[portugal]

If this(3E,6S,7S,8S)-8-(2'-cyclohexenyl)-6,7-epoxy-6-ethyl-1,3-nonadiene is reacted with methanol and sulfuric acid then its going to be an acid cataluysed epoxide opening and since its a weak nucleophile it open from the more substituted cyclohexenyl side and thus resuls in OH being on carbon 6. Thus the product has to be (3E,6R,8R)-8-(2'-cyclohexenyl)-6-ethyl-7-methoxy-1,3-nonadien-6-ol with racemisation at C7.

Is that right?

also if you react a disubstitutedcyclobutene with cold KMNO4 then you are going to get pairs of diastereomers and enatiomers as the product yea?

[sjb]

If this(3E,6S,7S,8S)-8-(2'-cyclohexenyl)-6,7-epoxy-6-ethyl-1,3-nonadiene is reacted with methanol and sulfuric acid then its going to be an acid cataluysed epoxide opening and since its a weak nucleophile it open from the more substituted cyclohexenyl side and thus resuls in OH being on carbon 6. Thus the product has to be (3E,6R,8R)-8-(2'-cyclohexenyl)-6-ethyl-7-methoxy-1,3-nonadien-6-ol with racemisation at C7.

Is that right?

In my experience ring-opening epoxides tends to be done with little loss of stereointegrity. Can you post an image of the epoxide, as I have gotten lost drawing it out to try and answer?

also if you react a disubstitutedcyclobutene with cold KMNO4 then you are going to get pairs of diastereomers and enatiomers as the product yea?

Private message, received Mon, 17 Mar 2008 11:38:48 +0000

the precise question is this:
What is the expected major organic product expected from reaction of one enantiomer of 2,3-dimethylcyclobutene with cold potassium permanganate under basic conditions?

thats why i say a racemic mixture

This is similar to the situation I assumed.

OK, let's simplify a bit.

What would be the expected product from cyclobutene?
 - are there enantiomeric forms there? Perhaps name the compounds formed?

And from 1-methylcyclobutene?
 - similarly a good way to see if the diols formed are enantiomers is to name these, too.

And finally from 2,3-dimethylcyclobut-1-ene?
 - As you have "an enantiomer of a particular cycloalkene" presumably there's a chiral centre in your molecule - where?

So let's arbitrarily assign that centre, for instance, (R)

Does the configuration of that centre change in the oxidation?

What are the names of the compounds formed?

What is the definition of enantiomeric pairs?

Are these forms enantiomers?

Consider the mechanism of the oxidation - can you see anything that may favour (or disfavour) one pathway over another?

Does this mean the oxidation forms both isomers in equal amounts? Does this mean a racemic mixture?

S

(From topic http://www.chemicalforums.com/index.php?topic=24428.0)

Have you worked through these? If so, then the answer should be reasonably obvious, it may be different if it's a different cyclobutene though.

[portugal]

its a bit hard to draw it out on here but i think i am right anyway with that 1st question.

yes i have answered that and the my first answer was wrong for the post a few days ago. it wasnt a racemic mixture. but for this one they dont tell you wheter its trans or cis and therefire you would that both are going to exist in the reaction thereofre you will get 4 products overall in which you get 2 paris of enationmers and 2 pairs of diastereomers. I dont think meso compounds would have to do with these typoe of questions so i disregarded that in my consideration of product formation. still not 100% sure thats why i came on here to get chemistry expert help

[Dan]

its going to be an acid cataluysed epoxide opening and since its a weak nucleophile it open from the more substituted cyclohexenyl side and thus resuls in OH being on carbon 6. Thus the product has to be (3E,6R,8R)-8-(2'-cyclohexenyl)-6-ethyl-7-methoxy-1,3-nonadien-6-ol with racemisation at C7.

Is that right?

No, draw the structure, C6 is a quaternary centre, C7 is tertiary - ie C6 is the more substituted side of the epoxide and partial positive charge will develop here more easily - so the nucleophile will attack at C6, and will proceed with inversion. This is sometimes called a "loose Sn2" mechanism.

[sjb]

Thinking on it, I'm not sure racemisation is the term you should use here.

It's true that position 7 is no longer chiral, but that's merely because the acid you used had the same isotope of hydrogen as the hydrogen already in place. If you had used D₂SO₄ and MeOD then you'd retain knowledge of the chirality. Sorry if I wasn't that clear.

As to the second part, it's not always the case that you get a pair of diastereomers or enantiomers when you react a disubstituted cyclobutene with KMnO₄. Assuming methyl substitution for simplicity's sake, what are the products, for instance when you oxidize (3R,4R)-3,4-dimethylcyclobutene, or 3,3-dimethylcyclobutene?

I make it 9 distinct disubstituted cyclobutenes, of which 2 could be meso, 4 have only one chiral centre, 2 have 2 chiral centres, and 1 is not chiral at all.

[portugal]

well for the disubstiuted cyclobutene its in particular a 2,3-diethylcyclobutene but its not saying which of the 2 possible enantiomers its a more general question and thats why i reckon diasteromic pairs and enantiomer pairs.

I definitely know that we were told in lectures by our professor that when its acid catalysed epoxide ring opening with a weak nucleophile like methanol, you get opening occuring from the more substituted side with like dan said loose sn2. But we were told that racemisation occurs at that carbon and hydroxyl carbon has same stereochemistry.
thereofore i change my answer to this then (3E,7S,8S)-8-(2'-cyclohexenyl)-6-ethyl-6-methoxy-1,3-nonadien-7-ol with racemisation at C6. Does anyone out there agree with me??

[sjb]

well for the disubstiuted cyclobutene its in particular a 2,3-diethylcyclobutene but its not saying which of the 2 possible enantiomers its a more general question and thats why i reckon diasteromic pairs and enantiomer pairs.

Fine. I'm not 100% sure what compound you mean here. The compound I call "2,3-diethylcyclobutene" would also be called "1,4-diethylcyclobutene", and has a four-membered ring, in the ring is a double bond, and on one sp² carbon you have a ethyl group, with a ethyl on the sp³ carbon next to it.

But you also mention meso forms, which makes me wonder if the compound in question is in fact 3,4-diethylcyclobutene. This has the ring as before, but ethyls on the 2 sp³ carbons.

Assuming the former, then there are only two chiral forms, making 4 diastereomeric diols. But the cyclobutene itself does not exist in cis / trans forms.

Here you get 4 cis-diols from 2 alkenes.

If you mean the latter, then you have 3 forms, the (3R,4R), the (3S,4S), and the (3R,4S). As there is a plane of symmetry running down the middle of the last molecule (between carbons 1 & 2, and 3 & 4), the (3S,4R) molecule is the same as the last mentioned.

Cis-dihydroxylation of the (3R,4R) form (and similarly the (3S,4S) form) will result in only one diol each, as there will always be the 2 hydroxyl groups on the same side as one ethyl, and opposite the other. If you cis-dihydroxylate the meso form then in principle you can get a form where the ethyls and hydroxyls are all on the same side, and one where the ethyls and hydroxyls are on different sides.

Here you get 4 cis-diols from 3 alkenes.

[miraculix]

@Dan: I'm sorry to say, but your reaction looks more than impossible. Did you ever try to perform an SN2-type reaction on the position you've mentioned? If substitution will occur on this center (which I doubt), than it will proceed via SN1 with complete loss of stereochemistry (if no strong ion-pair-formation is involvede). In my opinion, attack of the nucleophile will be at the less hindered carbon of the epoxide.


cheers

miraculix