00:02
In this problem, you have a molecular mass of 113.
00:06
So because this is odd, we know that we have a nitrogen.
00:09
So we're going to use the rule 13, fire molecular formula.
00:13
So we get c8, and our remainder here was 9.
00:22
So we get c8, each 17.
00:27
And we're told that we have a nitrogen.
00:31
So since that has a molecular mass of 14, we pull up a carbon and two hydrogens.
00:37
So we get c7, h15n.
00:41
And we have at least one oxygen.
00:45
So that's 16.
00:46
So we get ch4 pulled out.
00:49
So we'll have c6, h11, no.
00:55
And if you look at your carbon nmr, you only have five peaks.
00:59
So we probably have another oxygen here, which, considering how high our carbonyl peak is in our probably supports us having ester.
01:12
So we're going to pull it another ch4.
01:15
So we'll have c5, h7, no2.
01:22
And if we look at degrees of unsaturation, this tells us the number of rings and pyvons total.
01:28
So we multiply the number of carbons by 2, add 2, subtract the number of hydrogens, and add the number of nitrogen's, and divide by 2.
01:38
So you're going to have 12 minus 6 over 2, so you get three.
01:45
And with your other ir data, you had a peak at 2270.
01:49
So that's a night trial.
01:53
So between our two functional groups, we have three pie bonds...