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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

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    (Original post by ReTakingF324/5)
    Lone pairs. I don't know what you mean for the 2nd part?
    like for the reaction of chlorine with benzene, does the curly arrow go to the Cl or the + next to the Cl
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    (Original post by ranz)
    but how would u know its a cylic and not just 2 joined tigether in a striaght chain?


    Posted from TSR Mobile
    It's probably about alpha amino acids which form diemers that are cyclic peptides so the assumption is that is must be a cyclic compound. Otherwise, if its not mentioned that it's an alpha amino acids,then it could be a straight chain.
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    (Original post by tcameron)
    always wondered this, in the nucleophilic mechanisms does the curly arrow go from the lone pairs or the negative charge?
    And for electrophilic mechanism does the lone pair go to the positive charge or to the atom which has the positive charge on it?
    For nucleophilic mechanisms the curly arrow can go from the lone pair or the negative charge. The -ve charge represents the lone pair, so as long as you use one or the other I don't think it matters.
    For electrophilic mechanisms the curly arrow can just go to the electrophile, it doesn't have to be the +ve charge. This is what the mark scheme says anyway.
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    Can someone clarify that the splitting patterns in NMR are the adjacent protons on the adjacent CARBON atom?
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    (Original post by rory58824)
    Can someone clarify that the splitting patterns in NMR are the adjacent protons on the adjacent CARBON atom?
    Singlet = adjacent to no C bonded to a H
    Doublet = adjacent to CH
    Triplet = adjacent to CH2
    Quartet = adjacent to CH3
    Septet = adjacent to 2xCH3
    whatever else you can just call multiplet
    remember must only be protons which aren't in the same proton environment as the proton you're identifying so -CH2CH2- will not be split

    C13 nmr does not have splitting
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    (Original post by tcameron)
    Singlet = adjacent to no C bonded to a H
    Doublet = adjacent to CH
    Triplet = adjacent to CH2
    Quartet = adjacent to CH3
    Septet = adjacent to 2xCH3
    whatever else you can just call multiplet
    remember must only be protons which aren't in the same proton environment as the proton you're identifying so -CH2CH2- will not be split

    C13 nmr does not have splitting
    So the spltting patterns only include protons on adjacent CARBON atoms, not adjacent nitrogen or oxygen atoms?
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    (Original post by rory58824)
    So the spltting patterns only include protons on adjacent CARBON atoms, not adjacent nitrogen or oxygen atoms?
    Yeah it isn't to do with nitrogen or oxygen
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    Hi
    Just wanted to ask why an -O- bond is not shown at the end of the chain on the left when in other cases it is shown.
    And how is a resort unit for a hydroxypropanoic acid drawn?Name:  Screenshot_20160613-141857.png
Views: 98
Size:  73.1 KB
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    (Original post by tcameron)
    Yeah it isn't to do with nitrogen or oxygen
    Thanks, just wanted to clear that up

    Anyone predicting a tough paper tomorrow?
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    Hi guys, can anyone explain where the 2nd oxygen is coming from in this NMR question? In both the H and C diagrams, there is no peak showing the O in the ester but suddenly it appears in the final structure, and yes I saw that in the molecular formula there is supposed to be 2 O's but shouldn't it appear in the NMR diagrams?
    Molecular formula is C10H1202

    June 13, last question

    Attachment 548983548985Attachment 548983548985
    Attached Images
      
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    What are the steps for Gas Chromatography (and YES we do need to know this, expected as overall knowledge in the spec)
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    (Original post by underestimate)
    Hi
    Just wanted to ask why an -O- bond is not shown at the end of the chain on the left when in other cases it is shown.
    And how is a resort unit for a hydroxypropanoic acid drawn?Name:  Screenshot_20160613-141857.png
Views: 98
Size:  73.1 KB
    It could be shown on either side it does not matter as long as a full repeat unit is displayed
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    (Original post by Noura*)
    Hi guys, can anyone explain where the 2nd oxygen is coming from in this NMR question? In both the H and C diagrams, there is no peak showing the O in the ester but suddenly it appears in the final structure, and yes I saw that in the molecular formula there is supposed to be 2 O's but shouldn't it appear in the NMR diagrams?
    Molecular formula is C10H1202

    June 13, last question

    Attachment 548983548985Attachment 548983548985
    That's because there is no Hydrogens attached so there's no peak, you're expected to deduce that there has to be a =o attached to the benzene as there's a peak for c6h5c-
    (and obviously by looking at important peaks on the C spectrum)
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    (Original post by Noura*)
    Hi guys, can anyone explain where the 2nd oxygen is coming from in this NMR question? In both the H and C diagrams, there is no peak showing the O in the ester but suddenly it appears in the final structure, and yes I saw that in the molecular formula there is supposed to be 2 O's but shouldn't it appear in the NMR diagrams?
    Molecular formula is C10H1202

    June 13, last question

    Attachment 548983548985Attachment 548983548985
    If you look at the carbon-13 NMR, there is a peak at about 184ppm which indicates a carbon bonded to two oxygens, one with a double bond and one with a single bond. So this indicates that it's an ester.
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    (Original post by Noura*)
    Hi guys, can anyone explain where the 2nd oxygen is coming from in this NMR question? In both the H and C diagrams, there is no peak showing the O in the ester but suddenly it appears in the final structure, and yes I saw that in the molecular formula there is supposed to be 2 O's but shouldn't it appear in the NMR diagrams?
    Molecular formula is C10H1202

    June 13, last question

    Attachment 548983548985Attachment 548983548985
    The structure shows HC-C-O. A peak would be around 3.5-4ppm for HC-O, but that environment is not present in this structure.
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    Can anyone elaborate on what we are expected to know regarding this statement from the Spec:

    "describe and explain the increased use of esters of fatty acids as biodiesel."
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    (Original post by M.Branson98)
    If you look at the carbon-13 NMR, there is a peak at about 184ppm which indicates a carbon bonded to two oxygens, one with a double bond and one with a single bond. So this indicates that it's an ester.
    (Original post by qweening)
    That's because there is no Hydrogens attached so there's no peak, you're expected to deduce that there has to be a =o attached to the benzene as there's a peak for c6h5c-
    Oh rights yes, I get it, but shouldn't it appear in the carbon spec¿
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    You know when you have the polymer of a hydoxy carbonxlic acid (like CH2OHCOOH but with stuff in the middle) if it asks you to draw the repeat unit of the polymer which side do you put the O in the ester link?
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    (Original post by RetroSpectro)
    Can anyone elaborate on what we are expected to know regarding this statement from the Spec:

    "describe and explain the increased use of esters of fatty acids as biodiesel."
    triglycerides + methanol = biodisel + glycerol
    naoh catalyst
    methanol is excess to increase yield
    high atom economy and glycerol is a useful byproduct
    biggest thing is that methanol is renewable to used more as it doesn't release co2 or greenhouse gases for better for the environment etc
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    (Original post by qweening)
    What are the steps for Gas Chromatography (and YES we do need to know this, expected as overall knowledge in the spec)
    Component is injected into the column and separated through the movement of the mobile phase (an inert gas such as helium)
    Inside the column the components seperate by relative solubility if it is a liquid stationary phase on a solid support or adsorption if the stationary phase is a solid.
    The components which are most soluble will leave the column last and so have the longest retention time which is the time from injection to detection

    Not sure what else is needed?
 
 
 
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