ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: stonecold on May 28, 2010, 09:13:03 pm
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What are the exceptions which we have to know?
So far all I have is there is no peak splitting for hydrogen on hydroxyl and if the molecule is symmetrical then there is also no peak splitting.
e.g. Today I was doing checkpoints, and there was the high res NMR of 1,1,2-trichloropropane. For some unknown reason, there is like double splitting on the middle hydrogen or something. I was so baffled, it had 8 peaks! :o. How are we supposed to know this stuff, when all the text book tells us is 'n+1' rule which for many cases doesn't even apply for reasons 'beyond the scope of the course.'
Any tips would be great. :)
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Yes. Middle hydrogen is split by two different environments. For the sake of VCE, this isn't touched on much. But generally do n+1 on both environments, then multiply them. So (1+1)(3+1) = 8 splits.
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ty. does this only happen when there are atoms other than hydrogen present. because normally, you just add up the hydrogens from either side then add one yeah...
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As eloquently said by Mao, the n+1 rule is actually an exception to the general rule for these things in the real world, it just happens by some way or another, that the combinatorics bhind it make it simple for us VCE'ers. This rule I proposed isn't hard or fast, in the same way as n+1 isn't, it just works for the sake of VCE. In any case, you won't get questions like this.
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know why u use NMR. pros and cons...etc etc
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Yeah as i said somewhere before you you deffs wont get a question that will split up into more than 4 peaks. After multiplets appearing on stav 2010 our teacher panicked and frantically attempted to contact vcaa lol
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Also that the -H in the -CHCl group in CH3.CHCl.CHCl.CH3 is represented by a quartet rather than a quintet. For "n" in the "n+1" rule, "n" is the number of non-identical protons.
And hehe, 36 peaks for -CH in CH3.CH.(CH2OH).CH2NH2
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I shall reiterate myself, and clarify the problem with 'multiplicity'. I will use the molecule 4-hydroxybutanoic acid as an example
HO-CH2-CH2-CH2-COOH, the H environment discussed in detail is at 3C. (second carbon from the left). There are 5 H environments, the H on 3C split by two neighbouring CH2 groups of different environments.
Firstly, to all you mathematicians, n+1 rule distributes peak intensity according to Pascal's triangle
In multiplicity, when n+1 rule is applied more than once, the intensity of each split peak is split again according to Pascal's triangle. E.g. for a multiplicity of 3x3, the peak ratios are (1.3.1)(3.9.3)(1.3.1)
Good chemists do not call this splitting a nonet, it is referred to as a triplet of triplets.
For this case, the H is split by neighbouring H environments that are only slightly different. Direct application of n+1 ignoring multiplicity will give 5 peaks with peak ratio (1,4,10,4,1).
However, on the actual printout of the sheet, you will notice that the peak ratios are slightly different, more like (1,6,11,6,1). On zooming in, you will also notice that the three middle peaks have tiny splits like small split-ends in hair. This is caused by multiplicity, since the actual mechanism dictates three triplets of triplets are produced. However, each separation in each triplet is of very similar magnitudes, thus the differences are tiny.
To be pedantic, a triplet of triplets are produced. A normal organic chemist will slap you for being so pedantic. It's close enough to a quintet, don't be so anal, the environments are very similar (but ever so slightly different). So at the VCE level, do NOT worry about multiplicity. VCAA won't give you a question where multiplicity actually matters. Be aware, know how, but don't use it unless it's a bleedingly obvious MCQ.
Attached is the splitting diagram for anyone interested.
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Yeah, so if its bleedingly obvious like it asks for the environment of carbon 2 on: 2-methylpropane... then you can just do 4^3 = 64? I don't think this is right?
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Yeah, so if its bleedingly obvious like it asks for the environment of carbon 2 on: 2-methylpropane... then you can just do 4^3 = 64? I don't think this is right?
There are only two H environments on that thing. The H on 2C has 9 equivalent neighbours, split into a set of 10 peaks.
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eh so here the n+1 rule works. mmmmm.
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Can someone explain to me why symmetrical molecules don't peak split? for example, HO-CH2-CH2-CH2-OH
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Hmm I think its good to realise that the hydrogens attached to a carbon that is attached to like an ester like COOCH2 I think is does not split due to deshielding....(Please correct me)
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Can someone explain to me why symmetrical molecules don't peak split? for example, HO-CH2-CH2-CH2-OH
The H on the second carbon does split (into a quintet).
For another symmetrical molecule such as ethandiol, HO-CH2-CH2-OH, the H on the first and second carbon don't split. Splitting only occurs when neighbouring atoms are in a different environment. In this case, because of symmetry, the neighbouring environments are the same, thus no splitting.
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But there's still 2 H environments?
What about like OH-CH2-CH2-CH2-CH2-OH? Splitting of each carbon except the alcohol with 3 H enivornments?
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^Yeah, I think so. :)
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Thanks Mao, that is a really good explanation. I'm finally starting to get it now.
Do I have to worry about that multiplet stuff like in the question which I posted at the start?