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OCR A-level Chemistry A Paper 2 - 16th June 2025 [Exam Chat]

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Unofficial Mark Scheme

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Reply 580

Original post
by opiug
I got B it was 1:2 as it had 2 carboxylic acid groups so Na subbed twice to make the salt

Yeah I might be wrong idk. Do u remember what u got for mass of bromine gas question mcq?

Reply 581

Original post
by saumil2907
Yeah I might be wrong idk. Do u remember what u got for mass of bromine gas question mcq?

I think that one was B - 6.4g?

Reply 582

Original post
by Sophi3_123
Did anyone get a carboxylic acid for NMR?

I got an ester

Reply 583

has anyone got an UOMS? the answers aren’t showing up at the top for me anumore

Reply 584

Original post
by Lilia Bethany
I got an ester


believe it was isopropyl propanoate

Reply 585

Original post
by johndoe11111
How are people getting 8 stereoisomers?


you have to do it to the power of the number of chiral carbons i think

Reply 586

Does anybody have an unofficial mark scheme?

Reply 587

Original post
by saumil2907
I got 12.8 coz it said the starting compound was 0.02mol and it wanted it to react in a way that would make it saturated. So for that u need 2Br2 molecules and so the moles becomes 0.02 x 2 = 0.04mol as you have one mol of compound and 2 moles of bromine and then u do 0.04 x (79.9(4)) as it is br2 and u get 12.80 which is D. Idk if this is right method or not though

it was half of that cos u times it by the Mr of Br2 the 2 before the Br2 doesnt count u dont include it in Mr

Reply 588

Original post
by opiug
I got B it was 1:2 as it had 2 carboxylic acid groups so Na subbed twice to make the salt
What was the number you got? I think I put B but I don’t remember what B was?

Reply 589

Original post
by saumil2907
I got 12.8 coz it said the starting compound was 0.02mol and it wanted it to react in a way that would make it saturated. So for that u need 2Br2 molecules and so the moles becomes 0.02 x 2 = 0.04mol as you have one mol of compound and 2 moles of bromine and then u do 0.04 x (79.9(4)) as it is br2 and u get 12.80 which is D. Idk if this is right method or not though


U had 2Br2 but I swear the mole number before it doesn’t count when calculating moles?

Reply 590

guys am i the only one who got 5.04 for the percentage yield synthesis question... idek where i went wrong🤣

Reply 591

Original post
by xitz_jessix
U had 2Br2 but I swear the mole number before it doesn’t count when calculating moles?
Uhhh no… they do matter

Reply 592

Original post
by disapproving-blu
guys am i the only one who got 5.04 for the percentage yield synthesis question... idek where i went wrong🤣


I got 9.53 or something like that 😭

Reply 593

Original post
by Unrei
Uhhh no… they do matter

Nah they meant as in u had 2Br2 and so the moles would be 2 x 0.02 = 0.04. Then u calculate the mass of Br2 which is 0.02 x (79.9 x 2) which is 6.4

Reply 594

ChatGPT - The whole H–Br molecule, with a polar bond, is the initial electrophile.
The Hδ⁺ is the part of the molecule that accepts electrons, but it only acts electrophilically because of the bond’s polarity.

Reply 595

Original post
by johndoe11111
Nah they meant as in u had 2Br2 and so the moles would be 2 x 0.02 = 0.04. Then u calculate the mass of Br2 which is 0.02 x (79.9 x 2) which is 6.4

Oh sorry, I misunderstood, I even got the same answer as them

Reply 596

Original post
by xitz_jessix
ChatGPT - The whole H–Br molecule, with a polar bond, is the initial electrophile.
The Hδ⁺ is the part of the molecule that accepts electrons, but it only acts electrophilically because of the bond’s polarity.
So the answer was C?

Reply 597

Original post
by Unrei
So the answer was C?

It should be i mean all the others dont make sense

Reply 598

Original post
by Unrei
So the answer was C?


I think so cuz there was no Br+ as an answer like people said (it was either H+ or Br-)

Reply 599

Original post
by johndoe11111
It should be i mean all the others dont make sense
Nice that is what I thought too cause Br+ and H- doesnt even appear in the mechanism and the mechanism is called electrophilic addition so why would a dipoled HBr act as a nucleophile

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