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    (Original post by pav)
    that's what i thought too but apprently it's not. this question is from chemical ideas so im hoping it's a lot harder then what we would get in the exam, but there is no guarantee
    All the Hydrogen (protons) from the -COOH's are the 3. The 4 represents those that make up the CH2's and the 1 is for the Hydrogen that forms the Hydroxy group attached to the central carbon.

    Hence the 1:4:3. Me thinks :holmes:
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    When learning all of the reaction conditions is there a difference between reflux and heat under reflux?

    For example for Friedel-Craft reactions it only says reflux, but for Esterfication it says heat under reflux for some and just reflux for others??

    And do we need to know how to name amides?
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    1000g of benzene gives 1050g of phenol. Calculate the percentage yield.

    The answer is 87%

    Can someone please explain this to me?
    Thanks
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    (Original post by Tetanus)
    All the Hydrogen (protons) from the -COOH's are the 3. The 4 represents those that make up the CH2's and the 1 is for the Hydrogen that forms the Hydroxy group attached to the central carbon.

    Hence the 1:4:3. Me thinks :holmes:
    Yeah that makes sense. Thanks!
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    (Original post by ABCDemily)
    Kept below 10C to prevent a phenol forming. Also, the reagents have to be created in situ as they are very volatile and would otherwise bugger off and now work.

    Also, when you couple the diazonium salt with the phenol, has to be done over iceee.
    It's also done using alkaline conditions.
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    (Original post by Tetanus)
    All the Hydrogen (protons) from the -COOH's are the 3. The 4 represents those that make up the CH2's and the 1 is for the Hydrogen that forms the Hydroxy group attached to the central carbon.

    Hence the 1:4:3. Me thinks :holmes:
    Tetanus! I remember you from the biology thread!
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    AHHHHHHHHHHHHHH SOOOO panicky about this exam!!!!!! I feel like I have a rough idea of everything but will probably loose marks on defining it properly.. Argh!
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    (Original post by RayM)
    1000g of benzene gives 1050g of phenol. Calculate the percentage yield.

    The answer is 87%

    Can someone please explain this to me?
    Thanks
    C6H6 --------> C6H5OH

    1 mole of benzene makes 1 mole of phenol

    moles of benzene = 1000/78 = 12.82

    12.82 * 94 (Mr of phenol) = 1216.36

    1050g/1216.36 * 100 = 86.32%

    It's just actual mass over theoretical mass.
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    (Original post by RayM)
    1000g of benzene gives 1050g of phenol. Calculate the percentage yield.

    The answer is 87%

    Can someone please explain this to me?
    Thanks
    Ok ur in luck I actually did this question in a paper today.

    So % yield= (actual yield/ theoritcal yield) x100

    your given the actual yield which is 1050.

    So you have to work out the theoritcal yield.

    To do this you work out the moles of benzene= 1000/78= 12.8205
    (78 is the molecular mass of benzene)

    then you work out the molecular mass of phenol which is 94 and times it by 12.8205 to get 1205.1 (theoritcal yield)

    then you do (1050/1205.1) x100= 87%

    hope this helps.
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    I hate the Chemical ideas book. Whoevers great idea it was to bring that in needs shooting :]
    our college didn't do the F334 in Jan because we were doing the investigations... How did everyone do in them? :]
    it's odd that F335 is before F334... But just sitting with the revision book now :'D
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    (Original post by Boompw)
    When learning all of the reaction conditions is there a difference between reflux and heat under reflux?

    For example for Friedel-Craft reactions it only says reflux, but for Esterfication it says heat under reflux for some and just reflux for others??

    And do we need to know how to name amides?
    There's no difference. Refluxing is a technique with the vertical condenser and pear shaped flask? You need to be able to draw the apparatus for that, so I'd look that up
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    dreading this exam like helllllll
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    (Original post by I'mBadAtMaths)
    Tetanus! I remember you from the biology thread!
    :hi:
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    (Original post by Jhyzone)
    Someone explain to me how you work out chemical shifts from NMR in a simple way please.
    Thanks.
    How do work out as in how do you interpret the spectra?

    The shifts on the data sheet correspond to the position of each peak, as you probably know. Think of NMR as building a jigsaw puzzle, each peak tells you what, and how many of each piece. For example, a peak at 1.1 would show a C-CH3 proton, if the relative height of the peak was 2, then you would have two of these C-CH3 protons. You can then start to build up the shape of the molecule. in this case you would start with H3C-C-CH3. This is the most simply I can really explain it. The only other thing is the splitting of the peaks, say you have a peak at 4.2, corresponding to a C-OH proton, if this peak was split into 3, then it would tell you that adjacent carbon atom would have two hydrogen atoms. The number of peaks that the main peak is split into = n-1 H atoms on the adjacent carbon. This is also called the n+1 rule, where n hydrogen atoms = n+1 splits of a peak (thanks to the guy below for correcting me).

    Dos this help a bit?
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    (Original post by Conor 419)
    Why exactly do ligands cause orbitals to split?
    You don't need to know this! All you need to know is that they do, and that the splits are different energy levels which correspond to different frequencies of light, E=hv and all that jazz, etc
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    (Original post by DCRichards)
    How do work out as in how do you interpret the spectra?

    The shifts on the data sheet correspond to the position of each peak, as you probably know. Think of NMR as building a jigsaw puzzle, each peak tells you what, and how many of each piece. For example, a peak at 1.1 would show a C-CH3 proton, if the relative height of the peak was 2, then you would have two of these C-CH3 protons. You can then start to build up the shape of the molecule. in this case you would start with H3C-C-CH3. This is the most simply I can really explain it. The only other thing is the splitting of the peaks, say you have a peak at 4.2, corresponding to a C-OH proton, if this peak was split into 3, then it would tell you that adjacent carbon atom would have three hydrogen atoms. The number of peaks that the main peak is split into = number of H on the adjacent carbon.

    Dos this help a bit?
    Would in not be next to a carbon with 2 hydrogen atoms?
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    (Original post by mooniibuggy)
    Hmm, that's what I did for AS but my teacher said they're running out of questions to ask so they're using storylines more :/ also, i'm doing past papers and there's a bit of storylines not covered in the revision guide
    I did the 2011 january f335 past paper, and I did good on it from cgp alone .... You needed 80/120 for an A, and I got 99/120.

    However, you may be right if you are referring to the june 2010 paper, coz I have not seen that yet, so it might have a lot of storylines.

    I advise you, however, to ignore your teacher because this is only the third f335 exam since they changed the syllabus in 2009, so they will not be running out of things to ask for just yet, if ever.

    Good luck tomorrow
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    (Original post by zahre)
    Would in not be next to a carbon with 2 hydrogen atoms?
    Yes that is correct, good call. Excuse me while I go and relearn the n+1 rule
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    What are the main topics we really even need from chemical storylines? I read and its just a load of waffle
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    I know that fibre reactive dyes are the most colourfast, because the covalent bonds between fibre molecules and dye molecules are strong, permanent bonds.

    However, do we need to know the actual reactions that go on in fibre reactive dyes --- As in, do we need to know what reacts with the -OH or -NH groups to form the covalent bonds.


    The reason I ask is that in the cgp revision guide it shows the bonding (in diagrams) for hydrogen bonding between dyes and fibres, and for ionic interactions between dyes and fibres, so I know what groups interact.

    But for fibre reactive dyes, it just states they are formed from groups that react with the groups on the fibre. No diagrams are shown.

    Does this mean we do not need to know about what reacts with it, because as far as my experience goes, if its not in cgp revision guides, its not important.
 
 
 
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