NMR spectrum for octane
Started by Hemichordate
If you mean H-NMR, then I would say that,
1. CH2 -> CH3(most left position) looks like triplet to me
2. CH3->CH2<-CH2(second position) looks like quadroplet of triplets
3. CH2->CH2<-CH2(any mid position) looks like triplet of triplets.
So I am getting 3 in theory. In actual chart that I found there are only two multiplets visible, since 2) and 3) probably overlaps.
1. CH2 -> CH3(most left position) looks like triplet to me
2. CH3->CH2<-CH2(second position) looks like quadroplet of triplets
3. CH2->CH2<-CH2(any mid position) looks like triplet of triplets.
So I am getting 3 in theory. In actual chart that I found there are only two multiplets visible, since 2) and 3) probably overlaps.
I'm getting three as well. where'd you get the fourth one?
Wouldn't you have a signal for the innermost two CH2's?
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bump
CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Wouldn't the inner two CH2's both have 5 splits? I don't see why those two wouldn't count.
CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Wouldn't the inner two CH2's both have 5 splits? I don't see why those two wouldn't count.
How about this:
CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3
1) The black -CH2- groups have something in common: each -CH2- group is surrounded by -CH2-groups on both sides. That would give you a quintet.
2) The blue -CH2- groups are surrounded by 5 H's, and would give a sextet.
3) The red -CH3- groups would give a triplet due to the neighboring -CH2- groups.
The two quintets would have the same ppm value?
Oh, ok. I see now.
So if it had been like this:
Br-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Would the two central CH2's still be equivalent? The Br should affect how upfield/downfield the signal is, right?
So if it had been like this:
Br-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Would the two central CH2's still be equivalent? The Br should affect how upfield/downfield the signal is, right?
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I don't think there would be any appreciable difference in the peaks, that's why we'd only see three signals. If you were to put halogens on the ends of the alkane, then you'd probably see 4 signals.
Ok so the two CH2's in my original question would only count as one if the molecule were symmetrical. Got it.
Thanks pookie and seraph.
pookiez88,
I agree, except this one. Any comments?
Regarding #2). The three H on the left will split original line in four parts. We got quintet. Then the two H on the right will split quintet in three parts. totally we got 12. (or triplet of quintets). Am I wrong?
CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3
You have to count the neighboring hydrogens and follow the "n+1" rule for splitting. The blue -CH2- group has 5 neighboring hydrogens, 3 from the terminal -CH3- group, and 2 from the inner black -CH2- group for a total of 5 neighbors. 5+1=6, a sextet.
When dealing with non-equivalent neighboring hydrogens you can't simply use the n+1 to predict the number of peaks. Since the number of peaks corresponds to the number of possible spin states of neighboring hydrogens, having non-equivalent neighboring hydrogens actually increases the number of peaks. IIRC you have to multiply the n+1, so take (3+1)*(2+1) = 12 peaks in the form of a multiplet for the blue hydrogens.
Are you aware what the source of H-H splitting is? If yes skip to the next paragraph. Basically a hydrogen atom can have its spin aligned with the external magnetic field (and producing a signal downfield on the adjacent hydrogen I believe) or have its spin aligned against the field. For reasons I'm not sure of either alignment is equally likely. So if there is no adjacent hydrogen you get a singlet because the hydrogen is in its own environment. If there is 1 adjacent hydrogen its spin can be aligned with or against the external magnetic field, and hence produces a double, both of equal height. If there are 2 adjacent hydrogens, they can be both with, both against, or cancel each other out with regard to the magnetic field, so you get a triplet with the middle peak twice as large as the other two peaks. With 3 adjacent hydrogens, you have 4 possibilities and so on.
Now when you have hydrogens on both sides that are in different chemical environments, and lets say you have two on either side of the reference hydrogen (lets call it Hr) if both the hydrogens on one side are supporting the field, and the two hydrogens on the other side are against the field they don't cancel each other out because they are in different environments (whereas if they were equivalent they would cancel each other out), so you have to multiply the N+1s
Now when you have hydrogens on both sides that are in different chemical environments, and lets say you have two on either side of the reference hydrogen (lets call it Hr) if both the hydrogens on one side are supporting the field, and the two hydrogens on the other side are against the field they don't cancel each other out because they are in different environments (whereas if they were equivalent they would cancel each other out), so you have to multiply the N+1s
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi
You can use that link to pop up a spectrum for octane. Just type in octane in the "compound name" section. Can you show me what you mean by the 12 peaks there?
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Does the graph just show two peaks or is the third peak really small?
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi
That spectrum is really messy so I can't really see anything. In any case, the difference in the two sets hydrogens is small so that only high resolution NMR is likely to see the difference between those sets of hydrogens (which is why I only mentioned the H's on C-2, because in the middle of the chain the environments are so similar that for all practical purposes you can just use the regular N+1 rule). A better example would be something like the H on C2 in (or the H's on C-3) 2-pentanol, but the same principles work for other molecules.
