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    (Original post by Mango88)
    But the molecular formula only contained one oxygen
    Yes i'm not sure but I think it said suggest a molecular formula.
    And then I though it never said the sample was pure which means what you have there could be an impurity or me and others. And this would explain the strange splitting pattern in the NMR.
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    (Original post by ZAM-H)
    Look yh I'm sorry but that paper was disgusting, no two ways about it. You're coming on here and posting how well you think you did when ppl want to just forget about it. You need to know when to keep quiet.

    Posted from TSR Mobile
    Well I think I got about 72, so if you're going to hate on everyone then at least hate on me, it'll bounce off my hard exterior
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    (Original post by ZAM-H)
    Look yh I'm sorry but that paper was disgusting, no two ways about it. You're coming on here and posting how well you think you did when ppl want to just forget about it. You need to know when to keep quiet.

    Posted from TSR Mobile
    Mate, you need to leave..

    I didn't like the paper either but Im on this thread cause I want to discuss & so do others. If you want to forget about it, then don't come on here. Stop bashing others though, I'm not in the happiest mood after that exam but life goes on, so chill out.
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    (Original post by LELF)
    I Think you may be wrong:/ I had a aldehyde with a methyl group as someone else said further up - CH3CH(CH3)COH

    There were three peaks in total which suggested there was 3 hydrogen environments first of all which lead me away from it being a ketone as there was 4 carbons in the molecular formula so the compound must have had two environments which were equivalent and I couldn't think of a ketone which structure applied to those rules.

    I agree with you on the peak at 9.8 ppm indicating a aldehyde, but it had a doublet splitting pattern, and due to the n+1 rule that must mean its neighboring carbon bonded to one other hydrogen (CH).

    The best clue though is the peak split into 7. This must mean again due to n+1 it has neighboring carbons with a total of 6 hydrogen atoms, so that peak was referring to the carbon bonded to the two CH3 groups.

    Thats what i got anyway xD
    Not necessarily 3 enviroments doesn't always mean 3 carbons which is why I always stick to ppm because the data sheet confirms for sure a ketone with a CH3 or a Ch2.
    The other thing is I don't think the compound was pure so the molecular formula could have been an impure compound.
    TBH I have no clue.
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    (Original post by Chemistryxx77)
    That's fine i'm not here to hate I just want to figure it out.
    Ah but the aldehyde is a doublet because you have one H and the oxygen counts as 0 but the n+1 rule means one so therefore two peaks.
    I have no idea about the multi splitting peaks.
    There was no area of peak 6.
    They're not examples though those where the actual figures for the NMR given in ppm.
    Also for two environments to make six they both must be adjacent to a carbon
    e.g.
    (H3C)CH2(CH3) will give a peak area of 6.
    The NMR spectrum I'm pretty sure was a doublet with area 1 (the aldehyde porton), the multi split peak with area 1, and the area 6 doublet, there definitely was an area 6 peak. Yes that's correct and that's what the structure has, a (H3C)CH(CH3) central group with the aldehyde (CHO) also bonded to that central carbon (that carbon is bonded to 3 other carbons and a hydrogen). And it wasn't 2 environments I think, it was one environment split into a doublet (the area 6 doublet). This is certainly a very confusing question though!
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    4 or 5 carbon alkenes on the first page? as a result of the alcohol dehydration
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    (Original post by DragonRider17)
    The NMR spectrum I'm pretty sure was a doublet with area 1 (the aldehyde porton), the multi split peak with area 1, and the area 6 doublet, there definitely was an area 6 peak. Yes that's correct and that's what the structure has, a (H3C)CH(CH3) central group with the aldehyde (CHO) also bonded to that central carbon (that carbon is bonded to 3 other carbons and a hydrogen). And it wasn't 2 environments I think, it was one environment split into a doublet (the area 6 doublet). This is certainly a very confusing question though!
    It was a doublet for sure. There was a peak area six but that was for the C=O at around 2.5ppm but not 6 envirmonets.
    Oh wait are you taking about methyl propanal ?
    I know how many marks was it anyway 4 ?
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    (Original post by Chemistryxx77)
    It was a doublet for sure. There was a peak area six but that was for the C=O at around 2.5ppm but not 6 envirmonets.
    Oh wait are you taking about methyl propanal ?
    I know how many marks was it anyway 4 ?
    Yes that's right! Maybe my descriptions haven't been very clear, this is the molecule
    I think it was:


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    What did you put for the thalamide with hydrochloric acid? When they substituted half of the molecule with N
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    unofficial mark scheme?
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    (Original post by Mango88)
    What did you put for the thalamide with hydrochloric acid? When they substituted half of the molecule with N
    I hydrolysed both amide linkages opening the ring up, and put a carboxylic acid group on each end (I guess an ammonia molecule would have been lost)
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    (Original post by DragonRider17)
    I hydrolysed both amide linkages opening the ring up, and put a carboxylic acid group on each end (I guess an ammonia molecule would have been lost)
    Yeah, exactly what I did
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    For the yield question, first I calculated the moles of bezene (mass was 10g) so 10/(6x6+6x.101) = ~0.128. 1:1 mole ratio of bezene with benzoic acid so theoretical mass of benzoic acid = 0.128 x mr of benzoic acid = ~15.62 g

    Mass of benzoic acid obtained = ~6.2g (nvm it was 3.8 I misremembered)... So yield = mass obtained/theoritcal mass... 3.8/15.62 x100 = 24.3%

    I'm not sure whether this is correct, just putting how I answered it.
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    (Original post by wwwwwwwwwwwf)
    For the yield question, first I calculated the moles of bezene (mass was 10g) so 10/(6x6+6x.101) = ~0.128. 1:1 mole ratio of bezene with benzoic acid so theoretical mass of benzoic acid = 0.128 x mr of benzoic acid = ~15.62 g

    Mass of benzoic acid obtained = ~6.2g (cant remember exact value)... So yield = mass obtained/theoritcal mass... 6.2?/15.62 x100 = 39.7%

    I'm not sure whether this is correct, just putting how I answered it.
    I did exactly the same thing (I think the mass of bezoic acid was ~3.8g)
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    Was unsure how to answer the 7 marker too, so I think i just put conditions/reaction type/reactants and then I did the mechanism (including the making of the CH3+ electrophile for CH3Cl)
    I'm pretty sure the question said to Put reactants/reaction type/ conditions + 'describe the mechanism' (correct me if im wrong), so not 100% on whether i was supposed to write out the mechanism, but whatever
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    For the multiplet, if you count it you can see that its a septuplet, and from there you use the n-1 rule to find that the environment is surrounded by 6 protons, hence 2 -CH3 groups
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    (Original post by DragonRider17)
    I did exactly the same thing (I think the mass of bezoic acid was ~3.8g)
    Yeah that's it so the yield would be 24.3% (ill edit original post)
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    For the purity of benzoic acid, would they accept measuring the boiling temperature and comparing it to a literature value rather than melting temperature? Also why was the % yield so low?
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    (Original post by E15)
    For the purity of benzoic acid, would they accept measuring the boiling temperature and comparing it to a literature value rather than melting temperature? Also why was the % yield so low?
    I ran out of time before finishing the last question on why the yield was low, but my first reason was to do with the fact it was a two step reaction ~ so losses will be more or something like that
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    (Original post by E15)
    For the purity of benzoic acid, would they accept measuring the boiling temperature and comparing it to a literature value rather than melting temperature? Also why was the % yield so low?
    I think they would have to be pretty mean not to, you can obviously boil benzoic acid. I think you'd get the mark. And no idea, I just said some general stuff about not all the methylbenzene being oxidised and some being lost in the recrystallisation. I don't know if there is some major reason I've missed that explains the VERY low yeild.
 
 
 
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