How to Score Full Marks in the Board Exam
A complete bank of board-style questions with model answers for Alcohols, Phenols and Ethers. In the exam: write the IUPAC name carefully, draw clear structures with correct ortho/para positions, write reagents and conditions above the arrow, and give the NCERT-canonical reason (resonance stabilisation of phenoxide for acidity; hydrogen bonding for boiling point/solubility; carbocation stability for dehydration order). For “how will you convert” questions, show every reagent over each arrow. For distinguishing tests, name the reagent and the observation for each compound.
1-Mark Questions (Definitions & Direct)
Q1. Write the IUPAC name of (CH)C-OH. Answer: 2-methylpropan-2-ol (tert-butyl alcohol).
Q2. What is denatured alcohol? Answer: Ethanol made unfit for drinking by adding small amounts of poisonous substances such as methanol, pyridine and a dye, so that industrial alcohol can be used without excise duty.
Q3. Name the reagent used in the Lucas test. Answer: Lucas reagent — concentrated HCl with anhydrous ZnCl.
Q4. What is the product of the cumene process besides phenol? Answer: Acetone (propan-2-one).
Q5. Why is phenol more acidic than ethanol? (one line) Answer: The phenoxide ion is stabilised by resonance (delocalisation into the ring), whereas the ethoxide ion is not, so phenol loses its proton more readily.
Q6. Name the reaction: phenol + CHCl + NaOH → salicylaldehyde. Answer: Reimer-Tiemann reaction.
2-Mark Questions
Q7. Why are alcohols of low molecular mass soluble in water while higher alcohols are not? Answer: Lower alcohols form hydrogen bonds with water through their -OH group, making them miscible. As the alkyl (hydrocarbon) part grows, the non-polar portion dominates and disrupts H-bonding with water, so higher alcohols become only slightly soluble.
Q8. Explain why the boiling point of ethanol is higher than that of dimethyl ether though both have the formula CHO. Answer: Ethanol has an O-H group and forms intermolecular hydrogen bonds, which need extra energy to break, raising its boiling point. Dimethyl ether has no O-H and cannot hydrogen-bond between its molecules, so it boils much lower.
Q9. How will you distinguish between phenol and ethanol? Answer: Add neutral FeCl solution: phenol gives a characteristic violet colouration, while ethanol gives no colour. (Phenol also reacts with NaOH; ethanol does not.)
Q10. Why does phenol undergo electrophilic substitution more readily than benzene? Answer: The -OH group donates electron density into the ring through resonance (the oxygen lone pair), increasing electron density at the ortho and para positions and activating the ring toward electrophiles.
2-Mark Questions (continued)
Q11. Give the structures of the products formed when anisole is treated with HI. Answer: Phenol (CHOH) and methyl iodide (CHI). The aryl-oxygen bond does not cleave, so the methyl-oxygen bond breaks.
Q12. Why is the Williamson synthesis preferred for unsymmetrical ethers, and what type of halide is used? Answer: It cleanly couples a chosen sodium alkoxide with a chosen alkyl halide, allowing mixed ethers. A primary alkyl halide is used because the reaction is SN2; secondary/tertiary halides would give alkenes by elimination.
Q13. Arrange in increasing order of acidity: phenol, o-cresol, o-nitrophenol. Justify. Answer: o-cresol < phenol < o-nitrophenol. The -CH group (o-cresol) is electron-donating and lowers acidity; -NO (o-nitrophenol) is electron-withdrawing and raises acidity by stabilising the phenoxide.
Q14. What happens when phenol is treated with bromine water? Answer: A white precipitate of 2,4,6-tribromophenol forms, because the strongly activated ring undergoes trisubstitution at the two ortho and the para positions.
3-Mark Questions (Conversions)
Q15. How will you convert (i) propene to propan-2-ol, (ii) propene to propan-1-ol? Answer: (i) Treat propene with HO in the presence of dilute HSO (acid-catalysed Markovnikov hydration) → propan-2-ol. (ii) Treat propene with diborane (BH) followed by HO/OH- (hydroboration-oxidation, anti-Markovnikov) → propan-1-ol.
Q16. How will you convert (i) phenol to salicylic acid, (ii) phenol to picric acid? Answer: (i) Heat sodium phenoxide with CO under pressure (Kolbe's reaction) then acidify → salicylic acid. (ii) Treat phenol with concentrated HNO (nitration) → 2,4,6-trinitrophenol (picric acid).
Q17. How will you convert ethanol to diethyl ether and to ethene? State the conditions. Answer: Heat ethanol with conc. HSO at 413 K → diethyl ether (intermolecular dehydration). Heat ethanol with conc. HSO at 443 K → ethene (intramolecular dehydration). Temperature decides the product.
3-Mark Questions (Reasoning & Mechanism)
Q18. Write the mechanism of acid-catalysed hydration of ethene to ethanol. Answer: Step 1 — ethene's π electrons attack a proton (from HO+) to form a carbocation (ethyl cation). Step 2 — a water molecule attacks the carbocation to give a protonated alcohol (oxonium ion). Step 3 — loss of a proton gives ethanol. The acid is regenerated, so it acts as a catalyst.
Q19. Explain why p-nitrophenol is more acidic than phenol with resonance. Answer: In the p-nitrophenoxide ion, the negative charge is delocalised not only into the ring but also onto the oxygen atoms of the -NO group (an additional resonance structure). This extra stabilisation of the conjugate base makes p-nitrophenol a stronger acid than phenol.
Q20. Give reasons: (a) ortho-nitrophenol is more volatile than para-nitrophenol; (b) phenol does not undergo protonation easily. Answer: (a) o-Nitrophenol forms an intramolecular hydrogen bond (chelation), so its molecules do not associate and it is steam-volatile; p-nitrophenol forms intermolecular H-bonds and is less volatile. (b) In phenol the oxygen lone pair is delocalised into the ring, so it is less available for protonation.
3-Mark Questions (Tests & Distinctions)
Q21. How will you distinguish between primary, secondary and tertiary alcohols by the Lucas test? Answer: Add Lucas reagent (conc. HCl + anhydrous ZnCl) at room temperature. A tertiary alcohol gives immediate turbidity, a secondary alcohol gives turbidity in about 5 minutes, and a primary alcohol gives no turbidity at room temperature (only on heating). The order follows carbocation stability 3° > 2° > 1°.
Q22. Distinguish between (i) phenol and benzoic acid, (ii) methanol and ethanol. Answer: (i) Benzoic acid gives effervescence (CO) with NaHCO while phenol does not; also phenol gives a violet colour with neutral FeCl. (ii) Ethanol gives a positive iodoform test (yellow precipitate of CHI with I/NaOH) while methanol does not.
Q23. What is the Victor Meyer test? What colours do 1°, 2° and 3° alcohols give? Answer: A test that converts the alcohol to a nitroalkane and then to a nitrolic acid treated with NaOH. A primary alcohol gives a blood-red colour, a secondary alcohol gives a blue colour, and a tertiary alcohol gives no colour.
5-Mark Questions (Long Answer)
Q24. (a) Why is phenol more acidic than ethanol? (b) How is the acidity of phenol affected by an electron-withdrawing group and an electron-donating group? Give one example each. Answer: (a) When phenol loses its proton, the resulting phenoxide ion is stabilised by resonance — the negative charge is delocalised onto the ortho and para carbons of the ring. The ethoxide ion has no such delocalisation and is destabilised by the +I effect of the ethyl group, so phenol is much more acidic. (b) An electron-withdrawing group such as -NO stabilises the phenoxide and increases acidity (p-nitrophenol > phenol). An electron-donating group such as -CH destabilises the phenoxide and decreases acidity (p-cresol < phenol).
Q25. Describe the industrial preparation of phenol by the cumene process with equations. Answer: Benzene is alkylated with propene in the presence of acid to give cumene (isopropylbenzene). Cumene is oxidised by air (O) to cumene hydroperoxide. Treatment of cumene hydroperoxide with dilute acid cleaves it to phenol and acetone. Thus: benzene → cumene → cumene hydroperoxide → phenol + acetone. The co-product acetone makes the process economical.
5-Mark Questions (continued)
Q26. (a) Write the reactions of the Williamson synthesis of tert-butyl methyl ether choosing the correct reagents. (b) Why is the other combination unsuitable? Answer: (a) Use sodium tert-butoxide and methyl iodide: (CH)C-O-Na+ + CHI → (CH)C-O-CH + NaI. The methyl iodide (a primary/methyl halide) undergoes SN2. (b) The combination sodium methoxide + tert-butyl bromide is unsuitable because tert-butyl bromide is a 3° halide; with the strongly basic alkoxide it undergoes E2 elimination to 2-methylpropene instead of substitution, so no ether forms.
Q27. How are the following carried out? (i) Kolbe's reaction (ii) Reimer-Tiemann reaction (iii) reaction of phenol with conc. HNO. Answer: (i) Sodium phenoxide is heated with CO under pressure (about 400 K, 4-7 atm) and the product acidified to give salicylic acid. (ii) Phenol is treated with chloroform and aqueous NaOH; the resulting intermediate is hydrolysed (acidified) to give salicylaldehyde, with the -CHO group mainly at the ortho position. (iii) Phenol with concentrated HNO undergoes nitration at the two ortho and the para positions to give 2,4,6-trinitrophenol (picric acid).
Quick-Fire Board Favourites
Q28. Why is the C-O-H bond angle in alcohols slightly less than the tetrahedral angle? Answer: Because of the repulsion between the two lone pairs on oxygen, which compresses the bond angle (about 108.5°) slightly below the ideal tetrahedral value.
Q29. Why does tertiary butyl alcohol dehydrate most easily among isomeric butanols? Answer: Its dehydration proceeds through a stable tertiary carbocation, which forms most easily; hence 3° alcohols dehydrate fastest (order 3° > 2° > 1°).
Q30. Give one chemical test to distinguish ethanol from phenol. Answer: Add neutral FeCl: phenol gives a violet colour; ethanol gives no colour.
Q31. Why can't a mixed ether be prepared satisfactorily by dehydration of alcohols? Answer: Dehydration of a mixture of two alcohols gives a mixture of three ethers (two symmetrical and one mixed), so the desired mixed ether cannot be obtained pure; Williamson synthesis is used instead.
Q32. What is power alcohol? Answer: A blend of ethanol with petrol used as a motor fuel.
More Board Favourites
Q33. Write the product when propan-1-ol is oxidised by acidified KCrO. Answer: Propanoic acid (it is first oxidised to propanal and then to the carboxylic acid).
Q34. Why is the -OH group in phenol ortho/para-directing? Answer: The lone pair on oxygen is delocalised into the ring, raising electron density specifically at the ortho and para positions, so electrophiles attack there.
Q35. Give the IUPAC name and one use of CHOH; why is it dangerous? Answer: IUPAC name: methanol. It is used as a solvent and to make formaldehyde. It is dangerous because it is oxidised in the body to toxic products, causing blindness and even death if ingested.
Q36. How is salicylic acid prepared from phenol? Name the reaction. Answer: By Kolbe's reaction: sodium phenoxide is heated with CO under pressure and the product acidified to give salicylic acid (2-hydroxybenzoic acid).