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

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    (Original post by annaj97)
    Why do only O-H and N-H peaks disappear when you add D2O?
    Don't forget -COOH
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    (Original post by ranz)
    what about in standard isolectric point


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    At isoelectric point it exists as zwitterion with NO OVERALL CHARGE (in definition) hence its only COO- and NH3+ and the R group is NOT affected
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    (Original post by Magicalgeofray)
    D2O Undergoes Proton exchange with the Hydrogen atoms (protons). D atoms have an even number of nucleons so are undetected in the NMR spectra and results in no O-H or N-H peaks
    Why does this only happen with O-H and N-H peaks?
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    (Original post by CalistaJupiter)
    They don't all protonate… I did a past paper that had a 'zwitterion' with an extra cooh side group and the mark scheme said "reject if both cooh groups are shown to have donated a proton" because only groups directly involved in being an amino acid (eg RCH(NH2)COOH) actually protonate
    yes BUT if its then in acidic conditions or pH below isoelectric point, the ALL NH2s become NH3+ COOH stay COOH

    pH above IP, all COOH beccome COO- and NH2 stay NH2
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    (Original post by annaj97)
    Why does this only happen with O-H and N-H peaks?
    -OH, -NH and -COOH protons are labile and easily interchange with D2O protons
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    (Original post by Rust Cohle)
    Ooops, number 4 was typo; should be suggest why organic product in q3 cannot be hydrolyses.
    1, 2 and 3 three correct.
    still think its cos theres no amide/peptide/ester linkage hence cannot hydrolyse? :/
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    (Original post by annaj97)
    Why does this only happen with O-H and N-H peaks?
    It can to COOH and S-H as well and the reason is because their protons are easily exchangeable/'labile'
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    (Original post by tcameron)
    the isolectric point is when it exists as a zwitterion, not below the isoelectric point
    yep i know :s what did i say that was wrong?
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    But carboxylic acid can still go through reduction if added 2[H] e.g. Ethanoic acid to Ethanol
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    If anyone has ran out of Past papers i would recommend AQA Unit 4 papers from about half way onwards is very similar to our spec, however the questions deal with concepts our spec has not covered, yet we can apply our knowledge to questions such as reaction mechanisms, personally these were great for practicing those kinda awkward application based questions.
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    (Original post by sakuraton)
    But carboxylic acid can still go through reduction if added 2[H] e.g. Methanoic acid to Methanol
    because methanol is effectively an aldehyde
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    (Original post by Magicalgeofray)
    Attachment 549315Attachment 549307 This is what the mark scheme says on Zwitter ions for this question i was confused too so emailed my teacher, and this was her reply but I am still confused:
    [/font][/font][/font][/font]
    For zwitterions, its only the NH2 and COOH in the general formula that are affected - NH3+ and COO- respectively, giving no overall charge

    Break at the peptide linkage
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    (Original post by tcameron)
    because methanol is effectively an aldehyde
    Whoops wrong example, ethanoic acid to ethanol
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    (Original post by tcameron)
    because methanol is effectively an aldehyde
    Need to practise my spelling
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    https://en.wikipedia.org/wiki/Zwitterion
    In chemistry, a zwitterion is a neutral molecule with both a positive and a negative electrical charge. Multiple positive and negative charges can be present. Zwitterions are distinct from dipoles, at different locations within that molecule.

    if there is a molecule with n amines and n carboxyls, wouldn't it be all protonated as long as the charge stays netural? (n-n=0)
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    (Original post by sakuraton)
    Whoops wrong example, ethanoic acid to ethanol
    I even meant to say methanioic acid is an aldehyde not methanol

    And I don't think that reaction is possible
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    (Original post by lai812matthew)
    https://en.wikipedia.org/wiki/Zwitterion
    In chemistry, a zwitterion is a neutral molecule with both a positive and a negative electrical charge. Multiple positive and negative charges can be present. Zwitterions are distinct from dipoles, at different locations within that molecule.

    if there is a molecule with n amines and n carboxyls, wouldn't it be all protonated as long as the charge stays netural? (n-n=0)
    Nope nope nope in zwitterions it ist JUST the NH2 and COOH in the general formula that are affected, forget about the R group completely - if my understanding/method is correct.

    But if its put in acid/alkaline respectively, all COOH/NH3 react accordingly if that makes sense?
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    (Original post by tcameron)
    I even meant to say methanioic acid is an aldehyde not methanol

    And I don't think that reaction is possible
    Then can you explain this:

    http://www.ocr.org.uk/Images/79471-q...d-analysis.pdf

    http://www.ocr.org.uk/Images/61007-m...is-january.pdf

    January 2012, Question 3)b) Thank you!
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    does anyone have a copy of 2015 papers for F324??
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    (Original post by sakuraton)
    Then can you explain this:

    http://www.ocr.org.uk/Images/79471-q...d-analysis.pdf

    http://www.ocr.org.uk/Images/61007-m...is-january.pdf

    January 2012, Question 3)b) Thank you!
    I thought it wasn't possible too until I saw this o.o
 
 
 
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