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Edexcel AS Chemistry Unit 1 22/5/2015 Unofficial Mark Scheme

Completed Unit 2 mark scheme here: http://www.thestudentroom.co.uk/showthread.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.


20 Multiple Choice - marks



16 17 Question - Mass spectrometry marks



18 20 Question - Organic chemistry marks



10 19 Question - Sodium chloride marks



14 20 Question - Enthalpy change marks

(edited 8 years ago)

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:woo:
Cheers C0balt :biggrin:
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.
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).
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.


Yes but fragmentation? I dunno
Reply 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.


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.

Posted from TSR Mobile
Original post by sat_freak
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).


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 :smile:
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
(edited 8 years ago)
Reply 9
Original post by sat_freak
Q 18 (c) (ii) As the secondary carbocation is more stable than the primary (will form the major product).


It is tertiary carbocation

Posted from TSR Mobile
Reply 10
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 :smile:


I said both but it's tertiary I believe lol

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Original post by C0balt
I said both but it's tertiary I believe lol

Posted from TSR Mobile


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?
Reply 12
Thanks, have forgotten my answers though.


Didn't like that exam tbh :sad:
Reply 13
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?


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.


Posted from TSR Mobile
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.


Posted from TSR Mobile


Yeah. Hopefully I have got the mark since it does make sense!
Original post by C0balt
It is tertiary carbocation

Posted from TSR Mobile


Yeah sorry, it is tertiary. And tertiary is more stable than primary.
Original post by GrandMasti
Yeah. Hopefully I have got the mark since it does make sense!


Maybe they wanted to hear the word(s) : "inductive effect of the alkyl groups"

Or Markovnikov's rule.
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.

Posted from TSR Mobile


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, 1:2:1 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.
(edited 8 years ago)
Original post by sat_freak
Maybe they wanted to hear the word(s) : "inductive effect of the alkyl groups"

Or Markovnikov's rule.


Ah well, it is only 1 mark. Also, can you please upload the unit 1 and 2 biology papers onto the biology thread. That would be great! Thanks :biggrin:
Reply 19
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, 1:2:1 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.

I don't think relative abundance between the atomic peaks and molecular peaks would be a marking point. Two atomic peak having the same height can be a marking point because it said half 81 and half 79

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