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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student treated propene with sulphuric acid and water. He expected propan-1-ol but obtained propan-2-ol instead. Explain why this happened and state the rule governing this outcome.

Acid-catalysed hydration of propene follows Markovnikov's rule: the proton (electrophile) adds to the carbon with more hydrogen atoms (C-1), generating a secondary carbocation at C-2, which is more stable than a primary carbocation at C-1.
Water then attacks the secondary carbocation at C-2, and loss of a proton gives propan-2-ol; so the –OH group adds to the more-substituted carbon, which is C-2 in propene.
To obtain propan-1-ol (anti-Markovnikov product), the student should use hydroboration-oxidation with diborane followed by H2O2/NaOH, where boron adds to C-1 and –OH is ultimately placed at C-1.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

With reference to the reaction of 3-methylbutan-2-ol with HBr, explain why the product obtained is 2-bromo-2-methylbutane rather than a simple secondary bromide at C-2.

When 3-methylbutan-2-ol is protonated, the OH2+ leaves from C-2, generating a secondary carbocation at C-2.
The secondary carbocation at C-2 rearranges to a more stable tertiary carbocation at C-3 by a hydride shift from C-3 to C-2, as tertiary carbocations are more stable than secondary ones.
Bromide ion then attacks the tertiary carbocation at C-3, giving 2-bromo-2-methylbutane (a tertiary bromide); carbocation rearrangement via hydride shift is the key step explaining the non-Markovnikov product.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

In a laboratory, you are given butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. Describe how you would use the Lucas test to identify each compound.

Dissolve a few drops of each alcohol in Lucas reagent (conc. HCl + anhydrous ZnCl2) at room temperature and observe turbidity.
2-Methylpropan-2-ol (tertiary) will produce immediate turbidity at room temperature because it forms a stable tertiary carbocation rapidly via SN1; the insoluble tert-butyl chloride separates.
Butan-2-ol (secondary) will develop turbidity slowly (within 5 minutes), while butan-1-ol (primary) will show no turbidity at room temperature even after prolonged standing, allowing clear identification of all three.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A factory produces salicylic acid from phenol. Describe the Kolbe's reaction used for this purpose, explaining why phenoxide ion is used rather than phenol itself.

Phenol is first treated with NaOH to generate sodium phenoxide; the phenoxide ion has a full negative charge on oxygen that is delocalised into the ring, making the ring far more nucleophilic than in phenol itself.
Sodium phenoxide is then heated with CO2 under pressure (Kolbe's reaction); the highly activated ring undergoes electrophilic attack by CO2 at the ortho position, forming the carboxylated product.
After acidification, ortho-hydroxybenzoic acid (salicylic acid) is obtained; salicylic acid is a precursor to aspirin (acetylsalicylic acid), making this reaction industrially important for pharmaceutical manufacturing.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

When methanol-poisoned patients are treated with intravenous ethanol infusions, the poisoning effects are neutralised. Explain the biochemical reason for this treatment.

In the body, methanol is oxidised first to methanal (formaldehyde) and then to methanoic acid (formic acid); it is methanoic acid that causes blindness and death by disrupting cellular metabolism.
When ethanol is administered intravenously, it competes for the same oxidative enzyme (alcohol dehydrogenase) responsible for converting both methanol and ethanol to their aldehydes.
The enzyme preferentially oxidises ethanol; by saturating (swamping) the enzyme, the oxidation of methanol to the toxic methanal is slowed, allowing the kidneys sufficient time to excrete methanol unchanged before it can be converted to its toxic metabolites.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

In wine making, the alcohol content cannot exceed 14% by fermentation alone. Explain the reason, and describe what happens if air enters the fermentation mixture.

The enzyme zymase (present in yeast) catalyses the conversion of glucose to ethanol and CO2; however, its catalytic activity is inhibited (denatured) when the ethanol concentration in the mixture exceeds approximately 14%, so fermentation stops automatically beyond this concentration.
If air (oxygen) enters the anaerobic fermentation mixture, the oxygen oxidises the ethanol formed to ethanoic acid (acetic acid); ethanoic acid has a sour taste and destroys the flavour of the alcoholic beverage.
To obtain ethanol concentrations above 14% (as in spirits), additional distillation of the fermentation mixture is required, which concentrates the ethanol by fractional distillation.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A chemist wants to prepare diethyl ether from ethanol. She heats ethanol with conc. H2SO4 at 413 K. Explain the mechanism and state why the temperature must be kept at 413 K rather than 443 K.

At 413 K, protonation of ethanol by H2SO4 gives the ethyl oxonium ion; a second molecule of ethanol attacks the protonated ethanol (SN2 bimolecular mechanism), displacing water to give diethyl ether and water; this is an intermolecular dehydration.
At 443 K, the rate-determining step shifts towards intramolecular E2 elimination: the protonated ethanol loses a proton from the adjacent carbon to give ethene (alkene) instead of ether; higher temperature favours the formation of the more thermodynamically stable alkene product.
Therefore, maintaining the temperature at 413 K ensures the kinetically controlled SN2 ether-forming pathway predominates; exceeding this temperature reverses product selectivity towards alkene.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

Explain why o-nitrophenol can be separated from p-nitrophenol by steam distillation, while p-nitrophenol remains in the residue.

Ortho-nitrophenol forms an intramolecular hydrogen bond between the –OH and the adjacent nitro group oxygen within the same molecule; this internal H-bonding blocks the formation of intermolecular hydrogen bonds, resulting in a lower boiling point and making o-nitrophenol steam-volatile.
Para-nitrophenol cannot form intramolecular hydrogen bonds because the –OH and –NO2 groups are on opposite sides of the ring; instead, p-nitrophenol molecules associate strongly through intermolecular hydrogen bonding, raising the boiling point and making it non-volatile in steam.
During steam distillation, only o-nitrophenol distils over with steam, while p-nitrophenol remains in the flask; this allows clean separation of the two isomers produced in the nitration of phenol with dilute HNO3.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A student attempts to prepare t-butyl ethyl ether by reacting sodium ethoxide with t-butyl bromide. The reaction gives an alkene, not an ether. Explain the observation and suggest an alternative route to the ether.

Sodium ethoxide (C2H5ONa) is a strong base as well as a nucleophile; when it reacts with t-butyl bromide (a tertiary halide), the bulky tertiary carbon makes SN2 attack very difficult, while the base abstracts a β-hydrogen preferentially via E2 elimination.
The result is 2-methylpropene (isobutylene) and no ether is formed; tertiary alkyl halides invariably give elimination products when treated with strong bases/nucleophiles.
The correct approach uses the tertiary alkoxide: treat t-butanol with sodium metal to form sodium t-butoxide, then react it with ethyl bromide (primary halide) by SN2 — (CH3)3CONa + CH3CH2Br → (CH3)3C–O–CH2CH3 + NaBr; primary alkyl halides are susceptible to SN2 and do not undergo elimination readily.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

With reference to the reaction of anisole with bromine in ethanoic acid, explain why no Lewis acid catalyst is required, unlike in the bromination of benzene.

Bromination of benzene requires a Lewis acid catalyst (FeBr3) to polarise the Br–Br bond and generate the electrophile Br+; benzene's electron density alone is insufficient to polarise bromine.
In anisole, the methoxy group (–OCH3) donates electron density into the benzene ring through resonance of oxygen lone pairs; this activates the ring sufficiently so that even the weakly electrophilic Br2 (which is polarised to some extent) can react without any additional catalyst.
The highly electron-rich ortho and para positions in anisole are attacked preferentially, giving mainly para-bromoanisole (90% yield) and minor ortho-bromoanisole, demonstrating both the activating effect and ortho/para directing ability of the methoxy group.

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Chapter 7: Alcohols, Phenols and Ethers | Application Questions

A student treated propene with sulphuric acid and water. He expected propan-1-ol but obtained propan-2-ol instead. Explain why this happened and state the rule governing this outcome.

With reference to the reaction of 3-methylbutan-2-ol with HBr, explain why the product obtained is 2-bromo-2-methylbutane rather than a simple secondary bromide at C-2.

In a laboratory, you are given butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. Describe how you would use the Lucas test to identify each compound.

A factory produces salicylic acid from phenol. Describe the Kolbe's reaction used for this purpose, explaining why phenoxide ion is used rather than phenol itself.

When methanol-poisoned patients are treated with intravenous ethanol infusions, the poisoning effects are neutralised. Explain the biochemical reason for this treatment.

In wine making, the alcohol content cannot exceed 14% by fermentation alone. Explain the reason, and describe what happens if air enters the fermentation mixture.

A chemist wants to prepare diethyl ether from ethanol. She heats ethanol with conc. H2SO4 at 413 K. Explain the mechanism and state why the temperature must be kept at 413 K rather than 443 K.

Explain why o-nitrophenol can be separated from p-nitrophenol by steam distillation, while p-nitrophenol remains in the residue.

A student attempts to prepare t-butyl ethyl ether by reacting sodium ethoxide with t-butyl bromide. The reaction gives an alkene, not an ether. Explain the observation and suggest an alternative route to the ether.

With reference to the reaction of anisole with bromine in ethanoic acid, explain why no Lewis acid catalyst is required, unlike in the bromination of benzene.

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