Alcohols, Phenols and EthersNEET Chemistry · Class 12 · NCERT Chapter 7

Both compounds are isomers with the same molecular mass (46 g/mol). The difference is due to intermolecular hydrogen bonding. Ethanol (CH3CH2OH) has an O-H bond — the hydrogen is partially positive and the oxygen of a neighbouring molecule is partially negative. This O-H...O hydrogen bond is a relatively strong intermolecular force (20-40 kJ/mol) that must be broken when ethanol evaporates, requiring much more energy and therefore a higher boiling point. Dimethyl ether (CH3-O-CH3) has no O-H bond. It cannot form hydrogen bonds with itself — only weak van der Waals forces hold its molecules together. So it boils at -24°C, much lower than ethanol. This principle applies to all alcohol vs ether comparisons.

When phenol (C6H5OH) donates a proton, it forms the phenoxide ion (C6H5O-). The negative charge on oxygen is delocalised into the benzene ring through resonance — five resonance structures spread the charge, stabilising the phenoxide ion. This stabilisation makes it easier to form the phenoxide (more acidic, pKa ~10). Ethanol forms an ethoxide ion (C2H5O-) where the negative charge stays on oxygen with no resonance stabilisation — it is less stable, so ethanol is less acidic (pKa ~16). Electron-withdrawing groups (like -NO2) at ortho or para positions on the phenol ring further delocalise the negative charge of the phenoxide ion through resonance onto the EWG oxygens. This extra stabilisation makes phenol even more acidic. For example, 2,4,6-trinitrophenol (picric acid) is as strong as mineral acids!

Williamson ether synthesis makes an ether by reacting a sodium alkoxide (R-O-Na) with an alkyl halide (R'X): R-ONa + R'X → R-O-R' + NaX. The alkoxide is a strong base/nucleophile. The alkyl halide must be primary (or methyl) because the reaction proceeds by SN2 — the alkoxide attacks the back face of the alkyl carbon. If R'X is secondary or tertiary, the alkoxide will instead act as a base and cause E2 elimination to give an alkene, not an ether. So for making CH3-O-C(CH3)3 (methyl tert-butyl ether, MTBE): use (CH3)3C-ONa (tert-butoxide) + CH3I (methyl iodide), NOT CH3ONa + (CH3)3CBr (which would give isobutylene).

The Lucas test uses a mixture of anhydrous zinc chloride (ZnCl2) and concentrated hydrochloric acid. The alcohol reacts to form an alkyl chloride (insoluble in the reagent), which appears as turbidity (cloudiness). The test works by SN1: ZnCl2 coordinates with the -OH of the alcohol, making it a better leaving group (as Zn-OH complex). Then Cl- attacks. Tertiary alcohols form stable 3° carbocations immediately — turbidity appears in seconds. Secondary alcohols form 2° carbocations more slowly — turbidity appears in about 5 minutes. Primary alcohols form very unstable 1° carbocations — no turbidity at room temperature (reaction requires heating). The test is only useful for alcohols up to about C5 because larger alcohols are insoluble in water and cannot be tested this way.

Both reactions start with phenol (phenoxide) and introduce a group onto the ring, but they are completely different. Kolbe synthesis (Kolbe-Schmitt): sodium phenoxide + CO2 gas (125°C, 5 atm) → sodium salicylate (sodium 2-hydroxybenzoate) → salicylic acid on acidification. The product is a carboxylic acid at the ortho position. Salicylic acid is the starting material for aspirin. Reimer-Tiemann reaction: phenol + chloroform (CHCl3) + NaOH (warm) → 2-hydroxybenzaldehyde (salicylaldehyde) as major product, plus some 4-hydroxybenzaldehyde. The reactive intermediate is dichlorocarbene (:CCl2), generated from CHCl3 + NaOH. It reacts with the activated phenoxide ring (highly nucleophilic at ortho/para) to introduce a -CHO group. The product is an aldehyde at the ortho position.

When methyl tert-butyl ether (CH3-O-C(CH3)3) reacts with HI, the tert-butyl C-O bond breaks, not the methyl C-O bond. The mechanism is: (1) O is protonated by HI to give an oxonium ion. (2) I- now acts as nucleophile or the C-O bond breaks. The deciding factor is which carbocation is more stable: CH3+ (methyl, very unstable) vs (CH3)3C+ (tertiary, very stable). The system chooses to break the tert-butyl C-O bond and form the stable 3° carbocation (SN1). I- then attacks this carbocation to give (CH3)3CI (tert-butyl iodide), and CH3OH (methanol) is released. General rule for ether cleavage: I- attacks the more substituted carbon if it can form a stable carbocation (SN1), OR attacks the less hindered carbon if the substrate is primary (SN2).

You can expect 2-4 questions from this chapter each year in NEET. The most frequently tested topics are: (1) Acidity comparison — ordering phenol, substituted phenols, alcohols, water. Know that EWG increases acidity and EDG decreases it. (2) Oxidation of alcohols — what product forms with PCC vs KMnO4, and 1° vs 2° vs 3° alcohols. (3) Named reactions — Kolbe, Reimer-Tiemann, Williamson synthesis. Know products and conditions clearly. (4) Lucas test — which alcohol reacts when. (5) Ether cleavage by HI — which C-O bond breaks and why. Practise these specific question types and you will cover the bulk of what NEET tests.