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General Organic Chemistry (GOC)

CBSE Class 11 & JEE Mains • Module 01 of 20 • The Foundation of All Organic Chemistry

📍 Chapter Overview

General Organic Chemistry — Mind Map

This chapter establishes the conceptual foundation for every organic reaction you will ever study. Master bond fission, reactive intermediates, and electronic effects — everything else builds on this.

Topics Covered: Bond Fission (Homo/Heterolytic) · Carbocations · Carbanions · Free Radicals · Inductive Effect · Resonance · Hyperconjugation · Aromaticity (Hückel) · Acidity & Basicity of Organic Molecules · Types of Reactions

🤖 AI Image Prompt (Mind Map / Chapter Overview): A vibrant, colorful organic chemistry mind map on a dark navy background. Central node labeled "General Organic Chemistry (GOC)" with 6 branches radiating outward: (1) "Bond Fission" with sub-nodes Homolysis → Free Radicals and Heterolysis → Ions; (2) "Reactive Intermediates" with Carbocation (sp2, planar), Carbanion (sp3, pyramidal), Free Radical; (3) "Electronic Effects" with Inductive Effect, Resonance/Mesomeric, Hyperconjugation; (4) "Aromaticity" with Hückel rule 4n+2, benzene ring icon; (5) "Acidity & Basicity" with conjugate base stability factors; (6) "Types of Reactions" with SR, AE, AN, EAS. Use neon green, electric blue, and golden yellow colors. Chemistry textbook illustration style, educational poster quality, ultra detailed, high resolution.

1. Bond Fission — How Bonds Break

When a covalent bond breaks, the two electrons are distributed between atoms in two distinct ways:

A. Homolytic Fission — produces Free Radicals

Each atom gets one electron. Uses fish-hook (half-headed) arrows.

Conditions: UV light, high temperature, non-polar solvents.
Example: Cl–Cl → 2 Cl• (UV light)

B. Heterolytic Fission — produces Ions

Both electrons go to one atom. Uses full curved arrows.

Conditions: Polar solvents, ionic conditions.
Example: HCl → H⁺ + Cl⁻ in water

2. Carbocations (Carbenium Ions)

Definition: Carbon with a positive charge — only 6 electrons (3 bonds). sp² hybridized, planar, with an empty p orbital perpendicular to the plane.

Stability: 3° > 2° > 1° > Me > Vinyl ≈ Aryl (very unstable)

Why? Alkyl groups donate electrons via hyperconjugation (C–H σ bond overlaps with empty p orbital) AND +I inductive effect. More alkyl groups = more electron donation = more stable carbocation.

Special cases: Allylic (CH₂=CH–CH₂⁺) stabilized by 2 resonance structures ≈ 3°. Benzylic stabilized by 4+ resonance structures. Vinyl cations are sp hybridized — extremely unstable.

3. Carbanions & Free Radicals

Reactive Intermediates Comparison Diagram

Chemistry educational diagram comparing three reactive intermediates side-by-side on white background. Left: Carbocation (C+) — flat triangular sp2 geometry, empty p orbital shown as dashed lobe above/below plane, label "planar, 6 electrons". Center: Carbanion (C-) — pyramidal sp3 geometry, lone pair shown as filled oval, label "pyramidal, 8 electrons". Right: Free Radical (C•) — flat sp2 geometry, half-filled p orbital with single electron dot, label "planar, 7 electrons". Each structure shows 3 R groups attached. Use color coding: red for carbocation, blue for carbanion, green for free radical. Clean textbook illustration style, high resolution, no background clutter.

Carbanion (C⁻): sp³, pyramidal, 8 electrons. Stability: Me > 1° > 2° > 3° (opposite to carbocations). Alkyl groups destabilize (donate electrons → bad for negative charge). sp carbanion (HC≡C⁻) most stable — sp C is most electronegative.

Free Radical (C•): sp², planar, 7 electrons (1 unpaired). Stability: 3° > 2° > 1° > Me (same as carbocations — hyperconjugation stabilizes). Allylic and benzylic radicals are extra stable.

Intermediate	Charge	Hybridization	Shape	Stability Order
Carbocation C⁺	+1	sp²	Planar 120°	3° > 2° > 1° > Me
Carbanion C⁻	−1	sp³	Pyramidal 107°	Me > 1° > 2° > 3°
Free Radical C•	0	sp²	Planar	3° > 2° > 1° > Me

4. Benzene — Drawing the Aromatic Ring

Hückel's Rule for Aromaticity — 3 Conditions (ALL must be met):

Planar ring
Completely conjugated (p orbital on every atom)
(4n + 2) π electrons where n = 0, 1, 2, 3…

Benzene: 6π electrons, n=1 → Aromatic ✓

Aromatic vs Anti-aromatic Ring Systems

Chemistry diagram showing 6 ring structures side by side: (1) Benzene C6H6 — regular hexagon with inner circle, labeled "6π, aromatic" in green; (2) Pyridine — hexagon with N at top, labeled "6π, aromatic, lone pair NOT in π"; (3) Pyrrole — pentagon with NH, inner circle, labeled "6π, aromatic, lone pair IN π"; (4) Furan — pentagon with O, labeled "6π, aromatic"; (5) Cyclobutadiene — square, labeled "4π, anti-aromatic" in red; (6) Cyclopentadienyl anion C5H5- — pentagon with negative charge, inner circle, labeled "6π, aromatic". Each drawn with structural formula, color coded green for aromatic, red for anti-aromatic. Clean white background, educational style.

Compound	π electrons	n	Classification
Benzene C₆H₆	6	1	Aromatic ✓
Pyridine	6	1	Aromatic ✓ (N lone pair NOT in π)
Pyrrole	6	1	Aromatic ✓ (N lone pair IS in π)
Furan	6	1	Aromatic ✓
Cyclobutadiene	4	—	Anti-aromatic ✗ (4n, n=1)
Cyclopentadienyl anion C₅H₅⁻	6	1	Aromatic ✓
Tropylium cation C₇H₇⁺	6	1	Aromatic ✓

5. Inductive Effect

−I Groups (withdrawing, stronger first): −NR₃⁺ > −NO₂ > −CN > −COOH > −F > −Cl > −Br > −I > −OH > −OR

+I Groups (donating): −O⁻ > −NR₂ > −OR > −OH > −alkyl (3° > 2° > 1°)

Application: ClCH₂COOH is more acidic than CH₃COOH because Cl (−I) stabilizes COO⁻ by withdrawing electrons → stronger acid.

6. Resonance (Mesomeric Effect)

Rules for Valid Resonance Structures:

Same arrangement of atoms — only electrons move, NEVER atoms.
No structure can violate the octet rule for C, N, O (except carbocations on C).
More equivalent resonance structures → greater resonance energy → more stable.
Structure with −ve charge on more electronegative atom is more stable contributor.

+M Groups (donate via lone pairs into π): −NH₂ > −NHR > −OR > −OH

−M Groups (withdraw from π): −NO₂ > −CN > −CHO > −COR > −COOH

7. Hyperconjugation

Hyperconjugation Diagram — C–H σ bond donating into empty p orbital

Chemistry diagram illustrating hyperconjugation. Left structure: propyl carbocation CH3-CH2-C+(R2) showing the central carbocation with empty p orbital (two lobes above and below plane, shown in light purple dashed). Adjacent C-H sigma bond of methyl group shown with electron density arrow flowing into the empty p orbital. Right: Newman projection down the C-C bond showing overlap. Labels: "C-H σ bond", "empty p orbital", "electron donation", "hyperconjugation stabilization". Include the number of alpha-H atoms for methyl (3H), ethyl (6H), and isopropyl carbocation (6H). Color scheme: blue for bonds, purple for p orbital, red for electron flow. Clean white background, textbook educational illustration.

Hyperconjugation (No-Bond Resonance): The σ electrons of a C–H bond adjacent to a π bond or carbocation partially overlap into the π system, stabilizing it.

Hyperconjugating H atoms = total H atoms on ALL α-carbons

Propene CH₃–CH=CH₂: 3 hyperconjugating H atoms (on the CH₃)

2-methylpropene (CH₃)₂C=CH₂: 6 hyperconjugating H atoms

tert-butyl cation (CH₃)₃C⁺: 9 hyperconjugating H atoms — most stable!

8. Acidity and Basicity — Conjugate Base Method

4 Factors for Conjugate Base Stability (= acidity):

A. Resonance: More delocalization = more stable = stronger acid.
B. Inductive (−I): EWG near the negative charge = stabilizes it = more acidic.
C. Hybridization: sp > sp² > sp³ (sp most electronegative). HC≡CH more acidic than alkenes or alkanes.
D. Polarizability: Larger atom stabilizes charge better. HI > HBr > HCl > HF (despite F most electronegative!).

Worked Examples

Ex 1 M: Arrange in stability: CH₃⁺, (CH₃)₂CH⁺, (CH₃)₃C⁺, CH₂=CHCH₂⁺

Solution: Allylic (2 resonance structures) ≈ 3° > 2° > 1° (methyl).
Order: (CH₃)₃C⁺ ≈ CH₂=CHCH₂⁺ > (CH₃)₂CH⁺ > CH₃⁺

Ex 2 M: Why is phenol more acidic than ethanol?

Solution: PhO⁻ has 5 resonance structures — negative charge delocalized over ring. EtO⁻ has no resonance. More stable PhO⁻ → phenol is stronger acid (pKa ~10 vs ~16).

Ex 3 H: Is cyclopentadienyl anion C₅H₅⁻ aromatic? Justify.

Solution: Planar ring ✓. All 5 C are sp² ✓. 4 × C=C π bonds (4e) + lone pair on C⁻ (2e) = 6π total = 4(1)+2, n=1 ✓. YES, aromatic. This is why cyclopentadiene (pKa ~16) is unusually easy to deprotonate.

Practice Problems

E Q1. Classify: H₂O, BF₃, CN⁻, NO₂⁺, NH₃, AlCl₃ as electrophile or nucleophile.

E Q2. How many hyperconjugating α-H atoms in: (a) propene, (b) 2-methylpropene, (c) (CH₃)₃C⁺?

M Q3. Arrange carbanions by stability: CH₃⁻, (CH₃)₃C⁻, HC≡C⁻, CH₂=CH⁻. Justify.

M Q4. Arrange in acidity: CH₃COOH, CCl₃COOH, CF₃COOH, CBr₃COOH. Justify.

H Q5. Why is HF less acidic than HCl despite F being more electronegative? Which factor overrides?

H Q6. Is cyclopropenyl cation C₃H₃⁺ (2π electrons) aromatic? Check all 3 Hückel conditions.

Confusing +M and +I effects — halogens are −I but +M (weak). Never mix these up.
Applying resonance to σ bonds — resonance involves only π bonds and lone pairs in conjugation.
Acidity of HX: always use polarizability, NOT electronegativity → HI > HBr > HCl > HF.
Using full curved arrows for homolysis — fish-hook (half-headed) arrow for homolysis ONLY.
Counting N lone pair in pyridine as part of π system — it is NOT (it's in sp² orbital, in plane).

Concept	Key Point
Homolysis	Fish-hook arrows → free radicals
Heterolysis	Full curved arrow → ions
Carbocation stability	3° > 2° > 1° (hyperconjugation + +I)
Carbanion stability	Me > 1° > 2° > 3° (opposite!)
Aromaticity	Planar + conjugated + (4n+2)π
Acidity of HX	HI > HBr > HCl > HF (polarizability)
−I effect	Stabilizes −ve charge → increases acidity
Hyperconjugation	C–H σ electrons donate into adjacent π/p system
Pyrrole N	Lone pair in π (aromatic). Pyridine N lone pair NOT in π.