Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: portugal on February 27, 2008, 01:45:43 AM
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What would the mechanism look like for the ring opening of trans-2,3-dimethyloxirane with HBr look like?
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What is your nucleophile(s)? what is your electrophile(s)?
nucleophiles attack electrophiles.
hint - epoxides open by different mechanisms whether the reaction is run in acidic or basic conditions. which are you working under?
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Yea nucleophile is Br^-1 and electrophile is H^+1. Its acidic condition and in particualar very acidic as HBr is a strong acid.
Its the actual mechanism that i find difficult in this case because in 1st year organic we never saw nothing like this type.
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Good, we know we're in acid, so the epoxide opening mechanism will follow the acidic conditions.
H+ is electrophilic, Br- is nueclophilic. If that were the end of the story, there'd be no reaction, as Br- would attack H+ and that would be the end of it. There's one more nucleophile, can you find it? What electrophile might that attack? How does that change the epoxide? Does it make any other atoms more nucleophilic or electrophilic? If so which ones, and how might the mechanism proceed from there?
start here:
http://www.cem.msu.edu/~reusch/VirtTxtJml/alcohol2.htm#ethnom
http://www.vanderbilt.edu/AnS/Chemistry/Rizzo/Chem220b/Chapter_18.pdf
Bonus question: is trans-2,3-dimethyloxirane chiral? what about cis-2,3-dimethyloxirane - is it chiral?
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All i can think of is that other nucleophile is the oxygen because of its lone pair of electrtons and this attacks the hyfrogen resulting in epoxed being converted into alcholol. i dont know the rest especially which steps go were in the mechanism becuase of two nucleophiles etc.??
plus bonus question both are chiral as contain 4 diff groups on the stereogenic carbon centres.
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Good, the oxygen is the other nucleophile. You're exactly right, the oxygen lone pair will attack the proton as the first step of the mechanism, but it won't form an alcohol straight out. The lone pair does the attacking, not the electrons from one of the carbon-oxygen bonds. What will form, though, is an epoxide, where the oxygen atom is still attached to both carbon atoms along with a proton. What is special about this type of oxygen atom? How might that modulate the reactivity of the epoxide. That is, after the first step where you have correctly indicated the oxygen atom attacks a proton, what are the new nucleophile(s)? what are the new electrophile(s)? Nucleophiles attack electrophiles.
As per the bonus, why is cis-2,3-dimethyloxirane not a chiral molecule? what is unique about the enantiomer of cis-2,3-dimethyloxirane?
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I am really lost now i need to see what this mechansim looks like for my practical 2morrow could someone please do me a favour and draw one up
thank you
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No, I won't draw it all out for you, but I will draw up what you've figured out so far. See attached.
So far, we have activation of the epoxide oxygen with acid. Given the intermediate we have drawn, what are the nucleophiles? what are the electrophiles?
hint: we already listed 2 nucleophiles and 1 electrophile from the very first step of the mechanism. now, there is 1 nucleophile (was also a nucleophile before) and 2 electrophiles (1 was an electrophile before, and 1 new one). can you find them all? Remembering that nucleophiles attack electrophiles, can you predict what the next step of the mechanism is?
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the nucleophile is Br and electrophile is C-O bond and H ion. is that it . next step is transtion stae thhen breaking of ring????i am not sure how it goes from here
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Yes ... almost. Br isn't the nucleophile, Br- is. The 2 electrophiles are indeed the H of the O-H bond and the C of the C-O bond. In this particular example, both C-O bonds are identical, and the Br- will attack both equally. (do you know which carbon atom would be attacked if the carbons were different, say in 1,1-dimethyloxirane?)
Thus, the next productive step in our mechanism is indeed the opening of the epoxide ring by Br-. With careful minding of stereochemistry, can you describe, draw out, or name the product that will be formed?
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I have done all that see i know what i am talking about its just my mechanism sucks i need to see one that is good and its like 111am at moment and i have a prac 2morrow at 8 so i am rushing.
since i have answered all your questions could i please hav your mechanism chemistry genius
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Show this mechanism on the forum (use additional otions for graphic).
Azmanam will correct it if needed. These are forum rules
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i dont have program to do this mechanism creations. Plus i dont know what to do. Forum rules arent to great on here then
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You can get a free structure drawing program called ISISDraw. You'll need to register (free - http://www.mdli.com/my_account/register1.jsp), then choose MDL/ISISDraw from the dropdown on this page (http://www.mdli.com/downloads/downloadable/index.jsp). follow instructions to download MDL ISISDraw 2.5.
It's not the best structure drawing program, but it's the free-est.
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I will have to leave it now i will ask a demonstrator about it in prac tomorrow
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You can use free ISIS DRAW.
Look at the mechanism at:
http://depts.washington.edu/chemcrs/bulkdisk/chem237A_aut05/exams_Final%20Key-Pink.pdf
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I have worked the product (3-bromo-2-butanol) and the mechanism out now for both the trans(2R, 3R) and Trans (2S,3S), however i have come across 2 products for each of them becuase of the equla probability that the C-O bond will be broken on either side and i need to know which one of each two is the major product. How am i supposed to know that???
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trans(2R, 3R) and Trans (2S,3S)
wrong using trans - trans suggest double bond
i need to know which one of each two is the major product. How am i supposed to know that???
It is normal in the case of more symmetric compounds that both products are formed in exactly the same amounts
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@ AWK: agreed.
cis/trans is generally reserved for geometry about a double bond, and sometimes for relative positions in a cyclic system. syn/anti is also generally used for relative positions in a cyclic systems, but it can also be used for net creation of 2 new stereocenters.
And you are correct, you form a racemic mixture of two products, and both will be formed in equal amounts.
I didn't check your r/s nomenclature, but assuming you prioritized each substituent correctly, then a job very well done.