Class 12 Chemistry Ch 7 Alcohols Phenols Ethers Formula Sheet PDF

Ans. You can download the Alcohols, Phenols and Ethers Class 12 Chemistry Formula Sheet PDF directly from this Collegedunia page. Both the Normal and HD versions are available and free of cost.

Ans. Yes. This page reflects the current 2026-27 syllabus for Class 12 Chemistry. Alcohols, Phenols and Ethers is fully retained in the new edition with no formula cuts; every relation in Sections 7.1 to 7.6 of the NCERT remains examinable.

Ans. The Formula Sheet PDF runs approximately 8 to 9 pages and covers the master reaction table, symbol glossary, acidity / boiling-point reference, quick-fact MCQ cards, and four common numerical pattern templates.

Ans. The phenoxide ion (C₆H₅-O^-) is stabilised by resonance: the negative charge is delocalised onto the ortho and para carbons of the benzene ring, spreading it over multiple atoms. The ethoxide ion has no such delocalisation, so the negative charge stays fully on one oxygen. The resonance-stabilised phenoxide is roughly 10⁶ times more stable than ethoxide, which is why phenol has pK_a 10 and ethanol pK_a 16.

Ans. At 413 K (140 °C), conc. H₂SO₄ catalyses intermolecular dehydration of ethanol to give diethyl ether: 2 C₂H₅-OH → C₂H₅-O-C₂H₅ + H₂O. At 443 K (170 °C), the same acid catalyses intramolecular dehydration to give ethene: C₂H₅-OH → C₂H₄ + H₂O. The temperature is the only variable that flips the outcome.

Ans. Williamson synthesis is the S_N2 reaction of a sodium alkoxide with an alkyl halide: R-O^-Na⁺ + R'-X → R-O-R' + NaX. It is the standard route for unsymmetrical ethers. It fails when R'-X is a tertiary (3°) halide because steric crowding pushes the reaction to E2 elimination, giving an alkene instead. The fix is to choose the 1° halide as R'-X and put the bulkier group on the alkoxide.

Ans. When anisole (C₆H₅-O-CH₃) reacts with HI, the I^- nucleophile attacks the methyl carbon (S_N2 at sp³ C), not the aryl carbon (sp², locked in the aromatic ring). The aryl C-O bond has partial double-bond character from lone-pair donation into the ring, making it too strong to cleave. The products are therefore C₆H₅-OH (phenol) + CH₃-I (methyl iodide).

Ans. The Reimer-Tiemann reaction treats phenol with chloroform (CHCl₃) in the presence of aqueous NaOH to introduce a -CHO group at the ortho position of the ring, giving salicylaldehyde (2-hydroxybenzaldehyde). Mechanism: NaOH deprotonates CHCl₃ to dichlorocarbene (:CCl₂), which attacks the activated phenoxide at the ortho carbon; subsequent hydrolysis of -CCl₂H gives -CHO. The reaction is examinable in both CBSE and JEE Main.

Ans. Sodium phenoxide is heated with CO₂ at 400 K and 4 to 7 atm; the carboxylate intermediate is acidified to give salicylic acid (2-hydroxybenzoic acid). The mechanism involves electrophilic attack of CO₂ on the activated ortho carbon of the phenoxide. The Kolbe reaction is the industrial route to salicylic acid, the precursor of aspirin.

Ans. The cumene process oxidises cumene (isopropylbenzene) with atmospheric O₂ to cumene hydroperoxide, then acidifies to give phenol and acetone (valuable co-product). The Dow process hydrolyses chlorobenzene with NaOH at 623 K and 320 atm to give phenol. Cumene is the dominant industrial route today because acetone offsets the cost.

Ans. Picric acid is prepared by stepwise nitration of phenol: dilute HNO₃ gives ortho/para-nitrophenol; more concentrated HNO₃ gives 2,4-dinitrophenol; final nitration with conc. HNO₃ + H₂SO₄ gives picric acid (pK_a 0.4), stronger than acetic acid. Three -NO₂ groups stabilise the conjugate base by resonance and -I effects.

Ans. Hydroboration-oxidation uses B₂H₆ in THF followed by alkaline H₂O₂ on an alkene. The boron attaches to the less-substituted carbon (anti-Markovnikov), and oxidation replaces it with -OH without rearrangement. The reaction is concerted and syn-additive, so no carbocation forms and no Wagner-Meerwein rearrangement is possible. This makes it CBSE's preferred 3-mark answer when the question demands a rearrangement-free preparation.

Ans. PCC (pyridinium chlorochromate) in dichloromethane is a mild oxidant that stops at R-CHO; KMnO₄ is a strong aqueous oxidant that overshoots to R-COOH because the aldehyde hydrates in water and is oxidised further. For secondary alcohols, both reagents give the ketone (no further oxidation easily). Tertiary alcohols resist both because there is no alpha-H.

Ans. Saytzeff's rule says that in an E1 dehydration, the more-substituted (more stable) alkene is the major product. 2-Methylbutan-2-ol with conc. H₂SO₄ at 443 K gives 2-methylbut-2-ene (trisubstituted) as the major product over 2-methylbut-1-ene (disubstituted). Alkene stability follows hyperconjugation: more alpha-H atoms mean more hyperconjugative stabilisation.

Ans. In water, the phenol partially ionises to phenoxide, which is much more activated than phenol itself; the highly activated ring undergoes triple electrophilic substitution at the 2, 4 and 6 positions to give 2,4,6-tribromophenol (white precipitate) very fast. In CS₂ (a non-polar solvent) at low temperature, ionisation is suppressed and the reaction stops at the mono-bromo stage, giving a mixture of o- and p-bromophenol.

Ans. The Lucas test mixes the alcohol with Lucas reagent (concentrated HCl + anhydrous ZnCl₂) at room temperature. Tertiary alcohols give immediate turbidity because the 3° carbocation forms fast; secondary alcohols give turbidity in 5 to 10 minutes; primary alcohols give no turbidity at room temperature and need heating. The Lucas test is the standard CBSE / JEE Main / NEET distinction question for the chapter.