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Discussion in 'DAT Discussions' started by nt4reall, Jun 19, 2008.
Can anyone help me to solve this problem , How may signals chlorocyclohexane would produce? Thanks
Two triplets, one quartet, and one quintet
Bruin is right. 1 Quartet and 3 quintets.
for NMR do u count the number of H surrounding and the peak would be n+1?
Simiam, how many types of hydrogens are in this molecule? The answer is 4, and not 6. Remember, you don't count the similar hydrogens again. I guess you are counting the similar type of hydrogens again, and that is why you end up with so many quintets. the Hydrogens at C4 have 4 surrounding hydrogens, but that is not gonna give you a quintet, because the two Hydrogens on the C3 are similar hydrogens to the ones on C5. So you count only two of them [not 4!], and that gives you a triplet. Use the same rule for the rest.
Correct me if I'm wrong and Brian is right...
So, based on what you have said, propane, CH3CH2CH3, must have a quartet and a triplet. An interesting theory...
I'm still confuse about this but I think I've got 3(5) --> quintets and 1(4)-->quartet.may be I can scan it so we can confirm this.
C1 H = triplet (split by the C2 hydrogens)
C2 H2 =sextet (split into triplet by C3 H2 and each triplet is split into a doublet by C1 H)
C3 H2 = triplet (C2 and C4 have equivalent hydrogens)
C4 H2 = triplet (C3 and C5 have equivalent hydrogens)
See correction in a post below.
Bruin, refer to the answer by doc toothache. That is what I'm talking about. I know there is a rule like that. I sent you a message about this. I will get back to you once I review this session.
I think your answer and reasoning are somewhat different from doc toothache's?
So doc toothache, based on what you have said, 2-methyl-butanol, CH3CH2CHCH3CH2OH, should give octet on C3?
This spectrum shows a heptet and a triplet. It is exactly proving my point.
In 2- methylbutanol, the hydrogen on C2 is expected to be split by the -CH3 hydrogens into a doublet. Since the hydrogens on C1 and C3 are in slightly different environment and are, therefore, non equivalent, further spliting is expected.
(click on the spectrum to see a larger image with the hydrogen assignments)
Theoretically, the C2 hydrogens in propane should be a quartet since the two methyl groups are equivalent. Why in real life, the NMR spectra may be different than expected, you will have to ask a chem prof. In the meantime, we may be making this more complicated than it needs to be. For example, in the case of chlorocyclopropane, the C 1 hydrogen can be in either axial or equatorial position, and depending on the experimental conditions we may or may not have a 50:50 mixture. You can pretty much bet that the DAT material will stick to basics since otherwise it would require some experience in both advanced organic chemistry and NMR spectroscopy.
Correction: (See post below)
The C 2 hydrogens in propane should be a heptet since there are 6 equivalent hydrogens on the 2 methyl groups.
Very interesting. I will have to read more of O.chem textbook to fully understand your reasoning. Thank you for enlightening me.
Bruin, my point is that you were missing something there. I'm not claiming to have the correct answer at all. All I'm saying is that there are some cases where you don't count similar hydrogens. As I said, I will review them and get back to you.
Propane should have a heptet. While the methyl group hydrogens are all equivalent, there are six of them, from the n +1 rule that would be 6+1=7 peaks.
I agree with Bruinlove...
C1 - quintet
C2 - quartet
C3 - quintet
C4 - quintet
C2-sextet- in this case the hydrogens from C 1 and C3 are non equivalent-the hydrogens from C1 will cause split into a doublet and each peak of the doublet will again be split into triplet (most upfield signal of the -CH2 hydrogens)
C3-multiplet- if C2 and C4 are non equivalent we should see a triplet of a triplet
C4-quintet-in the furthest downfield position.
"Whenever two (or three) sets of adjacent protons are equivalent to each other, use the n+1 rule to determine the splitting pattern."
"When two sets of adjacent protons are different from each other (n protons on one adjacent carbon and m protons on the other), the number of peaks in an NMR signal=(n+1)(m+1)."