Ethers: Describe the Williamson synthesis of ethers and its limitations with tertiary halides

Chapter: Alcohols, Phenols and Ethers

Topic: Williamson Ether Synthesis and Its Limitations

Williamson Synthesis of Ethers

The Williamson ether synthesis is an important method for preparing ethers by reacting a suitable sodium alkoxide or sodium phenoxide with an alkyl halide.

R–ONa + R′–X → R–O–R′ + NaX

The reaction generally proceeds through an SN2 mechanism.

Alkoxide / Phenoxide + Alkyl halide
↓
SN2 substitution
↓
Ether

An alcohol reacts with sodium metal to form sodium alkoxide.

2ROH + 2Na → 2RONa + H₂↑

The resulting sodium alkoxide acts as the nucleophile in the Williamson synthesis.

ROH → RONa → Nucleophile

R–ONa + R′–X → R–O–R′ + NaX

Where:

• R–ONa = sodium alkoxide
• R′–X = alkyl halide
• R–O–R′ = ether
• NaX = sodium halide

The alkyl halide should preferably be a primary alkyl halide for a good yield.

The alkoxide ion acts as a nucleophile and attacks the carbon atom attached to the leaving group.

RO⁻ + R′–X
↓
Back-side attack
↓
R–O–R′ + X⁻

The mechanism is a single-step SN2 nucleophilic substitution.

SN2 → One-step reaction → Ether formation

Sodium ethoxide reacts with bromoethane to form ethoxyethane (diethyl ether).

C₂H₅ONa + C₂H₅Br
↓
C₂H₅–O–C₂H₅ + NaBr

Here:

• C₂H₅ONa = sodium ethoxide
• C₂H₅Br = bromoethane
• C₂H₅–O–C₂H₅ = ethoxyethane
• NaBr = sodium bromide

Bromoethane + Sodium ethoxide → Diethyl ether

Williamson synthesis is particularly useful for preparing unsymmetrical ethers.

R–ONa + R′–X
↓
R–O–R′

For example, sodium ethoxide and methyl iodide give methoxyethane.

C₂H₅ONa + CH₃I
↓
CH₃OC₂H₅ + NaI

Methyl/primary alkyl halide + Alkoxide → Good Williamson synthesis

Tertiary alkyl halides are generally unsuitable for Williamson ether synthesis because the reaction proceeds through an SN2 mechanism.

In a tertiary alkyl halide, the carbon bearing the halogen is highly sterically hindered. Therefore, backside attack by the alkoxide ion becomes difficult.

3° R–X
↓
Steric hindrance
↓
SN2 attack difficult
↓
Elimination (E2) favoured
↓
Alkene

Tertiary halide + Alkoxide
→ E2 elimination predominates
→ Alkene forms instead of ether

Consider tert-butyl bromide with sodium ethoxide.

(CH₃)₃C–Br + C₂H₅ONa
↓
E2 elimination
↓
(CH₃)₂C=CH₂

Instead of ether formation, the strong alkoxide base removes a β-hydrogen and an alkene is formed.

Exam Point: Tertiary alkyl halides favour elimination because SN2 backside attack is sterically hindered.

Halide	Williamson Reaction	Reason
Methyl	Excellent	Very low steric hindrance
Primary	Very good	SN2 favoured
Secondary	Less suitable	Some elimination may occur
Tertiary	Not suitable	E2 predominates

1. The reaction works best with methyl and primary alkyl halides.
2. Secondary alkyl halides may give a mixture of substitution and elimination products.
3. Tertiary alkyl halides generally undergo E2 elimination instead of SN2 substitution.
4. Strongly hindered substrates are poor choices because SN2 attack is sterically difficult.
5. Aryl halides such as chlorobenzene generally do not undergo ordinary SN2 Williamson substitution.

For Williamson synthesis:

Choose a methyl or primary alkyl halide.

Q. What is Williamson ether synthesis?

Answer:

Williamson ether synthesis is the preparation of ethers by reacting sodium alkoxide or sodium phenoxide with a suitable alkyl halide. The reaction generally follows the SN2 mechanism.

RONa + R′X → ROR′ + NaX

Q. Explain Williamson ether synthesis with an example.

Answer:

Sodium alkoxide reacts with an alkyl halide through an SN2 nucleophilic substitution reaction to form an ether.

C₂H₅ONa + CH₃I → CH₃OC₂H₅ + NaI

Here sodium ethoxide acts as the nucleophile and methyl iodide acts as the substrate.

Q. Why are tertiary alkyl halides not suitable for Williamson ether synthesis?

Answer:

1. Williamson synthesis proceeds by an SN2 mechanism.
2. SN2 requires backside attack on the carbon attached to the leaving group.
3. In tertiary alkyl halides, this carbon is highly sterically hindered.
4. Therefore SN2 substitution is difficult and the strong alkoxide ion acts as a base, causing E2 elimination to form an alkene.

Tertiary halide → SN2 difficult → E2 elimination favoured

Q. Describe Williamson ether synthesis and discuss its limitations.

Williamson Reaction

R–ONa + R′–X → R–O–R′ + NaX

It is an SN2 reaction in which an alkoxide ion attacks an alkyl halide.

Limitations

1. Methyl and primary alkyl halides give the best results.
2. Secondary halides may undergo competing elimination.
3. Tertiary halides do not favour SN2 because of steric hindrance.
4. Tertiary halides generally give alkenes by E2 elimination.
5. Aryl halides are not suitable for ordinary SN2 Williamson synthesis.

Williamson synthesis is most effective when the alkyl halide is methyl or primary.

Q. Explain the Williamson synthesis of ethers, its mechanism, examples and limitations with tertiary alkyl halides.

1. Definition

Williamson ether synthesis is a method for preparing ethers by reacting a sodium alkoxide or sodium phenoxide with an alkyl halide.

R–ONa + R′–X → R–O–R′ + NaX

2. Mechanism

The alkoxide ion attacks the carbon atom of the alkyl halide from the back side and displaces the halide ion.

RO⁻ + R′–X
↓
Back-side SN2 attack
↓
R–O–R′ + X⁻

3. Example

C₂H₅ONa + C₂H₅Br → C₂H₅OC₂H₅ + NaBr

This produces ethoxyethane (diethyl ether).

4. Limitation with Tertiary Halides

Tertiary alkyl halides are highly sterically hindered. Therefore, backside attack required for SN2 is difficult.

3° R–X + RO⁻
↓
SN2 hindered
↓
E2 elimination
↓
Alkene

5. Final Conclusion

Methyl / Primary alkyl halide → SN2 → Ether formation

Tertiary alkyl halide → SN2 hindered → E2 elimination → Alkene formation

1. Williamson synthesis is used for the preparation of:

A) Aldehydes

B) Ethers

C) Ketones only

D) Amines

✅ Answer

B) Ethers

---

2. The nucleophile in Williamson synthesis is generally:

A) Alkoxide ion

B) H⁺

C) Cl⁻ only

D) Water

✅ Answer

A) Alkoxide ion

---

3. Williamson synthesis generally follows:

A) SN1

B) SN2

C) E1 only

D) Addition

✅ Answer

B) SN2

---

4. General reaction of Williamson synthesis is:

A) RONa + R′X → ROR′ + NaX

B) ROH → RCHO

C) RX → RH

D) RCOOH → RCHO

✅ Answer

A

---

5. Which substrate gives the best Williamson reaction?

A) Tertiary alkyl halide

B) Primary alkyl halide

C) Highly hindered tertiary alcohol

D) Quaternary carbon compound

✅ Answer

B) Primary alkyl halide

---

6. Methyl halides are:

A) Very suitable

B) Unsuitable

C) Always eliminated

D) Radical substrates

✅ Answer

A) Very suitable

---

7. Tertiary alkyl halides are unsuitable mainly because:

A) SN2 is sterically hindered

B) Oxygen is absent

C) They cannot contain halogen

D) They are gases

✅ Answer

A

---

8. With tertiary halides, alkoxide ions generally favour:

A) E2 elimination

B) SN2 strongly

C) Addition

D) Reduction

✅ Answer

A) E2 elimination

---

9. The product of E2 reaction of a tertiary halide is generally:

A) Alkene

B) Ether

C) Alcohol only

D) Aldehyde

✅ Answer

A) Alkene

---

10. SN2 reaction involves:

A) Backside attack

B) Carbocation only

C) Free radical

D) Electrophilic attack only

✅ Answer

A

---

11. Sodium ethoxide + bromoethane gives:

A) Ethoxyethane

B) Ethene only

C) Ethanal

D) Ethanoic acid

✅ Answer

A) Ethoxyethane

---

12. The leaving group in an alkyl bromide is:

A) Br⁻

B) Na⁺

C) O⁻

D) H⁺

✅ Answer

A) Br⁻

---

13. Williamson synthesis can prepare:

A) Symmetrical ethers

B) Unsymmetrical ethers

C) Both

D) Neither

✅ Answer

C) Both

---

14. For unsymmetrical ethers, the less hindered alkyl halide is usually preferred because:

A) SN2 is easier

B) SN1 is always required

C) E1 is required

D) No reaction occurs

✅ Answer

A

---

15. Secondary alkyl halides in Williamson synthesis may undergo:

A) Only substitution

B) Competing elimination

C) Only oxidation

D) No reaction

✅ Answer

B

---

16. Which is most suitable for Williamson synthesis?

A) CH₃I

B) (CH₃)₃CBr

C) Highly hindered tertiary bromide

D) Aryl halide

✅ Answer

A) CH₃I

---

17. Sodium phenoxide can react with a suitable alkyl halide to form:

A) Aryl ether

B) Alkane only

C) Ketone only

D) Amide

✅ Answer

A) Aryl ether

---

18. Williamson synthesis is essentially a:

A) Nucleophilic substitution

B) Electrophilic addition

C) Free-radical polymerisation

D) Oxidation

✅ Answer

A

---

19. Tertiary halide + strong alkoxide generally produces:

A) Ether as major product

B) Alkene by elimination

C) Aldehyde

D) Acid

✅ Answer

B

---

20. The key requirement for a successful Williamson synthesis is:

A) Suitable alkyl halide capable of SN2

B) Tertiary halide only

C) Carbocation formation

D) Aromatic halide only

✅ Answer

A

🎯 Quick Revision

Williamson synthesis: RONa + R′X → ROR′ + NaX
Mechanism: SN2
Nucleophile: Alkoxide / Phenoxide ion
Best substrate: Methyl or Primary alkyl halide
Secondary halide: Substitution + elimination may compete
Tertiary halide: SN2 strongly hindered
Tertiary halide + alkoxide: E2 elimination favoured
Main product with tertiary halide: Alkene

Alcohols/Phenols: Explain why Phenol is more acidic than Ethanol using resonance stabilisation

Chapter: Alcohols, Phenols and Ethers

Topic: Acidity of Phenol and Ethanol – Resonance Stabilisation

Why is Phenol More Acidic than Ethanol?

Phenol (C₆H₅OH) is considerably more acidic than ethanol (C₂H₅OH). The main reason is the greater stability of the conjugate base formed after loss of H⁺.

Phenol → H⁺ + Phenoxide ion

Ethanol → H⁺ + Ethoxide ion

Phenoxide ion is resonance-stabilised
↓
More stable conjugate base
↓
Phenol is more acidic

Phenol

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

Phenol loses H⁺ to form the phenoxide ion.

Ethanol

C₂H₅OH ⇌ C₂H₅O⁻ + H⁺

Ethanol loses H⁺ to form the ethoxide ion.

The stability of the conjugate base determines the relative acidity.

Phenoxide Ion

In the phenoxide ion, the negative charge initially resides on oxygen. However, the lone pair/negative charge can be delocalised into the benzene ring through resonance.

C₆H₅O⁻
↓
Negative charge delocalisation
↓
Ortho and para positions of the ring

Thus, the negative charge is not confined completely to oxygen. It is delocalised over the oxygen atom and the aromatic ring.

Important: Resonance does not mean that the negative charge permanently moves from one atom to another. The actual phenoxide ion is a resonance hybrid.

Delocalisation of negative charge → Greater stability of phenoxide ion

Ethoxide Ion

C₂H₅O⁻

In ethoxide ion, the negative charge remains essentially localised on the oxygen atom. There is no benzene π-system available for resonance delocalisation.

C₂H₅O⁻
↓
Negative charge mainly on O
↓
No resonance stabilisation

Therefore, ethoxide ion is less stable than phenoxide ion.

Ethoxide ion → Localised negative charge → Less stable

The ethyl group in ethanol has a +I (electron-releasing) effect.

C₂H₅ → O⁻
Electron-releasing effect
↓
Negative charge becomes less stable

The electron-releasing alkyl group increases electron density around the oxygen atom and destabilises the ethoxide ion.

This further decreases the acidity of ethanol.

+I effect of alkyl group → Destabilises ethoxide ion

Property	Phenol	Ethanol
Formula	C₆H₅OH	C₂H₅OH
Conjugate base	Phenoxide ion	Ethoxide ion
Resonance stabilisation	Present	Absent
Negative charge	Delocalised	Mainly localised on O
Conjugate-base stability	Higher	Lower
Acidity	Higher	Lower

Acid Strength
↓
Stability of Conjugate Base
↓
More Stable Conjugate Base
↓
Stronger Acid

Phenoxide ion more stable than Ethoxide ion
↓
Phenol more acidic than Ethanol

Q. Why is phenol more acidic than ethanol?

Answer:

Phenol forms a resonance-stabilised phenoxide ion after loss of H⁺. The negative charge in phenoxide ion is delocalised over the oxygen atom and the aromatic ring.

Ethoxide ion does not have resonance stabilisation. Hence phenol is more acidic than ethanol.

Q. Explain the role of resonance in the greater acidity of phenol.

Answer:

1. Phenol loses H⁺ to form the phenoxide ion.
2. The negative charge of phenoxide ion is delocalised through resonance into the benzene ring.
3. This delocalisation stabilises phenoxide ion and therefore makes phenol more acidic than ethanol.

Q. Compare the acidity of phenol and ethanol on the basis of stability of their conjugate bases.

Answer:

Phenol ionises as:

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

The phenoxide ion is resonance-stabilised because the negative charge can be delocalised over the aromatic ring.

Ethanol ionises as:

C₂H₅OH ⇌ C₂H₅O⁻ + H⁺

The ethoxide ion has no resonance stabilisation and the ethyl group also shows a +I effect, which destabilises the negative charge.

Phenoxide ion > Ethoxide ion in stability
Therefore:
Phenol > Ethanol in acidity

Q. Explain with resonance why phenol is more acidic than ethanol.

Step 1 – Ionisation of Phenol

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

Step 2 – Resonance Stabilisation

The negative charge of the phenoxide ion is delocalised into the benzene ring through resonance. The resonance structures involve mainly the ortho and para positions.

Phenoxide ion
↓
Negative charge delocalisation
↓
Resonance stabilisation

Step 3 – Ethanol

C₂H₅OH ⇌ C₂H₅O⁻ + H⁺

Ethoxide ion has no resonance stabilisation. Moreover, the ethyl group has a +I effect and destabilises the negative charge.

Greater stability of phenoxide ion
↓
Greater tendency of phenol to lose H⁺
↓
Phenol is more acidic than ethanol.

Q. Explain in detail why phenol is more acidic than ethanol using resonance stabilisation and inductive effects.

1. Ionisation of Phenol

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

Phenol produces the phenoxide ion after losing a proton.

2. Resonance Stabilisation

In phenoxide ion, the negative charge is delocalised from oxygen into the aromatic π-system. This produces several important resonance contributors involving the ortho and para positions.

Negative charge on O
↓
Delocalisation into benzene ring
↓
Resonance-stabilised phenoxide ion

3. Ionisation of Ethanol

C₂H₅OH ⇌ C₂H₅O⁻ + H⁺

The ethoxide ion has no aromatic π-system and therefore cannot delocalise its negative charge by resonance.

4. +I Effect of Ethyl Group

The ethyl group is electron-releasing. Its +I effect increases the electron density on the oxygen atom of ethoxide ion and makes the negative charge less stable.

5. Final Conclusion

Phenoxide ion → Resonance stabilised
Ethoxide ion → No resonance stabilisation + +I destabilisation

Phenoxide ion is more stable
↓
Phenol is more acidic than ethanol

1. Which is more acidic?

A) Ethanol

B) Phenol

C) Both equally

D) Ether

✅ Answer

B) Phenol

---

2. The conjugate base of phenol is:

A) Ethoxide ion

B) Phenoxide ion

C) Hydroxide ion

D) Oxide ion

✅ Answer

B) Phenoxide ion

---

3. Phenol is more acidic mainly because:

A) Phenoxide ion is resonance-stabilised

B) Phenol has no oxygen

C) Ethanol is aromatic

D) Phenol is ionic

✅ Answer

A

---

4. The negative charge in phenoxide ion is:

A) Completely localised on carbon

B) Delocalised

C) Absent

D) Only on hydrogen

✅ Answer

B) Delocalised

---

5. Ethoxide ion is less stable because:

A) It is resonance-stabilised

B) It has no resonance stabilisation

C) It is aromatic

D) It has no oxygen

✅ Answer

B

---

6. Resonance in phenoxide ion involves:

A) Benzene π-system

B) Alkane σ-system only

C) Hydrogen bond only

D) Metal orbitals

✅ Answer

A

---

7. The more stable conjugate base corresponds to:

A) Weaker acid

B) Stronger acid

C) No acid

D) Always a base

✅ Answer

B

---

8. Which ion is more stable?

A) Ethoxide

B) Phenoxide

C) Both equally

D) Neither

✅ Answer

B) Phenoxide

---

9. The ethyl group exhibits:

A) +I effect

B) –I effect

C) No inductive effect

D) Only resonance effect

✅ Answer

A) +I effect

---

10. The +I effect of ethyl group:

A) Stabilises ethoxide ion

B) Destabilises ethoxide ion

C) Has no effect

D) Removes oxygen

✅ Answer

B

---

11. Phenol ionises to give:

A) C₆H₅O⁻ + H⁺

B) C₂H₅O⁻ + H⁺

C) C₆H₆ + OH⁻

D) CH₄ + O⁻

✅ Answer

A

---

12. Ethanol ionises to give:

A) Phenoxide ion

B) Ethoxide ion

C) Phenyl ion

D) Carbonate ion

✅ Answer

B

---

13. Resonance stabilisation generally makes an ion:

A) Less stable

B) More stable

C) Neutral

D) Radioactive

✅ Answer

B

---

14. Phenoxide ion is stabilised by delocalisation of:

A) Positive charge

B) Negative charge

C) Neutrons

D) Protons

✅ Answer

B

---

15. Which statement is correct?

A) Phenol forms a less stable conjugate base

B) Phenol forms a more stable conjugate base

C) Ethanol is aromatic

D) Ethoxide is resonance-stabilised by benzene

✅ Answer

B

---

16. The acidity of phenol is related directly to the stability of:

A) Phenoxide ion

B) Benzene

C) Ethane

D) Ethyl radical

✅ Answer

A

---

17. Which ion lacks resonance stabilisation?

A) Phenoxide

B) Ethoxide

C) Both

D) Neither

✅ Answer

B) Ethoxide

---

18. Phenol is more acidic than ethanol because:

A) Phenoxide is more stable

B) Ethanol is aromatic

C) Ethoxide has resonance

D) Phenol has no oxygen

✅ Answer

A

---

19. Greater conjugate-base stability means:

A) Greater acidity

B) Lower acidity

C) No acidity

D) No relation

✅ Answer

A

---

20. Correct order of acidity is:

A) Ethanol > Phenol

B) Phenol > Ethanol

C) Ethanol = Phenol

D) Neither is acidic

✅ Answer

B) Phenol > Ethanol

🎯 Quick Revision

Phenol: C₆H₅OH
Ethanol: C₂H₅OH
Phenol conjugate base: Phenoxide ion
Ethanol conjugate base: Ethoxide ion
Phenoxide: Resonance stabilised
Ethoxide: No resonance stabilisation
Ethyl group: +I effect
Final order: Phenol > Ethanol in acidity

Haloarenes: Why are haloarenes less reactive towards nucleophilic substitution than haloalkanes

Chapter: Haloalkanes and Haloarenes

Topic: Reactivity of Haloarenes towards Nucleophilic Substitution

Why are Haloarenes Less Reactive?

Haloarenes such as chlorobenzene (C₆H₅Cl) are much less reactive towards nucleophilic substitution than haloalkanes such as chloroethane (C₂H₅Cl).

Haloalkanes → More reactive

Haloarenes → Less reactive

The main reason is the special nature of the C–X bond in haloarenes, caused mainly by resonance.

Most Important Reason

In haloarenes, the lone pair of electrons on the halogen atom participates in resonance with the benzene ring.

C₆H₅–Cl
↕
Resonance interaction
↕
Partial C=Cl character

Due to resonance, the carbon–halogen bond acquires partial double-bond character.

C–X bond in haloarenes → Partial double bond character

Therefore, this bond is stronger and more difficult to break than the ordinary C–X bond in haloalkanes.

In haloarenes, the carbon attached to halogen is sp² hybridised.

In haloalkanes, the carbon attached to halogen is generally sp³ hybridised.

Haloarene: C(sp²)–X

Haloalkane: C(sp³)–X

The sp² carbon has greater s-character than sp³ carbon. Consequently, the C–X bond in haloarenes is shorter and stronger.

Haloarene C–X bond → Shorter + Stronger

For an SN1 reaction, the C–X bond must break to form a carbocation.

C₆H₅–X
↓
C₆H₅⁺ + X⁻

Formation of a phenyl carbocation is highly unstable. Therefore, the SN1 mechanism is not favoured.

Important: The phenyl carbocation is much less stable than ordinary alkyl carbocations.

SN1 mechanism → Not favoured in haloarenes

SN2 reactions require a nucleophile to attack the carbon atom from the backside.

Nu⁻ → C–X
Back-side attack

In haloarenes, the carbon attached to halogen is part of the planar aromatic ring. The ring structure and stronger C–X bond make the required backside displacement difficult.

Thus, ordinary SN2 substitution is strongly disfavoured for aryl halides.

SN2 mechanism → Strongly hindered / disfavoured

The halogen lone pair overlaps with the π-system of the benzene ring. This resonance gives the C–X bond additional stability.

Lone pair on X
↓
Benzene π-system
↓
Resonance
↓
Stronger C–X bond

Therefore, breaking the C–X bond requires more energy.

1. The C–X bond in haloarenes has partial double-bond character due to resonance.
2. The carbon attached to halogen is sp² hybridised.
3. The C–X bond is shorter and stronger.
4. SN1 is difficult because formation of a phenyl carbocation is highly unfavourable.
5. SN2 is difficult because backside displacement at the aryl carbon is strongly disfavoured.
6. The aromatic ring and resonance provide additional stability to the haloarene.

Therefore, haloarenes are much less reactive towards nucleophilic substitution than haloalkanes.

Property	Haloalkanes	Haloarenes
Carbon hybridisation	sp³	sp²
C–X bond	Single bond	Partial double-bond character
Bond strength	Relatively lower	Relatively higher
SN1	Can occur depending on substrate	Not favoured
SN2	Common pathway	Strongly disfavoured
Nucleophilic substitution	More reactive	Less reactive

Chlorobenzene

C₆H₅–Cl

The C–Cl bond is strengthened by resonance and has partial double-bond character.

Chloroethane

CH₃CH₂–Cl

The C–Cl bond is a normal σ bond and does not have the same resonance interaction with an aromatic ring.

Chloroethane → More reactive
Chlorobenzene → Less reactive

Q. Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?

Answer:

In haloarenes, the lone pair of halogen participates in resonance with the benzene ring. Hence, the C–X bond acquires partial double-bond character and becomes stronger.

Therefore, haloarenes are less reactive towards nucleophilic substitution.

Q. Give three reasons why chlorobenzene is less reactive than chloroethane towards nucleophilic substitution.

1. The C–Cl bond in chlorobenzene has partial double-bond character due to resonance.
2. The carbon attached to chlorine is sp² hybridised, making the bond shorter and stronger.
3. Both SN1 and ordinary SN2 pathways are unfavourable for an aryl halide.

Q. Explain why haloarenes are less reactive towards nucleophilic substitution than haloalkanes.

Answer:

1. The halogen lone pair is delocalised into the aromatic ring.
2. This produces partial double-bond character in the C–X bond.
3. The carbon attached to halogen is sp² hybridised, giving a shorter and stronger C–X bond.
4. Formation of a phenyl carbocation is highly unfavourable, so SN1 is not favoured; backside substitution at the aryl carbon is also strongly disfavoured.

Hence haloarenes are less reactive than haloalkanes.

Q. Discuss the factors responsible for the low reactivity of haloarenes towards nucleophilic substitution.

1. Resonance

The lone pair of the halogen participates in resonance with the benzene ring.

C₆H₅–X
↓
Resonance
↓
Partial C=X character

2. Strong C–X Bond

The carbon attached to halogen is sp² hybridised. The C–X bond is shorter and stronger than a typical C–X bond in haloalkanes.

3. SN1 is Unfavourable

SN1 requires carbocation formation. Formation of a phenyl carbocation is highly unstable.

4. SN2 is Unfavourable

Backside displacement at the aromatic carbon is strongly disfavoured.

Resonance + Strong C–X bond + Unfavourable SN1/SN2
↓
Low reactivity of haloarenes

Q. Explain in detail why haloarenes are less reactive towards nucleophilic substitution than haloalkanes.

1. Resonance Effect

The lone pair present on the halogen atom overlaps with the π-electron system of the benzene ring.

As a result, the C–X bond acquires partial double-bond character.

Halogen lone pair
↓
π-system of benzene
↓
Resonance
↓
Partial double-bond character of C–X

2. Shorter and Stronger C–X Bond

The carbon attached to the halogen is sp² hybridised. Since sp² hybridisation has greater s-character than sp³ hybridisation, the C–X bond is shorter and stronger.

3. SN1 Mechanism is Unfavourable

SN1 requires formation of a carbocation. Formation of a phenyl carbocation is highly unstable and therefore the pathway is unfavourable.

C₆H₅–X → C₆H₅⁺ + X⁻
Highly unfavourable

4. SN2 Mechanism is Unfavourable

SN2 requires backside attack by the nucleophile. In an aryl halide, the carbon–halogen bond is part of the planar aromatic system and is strengthened by resonance, making ordinary backside displacement strongly disfavoured.

5. Final Result

Haloarenes:
Resonance → Strong C–X bond
+
SN1 unfavourable
+
SN2 strongly disfavoured

Therefore → Haloarenes are less reactive towards nucleophilic substitution.

1. Haloarenes are generally ______ reactive than haloalkanes towards nucleophilic substitution.

A) More

B) Less

C) Equally

D) Infinitely more

✅ Answer

B) Less

---

2. The carbon bonded to halogen in haloarenes is generally:

A) sp³ hybridised

B) sp² hybridised

C) sp hybridised

D) Unhybridised

✅ Answer

B) sp²

---

3. The C–X bond in haloarenes has:

A) Pure ionic character

B) Partial double-bond character

C) Triple-bond character

D) No bond character

✅ Answer

B

---

4. Partial double-bond character arises mainly due to:

A) Hyperconjugation only

B) Resonance

C) Hydrogen bonding

D) Ionic dissociation

✅ Answer

B) Resonance

---

5. In haloarenes, the lone pair of halogen interacts with:

A) σ-system only

B) π-system of benzene

C) Nucleus

D) Solvent only

✅ Answer

B

---

6. The C–X bond in chlorobenzene is:

A) Longer and weaker than in chloroethane

B) Shorter and stronger than in chloroethane

C) Identical in all respects

D) Ionic

✅ Answer

B

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7. SN1 reaction of chlorobenzene is difficult mainly because:

A) Chlorine is absent

B) Phenyl carbocation is highly unstable

C) Benzene is saturated

D) Chlorobenzene is ionic

✅ Answer

B

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8. SN2 reactions require:

A) Backside attack

B) Carbocation formation

C) Radical formation

D) Proton transfer only

✅ Answer

A

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9. Ordinary SN2 substitution is strongly disfavoured in:

A) Methyl halides

B) Primary haloalkanes

C) Aryl halides

D) Some primary alkyl systems

✅ Answer

C) Aryl halides

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10. Which compound is a haloarene?

A) CH₃Cl

B) C₂H₅Br

C) C₆H₅Cl

D) CH₃CH₂I

✅ Answer

C) C₆H₅Cl

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11. Chlorobenzene is less reactive than chloroethane because:

A) C–Cl bond is strengthened by resonance

B) It has no carbon

C) It is an ionic compound

D) Chlorine is absent

✅ Answer

A

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12. Greater s-character in sp² carbon makes the C–X bond:

A) Weaker

B) Stronger

C) Ionic only

D) Non-existent

✅ Answer

B) Stronger

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13. Which effect is particularly important in chlorobenzene?

A) Resonance

B) Nuclear fission

C) Hydrogen bonding only

D) Metallic bonding

✅ Answer

A

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14. The aromatic ring in chlorobenzene is:

A) Planar

B) Tetrahedral

C) Linear

D) Octahedral

✅ Answer

A

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15. Which statement is correct?

A) Haloarenes have a weaker C–X bond due to resonance

B) Haloarenes have a stronger C–X bond due to resonance

C) Haloarenes do not contain halogen

D) Haloarenes are always ionic

✅ Answer

B

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16. Phenyl carbocation formation in an SN1 pathway is:

A) Highly favourable

B) Highly unfavourable

C) Always spontaneous

D) Required for SN2

✅ Answer

B

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17. Compared with haloalkanes, the C–X bond in haloarenes is generally:

A) Shorter

B) Longer

C) Absent

D) Metallic

✅ Answer

A) Shorter

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18. Which combination explains low reactivity of haloarenes?

A) Resonance + strong C–X bond

B) Weak C–X bond only

C) Ionic bonding only

D) Hydrogen bonding only

✅ Answer

A

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19. Which carbon is involved in nucleophilic substitution of chlorobenzene?

A) sp³ carbon of alkyl chain

B) sp² carbon of aromatic ring

C) sp carbon only

D) No carbon

✅ Answer

B

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20. The best summary for haloarene reactivity is:

A) Haloarenes are more reactive because of resonance

B) Haloarenes are less reactive because resonance strengthens the C–X bond

C) Haloarenes are always ionic

D) Haloarenes cannot contain chlorine

✅ Answer

B

🎯 Quick Revision

Haloarenes: Less reactive towards nucleophilic substitution
Carbon attached to X: sp² hybridised
Main reason: Resonance
C–X bond: Partial double-bond character
C–X bond: Shorter and stronger
SN1: Unfavourable due to unstable phenyl carbocation
SN2: Ordinary backside displacement at aryl carbon is strongly disfavoured
Example: Chlorobenzene (C₆H₅Cl)