00:01
We've got some nmr problems that we're doing here.
00:07
And so the first part is about carbon nmr, where we have a few hydrocarbons, namely n -pentane, 3 -methyl n -pentane, and sys 1 -3 -dymethyl cyclohexin.
00:30
And we are looking at how many carbon -an -mortemortyxene.
00:36
Signals that will be for each one.
00:38
And so for doing that, we really just have to count the carbons and account for symmetry.
00:44
So we see here that there's a line, that there's a plane of symmetry for this compound, meaning that you will have this one methylene, which will be a quintet, or actually, never mind.
00:57
We're looking at carbon, not proton.
00:59
But anyway, we'll have, we'll have this carbon, and then this carbon, which is going to be equivalent to this one, as well as this methyl, which is going to be equivalent to this one.
01:08
So we're going to have three signals for that.
01:11
Now for our hexane right here, we are again going to have a plane symmetry, and we will have the same three that we have for the previous compound, as well as a fourth one for this methyl group here, which is on the central carbon.
01:32
So that'll be four.
01:33
And now for our, for our dimethyclohexane, we will have our methylene here, our two methines that are equivalent, our two methylines that are equivalent, and then our methyline that is going to be non -equivalent, as well as our two methyl groups that are also going to be equivalent to one another because we have our plane of symmetry.
01:57
And so that'll be one, two, three, four, five distinct signals.
02:03
And so in the choices for this question, that would be answer c, 3, 4, and 5.
02:12
And now we have another set of compounds that we're looking at proton signals from.
02:22
And let's just go ahead and do this...