Alcohols, Phenols and Ethers — Class 12 Chemistry practice questions

Question 1

A quality-control chemist in a pharmaceutical laboratory has four unlabelled bottles, each containing one of the following compounds: ethanol (A), phenol (B), cyclohexanol (C), and benzyl alcohol (D). She needs to identify each compound using simple chemical tests and also understands why certain reactions behave differently with these compounds.

(i) She adds a small amount of neutral FeCl₃ solution to each sample. Only one bottle gives a characteristic colour. Identify the compound and name the colour produced. (1 mark)

(ii) She then treats samples A and C separately with Lucas reagent (conc. HCl + anhydrous ZnCl₂) at room temperature. Describe what she observes in each case and explain the difference in reactivity. (2 marks)

(iii) When phenol is treated with Br₂ water (not Br₂/CS₂), a white precipitate forms immediately even without a Lewis acid catalyst. Name the product and give a reason why this reaction is so facile compared to bromination of benzene. (1 mark)

Accepted Answer

(i) Compound B (phenol) gives a characteristic violet / purple colour with neutral FeCl₃ solution.

[Value point: identification of phenol — ½ mark; colour — ½ mark]

(ii) Lucas test observations:

• Ethanol (A) — primary alcohol: no turbidity / no cloudiness is observed at room temperature. The solution remains clear. (The reaction with Lucas reagent is too slow at room temperature for a primary alcohol.)

• Cyclohexanol (C) — secondary alcohol: turbidity / cloudiness appears after approximately 5 minutes, because the reaction proceeds via the SN1 mechanism with moderate rate for a secondary carbocation intermediate.

Reason for the difference: The rate of reaction with Lucas reagent depends on the stability of the carbocation intermediate formed. Cyclohexanol gives a secondary carbocation, which is more stable than the primary carbocation that would form from ethanol. Therefore, cyclohexanol reacts faster, producing the insoluble alkyl chloride (cloudiness) within a few minutes, while ethanol shows no observable reaction at room temperature.

[Value point: correct observation for ethanol — ½ mark; correct observation for cyclohexanol — ½ mark; correct reason based on carbocation stability / mechanism — 1 mark]

(iii) Product: 2,4,6-tribromophenol (white precipitate ↓)

Reaction:

C₆H₅OH + 3Br₂(aq) → C₆H₂Br₃OH↓ (2,4,6-tribromophenol) + 3HBr

Reason: Due to the strong electron-donating effect of the —OH group through resonance, the electron density at the ortho and para positions of the benzene ring is greatly increased. This makes the ring highly activated towards electrophilic aromatic substitution. As a result, Br₂ (a weak electrophile) can react directly without a Lewis acid catalyst, and all three activated positions (two ortho + one para) are substituted in a single step.

[Value point: correct product name / structure — ½ mark; correct reason (ring activation by —OH through resonance / increased electron density at o- and p-positions) — ½ mark]

Question 2

A chemistry laboratory has four unlabelled bottles, each containing one of the following compounds: ethanol (CH₃CH₂OH), phenol (C₆H₅OH), ethoxyethane (CH₃CH₂OCH₂CH₃), and propan-2-ol ((CH₃)₂CHOH). A student performs the following tests to identify each compound:

(a) Which compound gives a violet/purple colouration with neutral FeCl₃ solution? Name the test and state the observation. (1 mark)

(b) Which compound among the two alcohols (ethanol and propan-2-ol) gives an immediate turbidity with Lucas reagent (conc. HCl + anhydrous ZnCl₂) at room temperature? Give the chemical reaction involved. (1 mark)

(c) Which compound does NOT give any visible reaction with sodium metal, and why? (1 mark)

(d) Ethanol can be converted into ethoxyethane (diethyl ether) in the laboratory. Name the method used and write the chemical equation with reagents and conditions. (1 mark)

Accepted Answer

(a) Phenol (C₆H₅OH) gives a violet/purple colouration with neutral FeCl₃ solution.
Test: FeCl₃ test (Ferric chloride test).
Observation: Phenol reacts with neutral FeCl₃ to give a characteristic violet/purple colour. Ethanol, propan-2-ol and ethoxyethane do not give this colour.

(b) Propan-2-ol gives an immediate turbidity (cloudiness) with Lucas reagent at room temperature.
Due to the Lucas test: tertiary and secondary alcohols react with Lucas reagent (conc. HCl + anhydrous ZnCl₂) at room temperature, but propan-2-ol (a secondary alcohol) reacts within about 5 minutes giving turbidity, while ethanol (a primary alcohol) shows no turbidity at room temperature.

Chemical reaction:

(CH₃)₂CHOH + HCl —[anhydrous ZnCl₂, room temp.]→ (CH₃)₂CHCl + H₂O

∴ Propan-2-ol gives turbidity due to formation of insoluble (CH₃)₂CHCl (2-chloropropane).

(c) Ethoxyethane (CH₃CH₂OCH₂CH₃) does NOT give any visible reaction with sodium metal.
Reason: Due to the absence of an active O–H bond (hydroxyl group), ethers do not react with sodium metal. Alcohols and phenol react with Na to liberate H₂ gas, but ethers have no replaceable hydrogen on oxygen; hence, no reaction is observed.

Question 3

Meera is working in an organic chemistry lab. She has four unlabelled bottles, each containing one of the following compounds: ethanol (CH₃CH₂OH), phenol (C₆H₅OH), diethyl ether (C₂H₅OC₂H₅), and cyclohexanol (C₆H₁₁OH). She needs to identify each compound using simple chemical tests and also carry out a specific conversion.

Answer the following questions based on the above scenario:

(a) Meera adds a few drops of neutral FeCl₃ solution to samples from two bottles. One sample gives a violet/purple colouration while the other gives no colour. Identify the two compounds and explain the observation. (2 marks)

(b) Meera performs the Lucas test (conc. HCl + anhydrous ZnCl₂) at room temperature on ethanol and cyclohexanol. State the observation for each compound and identify which reacts faster. Give one reason. (1 mark)

(c) Meera wants to convert phenol to salicylaldehyde. Name the reaction involved and write the reagents/conditions required. (1 mark)

Accepted Answer

(a) The compound that gives a violet/purple colouration with neutral FeCl₃ solution is phenol (C₆H₅OH). Phenol forms a coloured complex (iron(III) phenoxide) with FeCl₃ due to the presence of the –OH group directly attached to the benzene ring, which gives a characteristic violet/purple colour.

The compound that gives no colour with FeCl₃ is diethyl ether (C₂H₅OC₂H₅). Diethyl ether has no –OH group; hence it does not react with FeCl₃ and produces no colouration.

∴ Phenol → violet/purple colour with FeCl₃; diethyl ether → no colour.

(b) Lucas test observation:
• Ethanol (primary alcohol): No cloudiness/turbidity at room temperature. Primary alcohols do not react with Lucas reagent at room temperature.
• Cyclohexanol (secondary alcohol): Cloudiness/turbidity appears after approximately 5 minutes at room temperature.

Cyclohexanol reacts faster than ethanol. This is because cyclohexanol is a secondary alcohol and forms a more stable secondary carbocation intermediate under SN1 conditions, whereas ethanol (primary alcohol) forms a less stable primary carbocation and requires heating.

(c) The reaction involved is the Reimer–Tiemann reaction.

Reagents/Conditions: Phenol is treated with CHCl₃ (chloroform) and aqueous NaOH (aq. NaOH), followed by acidification.

C₆H₅OH + CHCl₃ → (aq. NaOH, Δ, then H₃O⁺) → 2-hydroxybenzaldehyde (salicylaldehyde)

∴ Reimer–Tiemann reaction; reagents: CHCl₃ + aq. NaOH, followed by acidification.

Question 4

A chemist working in a pharmaceutical laboratory has four unlabelled bottles, each containing one of the following compounds: ethanol (CH₃CH₂OH), phenol (C₆H₅OH), cyclohexanol (C₆H₁₁OH), and ethoxyethane (CH₃CH₂OCH₂CH₃). To identify each compound, she performs a series of chemical tests and also needs to arrange them in increasing order of boiling point for a separation procedure.

(a) She treats each compound with neutral FeCl₃ solution. Which compound gives a characteristic violet/purple colouration? Write the equation for this reaction.  (2 marks)

(b) She then applies the Lucas test (conc. HCl + anhydrous ZnCl₂) to distinguish between the two remaining alcohols. Predict the observation for each alcohol and identify which is which.  (1 mark)

(c) Arrange the four compounds in increasing order of boiling point and give one reason for the position of ethoxyethane in your order.  (1 mark)

Accepted Answer

(a) FeCl₃ Test — Identification of Phenol (2 marks)

Phenol (C₆H₅OH) gives a characteristic violet/purple colouration with neutral FeCl₃ solution. The other three compounds — ethanol, cyclohexanol and ethoxyethane — do not give this colouration.

The reaction is:

6 C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ + 3 HCl + 3 H⁺

(or acceptably written as the formation of the iron(III) phenoxide complex giving a violet/purple colour)

This test confirms the bottle containing phenol.

(b) Lucas Test — Distinguishing Ethanol from Cyclohexanol (1 mark)

Reagent: conc. HCl + anhydrous ZnCl₂ at room temperature.

• Cyclohexanol (secondary alcohol, 2°): develops turbidity (cloudiness) within approximately 5 minutes, as the secondary carbocation is formed at a moderate rate.
  C₆H₁₁OH + HCl →(anhydrous ZnCl₂) C₆H₁₁Cl (turbid) + H₂O

• Ethanol (primary alcohol, 1°): no turbidity / remains clear at room temperature (reacts only on heating).

∴ The bottle showing turbidity within 5 min = cyclohexanol; the bottle showing no turbidity = ethanol. The fourth bottle (no reaction with FeCl₃, no reaction in Lucas test) = ethoxyethane.

(c) Increasing Order of Boiling Point (1 mark)

Increasing order:
ethoxyethane (34.6°C) < ethanol (78.4°C) < cyclohexanol (161°C) < phenol (182°C)

Reason for position of ethoxyethane: Ethoxyethane has the lowest boiling point because it cannot form intermolecular hydrogen bonds (the oxygen is flanked by two alkyl groups and there is no O–H bond), so only weak van der Waals/dipole–dipole forces operate, resulting in much lower energy needed for vaporisation compared to the alcohols and phenol which undergo intermolecular hydrogen bonding.

Question 5

A chemistry student is working in a laboratory and has three unlabelled bottles, each containing one of the following compounds: cyclohexanol, phenol, and diethyl ether. The student needs to identify each compound using simple chemical tests. She also notices that phenol is significantly more acidic than cyclohexanol, even though both contain an –OH group.

(a) Suggest ONE chemical test (reagent and expected observation for each compound) that would distinguish phenol from cyclohexanol.  (2 marks)

(b) Account for the fact that phenol is a much stronger acid than cyclohexanol.  (1 mark)

(c) When the student treats phenol with bromine water, a white precipitate is obtained immediately, whereas cyclohexanol gives no such precipitate under the same conditions. Identify the white precipitate and write the balanced chemical equation for the reaction of phenol with bromine water.  (1 mark)

Accepted Answer

(a) Neutral FeCl₃ test:
Add a few drops of neutral FeCl₃ solution to each compound.
• Phenol → gives a distinct violet/purple colouration.
• Cyclohexanol → gives no colour (solution remains pale yellow/colourless).
∴ The compound producing violet/purple colouration with neutral FeCl₃ is phenol; the compound showing no colour change is cyclohexanol.

[Award 1 mark for correct reagent (neutral FeCl₃); 1 mark for correct observation stated for BOTH phenol (violet/purple colour) and cyclohexanol (no colour change).]

(b) Due to resonance stabilisation of the phenoxide ion (C₆H₅O⁻): the negative charge on the oxygen is delocalised over the benzene ring through five resonance structures, dispersing the charge and greatly stabilising the conjugate base. In cyclohexanol, the cyclohexoxide ion (C₆H₁₁O⁻) has no such resonance stabilisation — the negative charge remains localised on oxygen. The greater stability of the conjugate base makes phenol a much stronger acid than cyclohexanol.

[Award 1 mark for resonance stabilisation of phenoxide ion / delocalisation of negative charge over the benzene ring.]

(c) The white precipitate is 2,4,6-tribromophenol.

Balanced equation:

C₆H₅OH + 3Br₂(aq) → C₆H₂Br₃OH↓ + 3HBr

(i.e. phenol + 3Br₂ → 2,4,6-tribromophenol↓ (white ppt) + 3HBr)

∴ The white precipitate formed is 2,4,6-tribromophenol.

[Award ½ mark for correctly identifying the precipitate as 2,4,6-tribromophenol; ½ mark for the correctly balanced equation.]

Question 6

A student in an organic chemistry lab is given four unlabelled bottles, each containing one of the following compounds: (i) ethanol, (ii) phenol, (iii) cyclohexanol, (iv) tert-butanol. The student is asked to identify all four using simple chemical tests available in the lab: FeCl₃ solution, Lucas reagent (conc. HCl + anhyd. ZnCl₂), and chromic anhydride (CrO₃ in H₂SO₄, Jones reagent).

(a) Which compound will give a characteristic violet/purple colour with FeCl₃ solution? Name the compound and write the chemical equation for the reaction. (2 marks)

(b) The remaining three compounds are tested with Lucas reagent. Arrange them in the order of reactivity (fastest to slowest) and justify the order. (1 mark)

(c) Among ethanol and cyclohexanol, which one will show a colour change when treated with CrO₃/H₂SO₄, and what is the product formed? (1 mark)

Accepted Answer

Part (a) — 2 marks

Phenol gives a characteristic violet/purple colour with neutral FeCl₃ solution due to the formation of an iron–phenol complex. Alcohols (ethanol, cyclohexanol, tert-butanol) do not give this colour change.

Chemical equation:

6 C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]³⁻ + 3 HCl + 3 H⁺

(Accepted simplified form: FeCl₃ + excess phenol → violet/purple iron–phenoxide complex)

∴ Compound (ii) phenol is identified by the violet/purple colouration with FeCl₃ solution.

[Value points: identification of phenol — ½ mark; reason/observation (violet/purple colour, FeCl₃ complex) — ½ mark; chemical equation written — 1 mark]

---

Part (b) — 1 mark

Lucas test reactivity order (fastest to slowest):

tert-butanol > cyclohexanol > ethanol

Justification: Lucas reagent works via an SN1 mechanism — the alcohol is first protonated, then loses water to form a carbocation intermediate. The ease of carbocation formation determines the rate.

• tert-butanol forms a stable 3° carbocation → reacts immediately (turbidity/cloudiness at once).
• cyclohexanol forms a 2° carbocation → reacts within ~5 minutes (turbidity after some time).
• ethanol forms an unstable 1° carbocation → no observable reaction at room temperature.

∴ Order: tert-butanol > cyclohexanol > ethanol (3° > 2° > 1°), based on stability of the carbocation intermediate.

[Value points: correct order — ½ mark; correct reason (carbocation stability / SN1) — ½ mark]

---

Part (c) — 1 mark

Both ethanol (1° alcohol) and cyclohexanol (2° alcohol) are oxidised by CrO₃/H₂SO₄ (Jones reagent), shown by a colour change from orange to green (Cr⁶⁺ → Cr³⁺). However, they give different products:

• Ethanol (1° alcohol) is oxidised to acetaldehyde (ethanal), and on further oxidation to acetic acid (ethanoic acid).

CH₃CH₂OH + [O] →(CrO₃/H₂SO₄) CH₃CHO →(further [O]) CH₃COOH

• Cyclohexanol (2° alcohol) is oxidised to cyclohexanone (a ketone).

C₆H₁₁OH + [O] →(CrO₃/H₂SO₄) C₆H₁₀O (cyclohexanone)

Since the question asks which shows a colour change: both do. The product from cyclohexanol is cyclohexanone (ketone); the product from ethanol is ethanal/acetic acid (aldehyde/carboxylic acid).

∴ Both ethanol and cyclohexanol show the orange → green colour change with CrO₃/H₂SO₄. The product formed from cyclohexanol is cyclohexanone; from ethanol, the primary product is ethanal (acetaldehyde).

[Value points: correct identification that both react (or specifying cyclohexanol → cyclohexanone as the ketone product) — ½ mark; correct product named — ½ mark]

Question 7

A pharmaceutical chemist is working on synthesising two medicinal compounds from phenol. In Route A, she reacts sodium phenoxide with CO₂ under pressure at 400 K to obtain compound X (an antiseptic). In Route B, she reacts sodium phenoxide with chloroethane in dry ether to obtain compound Y (an anaesthetic). She also notices that compound Y shows lower reactivity towards electrophilic substitution compared to phenol itself.

(a) Identify compounds X and Y and write a balanced chemical equation for each synthesis. (2)
(b) Give one reason why compound Y shows lower reactivity towards electrophilic substitution than phenol. (1)
(c) State one test, with observation, to distinguish phenol from compound Y. (1)

Accepted Answer

(a)

Compound X is salicylic acid (2-hydroxybenzoic acid), formed by Kolbe's reaction.
Compound Y is ethoxybenzene (phenetole), formed by Williamson synthesis.

Route A — Kolbe's reaction:

C₆H₅ONa + CO₂ → [400 K, pressure] → C₆H₄(OH)COONa

C₆H₄(OH)COONa + HCl → C₆H₄(OH)COOH + NaCl

(Sodium phenoxide reacts with CO₂ under pressure at 400 K to give sodium salicylate; acidification gives salicylic acid — Compound X.)

Route B — Williamson synthesis:

C₆H₅ONa + C₂H₅Cl → [dry ether] → C₆H₅OC₂H₅ + NaCl

(Sodium phenoxide undergoes SN2 reaction with chloroethane to give ethoxybenzene — Compound Y.)

(b)

Due to the absence of the free –OH group in compound Y (ethoxybenzene), the –OC₂H₅ group donates electrons to the ring less effectively than the –OH group in phenol (the lone pairs on oxygen are less available for resonance with the ring because of the electron-withdrawing inductive effect of the ethyl group), making the ring slightly less electron-rich and hence less reactive towards electrophilic substitution than phenol.

(c)

FeCl₃ test:
— Phenol gives a violet/purple colouration with neutral FeCl₃ solution (due to formation of a coloured iron–phenolate complex).
— Ethoxybenzene (compound Y) gives no colour change with FeCl₃ solution.
∴ The compound that gives a violet/purple colour is phenol; the compound that gives no colour is compound Y (ethoxybenzene).

Question 8

A chemistry student is working in a laboratory and has three unlabelled bottles, each containing one of the following compounds: ethanol (CH₃CH₂OH), phenol (C₆H₅OH), and diethyl ether (C₂H₅OC₂H₅). She has access to the following reagents: neutral FeCl₃ solution, sodium metal, and bromine water (Br₂/H₂O).

(a) Which reagent will help her distinguish phenol from both ethanol and diethyl ether in a single test? State the observation for ALL THREE compounds with this reagent.  (2 marks)

(b) After identifying phenol, she adds sodium metal to the remaining two bottles separately. Which compound — ethanol or diethyl ether — reacts with sodium metal, and what observation confirms the reaction?  (1 mark)

(c) Write the balanced chemical equation for the reaction of phenol with sodium metal.  (1 mark)

Accepted Answer

CBSE Marking Scheme — Total: 4 marks

(a) Reagent: Neutral FeCl₃ solution  [½ mark for naming the reagent]

Observations with neutral FeCl₃:
• Phenol → gives a characteristic violet / purple colour.  [½ mark]
• Ethanol → gives no colour / no characteristic change (solution remains pale yellow-brown).  [½ mark]
• Diethyl ether → gives no colour / no characteristic change.  [½ mark]

∴ Neutral FeCl₃ distinguishes phenol (violet colour) from both ethanol and diethyl ether (no colour).

[Award full 2 marks only if observations for ALL THREE compounds are stated.]

(b) Ethanol reacts with sodium metal.  [½ mark]

Observation: Brisk effervescence / bubbles of hydrogen gas (H₂↑) are produced.  [½ mark]

(Diethyl ether does NOT react with sodium metal under ordinary conditions because the C–O–C oxygen is much less acidic than the O–H of ethanol; there is no O–H bond to donate a proton.)

∴ The bottle that produces brisk effervescence with sodium metal is ethanol; the one that shows no reaction is diethyl ether.

(c) Balanced equation for the reaction of phenol with sodium metal:

C₆H₅OH  +  Na  →  C₆H₅O⁻Na⁺  +  ½ H₂↑

or, written as the molecular equation:

2 C₆H₅OH  +  2 Na  →  2 C₆H₅ONa  +  H₂↑

[1 mark — award for a balanced equation with correct product sodium phenoxide (C₆H₅ONa) and H₂↑; ½ mark if product is correct but equation is not balanced or H₂↑ is missing.]

Question 9

Which of the following reagents is used to distinguish ethanol from phenol?

Accepted Answer

(B) FeCl₃ solution

Explanation: FeCl₃ solution gives a characteristic violet/purple colour with phenol due to formation of a coloured complex, whereas ethanol does not give any colour with FeCl₃. Hence FeCl₃ solution distinguishes ethanol from phenol.

Question 10

Assertion (A): Phenol is more acidic than ethanol.
Reason (R): The phenoxide ion formed after loss of a proton from phenol is stabilised by resonance with the benzene ring.

Accepted Answer

Option (A)

Explanation: Phenol is indeed more acidic than ethanol — Assertion (A) is TRUE. The lone pair on oxygen of phenoxide ion delocalises into the benzene ring through five resonance structures, distributing the negative charge over the ortho and para carbon atoms; this resonance stabilisation lowers the energy of the conjugate base and makes proton release more favourable. The ethoxide ion (from ethanol) has no such stabilisation. Since Reason (R) correctly identifies resonance stabilisation of the phenoxide ion as the cause, it is TRUE and is the correct explanation of (A).

Question 11

Assertion (A) : Phenol is more acidic than ethanol.
Reason (R) : The phenoxide ion is stabilised by resonance, whereas the ethoxide ion is not.

Accepted Answer

(A) Both A and R are true, and R is the correct explanation of A.

Explanation: Phenol is a stronger acid than ethanol because the conjugate base of phenol — the phenoxide ion — is stabilised by delocalisation of the negative charge over the benzene ring through resonance (five resonance structures). The ethoxide ion, by contrast, has no such delocalisation; the negative charge remains entirely on oxygen. This greater stability of the phenoxide ion shifts the equilibrium towards ionisation, making phenol significantly more acidic than ethanol. Since R directly and correctly accounts for the observation stated in A, option (A) is correct.

Question 12

Which reagent is used in the Lucas test to distinguish between primary, secondary and tertiary alcohols?

Accepted Answer

(B) Conc. HCl + anhydrous ZnCl₂

Explanation: The Lucas test uses a mixture of concentrated HCl and anhydrous ZnCl₂ (Lucas reagent). Tertiary alcohols react immediately giving a turbid (cloudy) mixture due to formation of an insoluble alkyl chloride; secondary alcohols react in about 5 minutes; primary alcohols show no turbidity at room temperature.

Question 13

Which reagent is used to distinguish phenol from ethanol?

Accepted Answer

Option (B) — Neutral FeCl₃ solution.

Neutral FeCl₃ solution gives a characteristic violet/purple colouration with phenol due to the formation of a phenol–iron complex, whereas ethanol does not produce any colour with FeCl₃. This colour difference allows phenol to be distinguished from ethanol.

Question 14

Account for the following:
(i) Phenol is a stronger acid than cyclohexanol.
(ii) The boiling point of ethanol (C₂H₅OH) is much higher than that of its isomer dimethyl ether (CH₃OCH₃).

Accepted Answer

(i) Phenol is a stronger acid than cyclohexanol.

Due to resonance stabilisation of the phenoxide ion — the negative charge on the oxygen is delocalised over the benzene ring through five resonance structures. No such stabilisation is possible in the cyclohexanoxide ion, which bears the entire negative charge on oxygen alone. ∴ Phenol readily loses a proton and is more acidic than cyclohexanol. [1 mark]

(ii) Ethanol has a much higher boiling point than dimethyl ether.

Due to the presence of an –OH group, ethanol molecules are associated through strong intermolecular hydrogen bonding (O–H···O). Dimethyl ether has no O–H bond and therefore cannot form hydrogen bonds with other ether molecules; only weak van der Waals forces act between them. Extra energy is required to break the hydrogen bonds in ethanol, raising its boiling point significantly. [1 mark]

Question 15

Why is phenol more acidic than ethanol?

Accepted Answer

Due to resonance stabilisation of the phenoxide ion (C₆H₅O⁻), phenol is more acidic than ethanol.

When phenol loses a proton, the phenoxide ion formed is stabilised by delocalisation of the negative charge over the benzene ring through resonance — the lone pair on oxygen is in conjugation with the π system, spreading the charge onto the ring carbon atoms (ortho and para positions). This makes the phenoxide ion relatively stable and phenol a stronger acid.

In contrast, when ethanol loses a proton, the ethoxide ion (C₂H₅O⁻) formed carries the full negative charge localised on the oxygen atom alone — there is no resonance stabilisation. Hence ethoxide is a much stronger base (less stable conjugate base), and ethanol is a weaker acid than phenol.

∴ Phenol (pKa ≈ 10) is significantly more acidic than ethanol (pKa ≈ 16).

Question 16

A research chemist is studying the reactivity and acidity of three compounds isolated from a plant extract:

Compound P: 2,4,6-trinitrophenol (picric acid)
Compound Q: 4-methylphenol (p-cresol)
Compound R: Cyclohexanol

(a) Arrange compounds P, Q and R in increasing order of acid strength. Give a reason for the position of compound P in the order. (2 marks)

(b) The chemist treats compound Q with CHCl₃ in the presence of aqueous NaOH followed by acidification with dilute HCl. Name the reaction taking place, identify the organic product formed, and state the position at which the new functional group is introduced on the ring. (2 marks)

Accepted Answer

PART (a) [2 marks]

Increasing order of acid strength:

R < Q < P

i.e., Cyclohexanol < 4-methylphenol < 2,4,6-trinitrophenol

Reason for the position of Compound P:

The three –NO₂ groups at the ortho and para positions are strongly electron-withdrawing groups (–I effect and resonance withdrawal / −M effect). They withdraw electron density from the phenoxide ion formed after loss of H⁺, thereby stabilising the phenoxide ion extensively through dispersal of negative charge. This makes the O–H bond weaker and proton donation much easier, making 2,4,6-trinitrophenol (picric acid) far more acidic than simple phenol or p-cresol.

∴ Position of P: Strongest acid in the series.

[Note on Q vs R: Phenol (and its alkyl derivatives such as p-cresol) is more acidic than cyclohexanol because the phenoxide ion is stabilised by resonance (delocalisation of the negative charge over the ring), whereas no such stabilisation exists for the cyclohexoxide ion. The –CH₃ group in p-cresol is an electron-donating group (+I effect) which slightly destabilises the phenoxide ion relative to phenoxide itself, but p-cresol remains far more acidic than cyclohexanol.]

---

PART (b) [2 marks]

Name of the reaction: Reimer–Tiemann Reaction

Reagent and conditions: CHCl₃ + aq. NaOH → (heat, Δ) → followed by acidification with dil. HCl

Product formed: 2-hydroxy-4-methylbenzaldehyde (5-methylsalicylaldehyde)

The electrophilic species :CCl₂ (dichlorocarbene), generated from CHCl₃ and NaOH, attacks the ring ortho to the –OH group. The intermediate is hydrolysed and then acidified to give an aldehyde group (–CHO) at the ortho position.

Position of introduction: ortho to the –OH group (C-2 position relative to –OH; para is blocked by –CH₃, so ortho product is the major product).

Reaction:

4-methylphenol + CHCl₃ → (aq. NaOH, Δ, then dil. HCl) → 2-hydroxy-4-methylbenzaldehyde

∴ The Reimer–Tiemann reaction introduces a –CHO group at the ortho position of the ring relative to the phenolic –OH group.

Question 17

Give reasons for each of the following observations:
(a) Phenol is a stronger acid than ethanol.
(b) The boiling point of propan-1-ol (b.p. 97 °C) is much higher than that of methoxyethane (b.p. 11 °C), although both have the same molecular formula C₃H₈O.
(c) In the reaction of phenol with bromine water, tribromination occurs readily even without a Lewis acid catalyst, whereas bromination of benzene requires anhydrous AlCl₃ as catalyst.

Accepted Answer

(a) Phenol is a stronger acid than ethanol.
Due to resonance stabilisation of the phenoxide ion (C₆H₅O⁻), phenol is a much stronger acid than ethanol.
When phenol loses a proton, the negative charge on oxygen in the phenoxide ion is delocalised over the benzene ring through five resonance structures — the charge is distributed on the oxygen and the ortho and para carbon atoms. This extensive delocalisation stabilises the phenoxide ion, making it a weaker conjugate base and shifting the equilibrium towards ionisation.
In ethanol, the ethoxide ion (C₂H₅O⁻) carries the negative charge entirely on oxygen with no such resonance delocalisation, so it is far less stable. Hence phenol (pKa ≈ 10) is a much stronger acid than ethanol (pKa ≈ 16).

(b) Propan-1-ol has a much higher boiling point than methoxyethane (same molecular formula C₃H₈O).
Due to the presence of an –OH group, propan-1-ol molecules undergo strong intermolecular hydrogen bonding with one another (O–H···O), which requires considerable energy to break. Methoxyethane has no O–H bond and therefore cannot form intermolecular hydrogen bonds — only weaker van der Waals (dipole–dipole and London dispersion) forces operate between its molecules. The much greater intermolecular association in propan-1-ol raises its boiling point significantly compared to methoxyethane.

(c) Phenol undergoes electrophilic bromination without a Lewis acid catalyst, whereas benzene requires anhydrous AlCl₃.
The –OH group attached to the benzene ring in phenol is a powerful electron-donating group. The lone pair on oxygen is donated into the ring by resonance, greatly increasing the electron density at the ortho and para positions. This makes the ring highly activated towards electrophilic aromatic substitution, so even the weakly electrophilic Br₂ molecule (without activation by a Lewis acid) reacts readily. The high electron density facilitates rapid successive bromination at all three activated positions (2, 4 and 6), giving 2,4,6-tribromophenol directly.
In benzene, the ring is not activated; Br₂ alone is too weak an electrophile. Anhydrous AlCl₃ is needed to polarise the Br–Br bond and generate a more powerful electrophile (Br⁺ or Br···AlCl₃) to overcome the energy barrier.

Question 18

Account for the following:
(a) Phenol is a stronger acid than ethanol.
(b) The boiling point of ethanol (351 K) is much higher than that of its isomer dimethyl ether (249 K), although both have the same molecular formula C₂H₆O.
(c) Tertiary alcohols undergo dehydration most readily among primary, secondary and tertiary alcohols.

Accepted Answer

(a) Due to resonance stabilisation of the phenoxide ion formed after loss of a proton, phenol is a stronger acid than ethanol.

When phenol loses a proton, the phenoxide ion is formed. The negative charge on the oxygen is delocalised over the benzene ring through five resonance structures, thereby stabilising the conjugate base. In ethanol, the ethoxide ion formed on deprotonation carries the negative charge localised on the oxygen alone — no such resonance stabilisation is possible. Hence, the equilibrium for ionisation lies further to the right for phenol, making it a stronger acid than ethanol.

(b) Due to intermolecular hydrogen bonding, ethanol has a much higher boiling point than dimethyl ether.

Ethanol (C₂H₅OH) contains an O–H bond; the oxygen atom is highly electronegative and the hydrogen is sufficiently positive to form strong intermolecular hydrogen bonds between molecules. A large amount of energy is required to break these associations before the liquid can vaporise, so the boiling point is high (351 K). Dimethyl ether (CH₃OCH₃) has no O–H bond and therefore cannot form intermolecular hydrogen bonds; only weak van der Waals forces operate between its molecules, resulting in a much lower boiling point (249 K).

(c) Due to the greater stability of the carbocation intermediate formed during dehydration, tertiary alcohols undergo dehydration most readily.

Dehydration of alcohols proceeds via an E1 mechanism involving a carbocation intermediate. A tertiary carbocation is stabilised by three alkyl groups through hyperconjugation and inductive effects, making it far more stable (and more readily formed) than a secondary or primary carbocation. Since the rate-determining step is the formation of this carbocation, tertiary alcohols require the least activation energy and therefore dehydrate most readily.

Question 19

A school chemistry club is investigating the properties of three organic compounds found in household products. They collect the following observations:

• Compound P is a colourless liquid present in hand sanitisers. Its molecular formula is C₂H₅OH. It turns orange K₂Cr₂O₇ solution green on warming with acidified K₂Cr₂O₇.
• Compound Q is a colourless liquid with a distinctive odour, used as an antiseptic. Its molecular formula is C₆H₅OH. It gives a violet/purple colouration with neutral FeCl₃ solution.
• Compound R is obtained when Compound P is heated with excess Compound Q in the presence of conc. H₂SO₄. It is used as a solvent.

On the basis of this information, answer the following questions:

(a) Identify compounds P, Q and R. (1 mark)
(b) Write the chemical equation for the reaction of Compound P with acidified K₂Cr₂O₇, clearly showing the reagent and conditions over the arrow. (1 mark)
(c) Compound Q is a stronger acid than Compound P. Give ONE reason to justify this. (1 mark)
(d) When Compound Q is treated with Br₂ water, a white precipitate is formed. Write the chemical equation for this reaction and name the product. (1 mark)

Accepted Answer

(a) Identification of compounds:

Compound P: Ethanol (CH₃CH₂OH) — a primary alcohol, present in hand sanitisers; oxidised by acidified K₂Cr₂O₇.

Compound Q: Phenol (C₆H₅OH) — an aromatic alcohol/phenol, used as antiseptic; gives violet/purple colour with FeCl₃.

Compound R: Ethyl phenyl ether / Phenetole (C₆H₅OC₂H₅) — formed by Williamson-type acid-catalysed etherification of phenol with ethanol; used as a solvent.

(All three correct — 1 mark; any two correct — ½ mark)

(b) Reaction of Compound P (ethanol) with acidified K₂Cr₂O₇:

The oxidation of a primary alcohol by acidified K₂Cr₂O₇ proceeds as:

3 C₂H₅OH + K₂Cr₂O₇ + 4 H₂SO₄ → 3 CH₃CHO + K₂SO₄ + Cr₂(SO₄)₃ + 7 H₂O

[Condition over arrow: acidified K₂Cr₂O₇ / H⁺, Δ]

Note: The primary alcohol is first oxidised to acetaldehyde (ethanal); with excess oxidant it may proceed further to acetic acid (ethanoic acid). Both products are acceptable.

[Correct balanced equation with condition — 1 mark; equation without condition — ½ mark]

(c) Justification — Phenol (Q) is a stronger acid than ethanol (P):

Due to resonance stabilisation of the phenoxide ion (C₆H₅O⁻), the negative charge is delocalised over the oxygen atom and the ortho and para positions of the benzene ring through five resonance structures. This makes the conjugate base (phenoxide ion) far more stable than the ethoxide ion (C₂H₅O⁻), which has no such resonance stabilisation. Greater stability of the conjugate base means phenol loses its proton more readily, making it a stronger acid.

[Resonance stabilisation of phenoxide ion stated — 1 mark; any equivalent correct reason accepted]

(d) Reaction of Compound Q (phenol) with Br₂ water:

Phenol undergoes electrophilic aromatic substitution with bromine water. The –OH group is a strong activating group and directs bromine to the ortho and para positions. All three positions (two ortho, one para) are substituted simultaneously even without a Lewis acid catalyst:

C₆H₅OH + 3 Br₂ → 2,4,6-tribromophenol↓ + 3 HBr

Product: 2,4,6-tribromophenol (white precipitate)

[Correct balanced equation — ½ mark; correct name of product — ½ mark]

Question 20

A chemistry student is studying the properties of two colourless liquids labelled P and Q in the laboratory. Liquid P turns blue litmus red and gives a violet colouration with neutral FeCl₃ solution. Liquid Q does not affect litmus and gives an orange-yellow precipitate with 2,4-dinitrophenylhydrazine (2,4-DNP) reagent but does not give a silver mirror with Tollens' reagent. The student also notes that Liquid P reacts with bromine water immediately, giving a white precipitate, while Liquid Q does not react with bromine water under the same conditions.

(a) Identify Liquids P and Q. Give one reason for each identification.          (2 marks)
(b) Which liquid — P or Q — will react with sodium metal to liberate hydrogen gas? Write the balanced chemical equation for the reaction.          (1 mark)
(c) The student wishes to convert Liquid P into its acetyl derivative using acetic anhydride. Name the reaction and write the balanced chemical equation for it.          (1 mark)

Accepted Answer

(a) Identification of P and Q:

Liquid P is Phenol (C₆H₅OH).
Reason: Phenol is acidic (turns blue litmus red), gives a characteristic violet/purple colouration with neutral FeCl₃ solution, and reacts with bromine water immediately to give a white precipitate of 2,4,6-tribromophenol (trisubstitution due to the strongly activating –OH group).

Liquid Q is Acetone (CH₃COCH₃) / a ketone.
Reason: Q gives an orange-yellow precipitate with 2,4-DNP reagent, confirming the presence of a carbonyl (C=O) group. It does not give a silver mirror with Tollens' reagent, ruling out an aldehyde and confirming it is a ketone. It does not react with bromine water under normal conditions, consistent with a simple ketone.

(b) Liquid P (Phenol) will react with sodium metal to liberate hydrogen gas.
Phenol is acidic (stronger acid than alcohol) and reacts with sodium to release H₂.
Balanced equation:

2 C₆H₅OH + 2 Na → 2 C₆H₅O⁻Na⁺ + H₂↑
(Sodium phenoxide)

Note: Liquid Q (acetone/ketone) does not have an O–H bond and does not react with sodium metal to liberate H₂.

(c) The reaction of phenol with acetic anhydride to form its acetyl derivative is called Acetylation (Esterification/Acylation).

Balanced equation:

C₆H₅OH + (CH₃CO)₂O → C₆H₅OOCCH₃ + CH₃COOH

(Phenol)   (Acetic anhydride)   (Phenyl acetate)   (Acetic acid)

The –OH group of phenol is acetylated by acetic anhydride, forming phenyl acetate (an ester) and acetic acid as the by-product.

Alcohols, Phenols and Ethers: Class 12 Chemistry Practice Questions

Want unlimited practice on Alcohols, Phenols and Ethers?