00:01
This question asks us to explain why only a substitution product is formed in this reaction with methoxide and no elimination product.
00:10
So if we think about what type of substitution and elimination would be happening here, we have a secondary leaving group and a good nucleophile good base.
00:19
So we would have sn2 and e2.
00:23
And so sn2 doesn't have a whole lot of requirements other than a primary or secondary leaving group and a good nucleophile, which we have.
00:30
So that can happen.
00:32
But the e2 has some more requirements than that.
00:36
It needs, and most importantly, it needs an anti -perioplaneer hydrogen to the leaving group.
00:40
So what that means is if we were to draw this in chair confirmation, what am i doing? that's a really bad chair.
00:49
If we were to draw this in chair confirmation, i'm going to put the methyl group up here, our bromine here, and our methyl here.
00:58
And so the hydrogens all would be equatorial on these carbons.
01:05
Now, for an e2 reaction, we need an anti -peri -planar hydrogen...