.15 g gave 0.12 g AgBr (M = 188) → %Br = (80 × 0.12 × 100)/(188 × 0.15) = 34.04%.
• Prob 8.24 (Carius S): 0.157 g gave 0.4813 g BaSO4 (M = 233) → %S = (32 × 0.4813 × 100)/(233 × 0.157) = 42.10%.
Four classic NCERT & Kerala DHSE exam "WHY?" questions:
1. Why boil Lassaigne's extract with conc. HNO3 before AgNO3? To decompose interfering NaCN and Na2S (otherwise they form AgCN/Ag2S precipitates)!
2. Why use acetic acid (NOT H2SO4) in the lead acetate sulphur test? Sulphuric acid would react with Pb2+ to form white PbSO4!
3. Why doesn't CCl4 give AgCl directly? Covalent C–Cl bonds produce no free Cl− ions!
4. Which N-compounds FAIL Kjeldahl's method? Nitro (–NO2), Azo (–N=N–), and ring N (pyridine)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 2 Marks): What is Lassaigne's test? Why is an organic compound fused with metallic sodium before testing for nitrogen, sulphur, and halogens?

Lassaigne's Test: A qualitative method to detect N, S, and halogens (Cl, Br, I) in organic compounds.

Reason for Sodium Fusion: Elements in organic compounds are bonded covalently and do not give free ions in water. Fusing with metallic sodium converts them into water-soluble ionic compounds: Na + C + N → NaCN; 2Na + S → Na2S; Na + X → NaX, which readily undergo ionic precipitation/colour tests.

Q2 (Kerala HSE SAY / NCERT Ex 8.25 & 8.29 • 2 Marks): (a) Why is Lassaigne's extract boiled with concentrated HNO3 before adding AgNO3 while testing for halogens? (b) Will CCl4 give a white precipitate of AgCl on heating with AgNO3? Give reason.

(a) If N or S is present, Lassaigne's extract contains NaCN or Na2S, which would react with AgNO3 to form interfering precipitates (AgCN or Ag2S). Boiling with conc. HNO3 decomposes NaCN and Na2S into volatile HCN and H2S gases.

(b) No—CCl4 is purely covalent and does not ionise to give free Cl− ions needed to precipitate AgCl.

Q3 (Kerala HSE March • 2 Marks): Arrange the following carbocations in increasing order of stability and explain the reason: CH3+, (CH3)2CH+, CH3CH2+, (CH3)3C+.

Increasing Stability Order: CH3+ < CH3CH2+ < (CH3)2CH+ < (CH3)3C+.

Explanation: Alkyl groups directly attached to the positively charged sp2 carbon disperse positive charge through both +I (inductive) effect and hyperconjugation (9 α-H in 3°, 6 in 2°, 3 in 1°, 0 in CH3+). Greater hyperconjugation gives greater stability.

Q4 (Kerala HSE Model / NCERT 8.8 • 3 Marks): How will you separate the following mixtures? State the principle in each case: (a) Chloroform (b.p. 334 K) and Aniline (b.p. 457 K) (b) Glycerol from spent-lye (c) Camphor and Calcium sulphate.

(a) Simple Distillation: Large difference in boiling points (123 K); lower-boiling chloroform vaporises and condenses first.

(b) Distillation under Reduced Pressure: Glycerol has a high b.p. and decomposes before boiling at atmospheric pressure; lowering external pressure allows it to boil at a lower temperature without decomposition.

(c) Sublimation: Camphor sublimes directly from solid to vapour on heating while calcium sulphate remains behind as non-sublimable residue.

Q5 (Kerala HSE March (Repeated) • 3 Marks): Write the IUPAC names of: (a) CH3–CH2–CH(OH)–CH2–CH2–CH(CH3)–CH2–CH3 (b) CH3–CO–CH2–CH2–CH2–COOH (c) HC≡C–CH=CH–CH=CH2.

(a) 6-Methyloctan-3-ol (longest 8-C chain; –OH gets lowest locant 3).

(b) 5-Oxohexanoic acid (–COOH is principal functional group at C-1; ketone at C-5 is prefix oxo-).

(c) Hexa-1,3-dien-5-yne (double bonds receive lower locants 1,3 than triple bond 5; 'a' retained in hexa).

Q6 (Kerala HSE SAY • 2 Marks): What colour is obtained in Lassaigne's test when an organic compound contains: (a) Nitrogen alone (b) Both Nitrogen and Sulphur together? Give the chemical species responsible for each colour.

(a) Nitrogen alone: Prussian blue due to iron(III) hexacyanidoferrate(II), Fe4[Fe(CN)6]3·xH2O.

(b) Both N and S together: Blood-red colour due to formation of sodium thiocyanate (NaSCN) giving [Fe(SCN)]2+ with Fe3+ ions.

Q7 (Kerala HSE March (Standard Numerical) • 3 Marks): In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find the percentage of bromine in the compound. [Atomic mass: Ag = 108, Br = 80]

Molar mass of AgBr = 108 + 80 = 188 g mol-1.
188 g AgBr contains 80 g Br.
Percentage of Br = (80 × 0.12 × 100) / (188 × 0.15) = 34.04%.

Q8 (Kerala HSE / NCERT Problem 8.21 • 4 Marks): In Dumas method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)

Actual pressure of dry N2, p1 = 715 − 15 = 700 mm Hg.
Volume at STP, V = (p1V1 × 273) / (760 × T1) = (700 × 50 × 273) / (760 × 300) = 41.9 mL.
22,400 mL N2 at STP weighs 28 g.
% of Nitrogen = (28 × 41.9 × 100) / (22400 × 0.3) = 17.46%.

Q9 (Kerala HSE Improvement • 2 Marks): Distinguish between Inductive effect and Electromeric effect (any two points). Identify which explains: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH.

Inductive Effect: (1) Permanent effect operating in ground state. (2) Involves slight displacement of σ-bond electrons along a saturated chain without reagent.

Electromeric Effect: (1) Temporary effect operating only in presence of attacking reagent. (2) Involves complete transfer of π-electron pair in multiple bonds.

The carboxylic acid acidity order is explained by the −I (Inductive) effect of chlorine atoms.

Q10 (Kerala HSE March • 2 Marks): Explain the principle of Paper Chromatography. Define Retardation Factor (Rf).

Principle: Paper chromatography is a type of partition chromatography based on continuous differential partitioning of mixture components between a stationary phase (water trapped inside special chromatography paper) and a moving mobile phase (solvent rising by capillary action).

Retardation Factor (Rf): Rf = (Distance moved by the substance from base line, x) / (Distance moved by the solvent front from base line, y).

Q11 (Kerala HSE Model • 1 Marks): Name two classes of nitrogen-containing organic compounds for which Kjeldahl's method is NOT applicable.

Kjeldahl's method is not applicable to compounds containing nitrogen in nitro groups (–NO2), azo groups (–N=N–), or nitrogen present inside a ring (e.g., pyridine), because their nitrogen does not convert to ammonium sulphate with conc. H2SO4.

Q12 (Kerala HSE SAY • 3 Marks): A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M NaOH for neutralisation. Find the percentage of nitrogen. (NCERT Ex 8.33)

60 mL of 0.5 M NaOH neutralises 30 mL of 0.5 M H2SO4 (since H2SO4 is dibasic: 2 NaOH ≡ 1 H2SO4).
Acid consumed by NH3 = 50 − 30 = 20 mL of 0.5 M H2SO4 = 40 mL of 0.5 M NH3 = 20 mL of 1 M NH3.
Mass of N = (14 × 20) / 1000 = 0.28 g.
% of Nitrogen = (0.28 / 0.50) × 100 = 56.0%.