The Hook: Anesthesia and the Sleepy Molecule
Imagine you’re in a hospital scene from 3 Idiots (2009) - remember the delivery scene? Before modern anesthesia, doctors used diethyl ether to put patients to sleep during surgery. One whiff of “ether” and you’re out cold!
But here’s the chemistry puzzle: Ethers have the same functional group as alcohols (oxygen!), but they behave completely differently:
- Alcohols are reactive (oxidation, substitution, elimination)
- Ethers are super stable (barely react with anything!)
The JEE Question: Why is ether so unreactive? How do you make it? And when it DOES react (HI cleavage), which C-O bond breaks first? Get this right, and you’ll crack the 3-4 marks JEE loves to test on ethers!
The Core Concept: The Oxygen Bridge
General Structure
$$\boxed{\text{R-O-R' (oxygen connecting two carbon groups)}}$$Types:
Symmetrical (Simple) Ether: R-O-R (same groups on both sides)
- Example: CH₃-O-CH₃ (dimethyl ether)
- Example: C₂H₅-O-C₂H₅ (diethyl ether)
Unsymmetrical (Mixed) Ether: R-O-R’ (different groups)
- Example: CH₃-O-C₂H₅ (ethyl methyl ether)
- Example: C₆H₅-O-CH₃ (methoxybenzene, anisole)
In simple terms: Ethers are like a bridge where oxygen connects two carbon “islands.” Unlike alcohols (R-OH), there’s no H on oxygen, which makes ethers much less reactive!
Nomenclature
IUPAC System:
- Smaller group → alkoxy
- Larger group → alkane
Examples:
| Structure | IUPAC Name | Common Name |
|---|---|---|
| CH₃-O-CH₃ | Methoxymethane | Dimethyl ether |
| C₂H₅-O-C₂H₅ | Ethoxyethane | Diethyl ether |
| CH₃-O-C₂H₅ | Methoxyethane | Ethyl methyl ether |
| C₆H₅-O-CH₃ | Methoxybenzene | Anisole |
| C₆H₅-O-C₆H₅ | Diphenyl ether | Phenyl ether |
Memory Trick: “Smaller becomes Suffix (-oxy), Larger stays as main chain name!”
Interactive Demo: Visualize Ether Structure
Explore the molecular structure of ethers and understand the oxygen bridge.
Physical Properties: Why Ethers are Different
Comparison with Alcohols and Alkanes
| Property | Alkane (R-R') | Ether (R-O-R') | Alcohol (R-OH) |
|---|---|---|---|
| Boiling Point | Lowest | Intermediate | Highest |
| H-bonding | No | No | Yes (O-H···O) |
| Polarity | Non-polar | Polar | Polar |
| Water solubility | Insoluble | Slightly soluble | Soluble (small) |
Boiling Point Order (for similar molecular mass):
$$\boxed{\text{Alcohol} > \text{Ether} > \text{Alkane}}$$Example:
- CH₃-CH₂-OH (ethanol): BP = 78°C
- CH₃-O-CH₃ (dimethyl ether): BP = -24°C
- CH₃-CH₂-CH₃ (propane): BP = -42°C
Why?
- Alcohols: Strong intermolecular H-bonding (O-H···O-H) → high BP
- Ethers: Only dipole-dipole interactions (C-O dipole, but no H-bonding) → medium BP
- Alkanes: Only weak van der Waals forces → low BP
Memory Trick: “Alcohols have Amazing H-bonds - highest BP! Ethers are Exactly in the middle, alkanes are Abysmal (lowest)!”
Water Solubility
Small ethers (like dimethyl ether, diethyl ether) are slightly soluble in water.
Why?
- Ether oxygen can accept H-bonds from water (O···H-O-H)
- But ether cannot donate H-bonds (no H on oxygen)
- Result: Partial solubility
Pattern:
- Dimethyl ether (CH₃-O-CH₃): Soluble
- Diethyl ether (C₂H₅-O-C₂H₅): Slightly soluble
- Di-n-butyl ether: Insoluble (too large, hydrocarbon part dominates)
Preparation of Ethers: The Synthesis Toolkit
Method 1: Williamson Synthesis (THE MOST IMPORTANT!)
General Reaction:
R-O⁻Na⁺ + R'-X → R-O-R' + NaX
(Sodium alkoxide) (Alkyl halide) (Ether)
Full Sequence:
Step 1: Prepare sodium alkoxide
2 R-OH + 2 Na → 2 R-O⁻Na⁺ + H₂↑
Step 2: Williamson synthesis
R-O⁻Na⁺ + R'-X → R-O-R' + NaX
Mechanism (SN2):
δ- δ+
R-O⁻ + R'-X → R-O-R' + X⁻
(Nucleophile) (Electrophile)
Backside attack (SN2 mechanism)
Example:
Step 1:
2 CH₃-OH + 2 Na → 2 CH₃-O⁻Na⁺ + H₂↑
Step 2:
CH₃-O⁻Na⁺ + CH₃-I → CH₃-O-CH₃ + NaI
(Dimethyl ether)
Key Points:
- SN2 mechanism → works best with primary alkyl halides (R’-X should be 1°)
- Alkoxide acts as nucleophile, alkyl halide as electrophile
- For unsymmetrical ethers: Choose reagents carefully!
Rule: For best yield, use the less hindered alkyl halide!
Example: To make C₂H₅-O-C(CH₃)₃ (ethyl tert-butyl ether)
Option 1: C₂H₅-O⁻Na⁺ + (CH₃)₃C-Br
- Problem: 3° alkyl halide → E2 elimination (not SN2!) ✗
Option 2: (CH₃)₃C-O⁻Na⁺ + C₂H₅-Br
- 1° alkyl halide → SN2 works! ✓
Correct Answer: Use Option 2 (1° alkyl halide)
Memory Trick: “Williamson Wants 1° alkyl halide - otherwise elimination Wins!”
For Aromatic Ethers (like Anisole):
C₆H₅-O⁻Na⁺ + CH₃-I → C₆H₅-O-CH₃ + NaI
(Sodium phenoxide) (Anisole/Methoxybenzene)
Cannot do reverse: CH₃-O⁻Na⁺ + C₆H₅-X doesn’t work (aromatic halides unreactive in SN2!)
Method 2: Dehydration of Alcohols (for Symmetrical Ethers)
Reaction:
conc. H₂SO₄
2 R-OH ──────────────→ R-O-R + H₂O
140°C
Example:
conc. H₂SO₄
2 CH₃CH₂-OH ──────────→ CH₃CH₂-O-CH₂CH₃ + H₂O
140°C (Diethyl ether)
Mechanism (SN2):
Step 1: Protonation
CH₃CH₂-OH + H₂SO₄ → CH₃CH₂-OH₂⁺
Step 2: SN2 attack by another alcohol
CH₃CH₂-OH + CH₃CH₂-OH₂⁺ → CH₃CH₂-O⁺H-CH₂CH₃ + H₂O
|
H
Step 3: Deprotonation
CH₃CH₂-O⁺H-CH₂CH₃ -H⁺→ CH₃CH₂-O-CH₂CH₃
|
H
Limitation: Only makes symmetrical ethers (R-O-R)!
Temperature Matters:
| Temperature | Product | Reaction Type |
|---|---|---|
| 140°C | Ether (R-O-R) | Intermolecular dehydration |
| 170°C | Alkene (R-CH=CH₂) | Intramolecular dehydration (E1) |
Memory Trick: “140° → ether (lower temp), 170° → alkene (higher temp) - Elimination needs Extra heat!”
Common Mistake: Mixing up temperatures for ether vs alkene formation
Remember:
- 140°C + conc. H₂SO₄ → ether (two molecules of alcohol react)
- 170°C + conc. H₂SO₄ → alkene (one molecule of alcohol loses water)
JEE Pattern: They give you alcohol + H₂SO₄ and ask for product at different temperatures!
Reactions of Ethers: The Unreactive Champion
Why Ethers are Unreactive
Reasons:
- No acidic H: Unlike alcohols (R-OH), ethers (R-O-R) have no H on oxygen
- Strong C-O bonds: C-O single bonds are strong, difficult to break
- Low polarity: Less polar than alcohols → weaker interactions
Result: Ethers are excellent solvents for reactions (don’t interfere!)
Common Uses:
- Diethyl ether: Anesthetic, organic solvent
- THF (tetrahydrofuran): Solvent for Grignard reactions
- DME (dimethoxyethane): Solvent for organometallic chemistry
Reaction 1: Cleavage by HX (The ONLY Important Reaction!)
General Reaction:
Excess HI or HBr
R-O-R' ────────────────→ R-I + R'-I + H₂O
Heat
Reactivity Order of Acids:
$$\boxed{\text{HI} > \text{HBr} >> \text{HCl (very slow)}}$$Why? I⁻ is best nucleophile (large, polarizable)
Case 1: Symmetrical Ether
Example:
Excess HI
C₂H₅-O-C₂H₅ ────────→ 2 C₂H₅-I + H₂O
Heat
Mechanism:
Step 1: Protonation of ether oxygen
C₂H₅-O-C₂H₅ + HI → C₂H₅-O⁺H-C₂H₅ + I⁻
|
H
Step 2: SN2 attack by I⁻ (cleavage of C-O bond)
H
|
C₂H₅-O⁺-C₂H₅ + I⁻ → C₂H₅-OH + C₂H₅-I
|
H
Step 3: Conversion of alcohol to alkyl iodide
C₂H₅-OH + HI → C₂H₅-I + H₂O
Net Result: Both C-O bonds break → 2 alkyl halides
Case 2: Unsymmetrical Ether (Alkyl-Alkyl)
Example:
Excess HI
CH₃-O-C₂H₅ ────────→ CH₃-I + C₂H₅-I + H₂O
Question: Which C-O bond breaks first?
Answer: The bond that forms the more stable carbocation (SN1 pathway) OR the less hindered side (SN2 pathway)
For small alkyl groups (1°): SN2 mechanism dominates
- I⁻ attacks less hindered carbon
- Both bonds eventually break → both alkyl iodides form
Case 3: Aryl-Alkyl Ether (like Anisole)
Example:
Excess HI
C₆H₅-O-CH₃ ────────→ C₆H₅-OH + CH₃-I
Heat (Phenol)
Mechanism:
Step 1: Protonation
C₆H₅-O-CH₃ + HI → C₆H₅-O⁺H-CH₃ + I⁻
|
H
Step 2: Nucleophilic attack (SN2)
C₆H₅-O⁺H-CH₃ + I⁻ → C₆H₅-OH + CH₃-I
|
H
Key Point: Only the C-O bond attached to alkyl group breaks!
Why?
- C₆H₅-O bond has partial double-bond character (resonance with benzene ring)
- Aryl C-O bond is stronger than alkyl C-O bond
- I⁻ attacks the weaker, more accessible alkyl carbon
Product: Phenol + alkyl halide (NOT aryl halide!)
Memory Trick: “Aryl-Alkyl ethers → Alkyl halide + phenol (Aryl C-O bond never breaks!)”
Question Type: “C₆H₅-O-C₂H₅ + excess HI → ?”
Strategy:
- Identify ether type (aryl-alkyl, alkyl-alkyl, or symmetrical)
- For aryl-alkyl: Alkyl C-O breaks → phenol + alkyl halide
- For alkyl-alkyl (unsymmetrical): Both C-O bonds break → two alkyl halides
- For symmetrical: Both identical alkyl halides
Answer for above: C₆H₅-OH + C₂H₅-I
Common Trap: Students write C₆H₅-I (wrong! aryl C-O doesn’t break)
Reaction 2: Reaction with Acetyl Chloride/Acetic Anhydride
Only for phenolic ethers!
Example (Anisole):
CH₃COCl or (CH₃CO)₂O
C₆H₅-O-CH₃ ─────────────────────→
AlCl₃ (Lewis acid)
OCH₃ OCH₃
| |
/═\\ /═\\
| | COCH₃ + | |
\═/ \═/
COCH₃
(o-Methoxyacetophenone) (p-Methoxyacetophenone)
This is electrophilic aromatic substitution (Friedel-Crafts acylation)!
Key Point: -OCH₃ group is activating and ortho/para directing
Reaction 3: Electrophilic Aromatic Substitution (Aromatic Ethers)
For anisole (C₆H₅-O-CH₃):
Nitration:
HNO₃/H₂SO₄
C₆H₅-O-CH₃ ──────────→ o-NO₂-C₆H₄-O-CH₃ + p-NO₂-C₆H₄-O-CH₃
(ortho) (para, major)
Halogenation:
Br₂/FeBr₃
C₆H₅-O-CH₃ ──────────→ o-Br-C₆H₄-O-CH₃ + p-Br-C₆H₄-O-CH₃
Activating Effect: -OCH₃ (methoxy) activates the benzene ring via +M effect
- Electron donation through resonance
- ortho/para directing
Reactivity: Anisole is more reactive than benzene (like phenol!)
Special Ethers: Crown Ethers and Epoxides
Crown Ethers (Brief)
Structure: Cyclic polyethers with multiple -O- groups
Example: 18-Crown-6 (18 atoms in ring, 6 oxygens)
Use:
- Complexes with metal cations (K⁺, Na⁺)
- Makes metal salts soluble in organic solvents
- Used in phase-transfer catalysis
JEE Relevance: Low (mostly mentioned in organic chemistry applications)
Epoxides (Oxiranes)
Structure: Three-membered cyclic ether
General Structure:
O
/ \
R-CH--CH₂
Most Common: Ethylene oxide (oxirane)
O
/ \
H₂C--CH₂
Properties:
- Highly reactive (ring strain!)
- Much more reactive than regular ethers
- Undergo ring-opening reactions easily
Preparation:
mCPBA or RCO₃H
R-CH=CH₂ ──────────────→ R-CH--CH₂
\ /
O
Reactions (Ring Opening):
(a) Acid-catalyzed:
O H⁺/H₂O OH
/ \ ──────────→ |
R-CH--CH₂ R-CH-CH₂-OH
(b) Base-catalyzed:
O NaOH/H₂O O⁻
/ \ ──────────→ |
R-CH--CH₂ R-CH-CH₂-OH
JEE Focus: Epoxide reactions are covered in detail under Aldehydes & Ketones or advanced organic chemistry.
Memory Tricks & Patterns
The Master Mnemonic: “ETHERS WATCH CALMLY”
Ether boiling point: Alcohol > Ether > Alkane
Temperature: 140°C → ether, 170°C → alkene
HI > HBr » HCl (reactivity for cleavage)
Ether oxygen can accept H-bonds (not donate)
R-O-R structure (no H on oxygen)
SN2 mechanism (Williamson synthesis)
Williamson uses 1° alkyl halide (best yield)
Aryl-alkyl ether → phenol + alkyl halide (aryl C-O doesn’t break)
Two products from symmetrical ether cleavage
Cleaving requires excess HI/HBr + heat
Heat with H₂SO₄: temp determines product (ether vs alkene)
Crown ethers complex with metal ions
Anisole (C₆H₅-O-CH₃) undergoes electrophilic substitution (ortho/para)
Less reactive than alcohols (no -OH)
Methoxy (-OCH₃) is activating and ortho/para directing
Low reactivity → excellent solvent
Yields from Williamson: use less hindered alkyl halide
Williamson Synthesis Decision Tree
Question: Make R-O-R’ ether
Step 1: Identify if symmetrical or unsymmetrical
- Symmetrical (R-O-R): Use dehydration (2 R-OH + H₂SO₄, 140°C) OR Williamson
- Unsymmetrical (R-O-R’): Use Williamson only
Step 2 (for Williamson): Choose reagents
- Check which groups are primary, secondary, or tertiary
- Use less hindered group as alkyl halide (R-X)
- Use more hindered/aromatic group as alkoxide (R’-O⁻Na⁺)
Example: Make (CH₃)₃C-O-CH₃
- Option 1: (CH₃)₃C-O⁻Na⁺ + CH₃-I → ✓ (CH₃-I is 1°, good for SN2)
- Option 2: CH₃-O⁻Na⁺ + (CH₃)₃C-Br → ✗ ((CH₃)₃C-Br is 3°, gives elimination!)
Choose Option 1!
Common Mistakes to Avoid
Wrong: Thinking 140°C gives alkene, 170°C gives ether Right:
- 140°C + conc. H₂SO₄ → ether (intermolecular dehydration)
- 170°C + conc. H₂SO₄ → alkene (intramolecular dehydration)
Memory: Lower temp (140°C) → ether (two molecules combine); Higher temp (170°C) → alkene (one molecule eliminates)
Wrong: Predicting C₆H₅-I as product from C₆H₅-O-CH₃ + HI Right: Product is C₆H₅-OH + CH₃-I (aryl C-O bond doesn’t break!)
Reason: Aryl C-O bond has partial double-bond character (resonance), much stronger than alkyl C-O bond
JEE Trap: This is tested EVERY year in some form!
Wrong: Using 3° alkyl halide in Williamson synthesis Right: 3° alkyl halides undergo E2 elimination, not SN2!
Example:
CH₃-O⁻Na⁺ + (CH₃)₃C-Br → CH₂=C(CH₃)₂ + CH₃-OH + NaBr
(Alkene, NOT ether!)
Strategy: Always use 1° alkyl halide for best results; 2° can work but gives mixed products
Wrong: Expecting ethers to undergo oxidation/reduction like alcohols Right: Ethers are very unreactive - only HI/HBr cleavage works!
Why?
- No acidic H (can’t be oxidized like R-OH)
- Strong C-O bonds (don’t break easily)
- No electrophilic center (unlike carbonyl groups)
Exception: Epoxides (three-membered ring ethers) are highly reactive due to ring strain!
Wrong: Thinking ethers have higher BP than alcohols (both have oxygen!) Right: Alcohol > Ether > Alkane (for similar molecular mass)
Reason:
- Alcohols: Strong H-bonding (O-H···O-H)
- Ethers: Only dipole-dipole (no H on oxygen to donate!)
- Alkanes: Only van der Waals
JEE Pattern: They give compounds with similar molecular mass and ask to rank BP!
Practice Problems
Level 1: Foundation (NCERT Style)
Question: Write IUPAC names for: (a) CH₃-O-C₂H₅ (b) C₂H₅-O-C₂H₅ (c) C₆H₅-O-CH₃
Solution:
(a) CH₃-O-C₂H₅
- Smaller group: CH₃- → methoxy
- Larger group: C₂H₅- → ethane
- IUPAC Name: Methoxyethane
- Common name: Ethyl methyl ether
(b) C₂H₅-O-C₂H₅
- Both groups same: C₂H₅- → ethoxy + ethane
- IUPAC Name: Ethoxyethane
- Common name: Diethyl ether
(c) C₆H₅-O-CH₃
- Smaller group: CH₃- → methoxy
- Larger group: C₆H₅- (benzene) → benzene
- IUPAC Name: Methoxybenzene
- Common name: Anisole
Question: Why is the boiling point of ethanol (78°C) higher than that of dimethyl ether (-24°C), despite having the same molecular formula C₂H₆O?
Solution:
Reason: Intermolecular hydrogen bonding
Ethanol (CH₃-CH₂-OH):
- Has -OH group
- Forms strong intermolecular H-bonds: O-H···O-H
- Requires more energy to break these bonds → higher boiling point (78°C)
Dimethyl ether (CH₃-O-CH₃):
- Has no H on oxygen
- Cannot form H-bonds (only weaker dipole-dipole interactions)
- Less energy needed to vaporize → lower boiling point (-24°C)
Key: Presence of H-bonding dramatically increases BP!
Question: Complete the reaction:
2 CH₃CH₂-OH + conc. H₂SO₄ (140°C) → ?
Solution:
2 CH₃CH₂-OH + conc. H₂SO₄ (140°C) → CH₃CH₂-O-CH₂CH₃ + H₂O
(Ethoxyethane/Diethyl ether)
Reaction Type: Intermolecular dehydration of alcohol
Note: If temperature were 170°C, product would be alkene (CH₂=CH₂)!
Level 2: JEE Main Style
Question: Which of the following reagent combinations will give the best yield for preparing CH₃-O-C(CH₃)₃?
(A) CH₃-O⁻Na⁺ + (CH₃)₃C-Br (B) (CH₃)₃C-O⁻Na⁺ + CH₃-Br (C) Both give equal yield (D) Neither will work
Solution: (B)
Analysis:
Option A: CH₃-O⁻Na⁺ + (CH₃)₃C-Br
- (CH₃)₃C-Br is a 3° alkyl halide
- 3° halides favor E2 elimination over SN2!
- Expected product: (CH₃)₂C=CH₂ (alkene) + CH₃-OH
- Poor yield of ether ✗
Option B: (CH₃)₃C-O⁻Na⁺ + CH₃-Br
- CH₃-Br is a 1° alkyl halide (actually methyl halide, even better!)
- 1° halides excellent for SN2 mechanism
- Expected product: CH₃-O-C(CH₃)₃ (ether)
- Good yield ✓
Williamson Strategy: Use less hindered halide (1° > 2° » 3°)
Answer: (B)
Question: Identify the products A and B:
C₆H₅-O-CH₂-CH₃ + excess HI Heat→ A + B
(A) A = C₆H₅-I, B = CH₃CH₂-OH (B) A = C₆H₅-OH, B = CH₃CH₂-I (C) A = C₆H₅-I, B = CH₃CH₂-I (D) A = C₆H₅-OH, B = CH₃CH₂-OH
Solution: (B)
Analysis: This is an aryl-alkyl ether (phenyl ethyl ether).
Rule: In aryl-alkyl ethers, only the alkyl C-O bond breaks (aryl C-O bond has partial double-bond character, too strong!).
Mechanism:
Step 1: Protonation
C₆H₅-O-CH₂CH₃ + HI → C₆H₅-O⁺H-CH₂CH₃
|
H
Step 2: SN2 cleavage (alkyl C-O breaks)
C₆H₅-O⁺H-CH₂CH₃ + I⁻ → C₆H₅-OH + CH₃CH₂-I
|
H
Products:
- A = C₆H₅-OH (phenol)
- B = CH₃CH₂-I (ethyl iodide)
Answer: (B)
Common Mistake: Predicting C₆H₅-I (aryl C-O bond doesn’t break!)
Question: Arrange the following in increasing order of boiling point: I. CH₃-CH₂-CH₂-CH₃ (Butane) II. CH₃-CH₂-O-CH₃ (Methoxyethane) III. CH₃-CH₂-CH₂-OH (Propan-1-ol)
(A) I < II < III (B) III < II < I (C) II < I < III (D) I < III < II
Solution: (A) I < II < III
Analysis: All three have similar molecular mass (~60 amu).
I. Butane (CH₃-CH₂-CH₂-CH₃):
- Alkane, no polarity
- Only van der Waals forces
- Lowest BP (~0°C)
II. Methoxyethane (CH₃-CH₂-O-CH₃):
- Ether, polar C-O bond
- Dipole-dipole interactions (no H-bonding!)
- Intermediate BP (~8°C)
III. Propan-1-ol (CH₃-CH₂-CH₂-OH):
- Alcohol, has -OH
- Strong intermolecular H-bonding
- Highest BP (~97°C)
Increasing order: I < II < III ✓
Memory: Alkane < Ether < Alcohol (for similar MW)
Level 3: JEE Advanced Style
Question: An ether (A) with molecular formula C₇H₈O reacts with HI to give phenol and methyl iodide. (A) also undergoes nitration to give a mixture of ortho and para products. Identify (A) and write all reactions.
Solution:
Analysis:
- Reacts with HI → phenol + CH₃I → (A) must be aryl-alkyl ether with CH₃- group
- Molecular formula C₇H₈O → C₆H₅-O-CH₃ (6C from benzene + 1C from CH₃ + 1O = C₇H₈O) ✓
- Nitration gives ortho/para products → -OCH₃ is ortho/para directing (activating group) ✓
Compound A = Anisole (Methoxybenzene, C₆H₅-O-CH₃)
Reactions:
(1) Cleavage with HI:
Excess HI
C₆H₅-O-CH₃ ────────→ C₆H₅-OH + CH₃-I
Heat (Phenol) (Methyl iodide)
(2) Nitration:
HNO₃/H₂SO₄
C₆H₅-O-CH₃ ──────────→
OCH₃ OCH₃
| |
/═\\ /═\\
| | NO₂ + | |
\═/ \═/
NO₂
(o-Nitroanisole) (p-Nitroanisole, major)
Explanation:
- -OCH₃ (methoxy) is an electron-donating group (+M effect)
- Activates benzene ring (makes it more reactive than benzene)
- Directs incoming electrophile (NO₂⁺) to ortho/para positions
Question: How would you synthesize the following ethers using Williamson synthesis? (a) (CH₃)₂CH-O-CH₃ (b) C₆H₅-O-C₂H₅
Solution:
(a) (CH₃)₂CH-O-CH₃ (Isopropyl methyl ether)
Analysis:
- Unsymmetrical ether
- One side: (CH₃)₂CH- (2°)
- Other side: CH₃- (methyl)
Options:
Option 1: (CH₃)₂CH-O⁻Na⁺ + CH₃-I
- CH₃-I is methyl halide (best for SN2!)
- This is the BEST route ✓
Option 2: CH₃-O⁻Na⁺ + (CH₃)₂CH-Br
- (CH₃)₂CH-Br is 2° halide
- Will give competition between SN2 (ether) and E2 (alkene)
- Lower yield ✗
Best Synthesis:
Step 1: Prepare sodium isopropoxide
2 (CH₃)₂CH-OH + 2 Na → 2 (CH₃)₂CH-O⁻Na⁺ + H₂↑
Step 2: Williamson synthesis
(CH₃)₂CH-O⁻Na⁺ + CH₃-I → (CH₃)₂CH-O-CH₃ + NaI
(b) C₆H₅-O-C₂H₅ (Phenyl ethyl ether)
Analysis:
- Aryl-alkyl ether
- Aryl halides (C₆H₅-X) don’t undergo SN2!
- Must use phenoxide as nucleophile
Only One Option:
Step 1: Prepare sodium phenoxide
C₆H₅-OH + NaOH → C₆H₅-O⁻Na⁺ + H₂O
Step 2: Williamson synthesis
C₆H₅-O⁻Na⁺ + C₂H₅-Br → C₆H₅-O-C₂H₅ + NaBr
Cannot do reverse: C₂H₅-O⁻Na⁺ + C₆H₅-Br ✗ (aryl halides unreactive!)
Key Point: For aryl-alkyl ethers, aryl group must be the alkoxide (nucleophile)!
Question: Predict the major products and explain the mechanism:
(a) CH₃-O-CH₂-CH₃ + excess HI Heat→ ?
(b) CH₃-O-CH₃ + excess HBr Heat→ ?
Solution:
(a) CH₃-O-CH₂-CH₃ + excess HI:
Products: CH₃-I + C₂H₅-I + H₂O
Mechanism:
Step 1: Protonation
CH₃-O-CH₂CH₃ + HI → CH₃-O⁺H-CH₂CH₃ + I⁻
|
H
Step 2: SN2 cleavage (I⁻ attacks less hindered CH₃)
CH₃-O⁺H-CH₂CH₃ + I⁻ → CH₃-I + C₂H₅-OH
|
H
Step 3: Conversion of alcohol to alkyl iodide
C₂H₅-OH + HI → C₂H₅-I + H₂O
Net: Both C-O bonds break → CH₃-I + C₂H₅-I
(b) CH₃-O-CH₃ + excess HBr:
Products: 2 CH₃-Br + H₂O
Mechanism:
Step 1: Protonation
CH₃-O-CH₃ + HBr → CH₃-O⁺H-CH₃ + Br⁻
|
H
Step 2: SN2 cleavage
CH₃-O⁺H-CH₃ + Br⁻ → CH₃-Br + CH₃-OH
|
H
Step 3: Conversion
CH₃-OH + HBr → CH₃-Br + H₂O
Net: 2 CH₃-Br + H₂O
Note: Symmetrical ether → both C-O bonds break to give same product!
Quick Revision Box
| Situation | Formula/Approach | Key Point |
|---|---|---|
| Make symmetrical ether (R-O-R) | 2 R-OH + H₂SO₄, 140°C | Intermolecular dehydration |
| Make unsymmetrical ether (R-O-R’) | Williamson synthesis | R-O⁻Na⁺ + R’-X (R’-X must be 1°!) |
| Williamson with 3° halide | Elimination, NOT substitution! | Always use 1° halide for best yield |
| Cleave ether | Excess HI or HBr, heat | HI > HBr » HCl |
| Aryl-alkyl ether cleavage | Phenol + alkyl halide | Aryl C-O bond doesn’t break! |
| Symmetrical ether cleavage | Two same alkyl halides | Both C-O bonds break |
| Temperature for ether (dehydration) | 140°C | 170°C gives alkene! |
| Boiling point order | Alcohol > Ether > Alkane | H-bonding vs dipole vs van der Waals |
| Water solubility | Small ethers slightly soluble | Can accept H-bonds, can’t donate |
| Electrophilic substitution (anisole) | ortho/para products | -OCH₃ is activating, ortho/para directing |
Connection to Other Topics
Prerequisites (Review these first):
- Alcohols - dehydration to ethers, comparison of properties
- Organic Principles: SN1 and SN2 Mechanisms - for Williamson synthesis understanding
- Hydrocarbons: Alkyl Halides - reactivity in Williamson synthesis
Related Topics:
- Phenols - product of aryl-alkyl ether cleavage
- Aldehydes & Ketones - epoxide reactions, Grignard in ether solvent
- Carboxylic Acids - esterification comparison
Advanced Applications:
- Crown ethers (complexation chemistry)
- Epoxides (ring-opening reactions, polymer chemistry)
- THF as Grignard solvent (organometallic chemistry)
- Williamson synthesis in natural product synthesis
Teacher’s Summary
1. Structure Determines Reactivity:
- Ethers (R-O-R) have no H on oxygen → very unreactive
- Only significant reaction: HI/HBr cleavage (C-O bond breaking)
- Excellent solvents (don’t interfere with reactions)
2. Williamson Synthesis is King:
- General: R-O⁻Na⁺ + R’-X → R-O-R’ (SN2 mechanism)
- Critical Rule: Use 1° alkyl halide for best yield (2° gives mixed products, 3° gives elimination!)
- For aryl ethers: Aryl group MUST be alkoxide (C₆H₅-O⁻ + R-X), can’t reverse!
3. Dehydration Temperature:
- 140°C + conc. H₂SO₄ → ether (intermolecular: 2 R-OH → R-O-R)
- 170°C + conc. H₂SO₄ → alkene (intramolecular: R-OH → R-CH=CH₂)
4. Ether Cleavage Pattern:
- HI > HBr » HCl (reactivity)
- Aryl-alkyl ether: Only alkyl C-O breaks → phenol + alkyl halide
- Alkyl-alkyl ether: Both C-O bonds break → two alkyl halides
- Symmetrical ether: Two identical alkyl halides
5. Physical Properties:
- Boiling Point: Alcohol > Ether > Alkane (H-bonding > dipole-dipole > van der Waals)
- Solubility: Small ethers slightly soluble (accept H-bonds, don’t donate)
- Polarity: Ethers polar (C-O dipole) but less than alcohols
6. JEE Strategy:
- Master Williamson synthesis reagent selection (1° halide rule!)
- Know aryl-alkyl ether cleavage pattern (phenol + alkyl halide, NOT aryl halide!)
- Remember temperature distinction (140°C vs 170°C)
- Practice boiling point comparisons (alcohol vs ether vs alkane)
- Understand why ethers are unreactive (no acidic H, strong C-O bonds)
“Ethers are the silent partners of organic chemistry - unreactive by nature, but essential as solvents. Their only weakness? The C-O bond breaking under harsh acidic conditions. Master Williamson synthesis and HI cleavage, and you’ve conquered ethers for JEE!”
Previous Topic: ← Phenols - The acidic aromatic cousins Next Topic: Aldehydes & Ketones → - The carbonyl powerhouses!