00:01
This problem, we're asked to choose which of the four substances we've written here is.
00:10
Let me see.
00:15
Has a chiral exit? i'm sorry, it was octively active.
00:26
In order to be optically active, we have to have a carbon with four different things attached to it, if you will.
00:41
There we go.
00:46
So very quickly, let's just sketch a couple of these.
00:49
So this one is c, c, c, c, c, b .r.
00:58
So none of these has four different things attached to it.
01:04
And the bromobuteric acid is going to be c, c, c, with an h, and a br, and then a c -h2, and a c -o -o -h.
01:19
So this one has four different groups attached to it.
01:31
Carbon 1, 2, 3 has this group attached to it.
01:39
This, this, this.
01:40
This is optically active.
01:49
And this is 2 -bromo, 3 -metopropane...