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

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

Long Answer

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Hard

Question 1

Long Form

Compare the acidity of alcohols and phenols, explaining how the structure of the conjugate base and the nature of substituents influence acid strength.

Alcohols are weak acids (pKa ~16) because the alkoxide ion formed on ionisation bears the negative charge localised on oxygen, with no stabilisation by delocalisation; additionally, electron-donating alkyl groups destabilise the ion.
Phenols are far stronger acids (pKa ~10) because the phenoxide ion has its negative charge extensively delocalised over the aromatic ring through five resonance structures, providing exceptional stability to the conjugate base.
Electron-withdrawing groups (e.g., –NO2) at ortho and para positions further stabilise the phenoxide ion by delocalisation, making nitrophenols more acidic than phenol (e.g., o-nitrophenol pKa 7.2, p-nitrophenol pKa 7.1).
Electron-releasing groups (e.g., –CH3) in cresols destabilise the phenoxide ion, making cresols (pKa ~10.2) slightly less acidic than phenol.
Phenol reacts with aqueous NaOH to give sodium phenoxide, confirming its greater acidity over ethanol which does not react with aqueous NaOH.

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Question 2

Long Form

Describe the mechanism of acid-catalysed hydration of an alkene to form an alcohol, and contrast this with the hydroboration-oxidation method in terms of regiochemistry and mechanism.

Acid-catalysed hydration: the alkene is protonated by H3O+ to form a carbocation at the more-substituted carbon (Markovnikov's rule); water then acts as a nucleophile and attacks the carbocation; deprotonation gives the Markovnikov alcohol.
The mechanism involves three steps: protonation of alkene (fast), nucleophilic attack of water on carbocation (gives oxonium ion), then deprotonation to yield the alcohol; the carbocation formation is the slowest (rate-determining) step.
Hydroboration-oxidation: boron of diborane adds to the less-substituted (more hydrogen-bearing) sp2 carbon of the alkene, giving an organoborane; this gives anti-Markovnikov regiochemistry.
The trialkylborane intermediate is oxidised by H2O2/NaOH to liberate the alcohol; the boron is replaced by –OH with retention of configuration; the reaction produces the alcohol in excellent yield.
Key contrast: acid hydration gives Markovnikov product (OH on more-substituted C) while hydroboration-oxidation gives anti-Markovnikov product (OH on less-substituted C), making the two methods complementary for synthesis.

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Long Answer

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Hard

Question 3

Long Form

Explain the mechanism of acid-catalysed dehydration of ethanol to yield ethene, and justify why the ease of dehydration follows the order tertiary > secondary > primary.

Step 1 — Formation of protonated alcohol (ethyl oxonium ion): H+ from conc. H2SO4 adds to the oxygen of ethanol to form CH3CH2–OH2+ (fast and reversible equilibrium).
Step 2 — Formation of carbocation: the protonated alcohol slowly loses water to generate a carbocation; this is the slowest (rate-determining) step, forming CH3CH2+.
Step 3 — Elimination: a base (water or HSO4–) abstracts a proton from the carbon adjacent to the carbocation, giving ethene; the acid catalyst is regenerated; ethene is removed to drive equilibrium forward.
Ease of dehydration is tertiary > secondary > primary because the rate-determining step involves carbocation formation; tertiary carbocations are most stable due to hyperconjugation and inductive stabilisation by three alkyl groups, so they form most readily.
Therefore, tertiary alcohols dehydrate at lower temperatures (e.g., tert-butanol at 358 K) and under milder conditions than primary alcohols (e.g., ethanol at 443 K).

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Medium

Question 4

Long Form

Discuss the oxidation reactions of primary, secondary and tertiary alcohols, naming the reagents and products in each case.

Primary alcohols are oxidised to aldehydes by mild oxidising agents such as CrO3 in anhydrous conditions or PCC (pyridinium chlorochromate); PCC stops oxidation at the aldehyde stage and gives excellent yields.
The same primary alcohols are oxidised directly to carboxylic acids by strong oxidising agents such as acidified KMnO4 or K2Cr2O7, because the intermediate aldehyde is also oxidised.
Secondary alcohols are oxidised to ketones by CrO3 or chromic anhydride; the reaction involves C=O bond formation; ketones are not further oxidised under mild conditions.
Tertiary alcohols have no C–H bond on the carbon bearing –OH and therefore resist oxidation under mild conditions; only under very harsh conditions (hot KMnO4) do they undergo C–C bond cleavage to give a mixture of carboxylic acids.
When vapours of primary or secondary alcohols are passed over heated copper at 573 K, dehydrogenation occurs giving aldehyde or ketone respectively; tertiary alcohols undergo dehydration to alkenes under the same conditions.

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Medium

Question 5

Long Form

Describe four methods for the preparation of phenol from benzene derivatives, giving reagents, conditions and equations in outline.

From haloarenes: chlorobenzene is fused with NaOH at 623 K and 300 atm pressure; sodium phenoxide formed is acidified with HCl to give phenol; this requires drastic conditions due to the stability of the aryl C–Cl bond.
From benzenesulphonic acid: benzene is sulphonated with oleum; the sulphonic acid is heated with molten NaOH to give sodium phenoxide; acidification yields phenol.
From diazonium salts: aniline is treated with NaNO2/HCl at 273–278 K to give benzene diazonium chloride; warming with water or dilute acid hydrolyses the diazonium salt to phenol, releasing N2 and HCl.
From cumene (industrial): cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide; treatment with dilute H2SO4 gives phenol and acetone as co-products; this is the most important commercial source of phenol worldwide.
All four methods convert benzene derivatives to phenol but differ in atom economy, cost and scale; the cumene process is favoured industrially because acetone, a valuable by-product, is also obtained.

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Question 6

Long Form

Analyse the electrophilic aromatic substitution reactions of phenol, explaining the role of the –OH group in directing the incoming electrophile.

The –OH group attached to the benzene ring is an ortho, para director; it donates electron density to the ring through resonance, making ortho and para positions electron-rich and more susceptible to electrophilic attack.
In nitration with dilute HNO3 at 298 K, a mixture of ortho-nitrophenol and para-nitrophenol is formed; concentrated HNO3 gives 2,4,6-trinitrophenol (picric acid).
In halogenation with bromine in low-polarity solvents (CS2, CHCl3) at 273 K, monobromophenols (ortho and para) are formed without needing a Lewis acid catalyst, because the –OH group activates the ring sufficiently to polarise Br2; bromine water gives 2,4,6-tribromophenol (white precipitate).
In Kolbe's reaction, phenoxide ion (even more reactive than phenol) undergoes substitution with CO2 at ortho position to give salicylic acid; in Reimer-Tiemann reaction, CHCl3/NaOH introduces –CHO at ortho position to give salicylaldehyde.
The phenoxide ion generated in alkaline medium is more reactive than phenol because the negative charge on oxygen makes the ring even more electron-rich, facilitating reactions with weak electrophiles like CO2.

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Long Answer

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Question 7

Long Form

Describe the Williamson synthesis of ethers, explaining the mechanism, limitations and how these limitations are overcome with suitable examples.

In Williamson synthesis, a sodium alkoxide (R–ONa) reacts with a primary alkyl halide (R'–X) in an SN2 reaction; the alkoxide acts as a nucleophile attacking the less hindered carbon of the primary halide, displacing the halide ion: R–ONa + R'–X → R–O–R' + NaX.
The method can prepare both symmetrical ethers (same alkoxide and halide) and unsymmetrical ethers (different groups), and extends to phenoxide ions for preparation of aryl ethers.
Limitation: if secondary or tertiary alkyl halides are used, elimination predominates over substitution because alkoxides are also strong bases; for example, C2H5ONa with (CH3)3CBr gives 2-methylpropene, not the ether.
Overcoming limitation: to prepare an ether with a tertiary alkyl group, the tertiary group should be introduced as the alkoxide (from the corresponding alcohol treated with Na), while the primary alkyl halide is used as the other partner; e.g., (CH3)3CONa + C2H5Cl → (CH3)3C–O–C2H5.
Phenols, being more acidic than alcohols, readily form sodium phenoxides with NaOH; these phenoxides react with primary alkyl halides to give aryl alkyl ethers via Williamson synthesis.

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Medium

Question 8

Long Form

Compare the physical properties (boiling point, solubility) of alcohols, ethers and alkanes of comparable molecular mass, accounting for the differences in terms of intermolecular forces.

Boiling points: alcohols have the highest boiling points of the three classes because they form strong intermolecular hydrogen bonds via their –OH groups; e.g., ethanol (MW 46) boils at 351 K, much higher than methoxymethane (MW 46, b.p. 248 K) and propane (MW 44, b.p. 231 K).
Ethers have boiling points comparable to alkanes of similar mass because they lack O–H bonds and thus cannot self-associate by hydrogen bonding; only weak van der Waals forces operate between ether molecules.
Solubility in water: lower molecular mass alcohols are miscible with water in all proportions because –OH groups form strong hydrogen bonds with water; solubility decreases as the alkyl chain grows (greater hydrophobic contribution).
Ethers can also form hydrogen bonds with water through the lone pairs on oxygen; hence their water solubility is similar to that of alcohols of the same molecular mass; for example, ethoxyethane and butan-1-ol are both miscible to approximately 7.5–9 g/100 mL.
Alkanes are essentially insoluble in water because they are non-polar and cannot interact with water molecules through hydrogen bonding; like dissolves like — alkanes dissolve in non-polar solvents.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 9

Long Form

Discuss the preparation of ethers by acid-catalysed dehydration of alcohols and the Williamson synthesis, comparing their scope and mechanism.

Acid-catalysed dehydration: a primary alcohol is heated with conc. H2SO4 at relatively low temperature (e.g., 413 K for ethanol); protonation of alcohol oxygen gives an oxonium ion, which is attacked by a second alcohol molecule (SN2); loss of a proton gives the ether and water.
Scope of dehydration: limited to symmetrical ethers from primary unhindered alcohols; secondary and tertiary alcohols preferentially form alkenes (elimination competes); mixed ethers cannot be easily made by this route.
Williamson synthesis: sodium alkoxide plus primary alkyl halide (SN2); highly versatile for both symmetrical and unsymmetrical ethers; also applicable to aryl ethers using sodium phenoxide.
Scope of Williamson: superior for mixed ethers; however, tertiary alkyl halides still give elimination; the method is best when the halide is primary and the alkoxide provides the more complex group.
Overall, dehydration is a simple, inexpensive method for symmetric ethers from primary alcohols, while Williamson synthesis is the preferred laboratory method for its generality, regioselectivity and applicability to aryl ethers.

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Long Answer

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Hard

Question 10

Long Form

Explain the cleavage of C–O bonds in ethers by hydrogen halides, describing the mechanism for both dialkyl ethers and alkyl aryl ethers with examples.

Ethers are generally inert but react with concentrated HI or HBr at high temperature; HI is most reactive (HI > HBr > HCl) because iodide is the best nucleophile and HI is most acidic.
Mechanism (dialkyl ether): Step 1 — ether oxygen is protonated by HI to form an oxonium ion; Step 2 — iodide ion (nucleophile) attacks the less substituted carbon via SN2, displacing an alcohol molecule; the alcohol from Step 2 then reacts with excess HI to give the second alkyl iodide.
When one alkyl group is tertiary, the carbocation formed by C–O cleavage in Step 2 is a stable tertiary carbocation; the reaction follows SN1 pathway, and iodide attacks the tertiary carbocation to give tertiary alkyl iodide and the alcohol from the other group.
Alkyl aryl ether cleavage: protonation gives a methylphenyl oxonium ion; since the sp2 aryl carbon cannot undergo nucleophilic substitution, iodide breaks only the alkyl–O bond; products are phenol and alkyl iodide; phenol does not react further under these conditions.
The relative stability of the aryl–O bond (partial double bond character due to resonance with the ring) versus the alkyl–O bond (pure single bond) ensures exclusive cleavage at the alkyl–O bond in aryl alkyl ethers.

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Chapter 7: Alcohols, Phenols and Ethers | Long Questions

Compare the acidity of alcohols and phenols, explaining how the structure of the conjugate base and the nature of substituents influence acid strength.

Describe the mechanism of acid-catalysed hydration of an alkene to form an alcohol, and contrast this with the hydroboration-oxidation method in terms of regiochemistry and mechanism.

Explain the mechanism of acid-catalysed dehydration of ethanol to yield ethene, and justify why the ease of dehydration follows the order tertiary > secondary > primary.

Discuss the oxidation reactions of primary, secondary and tertiary alcohols, naming the reagents and products in each case.

Describe four methods for the preparation of phenol from benzene derivatives, giving reagents, conditions and equations in outline.

Analyse the electrophilic aromatic substitution reactions of phenol, explaining the role of the –OH group in directing the incoming electrophile.

Describe the Williamson synthesis of ethers, explaining the mechanism, limitations and how these limitations are overcome with suitable examples.

Compare the physical properties (boiling point, solubility) of alcohols, ethers and alkanes of comparable molecular mass, accounting for the differences in terms of intermolecular forces.

Discuss the preparation of ethers by acid-catalysed dehydration of alcohols and the Williamson synthesis, comparing their scope and mechanism.

Explain the cleavage of C–O bonds in ethers by hydrogen halides, describing the mechanism for both dialkyl ethers and alkyl aryl ethers with examples.

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