ChemistryCBSE Class 11 · Organic Chemistry – Some Basic Principles and Techniques

Electronic Effects, Reactive Intermediates and Types of Organic Reactions Class 11

Updated 17 October 2026 · Fact-checked

Electron displacement in organic molecules occurs by the inductive effect (permanent shift through σ bonds, falling off with distance), the resonance or mesomeric effect (delocalisation of π or lone-pair electrons), hyperconjugation (σ(C–H) electrons interacting with an adjacent π bond or empty p orbital) and the temporary electromeric effect. These effects fix the stability of the reactive intermediates formed by bond fission: carbocations (3° > 2° > 1° > methyl), free radicals (same order) and carbanions (methyl > 1° > 2° > 3°). Reactions are classified as substitution, addition, elimination and rearrangement.

Understand Electronic Effects, Reactive Intermediates and Types of Organic Reactions

Bond fission. *Homolytic* fission gives two free radicals (one electron to each atom, shown by half-arrows, typical in non-polar conditions or with light/heat). *Heterolytic* fission gives a cation and an anion (both bonding electrons go to the more electronegative atom, shown by a full curved arrow).

Intermediates. A carbocation has a positive carbon with six electrons: $s p^{2}$ , planar, empty p orbital. A carbanion has a negative carbon with a lone pair: $s p^{3}$ , pyramidal. A free radical has an unpaired electron on carbon: $s p^{2}$ (planar) or nearly so.

Inductive effect. The permanent polarisation of a σ bond caused by a difference in electronegativity, transmitted along the chain and falling off rapidly with distance. Groups such as $- N O_{2}$ , $- C N$ , $- F$ , $- C l$ , $- C O O H$ show the $- I$ effect (electron-withdrawing); alkyl groups show a $+ I$ effect (electron-releasing, relative to hydrogen).

Resonance (mesomeric) effect. Delocalisation of π electrons or lone pairs through conjugation. $+ R$ : groups that donate a lone pair (–OH, –NH₂, –OR, –Cl); $- R$ : groups that withdraw π electrons ( $- N O_{2}$ , $- C H O$ , $- C N$ , $- C O O H$ ). The resonance hybrid is more stable than any contributing structure, and the energy difference is the resonance energy.

Hyperconjugation. The interaction of the σ electrons of C–H bonds on a carbon next to a double bond, a carbocation or a radical with the neighbouring p orbital. The number of no-bond (hyperconjugative) structures equals the number of α-hydrogens: 3 for ethyl cation, 6 for isopropyl, 9 for tert-butyl.

Electromeric effect. A temporary shift of the π electrons of a multiple bond towards one atom when an attacking reagent approaches; it vanishes when the reagent is removed.

Stability orders. Carbocations: $3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{+}$ (+I and hyperconjugation). Free radicals: $3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{∙}$ . Carbanions: $C H_{3}^{-} > 1^{\circ} > 2^{\circ} > 3^{\circ}$ (alkyl groups push electrons onto a charged carbon and destabilise it). Resonance-stabilised ions (allyl, benzyl) are more stable than simple alkyl ions.

Types of organic reactions. Substitution (one atom or group replaced), addition (to a multiple bond), elimination (a small molecule such as HX removed to give a double bond), and rearrangement (a change in the skeleton or functional group within a molecule). By attacking species: nucleophilic or electrophilic, or free radical.

Electronic Effects, Reactive Intermediates and Types of Organic Reactions: formulas and key facts

Hyperconjugative structures: $number of no-bond structures = number of α -H$; Ethyl cation 3, isopropyl 6, tert-butyl 9.
Carbocation stability: $3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{+}$; +I effect and hyperconjugation.
Carbanion stability: $C H_{3}^{-} > 1^{\circ} > 2^{\circ} > 3^{\circ}$; Reverse of carbocations.
Acid strength and −I: $C l_{3} C C O O H > C l_{2} C H C O O H > C l C H_{2} C O O H > C H_{3} C O O H$; More −I groups stabilise the carboxylate.

How to solve Electronic Effects, Reactive Intermediates and Types of Organic Reactions questions

For stability and strength questions, find the group attached to the charged or reactive centre and decide whether it donates or withdraws electrons by induction, resonance or hyperconjugation.

1Draw the species and mark the charged or reactive atom.
2Identify groups attached to it and classify each as +I/−I and +R/−R.
3Count α-hydrogens for hyperconjugation and write resonance structures if a π bond or lone pair is adjacent.
4Rank by the extent of charge dispersal: more dispersal means more stability for an ion.
5Classify a reaction by comparing the reactants and products (substituted, added, removed, rearranged) and name the attacking species.

Electronic Effects, Reactive Intermediates and Types of Organic Reactions 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.

Choosing a stronger acid for a cleaning product

Suppose a lab compares ethanoic acid and chloroethanoic acid solutions of the same concentration.

Which is the stronger acid and what effect explains it?

Try it first, then show the solutionSolution

Chlorine is electronegative and pulls electrons through the σ bonds (−I effect).
That disperses the negative charge of the carboxylate ion, stabilising it, so chloroethanoic acid ionises more.

Answer: Chloroethanoic acid is stronger because of the −I effect of chlorine.

Memory hook: Electron-withdrawing groups strengthen acids.

Choosing the faster alkyl halide

Suppose a lab must choose which of 1-bromobutane and 2-bromo-2-methylpropane will react faster with water by a path that goes through a carbocation.

Which one reacts faster and why?

Try it first, then show the solutionSolution

2-Bromo-2-methylpropane loses Br⁻ to give the tertiary carbocation (CH₃)₃C⁺, with nine α-hydrogens and three +I alkyl groups.
1-Bromobutane would give only a primary carbocation, which is far less stable, so it forms more slowly.

Answer: 2-Bromo-2-methylpropane, through the more stable tertiary carbocation.

Memory hook: The more stable the carbocation, the faster it forms.

Tricks to remember Electronic Effects, Reactive Intermediates and Types of Organic Reactions

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

Alkyl groups push
Alkyl groups are +I and give hyperconjugation: they help cations and radicals but hurt anions.
Use it when: Stability orders.
Opposite for carbanions
The carbanion order is the reverse of the carbocation order.
Use it when: Quick ranking.
Nine hydrogens, nine structures
tert-Butyl cation has nine α-H, so nine hyperconjugative structures.
Use it when: Counting.
Inductive fades fast
The effect falls off with each carbon: nearly gone after the third bond.
Use it when: Position of a halogen in an acid.

Electronic Effects, Reactive Intermediates and Types of Organic Reactions: solved examples

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

Example 1: Types of bond fission · 2 marks · Very short answerBoard level

Write the homolytic and heterolytic fission of the C–Br bond in CH₃–Br and name the species formed in each case.

Try it first, then show the solutionSolution

Homolytic: CH₃–Br → CH₃• + Br•, a methyl free radical and a bromine radical (each atom takes one electron). (1)
Heterolytic: CH₃–Br → CH₃⁺ + Br⁻, a methyl carbocation and a bromide ion (bromine takes both electrons because it is more electronegative). (1)

Answer: Radicals CH₃• and Br•; ions CH₃⁺ and Br⁻.

Example 2: Stability of carbocations · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

The number of α-hydrogens is 0, 3, 6 and 9, so the number of hyperconjugative structures is 0, 3, 6 and 9. (1)
Each additional alkyl group also releases electrons (+I) to the positive carbon, dispersing the charge. (1)
Order: CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺. (1)

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

Example 3: Inductive effect and acidity · 3 marks · Short answerHarder board level

Arrange CH₃COOH, ClCH₂COOH, Cl₂CHCOOH and Cl₃CCOOH in decreasing order of acid strength, and explain why HCOOH is a stronger acid than CH₃COOH.

Try it first, then show the solutionSolution

Each Cl has a −I effect that stabilises the carboxylate anion by dispersing the negative charge; more chlorines give more stabilisation. (1)
Order: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH. (1)
In CH₃COOH the methyl group is +I and pushes electrons onto the carboxylate, destabilising it; HCOOH has no such group, so it is the stronger acid. (1)

Answer: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH; HCOOH > CH₃COOH because methyl is +I.

Example 4: Resonance and acidity or basicity · 4 marks · Long answerHarder board level

Explain with resonance why (a) phenol is a stronger acid than ethanol, (b) aniline is a weaker base than methylamine, and (c) p-nitrophenol is a stronger acid than phenol.

Try it first, then show the solutionSolution

(a) The phenoxide ion's negative charge is delocalised into the ring over several resonance structures; the ethoxide ion's charge is localised on O and further pushed by the +I of the ethyl group. (1)
(b) The lone pair on nitrogen in aniline is delocalised into the ring (+R of –NH₂), so it is less available to accept a proton; in methylamine the +I methyl group makes the lone pair more available. (1)
(c) The –NO₂ group at the para position withdraws electron density (−R and −I), stabilising the p-nitrophenoxide ion further. (1)
In every case the more stable the conjugate base (for acids) or the more available the lone pair (for bases), the stronger the acid or base. (1)

Answer: (a), (c) the conjugate base is stabilised by resonance; (b) the lone pair is delocalised in aniline.

Example 5: Classification of reactions · 4 marks · Long answerHOTS / competency

Classify each reaction as substitution, addition, elimination or rearrangement, and name the attacking species or the intermediate: (i) CH₃CH₂Br + OH⁻(aq) → CH₃CH₂OH + Br⁻; (ii) CH₃CH₂Br + alcoholic KOH → CH₂=CH₂ + KBr + H₂O; (iii) CH₂=CH₂ + Br₂ → BrCH₂CH₂Br; (iv) (CH₃)₃CBr + H₂O → (CH₃)₃COH + HBr through a carbocation.

Try it first, then show the solutionSolution

(i) Substitution: the nucleophile OH⁻ replaces bromine. (1)
(ii) Elimination: HBr is removed from adjacent carbons giving a double bond. (1)
(iii) Addition: the π bond adds Br₂ (electrophilic addition, the attacking species is the polarised bromine, Br⁺ end). (1)
(iv) Nucleophilic substitution proceeding through the tertiary carbocation (CH₃)₃C⁺, stable because of +I and nine α-hydrogens. (1)

Answer: (i) substitution; (ii) elimination; (iii) addition; (iv) substitution via a carbocation.

Where students lose marks in Electronic Effects, Reactive Intermediates and Types of Organic Reactions

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

Mistake: Writing the carbanion stability order the same as carbocations.
Applying one rule to both.
Do this: Alkyl groups stabilise cations and radicals but destabilise carbanions.
Usually costs: The ordering.
Mistake: Calling the electromeric effect permanent.
Confusing it with the inductive effect.
Do this: Electromeric is temporary and appears only in the presence of an attacking reagent.
Usually costs: Definition marks.
Mistake: Stating that the inductive effect operates through π bonds.
Mixing it with resonance.
Do this: Inductive: σ bonds; resonance: π bonds or lone pairs.
Usually costs: Concept marks.
Mistake: Using half-arrows for heterolytic fission, or full arrows for homolytic fission.
Wrong curved-arrow convention.
Do this: A full arrow moves an electron pair; a half-arrow (fishhook) moves one electron.
Usually costs: Presentation marks.

Exam tips

Name the effect and the direction of electron flow in every explanation.
Count α-hydrogens when you use hyperconjugation.
Draw curved arrows from the electron source to the destination.
Write the structure of the intermediate with its charge and hybridisation.
For acid-base strength, argue through the stability of the conjugate base or the availability of the lone pair.

Practice questions on Electronic Effects, Reactive Intermediates and Types of Organic Reactions

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 carbocation has the maximum number of hyperconjugative structures?
(a) CH₃CH₂⁺ (b) (CH₃)₂CH⁺ (c) (CH₃)₃C⁺ (d) CH₃⁺
Hint
Count α-hydrogens.
Answer
(c): nine α-hydrogens.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The carbon of a carbocation is sp² hybridised and the ion is planar.
Reason (R): A carbocation has three bond pairs around the central carbon and an empty p orbital.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Three regions of electron density.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
State two differences between the inductive effect and the resonance effect.
Hint
Think of the bonds involved and the distance.
Answer
The inductive effect operates through σ bonds and weakens rapidly with distance; the resonance effect operates through π electrons or lone pairs in conjugation and is transmitted over the whole conjugated system. The inductive effect is a permanent polarisation without electron delocalisation, while resonance involves delocalisation and gives a stabilised hybrid.
Q4 · 3 marks · Short answerHarder board level
Identify the nature (+I/−I and +R/−R) of each group: (a) –NO₂, (b) –OH, (c) –Cl, (d) –CHO. Which of these is deactivating yet an ortho-para director in electrophilic substitution of benzene?
Hint
Halogens show both effects.
Answer
(a) –NO₂: −I and −R. (b) –OH: −I and +R (+R dominates). (c) –Cl: −I and +R (−I dominates the reactivity). (d) –CHO: −I and −R. The halogen –Cl is deactivating (−I) but ortho-para directing (+R).
Q5 · 4 marks · Long answerHOTS / competency
(a) Arrange CH₃⁻, (CH₃)₂CH⁻ and CH₃CH₂⁻ in decreasing order of stability and explain. (b) Arrange the corresponding free radicals in order of stability. (c) Explain why the benzyl cation, C₆H₅CH₂⁺, is more stable than the ethyl cation.
Hint
Alkyl groups destabilise anions.
Answer
(a) CH₃⁻ > CH₃CH₂⁻ > (CH₃)₂CH⁻: the +I effect of alkyl groups increases the electron density on a carbon that is already negative, destabilising it. (b) Radicals: (CH₃)₂CH• > CH₃CH₂• > CH₃• (alkyl groups stabilise by hyperconjugation). (c) The positive charge of the benzyl cation is delocalised into the ring by resonance over several structures, whereas the ethyl cation has only three hyperconjugative structures.

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.Hybridisation of a carbocation carbon?Show answer
sp² (planar).
2.Which fission gives free radicals?Show answer
Homolytic.
3.Hyperconjugative structures in the ethyl cation?Show answer
Three.
4.Is the electromeric effect temporary?Show answer
Yes.
5.Stability order of carbanions?Show answer
Methyl > 1° > 2° > 3°.
6.Name the group that is −I and +R.Show answer
A halogen such as –Cl.

Electronic Effects, Reactive Intermediates and Types of Organic Reactions: frequently asked questions

Why is a tertiary carbocation more stable than a primary one?

It has more alkyl groups releasing electrons (+I) and more α-hydrogens for hyperconjugation, so the charge is better dispersed.

What is the difference between a nucleophile and an electrophile?

A nucleophile donates an electron pair to attack a positive centre; an electrophile accepts a pair.

Is hyperconjugation also called no-bond resonance?

Yes, because the contributing structures show no bond between the carbon and the hydrogen.

Does the inductive effect change the actual bond lengths?

It slightly polarises the bond, but it does not involve delocalisation of electrons.

What is a rearrangement reaction?

A reaction in which the carbon skeleton or functional group of a molecule changes within the same molecule, with no net loss or gain of atoms.