Current Affairs updated daily
KalviGavaniKalviGavani

Organic Reactions Concepts Solutions | KalviGavani

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions exercise solutions — reaction mechanisms, electrophile nucleophile, inductive and resonance effects, MCQs, full answers.

Updated 15 June 202618 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

Welcome to the Basic Concepts of Organic Reactions exercise solutions for Class 11 Chemistry (Tamil Nadu State Board). Here is how to get the most from this page:

1. Attempt every question on your own first — do not jump straight to the answers. 2. Compare your answer with the solution provided. If they differ, trace through the explanation step by step. 3. Re-work mistakes — write the corrected answer in your own words; this builds deeper understanding. 4. Use the Key Concepts section for a quick recap before exams.


Key Concepts (Quick Recap)

Types of Organic Reactions

Reaction TypeDescriptionExample
SubstitutionOne atom/group is replaced by anotherCH₄ + Cl₂ → CH₃Cl + HCl
AdditionAtoms/groups are added across a double/triple bondCH₂=CH₂ + H₂ → CH₃–CH₃
EliminationAtoms/groups are removed to form a double/triple bondC₂H₅OH → CH₂=CH₂ + H₂O
RearrangementMolecule rearranges to form an isomerCH₃CH₂CH₂CH₃ → (CH₃)₂CHCH₃ (rare, specific conditions)

Electrophile vs Nucleophile

FeatureElectrophileNucleophile
DefinitionElectron-seeking (electron-poor) speciesElectron-donating (electron-rich) species
ChargePositively charged or neutral with empty orbitalNegatively charged or neutral with lone pair
RoleAccepts an electron pairDonates an electron pair
ExamplesH⁺, NO₂⁺, SO₃, BF₃, AlCl₃OH⁻, CN⁻, NH₃, H₂O, CH₃O⁻
BehaviourAttacks electron-rich centresAttacks electron-deficient centres

Inductive Effect vs Resonance (Mesomeric) Effect

FeatureInductive Effect (I)Resonance/Mesomeric Effect (M)
Bonds involvedSigma (σ) bonds onlyPi (π) bonds and conjugated systems
NaturePermanent polarizationDelocalisation of π-electrons
RangeShort-range (dies out after 3-4 carbons)Long-range (through entire conjugated system)
+I effectElectron-donating groups (alkyl, –CH₃)+M: Electron-donating by resonance (–OH, –NH₂, –OR)
–I effectElectron-withdrawing groups (–NO₂, –CN, –Cl)–M: Electron-withdrawing by resonance (–NO₂, –CN, –CHO)
Effect on acidity–I groups increase acidity; +I groups decrease acidity–M groups increase acidity; +M groups decrease acidity

Free Radical Mechanism — Steps

StepDescriptionExample (Chlorination of Methane)
InitiationBond breaks homolytically to form free radicals (requires UV light or heat)Cl₂ → 2 Cl· (homolytic fission)
PropagationFree radical attacks a molecule to form a new radical (chain reaction)Cl· + CH₄ → HCl + ·CH₃; ·CH₃ + Cl₂ → CH₃Cl + Cl·
TerminationTwo radicals combine to form a stable molecule (chain ends)Cl· + Cl· → Cl₂; ·CH₃ + Cl· → CH₃Cl; ·CH₃ + ·CH₃ → C₂H₆

Electron-Displacing Effects Summary

EffectSymbolDirectionGroupsImpact on Reactivity
+Inductive+IPushes e⁻ away from itself–CH₃, –C₂H₅, –CH(CH₃)₂Increases nucleophilicity at ortho/para positions
–Inductive–IPulls e⁻ towards itself–Cl, –NO₂, –CN, –COOHDecreases nucleophilicity; increases acidity
+Mesomeric+MDonates e⁻ via resonance–OH, –NH₂, –OR, –NHRActivates ring for electrophilic substitution
–Mesomeric–MWithdraws e⁻ via resonance–NO₂, –CN, –CHO, –COOHDeactivates ring for electrophilic substitution

Homolytic vs Heterolytic Fission

FeatureHomolytic FissionHeterolytic Fission
DefinitionBond breaks equally; each atom gets one electronBond breaks unequally; both electrons go to one atom
ProductsTwo free radicalsA cation and an anion
ExampleCl₂ → 2 Cl·CH₃–Cl → CH₃⁺ + Cl⁻
ConditionsUV light, heat, peroxidesPolar solvents, ionic conditions

I. Choose the Best Answer (MCQs)

1. Which of the following is an electrophile?
Answer

(a) OH⁻ (b) NH₃ (c) NO₂⁺ (d) CN⁻

Answer(c) NO₂⁺
ExplanationAn electrophile is an electron-deficient species that seeks electrons. NO₂⁺ (nitronium ion) is positively charged and electron-deficient, making it a strong electrophile. OH⁻, NH₃, and CN⁻ are nucleophiles — they are electron-rich species that donate electron pairs.

2. The inductive effect is transmitted through:
Answer

(a) π-bonds (b) σ-bonds (c) Both π and σ-bonds (d) Neither π nor σ-bonds

Answer(b) σ-bonds
ExplanationThe inductive effect involves the permanent polarization of sigma (σ) bonds due to electronegativity differences. It operates through the sigma-bonded framework of a molecule and diminishes rapidly with distance (becomes negligible after 3-4 carbons). The resonance/mesomeric effect, on the other hand, operates through π-bonds.

3. Which of the following groups exhibits a +I effect?
Answer

(a) –NO₂ (b) –CN (c) –CH₃ (d) –Cl

Answer(c) –CH₃
ExplanationThe +I (positive inductive) effect is shown by electron-donating groups that push electron density away from themselves. Alkyl groups (–CH₃, –C₂H₅, etc.) are electron-donating due to hyperconjugation. –NO₂, –CN, and –Cl are electron-withdrawing groups that exhibit the –I effect.

4. In the free radical chlorination of methane, the initiation step is:
Answer

(a) CH₄ + Cl· → CH₃Cl + H· (b) Cl₂ → 2 Cl· (c) Cl· + CH₄ → ·CH₃ + HCl (d) ·CH₃ + ·CH₃ → C₂H₆

Answer(b) Cl₂ → 2 Cl·
ExplanationThe initiation step is the first step in a free radical mechanism where a bond breaks homolytically to generate free radicals. In the chlorination of methane, the Cl–Cl bond breaks under UV light to produce two chlorine radicals. Options (a) and (c) are propagation steps. Option (d) is a termination step.

5. Which of the following is NOT a nucleophile?
Answer

(a) H₂O (b) NH₃ (c) BF₃ (d) OH⁻

Answer(c) BF₃
ExplanationBF₃ (boron trifluoride) is an electrophile because boron has an incomplete octet (only 6 electrons) and seeks electrons. H₂O, NH₃, and OH⁻ all have lone pairs of electrons and can donate them, making them nucleophiles.

6. Hyperconjugation is also known as:
Answer

(a) No-bond resonance (b) Inductive effect (c) Electromeric effect (d) Resonance

Answer(a) No-bond resonance
ExplanationHyperconjugation involves the delocalisation of σ-electrons from an adjacent C–H bond into an empty or partially filled p-orbital or π-system. It is called "no-bond resonance" because the contributing structures show a C–H bond that appears to have been broken (no bond between C and H) while the hydrogen remains in place.

7. The resonance effect operates through:
Answer

(a) Only sigma bonds (b) Only pi bonds and conjugated systems (c) Only ionic bonds (d) Only hydrogen bonds

Answer(b) Only pi bonds and conjugated systems
ExplanationThe resonance (mesomeric) effect involves the delocalisation of π-electrons through conjugated systems — alternating single and double bonds, or systems with a lone pair adjacent to a double bond. It does not operate through sigma bonds alone (that is the inductive effect).

II. Fill in the Blanks

1. A species that seeks electrons is called an electrophile.

2. The permanent polarization of a sigma bond due to electronegativity difference is called the inductive effect.

3. In heterolytic fission, the bond breaks unequally, producing a cation and an anion.

4. The electron-donating groups that push electron density away from themselves exhibit the +I effect.

5. The delocalisation of π-electrons through conjugated systems is called the resonance or mesomeric effect.

6. In a free radical mechanism, the step where two radicals combine to form a stable molecule is called termination.

7. NH₃ acts as a nucleophile because it has a lone pair on nitrogen.

8. The –NO₂ group exhibits the –I and –M effect, withdrawing electrons both inductively and by resonance.

9. In the chlorination of methane, the propagation step produces HCl and methyl radical (·CH₃).

10. Hyperconjugation involves the delocalisation of sigma (σ) electrons from adjacent C–H bonds.

11. The stability order of carbocations is: 3° > 2° > 1° (tertiary > secondary > primary).

12. The order of reactivity for nucleophilic substitution is: 1° > 2° > 3° for SN2 mechanism.

13. The electromeric effect is a temporary polarization of π-electrons in the presence of a reagent.

14. BF₃ is an electrophile because boron has an incomplete octet.

15. The species formed by homolytic fission of a bond are called free radicals.


III. True or False (with Corrections)

1. Statement: Nucleophiles are electron-deficient species.
AnswerFalse. Nucleophiles are electron-rich species that donate an electron pair. Electrophiles are the electron-deficient species that accept an electron pair.
2. Statement: The inductive effect is permanent and operates through sigma bonds.
AnswerTrue. The inductive effect is a permanent electron displacement through sigma bonds caused by electronegativity differences. It is always present and does not depend on the approach of a reagent.
3. Statement: The resonance effect and the inductive effect are the same.
AnswerFalse. The inductive effect operates through sigma bonds and involves permanent polarization due to electronegativity. The resonance effect operates through pi bonds and conjugated systems and involves delocalisation of π-electrons. They are distinct electronic effects.
4. Statement: A tertiary carbocation is less stable than a primary carbocation.
AnswerFalse. A tertiary carbocation is MORE stable than a primary carbocation due to hyperconjugation and the +I effect of three alkyl groups that stabilise the positive charge. The stability order is 3° > 2° > 1°.
5. Statement: In homolytic fission, one atom gets both bonding electrons.
AnswerFalse. In homolytic fission, the bond breaks equally — each atom gets one electron, producing two free radicals. It is in HETEROLYTIC fission that one atom gets both bonding electrons, producing a cation and an anion.

IV. Match the Following

Set 1: Reagents and Types

Column AColumn B
1. OH⁻(a) Electrophile
2. NO₂⁺(b) Nucleophile
3. BF₃(c) Electrophile
4. NH₃(d) Nucleophile
5. AlCl₃(e) Electrophile
Answer

1. OH⁻ → (b) Nucleophile 2. NO₂⁺ → (a) Electrophile 3. BF₃ → (c) Electrophile 4. NH₃ → (d) Nucleophile 5. AlCl₃ → (e) Electrophile


Set 2: Effects and Groups

Column AColumn B
1. +I effect(a) –NO₂, –CN
2. –I effect(b) –CH₃, –C₂H₅
3. +M effect(c) –NO₂, –CHO
4. –M effect(d) –OH, –NH₂
Answer

1. +I effect → (b) –CH₃, –C₂H₅ 2. –I effect → (a) –NO₂, –CN 3. +M effect → (d) –OH, –NH₂ 4. –M effect → (c) –NO₂, –CHO


V. Answer Briefly (2-mark)

1. Define electrophile and nucleophile with two examples each.
Answer

An electrophile is an electron-deficient species that accepts an electron pair to form a new bond. Electrophiles are positively charged or neutral species with an empty orbital.

  • Example 1: H⁺ (proton) — seeks electrons to form a bond
  • Example 2: NO₂⁺ (nitronium ion) — used in nitration reactions

A nucleophile is an electron-rich species that donates an electron pair to form a new bond. Nucleophiles are negatively charged or neutral species with lone pairs.

  • Example 1: OH⁻ (hydroxide ion) — donates lone pair
  • Example 2: NH₃ (ammonia) — nitrogen lone pair acts as donor

2. Differentiate between homolytic and heterolytic fission.
Answer
FeatureHomolytic FissionHeterolytic Fission
Bond breakEqual — each atom gets one electronUnequal — one atom gets both electrons
ProductsTwo free radicalsA cation and an anion
ConditionsUV light, heat, peroxidesPolar solvents
ExampleCl₂ → 2 Cl·CH₃–Cl → CH₃⁺ + Cl⁻
EnergyRequires less energyRequires more energy

3. What is hyperconjugation? Give one example.
Answer

Hyperconjugation (also called no-bond resonance) is the delocalisation of σ-electrons of an adjacent C–H bond into an empty or partially filled p-orbital or π-system. It stabilises carbocations, alkenes, and free radicals.

Example: In propene (CH₃–CH=CH₂), the σ-electrons of the C–H bonds of the methyl group overlap with the π-orbital of the double bond, providing stability through hyperconjugation.

H H H H \ / \ / CH₃–CH=CH₂ ←→ ·CH₂–CH=CH₂ (no-bond resonance form)

This is why more substituted alkenes are more stable.


4. State the order of stability of carbocations and explain why.
Answer

The order of stability is: 3° > 2° > 1° > CH₃⁺

This is because:

  • Alkyl groups exert a +I effect, pushing electron density towards the positively charged carbon.
  • More alkyl groups mean more electron donation, stabilising the positive charge.
  • Additionally, hyperconjugation from adjacent C–H bonds further stabilises the carbocation.
  • A tertiary carbocation has 3 alkyl groups providing +I effect and 9 hyperconjugative C–H bonds, making it the most stable.

VI. Answer in Detail (5-mark)

1. Explain the inductive effect and resonance effect with examples and compare them.

Inductive Effect

The inductive effect is the permanent polarization of a sigma bond caused by the electronegativity difference between two atoms. The electron density shifts along the chain of sigma bonds, but the effect diminishes rapidly with distance.

Types:

  • +I Effect (Electron-donating): Groups that push electron density away from themselves. Alkyl groups (–CH₃, –C₂H₅, –CH(CH₃)₂) show the +I effect.
In CH₃–CH₂–Cl: The Cl atom withdraws electron density (–I effect) The CH₃ group donates electron density (+I effect) Net effect: polarization of C–Cl bond towards Cl
  • –I Effect (Electron-withdrawing): Groups that pull electron density towards themselves. Examples: –Cl, –NO₂, –CN, –COOH, –F.

Application — Effect on Acidity:

+I groups decrease acidity: CH₃–COOH (ethanoic acid, pKa ≈ 4.76) is weaker than H–COOH (methanoic acid, pKa ≈ 3.75) because –CH₃ pushes e⁻ towards –COOH –I groups increase acidity: F–CH₂–COOH (fluoroacetic acid) is stronger than CH₃–COOH because –F pulls e⁻ away from –COOH

Resonance (Mesomeric) Effect

The resonance effect is the delocalisation of π-electrons through conjugated systems (alternating single and double bonds, or a lone pair adjacent to a π-bond). It stabilises the molecule by distributing electron density over multiple atoms.

Types:

  • +M Effect (Electron-donating by resonance): Groups that donate electrons through resonance into the conjugated system. Examples: –OH, –NH₂, –OR, –NHR.
Phenol (C₆H₅–OH): The lone pair on oxygen delocalises into the benzene ring, making the ortho and para positions electron-rich. This activates the ring for electrophilic substitution.
  • –M Effect (Electron-withdrawing by resonance): Groups that withdraw electrons through resonance from the conjugated system. Examples: –NO₂, –CN, –CHO, –COOH.
Nitrobenzene (C₆H₅–NO₂): The –NO₂ group withdraws electron density from the ring, deactivating it for electrophilic substitution.

Comparison Table

FeatureInductive EffectResonance Effect
Bonds involvedSigma (σ)Pi (π) and conjugated systems
NaturePermanent polarizationDelocalisation of π-electrons
RangeShort (3-4 carbons)Long (entire conjugated system)
Depends onElectronegativity differencePresence of conjugation
Effect type+I or –I+M or –M

2. Describe the free radical mechanism of chlorination of methane with all steps.
Answer

The chlorination of methane proceeds via a free radical chain mechanism involving three types of steps: initiation, propagation, and termination.

Step 1: Initiation

The reaction begins when UV light (or heat) provides enough energy to break the Cl–Cl bond homolytically. Each chlorine atom retains one electron, forming two chlorine free radicals.

Cl₂ →(UV light)→ 2 Cl·

This is the energy-absorbing step that starts the chain reaction. The bond dissociation energy of Cl₂ is about 242 kJ/mol.

Step 2: Propagation

The chlorine radical attacks a methane molecule, abstracting a hydrogen atom to form HCl and a methyl radical.

Cl· + CH₄ → ·CH₃ + HCl

The highly reactive methyl radical then attacks a chlorine molecule, producing chloromethane and regenerating a chlorine radical.

·CH₃ + Cl₂ → CH₃Cl + Cl·

The regenerated Cl· radical can attack another CH₄ molecule, continuing the chain. This is why it is called a chain reaction — one initiation event leads to many product molecules.

Step 3: Further Chlorination (if excess Cl₂ is present)

The process can repeat with CH₃Cl:

Cl· + CH₃Cl → ·CH₂Cl + HCl ·CH₂Cl + Cl₂ → CH₂Cl₂ + Cl·

And further:

Cl· + CH₂Cl₂ → ·CHCl₂ + HCl ·CHCl₂ + Cl₂ → CHCl₃ + Cl·

And even further:

Cl· + CHCl₃ → ·CCl₃ + HCl ·CCl₃ + Cl₂ → CCl₄ + Cl·

Step 4: Termination

The chain reaction ends when two free radicals combine to form a stable molecule:

Cl· + Cl· → Cl₂ ·CH₃ + Cl· → CH₃Cl ·CH₃ + ·CH₃ → C₂H₆ (ethane)

Termination steps remove free radicals from the system, stopping the chain. Since radicals are present in very low concentrations, termination events are rare compared to propagation steps.

Energy Considerations

The overall reaction is exothermic because the bonds formed (C–Cl and H–Cl) release more energy than the bonds broken (C–H and Cl–Cl). However, the initiation step requires energy input (UV light) to break the Cl–Cl bond.


3. Explain substitution, addition, and elimination reactions with examples.

Substitution Reactions

In a substitution reaction, one atom or group in a molecule is replaced by another atom or group. There are two main types:

Nucleophilic Substitution (SN): A nucleophile replaces a leaving group.

CH₃Br + OH⁻ → CH₃OH + Br⁻

Here, OH⁻ (nucleophile) replaces Br⁻ (leaving group).

Electrophilic Substitution (SE): An electrophile replaces a hydrogen atom on an aromatic ring.

C₆H₆ + Br₂ →(FeBr₃)→ C₆H₅Br + HBr

Here, Br⁺ (electrophile) replaces a hydrogen on benzene.

Free Radical Substitution: A free radical replaces an atom.

CH₄ + Cl₂ →(UV)→ CH₃Cl + HCl

Addition Reactions

In an addition reaction, atoms or groups are added across a double or triple bond, converting it to a single bond. The unsaturation decreases.

Electrophilic Addition: An electrophile adds to an unsaturated bond.

CH₂=CH₂ + HBr → CH₃–CH₂Br

Catalytic Hydrogenation: Hydrogen adds across a double bond using a metal catalyst.

CH₂=CH₂ + H₂ →(Ni)→ CH₃–CH₃

Addition of Halogens:

CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br

Elimination Reactions

In an elimination reaction, atoms or groups are removed from adjacent carbons to form a double bond. Unsaturation increases.

Dehydration of Alcohols (E1):

C₂H₅OH →(H₂SO₄, 170°C)→ CH₂=CH₂ + H₂O

Dehydrohalogenation (E2):

CH₃CH₂CH₂Br + KOH(alc) → CH₃CH=CH₂ + KBr + H₂O

Saytzeff's Rule: In elimination reactions, the more substituted alkene (with more alkyl groups on the double-bonded carbons) is the major product.

CH₃–CHBr–CH₂–CH₃ + OH⁻(alc) → Major: CH₃–CH=CH–CH₃ (2-butene, more substituted) Minor: CH₂=CH–CH₂–CH₃ (1-butene, less substituted)

Comparison Table

FeatureSubstitutionAdditionElimination
SaturationNo changeIncreases saturationIncreases unsaturation
Bonds broken1 σ-bond1 π-bond2 σ-bonds
Bonds formed1 σ-bond2 σ-bonds1 π-bond
ExampleCH₄ + Cl₂ → CH₃Cl + HClCH₂=CH₂ + HBr → CH₃CH₂BrC₂H₅OH → CH₂=CH₂ + H₂O

Common Mistakes to Avoid

1. Confusing electrophiles and nucleophiles: Electrophiles SEEK electrons (electron-poor); nucleophiles DONATE electrons (electron-rich). Remember: "Electrophile = electron-loving = electron-acceptor."

2. Mixing up homolytic and heterolytic fission: In homolytic fission, each atom gets ONE electron (produces radicals). In heterolytic fission, one atom gets BOTH electrons (produces ions).

3. Confusing +I/–I with +M/–M effects: The inductive effect operates through sigma bonds and is short-range. The resonance effect operates through pi bonds and conjugated systems. Some groups (like –Cl) show –I effect but can show +M effect in certain contexts.

4. Forgetting that the inductive effect diminishes with distance: The inductive effect becomes negligible beyond 3-4 carbon atoms. It is NOT transmitted through the entire molecule like the resonance effect.

5. Mixing up the order of carbocation stability: The correct order is 3° > 2° > 1° > CH₃⁺. More alkyl groups = more +I effect = more stable carbocation. Do NOT reverse this.

6. Omitting the initiation step in free radical mechanisms: Every free radical mechanism must start with an initiation step (bond homolysis). Without it, there are no radicals to begin the chain.

7. Confusing substitution with addition reactions: Substitution REPLACES one group with another (saturation unchanged). Addition ADDS groups across a double bond (saturation increases). The number of products differs.

8. Ignoring Saytzeff's rule in elimination reactions: The more substituted alkene is the major product in elimination. Always check which alkene has more alkyl groups on the double-bonded carbons.


Exam Checklist

  • [ ] I can define and identify electrophiles and nucleophiles with examples
  • [ ] I can differentiate between homolytic and heterolytic fission
  • [ ] I can explain the inductive effect (+I and –I) with examples
  • [ ] I can explain the resonance/mesomeric effect (+M and –M) with examples
  • [ ] I can compare inductive and resonance effects
  • [ ] I can describe the free radical mechanism (initiation, propagation, termination)
  • [ ] I can explain hyperconjugation and its role in stability
  • [ ] I can write the stability order of carbocations, carbanions, and free radicals
  • [ ] I can distinguish substitution, addition, and elimination reactions with examples
  • [ ] I can explain the electromeric effect and its temporary nature

Frequently Asked Questions

The inductive effect is the permanent polarization of a sigma bond due to the electronegativity difference between atoms, transmitting electron density through the chain.