Previous year questions: Alcohols, Phenols and Ethers

The questions below come from official CBSE board papers and from official Sample and Additional Practice Question Papers. The year label in each heading is the one printed on the source, and the marks are the marks printed for that question or part.

Q1 (CBSE Board 2023, 2 marks)

For the pair phenol and cyclohexanol, answer the following. (a) Why is phenol more acidic than cyclohexanol? (b) Give one chemical test to distinguish between the two.

Answer: (a) The phenoxide ion left behind after the proton leaves has its negative charge delocalised over the ring and the oxygen through resonance, so it is much more stable than an alkoxide where the charge stays on one oxygen. Phenol is therefore the stronger acid. (b) Add sodium hydroxide solution. Phenol dissolves because it forms sodium phenoxide, while cyclohexanol does not react and stays in a separate layer. Marking: one mark for the phenoxide resonance stabilisation and one for a workable test such as sodium hydroxide, with the observation stated clearly.

Q2 (CBSE Board 2023, 3 marks)

(a) Why is the carbon oxygen bond length in phenols less than in methanol? (b) Arrange in order of increasing boiling point: ethoxyethane, butanal, butanol, n-butane. (c) How can phenol be prepared from anisole? Give the reaction.

Answer: (a) The oxygen lone pair is conjugated with the aromatic ring, so the carbon oxygen bond gains partial double bond character, and the carbon is sp2 hybridised. Both effects shorten the bond. (b) n-Butane, then ethoxyethane, then butanal, then butanol. Butanol hydrogen bonds strongly, butanal has a polar carbonyl group, the ether has only weak polarity and butane has none. (c) Heat anisole with concentrated HI. The alkyl oxygen bond breaks and the products are phenol and iodomethane. Part (a) needs the resonance and sp2 points, part (b) needs the complete order, and part (c) needs the reagent with both products named.

Q3 (CBSE Board 2024, 1 mark)

The reaction of an alkyl halide with sodium alkoxide forming an ether is known as (A) Wurtz reaction (B) Reimer Tiemann reaction (C) Williamson synthesis (D) Kolbe reaction

Answer: (C) Williamson synthesis. The alkoxide oxygen attacks the alkyl halide and displaces the halide in an SN2 step. It is the standard laboratory route to symmetrical and unsymmetrical ethers. The mark comes from the name of the reaction, so write Williamson synthesis clearly and add the SN2 attack of the alkoxide oxygen.

Q4 (CBSE Board 2024, 1 mark)

Which of the following compounds gives a ketone on oxidation with chromic anhydride? (A) (CH3)2CHCH2OH (B) CH3CH2CH2OH (C) (CH3)3COH (D) CH3CH2CH(OH)CH3

Answer: (D) CH3CH2CH(OH)CH3, butan-2-ol. Only a secondary alcohol has a hydrogen on the carbinol carbon and a hydrogen on a neighbouring carbon to lose, so it oxidises to a ketone. The two primary alcohols give aldehydes and then acids, and the tertiary alcohol resists oxidation. One mark for the correct product and one for naming the secondary alcohol that gives a ketone on oxidation.

Q5 (CBSE Board 2024, 1 mark)

The correct order of the ease of dehydration of alcohols by concentrated sulphuric acid is (A) (CH3)3COH > (CH3)2CHOH > CH3CH2OH (B) (CH3)2CHOH > CH3CH2OH > (CH3)3COH (C) CH3CH2OH > (CH3)2CHOH > (CH3)3COH (D) (CH3)2CHOH > (CH3)3COH > CH3CH2OH

Answer: (A). Dehydration goes through a carbocation, so ease follows cation stability: tertiary above secondary above primary. The tertiary alcohol forms the most stable carbocation and is easiest to dehydrate. Explain through the carbocation stability order, because the dehydration route through the cation is why the tertiary alcohol comes first.

Q6 (CBSE Board 2025, 3 marks)

(a) Arrange in increasing order of acidic strength: 3,5-dinitrophenol, 4-methylphenol, phenol, 2,4,6-trinitrophenol. (b) What happens when (write equations) (i) phenol is distilled with zinc dust, (ii) anisole is treated with HBr?

Answer: (a) 4-Methylphenol, then phenol, then 3,5-dinitrophenol, then 2,4,6-trinitrophenol. Nitro groups withdraw electron density and stabilise the phenoxide ion, and three of them act the most strongly, while the methyl group donates electron density and weakens the acid. (b)(i) Zinc dust removes the oxygen and phenol becomes benzene. (ii) HBr cleaves the ether at the alkyl oxygen bond, giving phenol and bromomethane. Part (a) carries one mark for the order and one for the nitro and methyl effects; part (b) carries one mark for each equation with zinc dust and with HBr.

Q7 (CBSE SQP 2022-23, 1 mark)

When 1-methylcyclohexanol is heated with acid, the alkene obtained as the main product is (a) 1-methylcyclohexane (b) 1-methylcyclohexene (c) 1-cyclohexylmethanol (d) 1-methylenecyclohexane

Answer: (b) 1-methylcyclohexene. Protonation of the OH group is followed by loss of water to give the tertiary carbocation, which is stable. Then a hydrogen is lost from a neighbouring ring carbon and the double bond forms, and Zaitsev's rule favours the more substituted alkene. The product follows from the tertiary carbocation and from Zaitsev's rule, so write both statements to protect the mark.

Q8 (CBSE SQP 2023-24, 3 marks)

Write the name of the reaction, the structure and the IUPAC name of the product for any two of the following. (a) Phenol reacts with CHCl3 in the presence of NaOH, followed by hydrolysis. (b) CH3CH2CH(CH3)CH(CH3)ONa reacts with C2H5Br.

Answer: (a) Reimer Tiemann reaction. A formyl group enters at the ortho position, so the product is 2-hydroxybenzaldehyde, also called salicylaldehyde. (b) Williamson ether synthesis. The alkoxide oxygen attacks the ethyl carbon and displaces bromide, giving 2-ethoxy-3-methylpentane, CH3CH2CH(CH3)CH(CH3)-O-C2H5. Each part needs the reaction name, the structure and the IUPAC name, so all three pieces must appear for full credit.

Q9 (CBSE SQP 2024-25, 3 marks)

Propan-1-ol is heated with an acid catalyst at different temperatures. (a) Identify the major product at 140 degrees celsius and the substitution mechanism followed. (b) Identify the major product at 180 degrees celsius and the substitution mechanism followed.

Answer: (a) A lower temperature favours the ether, so the product is dipropyl ether. The mechanism is SN2: one alcohol molecule is protonated, and a second alcohol molecule attacks its carbon and displaces water. (b) A higher temperature favours elimination, so the product is propene. The mechanism is E1-like: protonation, loss of water to give the carbocation in the slow step, then loss of a proton to form the double bond. At 140 degrees the ether is the product and SN2 is the mechanism; at 180 degrees the alkene is the product and the mechanism is E1.

Q10 (CBSE Additional PQ 2022-23, 1 mark)

Which of the following compounds could be present in the de-icing fluid used on a car window during extreme cold? (a) formaldehyde (b) phenol (c) propan-2-ol (d) acetic acid

Answer: (c) propan-2-ol. An alcohol can be used as an antifreeze and de-icing agent, and it mixes with water in all proportions through hydrogen bonding, which lowers the freezing point of the liquid film on the glass. The mark is for choosing an alcohol and giving the hydrogen bonding reason, so name propan-2-ol and explain the freezing point depression.

How marks are usually split

A two mark question in this chapter usually wants one reason and one observation, so pair every claim with evidence. For an acidity comparison, write the conjugate base, its stabilisation and the pKa value in that order. For a preparation, name the reagent, the condition and the product. A three mark question on reactions normally carries one mark per equation, so balance each equation, write the condition above the arrow and name the product underneath.

For a five mark question on phenol chemistry, the usual split is one mark for the structure or naming, two for the reaction equations and two for the explanation. Do not leave the explanation for the end in one line; attach a short reason to each equation so the examiner can award the reasoning marks even if the final product is wrong.

Practical questions follow the same logic. Where the question asks why a particular reagent is chosen, name the competing reagent and say what goes wrong with it. The thionyl chloride question is the standard example: the answer is the gaseous by products that escape and leave pure alkyl chloride behind. That is the shape of answer that earns the reasoning mark. One more habit pays off in the exam: whenever a question pairs an alcohol with an acid or a halogen acid, write the class of the alcohol first, because the mechanism and the reaction conditions both follow from that class.