Alcohols Phenols and Ethers Questions and Answers

Q: The order of acidity of the following compounds is(i) o-Nitrophenol (ii) Phenol (iii) o-Cresol (iv) Ethanol

The order is (iv) < (iii) < (ii) < (i).Phenols are more acidic than alcohols because the phenoxide ion is stabilised by resonance with the ring, so ethanol (iv) is the weakest acid.Among the phenols, an electron-withdrawing nitro group stabilises the phenoxide further, making o-nitrophenol (i) the strongest; an electron-releasing methyl group destabilises it, making o-cresol (iii) weaker than phenol (ii). Options placing ethanol above any phenol, or cresol above phenol, reverse these effects.

Q: Identify the allylic alcohol in the following.

The allylic alcohol is \(\mathrm{C_{6}H_{5}{-}CH{=}CH{-}C(CH_{3})_{2}{-}OH}\).In an allylic alcohol the -OH group is on an sp3 carbon directly next to a C=C double bond. Here the carbon carrying -OH, C(CH3)2, is attached to the CH=CH group, so it is allylic (a tertiary allylic alcohol).C6H5CH2OH is benzylic (next to a benzene ring). CH2=CH-OH is vinylic (the -OH is on the double-bond carbon itself). In C6H5-CH2-CH2-OH and CH2=CH-CH2-CH2-OH the -OH carbon is one position too far from the ring or double bond.

Q: p-Bromophenol is the major product formed when phenol is treated with

Phenol gives mainly p-bromophenol with Br2 in CS2 at 273K.The -OH group strongly activates the ring. In a non-polar solvent such as CS2 (or CHCl3) at low temperature, bromine is not polarised much and the reaction stops at monobromination, giving mainly the para product with some ortho.With bromine water, a polar solvent, phenol ionises partly to the even more reactive phenoxide and all three positions are substituted at once, giving a white precipitate of 2,4,6-tribromophenol. That is why bromine water is the trap.

Q: Reimer-Tiemann reaction using CHCl3 and aq.NaOH involves the conversion of phenol into

The Reimer-Tiemann reaction converts phenol into salicylaldehyde.Phenol treated with chloroform and aqueous sodium hydroxide gains a -CHO group at the ortho position. The reactive species is dichlorocarbene, formed from chloroform by the base; the dichloromethyl intermediate is hydrolysed to the aldehyde, giving 2-hydroxybenzaldehyde (salicylaldehyde).Salicylic acid is made from phenol by Kolbe's reaction, with CO2 and NaOH. Anisole is made by methylating phenol, and phenol is reduced to benzene by heating with zinc dust. Chlorobenzene is not formed in this reaction.

Q: When a mixture of CO and H2 is heated at 573 K-673 K under 200-300 atmospheric pressure in the presence of a catalyst, methanol is produced. The catalyst used is

The catalyst is ZnO-Cr2O3.Methanol is manufactured from synthesis gas (CO and H2). Under the high-pressure conditions given, 573-673 K and 200-300 atm, the textbook catalyst is a mixture of zinc oxide and chromium(III) oxide: CO + 2H2 → CH3OH.Pd-BaSO4 is the poisoned catalyst of the Rosenmund reduction (acid chlorides to aldehydes), and Pt-BaSO4 is a similar supported metal. Nickel and copper chromite catalysts are used for other hydrogenations, not for this high-pressure methanol process.

Q: Chloroethane is prepared by the action of PCl3 on ethanol. What is the inorganic product formed in the reaction?

The inorganic product is \(\text{H}_{3}\text{PO}_{3}\), phosphorous acid.\[ \begin{aligned} &3\text{C}_{2}\text{H}_{5}\text{OH} + \text{PCl}_{3} \\ &\rightarrow 3\text{C}_{2}\text{H}_{5}\text{Cl} + \text{H}_{3}\text{PO}_{3} \end{aligned} \]Each P-Cl bond is replaced by P-OH, so the three chlorines go to the three ethyl groups and phosphorus keeps its +3 state as \(\text{H}_{3}\text{PO}_{3}\).\(\text{POCl}_{3}\) (with HCl) is the by-product when \(\text{PCl}_{5}\) is used instead. \(\text{H}_{3}\text{PO}_{4}\) and \(\text{HPO}_{3}\) contain phosphorus in the +5 state, which \(\text{PCl}_{3}\) does not reach in this reaction.

Q: Which of the following oxidising agent is used to convert ethanol to ethanal?

The reagent is \(\text{CrO}_{3}\) in anhydrous medium.To stop the oxidation of a primary alcohol at the aldehyde, a mild reagent is needed under conditions with no water: chromium trioxide in an anhydrous medium (or pyridinium chlorochromate) gives ethanal and goes no further.Acidified and alkaline \(\text{KMnO}_{4}\) and acidified \(\text{K}_{2}\text{Cr}_{2}\text{O}_{7}\) are strong oxidants in water and carry the oxidation on to ethanoic acid. \(\text{H}_{2}\text{O}_{2}\) is not the reagent used for this conversion.

Q: The compound with the lowest pKa value is

The compound with the lowest pKa (strongest acid) is p-nitrophenol.The nitro group is strongly electron-withdrawing. In the para position it pulls charge out of the ring by both inductive and resonance effects and stabilises the phenoxide ion, so p-nitrophenol loses its proton most easily (pKa about 7.1).Phenol is weaker (about 10.0). Methyl groups donate electrons, so o- and p-cresol are weaker still, and ethanol (about 16) is the weakest acid of all here.

Q: 2-Bromopropane reacts with Mg in dry ether to give a product which on treatment with HCHO forms an addition product. The addition product on hydrolysis forms

The final product is 2-methyl-1-propanol.2-Bromopropane and magnesium give the Grignard reagent (CH3)2CHMgBr. It adds to formaldehyde (HCHO), and hydrolysis of the adduct gives a primary alcohol with one more carbon:(CH3)2CH-MgBr + HCHO → (CH3)2CH-CH2OMgBr → (CH3)2CH-CH2OH(CH3)2CHCH2OH is 2-methyl-1-propanol. Formaldehyde always gives a primary alcohol, which rules out the secondary and tertiary alcohols; 1-butanol has a straight chain, but the isopropyl group keeps the branch.

Q: The reagent used for the conversion of decanol into decanoic acid is

The reagent is Jones reagent.Jones reagent, chromium trioxide in aqueous sulphuric acid with acetone, is a strong oxidant that takes a primary alcohol such as decan-1-ol all the way to the carboxylic acid, decanoic acid.Tollens' and Fehling's reagents oxidise aldehydes, not alcohols. A Grignard reagent is a nucleophile used to make new carbon-carbon bonds, and DIBAL-H is a reducing agent, the opposite of what is needed.

21.

The order of acidity of the following compounds is
(i) o-Nitrophenol (ii) Phenol (iii) o-Cresol (iv) Ethanol

A.(i) < (iii) < (ii) < (iv)
B.(iii) < (i) < (ii) < (iv)
C.(i) < (ii) < (iii) < (iv)
D.(iv) < (iii) < (ii) < (i)
E.(iii) < (ii) < (i) < (iv)

Exam Year(s)

KEAM - 2025,

Answer: D

The order is (iv) < (iii) < (ii) < (i).

Phenols are more acidic than alcohols because the phenoxide ion is stabilised by resonance with the ring, so ethanol (iv) is the weakest acid.

Among the phenols, an electron-withdrawing nitro group stabilises the phenoxide further, making o-nitrophenol (i) the strongest; an electron-releasing methyl group destabilises it, making o-cresol (iii) weaker than phenol (ii). Options placing ethanol above any phenol, or cresol above phenol, reverse these effects.

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22.

Identify the allylic alcohol in the following.

A.\(\mathrm{C_{6}H_{5}CH_{2}OH}\)
B.\(\mathrm{C_{6}H_{5}{-}CH{=}CH{-}C(CH_{3})_{2}{-}OH}\)
C.\(\mathrm{CH_{2}{=}CH{-}OH}\)
D.\(\mathrm{C_{6}H_{5}{-}CH_{2}{-}CH_{2}{-}OH}\)
E.\(\mathrm{CH_{2}{=}CH{-}CH_{2}{-}CH_{2}{-}OH}\)

Exam Year(s)

KEAM - 2026,

Answer: B

The allylic alcohol is \(\mathrm{C_{6}H_{5}{-}CH{=}CH{-}C(CH_{3})_{2}{-}OH}\).

In an allylic alcohol the -OH group is on an sp³ carbon directly next to a C=C double bond. Here the carbon carrying -OH, C(CH₃)₂, is attached to the CH=CH group, so it is allylic (a tertiary allylic alcohol).

C₆H₅CH₂OH is benzylic (next to a benzene ring). CH₂=CH-OH is vinylic (the -OH is on the double-bond carbon itself). In C₆H₅-CH₂-CH₂-OH and CH₂=CH-CH₂-CH₂-OH the -OH carbon is one position too far from the ring or double bond.

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23.

p-Bromophenol is the major product formed when phenol is treated with

A.Bromine water
B.\(\mathrm{Br_{2}}\) in acetic acid at 300K
C.\(\mathrm{Br_{2}}\) in \(\mathrm{CCl_{4}}\) at 300K
D.\(\mathrm{Br_{2}}\) in \(\mathrm{CS_{2}}\) at 273K
E.\(\mathrm{Br_{2}}\) in acetone at 273K

Exam Year(s)

KEAM - 2026,

Answer: D

Phenol gives mainly p-bromophenol with Br₂ in CS₂ at 273K.

The -OH group strongly activates the ring. In a non-polar solvent such as CS₂ (or CHCl₃) at low temperature, bromine is not polarised much and the reaction stops at monobromination, giving mainly the para product with some ortho.

With bromine water, a polar solvent, phenol ionises partly to the even more reactive phenoxide and all three positions are substituted at once, giving a white precipitate of 2,4,6-tribromophenol. That is why bromine water is the trap.

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24.

Reimer-Tiemann reaction using CHCl₃ and aq.NaOH involves the conversion of phenol into

A.benzene
B.salicylic acid
C.anisole
D.chlorobenzene
E.salicylaldehyde

Exam Year(s)

KEAM - 2026,

Answer: E

The Reimer-Tiemann reaction converts phenol into salicylaldehyde.

Phenol treated with chloroform and aqueous sodium hydroxide gains a -CHO group at the ortho position. The reactive species is dichlorocarbene, formed from chloroform by the base; the dichloromethyl intermediate is hydrolysed to the aldehyde, giving 2-hydroxybenzaldehyde (salicylaldehyde).

Salicylic acid is made from phenol by Kolbe's reaction, with CO₂ and NaOH. Anisole is made by methylating phenol, and phenol is reduced to benzene by heating with zinc dust. Chlorobenzene is not formed in this reaction.

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25.

When a mixture of CO and H₂ is heated at 573 K-673 K under 200-300 atmospheric pressure in the presence of a catalyst, methanol is produced. The catalyst used is

A.Pt-\(\mathrm{BaSO_{4}}\)
B.ZnO- \(\mathrm{Cr_{2}O_{3}}\)
C.Ni-\(\mathrm{Cr_{2}O_{3}}\)
D.Pd-\(\mathrm{BaSO_{4}}\)
E.CuO-\(\mathrm{Cr_{2}O_{3}}\)

Exam Year(s)

KEAM - 2026,

Answer: B

The catalyst is ZnO-Cr₂O₃.

Methanol is manufactured from synthesis gas (CO and H₂). Under the high-pressure conditions given, 573-673 K and 200-300 atm, the textbook catalyst is a mixture of zinc oxide and chromium(III) oxide: CO + 2H₂ → CH₃OH.

Pd-BaSO₄ is the poisoned catalyst of the Rosenmund reduction (acid chlorides to aldehydes), and Pt-BaSO₄ is a similar supported metal. Nickel and copper chromite catalysts are used for other hydrogenations, not for this high-pressure methanol process.

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26.

Chloroethane is prepared by the action of PCl₃ on ethanol. What is the inorganic product formed in the reaction?

A.\(\text{POCl}_{3}\)
B.\(\text{HPO}_{3}\)
C.\(\text{H}_{3}\text{PO}_{3}\)
D.\(\text{H}_{2}\text{PO}_{2}\)
E.\(\text{H}_{3}\text{PO}_{4}\)

Exam Year(s)

KEAM - 2026,

Answer: C

The inorganic product is \(\text{H}_{3}\text{PO}_{3}\), phosphorous acid.

\[ \begin{aligned} &3\text{C}_{2}\text{H}_{5}\text{OH} + \text{PCl}_{3} \\ &\rightarrow 3\text{C}_{2}\text{H}_{5}\text{Cl} + \text{H}_{3}\text{PO}_{3} \end{aligned} \]

Each P-Cl bond is replaced by P-OH, so the three chlorines go to the three ethyl groups and phosphorus keeps its +3 state as \(\text{H}_{3}\text{PO}_{3}\).

\(\text{POCl}_{3}\) (with HCl) is the by-product when \(\text{PCl}_{5}\) is used instead. \(\text{H}_{3}\text{PO}_{4}\) and \(\text{HPO}_{3}\) contain phosphorus in the +5 state, which \(\text{PCl}_{3}\) does not reach in this reaction.

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27.

Which of the following oxidising agent is used to convert ethanol to ethanal?

A.Acidified \(\text{KMnO}_{4}\)
B.Alkaline \(\text{KMnO}_{4}\)
C.Acidified \(\text{K}_{2}\text{Cr}_{2}\text{O}_{7}\)
D.\(\text{H}_{2}\text{O}_{2}\) in anhydrous medium
E.\(\text{CrO}_{3}\) in anhydrous medium

Exam Year(s)

KEAM - 2026,

Answer: E

The reagent is \(\text{CrO}_{3}\) in anhydrous medium.

To stop the oxidation of a primary alcohol at the aldehyde, a mild reagent is needed under conditions with no water: chromium trioxide in an anhydrous medium (or pyridinium chlorochromate) gives ethanal and goes no further.

Acidified and alkaline \(\text{KMnO}_{4}\) and acidified \(\text{K}_{2}\text{Cr}_{2}\text{O}_{7}\) are strong oxidants in water and carry the oxidation on to ethanoic acid. \(\text{H}_{2}\text{O}_{2}\) is not the reagent used for this conversion.

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28.

The compound with the lowest pKa value is

A.ethanol
B.phenol
C.p-nitrophenol
D.o-cresol
E.p-cresol

Exam Year(s)

KEAM - 2026,

Answer: C

The compound with the lowest pK_a (strongest acid) is p-nitrophenol.

The nitro group is strongly electron-withdrawing. In the para position it pulls charge out of the ring by both inductive and resonance effects and stabilises the phenoxide ion, so p-nitrophenol loses its proton most easily (pK_a about 7.1).

Phenol is weaker (about 10.0). Methyl groups donate electrons, so o- and p-cresol are weaker still, and ethanol (about 16) is the weakest acid of all here.

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29.

2-Bromopropane reacts with Mg in dry ether to give a product which on treatment with HCHO forms an addition product. The addition product on hydrolysis forms

A.1-butanol
B.2-butanol
C.2-propanol
D.2-methyl-1-propanol
E.2-methyl-2-propanol

Exam Year(s)

KEAM - 2026,

Answer: D

The final product is 2-methyl-1-propanol.

2-Bromopropane and magnesium give the Grignard reagent (CH₃)₂CHMgBr. It adds to formaldehyde (HCHO), and hydrolysis of the adduct gives a primary alcohol with one more carbon:

(CH₃)₂CH-MgBr + HCHO → (CH₃)₂CH-CH₂OMgBr → (CH₃)₂CH-CH₂OH

(CH₃)₂CHCH₂OH is 2-methyl-1-propanol. Formaldehyde always gives a primary alcohol, which rules out the secondary and tertiary alcohols; 1-butanol has a straight chain, but the isopropyl group keeps the branch.

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30.

The reagent used for the conversion of decanol into decanoic acid is

A.Tollens's reagent
B.Jones reagent
C.Grignard reagent
D.Fehling's reagent
E.DIBAL-H

Exam Year(s)

KEAM - 2026,

Answer: B

The reagent is Jones reagent.

Jones reagent, chromium trioxide in aqueous sulphuric acid with acetone, is a strong oxidant that takes a primary alcohol such as decan-1-ol all the way to the carboxylic acid, decanoic acid.

Tollens' and Fehling's reagents oxidise aldehydes, not alcohols. A Grignard reagent is a nucleophile used to make new carbon-carbon bonds, and DIBAL-H is a reducing agent, the opposite of what is needed.

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