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OCR Salters Chemistry F335 12th June 2013 Exam revision thread Watch

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    (Original post by abzy1234)
    No problem! Tell me about it; I can't believe the exam is nearly 48 hours away! :eek:

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    Please don't remind me, I'm doing F334 as well which I haven't even started revising for as I have 4 other exams before it!
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    There was a question in the 2012 Jan paper about heating a dicarboxylic acid (fumaric acid). The acid loses water on heating to form compound X. In the question it asks you to draw the structural formula for compound X. Apparently the two carboxyl groups in the dicarboxylic acid react with each other on heating with the elimination of a water molecule. Is this reaction in the syllabus or is something we are supposed to be able to guess in the exam?
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    (Original post by abzy1234)
    No, a peak represents only one type of proton environment. I think you might be getting confused over splitting patterns?

    Remember, the size of the peaks tells you how much protons there are in that particular environment

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    Here the peak on the left represents the two protons between the oxygens but this peak is larger then the one on the right which represents 6 protons

    Is it the height of the peak which represents the amount of protons in the environment or the area underneath ?

    (the compound is diethoxymethane http://www.chemicalbook.com/Chemical..._CB4128049.htm)
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    Hi guys,
    I have a bit of a worry and I was wondering if others have experienced the same problem. Basically I finish a paper in 1 hour and 30 mins! I am worried that is too quick!
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    Oh yeah could anyone help with this question from the Jan 2013 paper, I've done the paper before but I missed out the whole dilution effect which confused the hell out of me.

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    (Original post by LRC4321)
    Hi guys,
    I have a bit of a worry and I was wondering if others have experienced the same problem. Basically I finish a paper in 1 hour and 30 mins! I am worried that is too quick!
    Yea, I never use up the 2 hours either.
    I wouldn't say that's a problem, it means you have time to check it a few times :P


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    (Original post by martynsteel)
    Oh yeah could anyone help with this question from the Jan 2013 paper, I've done the paper before but I missed out the whole dilution effect which confused the hell out of me.

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    Harantony explained it really well a few pages back:

    (Original post by Harantony)
    There is 20cm^3 of 0.015moldm^-3 HNO3 in a flask let's say.

    When you add the NaOH:
    You have now added 10cm^3 of 0.015moldm^-3 NaOH, so, before the reaction, there is 30cm^3 in the flask overall.
    20cm^3 of this is 0.015moldm^-3 HNO3
    10cm^3 of this is 0.015moldm^-3 NaOH

    The NaOH now reacts with the HNO3 because it's an acid-base reaction. Because they are of equal concentrations, you can deduce that, after the reaction:
    10cm^3 of 30cm^3 is 0.015moldm^-3 HNO3
    10cm^3 of 30cm^3 is 0.015moldm^-3 NaNO3
    10cm^3 is water.

    So, overall, there is 30cm^3 of liquid in the flask, with 10cm^3 of this being acid. Let's work out how many moles of acid are in there:
    10cm^3 = 0.01dm^3
    m = c x v, so:
    0.015moldm^-3 x 0.01dm^3 = 1.5x10^-4 moles of HNO3

    Now, we can work out the concentration of HNO3. Don't forget that there is 30cm^3 of liquid, and because we know that there are 1.5x10^-4 moles of HNO3 and that c=m/v:
    [1.5x10^-4]/0.03 = 5x10^-3 moldm^-3 of HNO3.

    Finding pH:
    We know that one mole HNO3 dissociates into one mole of H+ and one mole of NO3-. So there will be 5x10^-3 moles of H+ from 5x10^-3 moles of HNO3.
    pH = -log[H+], so
    pH = -log(5x10^-3)
    pH = 2.30 (3 s.f.)

    It's a lot of work for 2 marks!
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    (Original post by AmirHabeeb)
    Yes that's right but there are 3 different proton environments around that ring so each would give rise to a different peak (in the same range obviously). It's from quite an old legacy paper I believe.
    I think that the reason the integration of the peaks is 3:4 is because one of the H2C groups attached in that ring is bonded to a nitrogen which would downshift the peak arising to a different peak
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    (Original post by AmirHabeeb)
    Harantony explained it really well a few pages back:
    Yeah I posted my question without checking back haha!
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    (Original post by josephtsui)
    There was a question in the 2012 Jan paper about heating a dicarboxylic acid (fumaric acid). The acid loses water on heating to form compound X. In the question it asks you to draw the structural formula for compound X. Apparently the two carboxyl groups in the dicarboxylic acid react with each other on heating with the elimination of a water molecule. Is this reaction in the syllabus or is something we are supposed to be able to guess in the exam?
    i think you just had to be able to put it together as if you drew out the molecule without water you could see an o with no attachment which doesn't normally occur in non salt long chain molecules so the joining to make a cyclo chain should seem the reasonable solution
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    (Original post by super121)
    Please don't remind me, I'm doing F334 as well which I haven't even started revising for as I have 4 other exams before it!
    Ouch dude... Just gotta put your head down I guess; at least by next Wednesday it'll be all over

    (Original post by Tikara)
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    Here the peak on the left represents the two protons between the oxygens but this peak is larger then the one on the right which represents 6 protons

    Is it the height of the peak which represents the amount of protons in the environment or the area underneath ?

    (the compound is diethoxymethane http://www.chemicalbook.com/Chemical..._CB4128049.htm)
    Yes it's the height of the peak which represents the ratio/amount of protons. So the ratio of 6 should represent 6 protons in the CH2 environment. It's quite a confusing question as its diethoxymethane, so you'll have to watch out for the CH2-O environment.

    That's a horrible nmr question regardless!
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    (Original post by abzy1234)
    Ouch dude... Just gotta put your head down I guess; at least by next Wednesday it'll be all over



    Yes it's the height of the peak which represents the ratio/amount of protons. So the ratio of 6 should represent 6 protons in the CH2 environment. It's quite a confusing question as its diethoxymethane, so you'll have to watch out for the CH2-O environment.

    That's a horrible nmr question regardless!
    well I just asked my teacher and he stressed that it's the cumulative area under the peak (and its splitting) and as it's a singlet on the left it is in the end a lower area then the triplet on the right
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    (Original post by Tikara)
    well I just asked my teacher and he stressed that it's the cumulative area under the peak (and its splitting) and as it's a singlet on the left it is in the end a lower area then the triplet on the right
    They are interchangeable. The relative areas of the peaks is proportional to the number of protons in that proton type.
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    (Original post by AmirHabeeb)
    They are interchangeable. The relative areas of the peaks is proportional to the number of protons in that proton type.
    what's interchangeable ? sorry this nmr stuff is doing my head in xD
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    (Original post by gkweic)
    i think you just had to be able to put it together as if you drew out the molecule without water you could see an o with no attachment which doesn't normally occur in non salt long chain molecules so the joining to make a cyclo chain should seem the reasonable solution
    Not entirely sure if it's on the syllabus but the dehydration of two carboxylic acids forms an acid anhydride
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    (Original post by Tikara)
    well I just asked my teacher and he stressed that it's the cumulative area under the peak (and its splitting) and as it's a singlet on the left it is in the end a lower area then the triplet on the right
    Your teachers right surely? The integration of peaks in NMR is relative to the number of protons within those environments?
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    (Original post by martynsteel)
    Not entirely sure if it's on the syllabus but the dehydration of two carboxylic acids forms an acid anhydride
    i think thats on the medicine topic syllabus for f334 but as this is synoptic i guess we have to know that
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    Could anyone link me the F335 January 2013 question paper? x
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    Could someone please explain to me 2b(i) in Jan 2012, thanks
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    Hi, I have the January 2013 paper, but I don't have the mark scheme :/
    Could someone post on here please, be really grateful
 
 
 
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