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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

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Original post by itsConnor_
Sorry where is the presentation? :smile:


It's from https://www.cpdhub.ocr.org.uk/ which requires a (free) registration, however I've also uploaded a copy here: https://my.mixtape.moe/gfofsl.pdf
Reply 901
Original post by suibster
Assume the chemical shift is at where it indicates.


so if theres a peak at 2 then i should assume they mean at 1 ppm?


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Original post by marioman
It's from https://www.cpdhub.ocr.org.uk/ which requires a (free) registration, however I've also uploaded a copy here: https://my.mixtape.moe/gfofsl.pdf


WOW thanks so much
Reply 903
Original post by ImNervous
I think the height of the peak at 122 m/e would be greater as it does not contain the isotope so is more abundant? Not really sure though

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I believe the answer is 1:1 so not sure..
Reply 904
Original post by AqsaMx
I believe the answer is 1:1 so not sure..


i dont think we need to know this though? or am i wrong


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Reply 905
Original post by ranz
i dont think we need to know this though? or am i wrong


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It's to do with fragments ( mass spec) and its synoptic from as I believe
Reply 906
guys if theres an aldehyde for example, would there be a peak at both 9.5 ppm ( -CHO) and 3.5 ppm (HC-O) for thr aldehyde


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Original post by itsConnor_
But surely if stationary phase is solid, it separates by adsorption and if the stationary phase is a liquid, it separates by relative solubilities?


Original post by AqsaMx
I don't have the paper, it was given to me as a pack but the answer was adsorption..


If the stationary phase is liquid, then the separation occurs by relative solubility.
If the stationary phase is a solid, then the separation occurs by adsorption
Reply 908
Original post by RayMasterio
If the stationary phase is liquid, then the separation occurs by relative solubility.
If the stationary phase is a solid, then the separation occurs by adsorption


isnt gc always relative solubility and tlc adsorption?


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Original post by ranz
isnt gc always relative solubility and tlc adsorption?


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If the stationary phase in GC is a liquid, then the separation is done by relative solubility. If it was a solid, then it will be adsorption
With NMR I go like this...

Let's just say

Triplet peak at 0.8ppm with peak area of 2 shows 2xH in CH2, adjacent to CH2

And do the same for every peak in the spectra...

Isn't that just as fine?
Original post by Serine Soul
With NMR I go like this...

Let's just say

Triplet peak at 0.8ppm with peak area of 2 shows 2xH in CH2, adjacent to CH2

And do the same for every peak in the spectra...

Isn't that just as fine?


That's fine. There's no specific format that you must follow, as long as you put the info in you will get the marks!
Original post by ranz
guys if theres an aldehyde for example, would there be a peak at both 9.5 ppm ( -CHO) and 3.5 ppm (HC-O) for thr aldehyde


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No...
Aldehydes consist of C=O bonds, not C-O bonds, so only a peak between 9 - 10 ppm, and 2-3 ppm if there's another CH attached to the C in C=O
(edited 7 years ago)
Original post by marioman
That's fine. There's no specific format that you must follow, as long as you put the info in you will get the marks!


Yeah, the table looks really tidy though! (I'm a messy person) just wish I had been told to use it earlier :frown:
Reply 914
Original post by Serine Soul
No...
Aldehydes consist of C=O bonds, not C-O bonds, so only a peak between 9 - 10 ppm, and 2-3 ppm if there's another CH attached to the C in C=O


so if theres a carbon attached to the aldehyde it would show up as HC-O 2-3 ppm?


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can someone please tell me how to approach these questions? the method for them changes everytime and I never get full marks even though these are supposed to be easy marks
Reply 916
Original post by RayMasterio
can someone please tell me how to approach these questions? the method for them changes everytime and I never get full marks even though these are supposed to be easy marks


( actual yield/ theoretical yield ) x100 = % yield
and just rearrange to find different parts when needed.



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I'm going to do NMR like this
Peak at 0.8 ppm due to R-CH
Triplet: adjacent to CH2
Peak area of 3: CH3
-CH2CH3
Reply 918
Original post by yoda123
I'm going to do NMR like this
Peak at 0.8 ppm due to R-CH
Triplet: adjacent to CH2
Peak area of 3: CH3
-CH2CH3


yh thats s good layout i think table can get confusing and u might grt some examiner who doesnt think table is enough.


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Hi,
Does anyone know the conditions for the acid hydrolysis of polyesters and polyamides?


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