Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

Hey all, this is what I remember from the multiple choice:

Type of reaction: Transesterification
The hydrolysis of the palm oil would be: Triol (there was only 1)
Radiation leading to bond breaking: UV (free radical formation)
What the 2 mechanisms had in common: Both had bonds broken and formed
Only 1 leads to: Elimination
The one with the optical isomer: Aldehyde which was X
The one with geometric isomer: Alkene Y (had C=C with E/Z isomerism)
Which produced iodoform: Both W+X which is aldehyde and ketone
Which is oxidised to form a carboxylic acid: Aldehyde X
Which reacted with 2,4-DNP: W+X the carbonyls
Gas used for carrier gas chromatography: Nitrogen
To separate you use the: Stationary phase
The repeat unit for the diol and dicarboxylic acid: A
The monomer used for that polymer was: 2-hydroxybutanoic acid
Ammonia and propanoyl chloride: C2H5CONH2
The acid and base conjugate pairs was: A
The pH of the buffer solution: 6.21
The yield will increase by: reducing pressure and temperature
Changes affecting Kp: Temperature change
Iodine: High concentration of hexane, top layer

Please correct me if I'm wrong

Don't think any of them led to elimination. the cyanide one was addition and the bromine was substitution.

I thought it was 3-hydroxypropanoic acid.

What the 2 mechanisms had in common: Both had bonds broken and formed
Only 1 leads to: Elimination
Which is oxidised to form a carboxylic acid: Aldehyde X

The monomer used for that polymer was: 2-hydroxybutanoic acid


The pH of the buffer solution: 6.21

i think you're right actually. I just realised elimination means a double bond is formed, the whole time i was thinking one molecule gets eliminated lol. well thats one extra mark gone

nope deffo 2-hydroxypropanoic acid

I believe only 1 leads to racemic
I think there was secondary alcohol which would've been oxidised to carboxylic acid too
And it was 3-hydroxy because COOH carbon is the priority.
pH I got as 4.03 or 4.02 around that



Yeah there was a question in IAL paper or something where we had to deduce the mechanism for elimination with OH but the step in the MCQ didn't look like that. Well I wouldn't worry now, it won't change anything, and the paper was tricky so gb will be forgiving. Gl with unit 5 tho

I don't think so...Not sure xD



Secondary alcohols can't oxidise to carb acids

I don't think so...Not sure xD



Secondary alcohols can't oxidise to carb acids

Its 3 hydroxy look at the photo

And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

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yh that's what I got ! Pretty sure I got 19/18 on MC

Yeah there was a question in IAL paper or something where we had to deduce the mechanism for elimination with OH but the step in the MCQ didn't look like that. Well I wouldn't worry now, it won't change anything, and the paper was tricky so gb will be forgiving. Gl with unit 5 tho

Its 3 hydroxy look at the photo

And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

Posted from TSR Mobile

Yeah it was substitution for the other one by SN2. I also put that a racemic mixture forms but I can't remember if the molecule was symmetric (in which case it wouldn't form a racemic mixture obvs) :/

Its 3 hydroxy look at the photo

And oh my god that is right, flipping hell what happened to me lol it forms ketone of course

Posted from TSR Mobile

No I think you're right, I don't think it was a secondary alcohol I think it was primary just the OH was on the right and the carbon had 2 other hydrogen atoms bound to it

the molecule was indeed symmetric, there was a CH3 on either side of the C-Br, so I went for elimination. It asked what it leads to in that step, so i would say that it is true to say that in that step a Br- is eliminated......

But there was also a secondary alcohol molecule with a methyl group wasn't there

the molecule was indeed symmetric, there was a CH3 on either side of the C-Br, so I went for elimination. It asked what it leads to in that step, so i would say that it is true to say that in that step a Br- is eliminated......

I can't remember, there was 2 which would be oxidised to a carboxylic acid though wasn't there?

I don't think it was elimination
AFAIK in elimination there should also be a double headed arrow from one of C-H bond to form C=C