00:01
In this problem, we're given the molecule formula c6 h15n.
00:07
And we want to know the structure given proton nmr data.
00:13
So we have a singlet with one hydrogen at 0 .9 parts per million.
00:20
We have a triplet for three hydrogens.
00:25
At 1 .1.
00:32
We have a singlet for nine hydrogens at 1 .15, and we have a quartet for two hydrogens at 2 .6.
00:50
So, first of all, when you see a triplet with three hydrogens and a quartet with two hydrogens, this tells us that we have a ch3 group, that's our three hydrogens, and it's a tributl it has two neighbors, so it's next to a ch2 group, and our ch2 group is our two hydrogens, and it has three neighbors, so it's a quartet, and that means next to it.
01:21
We either have a carbon with no hydrogens, or in this case, notice how we're further down the field for this.
01:28
That usually means you're attached to an electronegative atom, so nitrogen in this case, and here, and here, then once we figured out that piece, let's real quick come back up here.
01:49
So if we look at degrees of unsaturation, this gives us the number of pie bonds and rings.
01:59
So we multiply the number of carbons by two and add two.
02:04
Subtract the number of hydrogens, add the number of nitrogen and divide by two.
02:11
So here we get zero, so we don't have any pie bonds or rings.
02:15
So then if we look at our options here, well, notice we have six carbons...