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NCERT Exemplar · Q57

Q.Which of the following reaction schemes will yield phenol? (Two or more options may be correct.) Scheme (a): chlorobenzene is fused with NaOH at high temperature and about 300 atm pressure, then treated with H2O/H+. Scheme (b): aniline (C6H5NH2) is treated first with NaNO2/HCl and then with H2O on warming. Scheme (c): benzene is treated with oleum, then with NaOH on heating, then with H+. Scheme (d): chlorobenzene is treated with aqueous NaOH at 298 K and 1 atm, then with HCl.

(i) scheme
(a) - chlorobenzene, fused NaOH at ~300 atm, then H2O/H+
(ii) scheme
(b) - aniline, NaNO2/HCl, then H2O (warming)
(iii) scheme
(c) - benzene, oleum, then NaOH (heating), then H+
(iv) scheme
(d) - chlorobenzene, aqueous NaOH at 298 K/1 atm, then HCl
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Three classic routes to phenol are the Dow process from chlorobenzene (fused NaOH under high temperature/pressure, then acid), diazonium-salt hydrolysis from aniline, and the sulphonation route from benzene (oleum -> alkali fusion -> acid). All of (a), (b) and (c) succeed; (d) fails because the aryl C-Cl bond is inert to aqueous NaOH under mild (298 K, 1 atm) conditions.

Scheme (a) - Dow process (yields phenol)

Chlorobenzene fused with NaOH at high temperature and high pressure gives sodium phenoxide; acidification (H2O/H+) liberates phenol. The forcing conditions are what make nucleophilic aromatic substitution possible.

Scheme (b) - via benzenediazonium salt (yields phenol)

Aniline + NaNO2/HCl (cold) gives benzenediazonium chloride; warming with water hydrolyses it to phenol (with loss of N2). Correct.

Scheme (c) - sulphonation route (yields phenol)

Benzene + oleum -> benzenesulphonic acid; fusion with NaOH (heating) -> sodium phenoxide; acidification (H+) -> phenol. Correct. …

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