Alcohols, Phenols and Ethers, on the exam screen.

Cumene gives phenol by air oxidation ($\mathrm{O}_{2}$) to cumene hydroperoxide then acid hydrolysis ($\mathrm{H}_{2}\mathrm{O}$/$\mathrm{H}^{+}$), so A-II. Acetic acid is converted to ethanol by esterification ($\mathrm{CH}_{3}\mathrm{OH}$/$\mathrm{H}^{+}$) followed by reduction ($\mathrm{H}_{2}$/catalyst), so B-III. Propan-1-ol is dehydrated (conc. $\mathrm{H}_{2}\mathrm{SO}_{4}$, heat) to propene, which on Markovnikov hydration ($\mathrm{H}^{+}$/$\mathrm{H}_{2}\mathrm{O}$) gives propan-2-ol, so C-IV. Benzene is converted to phenol by sulphonation (oleum), NaOH fusion, then acidification, so D-I. Hence A-II, B-III, C-IV, D-I.

In the phthalein dye test a phenol is heated with phthalic anhydride and concentrated H$_{2}$SO$_{4}$ to form phenolphthalein, which turns pink on addition of NaOH; this confirms a phenolic -OH group. Aldehydes are detected by Tollens'/Fehling's tests, carboxylic acids by effervescence with NaHCO$_{3}$, and alcohols by the ceric ammonium nitrate or Lucas tests.

HI protonates anisole at oxygen, then $\mathrm{I}^{-}$ attacks the methyl group by substitution nucleophilic bimolecular (S$_{\text{N}}$2) cleavage. The products are phenol and methyl iodide (CH$_{3}$I); the aryl C-O bond is not cleaved by S$_{\text{N}}$2. Chapter: Alcohols, Phenols and Ethers.

Why the other options are wrong:

- (1) Iodobenzene and $\mathrm{CH}_{3}\mathrm{OH}$: this needs $\mathrm{I}^{-}$ to attack the ring carbon, but S$_{\text{N}}$2 cannot occur at an sp$^{2}$ aryl carbon, so the aryl C-O bond stays intact.
- (2) Phenol and $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{I}$: anisole is a methyl ether, so the alkyl iodide formed is $\mathrm{CH}_{3}\mathrm{I}$; ethyl iodide would come from phenetole.
- (3) Iodobenzene and $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$: the aryl C-O bond is not broken, and anisole has no ethyl group to give ethanol.