00:01
In this problem, we see an alkenes, so double bonds between carbons, and we have an alkened addition here with hydrochloric acid.
00:09
So this, we're asked for our product, and then also what our relationship between products would be.
00:20
So if we start here, this is a markovnikov addition, which means our chlorine is going to end up on the more substituted carbon of the double bonds.
00:29
So here we have a carbon that's tertiary.
00:32
It's connected to three other carbons.
00:34
Whereas this one's only connected to two.
00:37
So our more substituted one is where chlorine is going to end up.
00:44
And notice our stereochemistry on the other side of our ring doesn't change...