OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

Scroll to see replies

Damn guys, retaking this. I did the last years paper and got an D. This year I am going to nail it. I feel so much more ready. Just hope it ain't extremely hard. I have a massive feeling fats will come up.

Reply 821

bulkybros100

Original post by marioman

Read the question carefully - in June 2015 it tells you to "use your answer to (c)" (14 peaks) and "the chemical shifts and splitting patterns in the ¹H NMR spectrum". Although it doesn't say it outright then it's heavily hinting that the main clues to the structure are in the ¹H spectrum rather than the ¹³C spectrum.

ahhh that makes more sense now

Reply 822

marioman

Original post by RayMasterio

My problem with the June 2015 paper was which environment caused the peak. Like the NH bond never crossed my mind, i thought it was a carbonyl compound because it has the same chemical shifts.

How would you overcome this problem?

The key is using all the information that has been given. The structure of precipitate L given in the question has an N-H environment. There is no mention of D₂O in the question, so you should know to look out for an N-H peak in the spectrum. This is easy to find because you can see (again using the given structure of L) that this peak will be a singlet, and the spectrum only has one singlet peak.

Reply 823

RayMasterio

I'm feeling confident for the exam. The F324 was a challenging paper given the questions + the time constraints. But without the time constraints, it's relatively straight forward.

How is everyone revising? I've done all the possible revision there is

Reply 824

qweening

In the mock of F324 on Jan I scraped an A at like 44/60, June 2015 WAS hard, I walked out and remember going ''well if that was my uni place, it would be gone'' but I sorta got the NMR question as I realised there was another benzene peak and then the rest fell into place. The WHOLE reason any of us (if youre on here, I assume you've been working your ass off and you're wiling to ask for help) would fail is because of pressure. Exams suck.

Reply 825

Lularose83

Original post by AqsaMx

Can anyone explain this question? Like how do you know the position of the double bond in the third monomer?

This is what confused me as well!

Reply 826

Lularose83

Also, for F325, in some of the titration questions on the last few pages the mark scheme says to multiply by 40? How do you know whether or not to do this? (Sorry I can't word it better :smile:

)

Reply 827

mechanism

Original post by AqsaMx

Can anyone explain this question? Like how do you know the position of the double bond in the third monomer?

Draw each carbon out (so not skeletal formulae) and then work from there. There isn't a CH2 group coming off the second carbon on the chain. By placing the double bond there you've added an extra carbon. Sorry this is probably quite a terrible explanation, but draw each carbon out and you might see what I mean. Also the double bond can't be connecting to the benzene carbon because each carbon on benzene cannot have a fourth bond.

Reply 828

mechanism

Original post by Lularose83

Also, for F325, in some of the titration questions on the last few pages the mark scheme says to multiply by 40? How do you know whether or not to do this? (Sorry I can't word it better :smile:

)

Usually the question will say 250cm^3 of the solution was prepared, and then 25cm^3 titrated. so to get the moles of the original solution you must x10. Also the molar ratio, in the case of 'x40', would be 1:4 meaning you must x4 as well as x10 = x40. Does this make sense?
-Sometimes it might say x50 in the mark scheme if the ratio is 1:5 and the orig solution has 10x more volume.

Reply 829

Lularose83

Original post by mechanism

Yes I think I get it now! Thank you so much :smile:

Reply 830

BrownGuyIrfan

Can anyone here explain to me the Answer to question 3d Part 2 (ii) In the June 2013 F324 Paper? It's soooo confusing...and I feel like none of the teachers on youtube doing paper walkthroughs actually know why the answer is what it is...they just say the answer and move on to the next question. It's only 1 mark but every mark makes a difference!

http://www.ocr.org.uk/Images/175441-question-paper-unit-f324-01-rings-polymers-and-analysis.pdf

(edited 7 years ago)

Reply 831

drprocrastinator

anyone have a formula sheet for F325?

Reply 832

itsConnor_

Original post by drprocrastinator

anyone have a formula sheet for F325?

http://www.ocr.org.uk/Images/74947-datasheet.pdf
data sheet ?

Reply 833

drprocrastinator

Original post by itsConnor_

http://www.ocr.org.uk/Images/74947-datasheet.pdf
data sheet ?

no like a list of the equations like pka and ka and kc etc

Reply 834

itsConnor_

Original post by BrownGuyIrfan

I need help with this question too, but im thinking it's because of where the open-ended bond is on the part that's added on? Hence the double bond must be where the oppen bond is?

Reply 835

ImNervous

Original post by BrownGuyIrfan

Mechanism (the user) explained it a couple of posts ago.

Reply 836

CalistaJupiter

Original post by itsConnor_

I need help with this question too, but im thinking it's because of where the open-ended bond is on the part that's added on? Hence the double bond must be where the oppen bond is?

Original post by BrownGuyIrfan

I think all the confusion with the question is "how do you know the monomer didn't have a methyl group?"
For me what makes the most sense is the fact that the monomer joins the polymer unit at two points (eg there is a dotted line coming off the monomer on top of the fact that it is directly attached to the unit). So it can't ever have been a methyl group becasue a methyl group can't undergo addition polymerisation, only a double bond can. And it's not like there could be two double bonds as then the carbon would have 5 bonds which isn't possible.
So the best way to think of it is to go through the polymerisation and although it seems/looks like there should be an extra methyl group, if there was, there would be not dotted line coming off to the bottom. That was probably really badly explained sorry :colondollar:

(edited 7 years ago)

Reply 837

AqsaMx

What name is given to the process by which components are separated by gas/liquid chromatography?

I thought the answer was relative solubility, but the MS says Adsorption???

Reply 838

RayMasterio

If the Benzene ring did not have the circle in the middle, it would be c6h12 right?

Reply 839

itsConnor_

Sorry but in 4-aminophenol ( https://gyazo.com/28ae388c73b6a5f851cee6dd42f8d250 ) when it accepts a proton does it form a covalent or ionic bond? i was thinking about it forming a dative covalent bond with the hydrogen due to its lone pair, is this right? cos then again it goes to NH3+ doesnt it?

Also how does it act as a base (in the question) if the lone pair of electrons are delocalised into the ring?

Thanks. :smile: