Chapter 7: Alcohols, Phenols and Ethers Case Study Questions | chemistry Class 12 CBSE

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Chapter 7: Alcohols, Phenols and Ethers Case Study Questions | chemistry Class 12 CBSE | ByteNotes.in

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A chemistry student is studying the preparation of alcohols in the laboratory. She observes that when propene reacts with dilute sulphuric acid (acid-catalysed hydration), the product formed is propan-2-ol, not propan-1-ol. She also notes that when the same alkene undergoes hydroboration-oxidation using diborane followed by hydrogen peroxide in alkaline medium, propan-1-ol is obtained. The student realises that the two reactions follow different mechanisms and give different regiochemical outcomes. The acid-catalysed hydration proceeds via a carbocation intermediate, while hydroboration-oxidation is a concerted addition that follows anti-Markovnikov selectivity. Both methods are important in synthetic organic chemistry for the selective preparation of primary and secondary alcohols from alkenes.

Question 1: What is the product formed when propene undergoes acid-catalysed hydration? Name the rule that governs this addition.

The product is propan-2-ol (secondary alcohol).
The addition follows Markovnikov's rule, where the –OH group attaches to the carbon bearing the greater number of substituents (more substituted carbon).

Question 2: Explain why hydroboration-oxidation of propene gives propan-1-ol. What reagents are used in this reaction?

Hydroboration-oxidation gives propan-1-ol because boron attaches to the sp² carbon bearing more hydrogen atoms (less substituted carbon), placing –OH at the terminal carbon — opposite to Markovnikov's rule.
Reagents used: Step 1 — diborane (BH₃)₂; Step 2 — H₂O₂ in aqueous NaOH.

Question 3: Describe the three-step mechanism of acid-catalysed hydration of propene to form propan-2-ol.

Step 1: Protonation of propene by H₃O⁺ (electrophilic attack) forms a secondary carbocation (CH₃–⁺CH–CH₃) — the more stable carbocation — and water.
Step 2: Nucleophilic attack of water on the carbocation gives an oxonium ion intermediate (CH₃–CH(OH₂⁺)–CH₃).
Step 3: Deprotonation of the oxonium ion by water releases H₃O⁺ (regenerating the acid catalyst) and yields propan-2-ol.

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Case Set 2

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In an organic chemistry practical, students are tasked with preparing phenol using different starting materials. One group starts with chlorobenzene and fuses it with NaOH at 623 K and 320 atm pressure, followed by acidification to obtain phenol. Another group prepares phenol from cumene (isopropylbenzene) by oxidising it with air to form cumene hydroperoxide, which is then treated with dilute acid to yield phenol and acetone as a by-product. A third group uses the diazonium salt route — aniline is treated with NaNO₂ and HCl at 273–278 K to form benzene diazonium chloride, which on warming with water yields phenol. The teacher points out that the cumene process is the most commercially significant method worldwide.

Question 1: Write the chemical equation for the preparation of phenol from chlorobenzene.

Chlorobenzene + NaOH → Sodium phenoxide (623 K, 300 atm) → Phenol (on acidification with HCl).
C₆H₅Cl + NaOH → C₆H₅ONa + HCl → C₆H₅OH (Sodium phenoxide is acidified to get phenol).

Question 2: What is the by-product formed in the cumene process for phenol manufacture, and why is this process commercially preferred?

Acetone (CH₃COCH₃) is the valuable by-product formed along with phenol in the cumene process.
It is commercially preferred because both phenol and acetone are industrially important chemicals obtained simultaneously, making the process economically efficient. Most worldwide production of phenol uses this method.

Question 3: Describe the preparation of phenol from aniline via diazonium salt. Write equations and mention conditions carefully.

Step 1 (Diazotisation): Aniline (C₆H₅NH₂) is treated with NaNO₂ + HCl at 273–278 K to form benzene diazonium chloride (C₆H₅N₂⁺Cl⁻).
Step 2 (Hydrolysis): Benzene diazonium chloride is warmed with water (or treated with dilute acid), releasing N₂ gas and HCl, and yielding phenol (C₆H₅OH).
Overall: C₆H₅NH₂ → C₆H₅N₂⁺Cl⁻ → C₆H₅OH + N₂ + HCl. Temperature must be kept low (273–278 K) during diazotisation to prevent decomposition of the diazonium salt.

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Case Set 3

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A student compares the boiling points of compounds of similar molecular masses: ethanol (MW = 46, bp = 351 K), methoxymethane/dimethyl ether (MW = 46, bp = 248 K), and propane (MW = 44, bp = 231 K). She is puzzled that although all three have similar molecular masses, their boiling points differ significantly. Her teacher explains that the differences are due to intermolecular forces. Ethanol molecules form strong intermolecular hydrogen bonds through the O–H group, requiring more energy to overcome. Methoxymethane, lacking an O–H bond, cannot form hydrogen bonds with itself, though it can accept hydrogen bonds from water. Propane, being nonpolar, can only exert weak van der Waals forces. The teacher also explains the concept of solubility of alcohols in water and the role of alkyl group size.

Question 1: Why does ethanol have a much higher boiling point than methoxymethane despite having the same molecular mass?

Ethanol has an –OH group that participates in intermolecular hydrogen bonding (O–H···O), forming strong associations between molecules that require more energy (higher temperature) to break.
Methoxymethane lacks an O–H bond and cannot form hydrogen bonds with itself, so only weak van der Waals forces operate, resulting in a much lower boiling point.

Question 2: Explain why methoxymethane is miscible with water to almost the same extent as butan-1-ol, even though it cannot form hydrogen bonds with itself.

Methoxymethane has two lone pairs on the oxygen atom, which can form hydrogen bonds with water molecules (R–O:···H–OH), making it soluble in water.
Butan-1-ol also forms hydrogen bonds with water through its –OH group. Since both compounds interact with water molecules similarly, their water miscibility is comparable (~7.5 and 9 g per 100 mL respectively).

Question 3: How does branching in the carbon chain and increase in carbon chain length affect the boiling points and water solubility of alcohols? Explain with reasoning.

Effect of chain length: Boiling points of alcohols increase with increasing number of carbon atoms because longer chains have greater van der Waals (London dispersion) forces due to increased surface area.
Effect of branching: Boiling points decrease with branching because branched molecules have a more compact (spherical) shape, reducing the surface area available for van der Waals interactions.
Effect on solubility: Solubility in water decreases with increase in the size of the alkyl/aryl (hydrophobic) group, because the non-polar part of the molecule increasingly dominates over the polar –OH group, reducing the ability to form hydrogen bonds with water.

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During an experiment on acidity, a student tests phenol, ethanol, and water with sodium metal. All three liberate hydrogen gas. However, when phenol is tested with aqueous NaOH, it dissolves completely forming sodium phenoxide, while ethanol does not react with NaOH under normal conditions. The pKa of phenol is 10.0, that of ethanol is 15.9, and water is 15.7. The teacher draws resonance structures of the phenoxide ion to explain its exceptional stability. In substituted phenols, the student notices that o-nitrophenol and p-nitrophenol are more acidic than phenol itself (pKa ≈ 7.1–7.2), while o-cresol and p-cresol are less acidic (pKa ≈ 10.2).

Question 1: Why is phenol a stronger acid than ethanol? Give one piece of experimental evidence.

Phenol is stronger than ethanol because the phenoxide ion formed after losing a proton is stabilised by delocalisation of the negative charge over the benzene ring (resonance), while in ethoxide ion the negative charge is localised on oxygen only.
Experimental evidence: Phenol reacts with aqueous NaOH to form sodium phenoxide, whereas ethanol does not react with NaOH — showing phenol is acidic enough to neutralise a strong base.

Question 2: Why does the presence of a nitro group at ortho or para position increase the acidity of phenol, while a methyl group decreases it?

Nitro group is an electron-withdrawing group (EWG). At ortho/para positions, it further stabilises the phenoxide ion by effectively delocalising the negative charge through resonance — making the phenol easier to ionise (more acidic).
Methyl group is an electron-releasing group (ERG). It increases electron density on oxygen, stabilising the O–H bond and making it harder to release the proton — thus decreasing acidity. Cresols are therefore less acidic than phenol.

Question 3: Explain why alcohols are weaker acids than water, and also why alcohols act as Brønsted bases. Use electronic reasoning.

Alcohols are weaker acids than water: Alkyl groups (–CH₃, –C₂H₅) are electron-releasing groups. They increase electron density on oxygen, decreasing the polarity of the O–H bond and making it harder to release a proton. This is confirmed by the reaction: R–O⁻ + H₂O → R–OH + OH⁻, showing water is a better proton donor (stronger acid) than alcohol. The order of acidity is: primary > secondary > tertiary alcohol.
Alcohols as Brønsted bases: Oxygen in alcohols has two lone pairs of electrons, enabling it to accept a proton from strong acids — acting as a proton acceptor (base). For example, R–OH + H⁺ → R–OH₂⁺ (protonated alcohol/oxonium ion). This base character is essential in reactions like dehydration and substitution with HX.

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Case Set 5

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In an organic synthesis lab, students perform dehydration of alcohols. When ethanol is heated with concentrated H₂SO₄ at 443 K, ethene is produced. But when the same reaction is carried out at 413 K, ethoxyethane (diethyl ether) is the major product. A student notices that 2-methylpropan-2-ol (t-butanol) dehydrates very easily at 358 K with 20% H₃PO₄, while propan-1-ol requires harsher conditions. Another observation is that the dehydration of secondary and tertiary alcohols tends to give the more substituted alkene as the major product. The teacher notes that the mechanism involves carbocation formation and that tertiary carbocations are most stable, explaining the ease of dehydration.

Question 1: Why does the dehydration of alcohols follow the order: tertiary > secondary > primary? Explain.

Dehydration involves formation of a carbocation intermediate (rate-determining step). Tertiary carbocations (3°) are most stable due to hyperconjugation and inductive effect of three alkyl groups, and form most easily.
Therefore, tertiary alcohols dehydrate under the mildest conditions, followed by secondary, and primary alcohols require the most harsh conditions. The ease of dehydration follows: tertiary > secondary > primary.

Question 2: How does temperature control determine whether ether or alkene is obtained from ethanol with H₂SO₄? Explain the mechanism for ether formation.

At 443 K, elimination is favoured and ethene (alkene) is the product. At 413 K, substitution is favoured and ethoxyethane (ether) is the product.
Ether formation mechanism (SN2): Ethanol is first protonated by H⁺ to form ethyloxonium ion (CH₃CH₂–O⁺H₂). A second molecule of ethanol acts as a nucleophile and attacks the protonated ethanol in an SN2 reaction, displacing water and forming ethoxyethane, which then loses a proton.

Question 3: Write the three-step mechanism for the acid-catalysed dehydration of ethanol to ethene, and explain why the acid is regenerated.

Step 1 (Fast): Ethanol is protonated by H⁺ to form the ethyloxonium ion (CH₃CH₂–O⁺H₂). This step is fast and reversible.
Step 2 (Slow — rate determining): The C–O bond breaks heterolytically; water leaves as a leaving group and a carbocation (CH₃CH₂⁺) is formed. This is the slowest step.
Step 3 (Fast): A proton is eliminated from the carbocation by a base (water molecule), forming ethene (CH₂=CH₂). The H⁺ released in Step 3 is the same acid used in Step 1, so the acid is regenerated (catalytic cycle). To drive equilibrium to the right, ethene is removed as formed.

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Chapter 7: Alcohols, Phenols and Ethers | Case Study Questions

Passage

Question 1: What is the product formed when propene undergoes acid-catalysed hydration? Name the rule that governs this addition.

Question 2: Explain why hydroboration-oxidation of propene gives propan-1-ol. What reagents are used in this reaction?

Question 3: Describe the three-step mechanism of acid-catalysed hydration of propene to form propan-2-ol.

Passage

Question 1: Write the chemical equation for the preparation of phenol from chlorobenzene.

Question 2: What is the by-product formed in the cumene process for phenol manufacture, and why is this process commercially preferred?

Question 3: Describe the preparation of phenol from aniline via diazonium salt. Write equations and mention conditions carefully.

Passage

Question 1: Why does ethanol have a much higher boiling point than methoxymethane despite having the same molecular mass?

Question 2: Explain why methoxymethane is miscible with water to almost the same extent as butan-1-ol, even though it cannot form hydrogen bonds with itself.

Question 3: How does branching in the carbon chain and increase in carbon chain length affect the boiling points and water solubility of alcohols? Explain with reasoning.

Passage

Question 1: Why is phenol a stronger acid than ethanol? Give one piece of experimental evidence.

Question 2: Why does the presence of a nitro group at ortho or para position increase the acidity of phenol, while a methyl group decreases it?

Question 3: Explain why alcohols are weaker acids than water, and also why alcohols act as Brønsted bases. Use electronic reasoning.

Passage

Question 1: Why does the dehydration of alcohols follow the order: tertiary > secondary > primary? Explain.

Question 2: How does temperature control determine whether ether or alkene is obtained from ethanol with H₂SO₄? Explain the mechanism for ether formation.

Question 3: Write the three-step mechanism for the acid-catalysed dehydration of ethanol to ethene, and explain why the acid is regenerated.

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