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

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

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Question 1

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Classify alcohols as primary, secondary and tertiary with one example each.

Primary alcohol: –OH is attached to a carbon bearing two hydrogen atoms and one alkyl group; example: propan-1-ol (CH3CH2CH2OH).
Secondary alcohol: –OH is on a carbon bonded to two alkyl groups; example: propan-2-ol (CH3CHOHCH3).
Tertiary alcohol: –OH is on a carbon bonded to three alkyl groups; example: 2-methylpropan-2-ol.

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Question 2

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Distinguish between allylic and benzylic alcohols with one example each.

Allylic alcohol: the –OH group is attached to an sp3 carbon adjacent to a C=C double bond; example: prop-2-en-1-ol (CH2=CH–CH2OH).
Benzylic alcohol: the –OH group is attached to an sp3 carbon next to an aromatic ring; example: phenylmethanol (C6H5CH2OH).
Both allylic and benzylic alcohols may themselves be primary, secondary or tertiary depending on the substitution at the carbon bearing –OH.

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Question 3

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How are symmetrical and unsymmetrical ethers distinguished? Give one example of each.

Symmetrical (simple) ether: both alkyl or aryl groups attached to the oxygen atom are identical; example: diethyl ether (C2H5OC2H5).
Unsymmetrical (mixed) ether: the two groups on oxygen are different; example: ethylmethyl ether (C2H5OCH3).
The classification is based solely on the identity of substituents on oxygen and is not related to the size or branching of the groups.

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Question 4

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Write the IUPAC name of: (i) CH3CH2CH(OH)CH3 and (ii) HO–CH2–CH(OH)–CH2OH.

(i) The longest chain has 4 carbons; –OH is at C-2; IUPAC name: butan-2-ol.
(ii) The chain has 3 carbons with –OH at C-1, C-2 and C-3; IUPAC name: propane-1,2,3-triol (glycerol).
For polyhydric alcohols, the terminal 'e' of the parent alkane is retained before the suffix 'ol', and the number of –OH groups is indicated by di, tri, etc.

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Question 5

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How is phenol prepared from chlorobenzene? Write the reaction conditions.

Chlorobenzene is fused with NaOH at 623 K and 300 atm pressure to form sodium phenoxide.
The sodium phenoxide is then acidified with hydrochloric acid to liberate phenol.
This method involves nucleophilic substitution at the aryl carbon, which requires drastic conditions because aryl C–Cl bonds are strong.

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Question 6

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What is the role of diborane in the hydroboration-oxidation method of preparing alcohols?

Diborane (BH3)2 adds to the alkene such that boron attaches to the sp2 carbon carrying more hydrogen atoms, opposite to Markovnikov's rule.
The resulting trialkylborane is then oxidised by H2O2 in the presence of aqueous NaOH to give the alcohol.
The net result is anti-Markovnikov addition of water; the method provides excellent yield of the alcohol.

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Question 7

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State the product obtained when (i) an aldehyde and (ii) a ketone are treated with NaBH4.

Aldehydes are reduced by NaBH4 to give primary alcohols; for example, ethanal gives ethanol.
Ketones are reduced by NaBH4 to give secondary alcohols; for example, propanone gives propan-2-ol.
NaBH4 is a mild and selective reducing agent that reduces carbonyl compounds but does not reduce C=C bonds under normal conditions.

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Question 8

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What type of alcohol is produced when a Grignard reagent reacts with (i) methanal and (ii) a ketone?

Reaction of any Grignard reagent (RMgX) with methanal (HCHO) gives a primary alcohol after hydrolysis, as only one carbon from the Grignard adds to a one-carbon aldehyde.
Reaction of a Grignard reagent with a ketone (RCOR') gives a tertiary alcohol after hydrolysis, because the carbon bearing the carbonyl group already has two alkyl substituents.
The general mechanism involves nucleophilic addition of the carbanion (R–) of RMgX to the carbonyl carbon, forming a metal alkoxide adduct, which is then hydrolysed.

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Question 9

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What is the C–O bond length in phenol and how does it compare to methanol? Give reason.

The C–O bond length in phenol is 136 pm, shorter than the 142 pm value in methanol.
The shorter bond arises because the –OH group in phenol is attached to an sp2 hybridised carbon of the aromatic ring, and the unshared electron pair on oxygen conjugates with the ring, giving partial double bond character to the C–O bond.
The sp2 hybridised carbon in phenol is more electronegative than the sp3 carbon in methanol, which also contracts the bond.

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Question 10

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Why are boiling points of alcohols higher than those of ethers of comparable molecular mass?

Alcohols form strong intermolecular hydrogen bonds through their –OH groups, requiring considerable energy to break these associations during boiling.
Ethers have no O–H bond and therefore cannot form hydrogen bonds among themselves; only weak van der Waals interactions are present.
Consequently, alcohols have significantly higher boiling points; for example, ethanol (b.p. 351 K) vs. methoxymethane (b.p. 248 K) despite similar molecular masses.

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Question 11

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Explain the order of acid strength: tertiary alcohol < secondary alcohol < primary alcohol.

Acid strength of alcohols depends on the stability of the alkoxide ion formed after loss of a proton.
Alkyl groups are electron-donating; more alkyl groups on the carbon bearing –O– increase electron density on oxygen, destabilising the alkoxide ion and reducing the tendency to ionise.
Therefore, tertiary alcohols (three alkyl groups) are weakest acids, secondary alcohols are intermediate, and primary alcohols (one alkyl group) are the strongest acids among the three classes.

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Question 12

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Why is phenol a stronger acid than ethanol?

In ethanol, the alkoxide ion formed on ionisation bears the negative charge localised on oxygen, making it less stable.
In phenol, the phenoxide ion formed on ionisation has its negative charge delocalised over the benzene ring through resonance (five resonance structures), greatly stabilising the anion.
Greater stabilisation of the conjugate base phenoxide ion shifts the ionisation equilibrium to the right, making phenol (pKa ~10) far more acidic than ethanol (pKa ~16).

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Question 13

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What is the Lucas test? What is the observation for a tertiary alcohol?

The Lucas test uses Lucas reagent (concentrated HCl and anhydrous ZnCl2) to distinguish primary, secondary and tertiary alcohols based on their rate of reaction with HCl.
Alcohols are soluble in the reagent, while the alkyl chloride formed is immiscible, producing turbidity.
Tertiary alcohols react immediately to produce turbidity at room temperature because they form stable tertiary carbocations, whereas primary alcohols show no turbidity at room temperature.

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Chapter 7: Alcohols, Phenols and Ethers | Short Questions

Classify alcohols as primary, secondary and tertiary with one example each.

Distinguish between allylic and benzylic alcohols with one example each.

How are symmetrical and unsymmetrical ethers distinguished? Give one example of each.

Write the IUPAC name of: (i) CH3CH2CH(OH)CH3 and (ii) HO–CH2–CH(OH)–CH2OH.

How is phenol prepared from chlorobenzene? Write the reaction conditions.

What is the role of diborane in the hydroboration-oxidation method of preparing alcohols?

State the product obtained when (i) an aldehyde and (ii) a ketone are treated with NaBH4.

What type of alcohol is produced when a Grignard reagent reacts with (i) methanal and (ii) a ketone?

What is the C–O bond length in phenol and how does it compare to methanol? Give reason.

Why are boiling points of alcohols higher than those of ethers of comparable molecular mass?

Explain the order of acid strength: tertiary alcohol < secondary alcohol < primary alcohol.

Why is phenol a stronger acid than ethanol?

What is the Lucas test? What is the observation for a tertiary alcohol?

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