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 activation (rate-and-orientation view)

Step 1: Recognise the substituent. \(-OR\) (alkoxy, e.g. methoxy \(-OCH_3\)) is an electron-releasing group; the decisive point is that the oxygen lone pair conjugates with the \(\pi\) system of the ring.

Step 2: Consequence for reactivity and orientation. Because oxygen feeds electron density into the ring, the ortho and para carbons become electron rich, so an incoming electrophile is attracted there. The transition state for o / p attack is the most stabilised one, hence anisole undergoes electrophilic aromatic substitution more readily than benzene and yields chiefly o / p products.

Step 3: Four illustrations using anisole:
(a) Bromination without a Lewis-acid catalyst: anisole with Br₂ in acetic acid gives mainly p-bromoanisole and some o-bromoanisole; benzene by contrast needs FeBr₃, so this alone shows activation.
(b) Nitration: HNO₃ with H₂SO₄ converts anisole to p- and o-nitroanisole.
(c) Friedel-Crafts acetylation: CH₃COCl with AlCl₃ acylates the para position to 4-methoxyacetophenone.
(d) Sulphonation: hot conc. H₂SO₄ gives 4-methoxybenzenesulphonic acid (with some ortho).

Step 4: Reading the evidence. The ease and the o / p selectivity seen in all four reactions are a direct demonstration that the \(-OR\) group activates the benzene ring.

Option 2: Phenol - two more preparations, acidity, and the ethanol comparison

Step 1: Preparation from a diazonium salt. Benzenediazonium chloride is warmed with water; it hydrolyses to phenol and releases nitrogen:
C₆H₅N₂Cl + H₂O \(\xrightarrow{warm}\) C₆H₅OH + N₂ + HCl.

Step 2: Preparation from benzenesulphonic acid. Benzene is sulphonated to benzenesulphonic acid; its sodium salt is fused with NaOH to give sodium phenoxide, which on acidification gives phenol:
C₆H₅SO₃H \(\xrightarrow{NaOH,\,fuse}\) C₆H₅ONa \(\xrightarrow{H^{+}}\) C₆H₅OH.

Step 3: Acidity shown with a base. Phenol dissolves in NaOH solution forming sodium phenoxide but does NOT react with NaHCO₃, so it is a weak acid (weaker than carbonic acid) yet clearly acidic:
C₆H₅OH + NaOH \(\rightarrow\) C₆H₅ONa + H₂O.

Step 4: Acidity shown with a metal. With active metals phenol evolves hydrogen: 2C₆H₅OH + 2Na \(\rightarrow\) 2C₆H₅ONa + H₂.

Step 5: Compare the conjugate bases. In phenoxide the oxygen sits on an sp² ring carbon and its negative charge is dispersed by resonance across the ring, giving a stable, weak base. In ethoxide the oxygen sits on an sp³ alkyl carbon, there is no resonance, and the alkyl group donates electrons, so the charge stays concentrated on oxygen making ethoxide a strong, unstable base. Since the more stable conjugate base corresponds to the stronger acid, \[\boxed{\text{phenol is more acidic than ethanol}}\]