Stereochemical outcomes in opening of vinyl epoxidesRegioselectivity 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
Why aren't satellites disintegrated even though they orbit earth within earth's Roche Limits?
What's is the easiest way to purchase a stock and hold it
Can ThermodynamicData be used with NSolve?
Why are stats in Angband written as 18/** instead of 19, 20...?
Why does the U.S military use mercenaries?
Gambler's Fallacy Dice
Don't replace "|" with "(empty)" when generating slugs from title?
Why is the S-duct intake on the Tu-154 uniquely oblong?
how to create an executable file for an AppleScript?
Taylor series leads to two different functions - why?
Can 2 light bulbs of 120V in series be used on 230V AC?
What would be the game balance implications for using the Gygax method for applying falling damage?
pwaS eht tirsf dna tasl setterl fo hace dorw
Why using a variable as index of a list-item does not retrieve that item with clist_item:Nn?
Former Employer just sent me an IP Agreement
Should all adjustments be random effects in a mixed linear effect?
Why use a retrograde orbit?
Hotel booking: Why is Agoda much cheaper than booking.com?
Will this series of events work to drown a tarrasque?
Shortest amud or daf in Shas?
Gaussian kernel density estimation with data from file
How do you cope with rejection?
How does this piece of code determine array size without using sizeof( )?
Why is so much ransomware breakable?
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 May 6 at 0:27
orthocresol♦
41.2k7123254
asked May 5 at 22:13
J. DeansJ. Deans
505
$endgroup$
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 May 6 at 0:27
orthocresol♦
41.2k7123254
asked May 5 at 22:13
J. DeansJ. Deans
505
$endgroup$
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
May 5 at 22:21
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♦
May 5 at 23:22
$begingroup$
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
May 5 at 23:35
$begingroup$
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♦
May 6 at 0:11
$begingroup$
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♦
May 6 at 0:16
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 May 6 at 0:27
orthocresol♦
41.2k7123254
asked May 5 at 22:13
J. DeansJ. Deans
505
$endgroup$
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 May 6 at 0:27
orthocresol♦
41.2k7123254
asked May 5 at 22:13
J. DeansJ. Deans
505
edited May 6 at 0:27
orthocresol♦
41.2k7123254
asked May 5 at 22:13
J. DeansJ. Deans
505
edited May 6 at 0:27
orthocresol♦
41.2k7123254
edited May 6 at 0:27
orthocresol♦
41.2k7123254
edited May 6 at 0:27
orthocresol♦
41.2k7123254
41.2k7123254
asked May 5 at 22:13
J. DeansJ. Deans
505
asked May 5 at 22:13
J. DeansJ. Deans
505
asked May 5 at 22:13
J. DeansJ. Deans
505
505
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
May 5 at 22:21
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♦
May 5 at 23:22
$begingroup$
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
May 5 at 23:35
$begingroup$
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♦
May 6 at 0:11
$begingroup$
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♦
May 6 at 0:16
add a comment |
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
May 5 at 22:21
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♦
May 5 at 23:22
$begingroup$
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
May 5 at 23:35
$begingroup$
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♦
May 6 at 0:11
$begingroup$
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♦
May 6 at 0:16
2
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
May 5 at 22:21
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
May 5 at 22:21
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♦
May 5 at 23:22
$begingroup$
(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♦
May 5 at 23:22
$begingroup$
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
May 5 at 23:35
$begingroup$
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
May 5 at 23:35
$begingroup$
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♦
May 6 at 0:11
$begingroup$
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♦
May 6 at 0:11
$begingroup$
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♦
May 6 at 0:16
$begingroup$
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♦
May 6 at 0:16
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 May 6 at 1:52
user55119user55119
4,46711242
$endgroup$
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "431"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f114882%2fstereochemical-outcomes-in-opening-of-vinyl-epoxides%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 May 6 at 1:52
user55119user55119
4,46711242
$endgroup$
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
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 May 6 at 1:52
user55119user55119
4,46711242
$endgroup$
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
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 May 6 at 1:52
user55119user55119
4,46711242
$endgroup$
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 May 6 at 1:52
user55119user55119
4,46711242
edited May 6 at 2:06
answered May 6 at 1:52
user55119user55119
4,46711242
answered May 6 at 1:52
user55119user55119
4,46711242
answered May 6 at 1:52
user55119user55119
4,46711242
4,46711242
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
add a comment |
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
From 6 to 7 neighbouring group effect , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?
$endgroup$
– Chakravarthy Kalyan
May 7 at 4:35
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
$begingroup$
No, that is why it is reactive. 4-Membered rings do form.
$endgroup$
– user55119
May 7 at 11:37
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f114882%2fstereochemical-outcomes-in-opening-of-vinyl-epoxides%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.
$endgroup$
– Mathew Mahindaratne
May 5 at 22:21
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♦
May 5 at 23:22
$begingroup$
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
May 5 at 23:35
$begingroup$
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♦
May 6 at 0:11
$begingroup$
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♦
May 6 at 0:16