8a-methyl-1,2,3,4,4a,8a-hexahydronaphthalen-4a-ylium carbocation rearrangement
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In this reaction after the attack of lone pairs on $ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?
carbocation rearrangements
edited May 28 at 16:09
Loong♦
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asked May 28 at 13:24
Sameer ThakurSameer Thakur
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In this reaction after the attack of lone pairs on $ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?
carbocation rearrangements
edited May 28 at 16:09
Loong♦
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asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
$endgroup$
$begingroup$
Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried?
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– Michael Lautman
May 28 at 13:43
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Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.
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– Sameer Thakur
May 28 at 13:54
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chemistry.stackexchange.com/questions/98588/…
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– Mithoron
May 30 at 23:02
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$begingroup$
In this reaction after the attack of lone pairs on $ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?
carbocation rearrangements
edited May 28 at 16:09
Loong♦
37.4k9 gold badges93 silver badges198 bronze badges
asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
$endgroup$
In this reaction after the attack of lone pairs on $ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?
carbocation rearrangements
carbocation rearrangements
edited May 28 at 16:09
Loong♦
37.4k9 gold badges93 silver badges198 bronze badges
asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
edited May 28 at 16:09
Loong♦
37.4k9 gold badges93 silver badges198 bronze badges
asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
edited May 28 at 16:09
Loong♦
37.4k9 gold badges93 silver badges198 bronze badges
edited May 28 at 16:09
Loong♦
37.4k9 gold badges93 silver badges198 bronze badges
edited May 28 at 16:09
Loong♦
37.4k9 gold badges93 silver badges198 bronze badges
37.4k9 gold badges93 silver badges198 bronze badges
asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
asked May 28 at 13:24
Sameer ThakurSameer Thakur
1146 bronze badges
1146 bronze badges
$begingroup$
Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried?
$endgroup$
– Michael Lautman
May 28 at 13:43
$begingroup$
Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.
$endgroup$
– Sameer Thakur
May 28 at 13:54
$begingroup$
chemistry.stackexchange.com/questions/98588/…
$endgroup$
– Mithoron
May 30 at 23:02
add a comment
|
$begingroup$
Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried?
$endgroup$
– Michael Lautman
May 28 at 13:43
$begingroup$
Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.
$endgroup$
– Sameer Thakur
May 28 at 13:54
$begingroup$
chemistry.stackexchange.com/questions/98588/…
$endgroup$
– Mithoron
May 30 at 23:02
$begingroup$
Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried?
$endgroup$
– Michael Lautman
May 28 at 13:43
$begingroup$
Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried?
$endgroup$
– Michael Lautman
May 28 at 13:43
$begingroup$
Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.
$endgroup$
– Sameer Thakur
May 28 at 13:54
$begingroup$
Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.
$endgroup$
– Sameer Thakur
May 28 at 13:54
$begingroup$
chemistry.stackexchange.com/questions/98588/…
$endgroup$
– Mithoron
May 30 at 23:02
$begingroup$
chemistry.stackexchange.com/questions/98588/…
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– Mithoron
May 30 at 23:02
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4 Answers
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I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^circ$-carbocation, which is compatible with the initial $3^circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:
edited Jun 7 at 15:04
Glorfindel
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answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
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You are right. The hydride shift is redundant in my solution.
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– Sameer Thakur
May 28 at 19:03
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While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued 
Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.
answered May 28 at 22:43
user55119user55119
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It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land.
There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons. In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic. Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.
answered May 28 at 14:15
Oscar LanziOscar Lanzi
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Why do we need to move the methyl group in the lower ring?
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– Sameer Thakur
May 28 at 14:47
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Upper ring otherwise cannot be fully conjugated.
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– Oscar Lanzi
May 28 at 14:51
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Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.
answered May 28 at 15:11
Sameer ThakurSameer Thakur
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Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
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– Withnail
May 28 at 15:29
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You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
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– Withnail
May 28 at 15:34
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Not sure about this. I'm still curious about the possibility of a nonclassical ion.
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– Oscar Lanzi
May 28 at 15:47
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@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
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– Sameer Thakur
May 28 at 15:51
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@OscarLanzi what do you mean by a nonclassical ion?
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– Sameer Thakur
May 28 at 15:51
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4 Answers
4
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4 Answers
4
active
oldest
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$begingroup$
I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^circ$-carbocation, which is compatible with the initial $3^circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:
edited Jun 7 at 15:04
Glorfindel
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answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
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You are right. The hydride shift is redundant in my solution.
$endgroup$
– Sameer Thakur
May 28 at 19:03
add a comment
|
$begingroup$
I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^circ$-carbocation, which is compatible with the initial $3^circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:
edited Jun 7 at 15:04
Glorfindel
1,5374 gold badges11 silver badges20 bronze badges
answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
11.6k2 gold badges15 silver badges40 bronze badges
$endgroup$
$begingroup$
You are right. The hydride shift is redundant in my solution.
$endgroup$
– Sameer Thakur
May 28 at 19:03
add a comment
|
$begingroup$
I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^circ$-carbocation, which is compatible with the initial $3^circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:
edited Jun 7 at 15:04
Glorfindel
1,5374 gold badges11 silver badges20 bronze badges
answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
11.6k2 gold badges15 silver badges40 bronze badges
$endgroup$
I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^circ$-carbocation, which is compatible with the initial $3^circ$-carbocation given by elimination of water. Thus, I think the following mechanism is a very reliable one for gaining aromaticity:
edited Jun 7 at 15:04
Glorfindel
1,5374 gold badges11 silver badges20 bronze badges
answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
11.6k2 gold badges15 silver badges40 bronze badges
edited Jun 7 at 15:04
Glorfindel
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edited Jun 7 at 15:04
Glorfindel
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edited Jun 7 at 15:04
Glorfindel
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1,5374 gold badges11 silver badges20 bronze badges
answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
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answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
11.6k2 gold badges15 silver badges40 bronze badges
answered May 28 at 18:11
Mathew MahindaratneMathew Mahindaratne
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11.6k2 gold badges15 silver badges40 bronze badges
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You are right. The hydride shift is redundant in my solution.
$endgroup$
– Sameer Thakur
May 28 at 19:03
add a comment
|
$begingroup$
You are right. The hydride shift is redundant in my solution.
$endgroup$
– Sameer Thakur
May 28 at 19:03
$begingroup$
You are right. The hydride shift is redundant in my solution.
$endgroup$
– Sameer Thakur
May 28 at 19:03
$begingroup$
You are right. The hydride shift is redundant in my solution.
$endgroup$
– Sameer Thakur
May 28 at 19:03
add a comment
|
$begingroup$
While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued
Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.
answered May 28 at 22:43
user55119user55119
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$endgroup$
add a comment
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$begingroup$
While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued
Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.
answered May 28 at 22:43
user55119user55119
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$endgroup$
add a comment
|
$begingroup$
While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued
Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.
answered May 28 at 22:43
user55119user55119
5,4542 gold badges14 silver badges46 bronze badges
$endgroup$
While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued
Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.
answered May 28 at 22:43
user55119user55119
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answered May 28 at 22:43
user55119user55119
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answered May 28 at 22:43
user55119user55119
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answered May 28 at 22:43
user55119user55119
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add a comment
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add a comment
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$begingroup$
It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land.
There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons. In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic. Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.
answered May 28 at 14:15
Oscar LanziOscar Lanzi
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$endgroup$
$begingroup$
Why do we need to move the methyl group in the lower ring?
$endgroup$
– Sameer Thakur
May 28 at 14:47
$begingroup$
Upper ring otherwise cannot be fully conjugated.
$endgroup$
– Oscar Lanzi
May 28 at 14:51
add a comment
|
$begingroup$
It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land.
There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons. In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic. Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.
answered May 28 at 14:15
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
$endgroup$
$begingroup$
Why do we need to move the methyl group in the lower ring?
$endgroup$
– Sameer Thakur
May 28 at 14:47
$begingroup$
Upper ring otherwise cannot be fully conjugated.
$endgroup$
– Oscar Lanzi
May 28 at 14:51
add a comment
|
$begingroup$
It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land.
There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons. In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic. Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.
answered May 28 at 14:15
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
$endgroup$
It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land.
There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, the methyl group moves over the bridging bond but not all the way across so that the bond between the methyl group is delocalized between the two bridgehead carbons. In this structure, if it forms, the delocalized electrons are also conjugated across the bridge like an additional pi bonding pair so that the top ring (6 conjugated electrons) and the bridge structure (2 electrons) are both aromatic. Possibly other users more familiar with the system can indicate whether this structure forms, leading to a new question.
answered May 28 at 14:15
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
edited May 28 at 15:46
answered May 28 at 14:15
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
answered May 28 at 14:15
Oscar LanziOscar Lanzi
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answered May 28 at 14:15
Oscar LanziOscar Lanzi
19k2 gold badges32 silver badges58 bronze badges
19k2 gold badges32 silver badges58 bronze badges
$begingroup$
Why do we need to move the methyl group in the lower ring?
$endgroup$
– Sameer Thakur
May 28 at 14:47
$begingroup$
Upper ring otherwise cannot be fully conjugated.
$endgroup$
– Oscar Lanzi
May 28 at 14:51
add a comment
|
$begingroup$
Why do we need to move the methyl group in the lower ring?
$endgroup$
– Sameer Thakur
May 28 at 14:47
$begingroup$
Upper ring otherwise cannot be fully conjugated.
$endgroup$
– Oscar Lanzi
May 28 at 14:51
$begingroup$
Why do we need to move the methyl group in the lower ring?
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– Sameer Thakur
May 28 at 14:47
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Why do we need to move the methyl group in the lower ring?
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– Sameer Thakur
May 28 at 14:47
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Upper ring otherwise cannot be fully conjugated.
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– Oscar Lanzi
May 28 at 14:51
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Upper ring otherwise cannot be fully conjugated.
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– Oscar Lanzi
May 28 at 14:51
add a comment
|
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Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
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$begingroup$
Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
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– Withnail
May 28 at 15:29
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You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
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– Withnail
May 28 at 15:34
$begingroup$
Not sure about this. I'm still curious about the possibility of a nonclassical ion.
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– Oscar Lanzi
May 28 at 15:47
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@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
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– Sameer Thakur
May 28 at 15:51
$begingroup$
@OscarLanzi what do you mean by a nonclassical ion?
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– Sameer Thakur
May 28 at 15:51
|
show 1 more comment
$begingroup$
Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
$endgroup$
$begingroup$
Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
$endgroup$
– Withnail
May 28 at 15:29
$begingroup$
You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
$endgroup$
– Withnail
May 28 at 15:34
$begingroup$
Not sure about this. I'm still curious about the possibility of a nonclassical ion.
$endgroup$
– Oscar Lanzi
May 28 at 15:47
$begingroup$
@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
@OscarLanzi what do you mean by a nonclassical ion?
$endgroup$
– Sameer Thakur
May 28 at 15:51
|
show 1 more comment
$begingroup$
Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
$endgroup$
Okay I figured it out. First I drew the resonating structure for the carbocation and then performed a methyl shift followed by a hydride shift. Then the H+ ion leaves via the dehydration mechanism making the ring aromatic in the process.
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
answered May 28 at 15:11
Sameer ThakurSameer Thakur
1146 bronze badges
1146 bronze badges
$begingroup$
Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
$endgroup$
– Withnail
May 28 at 15:29
$begingroup$
You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
$endgroup$
– Withnail
May 28 at 15:34
$begingroup$
Not sure about this. I'm still curious about the possibility of a nonclassical ion.
$endgroup$
– Oscar Lanzi
May 28 at 15:47
$begingroup$
@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
@OscarLanzi what do you mean by a nonclassical ion?
$endgroup$
– Sameer Thakur
May 28 at 15:51
|
show 1 more comment
$begingroup$
Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
$endgroup$
– Withnail
May 28 at 15:29
$begingroup$
You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
$endgroup$
– Withnail
May 28 at 15:34
$begingroup$
Not sure about this. I'm still curious about the possibility of a nonclassical ion.
$endgroup$
– Oscar Lanzi
May 28 at 15:47
$begingroup$
@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
@OscarLanzi what do you mean by a nonclassical ion?
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
$endgroup$
– Withnail
May 28 at 15:29
$begingroup$
Your first two "resonance structures" are confusing. Have you migrated the methyl group (not resonance) or turned the molecule over?
$endgroup$
– Withnail
May 28 at 15:29
$begingroup$
You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
$endgroup$
– Withnail
May 28 at 15:34
$begingroup$
You can get the second structure, just by resonance among the 4electron, 5 center pi system on the top ring. Then turn the molecule over.
$endgroup$
– Withnail
May 28 at 15:34
$begingroup$
Not sure about this. I'm still curious about the possibility of a nonclassical ion.
$endgroup$
– Oscar Lanzi
May 28 at 15:47
$begingroup$
Not sure about this. I'm still curious about the possibility of a nonclassical ion.
$endgroup$
– Oscar Lanzi
May 28 at 15:47
$begingroup$
@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
@Withnail I moved the double bonds towards left through resonance and then turned the molecule over. It is confusing but it would proceed the same way even without turning it.
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
@OscarLanzi what do you mean by a nonclassical ion?
$endgroup$
– Sameer Thakur
May 28 at 15:51
$begingroup$
@OscarLanzi what do you mean by a nonclassical ion?
$endgroup$
– Sameer Thakur
May 28 at 15:51
|
show 1 more comment
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$begingroup$
Be careful with your thinking. H+ ions do not attack. The lone pair on the alcohol attacks (or accepts ) H+. You should always think of mechanisms in terms of the movement of electrons. Have you attempted a mechanism? Can you show us what you have tried?
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– Michael Lautman
May 28 at 13:43
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Yes I phrased it wrong. The lone pair on alcohol attacks on the H+ ions to convert it into a good leaving group ie. water and the water leaves to form a carbocation as shown.
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– Sameer Thakur
May 28 at 13:54
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chemistry.stackexchange.com/questions/98588/…
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– Mithoron
May 30 at 23:02