00:01
In this problem, we're looking at a structure with the mucleum formula c -10 h -13 -n -o -2.
00:08
So here, we are given ir and nmr data to find our structure.
00:15
So if we look at ir first, notice that around 3 ,300 we have a peak.
00:22
That's a nitrogen -hydrogen stretch.
00:24
And because it's a similar peak, it means this is going to be a secondary amine.
00:30
Or a mine and around 1600 we have a strong absorption.
00:39
Those are for carbon -carbon -carbon double bonds.
00:43
Around 1200 we see an absorption, that's a carbon oxygen single bond.
00:50
So we expect an ether or an ester or an alcohol, but because we don't see an alcohol peak, we can rule that out.
01:02
So then if we move to our nmr data, you'll notice in six to eight parts per million range we have two really nice doublets and we see that and we have a pair of substituted benzene ring so we have two hydrogens on each side each has one neighbor making them doublets and we also see a one hydrogen peak in between those this isn't aromatic it's actually the hydrogen on nitrogen so then if we go back to our nmr you'll notice at step 4, we have two hydrogens that choke as a quartet.
01:42
So usually when we see two hydrogens as a quartet, it means we have a ch2 group next to a h3 group...