00:01
This question asked for the products of each of these reactions, including stereochemistry, and we are only drawing substitution products here.
00:09
So starting with letter a, we have a secondary leaving group and a good nucleophile.
00:14
So this will be an sn2 reaction, which means we'll only have one product, because sn2 reactions give us complete inversion of stereochemistry.
00:23
So we'll put the methoxy on a dash, since the bromine was on a wedge here, and the hydrogen will be on a wedge now.
00:32
So there's our single product for part a.
00:36
For part b, we have r3 -bromote -3 methyl -heptane and methanol.
00:41
So now we have a tertiary leaving group and a weak nucleophile.
00:45
So this will be sn1 for substitution.
00:48
So we'll get two anantiomers here because if we're doing sn1, it goes through a carbocadion, so you can end up with the nucleophile on either side.
00:58
So we'll have one where it's on the same side that the bromine was on, that will look like this, and we'll have one where it's on the opposite side like this.
01:12
So there's our two stereo ismeris for part b for part c.
01:17
This is also an sn1 reaction because we have a primary benzylic carbocation there and a weak nucleophile.
01:24
There is no stereochemistry here because we have two hydrogens on that carbon, so it's not a stereo center.
01:30
So we'll just get one product that looks like this.
01:33
We'll just replace the chlorine with the athoxy group...