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

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    Damn guys, retaking this. I did the last years paper and got an D. This year I am going to nail it. I feel so much more ready. Just hope it ain't extremely hard. I have a massive feeling fats will come up.
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    (Original post by marioman)
    Read the question carefully - in June 2015 it tells you to "use your answer to (c)" (14 peaks) and "the chemical shifts and splitting patterns in the 1H NMR spectrum". Although it doesn't say it outright then it's heavily hinting that the main clues to the structure are in the 1H spectrum rather than the 13C spectrum.

    ahhh that makes more sense now
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    (Original post by RayMasterio)
    My problem with the June 2015 paper was which environment caused the peak. Like the NH bond never crossed my mind, i thought it was a carbonyl compound because it has the same chemical shifts.

    How would you overcome this problem?
    The key is using all the information that has been given. The structure of precipitate L given in the question has an N-H environment. There is no mention of D2O in the question, so you should know to look out for an N-H peak in the spectrum. This is easy to find because you can see (again using the given structure of L) that this peak will be a singlet, and the spectrum only has one singlet peak.
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    I'm feeling confident for the exam. The F324 was a challenging paper given the questions + the time constraints. But without the time constraints, it's relatively straight forward.

    How is everyone revising? I've done all the possible revision there is
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    In the mock of F324 on Jan I scraped an A at like 44/60, June 2015 WAS hard, I walked out and remember going ''well if that was my uni place, it would be gone'' but I sorta got the NMR question as I realised there was another benzene peak and then the rest fell into place. The WHOLE reason any of us (if youre on here, I assume you've been working your ass off and you're wiling to ask for help) would fail is because of pressure. Exams suck.
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    (Original post by AqsaMx)
    Attachment 548477

    Can anyone explain this question? Like how do you know the position of the double bond in the third monomer?
    This is what confused me as well!
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    Also, for F325, in some of the titration questions on the last few pages the mark scheme says to multiply by 40? How do you know whether or not to do this? (Sorry I can't word it better )
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    (Original post by AqsaMx)
    Attachment 548477

    Can anyone explain this question? Like how do you know the position of the double bond in the third monomer?
    Draw each carbon out (so not skeletal formulae) and then work from there. There isn't a CH2 group coming off the second carbon on the chain. By placing the double bond there you've added an extra carbon. Sorry this is probably quite a terrible explanation, but draw each carbon out and you might see what I mean. Also the double bond can't be connecting to the benzene carbon because each carbon on benzene cannot have a fourth bond.
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    (Original post by Lularose83)
    Also, for F325, in some of the titration questions on the last few pages the mark scheme says to multiply by 40? How do you know whether or not to do this? (Sorry I can't word it better )
    Usually the question will say 250cm^3 of the solution was prepared, and then 25cm^3 titrated. so to get the moles of the original solution you must x10. Also the molar ratio, in the case of 'x40', would be 1:4 meaning you must x4 as well as x10 = x40. Does this make sense?
    -Sometimes it might say x50 in the mark scheme if the ratio is 1:5 and the orig solution has 10x more volume.
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    On page 79 of the textbook, why does it say that if the chromotogram were to be repeated in same conditions and same solvent, all Rf values will be the same?
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    (Original post by mechanism)
    Usually the question will say 250cm^3 of the solution was prepared, and then 25cm^3 titrated. so to get the moles of the original solution you must x10. Also the molar ratio, in the case of 'x40', would be 1:4 meaning you must x4 as well as x10 = x40. Does this make sense?
    -Sometimes it might say x50 in the mark scheme if the ratio is 1:5 and the orig solution has 10x more volume.
    Yes I think I get it now! Thank you so much
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    Can anyone here explain to me the Answer to question 3d Part 2 (ii) In the June 2013 F324 Paper? It's soooo confusing...and I feel like none of the teachers on youtube doing paper walkthroughs actually know why the answer is what it is...they just say the answer and move on to the next question. It's only 1 mark but every mark makes a difference!

    http://www.ocr.org.uk/Images/175441-...d-analysis.pdf
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    anyone have a formula sheet for F325?
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    (Original post by drprocrastinator)
    anyone have a formula sheet for F325?
    http://www.ocr.org.uk/Images/74947-datasheet.pdf
    data sheet ?
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    no like a list of the equations like pka and ka and kc etc
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    (Original post by BrownGuyIrfan)
    Can anyone here explain to me the Answer to question 3d Part 2 (ii) In the June 2013 F324 Paper? It's soooo confusing...and I feel like none of the teachers on youtube doing paper walkthroughs actually know why the answer is what it is...they just say the answer and move on to the next question. It's only 1 mark but every mark makes a difference!

    http://www.ocr.org.uk/Images/175441-...d-analysis.pdf
    I need help with this question too, but im thinking it's because of where the open-ended bond is on the part that's added on? Hence the double bond must be where the oppen bond is?
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    (Original post by BrownGuyIrfan)
    Can anyone here explain to me the Answer to question 3d Part 2 (ii) In the June 2013 F324 Paper? It's soooo confusing...and I feel like none of the teachers on youtube doing paper walkthroughs actually know why the answer is what it is...they just say the answer and move on to the next question. It's only 1 mark but every mark makes a difference!

    http://www.ocr.org.uk/Images/175441-...d-analysis.pdf
    Mechanism (the user) explained it a couple of posts ago.
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    (Original post by itsConnor_)
    I need help with this question too, but im thinking it's because of where the open-ended bond is on the part that's added on? Hence the double bond must be where the oppen bond is?
    (Original post by BrownGuyIrfan)
    Can anyone here explain to me the Answer to question 3d Part 2 (ii) In the June 2013 F324 Paper? It's soooo confusing...and I feel like none of the teachers on youtube doing paper walkthroughs actually know why the answer is what it is...they just say the answer and move on to the next question. It's only 1 mark but every mark makes a difference!

    http://www.ocr.org.uk/Images/175441-...d-analysis.pdf
    I think all the confusion with the question is "how do you know the monomer didn't have a methyl group?"
    For me what makes the most sense is the fact that the monomer joins the polymer unit at two points (eg there is a dotted line coming off the monomer on top of the fact that it is directly attached to the unit). So it can't ever have been a methyl group becasue a methyl group can't undergo addition polymerisation, only a double bond can. And it's not like there could be two double bonds as then the carbon would have 5 bonds which isn't possible.
    So the best way to think of it is to go through the polymerisation and although it seems/looks like there should be an extra methyl group, if there was, there would be not dotted line coming off to the bottom. That was probably really badly explained sorry
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    What name is given to the process by which components are separated by gas/liquid chromatography?

    I thought the answer was relative solubility, but the MS says Adsorption???
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    If the Benzene ring did not have the circle in the middle, it would be c6h12 right?
 
 
 
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