Q.Predict the products of the following reactions:

Concept: Ether cleavage by HX (SN1/SN2 depending on alkyl group) and nitration of an activated aromatic ring.

(i) $CH_3-CH_2-CH_2-O-CH_3 + HBr \longrightarrow$

The alkoxy group with the smaller alkyl (methyl) leaves as $CH_3Br$ via SN2.

Products: $CH_3Br$ and $CH_3CH_2CH_2OH$.

(ii) $C_6H_5-OC_2H_5 + HBr \longrightarrow$

The aryl–O bond is strong; cleavage occurs at the alkyl–O bond. SN2 gives $C_2H_5Br$ and phenol.

Products: $C_2H_5Br$ and $C_6H_5OH$.

(iii) $C_6H_5-OC_2H_5 \xrightarrow[\text{Conc. } HNO_3]{\text{Conc. } H_2SO_4}$

The –OC$_2$H$_5$ group is strongly activating ortho/para-directing. Nitration occurs at the ortho and para positions.

Products: $o$- and $p$-nitrophenetole ($o$- and $p$–$O_2N$–$C_6H_4$–$OC_2H_5$). …

The Concept: Why Ethers Cleave with HX

Ethers are generally unreactive, but strong acids like HBr, HI, or hot concentrated HX protonate the oxygen, turning the poor leaving group (alkoxide) into an excellent one (alcohol). Once protonated, a halide ion attacks the carbon — but which carbon? That depends on the mechanism:

• SN2 pathway: Favoured when the alkyl group is primary or methyl. The halide attacks the less hindered carbon, giving an alcohol and an alkyl halide.
• SN1 pathway: Favoured when the alkyl group can form a stable carbocation (tertiary, benzylic, allylic). The C–O bond breaks first, then the halide attacks the carbocation.

Step-by-Step Solutions

(i) $CH_3-CH_2-CH_2-O-CH_3 + HBr \longrightarrow$

1. Identify the alkyl groups: One is propyl (primary), the other is methyl (primary). Both are primary — no stable carbocation possible. So the mechanism is SN2.
2. Protonation: The ether oxygen gets protonated by HBr, forming $CH_3CH_2CH_2-\overset{+}{O}(H)-CH_3$.
3. Nucleophilic attack: The bromide ion ($Br^-$) attacks the less hindered carbon. Between propyl and methyl, methyl is less hindered (only one carbon vs three). So $Br^-$ attacks the methyl carbon.
4. Products: Methyl bromide ($CH_3Br$) and propanol ($CH_3CH_2CH_2OH$) form. With a large excess of HBr the alcohol could slowly react further, but the cleavage products of this reaction are the bromide and the alcohol.

Final products: $CH_3Br + CH_3CH_2CH_2OH$.

(ii) $C_6H_5-OC_2H_5 + HBr \longrightarrow$

1. Identify the groups: One is phenyl (aryl), the other is ethyl (primary). The phenyl group is attached to oxygen via an $sp^2$ carbon — the C–O bond in aryl ethers is very strong due to partial double-bond character (resonance with the ring). So the aryl–O bond does not break under normal conditions.
2. Protonation: The oxygen gets protonated.
3. Which bond breaks? The aryl–O bond cannot break (partial double-bond character), and an ethyl carbocation would be far too unstable for SN1. The ethyl carbon is primary, so bromide attacks it directly: the ethyl–O bond cleaves via SN2, giving phenol and ethyl bromide.
4. Products: $C_6H_5OH$ (phenol) and $C_2H_5Br$ (ethyl bromide).

Final products: $C_6H_5OH + C_2H_5Br$.

(iii) $C_6H_5-OC_2H_5 \xrightarrow[\text{Conc. } HNO_3]{\text{Conc. } H_2SO_4}$ …

Method: Acid-Catalysed Cleavage of Ethers (SN1 / SN2 Mechanism)

This is not Williamson Ether Synthesis — it is the reverse reaction: cleavage of ethers by strong acids like HBr, HI, or hot concentrated $H_2SO_4$.

Key Principle

• Ethers are cleaved by strong HX acids (HI > HBr > HCl) via protonation of the oxygen, followed by nucleophilic attack by the halide ion.
• The mechanism (SN1 or SN2) depends on the alkyl group structure:
  • Primary alkyl group → SN2 (backside attack, inversion)
  • Tertiary / benzylic / allylic → SN1 (carbocation intermediate)
• Aryl alkyl ethers cleave at the alkyl–O bond (not the aryl–O bond, which is too strong).

(i) $CH_3-CH_2-CH_2-O-CH_3 + HBr \longrightarrow$

Method: SN2 cleavage (both groups are primary)

Steps:

1. Protonation of ether oxygen by HBr → forms oxonium ion.
2. Nucleophilic attack by Br⁻ on the less hindered carbon (the methyl group, $CH_3$–) via SN2.
3. Products:
   • Methyl bromide ($CH_3Br$)
   • Propan-1-ol ($CH_3CH_2CH_2OH$)

Result:

$CH_3CH_2CH_2OH + CH_3Br$

(ii) $C_6H_5-OC_2H_5 + HBr \longrightarrow$

Method: SN2 cleavage of the alkyl–O bond (aryl–O bond is too strong; an ethyl cation is far too unstable for SN1)

Steps:

1. Protonation of oxygen.
2. Nucleophilic attack by Br⁻ on the ethyl carbon (alkyl side) — proceeds via SN2 because ethyl is primary.
3. Products:
   • Phenol ($C_6H_5OH$)
   • Ethyl bromide ($C_2H_5Br$)

Result:

$C_6H_5OH + C_2H_5Br$

(iii) $C_6H_5-OC_2H_5 \xrightarrow[\text{Conc. } HNO_3]{\text{Conc. } H_2SO_4}$

Method: Electrophilic Aromatic Substitution (Nitration) — not ether cleavage

Steps:

1. Conc. $H_2SO_4$ + Conc. $HNO_3$ generates the nitronium ion ($NO_2^+$).
2. The –OC₂H₅ group is a strong activating and ortho/para-directing group.
3. Nitration occurs on the benzene ring at the ortho and para positions.
4. Products:
   • o-Nitrophenetole and p-Nitrophenetole (mixture) …

🧠 Core Concept: Ether Cleavage by HX

Ethers are generally unreactive, but strong acids like HBr, HI, or hot concentrated H₂SO₄ can cleave the C–O bond. The mechanism is SN1 or SN2, depending on the alkyl group.

Key rule:

• Primary alkyl group → SN2 (less substituted carbon gets attacked by halide)
• Tertiary alkyl group → SN1 (more substituted carbon forms carbocation)
• Aryl group (C₆H₅–) → does not undergo cleavage under normal conditions (resonance stabilisation)

✗ Common Mistake #1: Forgetting that aryl–O bonds do not break

Example reaction:

(ii) $C_6H_5-OC_2H_5 + HBr \longrightarrow$

Wrong answer:

$C_6H_5Br + C_2H_5OH$ (breaking the aryl–O bond)

Why it’s wrong:

The C₆H₅–O bond has partial double-bond character due to resonance. HBr cannot break it under normal conditions.

✓ Correct answer:

$C_6H_5OH + C_2H_5Br$

(Only the ethyl–O bond breaks via SN2)

How to avoid:

• Remember: Aryl–O bonds are inert to HX cleavage.
• The halide always attacks the alkyl side, not the aromatic ring.

✗ Common Mistake #2: Ignoring carbocation stability in mixed ethers

Example reaction:

(iv) $(CH_3)_3C-OC_2H_5 \xrightarrow{HI}$

Wrong answer:

$(CH_3)_3C-OH + C_2H_5I$ (assuming SN2 attack on the ethyl side)

Why it’s wrong:

The tert-butyl group forms a very stable tertiary carbocation, so the reaction follows SN1, not SN2: the C–O bond to the tertiary carbon breaks first, and $I^-$ then attacks that carbocation. The more substituted carbon gets the iodine, while the oxygen leaves with the ethyl group as ethanol.

✓ Correct answer:

$(CH_3)_3C-I + C_2H_5OH$

(tertiary carbocation captured by iodide; ethanol keeps the oxygen)

How to avoid:

• Identify the bulkier alkyl group — if it’s tertiary, expect SN1.
• In SN1, the more substituted carbon gets the halide (via carbocation).
• In SN2, the less substituted carbon gets the halide.

✗ Common Mistake #3: Forgetting that conc. H₂SO₄ + conc. HNO₃ is nitration, not cleavage

Example reaction:

(iii) $C_6H_5-OC_2H_5 \xrightarrow[\text{Conc. } HNO_3]{\text{Conc. } H_2SO_4}$

Wrong answer:

$C_6H_5OH + C_2H_5NO_3$ (thinking it’s an ether cleavage)

Why it’s wrong:

This is a nitrating mixture — it performs electrophilic aromatic substitution, not ether cleavage. The ether bond remains intact.

✓ Correct answer:

The product is a mixture of ortho- and para-nitrophenyl ethyl ether:

$C_6H_4(NO_2)-OC_2H_5$ (with –NO₂ at ortho/para positions)

How to avoid:

• Recognise the reagent: conc. H₂SO₄ + conc. HNO₃ = nitration.
• The ether oxygen is an activating group (ortho/para directing).
• Do not confuse with HX cleavage conditions.

✗ Common Mistake #4: Attacking the wrong carbon in S$_N$2 cleavage

Example reaction:

(i) $CH_3-CH_2-CH_2-O-CH_3 + HBr \longrightarrow$

Wrong answer:

$CH_3-CH_2-CH_2-Br + CH_3OH$ (bromide attacking the propyl side)

Why it’s wrong: …