The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Aqa chem 4/ chem 5 june 2016 thread

Scroll to see replies

Reply 700
Avatar for newt1
newt1
7
Original post by Dizem.Tekin
Please someone say that exam was hard for them too... I started having a panic attack in there... It went horribly and I revised so much for this exam


check twitter, we all found it horrendous
Original post by j.ethers
Ok so for the last question we know it has to be a primary/secondary alcohol because of the dichromate reaction, but the O-H nmr peak was a singlet, suggesting the adjacent carbon to the alcohol group is itself bonded to no hydrogens. Therefore how can it be a primary or secondary alcohol??


You don't look at the adjacent Carbons if you reach an O. And O is basically a dead end when talking about splitting.
Original post by BeckaH2
Yeah can't remember the exact number but definitely around 10%. Because it was something like (23 / 233) x 100.


**** i got 22/233
Original post by thehollowcrown
Before I forget these are some of my answers, not saying they're right by any means though

ph values were something like 2.08, 2.32, 3.33
for the salt/acid weird question i got 1.20
9.09x10-3 for one question can't rememer which one
9 peaks for the tick box question
Last question I got the structure: CH3-O-C(CH3)2-CH2-CH2-OH


I got all of them also😊


Posted from TSR Mobile
Original post by MarimaKauser
I got all of them also😊


Posted from TSR Mobile

yess mate
Original post by cuppycake3
no one laugh but could it be

(ch3)3-c-o-ch2-ch2-o-h

i came up with that in two mins :frown:((((((((


Haha I got that too, but I couldn't get the structure so I was just like 'screw this' and drew some structure which looked alright to me....
I wrote my numerical answers on my formula sheet and took it out, these are what I got...
pH values of 2.08, 2.32 and 3.33
1.204 for the salt/acid ratio
0.33 mol and 4.16 mol
9.09x10^-3 for something (kc?)
9.402% hydrogen
112 for the molecular ion peak
K=3.12x10^-5
(edited 7 years ago)
Reply 707
Avatar for tbnl
tbnl
Original post by Rabadon
rip me


Word
Reply 708
Avatar for Fibsy
Fibsy
9
Original post by Bacardigan
Anyone know what happens to the rate equation if the experiment was repeated with 100x the concentration of ethanal but same concentrations of HCl and Iodine? I put that it would be the same...


I put the same, and my explanation was that the mechanism and therefore the rate determining step stays the same?!

That 9 marker screwed me over

Posted from TSR Mobile
I never did chem 4 today but i heard it was very hard :frown: Idk why Aqa are doing this? Like do they want people to go Uni?
nitrogen atom b had more alkyl groups (inductve effect etc) but a had an N-H bond, making that part of the molecule more soluble therefore more likely to react... right?? urgh, worst paper ever
Original post by ishabennett
Can we complain.......Next year is the new spec not this year.


That's what I thought! It was like they'd got so carried away writing the new specification that most of the questions would be better suited to next years students since they will have learnt different stuff! I do wonder whether anyone from the exam boards actually looks at places like TSR to get a feel for what the students thought.
That was the first of all my A2 exams by AQA,safe to say I'm going to fail this year. See you later uni xx
Original post by High Stakes
Chemistry teachers looked at it and said it's relatively easy. Should be high grade boundaries.


Are you joking?😱
For the rate equation i put the same as the one they gave us i thought it had no effect :frown:
Did anyone else other than me find that unbearably difficult? Definitely the hardest unit 4 paper I have done :frown:
That wasn't too bad, just a bit unlike past papers and I was REALLY pushed for time. I finished as she said 'pens down'.

Did anyone else get the rate expression as exactly the same in the question?

Also, what was the reason for lidocaine hydrochloride? I said to prevent the N acting as a base by protonating it and so reduces side effects. I spent about 5 minutes looking for a chiral carbon in that compound to talk about enantiomers, but I couldn't see one!
Original post by j.ethers
Ok so for the last question we know it has to be a primary/secondary alcohol because of the dichromate reaction, but the O-H nmr peak was a singlet, suggesting the adjacent carbon to the alcohol group is itself bonded to no hydrogens. Therefore how can it be a primary or secondary alcohol??


H is bonded to O which is bonded to C. Alcohols are always a singlet.
Reply 718
Avatar for Fibsy
Fibsy
9
Original post by High Stakes
Chemistry teachers looked at it and said it's relatively easy. Should be high grade boundaries.


Oh, you again

Posted from TSR Mobile
Original post by jakeysimps11
Same mate, 99% sure its correct


i got that too but its a tertiary alcohol so doesn't turn green when dichromate is added to it :frown:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you