00:02
And this problem, we are drawing the substitution and elimination products for each reaction.
00:08
So starting with letter a, we have a primary leaving group here and a good nucleophile, good base.
00:14
And so since this is primary, the only thing we can do with it is sn2.
00:18
So our product will just be replacing that bromine with the methoxy group, and it will look like this.
00:25
For part b, now we have a secondary leaving group, and again, a good nucleophile and good base.
00:30
So here we can do either sn2 or e2.
00:34
So the sn2 product will look like this.
00:37
We'll flip that stereo chemistry completely.
00:40
And then the e2 product, there's actually two of them.
00:44
We can either put the double bond on this side or on this side.
00:49
For part c, now we have a tertiary leaving group and a good base.
00:54
And so the only thing we can do here is e2.
00:57
We can't do sn2 because that can't happen on tertiaries.
01:01
So our e2 product will look like this or like this.
01:08
The double bond can go either direction.
01:11
For part d, again, we have that tertiary leaving group, but now we have a weak nucleophile, weak base...