The major products formed by heating C6H5CH2-O-C6H5 with HI are

The products are C6H5CH2I and C6H5OH.HI cleaves ethers by protonating the oxygen and then letting iodide attack a carbon. The O-phenyl bond has partial double bond character (the oxygen's lone pair is delocalised into the ring), and iodide cannot attack an sp2 ring carbon, so that bond survives and phenol is formed. The iodide attacks the benzylic CH2 instead, giving benzyl iodide.Options giving iodobenzene, benzene or toluene would need the strong aryl C-O bond to break or the benzyl group to be reduced, which does not happen with HI under these conditions.

Match the following reactions with the corresponding reagentsReactionsReagents(a) Oxidation of secondary alcohols to ketones(i) 85% H3PO4, 440 K(b) Dehydration of secondary alcohols to alkenes(ii) Na2Cr2O7/H2SO4(c) Reduction of ketones to secondary alcohols(iii) Chromic anhydride(d) Oxidation of phenol to benzoquinone(iv) NaBH4

The correct matching is (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).Secondary alcohols are oxidised to ketones by chromic anhydride (CrO3).Secondary alcohols are dehydrated to alkenes by 85% phosphoric acid at about 440 K.Ketones are reduced to secondary alcohols by sodium borohydride (NaBH4).Phenol is oxidised to benzoquinone by acidified sodium dichromate (Na2Cr2O7/H2SO4).NaBH4 is the only reducing agent listed, so it must go with (c), and phosphoric acid is the only dehydrating agent, so it must go with (b); only one option has both.

Alcohols Phenols and Ethers Questions and Answers

Q: Which one of the following reaction is called Kolbe's method?

Kolbe's method is the electrolysis of an aqueous solution of potassium carboxylates.In Kolbe's electrolytic method, a solution of the sodium or potassium salt of a carboxylic acid is electrolysed. At the anode the carboxylate ions lose electrons and carbon dioxide, and the alkyl radicals join: for example potassium ethanoate gives ethane.Treating alkyl halides with sodium in dry ether is the Wurtz reaction. Hydrogenation of alkynes, chlorination of chloroform and isomerisation of n-hexane with anhydrous AlCl3 are other reactions, not Kolbe's.

Q: The O-H bond length in methanol is

The O-H bond length in methanol is 96 pm.In methanol the oxygen is sp3 hybridised. The textbook structure gives the O-H bond as 96 pm and the C-O bond as 142 pm, with a C-O-H angle of 108.9°. The O-H bond is short because hydrogen is a very small atom.142 pm and 141 pm are close to the C-O bond length, and 136 pm is close to the C-O bond in phenol. 109 pm is a typical C-H bond length. None of these is the O-H bond.

Q: The products A, B & C from the following reactions are respectivelyPhenol (C6H5OH) gives A with NaOH, then CO2, H+; gives B with Zn; and gives C with Na2Cr2O7, H2SO4.

A, B and C are 2-hydroxybenzoic acid, benzene and benzoquinone.A: phenol with NaOH gives sodium phenoxide, which reacts with CO2 (Kolbe's reaction), and acidification gives salicylic acid, that is 2-hydroxybenzoic acid. B: heating phenol with zinc dust removes the oxygen, giving benzene. C: oxidation of phenol with sodium dichromate and sulphuric acid gives benzoquinone.Aspirin needs a further acetylation of salicylic acid. Salicylaldehyde comes from the Reimer-Tiemann reaction (CHCl3 and NaOH), and quinol is the reduction product of benzoquinone, not its oxidation product. Cyclohexanol and cyclohexanone would need hydrogenation.

Q: When 2-methylbutan-2-ol is treated with Lucas reagent con.HCl and ZnCl2 at room temperature, the product obtained is

The product is 2-chloro-2-methylbutane.2-Methylbutan-2-ol is a tertiary alcohol. With Lucas reagent (concentrated HCl and anhydrous ZnCl2) it reacts at once at room temperature through a stable tertiary carbocation, and chloride simply replaces the -OH on the same carbon, so the carbon skeleton is unchanged.The chlorobutanes and 1-chloropentane have different carbon skeletons, and 2-chloro-2-methylbutene would need a double bond, which this substitution does not create.

Q: The relative ease of dehydration of the following alcohols is(i) Propan-2-ol (ii) Propan-1-ol (iii) 2-Methylpropan-2-ol

The order is (iii) \(\gt\) (i) \(\gt\) (ii).Acid-catalysed dehydration goes through a carbocation, so its ease follows carbocation stability: tertiary \(\gt\) secondary \(\gt\) primary. 2-Methylpropan-2-ol is tertiary, propan-2-ol is secondary and propan-1-ol is primary.The tertiary alcohol dehydrates with dilute acid at a lower temperature, while the primary alcohol needs concentrated acid and a higher temperature. Every other option puts the primary alcohol ahead of a more substituted one.

Q: Which of the following statement is incorrect with Williamson synthesis?

The incorrect statement is Secondary or tertiary alkyl halides gives better results of ether formation.Williamson synthesis is an SN2 reaction, which works best with primary alkyl halides. Alkoxides are strong bases as well as nucleophiles, so with secondary and especially tertiary halides elimination wins and alkenes form instead of ethers.The other statements are true: the method makes both symmetrical and unsymmetrical ethers from an alkyl halide and a sodium alkoxide, by SN2 attack, and alkoxides are both nucleophiles and strong bases.

Q: An organic liquid 'A' containing C, H and O with pKa value 15.9, used in polishing woods, on heating with concentrated sulphuric acid gives a gaseous product 'B' with the empirical formula, CH2. 'B' decolourises bromine water as well as dilute permanganate solution. The compound A and B are respectively

A and B are ethanol and ethene.A liquid alcohol with pKa about 15.9, used in varnishes and polishes, is ethanol. Heated with concentrated sulphuric acid it dehydrates to ethene, C2H4, whose empirical formula is CH2. As an alkene, ethene decolourises bromine water and dilute alkaline permanganate (Baeyer's test).Methane and ethane are saturated and do not decolourise these reagents, and ethane's empirical formula is CH3. Phenol is a solid, more acidic (pKa about 10), and does not dehydrate to an alkene.

Q: When phenol is treated with excess of bromine water, it gives

Phenol gives 2,4,6-tribromophenol.The OH group strongly activates the benzene ring at the ortho and para positions. In bromine water (a polar solvent) bromination is so fast that all three activated positions are substituted, giving a white precipitate of 2,4,6-tribromophenol.Mono-bromination to o- and p-bromophenol happens only in a non-polar solvent such as CS2 at low temperature. 1,3,5-tribromophenol is not a valid name for the product, and 2,4-dibromophenol stops short.

31.

Which one of the following reaction is called Kolbe's method?

A.Hydrogenation of propyne with Pt/Pd/Ni
B.Chlorination of chloroform.
C.Treatment of alkyl halides with sodium metal in dry ethereal solution.
D.Electrolysis of an aqueous solution of potassium carboxylates.
E.Isomerization of n-hexane to 2-methylpentane in presence of anhy.\(\mathrm{AlCl_{3}}\)/ HCl.

Exam Year(s)

KEAM - 2026,

Answer: D

Kolbe's method is the electrolysis of an aqueous solution of potassium carboxylates.

In Kolbe's electrolytic method, a solution of the sodium or potassium salt of a carboxylic acid is electrolysed. At the anode the carboxylate ions lose electrons and carbon dioxide, and the alkyl radicals join: for example potassium ethanoate gives ethane.

Treating alkyl halides with sodium in dry ether is the Wurtz reaction. Hydrogenation of alkynes, chlorination of chloroform and isomerisation of n-hexane with anhydrous AlCl₃ are other reactions, not Kolbe's.

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32.

The O-H bond length in methanol is

A.109 pm
B.136 pm
C.96 pm
D.141 pm
E.142 pm

Exam Year(s)

KEAM - 2026,

Answer: C

The O-H bond length in methanol is 96 pm.

In methanol the oxygen is sp³ hybridised. The textbook structure gives the O-H bond as 96 pm and the C-O bond as 142 pm, with a C-O-H angle of 108.9°. The O-H bond is short because hydrogen is a very small atom.

142 pm and 141 pm are close to the C-O bond length, and 136 pm is close to the C-O bond in phenol. 109 pm is a typical C-H bond length. None of these is the O-H bond.

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33.

The products A, B & C from the following reactions are respectively

Phenol (C₆H₅OH) gives A with NaOH, then CO₂, H⁺; gives B with Zn; and gives C with Na₂Cr₂O₇, H₂SO₄.

A.aspirin, salicylic acid and benzene
B.2-hydroxybezoic acid, salicylaldehyde and quinol
C.salicylic acid, quinone and cyclohexanone
D.salicylaldehyde, benzene and cyclohexanol
E.2-hydroxybezoic acid, benzene and benzoquinone

Exam Year(s)

KEAM - 2026,

Answer: E

A, B and C are 2-hydroxybenzoic acid, benzene and benzoquinone.

A: phenol with NaOH gives sodium phenoxide, which reacts with CO₂ (Kolbe's reaction), and acidification gives salicylic acid, that is 2-hydroxybenzoic acid. B: heating phenol with zinc dust removes the oxygen, giving benzene. C: oxidation of phenol with sodium dichromate and sulphuric acid gives benzoquinone.

Aspirin needs a further acetylation of salicylic acid. Salicylaldehyde comes from the Reimer-Tiemann reaction (CHCl₃ and NaOH), and quinol is the reduction product of benzoquinone, not its oxidation product. Cyclohexanol and cyclohexanone would need hydrogenation.

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34.

The major products formed by heating C₆H₅CH₂-O-C₆H₅ with HI are

A.\(\mathrm{C_{6}H_{5}CH_{2}I}\) and \(\mathrm{C_{6}H_{5}\text{-}CH_{2}OH}\)
B.\(\mathrm{C_{6}H_{5}CH_{3}}\) and \(\mathrm{C_{6}H_{5}OH}\)
C.\(\mathrm{C_{6}H_{5}CH_{2}I}\) and \(\mathrm{C_{6}H_{6}}\)
D.\(\mathrm{C_{6}H_{5}\text{-}CH_{2}OH}\) and \(\mathrm{C_{6}H_{5}I}\)
E.\(\mathrm{C_{6}H_{5}CH_{2}I}\) and \(\mathrm{C_{6}H_{5}OH}\)

Exam Year(s)

KEAM - 2026,

Answer: E

The products are C₆H₅CH₂I and C₆H₅OH.

HI cleaves ethers by protonating the oxygen and then letting iodide attack a carbon. The O-phenyl bond has partial double bond character (the oxygen's lone pair is delocalised into the ring), and iodide cannot attack an sp² ring carbon, so that bond survives and phenol is formed. The iodide attacks the benzylic CH₂ instead, giving benzyl iodide.

Options giving iodobenzene, benzene or toluene would need the strong aryl C-O bond to break or the benzyl group to be reduced, which does not happen with HI under these conditions.

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35.

Match the following reactions with the corresponding reagents

Reactions	Reagents
(a) Oxidation of secondary alcohols to ketones	(i) 85% H₃PO₄, 440 K
(b) Dehydration of secondary alcohols to alkenes	(ii) Na₂Cr₂O₇/H₂SO₄
(c) Reduction of ketones to secondary alcohols	(iii) Chromic anhydride
(d) Oxidation of phenol to benzoquinone	(iv) NaBH₄

A.(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
B.(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
C.(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
D.(a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
E.(a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)

Exam Year(s)

KEAM - 2026,

Answer: B

The correct matching is (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).

Secondary alcohols are oxidised to ketones by chromic anhydride (CrO₃).
Secondary alcohols are dehydrated to alkenes by 85% phosphoric acid at about 440 K.
Ketones are reduced to secondary alcohols by sodium borohydride (NaBH₄).
Phenol is oxidised to benzoquinone by acidified sodium dichromate (Na₂Cr₂O₇/H₂SO₄).

NaBH₄ is the only reducing agent listed, so it must go with (c), and phosphoric acid is the only dehydrating agent, so it must go with (b); only one option has both.

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36.

When 2-methylbutan-2-ol is treated with Lucas reagent con.HCl and ZnCl₂ at room temperature, the product obtained is

A.2-chlorobutane
B.2-chloro-2-methylbutane
C.1-chlorobutane
D.1-chloropentane
E.2-chloro-2-methylbutene

Exam Year(s)

KEAM - 2026,

Answer: B

The product is 2-chloro-2-methylbutane.

2-Methylbutan-2-ol is a tertiary alcohol. With Lucas reagent (concentrated HCl and anhydrous ZnCl₂) it reacts at once at room temperature through a stable tertiary carbocation, and chloride simply replaces the -OH on the same carbon, so the carbon skeleton is unchanged.

The chlorobutanes and 1-chloropentane have different carbon skeletons, and 2-chloro-2-methylbutene would need a double bond, which this substitution does not create.

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37.

The relative ease of dehydration of the following alcohols is

(i) Propan-2-ol (ii) Propan-1-ol (iii) 2-Methylpropan-2-ol

A.(iii) \(\gt\) (ii) \(\gt\) (i)
B.(ii) \(\gt\) (i) \(\gt\) (iii)
C.(ii) \(\gt\) (iii) \(\gt\) (i)
D.(i) \(\gt\) (ii) \(\gt\) (iii)
E.(iii) \(\gt\) (i) \(\gt\) (ii)

Exam Year(s)

KEAM - 2026,

Answer: E

The order is (iii) \(\gt\) (i) \(\gt\) (ii).

Acid-catalysed dehydration goes through a carbocation, so its ease follows carbocation stability: tertiary \(\gt\) secondary \(\gt\) primary. 2-Methylpropan-2-ol is tertiary, propan-2-ol is secondary and propan-1-ol is primary.

The tertiary alcohol dehydrates with dilute acid at a lower temperature, while the primary alcohol needs concentrated acid and a higher temperature. Every other option puts the primary alcohol ahead of a more substituted one.

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38.

Which of the following statement is incorrect with Williamson synthesis?

A.It is used to prepare symmetrical and unsymmetrical ethers.
B.It involves reaction between alkyl halide and sodium alkoxide.
C.This reaction involves \(\mathrm{S_{N}2}\) attack.
D.Secondary or tertiary alkyl halides gives better results of ether formation.
E.Alkoxides are not only nucleophiles but also strong bases.

Exam Year(s)

KEAM - 2026,

Answer: D

The incorrect statement is Secondary or tertiary alkyl halides gives better results of ether formation.

Williamson synthesis is an S_N2 reaction, which works best with primary alkyl halides. Alkoxides are strong bases as well as nucleophiles, so with secondary and especially tertiary halides elimination wins and alkenes form instead of ethers.

The other statements are true: the method makes both symmetrical and unsymmetrical ethers from an alkyl halide and a sodium alkoxide, by S_N2 attack, and alkoxides are both nucleophiles and strong bases.

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39.

An organic liquid 'A' containing C, H and O with pKa value 15.9, used in polishing woods, on heating with concentrated sulphuric acid gives a gaseous product 'B' with the empirical formula, CH₂. 'B' decolourises bromine water as well as dilute permanganate solution. The compound A and B are respectively

A.methanol and methane
B.ethanol and ethane
C.Phenol and ethanol
D.ethanol and ethene
E.ethanol and methane

Exam Year(s)

KEAM - 2026,

Answer: D

A and B are ethanol and ethene.

A liquid alcohol with pKa about 15.9, used in varnishes and polishes, is ethanol. Heated with concentrated sulphuric acid it dehydrates to ethene, C₂H₄, whose empirical formula is CH₂. As an alkene, ethene decolourises bromine water and dilute alkaline permanganate (Baeyer's test).

Methane and ethane are saturated and do not decolourise these reagents, and ethane's empirical formula is CH₃. Phenol is a solid, more acidic (pKa about 10), and does not dehydrate to an alkene.

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40.

When phenol is treated with excess of bromine water, it gives

A.o-bromophenol
B.o- and p-bromophenol
C.1,3,5-tribromophenol
D.2,4-dibromophenol
E.2,4,6-tribromophenol

Exam Year(s)

KEAM - 2026,

Answer: E

Phenol gives 2,4,6-tribromophenol.

The OH group strongly activates the benzene ring at the ortho and para positions. In bromine water (a polar solvent) bromination is so fast that all three activated positions are substituted, giving a white precipitate of 2,4,6-tribromophenol.

Mono-bromination to o- and p-bromophenol happens only in a non-polar solvent such as CS₂ at low temperature. 1,3,5-tribromophenol is not a valid name for the product, and 2,4-dibromophenol stops short.

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