What did you guys draw for the partial hydrolysis of the tripeptide with the residue? I couldnt figure it out because they all had C=O bonds but i drew the one on the left on its own and the two on the right still attached
There was no iodine question it was bromine numbers it got everyone confused. I've heard of iodine number before but not bromine. And correct me if I'm wrong but iodine numbers aren't F334 anyway. Either f331 or F335
What did you guys draw for the partial hydrolysis of the tripeptide with the residue? I couldnt figure it out because they all had C=O bonds but i drew the one on the left on its own and the two on the right still attached
I drew the two on the left still attached and the one on the right on its own.
What did you guys draw for the partial hydrolysis of the tripeptide with the residue? I couldnt figure it out because they all had C=O bonds but i drew the one on the left on its own and the two on the right still attached
Fe2+ is oxidised to Fe3+ by the airFe2+ => Fe3+ + e-This makes all the iron become Fe3+They all react with the -OH ion like so:Fe3+ + 3OH- => Fe(OH)3 (s)Iron(iii) hydroxide is a red brown precipitate.Something like that but with better English 😊
This is what I put! forgot to include the Fe3+ to Fe2+ equation though so will have lost a mark
This paper has pissed me off. I thought the majority of the questions were unfair and some of the questions that came up are not even defined on the syllabus! Come on OCR, I really want to go to university!
Apart from question 3 i think that was relatively OK i guess, could've been alot worse. anyway whats done is done, time to focus on this weeks exams then cram F335 from thursday onwards
Did anyone else think that this paper was overly full of content? Like there wasn't enough time? There were no easy one markers or anything really and loads of very wordy questions with very little space for the answers!
Bromine questions I got 390 for the first and 13 for the one later on. Membrane was to allow current (electrons) to flow between the two electrochemical cells whilst keeping the ion solutions separate (salt bridge could've got the mark there I suppose because thats basically the job of it). The peptide that should have broken is the one just to the right of the r group mentioned in the question, so you had two residues on the left and a single amino acid on the right.
For the 5 marker of bromine this is the method I used:
After working out the moles of BrO3- in the previous question (= 0.005 moles and so was in excess) and looking at the reaction equation given, you know that the ratio of BrO3- was 1:3 for Br2, so the moles of Br2 is 3x0.005 = 0.015
We also know that it was 0.015 moles of br2 in 200cm3, but the hexadiene molecule only reacts with 83, so 0.015/200x83? I think
So that's the moles of br2 reacted is worked out, so you multiply that by mass of bromine and then x2 because two bromine molecules.
Lastly you times that by 10 because for bromine number it is of 100g of hexadiene whereas the question said 10g.
I don't know if this is right, this is just what I did. Anyone do the same?
BrO3- was in excess, wasn't it? So You had to use the moles of Br- to calculate the moles of Br2 (by dividing by 5 and multiplying by 3). I'm not 100% sure but that is what I did.