Acids: Compare the acidity of carboxylic acids and phenols.

Chapter: Aldehydes, Ketones and Carboxylic Acids

Topic: Acidity of Carboxylic Acids and Phenols

Carboxylic Acids vs Phenols – Acidity

Both carboxylic acids and phenols contain an acidic hydrogen attached to oxygen. However, carboxylic acids are considerably stronger acids than phenols.

R–COOH ⇌ R–COO⁻ + H⁺

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

Acidity:

Carboxylic acids >> Phenols

The strength of an acid depends largely on the stability of its conjugate base.

Stronger acid
↓
More stable conjugate base

When a carboxylic acid loses H⁺, it forms a carboxylate ion. Its negative charge is stabilised very effectively by resonance.

Phenol forms a phenoxide ion, which is comparatively less stabilised.

More stable carboxylate ion → Stronger acidity of carboxylic acids

The carboxylate ion has two equivalent resonance structures.

R–C(=O)–O⁻
↔
R–C(–O⁻)=O

The negative charge is delocalised equally over the two oxygen atoms. Therefore, the carboxylate ion is highly stable.

Negative charge distributed over two equivalent oxygen atoms
→ High stability
→ High acidity of carboxylic acid

Phenol loses H⁺ to form the phenoxide ion. The negative charge can be delocalised into the benzene ring.

C₆H₅OH → C₆H₅O⁻ + H⁺

Although resonance occurs, some resonance structures place the negative charge on carbon atoms of the benzene ring.

These structures are less favourable because oxygen is more electronegative than carbon and stabilises negative charge better.

Therefore, phenoxide ion is less effectively stabilised than the carboxylate ion.

Carboxylate Ion

Negative charge
↓
Shared between TWO oxygen atoms

Both resonance structures are equivalent and highly stable.

Phenoxide Ion

Negative charge
↓
Delocalised over oxygen and carbon atoms

The resonance structures are not equivalent, and some place negative charge on carbon.

Carboxylate ion is more stable than phenoxide ion.
Therefore:
Carboxylic acid > Phenol in acidity

Electron-withdrawing groups increase acidity by stabilising the negative charge of the conjugate base.

–I group
↓
Stabilises conjugate base
↓
Acidity increases

Electron-releasing groups such as alkyl groups generally decrease the acidity of carboxylic acids by destabilising the carboxylate ion.

Electron-withdrawing group → Acidity ↑
Electron-releasing group → Acidity ↓

Property	Carboxylic Acid	Phenol
Acidic group	–COOH	–OH attached to benzene
Conjugate base	Carboxylate ion	Phenoxide ion
Resonance stabilisation	Very strong	Comparatively less effective
Negative charge	Delocalised over two O atoms	Delocalised into ring
Conjugate-base stability	Higher	Lower
Acidity	Higher	Lower

The acid strength is inversely related to pKa.

Lower pKa → Stronger acid
Higher pKa → Weaker acid

Typical approximate values in water are:

Carboxylic acid: pKa ≈ 4–5
Phenol: pKa ≈ 10

Carboxylic acids have much lower pKa values → therefore they are much stronger acids than phenols.

Q. Which is more acidic: carboxylic acid or phenol? Give reason.

Answer:

Carboxylic acids are more acidic than phenols because the carboxylate ion formed after loss of H⁺ is highly stabilised by resonance, with the negative charge delocalised over two oxygen atoms.

Q. Compare the acidity of carboxylic acids and phenols.

Answer:

1. Carboxylic acids are stronger acids than phenols.
2. Carboxylic acids form carboxylate ions in which the negative charge is equally delocalised over two oxygen atoms.
3. The phenoxide ion is less effectively stabilised because some resonance structures place the negative charge on carbon atoms.

Carboxylic acid > Phenol in acidity

Q. Explain why carboxylic acids are stronger acids than phenols.

Answer:

Acid strength depends on the stability of the conjugate base formed after removal of H⁺.

RCOOH → RCOO⁻ + H⁺

The carboxylate ion is stabilised by two equivalent resonance structures. The negative charge is distributed over two oxygen atoms.

RCOO⁻ ↔ RCOO⁻

In contrast, the phenoxide ion has less effective resonance stabilisation because the negative charge is delocalised into the benzene ring and may reside on carbon atoms.

More stable conjugate base → Stronger acid
Carboxylic acid > Phenol

Q. Compare the acidity of carboxylic acids and phenols on the basis of resonance and inductive effects.

1. Ionisation

RCOOH ⇌ RCOO⁻ + H⁺

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

2. Resonance

The carboxylate ion has two equivalent resonance structures, with the negative charge shared between two oxygen atoms.

Phenoxide ion also shows resonance, but some contributing structures place the negative charge on carbon atoms.

3. Conjugate-Base Stability

The carboxylate ion is more stable than the phenoxide ion. Therefore, carboxylic acids lose H⁺ more readily.

4. Inductive Effect

Electron-withdrawing groups increase acidity by stabilising the conjugate base, whereas electron-releasing groups decrease acidity.

Overall:

Carboxylic acids are much stronger acids than phenols.

Q. Give a detailed comparison of the acidity of carboxylic acids and phenols. Explain the role of resonance and inductive effects.

1. Acid Ionisation

RCOOH ⇌ RCOO⁻ + H⁺

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

2. Importance of Conjugate Base

A stronger acid forms a more stable conjugate base. Therefore, the relative stability of carboxylate and phenoxide ions determines their acidity.

3. Carboxylate Ion

The carboxylate ion is stabilised by resonance. The negative charge is equally distributed between two oxygen atoms.

R–C(=O)–O⁻
↔
R–C(–O⁻)=O

4. Phenoxide Ion

Phenoxide ion is resonance stabilised, but the delocalisation extends into the benzene ring. Some contributing structures place negative charge on carbon atoms, which is less favourable than placing it on oxygen.

5. Inductive Effect

Electron-withdrawing groups increase acidity by stabilising the negative charge, while electron-releasing groups decrease acidity.

6. Final Comparison

Carboxylate ion is more stable than phenoxide ion.

Therefore:

Carboxylic acids >> Phenols

in acidity.

1. Which is more acidic?

A) Phenol

B) Carboxylic acid

C) Alcohol

D) Ether

✅ Answer

B) Carboxylic acid

---

2. The conjugate base of a carboxylic acid is:

A) Carbanion

B) Carboxylate ion

C) Phenoxide ion

D) Alkoxide ion

✅ Answer

B) Carboxylate ion

---

3. The conjugate base of phenol is:

A) Phenoxide ion

B) Carboxylate ion

C) Alkoxide ion

D) Hydroxide ion

✅ Answer

A) Phenoxide ion

---

4. Acid strength mainly depends on:

A) Stability of conjugate base

B) Molecular colour

C) Boiling point only

D) Density only

✅ Answer

A

---

5. Carboxylate ion is stabilised by:

A) Resonance

B) Only hydrogen bonding

C) Polymerisation

D) Oxidation

✅ Answer

A) Resonance

---

6. Negative charge in carboxylate ion is delocalised over:

A) One carbon atom

B) Two oxygen atoms

C) Three carbon atoms

D) Hydrogen atoms

✅ Answer

B) Two oxygen atoms

---

7. Which conjugate base is more stable?

A) Phenoxide ion

B) Carboxylate ion

C) Alkoxide ion

D) Carbanion

✅ Answer

B) Carboxylate ion

---

8. Greater stability of conjugate base means:

A) Stronger acid

B) Weaker acid

C) No acid

D) Stronger base only

✅ Answer

A) Stronger acid

---

9. Phenoxide ion is less stable mainly because:

A) Some resonance structures place negative charge on carbon

B) It has no resonance

C) It contains no oxygen

D) It is neutral

✅ Answer

A

---

10. Electron-withdrawing groups generally:

A) Increase acidity

B) Decrease acidity

C) Destroy acidity

D) Have no effect

✅ Answer

A) Increase acidity

---

11. Electron-releasing groups generally:

A) Increase acidity

B) Decrease acidity

C) Always form acids

D) Have no effect

✅ Answer

B) Decrease acidity

---

12. The –COOH group is characteristic of:

A) Alcohols

B) Carboxylic acids

C) Ethers

D) Amines

✅ Answer

B

---

13. Phenol contains:

A) –OH directly attached to benzene

B) –COOH

C) –CHO

D) –CO–

✅ Answer

A

---

14. Which ion has two equivalent major resonance forms?

A) Carboxylate ion

B) Phenoxide ion only

C) Alkoxide ion

D) Hydroxide ion

✅ Answer

A) Carboxylate ion

---

15. Approximate pKa of a simple carboxylic acid is:

A) 4–5

B) 10

C) 15

D) 25

✅ Answer

A) 4–5

---

16. Approximate pKa of phenol is:

A) 2

B) 5

C) 10

D) 20

✅ Answer

C) 10

---

17. Lower pKa indicates:

A) Stronger acid

B) Weaker acid

C) Stronger base only

D) Neutral compound

✅ Answer

A) Stronger acid

---

18. Which conjugate base has negative charge mainly stabilised on oxygen atoms?

A) Carboxylate ion

B) Carbanion

C) Phenyl anion

D) Alkyl anion

✅ Answer

A) Carboxylate ion

---

19. Correct acidity order is:

A) Phenol > Carboxylic acid

B) Carboxylic acid > Phenol

C) Alcohol > Carboxylic acid

D) Ether > Phenol

✅ Answer

B

---

20. The best explanation for greater acidity of carboxylic acids is:

A) More stable carboxylate ion

B) Less stable conjugate base

C) Absence of oxygen

D) No resonance

✅ Answer

A) More stable carboxylate ion

🎯 Quick Revision

Carboxylic acid: R–COOH
Phenol: C₆H₅OH
Carboxylic acid conjugate base: R–COO⁻
Phenol conjugate base: C₆H₅O⁻
Carboxylate ion: Negative charge delocalised over two oxygen atoms
Phenoxide ion: Negative charge delocalised into benzene ring
Conjugate-base stability: Carboxylate > Phenoxide
Acidity: Carboxylic acids >> Phenols
pKa: Carboxylic acids ≈ 4–5; Phenol ≈ 10

Aldehydes/Ketones: Why do aldehydes undergo Nucleophilic Addition more readily than ketones

Chapter: Aldehydes, Ketones and Carboxylic Acids

Topic: Nucleophilic Addition – Reactivity of Aldehydes vs Ketones

Why are Aldehydes More Reactive than Ketones?

Aldehydes undergo nucleophilic addition reactions more readily than ketones because of two major factors:

1. Electronic Effect
+
2. Steric Effect

Aldehyde → Less steric hindrance + more electrophilic carbonyl carbon
↓
More readily undergoes nucleophilic addition

The carbonyl group contains a polar C=O bond because oxygen is more electronegative than carbon.

Cδ⁺ = Oδ⁻

Therefore, the carbonyl carbon carries a partial positive charge and acts as an electrophilic centre.

A nucleophile attacks this electron-deficient carbon atom.

Nu⁻ → Cδ⁺=Oδ⁻
↓
Nucleophilic Addition

More electrophilic carbonyl carbon → Faster nucleophilic attack

Aldehyde

R–CHO

An aldehyde has one alkyl group and one hydrogen attached to the carbonyl carbon.

Ketone

R–CO–R′

A ketone has two alkyl groups attached to the carbonyl carbon.

Alkyl groups show a +I (electron-releasing) effect. They push electron density towards the carbonyl carbon.

More alkyl groups
↓
Greater +I effect
↓
Carbonyl carbon becomes less δ⁺
↓
Nucleophilic attack becomes difficult

Ketones have greater +I effect than aldehydes → Ketones are less reactive towards nucleophilic addition.

Aldehyde

R–CHO
1 alkyl group + 1 H

Hydrogen is very small. Therefore, there is comparatively less crowding around the carbonyl carbon.

Ketone

R–CO–R′
2 alkyl groups

The two alkyl groups occupy more space around the carbonyl carbon. Therefore, the nucleophile faces greater steric hindrance while approaching the carbonyl carbon.

Aldehyde → Less steric hindrance
Ketone → Greater steric hindrance

R₂C=O + Nu⁻
↓
Nu⁻ attacks carbonyl carbon
↓
Tetrahedral intermediate
↓
R₂C(O⁻)(Nu)

The nucleophile attacks the electrophilic carbonyl carbon. The π bond of C=O breaks and the electron pair moves towards oxygen.

C=O π bond breaks → Nucleophile attaches to carbon → Tetrahedral product/intermediate forms

Factor	Aldehydes	Ketones
General structure	R–CHO	R–CO–R′
Alkyl groups attached to C=O	One	Two
Steric hindrance	Lower	Higher
+I effect	Lower	Higher
Electrophilicity of carbonyl C	Higher	Lower
Nucleophilic addition	More readily	Less readily

For simple aliphatic carbonyl compounds, the general order of reactivity towards nucleophilic addition is:

Formaldehyde
>
Other Aldehydes
>
Ketones

Formaldehyde is particularly reactive because its carbonyl carbon is attached to two hydrogen atoms, giving minimum steric hindrance and minimum electron-releasing effect.

HCHO > RCHO > RCOR′

Exam Point:
Aldehydes are more reactive than ketones towards nucleophilic addition because the carbonyl carbon in aldehydes is both less sterically hindered and more positively polarised/electrophilic.

Q. Why do aldehydes undergo nucleophilic addition more readily than ketones?

Answer:

Aldehydes have only one alkyl group attached to the carbonyl carbon, whereas ketones have two. Hence aldehydes have less steric hindrance and a smaller +I effect. Their carbonyl carbon is therefore more electrophilic and is attacked more readily by nucleophiles.

Q. Give two reasons why aldehydes are more reactive than ketones towards nucleophilic addition.

Answer:

1. Steric effect: Aldehydes contain one alkyl group and one small hydrogen atom, so the carbonyl carbon is less sterically hindered.
2. Electronic effect: Ketones have two electron-releasing alkyl groups, which reduce the positive character of the carbonyl carbon.
3. Hence, the carbonyl carbon of aldehydes is more accessible and more electrophilic.

Q. Explain the greater reactivity of aldehydes towards nucleophilic addition reactions on the basis of steric and electronic effects.

Answer:

The carbonyl carbon is electrophilic because the C=O bond is polar. A nucleophile therefore attacks the carbonyl carbon.

In aldehydes, the carbonyl carbon is attached to one alkyl group and one hydrogen atom. In ketones, it is attached to two alkyl groups.

The two alkyl groups in ketones cause greater steric hindrance and also exert a stronger +I effect. This decreases the positive character of the carbonyl carbon.

Less steric hindrance + Greater electrophilicity
→ Aldehydes undergo nucleophilic addition more readily.

Q. Explain in detail why aldehydes are more reactive than ketones towards nucleophilic addition reactions.

1. Polar Nature of C=O

Cδ⁺ = Oδ⁻

The carbonyl carbon is electron deficient and is therefore attacked by nucleophiles.

2. Steric Effect

Aldehydes contain one alkyl group and one hydrogen atom. Ketones contain two alkyl groups.

Aldehyde: R–CHO
↓
Less steric hindrance

Ketone: R–CO–R′
↓
Greater steric hindrance

3. Electronic Effect

Alkyl groups have a +I effect. Therefore, ketones have a greater electron-releasing effect and reduce the electrophilicity of the carbonyl carbon.

4. Result

Aldehydes:
More electrophilic + Less steric hindrance

↓
Greater reactivity towards nucleophilic addition

Q. Discuss the factors responsible for the greater reactivity of aldehydes than ketones towards nucleophilic addition reactions.

1. Carbonyl Group is Electrophilic

Because oxygen is more electronegative than carbon, the carbonyl bond is polar.

Cδ⁺ = Oδ⁻

The carbonyl carbon is therefore the site of nucleophilic attack.

2. Structural Difference

Aldehyde: R–CHO
Ketone: R–CO–R′

An aldehyde has one alkyl group and one hydrogen, while a ketone has two alkyl groups.

3. Steric Effect

The hydrogen atom in aldehyde is very small. Consequently, the nucleophile can approach the carbonyl carbon more easily. Two alkyl groups in a ketone create greater steric hindrance.

4. Electronic Effect

Alkyl groups exert a +I effect and release electron density towards the carbonyl carbon. Ketones have two alkyl groups and hence a greater +I effect than aldehydes.

Thus, the carbonyl carbon of a ketone has a smaller partial positive charge and is less electrophilic.

5. Overall Reactivity

HCHO > RCHO > RCOR′

Therefore, aldehydes undergo nucleophilic addition more readily than ketones because they have:

(i) less steric hindrance
(ii) greater electrophilicity of the carbonyl carbon.

1. Aldehydes generally undergo nucleophilic addition:

A) Less readily than ketones

B) More readily than ketones

C) At the same rate always

D) Never

✅ Answer

B) More readily than ketones

---

2. The carbonyl carbon is:

A) Nucleophilic

B) Electrophilic

C) Neutral only

D) Radical

✅ Answer

B) Electrophilic

---

3. The carbonyl bond is polar because:

A) C and O have different electronegativities

B) Both atoms are identical

C) Carbon is more electronegative than oxygen

D) There is no polarity

✅ Answer

A

---

4. General structure of an aldehyde is:

A) R–CO–R′

B) R–CHO

C) R–COOH

D) R–O–R′

✅ Answer

B) R–CHO

---

5. General structure of a ketone is:

A) R–CHO

B) R–CO–R′

C) R–OH

D) R–COOH

✅ Answer

B) R–CO–R′

---

6. An aldehyde contains how many alkyl groups attached to carbonyl carbon?

A) Two

B) One

C) Three

D) None always

✅ Answer

B) One

---

7. A ketone contains:

A) One alkyl group

B) Two alkyl groups

C) No carbon

D) One hydrogen only

✅ Answer

B) Two alkyl groups

---

8. Alkyl groups generally exhibit:

A) +I effect

B) –I effect

C) No electronic effect

D) Only resonance

✅ Answer

A) +I effect

---

9. Greater +I effect makes the carbonyl carbon:

A) More electrophilic

B) Less electrophilic

C) Positively charged completely

D) A radical

✅ Answer

B) Less electrophilic

---

10. Which has greater steric hindrance at the carbonyl carbon?

A) Aldehyde

B) Ketone

C) Formaldehyde

D) None

✅ Answer

B) Ketone

---

11. The small atom attached to carbonyl carbon in an aldehyde is:

A) Hydrogen

B) Oxygen

C) Nitrogen

D) Chlorine

✅ Answer

A) Hydrogen

---

12. Nucleophile attacks the:

A) Carbonyl carbon

B) Carbonyl oxygen only

C) Hydrogen only

D) Alkyl hydrogen

✅ Answer

A) Carbonyl carbon

---

13. Which is generally the most reactive towards nucleophilic addition?

A) Formaldehyde

B) Acetone

C) 2-butanone

D) Highly substituted ketone

✅ Answer

A) Formaldehyde

---

14. Correct general reactivity order is:

A) Ketone > Aldehyde > Formaldehyde

B) Formaldehyde > Aldehyde > Ketone

C) Ketone > Formaldehyde > Aldehyde

D) All equal

✅ Answer

B

---

15. Which factor favours nucleophilic attack on aldehydes?

A) Greater steric hindrance

B) Less steric hindrance

C) Greater +I effect

D) Electron-rich carbonyl carbon

✅ Answer

B) Less steric hindrance

---

16. Ketones have lower reactivity because:

A) They have two alkyl groups

B) They have no carbonyl group

C) They are always ionic

D) Oxygen is absent

✅ Answer

A

---

17. Nucleophilic addition involves attack on an:

A) Electrophilic centre

B) Nucleophilic centre

C) Neutral hydrogen

D) Alkyl radical

✅ Answer

A) Electrophilic centre

---

18. Which carbonyl compound has minimum steric hindrance?

A) Formaldehyde

B) Acetone

C) 2-butanone

D) Highly branched ketone

✅ Answer

A) Formaldehyde

---

19. In ketones, two alkyl groups:

A) Increase +I effect and steric hindrance

B) Decrease both always

C) Remove carbonyl group

D) Make the compound ionic

✅ Answer

A

---

20. The two principal reasons for greater aldehyde reactivity are:

A) Less steric hindrance and greater electrophilicity

B) Greater steric hindrance and +I effect

C) Resonance only

D) Hydrogen bonding only

✅ Answer

A

🎯 Quick Revision

Aldehyde: R–CHO
Ketone: R–CO–R′
Carbonyl carbon: Electrophilic
Nucleophile: Attacks carbonyl carbon
Aldehyde: One alkyl group + one H
Ketone: Two alkyl groups
Steric hindrance: Aldehyde < Ketone
+I effect: Aldehyde < Ketone
Reactivity: HCHO > RCHO > RCOR′
Conclusion: Aldehydes undergo nucleophilic addition more readily than ketones.

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

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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

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8. With tertiary halides, alkoxide ions generally favour:

A) E2 elimination

B) SN2 strongly

C) Addition

D) Reduction

✅ Answer

A) E2 elimination

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

A) Alkene

B) Ether

C) Alcohol only

D) Aldehyde

✅ Answer

A) Alkene

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10. SN2 reaction involves:

A) Backside attack

B) Carbocation only

C) Free radical

D) Electrophilic attack only

✅ Answer

A

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11. Sodium ethoxide + bromoethane gives:

A) Ethoxyethane

B) Ethene only

C) Ethanal

D) Ethanoic acid

✅ Answer

A) Ethoxyethane

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12. The leaving group in an alkyl bromide is:

A) Br⁻

B) Na⁺

C) O⁻

D) H⁺

✅ Answer

A) Br⁻

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13. Williamson synthesis can prepare:

A) Symmetrical ethers

B) Unsymmetrical ethers

C) Both

D) Neither

✅ Answer

C) Both

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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

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15. Secondary alkyl halides in Williamson synthesis may undergo:

A) Only substitution

B) Competing elimination

C) Only oxidation

D) No reaction

✅ Answer

B

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16. Which is most suitable for Williamson synthesis?

A) CH₃I

B) (CH₃)₃CBr

C) Highly hindered tertiary bromide

D) Aryl halide

✅ Answer

A) CH₃I

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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

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18. Williamson synthesis is essentially a:

A) Nucleophilic substitution

B) Electrophilic addition

C) Free-radical polymerisation

D) Oxidation

✅ Answer

A

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19. Tertiary halide + strong alkoxide generally produces:

A) Ether as major product

B) Alkene by elimination

C) Aldehyde

D) Acid

✅ Answer

B

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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