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Stereochemical outcomes in opening of vinyl epoxides
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Stereochemical outcomes in opening of vinyl epoxides
Regioselectivity of acid-catalyzed ring-opening of epoxidesHow can I work out what reaction will happen?Deciding the major product in this Wurtz reactionWhy (abundance of M+1)/(abundance of M)=(number of carbon atoms)*(1.1/100)?Why is the endo product the major product in a Diels-Alder reaction?Side chain formation in polymerisation of ethyleneRegioselectivity in addition of hydrogen iodide to vinyl chlorideHow to rationalise the major product formed in a nucleophile promoted epoxide cleavage reaction?Why is this β hydrogen in trans position to the leaving group?Explaining stereochemistry in a sigmatropic 1,3-alkyl migration
$begingroup$
Explain the outcomes of the following reactions. In each case the major product is shown.
Firstly, I don't understand why the chloride always attacks on the right hand side of the epoxide. Also, regarding the stereochemistry for (ii), I assume the chlorine can open the epoxide as it is antiperiplanar to the left hand side of the epoxide?
organic-chemistry nucleophilic-substitution regioselectivity stereoselectivity
edited 3 hours ago
orthocresol♦
40.7k7120252
asked 5 hours ago
J. DeansJ. Deans
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Explain the outcomes of the following reactions. In each case the major product is shown.
Firstly, I don't understand why the chloride always attacks on the right hand side of the epoxide. Also, regarding the stereochemistry for (ii), I assume the chlorine can open the epoxide as it is antiperiplanar to the left hand side of the epoxide?
organic-chemistry nucleophilic-substitution regioselectivity stereoselectivity
edited 3 hours ago
orthocresol♦
40.7k7120252
asked 5 hours ago
J. DeansJ. Deans
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
5 hours ago
1
$begingroup$
(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.
$endgroup$
– orthocresol♦
4 hours ago
$begingroup$
Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does
$endgroup$
– J. Deans
3 hours ago
$begingroup$
I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
Ugh, I can't believe I'm actually correct: pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.
$endgroup$
– orthocresol♦
3 hours ago
add a comment |
$begingroup$
Explain the outcomes of the following reactions. In each case the major product is shown.
Firstly, I don't understand why the chloride always attacks on the right hand side of the epoxide. Also, regarding the stereochemistry for (ii), I assume the chlorine can open the epoxide as it is antiperiplanar to the left hand side of the epoxide?
organic-chemistry nucleophilic-substitution regioselectivity stereoselectivity
edited 3 hours ago
orthocresol♦
40.7k7120252
asked 5 hours ago
J. DeansJ. Deans
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Explain the outcomes of the following reactions. In each case the major product is shown.
Firstly, I don't understand why the chloride always attacks on the right hand side of the epoxide. Also, regarding the stereochemistry for (ii), I assume the chlorine can open the epoxide as it is antiperiplanar to the left hand side of the epoxide?
organic-chemistry nucleophilic-substitution regioselectivity stereoselectivity
organic-chemistry nucleophilic-substitution regioselectivity stereoselectivity
edited 3 hours ago
orthocresol♦
40.7k7120252
asked 5 hours ago
J. DeansJ. Deans
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
orthocresol♦
40.7k7120252
asked 5 hours ago
J. DeansJ. Deans
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
orthocresol♦
40.7k7120252
edited 3 hours ago
orthocresol♦
40.7k7120252
edited 3 hours ago
orthocresol♦
40.7k7120252
40.7k7120252
asked 5 hours ago
J. DeansJ. Deans
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
J. DeansJ. Deans
483
asked 5 hours ago
J. DeansJ. Deans
483
483
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J. Deans is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
5 hours ago
1
$begingroup$
(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.
$endgroup$
– orthocresol♦
4 hours ago
$begingroup$
Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does
$endgroup$
– J. Deans
3 hours ago
$begingroup$
I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
Ugh, I can't believe I'm actually correct: pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.
$endgroup$
– orthocresol♦
3 hours ago
add a comment |
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
5 hours ago
1
$begingroup$
(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.
$endgroup$
– orthocresol♦
4 hours ago
$begingroup$
Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does
$endgroup$
– J. Deans
3 hours ago
$begingroup$
I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
Ugh, I can't believe I'm actually correct: pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.
$endgroup$
– orthocresol♦
3 hours ago
2
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
5 hours ago
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
5 hours ago
1
1
$begingroup$
(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.
$endgroup$
– orthocresol♦
4 hours ago
$begingroup$
(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.
$endgroup$
– orthocresol♦
4 hours ago
$begingroup$
Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does
$endgroup$
– J. Deans
3 hours ago
$begingroup$
Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does
$endgroup$
– J. Deans
3 hours ago
$begingroup$
I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
Ugh, I can't believe I'm actually correct: pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
Ugh, I can't believe I'm actually correct: pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.
$endgroup$
– orthocresol♦
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii). The role of water is to cleave the silylether.
answered 1 hour ago
user55119user55119
4,44211142
$endgroup$
add a comment |
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$begingroup$
While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii). The role of water is to cleave the silylether.
answered 1 hour ago
user55119user55119
4,44211142
$endgroup$
add a comment |
$begingroup$
While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii). The role of water is to cleave the silylether.
answered 1 hour ago
user55119user55119
4,44211142
$endgroup$
add a comment |
$begingroup$
While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii). The role of water is to cleave the silylether.
answered 1 hour ago
user55119user55119
4,44211142
$endgroup$
While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii). The role of water is to cleave the silylether.
answered 1 hour ago
user55119user55119
4,44211142
edited 1 hour ago
answered 1 hour ago
user55119user55119
4,44211142
answered 1 hour ago
user55119user55119
4,44211142
answered 1 hour ago
user55119user55119
4,44211142
4,44211142
add a comment |
add a comment |
J. Deans is a new contributor. Be nice, and check out our Code of Conduct.
J. Deans is a new contributor. Be nice, and check out our Code of Conduct.
J. Deans is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
5 hours ago
1
$begingroup$
(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.
$endgroup$
– orthocresol♦
4 hours ago
$begingroup$
Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does
$endgroup$
– J. Deans
3 hours ago
$begingroup$
I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.
$endgroup$
– orthocresol♦
3 hours ago
$begingroup$
Ugh, I can't believe I'm actually correct: pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.
$endgroup$
– orthocresol♦
3 hours ago