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    (Original post by ScottLangley)
    That equation is not balanced, should be 2H2O
    So it should, my bad
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    (Original post by Haza296)
    how were you meant to answer the beta-carotene question without knowing how many carbon atoms were in the two rings?

    im convinced it's impossible
    It is not impossible at all. A ring means there are two less hydrogens compared to a straight chain. Ring size is irrelevant.
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    (Original post by Toffo132)
    For a benzene ring you need Tin(Sn) and Conc. HCl
    i put followed by conc NaOH since old spec papers allowed it provided you mentioned its not done in one step
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    (Original post by callumabbott99)
    Unsaturated simply means no double bonds, benzene rings are saturated
    No, it doesn't. It means that the compound has all of its valence electrons in single covalent bonds. In carbon's case, it has to be sp3 hybridised.
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    (Original post by student2034)
    But it was a dicarboxylic acid so 2 waters would have been produced so wouldn't the moles be 0.2 making 3.01x10^22 the answer?


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    Hydroxide was limiting reagent
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    (Original post by TeachChemistry)
    Hydroxide was limiting reagent
    It was the 6.02x10^22 wasn't it?
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    (Original post by student2034)
    f***


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    Check the data sheet, the leftmost peak is HC-O, not HC-CO. Top one is correct for the correct identification!
    Most people i've seen have also said the Top one, who is saying bottom?
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    (Original post by TeachChemistry)
    It is not impossible at all. A ring means there are two less hydrogens compared to a straight chain. Ring size is irrelevant.
    We were told it has one branched chain, but there are branched chains on the benzene, so I've no idea what OCR wanted.
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    I got 108 for that one
    (Original post by Asiangirl101)
    I got 108 or 180 cm^3 I can't remember
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    (Original post by M.E.L)
    I got the top one as well.. pretty sure it was the three methyl groups closer to the C double O bond
    It's the bottom one. You will lose one mark only I expect if you put the top one.
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    (Original post by medhelp)
    I swear it said how would you purify and not how would you purify AND check the purity?
    it asks *you* to check the purity via % yield so hopefully not putting the melting point check should be ok
    Yes you had to put how to check purity. This question is basically from PAG 6.3
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    (Original post by Toffo132)
    was a diprotic acid with a monobasic base. therefore 1:2, so 0.1mol of NaOH reacts with 0.05 mol acid, and 1 mol of acid forms 1 mol of water, so 0.05mol water forms, which is 0.05*6.02*1023= 3.01*10^22
    No but it literally told you 0.1 moles NaOH reacts with 0.1 moles of acid. It was a diprotic acid but only one side reacted because it wasn't in excess. Therefore 0.1 mol water
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    (Original post by StA200)
    It was the 6.02x10^22 wasn't it?
    Yes
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    (Original post by TeachChemistry)
    Yes
    How did your students find the paper? Easier than last week?

    And there was this question on doing an equation with an alcohol (1 mark), if you've seen the paper, do you remember if it was unsaturated or a saturated alcohol?
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    Is there an unofficial mark scheme available?
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    (Original post by tamoni4)
    For the last NMR question, I've got a
    2,2-dimethylpropyl ethanoate
    It had 3 methyl groups due to a big singlet with relative peak area of 9
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    (Original post by TeachChemistry)
    It's the bottom one. You will lose one mark only I expect if you put the top one.
    Ah alright thanks a lot
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    (Original post by AnnaRainbows)
    Why is it AgNO3 to measure the rate of activity/reaction? I put KCl because to see which one reacts fastest you need a displacement reaction? Cl displaces both iodine and bromine and then you can look at the speed... oh gosh
    That does measure rate, that just tells you which one is most reactive. AgNO3 shows which precipitate forms the fastest. And as the C-halogen bond enthalpy decreases down the group its easier to break C-I than C-Cl and C-Br therefore the yellow precipitate forms faster than the white and cream precipitates.
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    Phenol with ethanone joined together. for C NMR.
    https://www.thestudentroom.co.uk/att...2&d=1497870262


    For this I think it's wrong because on this picture you have 8 carbon environments and you needed 6 because there were only 6 peaks so you needed to put OH where it would make 6 environments
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    (Original post by Boss987)
    71% yield.
    Did anyone get 95 repeat units?
    Ester for hydrogen NMR. This includes 3 CH3.
    At the end L was a Ester: CH3CH2COOCH2C(CH3)2CH3
    M was a carboxylic acid: CH3CH2COOH
    N was an alcohol : HOCH2C(CH3)2CH3 (Ashleigh)
    recrystallisation for the method.
    Phenol with ethanone joined together. for C NMR.
    https://www.thestudentroom.co.uk/att...2&d=1497870262
    Butan-1-ol + 2O] --> Butanoic acid and Water, using acidified potassium dichromate and under reflux
    Butan-1-ol + [O] --> Butanal and water, using acidified potassium dichromate and distillation technique (AnnaRainbows)
    Diol more soluble as it has two hydroxyl groups.
    Alcohol forms h bonds with water so soluble.
    https://www.google.co.uk/url?sa=i&rc...97957876618458
    Beta carotene Q
    C40H56 + 11H2 --> C40H78 (TuffyandTab)

    MCQ
    BDCBDCACCBCAACB (tamoni4)
    AgNo3
    C7H14 - molecular formula
    Addition reaction (economy)
    Trigonal planar - Atom no. 3
    3 peaks for NMR in MCQ
    3 or 4 Chiral carbons??
    Acyl chloride + primary amine = amide MCQ
    Alicyclic and saturated??

    Good Paper!!!
    I'm pretty sure the formula for β-carotene, according to OCR, was not C40H56. I think it was C40H44. If you want I'll explain my reasoning, but I've seen them do something like this before where they simplified what actually happens.

    It was a question where ammonia is added dropwise to a complex solution. We know that it causes a precipitate to form and they wanted a reaction. The actual reaction was the complex ion + ammonia but in the mark scheme they put OH- instead, even though it's almost impossible for the OH- ions to react with the complex solution rather than ammonia
 
 
 
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