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OCR Chemistry A - AS in Depth - 2015 (new spec) UNOFFICIAL MARK SCHEME

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Original post by revise_or_fail
why is this apparatus not suitable for oxidising a secondary alcohol?

Its because some of the reactants are volatile so may evaporate before reacting, which is why you need to use reflux apparatus.


The reactants aren't that volatile, the product will be significantly more volatile; since it will only oxidise once distillation apparatus should be used to purify the product.
Original post by alevelnerd123
Idk how that C-O-H question was worth 4 marks

104.5 (1)
2 bond pairs, 2 lone pairs (1)
Lone pair repels more than bond pair (1)

Then I put it should be 109.5 if it was bond pair but each lone pair reduces the angle by 2.5, so the 2 lone pairs 109.5-5=104.5


I think the 4th mark is for
'Then I put it should be 109.5 if it was bond pair but each lone pair reduces the angle by 2.5, so the 2 lone pairs 109.5-5=104.5'
did ppl get 2.2x10^-3 for rate of reaction question
Original post by DoubleDoors
The reactants aren't that volatile, the product will be significantly more volatile; since it will only oxidise once distillation apparatus should be used to purify the product.


Can't distill a secondary alcohol. Has to be refluxed
Original post by youreanutter
did ppl get 2.2x10^-3 for rate of reaction question

I got something x10-4 I think did you use a tangent ?
Original post by Chocolate Rain
I got something x10-4 I think did you use a tangent ?
yeah i did the change in mass(96.4-95.96)/200
Is it ok that I drew the methylpropanoic acid molecule but got the wrong name?? I drew it correctly and did all the correct identifications.
My rate of reaction was about -6x10^-4 ish. Its all really dependent on the way you drew your tangent. I doubt you would lose marks unless you drew it completely wrong and did the wrong calculations.
Original post by youreanutter
yeah i did the change in mass(96.4-95.96)/200


Oh I drew a tangent then worked it out probs got a silly mistake 🤔
did u guys do the change in mass over time by doing 96.4-(wherever ur tanget corrseponded to y axis)/200
if so u shud get marks
Reply 50
Was the question at the top of the butan-2-ol question set not why is it a secondary alcohol? I'm sure it said that. Don't remember that question being anything to do with carbocations and I thought you just had to say because the alpha carbon was bonded to two other carbons.
(edited 7 years ago)
Original post by jn998
Was the question at the top of the butan-2-ol question set not why is it a secondary alcohol? I'm sure it said that. Don't remember that question being anything to do with carbocations and I thought you just had to say because the alpha carbon was bonded to two other carbons.


No I think it was, why is it a secondary alcohol? Its because the OH group is attached to a carbon tat is attached to 2 alkyl groups and 1 hydrogen atom
Reply 52
I thought it was butanoic acid in the spectroscopy question but this didnt quite add up because they mentioned the peak at m/z = 44 was caused by fragmentation of a secondary carbocation. The compound is more likely to be methylpropanoic acid rather than butanoic acid tbh
Reply 53
Original post by blugg
I thought it was butanoic acid in the spectroscopy question but this didnt quite add up because they mentioned the peak at m/z = 44 was caused by fragmentation of a secondary carbocation. The compound is more likely to be methylpropanoic acid rather than butanoic acid tbh


What would I get out of 6 for just using the IR spec to identify it as a carboxylic acid, using the % compositions to get the empirical and then the 88 peak to get the molecular formula and naming it butanoic acid, without talking about fragmentation or any of the other Mass spec peaks?
Original post by blugg
I thought it was butanoic acid in the spectroscopy question but this didnt quite add up because they mentioned the peak at m/z = 44 was caused by fragmentation of a secondary carbocation. The compound is more likely to be methylpropanoic acid rather than butanoic acid tbh


No they mentioned that the peak that caused a a secondary carbocation was the one with the highest intensity which was at m/z = 43
Reply 55
Original post by 4nonymous
No they mentioned that the peak that caused a a secondary carbocation was the one with the highest intensity which was at m/z = 43


Ohh yes, my bad
Reply 56
Original post by jn998
What would I get out of 6 for just using the IR spec to identify it as a carboxylic acid, using the % compositions to get the empirical and then the 88 peak to get the molecular formula and naming it butanoic acid, without talking about fragmentation or any of the other Mass spec peaks?


Probably 4 marks, I didn't discuss the fragmentation peaks either. If you wrote the molecular formula as well as the empirical formula then you should get 4. This also depends on whether you annotated the C=O peak as well as the -OH
Reply 57
Original post by akpai22
My rate of reaction was about -6x10^-4 ish. Its all really dependent on the way you drew your tangent. I doubt you would lose marks unless you drew it completely wrong and did the wrong calculations.


same but seeing everyone elses answers I thought I did it wrong
(-0.006)

Original post by youreanutter
did u guys do the change in mass over time by doing 96.4-(wherever ur tanget corrseponded to y axis)/200
if so u shud get marks


thank frick

Original post by jn998
Was the question at the top of the butan-2-ol question set not why is it a secondary alcohol? I'm sure it said that. Don't remember that question being anything to do with carbocations and I thought you just had to say because the alpha carbon was bonded to two other carbons.


thank you

Original post by blugg
I thought it was butanoic acid in the spectroscopy question but this didnt quite add up because they mentioned the peak at m/z = 44 was caused by fragmentation of a secondary carbocation. The compound is more likely to be methylpropanoic acid rather than butanoic acid tbh


Original post by 4nonymous
No they mentioned that the peak that caused a a secondary carbocation was the one with the highest intensity which was at m/z = 43
wait so what is the answer?
(edited 7 years ago)
Original post by alkaline.
same but seeing everyone elses answers I thought I did it wrong
(-0.006)


How can you have a negative rate of reaction
Reply 59
Original post by 4nonymous
How can you have a negative rate of reaction

****
I thought cause the gradient was negative but rate describes how fast the reaction is changing not direction...oh well

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