ChemistryISC Class 11 · Organic Chemistry: Some Basic Principles and Techniques

Electronic Effects, Reactive Intermediates and Reaction Mechanisms

Updated 17 October 2026 · Fact-checked

The inductive effect (permanent shift along σ bonds), resonance (delocalisation of π electrons or lone pairs) and hyperconjugation (delocalisation of σ electrons of C–H) control the stability of carbocations, free radicals and carbanions and the strength of acids. Substitution may proceed by SN1 (carbocation, 3° > 2° > 1°) or SN2 (single step, 1° > 2° > 3°).

More alkyl groups mean more α-hydrogens, more hyperconjugation and more +I effect, so the carbocation becomes more stable: 3° > 2° > 1° > methyl.

Illustration drawn for QuestionBuddy.

Understand Electronic Effects, Intermediates and Reaction Mechanisms

Bond fission. Homolytic fission splits a bond so each atom gets one electron, giving free radicals (typically in light or at high temperature). Heterolytic fission gives a cation and an anion (in polar solvents).

Reagents. An electrophile is electron-deficient and accepts an electron pair (H⁺, NO₂⁺, carbocations, AlCl₃, BF₃). A nucleophile is electron-rich and donates a pair (OH⁻, CN⁻, NH₃, H₂O, carbanions).

Inductive effect. The permanent polarisation of a σ bond because of an electronegativity difference. Groups such as –NO₂, –CN, –F, –Cl, –COOH, –OH have −I (electron-withdrawing); alkyl groups have +I (electron-releasing). The effect weakens rapidly with distance. −I groups strengthen acids by stabilising the anion: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH, and FCH₂COOH > ClCH₂COOH > BrCH₂COOH.

Resonance effect. Delocalisation of π electrons or lone pairs through conjugation, represented by contributing structures. It stabilises the carboxylate ion, phenoxide ion and benzene, and makes phenol more acidic than ethanol.

Hyperconjugation. Delocalisation of electrons of the σ C–H bond adjacent (α-H) to a positive carbon, a free radical or a double bond. The number of α-hydrogens gives the number of hyperconjugative structures: ethyl cation 3, isopropyl cation 6, tert-butyl cation 9.

Stability. Carbocations and free radicals: 3° > 2° > 1° > CH₃ (+I and hyperconjugation); carbanions: CH₃⁻ > 1° > 2° > 3°.

Substitution mechanisms. SN2: one step, rate = k[RX][Nu], inversion of configuration, 1° > 2° > 3°. SN1: two steps via a carbocation, rate = k[RX], racemisation for a chiral centre, 3° > 2° > 1°. Elimination: E1 (carbocation) and E2 (single step, with a strong base).

Electronic Effects, Intermediates and Reaction Mechanisms: formulas and key facts

Stability of carbocations: $3° > 2° > 1° > C H_{3}^{+}$; Same order for free radicals.
Hyperconjugative structures: $n u mb er = n u mb er o f α - h y d r o g e n s$; CH₃CH₂⁺ 3 | (CH₃)₂CH⁺ 6 | (CH₃)₃C⁺ 9.
Rate laws: $S N 1 : r a t e = k [R X] ∣ S N 2 : r a t e = k [R X] [N u]$; First order and second order.
Acid strength: $K a = 1 0^{(} - p K a)$; A lower pKa is a stronger acid.

Electronic Effects, Intermediates and Reaction Mechanisms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why chloroacetic acid is stronger than vinegar

Suppose a chemist compares acetic acid (pKa ≈ 4.76) with chloroacetic acid (pKa ≈ 2.86).

Compare their Ka values and explain.

Try it first, then show the solutionSolution

Ka(CH₃COOH) = 10⁻⁴·⁷⁶ = 1.74 × 10⁻⁵; Ka(ClCH₂COOH) = 10⁻²·⁸⁶ = 1.38 × 10⁻³.
Ratio = 10^(4.76 − 2.86) = 10¹·⁹ ≈ 80.
The −I effect of Cl disperses the negative charge of ClCH₂COO⁻, stabilising it and making the acid stronger.

Answer: Chloroacetic acid is about 80 times stronger.

Memory hook: A −I group stabilises the conjugate base.

Tricks to remember Electronic Effects, Intermediates and Reaction Mechanisms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Alkyls push, halogens pull
+I for alkyl groups stabilises positive charge; −I for halogens stabilises negative charge.
Use it when: Carbocation and acid-strength comparisons.
SN1 = 3°, SN2 = 1°
Bulky tertiary halides favour SN1; unhindered primary halides favour SN2.
Use it when: Choosing a mechanism.

Electronic Effects, Intermediates and Reaction Mechanisms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Stability of carbocations · 1 mark · MCQBoard level

The most stable carbocation among the following is:
(a) CH₃⁺
(b) CH₃CH₂⁺
(c) (CH₃)₂CH⁺
(d) (CH₃)₃C⁺

Try it first, then show the solutionSolution

Three methyl groups give the most +I effect and 9 hyperconjugative structures.

Answer: (d) (CH₃)₃C⁺

Example 2: Order of stability · 3 marks · Short answerBoard level

Arrange CH₃⁺, CH₃CH₂⁺, (CH₃)₂CH⁺ and (CH₃)₃C⁺ in increasing order of stability, giving the number of hyperconjugative structures for each.

Try it first, then show the solutionSolution

Number of α-hydrogens: 0, 3, 6, 9 respectively. (1.5 marks)
More α-H and more +I alkyl groups spread the positive charge. (0.75 mark)
Order: CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺. (0.75 mark)

Answer: CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺

Example 3: Acid strength and the inductive effect · 4 marks · Short answerHarder board level

The pKa values (approximate) are CH₃COOH 4.76, ClCH₂COOH 2.86, Cl₂CHCOOH 1.29 and Cl₃CCOOH 0.65. Arrange the acids in order of decreasing strength, explain, and find how many times stronger Cl₃CCOOH is than CH₃COOH.

Try it first, then show the solutionSolution

Decreasing strength: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH. (1 mark)
Each Cl has a −I effect, pulling electrons away from the carboxylate and stabilising the anion; more Cl atoms give a greater effect. (1.5 marks)
Ratio of Ka = 10^(4.76 − 0.65) = 10⁴·¹¹. (1 mark)
= 1.3 × 10⁴ times stronger. (0.5 mark)

Answer: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH; Cl₃CCOOH is about 1.3 × 10⁴ times stronger.

Example 4: Reagents and bond fission · 4 marks · Short answerHarder board level

Classify as nucleophile or electrophile: OH⁻, NO₂⁺, NH₃, AlCl₃, CN⁻, H⁺. State whether the fission of the Cl–Cl bond in sunlight and of H–Br in water is homolytic or heterolytic.

Try it first, then show the solutionSolution

Nucleophiles (electron-rich, donate a pair): OH⁻, NH₃, CN⁻. (1.5 marks)
Electrophiles (electron-deficient): NO₂⁺, AlCl₃, H⁺. (1.5 marks)
Cl–Cl in sunlight: homolytic, giving two Cl· radicals. (0.5 mark)
H–Br in water: heterolytic, giving H⁺ and Br⁻. (0.5 mark)

Answer: Nucleophiles: OH⁻, NH₃, CN⁻. Electrophiles: NO₂⁺, AlCl₃, H⁺. Cl–Cl homolytic; H–Br heterolytic.

Example 5: SN1 versus SN2 · 5 marks · Long answerHOTS / competency

tert-Butyl bromide reacts with aqueous hydroxide by SN1, while methyl bromide reacts by SN2. (i) Write the rate law for each. (ii) What happens to the rate of each if [OH⁻] is doubled? (iii) What happens if both concentrations are doubled? (iv) What is the stereochemical outcome at a chiral carbon in each mechanism?

Try it first, then show the solutionSolution

SN1: rate = k[(CH₃)₃CBr] (rate-determining ionisation to a carbocation); SN2: rate = k[CH₃Br][OH⁻]. (1.5 marks)
Doubling [OH⁻]: SN1 rate unchanged; SN2 rate doubles. (1 mark)
Doubling both concentrations: SN1 rate doubles; SN2 rate becomes 4 times. (1 mark)
SN2: inversion of configuration (back-side attack). (0.75 mark)
SN1: the planar carbocation is attacked from both faces, giving a racemic mixture. (0.75 mark)

Answer: SN1 first order, racemisation; SN2 second order, inversion.

Where students lose marks in Electronic Effects, Intermediates and Reaction Mechanisms

The slips examiners see most often, and what to write instead.

Mistake: Reversing the carbanion order
Giving the same stability order as for carbocations.
Do this: Carbanions are destabilised by +I groups: CH₃⁻ is the most stable.
Usually costs: 2 marks.
Mistake: Confusing inductive and resonance effects
Explaining phenol acidity with the inductive effect only.
Do this: Resonance delocalises the negative charge in phenoxide.
Usually costs: 1 to 2 marks.
Mistake: SN1/SN2 stereochemistry reversed
Writing inversion for SN1.
Do this: SN2: inversion; SN1: racemisation.
Usually costs: 1 to 2 marks.

Exam tips

Name the effect (+I, −I, resonance, hyperconjugation) when you give a reason.
Draw resonance structures with curved arrows for the conjugate base.
For SN1 and SN2 give the rate law and the stereochemical result.
State the order of stability and the reason in the same sentence.

Practice questions on Electronic Effects, Intermediates and Reaction Mechanisms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following groups shows a −I effect?
(a) –CH₃
(b) –C₂H₅
(c) –NO₂
(d) –(CH₃)₃C
Hint
Alkyl groups release electrons.
Answer
(c) –NO₂
Q2 · 2 marks · Short answerBoard level
Define a nucleophile and give two examples.
Hint
It has a lone pair or a negative charge.
Answer
A nucleophile is an electron-rich species that donates a pair of electrons to an electron-deficient centre; for example OH⁻ and NH₃.
Q3 · 3 marks · Short answerHarder board level
Explain why FCH₂COOH is a stronger acid than ClCH₂COOH, and why HCOOH is stronger than CH₃COOH.
Hint
Think about the strength of −I and +I.
Answer
F is more electronegative than Cl, so its −I effect is greater and stabilises the anion more. The CH₃ in CH₃COOH has a +I effect, which destabilises the carboxylate; HCOOH has H instead, so it is the stronger acid.
Q4 · 3 marks · Short answerHarder board level
Why is phenol more acidic than ethanol? Use resonance.
Hint
Compare the stabilisation of the conjugate bases.
Answer
The phenoxide ion has its negative charge delocalised over the benzene ring through resonance, which stabilises it. The ethoxide ion has no such delocalisation, and its alkyl group has a +I effect that destabilises it.
Q5 · 3 marks · Short answerHarder board level
Give three differences between SN1 and SN2 reactions.
Hint
Compare steps, rate law, and substrate preference.
Answer
SN1: two steps through a carbocation, first-order rate, 3° > 2° > 1°, racemisation. SN2: single step, second-order rate, 1° > 2° > 3°, inversion of configuration.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student studies the stability of free radicals and carbanions in the series methyl, ethyl, isopropyl and tert-butyl.
(i) Arrange the free radicals in order of stability. (1 mark)
(ii) Arrange the carbanions in order of stability. (1 mark)
(iii) Give the reason for (i). (1 mark)
(iv) Give the reason for (ii). (1 mark)
Hint
Alkyl groups release electrons.
Answer
(i) tert-butyl > isopropyl > ethyl > methyl. (ii) methyl > ethyl > isopropyl > tert-butyl. (iii) The +I effect and hyperconjugation of the alkyl groups stabilise the electron-deficient radical. (iv) The +I effect of the alkyl groups intensifies the negative charge on the carbanion carbon, destabilising it.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Order of carbocation stability?Show answer
3° > 2° > 1° > methyl.
2.Which has more hyperconjugation, ethyl or isopropyl cation?Show answer
Isopropyl (6 α-H against 3).
3.Define the inductive effect.Show answer
Permanent polarisation of σ bonds due to electronegativity difference.
4.SN2 rate law?Show answer
rate = k[RX][Nu].

Electronic Effects, Intermediates and Reaction Mechanisms: frequently asked questions

Is hyperconjugation the same as resonance?

It is similar, but the delocalised electrons are the σ electrons of the C–H bond, so it is sometimes called "no-bond resonance".

Why do tertiary halides prefer SN1?

They form the most stable carbocation and the three alkyl groups hinder back-side attack, which is needed for SN2.

Is the inductive effect permanent?

Yes, it operates in the ground state of the molecule, unlike the electromeric effect, which appears only in the presence of an attacking reagent.