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    Some random answers:
    6.02*10^22
    Ch3Ch2COOCH2c(CH3)3
    Carotene saturated had 78 hydrogen and unsaturated 56 hydrogen
    But-3-ene-1-ol
    95 repeat units
    Phenylethanone with OH group opposite
    Silver nitrate solution
    HCL/H20 for hydrolysis, NaBH4 for reduction and sn/concentrated HCL for reduction of NO2 group
    (can anyone remember the naming Of The cyclic compound)
    4 chiral carbons
    Like 71% yield, you can recrystallise by dissolving in the solvent then filtration under reduced pressure, before testing for purity by taking the melting point and comparing to known values
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    (Original post by Lokikado)
    If i put "the impurities crystallize out" for the six marker but put everything else correctly, how many marks would i lose?
    1 mark at most, id say
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    (Original post by samwray)
    1 mark at most, id say
    Thank you!
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    (Original post by Toffo132)
    For a benzene ring you need Tin(Sn) and Conc. HCl
    I did that, but I mean the one where you're reducing a benzene ring with a KETONE on it to an alcohol. Can it be H2/nickel here?
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    (Original post by thursdaygirl)
    I did that, but I mean the one where you're reducing a benzene ring with a KETONE on it to an alcohol. Can it be H2/nickel here?

    You need NaBH4
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    (Original post by iamDev)
    It was weakly acidic as ph 5. So it's carboxylic acid or phenol.not c acid because it didn't react with carbonate. Must be phenol
    It said it for medium an orange precipitate and it didn't react with Tolles.

    I got a disubstituted benzene..money attached to a ketone and the other one bonded to ch3
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    (Original post by thursdaygirl)
    I did that, but I mean the one where you're reducing a benzene ring with a KETONE on it to an alcohol. Can it be H2/nickel here?
    It says in my revision guide that you can put H2 with nickel catalyst
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    (Original post by Dynamic_Vicz)
    For the IUPAC I had trouble naming the compound. I wrote hexene-3-ol. Would this suffice?
    I would say no because you have only identified the position of the hydroxyl group and not the alkene group. Slightly ambiguous. Then again who knows ahaha
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    I put 1hydroxyhexan3ene don't know if this is alright.
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    (Original post by iamDev)
    It was weakly acidic as ph 5. So it's carboxylic acid or phenol.not c acid because it didn't react with carbonate. Must be phenol
    I got a ketone as it didn't react with tollens reagent
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    (Original post by AliMoh4med)
    I got a ketone as it didn't react with tollens reagent
    Yh that's correct too. It was a phenol attached to C=O CH3
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    (Original post by tamoni4)
    I got 108 cm^3
    Gah I somehow got this wrong! I got like 127cm3, how did people work this out, has anyone made a list of the questions yet????
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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!!!
    If the answer for purity was in terms of separating funnel and anhydrous salt, does that get you any marks?
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    (Original post by Fasih98)
    If the answer for purity was in terms of separating funnel and anhydrous salt, does that get you any marks?
    A separating funnel and drying agent are used in the preparation of an organic liquid, so as the product you want to obtain is solid, unfortunately I imagine not - that's only about 3 marks however, the other two will be for % yield and determining the melting point
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    (Original post by TeachChemistry)
    You need NaBH4
    I wrote nabh4 but wonder why don't the mark schemes ask for it to be aqueous or with protons ? Would it also give a proton and a hydride ion?
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    (Original post by TLS1234)
    Highest ive seen in past papers is 92-A*, 87-A, 76-B but then again who knows how everyone found it
    That was last year's I think. But last year's was fairly straight forward tbh. I thought there was a lot more room for errors on this so I'm thinking 78ish for an A
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    (Original post by Paulina0)
    It had 3 methyl groups due to a big singlet with relative peak area of 9
    No, it had 2 methyl groups and the third one was a continuation of the chain. You cannot have a 2,2,2-trimethylethyl ethanoate.
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    (Original post by Jitesh)
    I wrote nabh4 but wonder why don't the mark schemes ask for it to be aqueous or with protons ? Would it also give a proton and a hydride ion?
    I think so because isn't it nucleophilic substitution so would involve that anyway haha idk lol
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    Guys for E and F with Markovnikov I swear you could have drawn E as major and F as minor OR the other way around like it didn't specify. When they asked what the major product was at the end i'm assuming it will be BASED on what we had drawn? My one just happened to be F because the hydrogen attached to a carbon with 6 H's compared to the other molecule where it attaches to 3 because I don't think you count the methyl on the CH2CH3. Can anyone vouch for this or am I just rubbish at chem??
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    What is an A* in general? Like 85%?
 
 
 
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