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OCR Chemistry A F324 Rings, Polymers and Analysis Tue 19 June 2012

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Reply 260
Hey, can anyone help please, I've had a sudden confusion over gas and thin layer chromatography. Do they both separate by adsorption? I thought one separated by relative solubility??


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Reply 261
give an example of one you don't understand and we could all help you go through it...lotsa brains together is better than 1 i guess:biggrin:

i might be able to help slightly....
i think the nicest one to understand is proton nmr
using the number of peaks you can work out number of hydrogen environments
then work out ratio of peaks- as in the height (it'll probably either tell you this- shown as numbers on top of each environment, or give you 'steps' to measure height of)
you may be given molecular formula: C8H16O2
using H+ nmr
say you find 4 environments
say the ratio is 2 : 2 : 3 : 9
from the number of mini peaks in each environment you find the splitting pattern of q : s : t : s

2 : 2 : 3 : 9
q : s : t : s

from this you know that 9 protons means there must be 3x CH3 in the same environment

the 2 singlet peaks tell us that they are not next to any carbons with any hydrogen's on them

the q tells us that it is adjacent to a CH2-CH2

the t tells us this must be adjacent to a CH3 group only

using the n+1 rule along with delta ppm values we can work out bonds/groups responsible for the peaks

so we know there are 8 carbons, 16 hydrogens and 2 oxygens....assuming from values we now know this is an ester

it must have a carbon with 3 CH3 groups attached, for the ester it must have a carbon double bonded to an oxygen and single bonded to another oxygen, there must be a CH3 at the end....

so structure must be C(CH3)3CH2OOCCH2CH3 AKA
CH3CH2C=OOCH2C(CH3)3

i know it seems really confusing the way i've tried to write about it, it's a lot easier just writing down the ratio and splitiing pattern and having lots of annotations linking to values and groups etc.

p.s. remember when your doing an nmr question that even if the question is out of 10 around 6-8 of the marks are for all of your working out and that includes just writing down any values from the data table and ratios which the question may have already mentioned, it's better to write down the same thing again just in case you get a mark for it even if you might not get any credit it's worth a try. only about 2-4 of the marks are for coming up with the correct structure: even if you get half of the structure right through guessing you could get almost full marks

i hope this helped a little:^_^:
Reply 262
Original post by Bright
Ahh bekhuda hoselam sar rafte, delam gereft! :frown:

And I keep thinking about my Eng Lit the day after this and bio!!!!.. :'(


akheysh me too! do you have bio f214??
mitarsam :frown:
look at inbox!
Reply 263
Original post by Aa234
Hey, can anyone help please, I've had a sudden confusion over gas and thin layer chromatography. Do they both separate by adsorption? I thought one separated by relative solubility??


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dont freak out!!!

this is all you need to remember:

Chromatography is an analytical technique that separates components in a mixture between a mobile phase and a stationary phase

GC- seperation by adsorption
TLC- seperation by relative solubility

MP(phase that does moves) can be liquid/gas
SP(phase that doesn’t move) can be solid/liquid on a solid support

For TLC the Rf = distance moved by component / distance moved by solvent front

Limitations for TLC: similar components have similar Rf values & unknown compounds have no value for comparison

For GC Retention Time = time taken for compound to pass from column inlet to detector, area under peak is proportional to amount

Limitations for GC: similar compounds have similar retention times, some substances may ‘hide’ under other peaks and unknowns have no ref: RT’s for comparison
Reply 264
anyone have any guesses about what they think might come up on tuesday
anyone wanna share thoughts???
Reply 265
Original post by browb003
Yeah that's correct :smile:

Arenes undergo electrophilic substitution rather than addition as substitution means that the delocalised ring of electrons is not permanently broken, and the delocalised ring makes benzene more stable so it would want to keep the delocalised ring

Substitution would permanently break the delocalised ring and so substitution is not energetically favourable for arenes :smile:


thankyouuuuuuuuuuuu!!!!!!!!!!!!!!!:awesome::banana::banana2:
Reply 266
Original post by hetty123
dont freak out!!!

this is all you need to remember:

Chromatography is an analytical technique that separates components in a mixture between a mobile phase and a stationary phase

GC- seperation by adsorption
TLC- seperation by relative solubility

MP(phase that does moves) can be liquid/gas
SP(phase that doesn’t move) can be solid/liquid on a solid support

For TLC the Rf = distance moved by component / distance moved by solvent front

Limitations for TLC: similar components have similar Rf values & unknown compounds have no value for comparison

For GC Retention Time = time taken for compound to pass from column inlet to detector, area under peak is proportional to amount

Limitations for GC: similar compounds have similar retention times, some substances may ‘hide’ under other peaks and unknowns have no ref: RT’s for comparison


Hey thank you for this, but I just looked at jan 2011 question 5 past paper and it says that tlc separates by adsorption?


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Reply 267
Original post by Aa234
Hey thank you for this, but I just looked at jan 2011 question 5 past paper and it says that tlc separates by adsorption?


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tlc does separate by adsorption, i think she made a mistake
gc separates by relative solubility and adsorption depending on the stationary phase
if the stationary phase is liquid then it separates via solubility if its solid its absorption
Original post by niaghez
SAME. Im just sitting here staring at a page lol.


Best way to revise for this paper is to just write all the reactions down in module 1 and 2. I did this, then go over them a couple hours later, just keep doing this and they will stick. Then for module 3 just past papers. Do the examples, its better then reading the book over and over again. :smile: good luck :smile:
Original post by Bright
Ummm I'm finding this quite difficult :/ It's A LOT of memorising to do and my brain is already saturated from F322 and F325 revision.. :/


Best way to revise for this paper is to just write all the reactions down in module 1 and 2. I did this, then go over them a couple hours later, just keep doing this and they will stick. Then for module 3 just past papers. Do the examples, its better then reading the book over and over again. good luck
Reply 270
Original post by lagoom
tlc does separate by adsorption, i think she made a mistake
gc separates by relative solubility and adsorption depending on the stationary phase
if the stationary phase is liquid then it separates via solubility if its solid its absorption


Oh right okay, thank you :smile:


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Reply 271
Original post by gerrard1892
Best way to revise for this paper is to just write all the reactions down in module 1 and 2. I did this, then go over them a couple hours later, just keep doing this and they will stick. Then for module 3 just past papers. Do the examples, its better then reading the book over and over again. good luck


Ok thanks!! :]
Reply 272
Anyone have the link for jan 2012 paper
Reply 273
Original post by kingam
Anyone have the link for jan 2012 paper


http://www.thestudentroom.co.uk/showthread.php?t=2013091
Reply 274
For mass spec is every fragment charged? Because ive got here that ch3 is neutral?? :s


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Reply 275
TIme to do F324...again.

Just hoping I remember some stuff :biggrin:
Reply 276
Original post by Aa234
For mass spec is every fragment charged? Because ive got here that ch3 is neutral?? :s


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I believe they always have a +?
Original post by tomkeys
It is not an arene, just a cyclic hydrocarbon with two alkene groups thrown in. As has been mentioned above, it undergoes electrophilic addition, rather than electrophilic substitution, which is what I believe you are describing. A good add-on question would be "Why does this compound react differently to an arene?" :tongue:


For this questions 'Why does this compound react differently to an arene?' would you write about cyclohexene having pi eledtrons which are localised between 2 carbon atoms so elctron density increases so molecules are polarised more easily(dipole induced)
whereas benzen is rsisiatnt to bromination because it is more stable it has pi electrons which are delocalised and so electron density is insuffincet, hence cannot induce a dipole in Benzene?? anything else..am i on the right track or no ?? :smile:
Reply 278
Original post by rasklatz
I believe they always have a +?


Oh okay thanks!


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Reply 279
Original post by Fatima0065
For this questions 'Why does this compound react differently to an arene?' would you write about cyclohexene having pi eledtrons which are localised between 2 carbon atoms so elctron density increases so molecules are polarised more easily(dipole induced)
whereas benzen is rsisiatnt to bromination because it is more stable it has pi electrons which are delocalised and so electron density is insuffincet, hence cannot induce a dipole in Benzene?? anything else..am i on the right track or no ?? :smile:


For A-Level, I think so:smile: I think there was a question on the specimen paper about comparing the reactivity of benzene, phenol and cyclohexene with bromine and you seem to have hit the marking points that were on there for the first and third compounds.

I imagine the reason why benzene undergoes substitution has something to do with its stability and this being better preserved by a substitution rather than an addition, but that's beyond A-Level as far as I know.

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