00:01
In this problem, we're thinking about the proton -enomr spectrum for 1 chlorubutane.
00:08
So we'll draw that here.
00:13
We have 2 -methal, 2 -propanol, and 2 -methyl -propane.
00:29
And we want to match these two.
00:33
So first of all, let's think about our 2 -methylopropanol.
00:43
And here notice that our methyl groups all have the exact same chemical environment.
00:52
So that means they're all going to show up in a single peak.
00:56
And their neighboring carbon here has zero hydrogens.
00:59
So we're expecting a single.
01:02
And we're going to see the same exact thing here.
01:06
But here we're going to have one extra peak for that hydrogen of our hydroxyl group.
01:13
So we're going to be.
01:17
Assign our first spectrum with the singlet right before 2 to our 2 -bromo 2 methyl propane and that means then our second spectrum is our 2 -methyl 2 propanol and then our last one is going to be 1 chlorobutane so if we think about assigning our peaks we have 2 hydrogen's here with two neighboring hydrogens.
02:01
So it's going to be a triplet.
02:04
And it's going to be further downfield because it's on the same carbon as this chlorine...