Wjec ch4 2016

But the molecular formula only contained one oxygen

Look yh I'm sorry but that paper was disgusting, no two ways about it. You're coming on here and posting how well you think you did when ppl want to just forget about it. You need to know when to keep quiet.

Posted from TSR Mobile

Look yh I'm sorry but that paper was disgusting, no two ways about it. You're coming on here and posting how well you think you did when ppl want to just forget about it. You need to know when to keep quiet.

Posted from TSR Mobile

I Think you may be wrong:/ I had a aldehyde with a methyl group as someone else said further up - CH3CH(CH3)COH

There were three peaks in total which suggested there was 3 hydrogen environments first of all which lead me away from it being a ketone as there was 4 carbons in the molecular formula so the compound must have had two environments which were equivalent and I couldn't think of a ketone which structure applied to those rules.

I agree with you on the peak at 9.8 ppm indicating a aldehyde, but it had a doublet splitting pattern, and due to the n+1 rule that must mean its neighboring carbon bonded to one other hydrogen (CH).

The best clue though is the peak split into 7. This must mean again due to n+1 it has neighboring carbons with a total of 6 hydrogen atoms, so that peak was referring to the carbon bonded to the two CH3 groups.

Thats what i got anyway xD

That's fine i'm not here to hate I just want to figure it out.
Ah but the aldehyde is a doublet because you have one H and the oxygen counts as 0 but the n+1 rule means one so therefore two peaks.
I have no idea about the multi splitting peaks.
There was no area of peak 6.
They're not examples though those where the actual figures for the NMR given in ppm.
Also for two environments to make six they both must be adjacent to a carbon
e.g.
(H3C)CH2(CH3) will give a peak area of 6.

The NMR spectrum I'm pretty sure was a doublet with area 1 (the aldehyde porton), the multi split peak with area 1, and the area 6 doublet, there definitely was an area 6 peak. Yes that's correct and that's what the structure has, a (H3C)CH(CH3) central group with the aldehyde (CHO) also bonded to that central carbon (that carbon is bonded to 3 other carbons and a hydrogen). And it wasn't 2 environments I think, it was one environment split into a doublet (the area 6 doublet). This is certainly a very confusing question though!

It was a doublet for sure. There was a peak area six but that was for the C=O at around 2.5ppm but not 6 envirmonets.
Oh wait are you taking about methyl propanal ?
I know how many marks was it anyway 4 ?

What did you put for the thalamide with hydrochloric acid? When they substituted half of the molecule with N

I hydrolysed both amide linkages opening the ring up, and put a carboxylic acid group on each end (I guess an ammonia molecule would have been lost)

For the yield question, first I calculated the moles of bezene (mass was 10g) so 10/(6x6+6x.101) = ~0.128. 1:1 mole ratio of bezene with benzoic acid so theoretical mass of benzoic acid = 0.128 x mr of benzoic acid = ~15.62 g

Mass of benzoic acid obtained = ~6.2g (cant remember exact value)... So yield = mass obtained/theoritcal mass... 6.2?/15.62 x100 = 39.7%

I'm not sure whether this is correct, just putting how I answered it.

I did exactly the same thing (I think the mass of bezoic acid was ~3.8g)

For the purity of benzoic acid, would they accept measuring the boiling temperature and comparing it to a literature value rather than melting temperature? Also why was the % yield so low?

For the purity of benzoic acid, would they accept measuring the boiling temperature and comparing it to a literature value rather than melting temperature? Also why was the % yield so low?