Question

Explain by giving four examples that the alkoxy group (-OR) activates the aromatic ring towards electrophilic substitution.
OR
Describe two methods of preparation of phenol. Give two reactions that demonstrate the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.

Hint:

The oxygen of -OR donates its lone pair into the ring (+R effect), so it is an activating ortho/para director; for the OR-alternative, phenol's acidity comes from resonance stabilisation of the phenoxide ion, absent in ethoxide.

Answer 1

Option 1: Alkoxy (-OR) group activates the ring

Step 1: Nature of the group. An alkoxy group such as methoxy (-OCH₃) has an oxygen atom carrying lone pairs and joined directly to the benzene ring. Through its lone pair the oxygen pushes electron density into the ring by resonance (+R / +M effect), which easily outweighs its weak -I effect. So -OR is an activating group and an ortho / para director.

Step 2: Why it activates. The oxygen lone pair delocalises into the ring and builds up extra negative charge chiefly at the ortho and para carbons. This raises the ring's electron density, so the ring attracts an electrophile \( E^{+} \) more strongly than plain benzene and reacts faster. The arenium-ion intermediate formed at o / p positions gets an extra resonance structure in which oxygen bears the positive charge (an oxonium form), lowering the activation energy.

Step 3: Four examples of electrophilic substitution on anisole (C₆H₅OCH₃):
(i) Halogenation: C₆H₅OCH₃ + Br₂ \(\xrightarrow{CH_3COOH}\) o- and p-bromoanisole (para major) + HBr. No catalyst is needed, unlike benzene.
(ii) Nitration: C₆H₅OCH₃ + HNO₃ \(\xrightarrow{conc.\,H_2SO_4}\) o- and p-nitroanisole + H₂O.
(iii) Sulphonation: C₆H₅OCH₃ + conc. H₂SO₄ \(\rightarrow\) o- and p-methoxybenzenesulphonic acid + H₂O.
(iv) Friedel-Crafts acylation: C₆H₅OCH₃ + CH₃COCl \(\xrightarrow{anhyd.\,AlCl_3}\) p-methoxyacetophenone + HCl.

Step 4: Conclusion. In each reaction substitution is faster than in benzene and occurs mainly at the ortho and para positions, which proves that the alkoxy group activates the aromatic ring and directs the electrophile to o / p.

Option 2: Phenol preparation, acidity and comparison with ethanol

Step 1: Preparation (Method 1) - from cumene. Cumene (isopropylbenzene) is oxidised by air and then treated with dilute acid to give phenol and acetone:
C₆H₅CH(CH₃)₂ \(\xrightarrow{O_2}\) C₆H₅C(CH₃)₂OOH \(\xrightarrow{H^{+}}\) C₆H₅OH + (CH₃)₂CO.

Step 2: Preparation (Method 2) - from chlorobenzene (Dow process). Chlorobenzene is heated with aqueous NaOH at 623 K and 320 atm to give sodium phenoxide, which on acidification gives phenol:
C₆H₅Cl + NaOH \(\xrightarrow{623\,K,\,320\,atm}\) C₆H₅ONa \(\xrightarrow{H^{+}}\) C₆H₅OH.

Step 3: Acidic nature (Reaction 1) - with NaOH. Phenol dissolves in aqueous NaOH forming a salt, showing it is acidic:
C₆H₅OH + NaOH \(\rightarrow\) C₆H₅ONa + H₂O.

Step 4: Acidic nature (Reaction 2) - with sodium metal. Phenol reacts with active metals liberating hydrogen gas, confirming the acidic O-H:
2C₆H₅OH + 2Na \(\rightarrow\) 2C₆H₅ONa + H₂↑.

Step 5: Phenol vs ethanol. Phenol is much more acidic than ethanol. When phenol loses H⁺, the resulting phenoxide ion has its negative charge spread over the ring by resonance (onto the ortho and para carbons), so it is well stabilised. When ethanol loses H⁺, ethoxide has no resonance and the electron-donating alkyl group (+I effect) pushes charge onto oxygen, intensifying it and destabilising the ion. A more stable conjugate base means a stronger acid, so \[\boxed{\text{acidity: phenol } (pK_a \approx 10) \gg \text{ ethanol } (pK_a \approx 16)}\]