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Alcohols Phenols and Ethers Questions and Answers

11.

Which of the following is the weakest acid?

Exam Year(s)

KEAM - 2025,

Answer: D

The weakest acid is Ethanol.

Acid strength depends on how stable the anion is after the proton leaves. When phenol loses a proton, the negative charge on the phenoxide ion is delocalised into the benzene ring by resonance. The ethoxide ion has no such resonance, and the ethyl group's +I effect concentrates the charge on oxygen further, so ethanol (pKa about 16) is a much weaker acid than any of the phenols (pKa about 7 to 10).

Among the phenols, p-nitrophenol is the strongest, since the nitro group stabilises the phenoxide further. The cresols carry an electron-donating methyl group and are slightly weaker than phenol itself, but they are still far stronger acids than ethanol.

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Answer: B

Lucas reagent is concentrated HCl with anhydrous \(\mathrm{ZnCl_{2}}\).

Lucas reagent distinguishes primary, secondary and tertiary alcohols. Zinc chloride, a Lewis acid, helps the alcohol ionise, and chloride converts it to an alkyl chloride, which is insoluble and makes the solution turbid. Tertiary alcohols turn cloudy at once, secondary alcohols within about five minutes, and primary alcohols not noticeably at room temperature.

The reagent needs hydrochloric acid because the product is an alkyl chloride. Nitric acid, sulphuric acid, acetic acid and oleum (printed 'Oleium') do not supply chloride ions, so none of those mixtures is Lucas reagent.

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13.

The major product obtained in the dehydration of ethanol in the presence of \(H_{2}SO_{4}\) at 413 K is

Exam Year(s)

KEAM - 2025,

Answer: D

The major product is ethoxyethane.

With concentrated sulphuric acid at 413 K, ethanol is in excess and dehydrates intermolecularly: one protonated ethanol molecule is attacked by another, giving diethyl ether:

\[ 2C_{2}H_{5}OH \to C_{2}H_{5}OC_{2}H_{5} + H_{2}O \]

At a higher temperature (about 443 K) intramolecular dehydration gives ethene instead. Ethanal and ethanoic acid are oxidation products, and ethane and ethyne are not formed by dehydration.

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14.

The product formed in the following reaction is
\(CH_{3} - CH_{2} - CH(CH_{3}) - CH(CH_{3}) - ONa + C_{2}H_{5}Br \to\)

Exam Year(s)

KEAM - 2025,

Answer: A

The product is 2-Ethoxy-3-methylpentane.

This is a Williamson ether synthesis: the alkoxide ion displaces bromide from bromoethane, attaching an ethyl group to the oxygen. The carbon skeleton of the alkoxide is unchanged: CH3CH2CH(CH3)CH(CH3)-O-C2H5.

The longest chain has five carbons; numbering from the end nearer the substituents puts the ethoxy group at C2 and the methyl at C3. '2-Ethoxy-4-methyl' numbers from the wrong end, and the other names describe different skeletons.

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15.

Which of the following compound has the highest boiling point?

Exam Year(s)

KEAM - 2025,

Answer: B

The highest boiling point is that of propan-1-ol.

These compounds have similar molar masses (44 to 60 g/mol), so the type of intermolecular force decides the order. Propan-1-ol molecules form hydrogen bonds through their O-H groups, the strongest attraction here, so it boils at 97°C.

Propanal (49°C) and acetone (56°C) have only dipole-dipole attractions, methoxymethane (−24°C) is a weakly polar ether, and n-butane (about 0°C) has only dispersion forces.

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16.

Phenol is treated with Con.H2SO4 to gives a product 'X" which on treatment with Con.HNO3 gives compound 'Y'. The compounds 'X' and 'Y' are respectively

Exam Year(s)

KEAM - 2025,

Answer: E

X and Y are Phenol-2,4-disulphonic acid and picric acid.

Concentrated sulphuric acid sulphonates the activated phenol ring at both the ortho and para positions, giving phenol-2,4-disulphonic acid. Treating this with concentrated nitric acid replaces the sulphonic acid groups by nitro groups and nitrates the remaining ortho position, giving 2,4,6-trinitrophenol, picric acid.

This route is used because direct nitration of phenol with concentrated nitric acid oxidises much of it. A mixture of 2- and 4-nitrophenol comes from dilute nitric acid on phenol itself, not from the sulphonic acid with concentrated acid.

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17.

Denatured alcohol with colour and foul smell is made now a days by mixing ethanol with

Exam Year(s)

KEAM - 2025,

Answer: C

Denatured alcohol is made by adding CuSO4 and pyridine.

Ethanol for industrial use is made unfit to drink by adding poisonous or unpleasant substances. Copper sulphate gives it a colour, and pyridine a foul smell, so it is easily recognised.

Methanol is also a denaturant (it makes methylated spirit), but it adds neither colour nor smell, which the question specifies. The other pairs are not used for this purpose.

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18.

Which of the following compound contains two primary alcoholic and one secondary alcoholic groups?

Exam Year(s)

KEAM - 2025,

Answer: D

The compound is glycerol.

Glycerol is propane-1,2,3-triol, HOCH2–CH(OH)–CH2OH. The two end –OH groups are on carbons attached to only one other carbon (primary), and the middle one is on a carbon attached to two (secondary).

Ethylene glycol has two primary –OH groups only, isopropyl and 2° butyl alcohol have one secondary –OH, and 3° butyl alcohol has one tertiary –OH.

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Answer: D

The correct order is C6H5OH > m-cresol > p-cresol > C2H5OH.

Phenols are far more acidic than ethanol, because the phenoxide ion is stabilised by resonance with the ring; ethanol comes last. A methyl group donates electrons and destabilises the phenoxide ion, so the cresols are weaker acids than phenol itself.

The methyl group donates most strongly from the para position, where it acts through hyperconjugation as well as induction on the carbon carrying the negative charge; from the meta position it acts by induction only. So p-cresol is weaker than m-cresol (pKa about 10.3 against 10.1, with phenol about 10.0). Options putting ethanol above any phenol, or a cresol above phenol, reverse these effects.

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20.

The products P1, P2, P3 and P4 of the following reactions are

(i) \(\mathrm{C_{6}H_{5}SO_{3}Na \xrightarrow{NaOH,\ H^{+}} P_{1}}\)
(ii) \(\mathrm{C_{6}H_{5}OH \xrightarrow{CO_{2} / NaOH / H^{+}} P_{2}}\)
(iii) \(\mathrm{C_{6}H_{5}OH \xrightarrow{NaOH,\ CH_{3}Cl} P_{3}}\)
(iv) \(\mathrm{o\text{-}HOC_{6}H_{4}COOH \xrightarrow{(CH_{3}CO)_{2}O,\ H^{+}} P_{4}}\)

Exam Year(s)

KEAM - 2025,

Answer: C

The products are P1 = Phenol, P2 = Salicylic acid, P3 = Anisole, P4 = Aspirin.

  1. Sodium benzenesulphonate fused with NaOH gives sodium phenoxide, and acid releases phenol.
  2. Kolbe's reaction: sodium phenoxide with CO2 under pressure, then acid, gives salicylic acid (2-hydroxybenzoic acid).
  3. Williamson synthesis: phenoxide with methyl chloride gives methyl phenyl ether, anisole.
  4. Acetylation of the phenolic OH of salicylic acid with acetic anhydride gives acetylsalicylic acid, aspirin.

Salicylaldehyde comes from the Reimer-Tiemann reaction (chloroform and alkali), not from CO2. Cyclohexanol would need hydrogenation of the ring, which none of these reagents does. Phenyl acetate is the acetate of phenol itself, but the substrate in (iv) is salicylic acid, so its acetate is aspirin.

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