# NMR spectrum for octane

Discussion in 'MCAT Study Question Q&A' started by Hemichordate, Jul 17, 2009.

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1. ### HemichordatePeds 7+ Year Member

May 5, 2008
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

3. ### kentavr 2+ Year Member

103
1
Nov 13, 2006
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.

4. ### thebillsfanUnseasoned Veteran 5+ Year Member

778
0
Dec 22, 2008
I'm getting three as well. where'd you get the fourth one?

5. ### HemichordatePeds 7+ Year Member

May 5, 2008
Wouldn't you have a signal for the innermost two CH2's?

6. ### HemichordatePeds 7+ Year Member

May 5, 2008
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.

7. ### Seraph 84

170
0
Jun 28, 2009
I'm see a triplet, sextet, and two quintets.

8. ### G1SG2 5+ Year Member

May 2, 2008

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.

9. ### HemichordatePeds 7+ Year Member

May 5, 2008
The two quintets would have the same ppm value?

10. ### G1SG2 5+ Year Member

May 2, 2008
It's just 1 quintet; they are equivalent hydrogens.

11. ### HemichordatePeds 7+ Year Member

May 5, 2008
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?

12. ### G1SG2 5+ Year Member

May 2, 2008
Right. The one closer to the Br would be further downfield.

13. ### Seraph 84

170
0
Jun 28, 2009
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.

14. ### HemichordatePeds 7+ Year Member

May 5, 2008
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.

15. ### kentavr 2+ Year Member

103
1
Nov 13, 2006
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?

16. ### G1SG2 5+ Year Member

May 2, 2008
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.

17. ### wanderer 5+ Year Member

1,980
28
Dec 14, 2008
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.

18. ### G1SG2 5+ Year Member

May 2, 2008
? I'm not sure I understand that. Can you please elaborate?

19. ### wanderer 5+ Year Member

1,980
28
Dec 14, 2008
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

20. ### G1SG2 5+ Year Member

May 2, 2008
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!

21. ### wanderer 5+ Year Member

1,980
28
Dec 14, 2008
So why would the blue hydrogens be split equally by the methyl and the other hydrogens (-CH2-) if those hydrogens are themselves not equivalent?

22. ### G1SG2 5+ Year Member

May 2, 2008
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?

May 5, 2008
24. ### wanderer 5+ Year Member

1,980
28
Dec 14, 2008
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.