[science123]

Do we get both product A and it's enantiomer (product B) or just A ? I was thinking that we would get only product A because the the epoxide ring is facing up and shielding the top side thus ethanol has to attack from bottom resulting in product A only and inverting the chiral center (by pushing the methyl up). Also is product C possible ? Thanks

[orgstudy]

In my opinion, you will get A and C ...
well i have no well proven and confirmed theory on this but just my own experience with organic over past 5 yrs ...

While Epoxide opening ... if any C attached to O is 3degree, the mechanism occurs with SN1 or else SN2 ...

and as you know ... 3 degree C na d give both d & l forms ... A and C are possible ... of course O will not change its plane, so B is not possible!!!!

[science123]

If we use H3O+ as a nucleophile, we would get both syn and trans diol but I have never seen anywhere that epoxide opening gives syn diol. My instructor used OsO4 to make syn diol and mCPBA followed by H3O+ to make trans diol. So, I am guessing that the mechanism is always Sn2 (nuc. attacking from opposite side of the epoxide resulting in trans product). What's confusing is that I heard Sn2 rarely happens on tertiary carbon which is the case here!

The part that's sn1 like is that the more substituted carbon is attacked because it bears more positive charge in the hybrid. I might be wrong though.
 
Also, is it possible for the nucleophile to attack from the top side in this case?

[orgstudy]

why would H₃O⁺ act as nucleophile?... its positively charged.

And i dont know what syn and trans we are talking about. syn and trans are used in case of double bonds i guess...

are we still talking about the same question?

[science123]

http://i.min.us/ibQyJw.JPG

My mistake, I was talking about opening the epoxide under acidic conditions with H2O as a nucleophile.

Product B is a syn diol (both OH are facing up) and I have never heard my instructor or text book talk about an epoxide opening with H3O+ give a syn diol. The method to make syn diols I learnt was from alkenes using OsO4 followed by H2O.

Based on that I think the mechanism for epoxide opening doesn't involve a planar carbocation where the nucleophile can attack from top and bottom resulting in product A and B. Rather, I think it's SN2 attack by the nucleophile from below the epoxide giving trans product A. Product B is not formed because the nucleophile cannot attack from the top (the epoxide is shielding the top side).

If we use the same concept for ethanol as nucleophile, then I am guessing Product A is the only possible product. My instructor didn't go into much detail regarding stereochem and that is confusing part.

[orgstudy]

Well what do you think of this product!!!

[science123]

both carbons of the epoxide are tertiary and hence there is no preference.

[sjb]

both carbons of the epoxide are tertiary and hence there is no preference.

Whilst you are correct that both are tertiary, that's not perhaps the real reason why there is no preference. If you had 1-ethyl-2-methylcyclohexene oxide they would still both be tertiary, but it might be expected that one product predominates over the other. Any thoughts why that may be the case?

[science123]

No idea..has to do with chair conformation perhaps??

[orgopete]

both carbons of the epoxide are tertiary and hence there is no preference.

The products are the chiral isomers of the reaction. Due to symmetry, one racemic product should result.