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Author Topic: jane1234's chemistry questions!!  (Read 12414 times)  Share 

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Re: jane1234's chemistry questions!!
« Reply #45 on: May 19, 2011, 09:51:54 pm »
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You are absolutely correct

In benzene they should only be 1 carbon and 1 hydrogen environment

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Re: jane1234's chemistry questions!!
« Reply #46 on: May 19, 2011, 10:06:23 pm »
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Just adding to Pixon's post, you apply multiplicity when the two adjacent hydrogen environments are not the same.

eg. propane- the two proton environments adjacent to the protons attached to the middle C are the same environment... you add them together 6+1=7 multiplet

propan-1-ol- the two proton environments adjacent to the protons attached to the middle C are NOT the same environment... use (3+1)(2+1)=12 multiplet

This came up in my school SAC today but I don't think it will be on exams.
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Re: jane1234's chemistry questions!!
« Reply #47 on: May 19, 2011, 10:12:51 pm »
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Not too sure about your propane. It should only have a quartet and a triplet

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Re: jane1234's chemistry questions!!
« Reply #48 on: May 19, 2011, 10:26:38 pm »
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Ok, a multiplet forms when there are very very similar environments. Although Benzene (C6H6) will only have one environment, add a couple of branches off this and the multiplet arises. This is because while being very similar, the branches cause a slight difference in the environments, but not enough create it's own set of peaks.

That, in a nutshell, is multiplicity.

Although Pixon's point is very valid, since VCE chemistry does not teach multiplicity it can't really be assessed, but once again - it wouldn't suprise me either if they put something up that everyone knew in a low res NMR - since they do, afterall, have the shift of Benzene in the Data book...

For anyone that's interested, this is the NMR of Aspirin:

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Pixon

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Re: jane1234's chemistry questions!!
« Reply #49 on: May 19, 2011, 10:53:21 pm »
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It's obvious, without knowing multiplicity that aspirin has more than 1 peak and to those who know about multiplicity, they will know that multiplets arise.

Ok, a multiplet forms when there are very very similar environments. Although Benzene (C6H6) will only have one environment, add a couple of branches off this and the multiplet arises. This is because while being very similar, the branches cause a slight difference in the environments, but not enough create it's own set of peaks.

That, in a nutshell, is multiplicity.

First of all...multiplets aren't something special. A triplet is still a multiplet. Or what do you think a multiplet is? From your explanation, I don't feel as though we're talking about the same thing.
But to the main issue: I think you're missing the point, or you're not explaining it very well. It's not the fact that you have "very very similar environments"...it's just simply that they are different and the branches are why you have this difference.
You then get strange multiplets because you have a proton environment being affected by different adjacent proton environments in different ways. But because the impact these adjacent protons have isn't great, the peaks come very close together in such a way based on the combination of possible alpha/beta spin states. You start to get doublet of doublets, etc.
« Last Edit: May 19, 2011, 10:55:09 pm by Pixon »
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Re: jane1234's chemistry questions!!
« Reply #50 on: May 19, 2011, 11:03:11 pm »
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Yes, I think we are on very different wavelengths here Pixon - I think we should just leave it as is - it's nothing worth worrying about!

Yours is probably a better explanation, so people read Pixon's post - it'll make a lot more sense.
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Re: jane1234's chemistry questions!!
« Reply #51 on: May 19, 2011, 11:23:16 pm »
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It's all good...this can return to being Jane's thread again. :)

p.s. As luken said, it's really nothing to worry about...only read if you're interested, it won't affect your VCE.
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Re: jane1234's chemistry questions!!
« Reply #52 on: May 20, 2011, 12:13:41 am »
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I think I should say something here.

Firstly, good work to Water and Pixon, Pixon especially. I mostly agree with what they have said about NMR.

It is crucial to understand that in NMR, two atoms are in the same environment ONLY IF their connectivity in the molecule is exactly the same. You need to consider the immediate neighbours, the second neighbours, the third neighbours, etc.

In another example, consider the nylon 6-6 dimer HOOC-CH2-CH2-CH2-CH2-CONH-CH2-CH2-CH2-CH2-CH2-CH2-NH2

In this example, each CH2 are slightly different to each other. In 13C NMR, you would expect 12 different peaks, in 1H NMR, you would expect 13 different peaks.


This blurb is beyond VCE level, but is useful if you want to get to the bottom of what is meant by 'multiplicity' and 'multiplet'.

On the topic of multiplicity, the term multiplicity (not to be confused with multiplet) refers to the peak-splitting effects of spin-spin coupling between neighbouring nuclei. Different neighbouring nuclei have different splitting frequencies (the separation between the split peaks, splitting frequency depends on the chemical environment). The n+1 rule is an approximation for calculating how many stable energy states there are in spin-spin coupling.

Consider 1-propanol: CH3-CH2-CH2-OH, the underlined H is split by two neighbouring groups at two slightly different frequencies. CH3 splits it into a 4-tet, and CH2 splits it into a 3-tet. Thus the underlined H is split into a 12-tet peak. However, the two splitting frequencies are very similar, and if you draw out the splitting diagram with very similar splitting frequencies, you will get a 6-tet. This is the n+1 approximation (which is valid most of the time).

Multiplet arise from overlap of peaks in similar chemical environments. This is the assignment we give to peak patterns that are too complex to assign. Going back to the nylon 6-6 dimer example, we won't actually see 13 distinct H peaks, because many of the CH2 groups are very similar to each other and they will overlap. This overlap creates a complex peak pattern that is usually unresolved, and cannot be assigned to simple n-tet peaks.

Multiplets also arise from complex splitting patterns that involve isotopes, 2J, 3J and 4J splitting, and various other complex interactions that makes it different to the simple n-tet peaks (n+1 approximation). This is seen a lot in aromatic compounds, since 4J splitting is significant (4J splitting is very weak in aliphatic compounds).



In summary for VCE students:
-The only time when two atoms are in the same environment is if they are exactly the same (i.e. have the same connectivity no matter how far you look). This means there must be some symmetry element. If you can't find symmetry, they are most probably in different environments.

-Use the n+1 rule, but understand that the actual physics of NMR is a lot more complicated, and you see many more complex patterns. The n+1 rule is an useful approximation, but an approximation nonetheless.
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Re: jane1234's chemistry questions!!
« Reply #53 on: May 20, 2011, 04:44:47 pm »
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If you think about it, complex splitting is unlikely to be examined because it would be way too difficult for us to work out the molecule with clusters with 12 or more peaks. While with triplets we can confirm the molecule contains a proton environment adjacent to CH2

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Re: jane1234's chemistry questions!!
« Reply #54 on: May 20, 2011, 05:01:07 pm »
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If you think about it, complex splitting is unlikely to be examined because it would be way too difficult for us to work out the molecule with clusters with 12 or more peaks. While with triplets we can confirm the molecule contains a proton environment adjacent to CH2
Exactly my point, but let's just leave it with that.
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Re: jane1234's chemistry questions!!
« Reply #55 on: May 22, 2011, 10:51:35 pm »
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From the TSSM 2009 Exam:

Note: Previous question deduced that the concentration of Cu2+ ions is 1370 ug/L in the 1mL sample and 13.7 ug/L in the 100mL sample.

Question 2:
It is found that the copper ions came from the compound copper chloride, CuCl2. The mass of copper chloride, in mg, in the original 1 mL sample of water is closest to:

A 0.0026
B 1.25
C 2.9
D 3.2

The answers say C.
n(Cu2+) = 0.00137/63.5 = 0.0000216
n(CuCl2) = 0.0000216
m(CuCl2) = 2.9

I get how they worked it out...  but if you got the concentration of the diluted sample (100mL) it is 13.7 ug/L and if you use this value you will get 0.029 mg of CuCl2 in the diluted sample, but shouldn't there be the same mass of CuCl2 in both samples? ???

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Re: jane1234's chemistry questions!!
« Reply #56 on: May 22, 2011, 11:25:22 pm »
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1370 ug/L in the 1mL sample   =/=    13.7 ug/L in the 100mL sample
137 ug / 100ml =/= 1.37 in 10ml sample
13.7 ug / 10ml
1.37 / 1ml

I'm not sure what you're asking, but if it's dilution well yeah, it should be the same amount of mass?
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jane1234

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Re: jane1234's chemistry questions!!
« Reply #57 on: May 23, 2011, 12:26:17 am »
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1370 ug/L in the 1mL sample   =/=    13.7 ug/L in the 100mL sample
137 ug / 100ml =/= 1.37 in 10ml sample
13.7 ug / 10ml
1.37 / 1ml

I'm not sure what you're asking, but if it's dilution well yeah, it should be the same amount of mass?

Sorry, not very clear. I was just wondering if their answer was right.
Because they said that there are 1370 ug of Cu2+ ions in the 1mL sample, even though it is 1370 ug per litre. I thought that you divided by 1000 to get 1.37 ug of Cu2+ ions in the 1mL sample.
And with the 100mL sample, it is 13.7 ug per litre. If you divided by 10, you would get 1.37 ug in the 100 mL sample, which makes sense that there is the same mass in both. If you do it by their method, you get two different masses in the two different solutions (but all you have added is water, which shouldn't affect the mass of Cu2+ ions).
Does that kind of make more sense?
Basically I don't think any of the answers are right, though I'm probably wrong...   

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Re: jane1234's chemistry questions!!
« Reply #58 on: May 23, 2011, 07:34:25 am »
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Nah, I'm pretty sure you're right. If I remember correctly, there are quite a few errors in their exams - this being one of them. Just follow on with the concentration/volume your worked out :)
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Re: jane1234's chemistry questions!!
« Reply #59 on: May 26, 2011, 11:51:25 pm »
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Okay just did Neap 2008:

1. When you have masses of a precipitate as 1.183g 1.137g 1.109g 1.110g and 1.108g, to find the average mass would you take all the values, or just the last three??? The answers say only the last three, but this doesn't even follow the titration +-0.05mL concordant results method. Would VCAA accept it if you used all 5?

2. Is an NH group considered an amine? Even though it only has one H?

Thanks :)