The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Aqa chem 4/ chem 5 june 2016 thread

Scroll to see replies

Original post by Gezza_O'Brien
Literally this is me. Watched all 12 seasons of greys anatomy in 2 months lmao and now look at me. Push one of Epi cuz dat exam nearly stopped my heart


Posted from TSR Mobile


get me an OR because i think i have a brain aneurism
Original post by terpanter
i think acyl chloride form ppt while haloalkane doesn't..?


yeah u are right I was wrong so feelsbadman for me
Original post by randoms132
I pretty sure it was 8 peaks for that c13 nmr question!


I could have sworn it was 9
Original post by Capsicle102
There was a question like this in a previous paper in which the mark scheme answer was:
- Chlorine is more electronegative
- Chlorine draws electron density towards itself
- OH bond is weaker


Is there any chance you can find this MS?

I said (which is in my notes/online):

Chlorine is a very electronegative atom which withdraws the electron density from the COO- ion, making the COO- ion less negative, more stable and more acidic.

I believe this was for two marks.
Original post by Aerosmith
It was 8 peaks for the carbon-13 n.m.r one right? The first box?


i got 9
Original post by Cadherin
It was 9 - did you miss out the CH3 groups on the benzene (both in one environment) as I did initially and discovered my mistake>


Yeah I got 9 too!! Pretty sure it's right because I checked it like 3 times
Original post by Scarly9
Yeah but in a previous pass paper silver nitrate was used to distinguish between a long chained haloalkane and acylchloride don't know why it works but it does >_<


I put AgNO3. If acidified, no marks.

Acidificed AgNO3 would work for the halogenoalkanes. However, if you don't acidify it, the acyl chloride would show a white precipitate and the Cl bonded directly to a methyl group wouldn't.
Original post by Stevesteve12
Could the last question be CH3-O-CH2-C(CH3)2-CH2OH???
I got that!!
Original post by HannahC-H
Ah sh*t, I said it was because the polyester would be broken down by hydrolysis in the body and so wouldn't have to be removed once the wound had healed as they'd naturally break down…..


That's what I said because the medical stuff through me off! But I think it's right though... maybe not what they want though!
Original post by Suits101
No, sorry.



D: zero
E: second

People are in agreeance with E, but D people are getting 1/0 so I don't know.


D is definetly 1 i had extra time and worked out k then put it into my own rate equation aha >_<
Original post by Cadherin
It was 9 - did you miss out the CH3 groups on the benzene (both in one environment) as I did initially and discovered my mistake>


Do you by any chance Remember the name of the molecule? I need to find out because i re counted at least 3 times and i was pretty sure :-(
Original post by randoms132
Do you by any chance Remember the name of the molecule? I need to find out because i re counted at least 3 times and i was pretty sure :-(


Lidocaine I seem to remember?
Someone post a mark scheme please. Also the last question was so bad I wanted to faint lol
Reply 1033
Avatar for cowie
cowie
7
ImageUploadedByStudent Room1465919833.201290.jpg This was the answer to the last question.. (I got it wrong and drew it as a tertiary alcohol:/
The integration value of 6 with the CH3 included the one on the left..)
How many marks would I loose for tertiary alcohol? I'd imagine 1 for not mentioning it having to be primary or secondary, 1 for wrong product and 2 probably for an error in my working that showed a segment of R-C(CH3)2OH..
So most likely -3/4 what do you think?


Posted from TSR Mobile
Desperately need to see a copy of the paper, I've blanked out that stitches question and can't remember what i wrote. Were there two questions about medical applications or just one? that paper has wrecked my head hahaha
Original post by Anam
Wait so i got a cl* radical in my fragmentation equation is that wrong as i used ch2ocl+ as my fragment ion as it adds up to 65 too


There may be different answers, but from my notes this should work:

[ClCH2COCl]+* -> [COCl]+ (37Cl) + [ClCH2]* (35 Cl)
(edited 7 years ago)
I highly doubt for the isomer question you had to put which isotope of Cl you used for that question on the equation for splitting the isotope. Questions like that have come up before - the only reason they specify is to allow you to work out the structure, which is what the question was assessing. Typically the three marks available would be fore 1) the structure of the molecular ion radical (whether you remembered to put the positive and radical signs), 2) For the structure of the fragment ion, and 3) For the structure of the remaining radical
Original post by Cadherin
I put AgNO3. If acidified, no marks.

Acidificed AgNO3 would work for the halogenoalkanes. However, if you don't acidify it, the acyl chloride would show a white precipitate and the Cl bonded directly to a methyl group wouldn't.
you can acidity it because acidifying it gets rid of in wanted ions if you put NaOH with AgNO3 that would of been no marks as they give the same the result
Original post by _newbe
I got that!!


no because then all your peaks would be singlets
Damn That Exam was solid

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you