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8.1 General Introduction

Organic compounds are vital for sustaining life on Earth — from the DNA that carries our genetic code to the proteins in our blood, muscles and skin. They also appear in fuels, polymers, dyes, medicines, and clothing.

Definition

Organic Chemistry is the branch of chemistry dealing with carbon compounds — primarily those formed with H, O, N, S, P, and halogens through covalent bonds.

Catenation: Carbon's unique ability to form covalent bonds with other carbon atoms — the root of organic chemistry's diversity.

Historical Milestones

Year Scientist Event
~1780 Berzilius (Sweden) Proposed Vital Force Theory — organic compounds can only be made inside living organisms by a mysterious "vital force"
1828 F. Wöhler Disproved Vital Force — synthesised urea (organic) from ammonium cyanate (inorganic) in a lab
1845 Kolbe Synthesised acetic acid from inorganic sources
1856 Berthelot Synthesised methane from inorganic sources
Later Electronic theory of covalent bonding → modern organic chemistry
Wöhler's Synthesis (1828)
NH4CNO  —(Heat)→  NH2CONH2
Ammonium cyanate  →  Urea (first organic compound synthesised from inorganic source)

8.2 Tetravalence of Carbon & Shapes of Organic Molecules

Carbon (atomic number 6, config 1s²2s²2p²) has 4 valence electrons and always forms 4 bonds. The shape depends on hybridisation.

8.2.1 Hybridisation & Molecular Shapes

sp³ Hybridisation
C H H H H
Methane (CH₄)
Tetrahedral  ·  109.5°
sp² Hybridisation
C C H H H H
Ethene (C₂H₄)
Trigonal Planar  ·  120°
sp Hybridisation
H C C H
Ethyne (C₂H₂)
Linear  ·  180°
Hybridisation s character Shape Bond Angle Example Bond character
sp³ 25% Tetrahedral 109.5° CH4 Longest, weakest, least electronegative C
sp² 33% Trigonal Planar 120° C2H4 Intermediate length & strength
sp 50% Linear 180° C2H2 Shortest, strongest; most electronegative C
Key Rule

Greater s-character → Greater electronegativity of C → Shorter & stronger bonds formed.

Electronegativity order: sp (50% s) > sp² (33% s) > sp³ (25% s)

Orbital Hybrid Diagrams
sp³, sp², sp Orbital Hybrid Diagrams

8.2.2 Characteristics of π Bonds

Wedge-and-Dash Representation of Methane
Wedge-and-Dash Representation of CH₄ (Methane)
Molecular Models
Three Types of Molecular Models: Framework, Ball-and-Stick, and Space-Filling

8.3 Structural Representations of Organic Compounds

Type Description Example – Ethane
Complete / Lewis Structure Every bond shown as a dash; all atoms written; lone pairs may be shown H–C(H)(H)–C(H)(H)–H (full expanded)
Condensed Formula Dashes omitted; identical groups shown with subscripts CH3CH3
Bond-line (Skeletal) Formula C and H not shown; C–C bonds as zig-zag lines; heteroatoms written A straight zig-zag line (2 carbons)
Bond-Line Rules
  • Terminal ends of lines = –CH3 groups (unless a functional group is shown there)
  • Line junctions = carbon atoms — add enough H's to complete 4 bonds
  • Atoms like O, N, Cl, Br ARE written explicitly
  • Carbon and hydrogen are NOT written — they are implied

Expanded Structural Formulas of Important Compounds

Methanol
H C H H O H
CH₃OH
Formaldehyde (Methanal)
H C H O
HCHO  ·  Methanal
Acetic Acid (Ethanoic Acid)
H C H H C O OH
CH₃COOH
Complete Structural Formulas
Complete Structural Formulas – Ethane, Ethene, Ethyne, Methanol

8.4 Classification of Organic Compounds

Classification Mind Map
Classification of Organic Compounds — Comprehensive Mind Map

Benzene Ring SVG Representations

Aromatic compounds are centred around the benzene ring. All substituents are shown in matching amber style, properly connected to the ring.

Benzene
C₆H₆
Toluene
CH₃
Methylbenzene
Aniline
NH₂
Aminobenzene
Chlorobenzene
Cl
Chlorobenzene
Nitrobenzene
NO₂
Nitrobenzene
Phenol
OH
Phenol (C₆H₅OH)

Ortho, Meta & Para Disubstituted Benzene

1,2-Dibromobenzene
1 2 Br Br
ortho (o-)
Positions 1,2
1,3-Dibromobenzene
1 3 Br Br
meta (m-)
Positions 1,3
1,4-Dibromobenzene
1 4 Br Br
para (p-)
Positions 1,4

8.4.1 Functional Group

Definition

An atom or group of atoms joined to the carbon chain that is responsible for the characteristic chemical properties of the compound.

Examples: –OH (hydroxyl), –CHO (aldehyde), –COOH (carboxylic acid), –NH2 (amino), C=C (alkene), C≡C (alkyne)

8.4.2 Homologous Series

Definition

A series of organic compounds each containing the same functional group; each successive member differs from the previous by –CH2– (14 mass units). Members are called homologues.


8.5 Nomenclature of Organic Compounds (IUPAC)

IUPAC = International Union of Pure and Applied Chemistry. In this system, names are correlated with structure so that the reader can deduce structure from name.

8.5.2 IUPAC Nomenclature of Alkanes

Suffix: –ane. Prefix = number of carbons.

Name Formula Name Formula
Methane CH4 Heptane C7H16
Ethane C2H6 Octane C8H18
Propane C3H8 Nonane C9H20
Butane C4H10 Decane C10H22
Pentane C5H12 Icosane C20H42
Hexane C6H14 Triacontane C30H62

Branched Alkyl Groups – SVG Structures

Isopropyl
CH₃ CH₃ CH (attach)
(CH₃)₂CH–
tert-Butyl
CH₃ C CH₃ CH₃ (attach)
(CH₃)₃C–
Neopentyl
CH₃ C CH₃ CH₃ CH₂ (attach)
–CH₂C(CH₃)₃

Rules for Naming Branched-Chain Alkanes

6 Golden Rules
  1. Longest chain = parent (root) alkane
  2. Lowest locants: number from end closest to first branch, so branched C-atoms get the lowest possible numbers
  3. Name branches alphabetically: different alkyl groups listed in alphabetical order; numbers separated by hyphens from group names
  4. Identical groups: use di-, tri-, tetra-, etc. (these prefixes are NOT considered in alphabetical ordering)
  5. Equivalent positions: the substituent coming first alphabetically gets the lower number
  6. Branched substituents: number the substituent from its attachment point (C1 of branch); name placed in parentheses
KEY CONCEPT

Functional Group Priority Order (Decreasing)

  • 1–COOH  → suffix: -oic acid
  • 2–SO3H  → suffix: sulphonic acid
  • 3–COOR  → suffix: -oate
  • 4–COX (acyl halide)  → suffix: -oyl halide
  • 5–CONH2 (amide)  → suffix: -amide
  • 6–CN (nitrile)  → suffix: nitrile
  • 7–CHO (aldehyde)  → suffix: -al
  • 8>C=O (ketone)  → suffix: -one
  • 9–OH (alcohol)  → suffix: -ol
  • 10–NH2 (amine)  → suffix: -amine
  • 11C=C (alkene)  → suffix: -ene
  • 12C≡C (alkyne)  → suffix: -yne

Always prefix (never suffix): –R, C6H5– (phenyl), –F, –Cl, –Br, –I, –NO2, alkoxy (–OR)


8.6 Isomerism

Definition

Isomerism: Two or more compounds with the same molecular formula but different properties. Such compounds are isomers.

Type Cause Example
Chain Isomerism Different carbon skeletons C5H12: Pentane, 2-Methylbutane (Isopentane), 2,2-Dimethylpropane (Neopentane)
Position Isomerism Same skeleton & functional group, different position C3H8O: Propan-1-ol vs Propan-2-ol
Functional Group Isomerism Same formula, different functional groups C3H6O: Propanone (ketone) vs Propanal (aldehyde)
Metamerism Different alkyl chains on either side of the same functional group C4H10O: CH3OC3H7 vs C2H5OC2H5
Stereoisomerism – Geometrical Same bonds, different spatial arrangement due to restricted rotation (C=C) cis-But-2-ene vs trans-But-2-ene
Stereoisomerism – Optical Same bonds, non-superimposable mirror images (chiral centre) L- and D-lactic acid
Cis-Trans Isomers
Geometrical Isomerism: cis and trans But-2-ene

8.7 Fundamental Concepts in Organic Reaction Mechanism

8.7.1 Fission of Covalent Bonds

Heterolytic Cleavage (Ionic)

Shared pair stays with one fragment → charged species.

CH3–Br  →  CH3+ + Br
  • Carbocation (R+): C with 6e, sp² hybridised, trigonal planar, very reactive
  • Carbanion (R): C gets the pair, sp³ hybridised, distorted tetrahedral
  • → Polar / Ionic reactions
Homolytic Cleavage (Free Radical)

Each fragment gets one electron → neutral radicals.

R–X  —(heat/light)→  R• + X•
  • Free radical (R•): neutral, one unpaired electron, very reactive
  • Shown with half-headed (fish hook) curved arrow
  • → Homopolar / Free radical reactions
Carbocation and Carbanion Shapes
Shapes of Methyl Carbocation (sp²) and Methyl Carbanion (sp³)

Carbocation Stability Order

·CH₃methyl
<
1° R⁺primary
<
2° R⁺secondary
<
3° R⁺tertiary

Reason: More alkyl groups → more hyperconjugation + inductive stabilisation → more stable carbocation.

8.7.2 Nucleophiles and Electrophiles

Nucleophile (Nu:)

= Nucleus-seeker. Electron-rich. Donates electron pair to electrophile.

Charged: HO, CN, HS, C2H5O, R3C:

Neutral with lone pair: H2O, NH3, R3N, ROH

Electrophile (E+)

= Electron-seeker. Electron-deficient. Accepts electron pair from nucleophile.

Positive ions: H+, NO2+, CH3+

Neutral: BF3, AlCl3, SO3; R–X (δ+ on C), >C=O

8.7.4 – 8.7.9 Electron Displacement Effects

Inductive Effect
Inductive Effect in Chloroethane
Effect Nature Transmission Key Feature
Inductive Permanent, ground state Through σ-bonds; decreases rapidly; almost zero after 3 bonds Electron withdrawal (–I: halogens, –NO2, –CN) or donation (+I: alkyl groups)
Resonance (Mesomeric) Permanent, ground state Through conjugated π system +R (electron away from group): –OH, –NH2, halogens  |  –R (electrons toward group): –COOH, –CHO, –NO2, –CN
Electromeric Temporary — only when reagent approaches Complete π electron transfer to one atom on demand +E: electrons go to atom where reagent attaches  |  –E: electrons go away from attachment point. Annulled when reagent removed.
Hyperconjugation Permanent σ(C–H) electrons → adjacent π or empty p orbital "No-bond resonance". More alkyl groups = more stabilisation of carbocation/free radical/alkene

Resonance in Benzene – Key Diagrams

Key Fact

Benzene's actual C–C bond length = 139 pm — intermediate between C–C single (154 pm) and C=C double (134 pm). This proves it is a resonance hybrid, not alternating single/double bonds.

Acetate Resonance
Resonance Structures of CH₃COO⁻ (Acetate Ion)

8.7.9 Hyperconjugation – Explained with SVG

Hyperconjugation = σ(C–H) electrons of an adjacent alkyl group delocalise into the empty p-orbital of a carbocation or π-system. This disperses the positive charge and stabilises the species.

H H H aligning C C⁺ empty p-orbital No-bond Resonance: H–C–C⁺ ↔ H⁺ + C=C σ(C–H) electrons delocalise into empty p orbital More C–H bonds adjacent to C⁺ = More stabilisation CH₃⁺ (0) < 1° (3 C–H) < 2° (6 C–H) < 3° (9 C–H) ∴ tert-butyl cation is the MOST stable
Hyperconjugation in ethyl cation — one σ(C–H) bond aligns with the empty p-orbital of C⁺, delocalising electrons and dispersing the positive charge.

8.8 Methods of Purification of Organic Compounds

Method Principle Used For Key Details
Sublimation Solid → vapour directly (bypasses liquid state) Sublimable compounds from non-sublimable impurities Camphor, iodine, naphthalene, benzoic acid
Crystallisation Difference in solubility at different temperatures Purification of solid organic compounds Dissolve at high T in minimum solvent → cool → filter crystals. Activated charcoal removes coloured impurities.
Simple Distillation Difference in boiling points Volatile liquid from non-volatile impurities; liquids with large bp difference CHCl3 (bp 334 K) from aniline (bp 457 K)
Fractional Distillation Same + fractionating column (multiple condensation-vaporisation steps) Liquids with close boiling points (e.g. petroleum) Each step = "theoretical plate". Commercially: hundreds of plates.
Vacuum Distillation Liquid boils when vapour pressure = reduced external pressure Liquids with very high bp or those that decompose at their bp Glycerol in soap industry. Uses water pump / vacuum pump.
Steam Distillation p(total) = p₁(organic) + p₂(water); boiling below 373 K Steam volatile, water-immiscible compounds Aniline separation. Since p₁ < p (atm), boiling occurs below 373 K.
Differential Extraction Organic compound more soluble in organic solvent than water Extracting organic compound from aqueous medium Separating funnel; two immiscible layers form; organic layer separated.
Chromatography Differential migration between stationary and mobile phases Separation, purification, identification Best modern technique. TLC, column, paper chromatography.
Simple Distillation Apparatus
Simple Distillation Apparatus
Fractional Distillation
Fractional Distillation Setup and Column Types
Distillation Under Reduced Pressure
Distillation Under Reduced Pressure (Vacuum Distillation) ·
Differential Extraction
Differential Extraction in a Separating Funnel

Rf Value in Chromatography

$$R_f = \frac{\text{Distance moved by substance from base line } (x)}{\text{Distance moved by solvent from base line } (y)}$$

Rf is characteristic of a compound in a given solvent system. Range: 0 to 1. Different compounds have different Rf values → basis of separation and identification.

Chromatography
Developing Chromatogram in TLC Jar and Developed Plate ·

8.9 Qualitative Analysis of Organic Compounds

8.9.1 Detection of Carbon and Hydrogen

Method

Heat compound with copper(II) oxide (CuO). Carbon → CO2 (tested with lime water); Hydrogen → H2O (tested with anhydrous CuSO4).

C + 2CuO  →  2Cu + CO2   (carbon burned)
2H + CuO  →  Cu + H2O   (hydrogen burned)
CO2 + Ca(OH)2  →  CaCO3↓ + H2O   (lime water turns milky)
5H2O + CuSO4  →  CuSO4·5H2O   (white anhydrous salt turns blue)

8.9.2 Lassaigne's Test (N, S, Halogens, P)

KEY CONCEPT

Principle: Organic compound fused with metallic sodium → converts covalent N, S, X into ionic (water-soluble) forms: NaCN, Na2S, NaX. The fused mass dissolved in distilled water = Sodium Fusion Extract (SFE).

Na + C + N  —(heat)→  NaCN
2Na + S  —(heat)→  Na2S
Na + X  —(heat)→  NaX  (X = Cl, Br, I)
Element Reagent Added to SFE Positive Test Observation Key Reaction
Nitrogen Boil with FeSO4 → acidify with conc. H2SO4 Prussian blue colour 6CN + Fe2+ → [Fe(CN)6]4– then 3[Fe(CN)6]4– + 4Fe3+ → Fe4[Fe(CN)6]3 (Prussian blue)
Sulphur (a) Acidify with acetic acid + lead acetate Black precipitate (PbS) S2– + Pb2+ → PbS↓ (black)
Sulphur (b) Sodium nitroprusside solution Violet colour S2– + [Fe(CN)5NO]2– → [Fe(CN)5NOS]4– (violet)
Chlorine Acidify with HNO3 + AgNO3 White ppt — soluble in NH4OH Cl + Ag+ → AgCl↓ (white)
Bromine Acidify with HNO3 + AgNO3 Pale yellow ppt — sparingly soluble in NH4OH Br + Ag+ → AgBr↓ (pale yellow)
Iodine Acidify with HNO3 + AgNO3 Yellow ppt — insoluble in NH4OH I + Ag+ → AgI↓ (yellow)
Phosphorus Oxidise with Na2O2 → boil with HNO3 → ammonium molybdate Yellow ppt (ammonium phosphomolybdate) H3PO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3PO4·12MoO3
⚠️ Special Cases

N + S both present: NaSCN forms → gives blood red colour with Fe3+ (no Prussian blue since no free CN). With excess Na: NaSCN decomposes → NaCN + Na2S → normal tests work.

Before halogen test: If N or S also present → first boil SFE with conc. HNO3 (to destroy NaCN and Na2S which would precipitate as AgCN/Ag2S and give false positives).

Why acetic acid, not H2SO4 in sulphur test? H2SO4 provides SO42– → PbSO4 precipitate (false positive). Acetic acid = inert, no interference.


8.10 Quantitative Analysis

8.10.1 Estimation of Carbon & Hydrogen (Liebig's Method)

Method

Known mass of compound burned in excess O2 with CuO. CO2 absorbed in KOH U-tube; H2O absorbed in CaCl2 U-tube. Increase in mass of each U-tube gives the amounts.

CxHy + (x + y/4) O2  →  x CO2 + (y/2) H2O
Let: m = mass of compound (g); m1 = mass of H2O (g); m2 = mass of CO2 (g) $$\% \text{ C} = \frac{12 \times m_2 \times 100}{44 \times m} \qquad \% \text{ H} = \frac{2 \times m_1 \times 100}{18 \times m}$$
📝 Solved Problem

0.246 g of organic compound gave 0.198 g CO2 and 0.1014 g H2O on combustion.

% C = (12 × 0.198 × 100) ÷ (44 × 0.246) = 21.95%

% H = (2 × 0.1014 × 100) ÷ (18 × 0.246) = 4.58%

Estimation of Carbon & Hydrogen
Estimation of Carbon & Hydrogen (Liebig's Combustion Apparatus)

8.10.2 Estimation of Nitrogen

Dumas Method

Compound + CuO in CO2 atmosphere → N2 gas collected over KOH. Measure volume.

CxHyNz + CuO → CO2 + H2O + N2

Convert volume to STP:   $V_{STP} = \dfrac{p_1 V_1 \times 273}{760 \times T_1}$

where p1 = atmospheric pressure − aqueous tension

$$\% N = \frac{28 \times V_{STP} \times 100}{22400 \times m}$$
Kjeldahl's Method

Compound + conc. H2SO4 → (NH4)2SO4; then + excess NaOH → NH3 gas absorbed in standard H2SO4; back-titrate unreacted acid with NaOH.

$$\% N = \frac{1.4 \times M \times 2(V - V_1/2)}{m}$$

M = molarity of H2SO4, V = vol of H2SO4 taken, V1 = vol of NaOH used, m = mass of compound.

NOT applicable to: –NO2 (nitro), azo groups, N in ring (e.g. pyridine).

📝 Solved Problem – Dumas Method

0.3 g compound → 50 mL N2 at 300 K and 715 mm pressure. Aqueous tension at 300 K = 15 mm.

Actual pressure = 715 – 15 = 700 mm

Volume at STP = (273 × 700 × 50) ÷ (300 × 760) = 41.9 mL

% N = (28 × 41.9 × 100) ÷ (22400 × 0.3) = 17.46%

Dumas Method
Dumas Method Apparatus for Nitrogen Estimation
Kjeldahl's Method
Kjeldahl's Method Apparatus for Nitrogen Estimation

8.10.3 Halogens – Carius Method

Method

Compound + fuming HNO3 + AgNO3 → sealed Carius tube in furnace → halogen forms silver halide (AgX). Filter, dry, weigh AgX.

$$\% X = \frac{\text{at. mass of X} \times m_1 \times 100}{\text{mol. mass of AgX} \times m}$$
📝 Solved Problem

0.15 g compound → 0.12 g AgBr.   M(AgBr) = 108 + 80 = 188 g/mol

% Br = (80 × 0.12 × 100) ÷ (188 × 0.15) = 34.04%

Carius Tube
Carius Tube for Halogen / Sulphur Estimation

8.10.4 Sulphur & Complete Summary

Element Method Product Measured Formula
C Liebig (CuO combustion) CO2 absorbed in KOH U-tube % C = 12×m2×100 / (44×m)
H Liebig (CuO combustion) H2O absorbed in CaCl2 U-tube % H = 2×m1×100 / (18×m)
N Dumas / Kjeldahl's N2 gas / NH3 absorbed in H2SO4 See formulas above
X (halogens) Carius (fuming HNO3 + AgNO3) AgX (silver halide) % X = at.mass(X)×m1×100 / (M(AgX)×m)
S Carius (fuming HNO3) BaSO4 precipitate % S = 32×m1×100 / (233×m)
P Fuming HNO3 + ammonium molybdate (NH4)3PO4·12MoO3 % P = 31×m1×100 / (1877×m)
O By difference % O = 100 − Σ(all other %)

✏️ Practice Questions

Q1
What are the hybridisation states of each carbon atom in the following compounds?
CH2=C=O,   CH3CH=CH2,   (CH3)2CO,   CH2=CHCN,   C6H6
Q2
How many σ and π bonds are present in HC≡C–CH=CH–CH3?
Q3
Write the IUPAC name of: HOCH2CH2CH2CH(CH3)CH(CH3)CH3
Q4
Which is more stable: O2NCH2CH2O or CH3CH2O? Give reason.
Q5
In Dumas' method for nitrogen estimation, 0.3 g of an organic compound gave 50 mL of N2 collected at 300 K and 715 mm pressure. Aqueous tension at 300 K = 15 mm. Find % nitrogen.
Q6
0.5 g organic compound → Kjeldahl's method → NH3 neutralised 10 mL of 1 M H2SO4. Find % N.
Q7
In Carius method, 0.15 g compound gave 0.12 g AgBr. Find % Br. [M(AgBr) = 188 g/mol]
Q8
Classify these reactions as substitution, addition, or elimination:
(a) CH3CH2Br + HS → CH3CH2SH + Br
(b) (CH3)2C=CH2 + HCl → (CH3)2ClC–CH3
(c) CH3CH2Br + HO → CH2=CH2 + H2O + Br
Q9
Why does benzene have equal C–C bond lengths of 139 pm, even though its Kekulé structures show alternating single and double bonds?
Q10
Identify the electrophilic centres in: CH3CH=O, CH3CN, CH3I
Q11
An organic compound contains 69% C and 4.8% H (remainder is O). Calculate masses of CO2 and H2O produced when 0.20 g of this compound undergoes complete combustion.
Q12
Will CCl4 give a white precipitate of AgCl when treated with AgNO3? Give reason.
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