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Edexcel AS Chemistry Unit 1 22/5/2015 Unofficial Mark Scheme
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Completed Unit 2 mark scheme here: http://www.thestudentroom.co.uk/show....php?t=3375301
Thanks to sat_freak I can now attempt to make a complete unofficial mark scheme for unit 1. However, my unit 2 performance was better than unit 1, so the quality of this mark scheme may be lower than unit 2.
Thanks to sat_freak I can now attempt to make a complete unofficial mark scheme for unit 1. However, my unit 2 performance was better than unit 1, so the quality of this mark scheme may be lower than unit 2.
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#4
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
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#5
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
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#6
(Original post by sat_freak)
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
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(Original post by sat_freak)
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
For Q 17 (c), as they specify that the sample is bromine gas, the peaks for 79 & 81 are probably not needed, as bromine is diatomic.
By your argument, just because propane is a gas, there would only be molecular ion peak, which obviously isn't the case.
State of the sample does not affect the final spectrum. Even if the bromine liquid was put into the mass spectrometer, the first step is vaporisation so it would become a gas anyways. Which is then bombarded by electrons which knocks electrons off and this can break the bond between the atoms. Putting gaseous state bromine will not prevent this from happening, and end result is same as putting bromine liquid or even solid.
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#8
(Original post by sat_freak)
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
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#9
Q17 (f) (i) Suggest two precautions that are necesary to ensure that the result of any analysis would be valid.
- Samples are fresh / not contaminated / collected properly with witnesses and all legal requirements.
- Proper ID / tagging / labelling of the samples is very crucial as the stakes are too high.
- Transport & storage iof samples.
- Equipment calibration / Use of control.
- Accuracy, precision, reliability, resolution of equipment
- Samples are fresh / not contaminated / collected properly with witnesses and all legal requirements.
- Proper ID / tagging / labelling of the samples is very crucial as the stakes are too high.
- Transport & storage iof samples.
- Equipment calibration / Use of control.
- Accuracy, precision, reliability, resolution of equipment
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(Original post by sat_freak)
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
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(Original post by GrandMasti)
Tbh that question was worded very weirdly, and so it could be interpreted as literally WHY is there a positive charge on that carbon, since it has 'lost an electron' or why that carbon and nothing else i.e. what you said
Tbh that question was worded very weirdly, and so it could be interpreted as literally WHY is there a positive charge on that carbon, since it has 'lost an electron' or why that carbon and nothing else i.e. what you said
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#12
So, I guess they may accept that the carbon atom has lost an electron in the formation of the C-Cl bond (since it is dative covalent). Although the tertiary argument is probably more correct. Is there any chance they may be lenient given the ambiguous wording of the question?
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(Original post by GrandMasti)
Wait, the wording of the question was: The structure of the carbocation is correctly drawn. Explain why the positive charge on the carbon atom shown.
So, I guess they may accept that the carbon atom has lost an electron in the formation of the C-Cl bond (since it is dative covalent). Although the tertiary argument is probably more correct. Is there any chance they may be lenient given the ambiguous wording of the question?
Wait, the wording of the question was: The structure of the carbocation is correctly drawn. Explain why the positive charge on the carbon atom shown.
So, I guess they may accept that the carbon atom has lost an electron in the formation of the C-Cl bond (since it is dative covalent). Although the tertiary argument is probably more correct. Is there any chance they may be lenient given the ambiguous wording of the question?
Honestly, primary can still form, just less likely, that's why I took the question as I said above. Because even if the student drew a primary carbonation it wouldn't have technically been incorrect.
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#15
(Original post by C0balt)
Yeah it was so vague. I took it as why that particular carbon rather than, say, the entire molecule, has a positive charge
Honestly, primary can still form, just less likely, that's why I took the question as I said above. Because even if the student drew a primary carbonation it wouldn't have technically been incorrect.
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Yeah it was so vague. I took it as why that particular carbon rather than, say, the entire molecule, has a positive charge
Honestly, primary can still form, just less likely, that's why I took the question as I said above. Because even if the student drew a primary carbonation it wouldn't have technically been incorrect.
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#17
(Original post by GrandMasti)
Yeah. Hopefully I have got the mark since it does make sense!
Yeah. Hopefully I have got the mark since it does make sense!
Or Markovnikov's rule.
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#18
(Original post by C0balt)
No, you do need it.
By your argument, just because propane is a gas, there would only be molecular ion peak, which obviously isn't the case.
State of the sample does not affect the final spectrum. Even if the bromine liquid was put into the mass spectrometer, the first step is vaporisation so it would become a gas anyways. Which is then bombarded by electrons which knocks electrons off and this can break the bond between the atoms. Putting gaseous state bromine will not prevent this from happening, and end result is same as putting bromine liquid or even solid.
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No, you do need it.
By your argument, just because propane is a gas, there would only be molecular ion peak, which obviously isn't the case.
State of the sample does not affect the final spectrum. Even if the bromine liquid was put into the mass spectrometer, the first step is vaporisation so it would become a gas anyways. Which is then bombarded by electrons which knocks electrons off and this can break the bond between the atoms. Putting gaseous state bromine will not prevent this from happening, and end result is same as putting bromine liquid or even solid.
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1 is evident. Gaseous bromine will still be mostly molecular.As the question has four marks, I believe 3 marks for three peaks and one for the correct proportion between them.
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#19
(Original post by sat_freak)
Maybe they wanted to hear the word(s) : "inductive effect of the alkyl groups"
Or Markovnikov's rule.
Maybe they wanted to hear the word(s) : "inductive effect of the alkyl groups"
Or Markovnikov's rule.
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(Original post by sat_freak)
Yeah that's correct but how would one figure out the size of the peak (relative intensity) of the atomic ions. For the molecular ions, as you said, 1
1 is evident. Gaseous bromine will still be mostly molecular.
As the question has four marks, I believe 3 marks for three peaks and one for the correct proportion between them.
Yeah that's correct but how would one figure out the size of the peak (relative intensity) of the atomic ions. For the molecular ions, as you said, 1
1 is evident. Gaseous bromine will still be mostly molecular.As the question has four marks, I believe 3 marks for three peaks and one for the correct proportion between them.
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