Questions to edexcel Re: U4 June 2011

Posted by: intechemistry on: February 17, 2012

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Here’s what I’ve just asked then (via e-mail)…

Dear Edxcel.

Some questions regarding Unit 4 June 2011.

Q11 (multiple choice) Answer stated = C
Correct me if I’m wrong, but it was my understanding that pure propanoic acid does not react with solid sodium carbonate. Aqueous propanoic acid, yes. (Unfortunately I have no pure propanoic acid with which to test my understanding). There is no indication that the propanoic acid is aqueous and the fact that _solid_ sodium carbonate is specified would suggest the questioner is trying to exclude the presence of water from this reaction. Hence neither propanoic acid or propan-1-ol would react. Also Fehlings solution has only been discussed in context of distinguishing aldehydes from ketones. There is no indication the the syllabus that Fehlings solution needs to be discussed with alcohols or in fact any other oxidisable species or that it’s oxidation strength is insufficient to oxidise alcohols and students have been specifically instructed that primary alcohols can undergo oxidation.
Could you please clear up those points so that I can best advise my students for this question.

Q14. Answer stated = C. There is a reasonable scientific answer in A i.e. the larger the mass, the more electrons and hence dispersion forces exist between the compound and the adsorbed liquid-stationary phase hence retention time would increase. Why would such understanding not gain credit?

Q19 d) ii) I suspect the simplified 1H NMR spectrum is incorrect with regards to the splitting of the peak at approx 1.8ppm. Could you please clarify the validity of the septet at 1prox 1.8ppm.

Thank you most kindly.

They replied: Sunday 19/Feb. 2012, and I felt compelled to send a follow-up:

Response (Anthony Ellison) 19/02/2012 08.06 AM
Dear Customer,

Thank you for your comments.

Re: June 2011 Unit 4

Q11 was in three parts. For Q11(a), the correct answer was C. Candidates scored highly on this response. All that this question was testing was that candidates realised that a carboxylic acid reacts with a carbonate, whereas an alcohol does not.

Fehling’s and Benedict’s solutions are mentioned in the Unit 4 specification only in the context of carbonyl chemistry – i.e. to distinguish between an aldehyde (test would be positive) and a ketone (test would be negative).

Re: Q14 – my answer would be D (as is on the published Edexcel mark scheme, in the secure area of the GCE Chemistry website) so I cannot agree with your comment “Q14. Answer stated = C”.

Re: Q19(d)(ii), The candidates were only required to identify the hydrogen environment in the molecule responsible for the singlet peak – which the majority of candidates did. The principal examiner presented the data, and the relevant question, in such a way as to be fair to the candidates.

Yours sincerely

Ask The Expert

My follow-up:

Dear Anthony. Thanks for your reply. I need to correspond further because I feel the points I raised have not been addressed.

Q11(a): I must reiterate, propoanoic acid will NOT react with solid sodium carbonate. However, aqueous propanoic acid will react with solid sodium carbonate.
There is no indication water is present – in fact, the opposite is true. Students with correct knowledge of this (non)reaction may therefore have not given C as the correct answer as it is an incorrect answer. The question is sub-standard.

Re: 11(b) My question regarding Fehlings solution relates to the compounds mentioned in (b), not part (c). I am unsure why you have taken time to inform me how aldehydes and ketones react with Fehlings reagent. Students are likely to have no knowledge of the oxidative potential for propan-1-ol or that of Fehlings solution. They do know that primary 1o alcohols can be oxidised. There will therefore be considerable doubt in their minds as to whether answer B will be correct.

Re: Q14. Apologies for the typo. Indeed the answer is D, but it is a trivial error and doesn’t address the initial query I raised. I said answer A has scientific merit. Why therefore will answer A get no credit? It resembles the error in 11 a) but from the opposite direction

Re: Q19(d)(ii) I asked about the validity of the spectrum at approx 1.8ppm, yet no mention of this was made. But the point about this enquiry is that the error will confuse students, likely causing many students to continually (and fruitlessly) modifying their structure until it completely agrees with the 1H NMR, only after which they will confidently circle the isolated methyl group.

Thanks again.

They replied:

Dear Customer,

I will endeavour to re-address one or two points by way of amplification of my original replies:

Q11(a) ORIGINAL REPLY WAS:

Q11 was in three parts. For Q11(a), the correct answer was C. Candidates scored highly on this response. All that this question was testing was that candidates realised that a carboxylic acid reacts with a carbonate, whereas an alcohol does not.

AMPLIFICATION:
This question was simply testing knowledge that acids react with carbonates. Solid sodium carbonate is usually used with weak acids so that there is a reasonable amount of fizzing. Although solutions / solvents were not specified in order to keep the question as simple as possible, candidates should have known that alcohols, aldehydes and ketones do not react with sodium carbonate. The high percentage of correct answers confirms this.

Q11(c) ORIGINAL REPLY WAS:

Fehling’s and Benedict’s solutions are mentioned in the Unit 4 specification only in the context of carbonyl chemistry – i.e. to distinguish between an aldehyde (test would be positive) and a ketone (test would be negative).

AMPLIFICATION:
Fehling’s and Benedict’s solutions are used to distinguish aldehydes from ketones, so answer B is correct in Q11(c), and, in fact, this was the highest scoring multiple-choice question (93.5% of candidates answered this item correctly).

Re: Q14: MY ORIGINAL REPLY WAS: – my answer would be D (as is on the published Edexcel mark scheme, in the secure area of the GCE 2008Chemistry website), so I cannot agree with your comment “Q14. Answer stated = C”.

Q14 AMPLIFICATION:
This question asks about separation of mixtures by gas chromatography in a long tube containing the stationary phase. At this point, the components have been vaporised, which is why the answer of volatility was NOT accepted.
There is a partition between the mobile phase and the stationary phase and components of the mixture will have different forces of attraction to the liquid stationary phase. Candidates are expected to know that there are different forces of attraction between each phase and the molecules in the mixture being separated, and that the stationary phase adsorbs chemicals to a greater or lesser extent.
The substances with larger molecular mass (answer A) spend longer in the column, as do those with lower volatility (answer C). However, the question asks for the reason, by use of the words “because the components differ in” and, therefore, the answer is D (which explains why substances spend different times in the column). So, to sum up:The only option which EXPLAINS the separation is response D, which was given by over 70% of the candidates.

Q19(d)(ii) ORIGINAL REPLY WAS:

Re: Q19(d)(ii), The candidates were only required to identify the hydrogen environment in the molecule responsible for the singlet peak – which the majority of candidates did. The principal examiner presented the data, and the relevant question, in such a way as to be fair to the candidates.

AMPLIFICATION:
The ester is CH3COOCH2CH(CH3)2, so if this was correct in (d)(i) it was easy to identify the singlet in d(ii) as coming from the first methyl group. The H in the CH with an adjacent CH2 and two adjacent CH3 groups would theoretically produce a nonet (i.e. the peak splitting into nine) . In the actual spectrum on the data base used, the end two lines of the nonet were so insignificant that they could hardly be seen, which is why in the very simplified spectrum it may have looked like a septet. This question did not trouble anyone who could draw the ester and, if the structure of the ester was incorrect, a transferred error mark was available for identifying an H in it which would give a singlet – so the candidates were treated very fairly indeed. We must bear in mind, too, the possibility that candidates in some Centres look at many spectra and may have come across the proton nmr spectrum of this compound by chance during their A2 chemistry studies – so we didn’t want to deviate from the ‘actual’ database spectrum to an unrealistic degree!

Yours sincerely,

Ask The Expert

and I responded once more:

I thank you once again for your reply in the way of ‘amplifications’, and I wish to respond.

Q11 a) Propanoic acid will NOT react with solid sodium carbonate. Aqueous propanoic acid will. To believe the two situations are equivolent isn’t something I can accept. In support of my position on this I point to the trusty Alka-Seltzer tablet. That most candidates gave the expected answer despite the answer being wrong is a credit to them for being able to make the best of a bad situation. It seems likely to me that in the process, that some students would have felt confusion and experience some time wastage. Well done to them for plummeting for the least worst answer available, It would nice to see Edexcel publish a comment on this to stop future students who will undertake revision using past years papers, saying that in retrospect, the omitted word ‘aqueous’ should have been included.

Q11 b) Unfortunately I have made a rather peculiar error, mixing up Q11 b) with Q11 c) thinking the question was to distinguish propan-1-ol from propanal. So your amplification has been useful, so thank you, and apologies for wasting your time on this. I trust in the future, Fehlings (or Tollens) reagent will not be proposed as a way to distinguish between a primary alcohol and an aldehyde, which would see the problem I mentioned arise.

Q14) I understand answer D is the ‘natural’ answer and agree with it being an answer, but my point is there is a reasonable arguement that says compounds of different molecular mass have different sizes, hence experience different degrees of steric retardation by the mobile phase. [For molecules that are similar in size it's only  matter of how close one examines the size before the size difference is detected]. In order to allay the concerns of students who has chosen answer A on the basis of steric consideration, I would now point out that the steric factor relates to the force of attraction to the liquid and that hence the force is more fundamental reason, which is why the answer is D.

Q
A quick look at Pascals triangle gives the peripheral peaks a relative height of 1/70 within the multiplet. I appreciate that to enlarge them would have introduced a secondary problem relating to analysis of relative peak heights. As a suggestion, if issues such as this arise in the future, perhaps informing the student in the text that the peak is indeed a nonet but that the peripheral peaks cannot be seen, would be helpful and prevent the possibility of students trying to fix the structure around the misleading multiplet. “This question did not trouble anyone who could draw the ester” This point is not sustainable as there is no way of knowing the time spent and frustration felt (if any) of a student who may have wanted to perfect the structure. After all, they were to circle the group in _their drawn structure_.

Thank you very much for your time, patience and assistance. It is reassuring to know edexcel provide this facility which helps with the quality of students knowledge and also the teaching of the subject.

Regards

I can’t believe I mixed up those two questions… Bizarre!! (musta’ been low on nasi lemak in my bloodstream)

Their last word on the matter:

 Response (Anthony Ellison)    26/02/2012 01.42 PM
Dear Customer,

In closing correspondence on this matter, I can assure you that all your comments have been noted and thanks for taking the trouble to submit them to Edexcel.

Yours sincerely,

Ask The Expert

So, that’s it. (until next time)

 

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