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    What did you guys draw for the partial hydrolysis of the tripeptide with the residue? I couldnt figure it out because they all had C=O bonds but i drew the one on the left on its own and the two on the right still attached
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    omg i got 48g for the bromine one.

    also, for the last question, for the cyanide ions, could the CN be on any part of the benzene ring?

    that was **** hard
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    Needed a very high A/A* in this exam as well LOOL.
    Full UMS in F335 here I come
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    (Original post by Zuzuvela)
    There was no iodine question it was bromine numbers it got everyone confused. I've heard of iodine number before but not bromine. And correct me if I'm wrong but iodine numbers aren't F334 anyway. Either f331 or F335

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    Thanks man, i was confused for a sec.
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    (Original post by yoyo.sat)
    What did you guys draw for the partial hydrolysis of the tripeptide with the residue? I couldnt figure it out because they all had C=O bonds but i drew the one on the left on its own and the two on the right still attached
    I drew the two on the left still attached and the one on the right on its own.


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    (Original post by yoyo.sat)
    What did you guys draw for the partial hydrolysis of the tripeptide with the residue? I couldnt figure it out because they all had C=O bonds but i drew the one on the left on its own and the two on the right still attached
    Yes that's what I did
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    (Original post by jack.lp.thompson)
    As the spec only says you need to know about [Fe(H20)6]2+ and [Fe(H20)6]3+ for iron complexes, I doubt it! But who knows!! 😄
    But it was talking about those haha
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    (Original post by jack.lp.thompson)
    Fe2+ is oxidised to Fe3+ by the airFe2+ => Fe3+ + e-This makes all the iron become Fe3+They all react with the -OH ion like so:Fe3+ + 3OH- => Fe(OH)3 (s)Iron(iii) hydroxide is a red brown precipitate.Something like that but with better English 😊
    This is what I put! forgot to include the Fe3+ to Fe2+ equation though so will have lost a mark
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    This paper has pissed me off. I thought the majority of the questions were unfair and some of the questions that came up are not even defined on the syllabus! Come on OCR, I really want to go to university!
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    (Original post by Kwaks)
    I drew the two on the left still attached and the one on the right on its own.


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    but there were two amide links so it split into 3 molecules
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    (Original post by Kwaks)
    I drew the two on the left still attached and the one on the right on its own.


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    I did it the other way round, did the two on the right but could easily be wrong
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    (Original post by dickbutt)
    Yeah I've seen everyone else got 7 too. How did you do it?
    I got 10 too!!?
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    (Original post by Daniel9998)
    but there were two amide links so it split into 3 molecules
    No it was a partial hydrolysis that only broke one amide bond
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    Apart from question 3 i think that was relatively OK i guess, could've been alot worse.
    anyway whats done is done, time to focus on this weeks exams then cram F335 from thursday onwards
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    (Original post by Daniel9998)
    but there were two amide links so it split into 3 molecules
    That's what I did as well! Think it was saying about partially because of the nh2 group at the bottom??
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    Did anyone else think that this paper was overly full of content? Like there wasn't enough time?
    There were no easy one markers or anything really and loads of very wordy questions with very little space for the answers!
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    (Original post by josephinemar25)
    No it was a partial hydrolysis that only broke one amide bond
    OH FFSSSS
    losing so many marks........................... ..........
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    (Original post by Daniel9998)
    but there were two amide links so it split into 3 molecules
    I made the same mistake as you but apparently because it was just partial hydrolysis, it only formed 2 molecules.
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    Bromine questions I got 390 for the first and 13 for the one later on.
    Membrane was to allow current (electrons) to flow between the two electrochemical cells whilst keeping the ion solutions separate (salt bridge could've got the mark there I suppose because thats basically the job of it).
    The peptide that should have broken is the one just to the right of the r group mentioned in the question, so you had two residues on the left and a single amino acid on the right.
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    (Original post by IBerrr)
    For the 5 marker of bromine this is the method I used:

    After working out the moles of BrO3- in the previous question (= 0.005 moles and so was in excess) and looking at the reaction equation given, you know that the ratio of BrO3- was 1:3 for Br2, so the moles of Br2 is 3x0.005 = 0.015

    We also know that it was 0.015 moles of br2 in 200cm3, but the hexadiene molecule only reacts with 83, so 0.015/200x83? I think

    So that's the moles of br2 reacted is worked out, so you multiply that by mass of bromine and then x2 because two bromine molecules.

    Lastly you times that by 10 because for bromine number it is of 100g of hexadiene whereas the question said 10g.

    I don't know if this is right, this is just what I did. Anyone do the same?
    BrO3- was in excess, wasn't it? So You had to use the moles of Br- to calculate the moles of Br2 (by dividing by 5 and multiplying by 3). I'm not 100% sure but that is what I did.
 
 
 
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