CBSE Important Questions Class 12 Chemistry Chapter 7 – Alcohols, Phenols and Ethers

Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers is part of the CBSE Class 12 Chemistry syllabus. These three families of compounds all contain oxygen. Students learn their classification and nomenclature, methods of preparation, physical properties and the reasons behind them, acidity of alcohols and phenols, the chemical reactions of each class, and important reactions such as Kolbe's reaction, Reimer-Tiemann reaction and Williamson synthesis.

These CBSE important questions for Class 12 Chemistry Chapter 7 cover why ethers have lower boiling points than alcohols, the acidic strength of phenol compared with alcohol, the effect of substituents on the acidity of phenols, dehydration of alcohols, Williamson ether synthesis, the Lucas test for distinguishing alcohols, and the electrophilic substitution reactions of phenol. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. State why ethers have lower boiling points than alcohols of comparable molecular mass.

Answer: The difference is due to hydrogen bonding.
Alcohols contain an −OH group. The hydrogen attached to the highly electronegative oxygen can form strong intermolecular hydrogen bonds with neighbouring alcohol molecules. A great deal of energy is needed to break these bonds, so the boiling point is high.
Ethers have the structure R−O−R. There is no hydrogen attached to oxygen, so ether molecules cannot hydrogen bond with each other. They are held together only by weak dipole-dipole forces.
Less energy is needed to separate them, so ethers boil at much lower temperatures. For example, ethanol boils at 351 K while dimethyl ether, of the same molecular mass, boils at just 248 K.

Q2. Why is phenol more acidic than ethanol?

Answer: The answer lies in the stability of the ion left behind after losing the proton.
Phenol gives the phenoxide ion. The negative charge on the oxygen is delocalised over the benzene ring through resonance, spreading it over several carbon atoms. A spread-out charge is much more stable.
Ethanol gives the ethoxide ion. Here the negative charge stays concentrated on the single oxygen atom. Worse, the ethyl group is electron releasing (+I effect), which pushes more electron density onto the already negative oxygen and destabilises it further.
Since the phenoxide ion is far more stable, phenol gives up its proton much more readily and is therefore the stronger acid.

Q3. Arrange the following in increasing order of acidic strength: phenol, p-nitrophenol, p-cresol (p-methylphenol).

Answer: p-cresol < phenol < p-nitrophenol
• p-nitrophenol is the strongest. The −NO₂ group is electron withdrawing, so it pulls electron density away and stabilises the phenoxide ion, making the proton easier to lose.
• Phenol sits in the middle as the reference.
• p-cresol is the weakest. The −CH₃ group is electron releasing, so it pushes electron density towards the oxygen and destabilises the phenoxide ion.
General rule: electron-withdrawing groups increase the acidity of phenols, while electron-releasing groups decrease it.

Q4. What happens when ethanol is heated with concentrated H₂SO₄ at 443 K and at 413 K?

Answer: The temperature decides the product.
At 443 K (higher temperature): Dehydration gives an alkene.
CH₃CH₂OH → CH₂=CH₂ + H₂O
Ethene is formed by removing a molecule of water from a single alcohol molecule (intramolecular dehydration).
At 413 K (lower temperature): Dehydration gives an ether.
2CH₃CH₂OH → CH₃CH₂-O-CH₂CH₃ + H₂O
Diethyl ether is formed by removing water from between two alcohol molecules (intermolecular dehydration).
So the same reactants give completely different products simply by changing the temperature.

Q5. What is the Williamson ether synthesis? Why does it fail with tertiary alkyl halides?

Answer: Williamson synthesis prepares an ether by heating an alkyl halide with sodium alkoxide:
R-ONa + R′-X → R-O-R′ + NaX
It proceeds by an SN2 mechanism, in which the alkoxide ion attacks the carbon of the alkyl halide from the back.
Why it fails with tertiary halides: A tertiary carbon is crowded by three bulky alkyl groups, which cause steric hindrance and block the approaching alkoxide.
Also, the alkoxide ion is a strong base as well as a nucleophile. Unable to attack the crowded carbon, it instead removes a β-hydrogen, and the reaction becomes an elimination giving an alkene instead of the ether.
Rule: for the best results use a primary alkyl halide with a bulky alkoxide, not the other way round.

Q6. How will you distinguish between primary, secondary and tertiary alcohols using the Lucas test?

Answer: Lucas reagent is a mixture of concentrated HCl and anhydrous ZnCl₂. The alcohol is shaken with it at room temperature and the appearance of turbidity (cloudiness, from the insoluble alkyl chloride) is timed.
• Tertiary alcohol: turbidity appears immediately, because it forms a very stable tertiary carbocation.
• Secondary alcohol: turbidity appears in about 5 to 10 minutes.
• Primary alcohol: no turbidity at room temperature; heating is needed.
The order simply follows the stability of the carbocation formed: 3° > 2° > 1°.

Q7. What is Kolbe's reaction? Write the equation.

Answer: In Kolbe's reaction, phenol is converted into salicylic acid.
Sodium phenoxide is heated with carbon dioxide at about 400 K under 4 to 7 atmospheres of pressure, and the product is then treated with acid:
C₆H₅ONa + CO₂ → sodium salicylate → (on acidification) salicylic acid
Why it works: The phenoxide ion is more reactive than phenol itself towards electrophilic substitution. CO₂ acts as a weak electrophile and attacks the ortho position.
Importance: Salicylic acid is the starting material for making aspirin, which is why this reaction matters in industry.

Q8. What is the Reimer-Tiemann reaction?

Answer: In the Reimer-Tiemann reaction, phenol is treated with chloroform (CHCl₃) in the presence of aqueous NaOH at 340 K. On acidifying the product, salicylaldehyde (2-hydroxybenzaldehyde) is obtained.
How it works: The chloroform reacts with NaOH to give dichlorocarbene (:CCl₂), which is the actual electrophile. It attacks the ortho position of the phenoxide ring. The intermediate formed is then hydrolysed to give the −CHO group.
So this reaction introduces a −CHO group at the ortho position of phenol.

Q9. Why is the C-O-C bond angle in ethers slightly greater than the tetrahedral angle?

Answer: The bond angle in dimethyl ether is about 111.7°, slightly more than the normal tetrahedral angle of 109.5°.
Reason: The two alkyl groups attached to the oxygen are bulky. There is repulsion between these two bulky groups, which pushes them apart and widens the angle between them.
This repulsion is stronger than the lone pair-bond pair repulsion that would otherwise squeeze the angle down, so the net effect is a slightly larger angle.

Q10. Why does phenol undergo electrophilic substitution more easily than benzene? Name the positions where substitution takes place.

Answer: In phenol, the −OH group has a lone pair of electrons on the oxygen. This lone pair goes into resonance with the benzene ring, pushing electron density into the ring.
A ring with more electron density is more attractive to electrophiles, which are electron-seeking species. So phenol is activated and reacts faster than benzene.
Positions: The resonance structures show that the extra electron density builds up mainly at the ortho and para positions. So the −OH group is ortho-para directing.
For example, phenol reacts with bromine water to give 2,4,6-tribromophenol as a white precipitate, even without a catalyst — a reaction benzene will not undergo at all under the same conditions.

Q11. How are alcohols prepared from alkenes? Name the two methods.

Answer: Method 1: Acid-catalysed hydration
The alkene is treated with water in the presence of dilute H₂SO₄. The addition follows Markovnikov's rule, so the −OH goes to the more substituted carbon.
CH₃-CH=CH₂ + H₂O → CH₃-CH(OH)-CH₃ (propan-2-ol)
Method 2: Hydroboration-oxidation
The alkene is treated with diborane (B₂H₆) and the product is then oxidised with alkaline H₂O₂. The addition is anti-Markovnikov, so the −OH goes to the less substituted carbon.
CH₃-CH=CH₂ → CH₃-CH₂-CH₂-OH (propan-1-ol)
So the two methods give different products from the same alkene, which makes both useful.

Q12. What happens when phenol is treated with (a) bromine water and (b) dilute nitric acid?

Answer: (a) With bromine water: A white precipitate of 2,4,6-tribromophenol is formed at once.
C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr
All three positions are brominated in one go, with no catalyst needed. This shows how strongly the −OH group activates the ring. This reaction is used as a test for phenol.
(With bromine in a non-polar solvent such as CS₂ at low temperature, only the mono-brominated product is obtained.)
(b) With dilute nitric acid: A mixture of o-nitrophenol and p-nitrophenol is formed.
These two can be separated by steam distillation. The ortho isomer forms an intramolecular hydrogen bond and so is volatile and distils over; the para isomer forms intermolecular hydrogen bonds and stays behind.

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