Alcohols Phenols and Ethers Questions and Answers

Q: In which of the following compounds, -OH funtional group is present?

The -OH functional group is present in ethyl alcohol (ethanol).Ethanol, \(C_{2}H_{5}OH\), is an alcohol: its functional group is the hydroxyl group (-OH) attached to a saturated carbon atom. This group makes alcohols able to form hydrogen bonds, which is why ethanol mixes with water.Acetone (\(CH_{3}COCH_{3}\)) is a ketone with a carbonyl group, and formaldehyde (HCHO) is an aldehyde, also with a carbonyl group. Acetic acid has a carboxyl group (-COOH), which contains an OH as part of the carboxyl unit but is classed as a separate functional group, not as a hydroxyl group.

Q: Denatured alcohol contains

Denatured alcohol contains CuSO4 + pyridine (among other additives).Denatured (methylated) spirit is ethanol made unfit for drinking, so that it can be sold for industrial use without excise duty. It is commonly denatured with methanol and pyridine, which give it a foul smell and taste, and coloured with copper sulfate (or a dye) so that it can be recognised.Formaldehyde, quinoline, zinc chloride and acetone are not the standard denaturant combination.

Q: Major product formed in the following reaction is :

The product is the azide D, with the azide on the same face as the original OH.With DEAD and PPh3 the 4-hydroxyl is activated as an oxyphosphonium ion. The nearest nucleophile is the molecule's own carboxylate, which displaces it with inversion at C4, forming a bridged bicyclic lactone (an intramolecular Mitsunobu reaction).Azide then opens this lactone by SN2 attack at C4, again with inversion, releasing the carboxylic acid. Two inversions give overall retention, so N3 is hashed like the starting OH, as in D. A has the azide with net inversion, as a direct intermolecular Mitsunobu would give; B is the epimeric alcohol and C an elimination product.

Q: The major product formed in the following reaction is :

The product is the 2,3-epoxy alcohol A, (2R,3R)-2,3-epoxygeraniol.Sharpless asymmetric epoxidation with Ti(OiPr)4, a tartrate and t-BuOOH works only on allylic alcohols, because the substrate's OH must bind to titanium. Geraniol has two trisubstituted alkenes, but only the C2=C3 bond is allylic to the OH, so only that bond is epoxidised and the remote 6,7-alkene is untouched. That rules out C and D.The tartrate decides the face. L-(+)-DET gives (2S,3S)-epoxygeraniol, and D-(−)-DET, used here, gives the enantiomer, (2R,3R), which is the structure with the epoxide oxygen wedged as drawn in A. B is the (2S,3S) epoxide from (+)-DET.

Q: The major product of the following reaction is :

The major product is Ph-CH(OMe)-CH2-OH, 2-methoxy-2-phenylethanol.Under acidic conditions the epoxide oxygen is protonated, and the C-O bond to the benzylic carbon weakens because the developing positive charge there is stabilised by the phenyl ring. Methanol therefore attacks the more substituted, benzylic carbon in an SN1-like opening, and the oxygen stays on the CH2 as a primary OH.Ph-CH(OH)-CH2-OMe is the product of basic conditions (methoxide attacks the less hindered CH2). The diol would need water as the nucleophile, and styrene would require deoxygenation, not ring opening.

Q: The major product in the following reaction is :

The major product is (C), the decalin bearing OH and CHO.This is the Barton reaction. NOCl converts the alcohol into a nitrite ester. On irradiation the O-NO bond breaks, and the alkoxy radical takes a hydrogen from the methyl group held near it across the ring fusion, a 1,5-hydrogen shift through a six-membered transition state. The carbon radical captures NO, the nitroso compound tautomerises to an oxime, and acid hydrolysis gives the aldehyde, while the oxygen returns to an OH.The ring-fusion carbon skeleton is unchanged. (A) and (B) would need oxidation of the alcohol to a ketone, and (D) moves the carbonyl to a different carbon.

Q: The major product in the following reaction is :

The product is (D), the cyclic p-methoxyphenyl acetal.DDQ oxidizes p-methoxybenzyl (PMB) ethers by removing a hydride from the benzylic CH2. The p-methoxy group stabilizes the resulting oxocarbenium ion, Ar-CH=O+-R. In wet solvent water would attack it and release the free alcohol. Here the solvent is dichloromethane and the molecule has its own free OH two carbons away, in a 1,3 relationship, so that hydroxyl captures the cation intramolecularly. The six-membered ring formed is a 1,3-dioxane bearing the p-methoxyphenyl group on the acetal carbon: a PMP acetal.(A) would need DDQ to oxidize an ordinary secondary alcohol to a ketone, which it does not do, since the alcohol is neither allylic nor benzylic. (B), an ester, would need a second oxidation at the benzylic carbon. (C) moves the PMB group to the other oxygen, which the mechanism gives no route for.

Q: Predict the major product of the following reactions :

The answer is (A): I is 2-methoxy-2-phenylethanol and II is 2-methoxy-1-phenylethanol.Acid-catalysed opening (MeOH, H3O+): the epoxide oxygen is protonated, and the C-O bond to the benzylic carbon weakens, because that carbon can carry the developing positive charge. Methanol attacks the more substituted, benzylic carbon, so I has OMe on the benzylic carbon and OH on the CH2.Base (MeOH, NaOMe): methoxide opens the neutral epoxide by an SN2 attack at the less hindered CH2, so II has OMe on the CH2 and OH on the benzylic carbon.(B) reverses the two, and (C) and (D) give the same product for both conditions, ignoring the change in regiochemistry.

Q: The major product of the following reaction is :(i) Me3SiCl (1.0 eq.) Et3N; CH2Cl2 (ii) PDC, CH2Cl2 (iii) dil. HCl

The product is (B): the hindered alcohol is oxidised to a ketone and the other OH is kept.With only 1.0 equivalent of Me3SiCl and Et3N, the less hindered secondary alcohol is silylated selectively. The OH in the plain cyclopentane ring is far more accessible than the OH next to the gem-dimethyl (neopentyl-like) carbon, so it becomes the TMS ether. PDC then oxidises the free, hindered OH to a ketone, and dilute HCl removes the TMS group, restoring the first OH.(A) is what you get if the hindered OH were protected instead. (C), the diketone, needs both alcohols free during oxidation, and (D) shows an elimination, which none of these mild steps causes.

Q: Predict the major product of the following reaction :

The product is (A), the trans-diaxial bromohydrin.Ph3P/Br2 supplies bromide, which opens the epoxide by backside attack (after activation of the oxygen), and aqueous work-up gives the bromohydrin. The tert-butyl group locks the cyclohexane in the chair with tBu equatorial. By the Furst-Plattner rule, nucleophilic opening of an epoxide on a locked cyclohexane proceeds through a chair-like transition state, so the nucleophile and the oxygen end up trans and both axial.Only (A) shows OH and Br both axial and anti. (B) has both equatorial, the product of the higher-energy twist-boat pathway, and (C) and (D) have one axial and one equatorial group, which would mean a cis arrangement or a ring flip that the tert-butyl group prevents.

41.

In which of the following compounds, -OH funtional group is present?

A.Ethyl alcohol or Ethanol
B.Acetone
C.Acetic acid
D.Formaldehyde

Exam Year(s)

MPPSC Forest Service - 2023,

Answer: A

The -OH functional group is present in ethyl alcohol (ethanol).

Ethanol, \(C_{2}H_{5}OH\), is an alcohol: its functional group is the hydroxyl group (-OH) attached to a saturated carbon atom. This group makes alcohols able to form hydrogen bonds, which is why ethanol mixes with water.

Acetone (\(CH_{3}COCH_{3}\)) is a ketone with a carbonyl group, and formaldehyde (HCHO) is an aldehyde, also with a carbonyl group. Acetic acid has a carboxyl group (-COOH), which contains an OH as part of the carboxyl unit but is classed as a separate functional group, not as a hydroxyl group.

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42.

Denatured alcohol contains

A.CuSO\(_{4}\) + Pyridine
B.HCHO + Quinoline
C.ZnCl\(_{2}\) + Pyridine
D.Acetone + Quinoline

Exam Year(s)

MPPSC Forest Service - 2024,

Answer: A

Denatured alcohol contains CuSO₄ + pyridine (among other additives).

Denatured (methylated) spirit is ethanol made unfit for drinking, so that it can be sold for industrial use without excise duty. It is commonly denatured with methanol and pyridine, which give it a foul smell and taste, and coloured with copper sulfate (or a dye) so that it can be recognised.

Formaldehyde, quinoline, zinc chloride and acetone are not the standard denaturant combination.

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43.

Major product formed in the following reaction is :

N-benzyl 4-hydroxyproline with the OH hashed and COOH wedged, treated with (i) DEAD, PPh3, CH2Cl2, 0 C and (ii) NaN3, DME, 0 C.

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The product is the azide D, with the azide on the same face as the original OH.

With DEAD and PPh₃ the 4-hydroxyl is activated as an oxyphosphonium ion. The nearest nucleophile is the molecule's own carboxylate, which displaces it with inversion at C4, forming a bridged bicyclic lactone (an intramolecular Mitsunobu reaction).

Azide then opens this lactone by S_N2 attack at C4, again with inversion, releasing the carboxylic acid. Two inversions give overall retention, so N₃ is hashed like the starting OH, as in D. A has the azide with net inversion, as a direct intermolecular Mitsunobu would give; B is the epimeric alcohol and C an elimination product.

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44.

The major product formed in the following reaction is :

Geraniol treated with (-)-DET, Ti(OiPr)4, t-BuOOH, CH2Cl2 and 4 A molecular sieves.

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The product is the 2,3-epoxy alcohol A, (2R,3R)-2,3-epoxygeraniol.

Sharpless asymmetric epoxidation with Ti(OiPr)₄, a tartrate and t-BuOOH works only on allylic alcohols, because the substrate's OH must bind to titanium. Geraniol has two trisubstituted alkenes, but only the C2=C3 bond is allylic to the OH, so only that bond is epoxidised and the remote 6,7-alkene is untouched. That rules out C and D.

The tartrate decides the face. L-(+)-DET gives (2S,3S)-epoxygeraniol, and D-(−)-DET, used here, gives the enantiomer, (2R,3R), which is the structure with the epoxide oxygen wedged as drawn in A. B is the (2S,3S) epoxide from (+)-DET.

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45.

The major product of the following reaction is :

Styrene oxide (phenyloxirane) treated with MeOH and acid

A.\(\mathrm{Ph-CH(OMe)-CH_{2}-OH}\)
B.\(\mathrm{Ph-CH(OH)-CH_{2}-OMe}\)
C.\(\mathrm{Ph-CH(OH)-CH_{2}-OH}\)
D.\(\mathrm{Ph-CH=CH_{2}}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The major product is Ph-CH(OMe)-CH₂-OH, 2-methoxy-2-phenylethanol.

Under acidic conditions the epoxide oxygen is protonated, and the C-O bond to the benzylic carbon weakens because the developing positive charge there is stabilised by the phenyl ring. Methanol therefore attacks the more substituted, benzylic carbon in an S_N1-like opening, and the oxygen stays on the CH₂ as a primary OH.

Ph-CH(OH)-CH₂-OMe is the product of basic conditions (methoxide attacks the less hindered CH₂). The diol would need water as the nucleophile, and styrene would require deoxygenation, not ring opening.

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46.

The major product in the following reaction is :

A decalin bearing an OH and, across the ring fusion, a methyl group, treated with (a) NOCl and light and (b) H3O+

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The major product is (C), the decalin bearing OH and CHO.

This is the Barton reaction. NOCl converts the alcohol into a nitrite ester. On irradiation the O-NO bond breaks, and the alkoxy radical takes a hydrogen from the methyl group held near it across the ring fusion, a 1,5-hydrogen shift through a six-membered transition state. The carbon radical captures NO, the nitroso compound tautomerises to an oxime, and acid hydrolysis gives the aldehyde, while the oxygen returns to an OH.

The ring-fusion carbon skeleton is unchanged. (A) and (B) would need oxidation of the alcohol to a ketone, and (D) moves the carbonyl to a different carbon.

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47.

The major product in the following reaction is :

A secondary alcohol two carbons from a p-methoxybenzyl ether on a branched carbon chain, treated with DDQ in CH2Cl2

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The product is (D), the cyclic p-methoxyphenyl acetal.

DDQ oxidizes p-methoxybenzyl (PMB) ethers by removing a hydride from the benzylic CH₂. The p-methoxy group stabilizes the resulting oxocarbenium ion, Ar-CH=O⁺-R. In wet solvent water would attack it and release the free alcohol. Here the solvent is dichloromethane and the molecule has its own free OH two carbons away, in a 1,3 relationship, so that hydroxyl captures the cation intramolecularly. The six-membered ring formed is a 1,3-dioxane bearing the p-methoxyphenyl group on the acetal carbon: a PMP acetal.

(A) would need DDQ to oxidize an ordinary secondary alcohol to a ketone, which it does not do, since the alcohol is neither allylic nor benzylic. (B), an ester, would need a second oxidation at the benzylic carbon. (C) moves the PMB group to the other oxygen, which the mechanism gives no route for.

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48.

Predict the major product of the following reactions :

Styrene oxide treated with MeOH, H3O+ to give I, and with MeOH, NaOMe (printed MaOMe) to give II

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The answer is (A): I is 2-methoxy-2-phenylethanol and II is 2-methoxy-1-phenylethanol.

Acid-catalysed opening (MeOH, H₃O⁺): the epoxide oxygen is protonated, and the C-O bond to the benzylic carbon weakens, because that carbon can carry the developing positive charge. Methanol attacks the more substituted, benzylic carbon, so I has OMe on the benzylic carbon and OH on the CH₂.

Base (MeOH, NaOMe): methoxide opens the neutral epoxide by an S_N2 attack at the less hindered CH₂, so II has OMe on the CH₂ and OH on the benzylic carbon.

(B) reverses the two, and (C) and (D) give the same product for both conditions, ignoring the change in regiochemistry.

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49.

The major product of the following reaction is :

Bicyclo[3.3.0]octane-diol: one secondary OH in the plain ring and one secondary OH next to a gem-dimethyl carbon in the other ring

(i) Me₃SiCl (1.0 eq.) Et₃N; CH₂Cl₂ (ii) PDC, CH₂Cl₂ (iii) dil. HCl

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The product is (B): the hindered alcohol is oxidised to a ketone and the other OH is kept.

With only 1.0 equivalent of Me₃SiCl and Et₃N, the less hindered secondary alcohol is silylated selectively. The OH in the plain cyclopentane ring is far more accessible than the OH next to the gem-dimethyl (neopentyl-like) carbon, so it becomes the TMS ether. PDC then oxidises the free, hindered OH to a ketone, and dilute HCl removes the TMS group, restoring the first OH.

(A) is what you get if the hindered OH were protected instead. (C), the diketone, needs both alcohols free during oxidation, and (D) shows an elimination, which none of these mild steps causes.

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50.

Predict the major product of the following reaction :

4-tert-Butylcyclohexene oxide, epoxide drawn on the same face as tBu, treated with 1. Ph3P, Br2, DCM; 2. H3O+

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The product is (A), the trans-diaxial bromohydrin.

Ph₃P/Br₂ supplies bromide, which opens the epoxide by backside attack (after activation of the oxygen), and aqueous work-up gives the bromohydrin. The tert-butyl group locks the cyclohexane in the chair with tBu equatorial. By the Furst-Plattner rule, nucleophilic opening of an epoxide on a locked cyclohexane proceeds through a chair-like transition state, so the nucleophile and the oxygen end up trans and both axial.

Only (A) shows OH and Br both axial and anti. (B) has both equatorial, the product of the higher-energy twist-boat pathway, and (C) and (D) have one axial and one equatorial group, which would mean a cis arrangement or a ring flip that the tert-butyl group prevents.

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