General Organic Chemistry — Electronic Effects & Reactive Intermediates

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Every “arrange in order” question in organic chemistry — acidity, basicity, carbocation stability, reactivity — is answered with the same four tools: inductive effect, resonance, hyperconjugation and aromaticity. This note is those four tools, and how to decide which one wins.

What you will be able to do

• Rank groups by inductive and resonance effect, and decide which effect dominates
• Order carbocations, carbanions and free radicals by stability, with reasons
• Arrange acids (carboxylic acids, phenols) and bases (amines) by strength, step by step
• Use hyperconjugation for alkene and carbocation stability
• Decide whether a ring or ion is aromatic, antiaromatic or non-aromatic

The big picture

One question behind every ranking: where does the charge go?

In one line: Electrons are pushed or pulled through σ bonds (inductive) and through π systems (resonance, hyperconjugation). Anything that spreads a charge over more atoms stabilises it — and that stability decides acidity, basicity and which intermediate forms.

Think of it like this: A charge is like a crowd in a room. Crammed into one corner, it’s unstable. Open more doors (resonance) or let people lean on the walls (inductive and hyperconjugative donation) and it settles down. Every ranking question asks: which structure gives the charge the most room?

Key idea: Priority when effects compete: aromaticity > resonance > hyperconjugation > inductive. Use the strongest effect that applies, then break ties with the next one.

Bond fission and reagents

How bonds break decides what species you get.

Fission	What happens	Gives	Favoured by
Homolytic	Each atom keeps one electron	Free radicals (R•)	Light, heat, peroxides, non-polar conditions
Heterolytic	One atom keeps both electrons	Ions: carbocations (R⁺) or carbanions (R⁻)	Polar solvents, polar bonds

Electrophiles (electron-seeking)	Nucleophiles (nucleus-seeking)
H⁺, NO₂⁺, Br⁺, R⁺, carbonyl carbon, BF₃, AlCl₃ (neutral but electron-deficient)	OH⁻, CN⁻, RO⁻, NH₃, H₂O, R–OH, R⁻ (lone pair or negative charge)

Exam trap: Neutral species can be electrophiles or nucleophiles. BF₃ and AlCl₃ are electrophiles (empty orbital); NH₃ and H₂O are nucleophiles (lone pair). Don’t decide by charge alone.

Inductive effect (I)

The σ-bond tug-of-war: permanent, but short-range.

• A permanent shift of σ electrons toward the more electronegative atom along a chain.
• Fades fast with distance: after about three bonds it is negligible.
• −I (withdrawing): −NO₂ > −CN > −COOH > −F > −Cl > −Br > −I > −OH > −OCH₃ > −C₆H₅
• +I (donating): alkyl groups, (CH₃)₃C− > (CH₃)₂CH− > CH₃CH₂− > CH₃−; also −COO⁻ and −O⁻

Solved example 01 — Inductive effect and distance

Arrange in decreasing acid strength: 2-chlorobutanoic acid, 3-chlorobutanoic acid, 4-chlorobutanoic acid, butanoic acid.

1. Each Cl withdraws electrons and stabilises the carboxylate ion (−I).
2. The effect weakens with every carbon between Cl and −COOH.

Answer: 2-chloro > 3-chloro > 4-chloro > butanoic acid.

Resonance (mesomeric) effect, R or M

Electrons spread through π systems — long-range, but only through conjugation.

+R (donate into the π system)	−R (withdraw from the π system)
−NH₂, −NHR, −OH, −OR, −NHCOR, −O⁻, halogens	−NO₂, −CN, −CHO, −COR, −COOH, −COOR, −SO₃H

Rules for valid resonance structures

1. Only electrons move (lone pairs and π electrons); atoms stay put.
2. Every structure has the same number of paired and unpaired electrons.
3. More covalent bonds, complete octets and negative charge on the more electronegative atom → more important structure.

Key idea: Halogens are the exception. Their −I outweighs their +R, so chlorobenzene reacts more slowly than benzene in electrophilic substitution — yet +R still directs the new group to ortho/para.

Hyperconjugation

σ electrons from C–H bonds next door lend a hand.

Delocalisation of σ electrons from C–H bonds on the carbon adjacent (α) to a carbocation, a radical or a double bond. Also called “no-bond resonance”. More α-hydrogens → more stabilisation.

Species	α-hydrogens	Stability
(CH₃)₃C⁺	9	Highest
(CH₃)₂CH⁺	6
CH₃CH₂⁺	3
CH₃⁺	0	Lowest

• Alkene stability: more substituted alkenes are more stable (lower heat of hydrogenation): tetrasubstituted > trisubstituted > disubstituted > monosubstituted. This is the basis of Saytzeff’s rule.
• Toluene: hyperconjugation of the CH₃ group activates the ring and directs ortho/para.
• Anomaly: alkyl groups donate by hyperconjugation in the order CH₃− > CH₃CH₂− > (CH₃)₂CH− > (CH₃)₃C− (most α-H first) — the opposite of their +I order.

Reactive intermediates

The short-lived species that every mechanism passes through.

Intermediate	Carbon	Shape	Stability order
Carbocation	sp², 6 electrons	Planar	Benzyl ≈ allyl ≈ 3° > 2° > 1° > CH₃⁺
Carbanion	sp³, lone pair	Pyramidal	CH₃⁻ > 1° > 2° > 3° (reverse)
Free radical	sp², 7 electrons	Nearly planar	Benzyl ≈ allyl > 3° > 2° > 1° > CH₃•
Carbene	6 electrons, neutral	Bent (singlet) or linear (triplet)	Very reactive; e.g. :CCl₂ in Reimer–Tiemann

Key idea: Carbanions are the mirror image. Alkyl groups push electrons onto a centre that already has too many, so they destabilise carbanions. Electron-withdrawing groups (−NO₂, −CN, C=O) and resonance stabilise them.

Solved example 02 — Carbocation stability

Arrange: CH₃CH₂⁺, (CH₃)₃C⁺, C₆H₅CH₂⁺, CH₃⁺, CH₂=CH–CH₂⁺.

1. Benzyl and allyl: positive charge spread by resonance.
2. (CH₃)₃C⁺: 9 α-H plus three +I groups.
3. Then ethyl (3 α-H), then methyl (none).

Answer: C₆H₅CH₂⁺ ≈ CH₂=CHCH₂⁺ ≈ (CH₃)₃C⁺ > CH₃CH₂⁺ > CH₃⁺. (Exam keys usually place benzyl first.)

Check yourself

• C1. Which is more stable: (CH₃)₃C⁻ or CH₃⁻? Why?
• C2. Which is the most stable radical: CH₃•, (CH₃)₃C•, C₆H₅CH₂•?

Answers at the end.

Acidity

An acid is strong when its conjugate base is stable.

Acid	pKa	Reason
CH₃COOH	4.76	Reference
HCOOH	3.75	No +I alkyl group
ICH₂COOH	3.18	Weakest −I halogen
BrCH₂COOH	2.90
ClCH₂COOH	2.86
FCH₂COOH	2.59	Strongest −I halogen
Cl₂CHCOOH	1.29	Two −I groups
Cl₃CCOOH	0.65	Three −I groups

Phenols

• Phenol (pKa ≈ 10) is acidic because phenoxide is stabilised by resonance into the ring; ethanol (≈ 16) is not.
• −NO₂ at ortho or para increases acidity strongly (−R and −I); at meta only by −I. p-Nitrophenol (7.15) > o (7.23) > m (8.4) > phenol (10.0).
• Electron donors decrease acidity: p-cresol (10.2) and p-methoxyphenol are weaker than phenol.
• 2,4,6-Trinitrophenol (picric acid) has pKa ≈ 0.4 — as strong as many mineral acids.

Benzoic acids and the ortho effect

Electron-withdrawing groups strengthen benzoic acid (p-nitrobenzoic, pKa 3.44, vs benzoic 4.20); donors weaken it. Almost any group at the ortho position increases acidity, even a donating CH₃ (o-toluic acid, 3.91), because it twists −COOH out of the ring’s plane — the ortho effect.

RCOOH > phenol > H₂O > ROH > HC≡CH > NH₃ > alkanes

General order of acidity

Solved example 03 — Substituted acetic acids

Arrange: CH₃COOH, ClCH₂COOH, Cl₂CHCOOH, FCH₂COOH.

1. More −I groups → more stable carboxylate.
2. Two Cl atoms withdraw more than one F.
3. F withdraws more than Cl.

Answer: Cl₂CHCOOH > FCH₂COOH > ClCH₂COOH > CH₃COOH.

Solved example 04 — Phenols

Arrange: phenol, p-nitrophenol, m-nitrophenol, p-cresol.

1. p-NO₂: −R and −I → most stabilised phenoxide.
2. m-NO₂: −I only.
3. p-CH₃: +I and hyperconjugation destabilise phenoxide.

Answer: p-nitrophenol > m-nitrophenol > phenol > p-cresol.

Basicity of amines

A base is strong when its lone pair is available.

Amine	pKb (water)
(C₂H₅)₂NH	3.00
(C₂H₅)₃N	3.25
(CH₃)₂NH	3.27
C₂H₅NH₂	3.29
CH₃NH₂	3.38
(CH₃)₃N	4.22
NH₃	4.75
C₆H₅NH₂	9.38

• Alkyl groups (+I) raise basicity above NH₃.
• In water, solvation also matters: the protonated amine is stabilised by H bonds to water — best for 1° (three N–H), worst for 3° (one N–H). Combined with +I and steric effects:
• Methyl series: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.
• Ethyl series: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.
• In the gas phase (no solvation): 3° > 2° > 1° > NH₃.
• Aniline is much weaker: its lone pair is delocalised into the ring. Amides are almost neutral: the lone pair is delocalised onto C=O.

Solved example 05 — Basicity in water

Arrange: NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂.

1. +I raises basicity above NH₃; solvation favours 1° and 2°.
2. 3° amine’s cation is poorly solvated and crowded.
3. Aniline: lone pair in resonance.

Answer: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ > C₆H₅NH₂.

Aromaticity

Hückel’s rule and the special stability it brings.

A species is aromatic if it is (1) cyclic, (2) planar, (3) fully conjugated (every ring atom sp² with a p orbital), and (4) has (4n + 2) π electrons: 2, 6, 10, 14 …

Species	π electrons	Verdict
Benzene	6	Aromatic
Cyclopentadienyl anion	6	Aromatic
Tropylium (cycloheptatrienyl) cation	6	Aromatic
Cyclopropenyl cation	2	Aromatic
Pyridine, pyrrole, furan, thiophene	6	Aromatic
Naphthalene	10	Aromatic
Cyclobutadiene	4	Antiaromatic (4n, planar)
Cyclopentadienyl cation	4	Antiaromatic
Cyclooctatetraene	8	Non-aromatic (tub-shaped, not planar)
Cyclopentadiene	4 (one sp³ C)	Non-aromatic

Key idea: Aromaticity explains odd acidities: cyclopentadiene (pKa ≈ 16) is far more acidic than other hydrocarbons because losing H⁺ gives an aromatic anion. Tropylium bromide is ionic for the same reason.

Solved example 06 — Aromatic or not?

Classify the cyclopentadienyl cation and anion.

1. Both are planar, cyclic and fully conjugated.
2. Cation: 4 π electrons (4n, n = 1) → antiaromatic.
3. Anion: 6 π electrons (4n + 2, n = 1) → aromatic.

Answer: The anion forms easily; the cation is very unstable.

Check yourself

• C3. Is pyrrole’s N lone pair part of the aromatic sextet? What does that do to its basicity?
• C4. Why is cyclooctatetraene not antiaromatic?

Answers at the end.

Exam trap gallery

The rankings students get wrong most often.

#	What students write	What is actually correct	Why
1	Chlorobenzene is activated because Cl donates by resonance.	Chlorobenzene is deactivated (−I > +R) but still ortho/para-directing.	Halogens are the exception to “resonance wins”.
2	Carbanion stability follows carbocation order: 3° > 2° > 1°.	Reversed: CH₃⁻ > 1° > 2° > 3°.	Alkyl groups push electrons onto an already-rich centre.
3	Trimethylamine is the strongest base among methylamines in water.	(CH₃)₂NH is strongest; (CH₃)₃N is below CH₃NH₂.	Poor solvation of the 3° ammonium ion.
4	FCH₂COOH is stronger than Cl₂CHCOOH.	Cl₂CHCOOH (1.29) is stronger than FCH₂COOH (2.59).	Two −I groups beat one stronger one here.
5	m-Nitrophenol is as acidic as p-nitrophenol.	p (7.15) is stronger than m (8.4).	At meta, −NO₂ can’t withdraw by resonance from the phenoxide oxygen.
6	Cyclooctatetraene is antiaromatic (8 π electrons).	It is non-aromatic: it folds into a tub.	Aromaticity rules need planarity.
7	+I order of alkyl groups = hyperconjugation order.	They’re opposite: hyperconjugation is CH₃ > ethyl > isopropyl > tert-butyl.	Hyperconjugation counts α-H atoms.
8	BF₃ is a nucleophile because it has no charge.	BF₃ is an electrophile (empty p orbital).	Electron-deficient, not charge, decides.

Memory hook: “Spread the charge, gain the peace” — delocalisation = stability. · “Halogens: slow but ortho-para” — deactivate, yet o/p direct. · “Carbanions hate alkyls” — reverse of carbocation order. · “Two-ish in water” — 2° amine is the strongest base in water. · “4n + 2, flat and through” — Hückel’s checklist.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Which is an electrophile? (Easy)

• (a) NH₃
• (b) H₂O
• (c) AlCl₃
• (d) OH⁻

Q02. Homolytic fission produces: (Easy)

• (a) carbocations
• (b) carbanions
• (c) free radicals
• (d) carbenes only

Q03. The most stable carbocation is: (Easy)

• (a) CH₃⁺
• (b) CH₃CH₂⁺
• (c) (CH₃)₂CH⁺
• (d) (CH₃)₃C⁺

Q04. Which group shows −R effect? (Easy)

• (a) −OH
• (b) −NH₂
• (c) −NO₂
• (d) −OCH₃

Q05. Which is aromatic? (Easy)

• (a) Cyclobutadiene
• (b) Cyclooctatetraene
• (c) Cyclopentadienyl anion
• (d) Cyclopentadienyl cation

Q06. The strongest acid is: (Moderate)

• (a) CH₃COOH
• (b) ClCH₂COOH
• (c) Cl₂CHCOOH
• (d) FCH₂COOH

Q07. Which is the strongest base in water? (Moderate)

• (a) NH₃
• (b) CH₃NH₂
• (c) (CH₃)₂NH
• (d) (CH₃)₃N

Q08. The most acidic phenol is: (Moderate)

• (a) p-cresol
• (b) phenol
• (c) m-nitrophenol
• (d) p-nitrophenol

Q09. The most stable carbanion is: (Moderate)

• (a) CH₃⁻
• (b) CH₃CH₂⁻
• (c) (CH₃)₂CH⁻
• (d) (CH₃)₃C⁻

Q10. Number of α-hydrogens available for hyperconjugation in (CH₃)₂CH⁺ is: (Moderate)

• (a) 3
• (b) 6
• (c) 7
• (d) 9

Q11. Chlorobenzene undergoes electrophilic substitution: (Moderate)

• (a) faster than benzene, ortho/para
• (b) slower than benzene, meta
• (c) slower than benzene, ortho/para
• (d) faster than benzene, meta

Q12. o-Toluic acid is stronger than benzoic acid because of: (Moderate)

• (a) +I of CH₃
• (b) hyperconjugation
• (c) the ortho effect
• (d) resonance

Q13. Which is antiaromatic? (Tough)

• (a) Tropylium cation
• (b) Cyclopropenyl cation
• (c) Cyclopentadienyl cation
• (d) Benzene

Q14. In the gas phase, the basicity order of methylamines is: (Tough)

• (a) 2° > 1° > 3°
• (b) 3° > 2° > 1°
• (c) 1° > 2° > 3°
• (d) 2° > 3° > 1°

Q15. Which has the lowest heat of hydrogenation? (Tough)

• (a) But-1-ene
• (b) cis-But-2-ene
• (c) trans-But-2-ene
• (d) 2-Methylpropene

Answer key with explanations

1. (c) AlCl₃ — Al has an empty orbital.
2. (c) free radicals — Each atom keeps one electron.
3. (d) (CH₃)₃C⁺ — Most +I and hyperconjugation (9 α-H).
4. (c) −NO₂ — −NO₂ withdraws π electrons.
5. (c) Cyclopentadienyl anion — 6 π electrons, planar.
6. (c) Cl₂CHCOOH — pKa 1.29; two −I groups.
7. (c) (CH₃)₂NH — +I plus good solvation.
8. (d) p-nitrophenol — −R and −I of para NO₂.
9. (a) CH₃⁻ — No electron-donating groups.
10. (b) 6 — Two CH₃ groups × 3.
11. (c) slower than benzene, ortho/para — −I deactivates; +R directs o/p.
12. (c) the ortho effect — Steric twisting of −COOH out of the ring plane.
13. (c) Cyclopentadienyl cation — 4 π electrons, planar, conjugated.
14. (b) 3° > 2° > 1° — No solvation: +I alone decides.
15. (c) trans-But-2-ene — trans-But-2-ene is the most stable of these (disubstituted, no steric strain between methyls); 2-methylpropene is also disubstituted but its heat of hydrogenation is slightly higher.

Check-yourself answers

• C1. CH₃⁻: the three +I methyl groups destabilise (CH₃)₃C⁻.
• C2. C₆H₅CH₂• (resonance with the ring).
• C3. Yes: the lone pair is part of the six π electrons, so it isn’t available for protonation; pyrrole is a very weak base.
• C4. It isn’t planar — it folds into a tub, so the p orbitals don’t overlap all the way round.

One-page revision

Read this page the night before the exam.

Effects

• Priority: aromaticity > resonance > hyperconjugation > inductive
• −I: NO₂ > CN > COOH > F > Cl > Br > I
• +I: 3° alkyl > 2° > 1° > CH₃
• Inductive fades after ~3 bonds

Resonance

• +R: NH₂, OH, OR, halogens; −R: NO₂, CN, CHO, COOH
• Halogens: −I > +R → deactivate, o/p direct
• Only electrons move between structures

Intermediates

• Carbocation: benzyl ≈ allyl ≈ 3° > 2° > 1° > CH₃⁺
• Carbanion: CH₃⁻ > 1° > 2° > 3°
• Radical: like carbocation
• Hyperconjugation counts α-H

Acidity

• Stable conjugate base = strong acid
• Cl₃CCOOH > Cl₂CHCOOH > FCH₂COOH > ClCH₂COOH > CH₃COOH
• p-NO₂ phenol > m > phenol > p-cresol
• Ortho effect strengthens benzoic acids

Basicity

• Water: 2° > 1° > 3° > NH₃ (methyl)
• Ethyl: 2° > 3° > 1° > NH₃
• Gas phase: 3° > 2° > 1°
• Aniline, amides weak: lone pair delocalised

Aromaticity

• Cyclic, planar, conjugated, 4n + 2 π e⁻
• Cp⁻, tropylium⁺, cyclopropenyl⁺ aromatic
• Cyclobutadiene, Cp⁺ antiaromatic
• COT non-aromatic (tub)