00:01
In this problem, you're given various information from ir, nmr, and mass spec.
00:10
So we need to figure out the structure that we have.
00:15
So the first thing we're going to look at is our ir.
00:18
We have a peak at 1692, and this corresponds to a carbonyl, so a double bond between carbon and oxygen.
00:34
And from nmr, we can tell that we have 10 hydrogens.
00:39
So then we can use masspec to find our molecular formula.
00:44
So 150 is our molecular weight, and we use a rule of 13.
00:50
So the whole number we get here is 11, and we have a remainder of 7.
00:58
So we add 11 and 7, and that's our hydrogens.
01:03
But we know we have at least one oxygen, so we take away a second.
01:07
Ch4.
01:09
So we get c10 h14o, but again we have 10 hydrogens.
01:16
So it means we're going to have another oxygen and we get c9 h10o2.
01:25
And if we look at degrees of unsaturation, this tells us the number of pybonds and rings.
01:31
So multiply the number of carbons by two.
01:39
Add two.
01:41
Subtract the number of hydrogens and divide by two.
01:45
So here we get five...