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Effect of substituents on arene oxide-phenol rearrangement
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Matt Walsh
Effect of substituents on arene oxide-phenol rearrangement
What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.
First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.
Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:
Hence, mechanism 1 should be the preferred pathway
I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:
However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.
Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.
o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well
Clarification
I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:
First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1
Now, the intermediate 1(a) will try go for rearrangements that will generate species which are more energetically favorable. Due to the electron donating(+M effect) of the hydroxyl present adjacent to the cationic part in 1(a), a 1,2 deuterium shift will occur leading to the formation of 1(b) as shown in the third step of Reaction Mechanism-1
Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all
What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.
First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.
Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:
Hence, mechanism 1 should be the preferred pathway
I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:
However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.
Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.
o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well
Clarification
I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:
First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1
Now, the intermediate 1(a) will try go for rearrangements that will generate species which are more energetically favorable. Due to the electron donating(+M effect) of the hydroxyl present adjacent to the cationic part in 1(a), a 1,2 deuterium shift will occur leading to the formation of 1(b) as shown in the third step of Reaction Mechanism-1
Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all
Thanks for the answer! Now, I understand why is 1(a) more stable than 2(a). Similarly, we can say that, 1(b) is less stable than 2(b). If the presence of deuterium doesnt matter, how to decide which of the two pathways is feasible? For the labelled intermediates, relative potential energies follows this order: 2(b)<1(a)<1(b)<2(a). Product (b)s are more stable than (a)s due to the electron donating nature of hydroxyl group. Im unable to figure out which reaction will be preferred from this data. Did you consider any other things to arrive at the conclusion - reaction 1 is the preferred one?More
What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.
First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.
Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:
Hence, mechanism 1 should be the preferred pathway
I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:
However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.
Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.
o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well
Clarification
I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:
First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1
Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all
What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.
First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.
Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:
Hence, mechanism 1 should be the preferred pathway
I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:
However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.
Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.
o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well
Clarification
I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:
First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1
Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all
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