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psyxwar

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Re: VCE Chemistry Question Thread
« Reply #555 on: April 08, 2014, 06:05:51 pm »
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How much of the theory behind instrumentation do we need to know? Or is it mainly application (reading spectrographs and interpreting them rather than explaining the theory behind it?)
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Re: VCE Chemistry Question Thread
« Reply #556 on: April 08, 2014, 06:23:25 pm »
+2
How much of the theory behind instrumentation do we need to know? Or is it mainly application (reading spectrographs and interpreting them rather than explaining the theory behind it?)

You need to have a rough idea of the principles, but not exactly how each apparatus works. For example, in mass spectroscopy, a magnetic field is used to differentiate between positively charged ions based on mass and charge. But, you won't need to know how the detector works, or how the electrons are fired to ionise molecules.
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psyxwar

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Re: VCE Chemistry Question Thread
« Reply #557 on: April 08, 2014, 06:34:28 pm »
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You need to have a rough idea of the principles, but not exactly how each apparatus works. For example, in mass spectroscopy, a magnetic field is used to differentiate between positively charged ions based on mass and charge. But, you won't need to know how the detector works, or how the electrons are fired to ionise molecules.
Thanks. How about things like degrees of freedom in IR?
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Re: VCE Chemistry Question Thread
« Reply #558 on: April 08, 2014, 06:38:13 pm »
+1
Thanks. How about things like degrees of freedom in IR?
I don't think so.
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Re: VCE Chemistry Question Thread
« Reply #559 on: April 08, 2014, 11:25:04 pm »
+2
How much of the theory behind instrumentation do we need to know? Or is it mainly application (reading spectrographs and interpreting them rather than explaining the theory behind it?)

I believe the study design states that you don't actually need to know anything about operations (e.g. how AAS works), but rather the interpretation of the qualitative and quantitative data produced by these instrumental analyses.

psyxwar

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Re: VCE Chemistry Question Thread
« Reply #560 on: April 09, 2014, 04:24:29 pm »
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I saw an analogy being made between covalent bonds and springs, and thus stronger bonds and lighter atoms mean a higher frequency of vibration. Is it sufficient to say that say, a C=C bond vibrates at a higher frequency than say a C-C bond and therefore absorbs IR radiation of a higher freq/wavenumber?

Also is the reason carboxyllic acids dont exhibit the peak at roughly 2900 wavenumber due to the sp3 C-H bond because the broad and intense OH peak in that region masks it?
« Last Edit: April 09, 2014, 04:29:07 pm by psyxwar »
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #561 on: April 09, 2014, 04:33:51 pm »
+3
I saw an analogy being made between covalent bonds and springs, and thus stronger bonds and lighter atoms mean a higher frequency of vibration. Is it sufficient to say that say, a C=C bond vibrates at a higher frequency than say a C-C bond and therefore absorbs IR radiation of a higher freq/wavenumber?

Great question:
Basically, there are 2 factors affecting the frequency of radiation absorbed:
- Strength of the bond- as strength of bond increases (single --> double bond), frequency/energy increases. So the double bond does absorb more IR Radiation
- Mass of atoms in bond- As the masses increase, the freqency decreases (it will take a lot more energy to move bigger atoms so this makes sense)
Hope this helps :)
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Re: VCE Chemistry Question Thread
« Reply #562 on: April 09, 2014, 09:21:13 pm »
+1
I saw an analogy being made between covalent bonds and springs, and thus stronger bonds and lighter atoms mean a higher frequency of vibration. Is it sufficient to say that say, a C=C bond vibrates at a higher frequency than say a C-C bond and therefore absorbs IR radiation of a higher freq/wavenumber?

Also is the reason carboxyllic acids dont exhibit the peak at roughly 2900 wavenumber due to the sp3 C-H bond because the broad and intense OH peak in that region masks it?

Yeah. Think of absorption as a bit like springs. Essentially, the molecule bonds vibrate and they absorb radiation at the same frequency as the vibration frequency.

Yes. The OH peak masks it; if you have an oxygen atom in your compound but you don't see that masking but see a separate CH3 and OH absorption, you probably have an alcohol.
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Re: VCE Chemistry Question Thread
« Reply #563 on: April 09, 2014, 09:37:57 pm »
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Thanks!

Can someone clarify the following (about NMR)? Correct to see if I've misunderstood anything etc. There probably is some irrelevant stuff; my teacher didn't teach it very well so I had to self learn a lot of it. Wall of text ahead.

General
  • NMR spectroscopy uses an NMR instrument to induce a magnetic field.
  • Protons, neutrons and electrons have spin, which can be +1/2 or -1/2. Spin is quantised; a nucleon or electron has to be one of the two aforementioned spin states.
  • For a nucleus to be able to be analysed by NMR, it must be NMR active; that is, it must have a net nuclear spin: ie. it has an odd mass number. This is because if it had an even amount of nucleons, there would be no net nuclear spin as the spins of nucleons cancel out, leaving it overall neutral.
  • NMR works by inducing resonance: nuclei in the lower energy state of being magnetically aligned with the magnetic field of the NMR instrument absorb energy and thus 'flip' and oppose the magnetic field. They then re-emit this energy and return to the lower energy state.
  • Roughly 50% of nuclei are aligned both with and against the magnetic field. A slightly higher amount are aligned against the field, as it is a lower energy state
  • No two NMR instruments (is this what they are called?) will have a magnetic field of exactly the same strength. Hence, we assign a chemical shift of 0 to tetramethylsilane (TMS) and measure it in concentration (ppm) rather than frequency as everything is standardised against TMS. TMS is symmetrical and thus produces only one peak, and its non-polar nature means it will appear on the far right (upfield/ on the shielded end) of the NMR spectrograph

Carbon 13 NMR
  • Uses the carbon-13 isotope. Carbon-13 is pretty rare, it's abundance is something like 1%
  • Spectrograph goes from 0-200ppm
  • Number of peaks in carbon-13 NMR = number of different carbon environments
  • Chemically equivalent carbon atoms are in the same carbon environment. This arises due to the symmetry of a molecule (CH3CH2CH3 has 2 carbon environments, one with 2 carbons and one with 1 carbon)
  • Area under peaks is not proportional to the amount of carbon atoms in that particular environment

Proton NMR
  • Uses the abundant hydrogen-1 isotope (something like 99% abundance).
  • Spectrograph is usually 0-10ppm
  • Number of peaks only indicative of number of hydrogen environments in low resolution proton NMR; in high resolution proton NMR peak splitting occurs, and a peak  may split into many subpeaks.
  • Ratio of the area under the 'peaks'(where 1 peak may equal many subpeaks; one 'peak'=one hydrogen environment) is directly proportional to the ratio of hydrogen atoms in each environment.
  • Chemical shift of protons determined by 3 effects:
    • Magnetic arisotropy: this has the biggest affect on the chemical shift of a proton. Put simply, the magnetic field of nearby pi electrons in double bonds (ie. alkenes and arenes, but not alkynes(?)) acts in tandem with the NMR instrument's magnetic field, strengthening it. The effect of this is that more energy is required to flip the spin states of the protons, and hence radio waves of a higher frequency are required -> higher chemical shift. This effect diminishes the further away the proton is from the double bond.
    • Inductive effects of nearby electronegative atoms: this has a smaller effect on the chemical shift of a proton. Electronegative elements hog electrons, and thus have a deshielding effect by diminishing the electron cloud around the proton. Hence, the proton is subject to more of the magnetic field's strength (no electrons to shield it) and thus more energy and thus a higher chemical shift is required for resonance
    • Magnetic fields of nearby protons: this has the smallest effect on the chemical shift of a proton. Protons themselves act as magnets; they may have positive or negative spin (up or down spin?). Hence, a proton will be affected by the spin states of neighboring protons: if a proton neighboring a particular proton of interest, proton P, has a spin state is aligned with the magnetic field of the NMR instrument, then proton P will experience a slightly stronger magnetic field and thus require more energy to induce resonance, thus having a higher chemical shift. Conversely, if the neighboring proton's spin state is opposing the magnetic field of the NMR instrument, then proton P will experience a slightly weaker magnetic field and thus have a smaller chemical shift. This is the cause of peak splitting
  • What is peak splitting? Peak splitting is when a peak is split into smaller subpeaks. Peak splitting occurs because protons themselves act as magnets. Hence, if a hydrogen nucleus has one neighboring hydrogen nucleus (in a different chemical environment), that neighboring hydrogen nucleus can either be aligned with or against the NMR's magnetic field. Hence, the peak splits into 2 subpeaks of roughly equal height (because the probability of that neighboring nucleus being aligned or against the NMR's magnetic field is approximately equal).
    • 2 neighboring protons = 3 subpeaks in a 1:2:1 ratio. This is because the two neighboring protons can either both be against the field, both be aligned with the field, or one can be aligned and one can be against (for which there are 2 possible combinations, hence the 1:2:1 ratio)
    • 3 neighboring protons = 4 subpeaks in a 1:3:3:1 ratio.
    • In fact, there's a pattern. n neighboring protons will produce n+1 subpeaks: this is known as the n+1 rule. Note that chemically equivalent proteins don't cause splitting, nor do hydrogen bonding protons (eg, O-H, O-N). The ratio of the areas of these subpeaks can be modeled by Pascal's triangle.
    • But why doesn't peak-splitting happen for carbon-13 NMR? The low abundance of carbon-13 means that it is very unlikely for carbon-13 atoms to be adjacent to eachother, hence they will not cause splitting.
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Re: VCE Chemistry Question Thread
« Reply #564 on: April 10, 2014, 07:28:07 pm »
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to what extent do we need to know about how the equipment works in chromatography? e.g. how the columns work in gas chromatography, method of operation
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Re: VCE Chemistry Question Thread
« Reply #565 on: April 10, 2014, 08:04:17 pm »
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to what extent do we need to know about how the equipment works in chromatography? e.g. how the columns work in gas chromatography, method of operation

Specifically mentions on the study design that you don't need to know this. :)

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Re: VCE Chemistry Question Thread
« Reply #566 on: April 11, 2014, 09:49:26 am »
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In q 5c of 2012 chem exam 1, when drawing the structure, how can we determine on what carbon, the OH group is located?
Thanks :D
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Re: VCE Chemistry Question Thread
« Reply #567 on: April 11, 2014, 10:43:19 am »
+1
A few ways to do this, one way, there are only 2 peaks in the carbon NMR, but 3 carbons in the molecule so there must be symmetry somewhere, only way symmetry can work with this molecule is 2-propanol.

Second way is using the 1H-NMR.  There is a septet with an area of 1, which means that there must be an environment with 1 hydrogen (CH) next to 6 neighbouring hydrogens (2x CH3).  The only way this can work is if it is 2-propanol.

CH3CH(OH)CH3

The H is the peak at 1.2ppm
The H is the peak at 3.6ppm

In q 5c of 2012 chem exam 1, when drawing the structure, how can we determine on what carbon, the OH group is located?
Thanks :D
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #568 on: April 11, 2014, 10:58:29 am »
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A few ways to do this, one way, there are only 2 peaks in the carbon NMR, but 3 carbons in the molecule so there must be symmetry somewhere, only way symmetry can work with this molecule is 2-propanol.

Second way is using the 1H-NMR.  There is a septet with an area of 1, which means that there must be an environment with 1 hydrogen (CH) next to 6 neighbouring hydrogens (2x CH3).  The only way this can work is if it is 2-propanol.

CH3CH(OH)CH3

The H is the peak at 1.2ppm

The H is the peak at 3.6ppm

Thanks heaps jgoudie...
btw, just wanted to say that all your videos are awesome and they have helped me sooo much. Just a quick question, on your biomolecules-fat video, I thought that the formula for an unsaturated fatty acid was CnH2n-2O2 not CnH2nO2. I'm sure that I'm making the mistake but could you please explain where.
Thanks
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Re: VCE Chemistry Question Thread
« Reply #569 on: April 11, 2014, 11:21:12 am »
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For the synthesis of drugs using organic molecules, I'm aware that we need to know about the synthesis of acetyl-salicylic acid (aspirin) from salicylic acid. Do we also need to know about the production of penicillin? How about gene cloning?

Some clarification would be much appreciated!


Thank you!