Bertybassett
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Hi, I am struggling with some questions on this paper and was wondering if anyone could help. For question 2f, how do we get this structure. For 6ci, how do we deduce the structures of this? Also, what is the difference between GC and GLC (chromatography) - I thought these were the same? Many thanks
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BobbJo
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You did not link the paper. Also this should probably be posted in the chemistry forum.

GC and GLC are the same
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Bertybassett
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Sorry it put it in this forum for some reason. Would you get the mark for saying GC then even though the MS only has GLC? Paper is here https://pmt.physicsandmathstutor.com...y%203%20QP.pdf
(Original post by BobbJo)
You did not link the paper. Also this should probably be posted in the chemistry forum.

GC and GLC are the same
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BobbJo
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To deduce the structure work out the answer to the earlier parts.
2
b) C=O
c) 3 different elements. Cannot be Cl as 35 or 37 leaves a mass of 8 or 6
must be C, O, H. Cannot be 2 O atoms
so 1 O atom leaves 43 - 16 = 27
27 = 2 C atoms + 3 H atoms
so C2H3O+
but from (b) we know a C=O is present
so CH3C+O
C4H7ClO+ -> CH3C+O + C2H4Cl. (radical dot on the C2H4Cl)

e) doublet -> 1 adjacent proton
singlet -> 0
quartet -> 3
hence
CH3COCH(Cl)CH3

6ci) work out which alcohol needs to react with that anhydride to form that ester (look at the ester linkage)
the alcohol is CH3CH2OH
equation is anhydride + 2CH3CH2OH -> ester + H2O

You would probably get the mark if you wrote GC instead of GLC [but check syllabus and marking schemes for accepted answers]
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