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Alkylation of 1-naphthol in trifluoroethanol versus in DMSO
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Khary Williams
Alkylation of 1-naphthol in trifluoroethanol versus in DMSO
The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:
Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.
In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.
EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."
The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:
Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.
In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.
EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."
.But there isnt any scope of O-alkylation in the protic medium, because being in a highly polar and protic medium, oxygen will be highly solvated. The upcoming electrophile can not approach the Oxygen atom because of steric effects by the solvent, and as far as breaking aromaticity is concerned, if reaction were just to be relied on that, How do we have electrophilic aromatic substitution on benzene ring. All intermediates involved there requires breaking of aromaticity which is regained in the final productMore
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These solvents should give identical products.
DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).
Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't believe there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral.
DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).
Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't believe there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral.
The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:
Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.
In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.
EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."
The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:
Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.
In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.
EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."
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These solvents should give identical products.
DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).
Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't believe there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral.
These solvents should give identical products.
DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).
Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't believe there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral.
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