At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Here's one of those cases where carbon is more electronegative than the atom it's attached to. remember lithium is a metal so it's probably going want to exist as Li+ because losing an electron fills it's octet, it looks like this a complex. |dw:1446938798972:dw|
|dw:1446938958438:dw| I feel that because the epoxide, that both sides are susceptible to this nucleophile attacking. what you really have is a carbanion in disguise I feel that it attacks the expoxide from the opposite side, and has a 50:50 chance of attacking either carbon because I feel that they both have the same number of substituents.
Sorry I should have drawn it like this: |dw:1446939271824:dw| Then the second step |dw:1446939141219:dw|
This is my best attempt, you may want to check this |dw:1446939335985:dw|
@jerrijones let's try to see why the other three might be wrong
I don't think it's A/c This is a carbanion so you cant dump any more electrons onto it. it's more likely that it's going to dump it's electrons onto the oxygen to generate an alkoxide ion, an oxygen with a negative charge. so this is out. |dw:1446939487526:dw| I think B/D depend on what face the carbanion attacks the molecule from. you see that the epoxide is facing up because of the wedges. the carbanion must attack from the opposite side, that was my logic behind this question which would lead to generating this. hence it comes in from the back side. |dw:1446939699999:dw| then the alkoxide does displaces the iodine in an SN2 like fashion. that was my reasoning behind that reaction. |dw:1446939746870:dw| Hope this helped you
That helps tremendously! Thank you again!
You got it man! no problem