[lithiumuranium]

Another question about this year's prep problems.
http://postimage.org/image/a1gnawhon/

In part (c), the final product suggests that NH3 adds to the more highly branched site of the (S,S) epoxide to give the intermediate Z. (I think that's the only way for intramolecular SN2 of Z to give the product).

If that reasoning is correct, then why does the base selectively add to the more sterically hindered carbon?

[Dan]

Hint: The ammonia will displace bromide first.

[lithiumuranium]

ah, ok thanks. Just making sure, that means the intramolecular ring opening occurs at the closest carbon.