Alcohols Phenols and Ethers Questions and Answers

Q: Suggest most suitable reagent/condition for the following conversion :

The reagent is TEMPO (2.0%); NaOCl (1.25 eq), CH2Cl2.The substrate has a primary and a secondary alcohol, and only the primary one must become an aldehyde. The Anelli oxidation, catalytic TEMPO with bleach as the stoichiometric oxidant, works through an oxoammonium ion that is bulky and reacts much faster with unhindered primary alcohols. It stops at the aldehyde without over-oxidation, and with a limited amount of NaOCl the secondary alcohol is left alone.MnO2 oxidises only allylic and benzylic alcohols, and neither alcohol here is activated. PDC oxidises primary and secondary alcohols alike, so it would also give the ketone. DMDO is mainly an epoxidising agent, and when it does oxidise alcohols it attacks secondary ones in preference to primary.

Q: The major product of the following reaction is :

The product is the 2-methoxytetrahydrofuran, option A.The substrate is a cyclic enol ether. Acid protonates the carbon of the double bond that is not attached to oxygen, which gives an oxocarbenium ion next to the ring oxygen, a far more stable cation than any alternative. Methanol adds to that carbon and loss of H+ gives the mixed acetal, 2-methoxy-4,4-dimethyltetrahydrofuran, the methoxy group on the carbon next to the ring oxygen.Options B and C put OMe on carbons that are not attached to the ring oxygen, which would need an unstabilised carbocation (and C would need a methyl shift); D is just an isomeric alkene and does not involve methanol.

Q: The major products of the following reaction is :

The answer is option A: P carries a trityl ether on the primary CH2O and a TBS ether on the ring oxygen, and Q is the primary iodide made from the reduced ester.Trityl chloride is very bulky, so it protects only the primary alcohol; TBSOTf then silylates the remaining secondary alcohol. LiBH4 reduces the methyl ester to the primary alcohol, leaving both silyl and trityl ethers intact, and Ph3P, I2 and imidazole convert that alcohol to the iodide (Appel-type reaction), so the CH2CO2Me side chain becomes CH2CH2I. None of these steps touches a stereocentre, so the configuration of the starting tetrahydrofuran is retained.B and D put TBS on the primary oxygen and Tr on the ring oxygen, which reverses the selectivity; C inverts the configuration of the side chain in Q.

Q: The correct order of reactivity of the following alcohols with p-nitrobenzoyl chloride in a esterification reaction is :

The order is III > II > I, option A.Acylation with p-nitrobenzoyl chloride is faster for an equatorial alcohol than for an axial one, because the equatorial OH is less hindered. Locating each OH relative to the angular methyl group in the trans-decalin: in III the hydroxyl is equatorial, so III reacts fastest. In I and II the hydroxyl is axial, so both are slower than III. Of the two axial alcohols, the one in I lies on the same face as the angular methyl and is blocked by its 1,3-diaxial interaction, so I is the slowest and II comes between.This agrees with the official key. The axial or equatorial assignment depends on the position of each OH relative to the angular methyl group as drawn.

51.

Suggest most suitable reagent/condition for the following conversion :

A diol with one secondary and one primary OH, oxidised at the primary OH only to an aldehyde, the secondary OH untouched

A.\(\mathrm{MnO}_{2}\); \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\)
B.TEMPO (2.0%); NaOCl (1.25 eq), \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\)
C.Pyridinium dichromate (PDC); \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\)
D.Dimethyldioxirane (DMDO); \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The reagent is TEMPO (2.0%); NaOCl (1.25 eq), CH₂Cl₂.

The substrate has a primary and a secondary alcohol, and only the primary one must become an aldehyde. The Anelli oxidation, catalytic TEMPO with bleach as the stoichiometric oxidant, works through an oxoammonium ion that is bulky and reacts much faster with unhindered primary alcohols. It stops at the aldehyde without over-oxidation, and with a limited amount of NaOCl the secondary alcohol is left alone.

MnO₂ oxidises only allylic and benzylic alcohols, and neither alcohol here is activated. PDC oxidises primary and secondary alcohols alike, so it would also give the ketone. DMDO is mainly an epoxidising agent, and when it does oxidise alcohols it attacks secondary ones in preference to primary.

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52.

The major product of the following reaction is :

4,4-dimethyl-4,5-dihydrofuran (a cyclic enol ether) with MeOH and acid

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The product is the 2-methoxytetrahydrofuran, option A.

The substrate is a cyclic enol ether. Acid protonates the carbon of the double bond that is not attached to oxygen, which gives an oxocarbenium ion next to the ring oxygen, a far more stable cation than any alternative. Methanol adds to that carbon and loss of H⁺ gives the mixed acetal, 2-methoxy-4,4-dimethyltetrahydrofuran, the methoxy group on the carbon next to the ring oxygen.

Options B and C put OMe on carbons that are not attached to the ring oxygen, which would need an unstabilised carbocation (and C would need a methyl shift); D is just an isomeric alkene and does not involve methanol.

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53.

The major products of the following reaction is :

A tetrahydrofuran with a hydroxymethyl group, a hydroxyl group and a CH2CO2Me chain, treated with TrCl, then TBSOTf to give P, then LiBH4 and PPh3, I2, imidazole to give Q

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The answer is option A: P carries a trityl ether on the primary CH₂O and a TBS ether on the ring oxygen, and Q is the primary iodide made from the reduced ester.

Trityl chloride is very bulky, so it protects only the primary alcohol; TBSOTf then silylates the remaining secondary alcohol. LiBH₄ reduces the methyl ester to the primary alcohol, leaving both silyl and trityl ethers intact, and Ph₃P, I₂ and imidazole convert that alcohol to the iodide (Appel-type reaction), so the CH₂CO₂Me side chain becomes CH₂CH₂I. None of these steps touches a stereocentre, so the configuration of the starting tetrahydrofuran is retained.

B and D put TBS on the primary oxygen and Tr on the ring oxygen, which reverses the selectivity; C inverts the configuration of the side chain in Q.

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54.

The correct order of reactivity of the following alcohols with p-nitrobenzoyl chloride in a esterification reaction is :

Three trans-decalin alcohols I, II and III with an angular methyl group and a hydroxyl on the left ring at different positions and orientations

A.III \(\gt\) II \(\gt\) I
B.I \(\gt\) III \(\gt\) II
C.II \(\gt\) III \(\gt\) I
D.II \(\gt\) I \(\gt\) III

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The order is III > II > I, option A.

Acylation with p-nitrobenzoyl chloride is faster for an equatorial alcohol than for an axial one, because the equatorial OH is less hindered. Locating each OH relative to the angular methyl group in the trans-decalin: in III the hydroxyl is equatorial, so III reacts fastest. In I and II the hydroxyl is axial, so both are slower than III. Of the two axial alcohols, the one in I lies on the same face as the angular methyl and is blocked by its 1,3-diaxial interaction, so I is the slowest and II comes between.

This agrees with the official key. The axial or equatorial assignment depends on the position of each OH relative to the angular methyl group as drawn.

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