NMR spectrum for octane

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Hemichordate

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The NMR spectrum for octane has 4 peaks, right? I just want to make sure because the solutions for Kaplan Orgo Discretes insists that it has only 3 peaks.

Ch3(CH2)6CH3
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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.
 
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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.
 
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Hemichordate said:
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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.
Click to expand...

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.
 
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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?
 
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Hemichordate said:
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?
Click to expand...

Right. The one closer to the Br would be further downfield.
 
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Hemichordate said:
The two quintets would have the same ppm value?
Click to expand...

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.
 
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pookiez88 said:
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.
Click to expand...

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?
 
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kentavr said:
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?
Click to expand...

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.
 
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pookiez88 said:
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.
Click to expand...
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.
 
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wanderer100 said:
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.
Click to expand...

? I'm not sure I understand that. Can you please elaborate?
 
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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
 
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wanderer100 said:
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
Click to expand...

Yeah, I know what the source if splitting is, but I never heard of multiplying the N+1's. I've never done that in orgo!
 
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So why would the blue hydrogens be split equally by the methyl and the other hydrogens (-CH2-) if those hydrogens are themselves not equivalent?
 
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pookiez88 said:
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?
Click to expand...
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.
 
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