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    Can someone help on a couple of questions please on this test paper: http://store.aqa.org.uk/qual/gce/pdf...4-W-SQP-07.PDF

    On question 7(a) can some explain the answer to me
    On question 8(a)(ii) :

    i worked out the theorectical yield first like this: 10/93 x 135 but the markscheme says that you do: 10/93 x 0.5 x 135 and i dont understand why. ?? can someone kindly help me please, thanks
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    For 7(a)(i) you get 3 peaks because you could draw a line of symmetry through the molecule running between the OH groups. By symmetry, the two carbon bonded to the OH groups (carbons 1 and 2) are exactly the same from the point of view of the NMR machine. Carbons 3 and 6 are also equivalent by symmetry, as are carbons 4 and 5. Equivalent carbons have exactly the same chemical shift so will appear as one peak. 3 unique carbon environments give you 3 peaks. Use this same principle for part (b).

    For 7(a)(ii) they are telling you that the factors which affect chemical shift in proton NMR also affect chemical shift in 13C-NMR. therefore the carbons closest to the electronegative oxygens will have the highest chemical shifts so must be the peak at around 76 ppm.

    For part 8(a)(ii) I'm not sure what the x 0.5 is for seeing as it's a 1:1 mole ratio. Maybe I'm missing something as well. Perhaps someone else can comfirm that's a mistake.
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    (Original post by klgyal)
    On question 8(a)(ii) :

    i worked out the theorectical yield first like this: 10/93 x 135 but the markscheme says that you do: 10/93 x 0.5 x 135 and i dont understand why. ?? can someone kindly help me please, thanks
    Look at the moles equivalent in the equation given and you will see that 2 moles of phenylamine produce 1 mole of phenylethanamide...
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    (Original post by charco)
    Look at the moles equivalent in the equation given and you will see that 2 moles of phenylamine produce 1 mole of phenylethanamide...
    Oh yes, you're right. I actually missed the equation underneath and was looking at the main scheme with the benzene rings drawn above and I suspect the original poster was as well. It seems they are using an extra equivalent of phenylamine to act as a base to mop up HCl rather than using a dedicated base which I assumed would be the case. Assuming is bad. That would have been silly marks lost in the exam!
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    thanks everyone
 
 
 
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