Wjec ch4 2016

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Original post by Jackielyngreen30

I think the molecular formula was C6H10O for the first part of that question i guessed because the there was a spectrum at 72m/z. C=12, 72÷12=6 so I guessed there was 6 carbons. Thats all I put for there I hope i do get the full 4 marks for giving that guess if its right :/. And the structural formula I deduced from the NMR spec there was an aldehyde present at around (~9ppm) so I put it as a displayed formula (stupid mistake) = (in structural formula) CH3CH(CH3)CH2CH2CHO

m/z 72 was the full molecular mass. All the carbons, hydrogens and oxygen. Not just the carbon. The correct molecular formula was C4H8O :smile:

Reply 21

Jackielyngreen30

Original post by sumeyyatontus

m/z 72 was the full molecular mass. All the carbons, hydrogens and oxygen. Not just the carbon. The correct molecular formula was C4H8O :smile:

Oh god damnnnnnnn 😢😢😢😢

Reply 22

aladdin818

Hmmm questions 1,2 and 5 were okay..
3 was RANDOM. The stuff about racemic mixture and the pregnant woman? What did everyone do for that? and also I went blank :frown:

for hydrolysis of Sucrose?

4 was just horrid. I think I lost almost all marks on that question cause I hate NMR and I hate chromatography.

Relying on my insurance to accept me for uni now D: I don't think Ill get an A in chem.. :[

Reply 23

Pixie1234

I didn't like that paper atall! Can anyone do an unofficial mark scheme?

Reply 24

BeepBoop

Correct me if I'm wrong but the WJEC definition of racemic mixture is: 'an equimolar mixture of two optical isomers'.
Yet the question asks what do you get when a racemic mixture is formed from one isomer.

Either the question was worded terribly or I'm missing something...

Reply 25

Naomi173

That was exactly my problem! Couldn't work that out all! Seemed contradictory

Reply 26

chemstudent314

Original post by SackSly

What the f*ck was question 4!? I'm actually so annoyed because I went through the effort of memorising all the different reactions and conditions and barely any even came up! Terrible paper :/

yes so did I, i knew all the reactions, I had lithium tetrahydridoaluminate (III) in my head all year round and they hardly asked for reagents and conditions:/ Also I just don't like nmr or chromatography :/ and the four mark questions on question 2 were ridiculous how much can you actually say to get all the marks

(edited 7 years ago)

Reply 27

chemstudent314

Original post by BeepBoop

i was stuck on that for ages too and didn't really know what they wanted me to draw :/

Reply 28

Lewildspleen

It was a weird paper. I feel like I guessed most of it!

Reply 29

sarsssbelike

I'm somewhat relieved that I wasnt the only one struggling. That paper was horrific, probably lost loads on proving how the statements were incorrect 😵

Reply 30

thatcooldude2.0

Felt like I was guessing all of section B :colonhash:

Anyone else get methyl propanal?

Reply 31

DrSum98

Original post by thatcooldude2.0

Felt like I was guessing all of section B :colonhash:

Anyone else get methyl propanal?

Yep think that's the right answer :smile:

Reply 32

Chemistryxx77

Original post by sumeyyatontus

Me too 👍

I don't want to bring either of you down but it was revealed in the high resolution NMR that the doublet at 9.8 indicated an aldehyde. And another peak said somewhere between 2.0 and 3.0 there was a multi something number of peaks which means either "CH3-C=O(- R)" or "-CH2-C=O(-R)" so it had to have a ketone in there no matter what and the C=O was not part of the aldehyde. So it had to be an aldehyde with a ketone or possibly an ester.
Correct me if i'm wrong though.

Reply 33

Faisalshamallakh

Hitler has reacted, I tried my best guys...

https://www.youtube.com/watch?v=FfAJvji4Bgs

Reply 34

Mango88

Original post by Chemistryxx77

So what should the structure have been?

Reply 35

Mango88

What value did everyone get for the percentage yield?

Reply 36

DragonRider17

Original post by Chemistryxx77

Apologies if I don't quite understand what you're saying, but I'll show how I think ours works. Using CH3-CH(CH3)-CHO, you have the one proton on the aldehyde, with chemical shift of about 9.8 and area 1, which is a doublet as it is next to the single proton on the CH group. Then this CH proton is next to several other protons (2 CH3 groups and the CHO group), so is multi-split and has area 1. Basically for a peak to be multi-split it needs to be next to several carbons with protons. Those examples you have given would have no splitting as those protons are not next to any other protons. Then the two methyl groups are in identical envronments so produce one peak with area 6. It is also a doublet as each CH3 group is next to the CH group in the centre of the molecule. This seems to match the NMR prefectly and has Mr 72, which the mass spec confirmed was the case.

Reply 37

DragonRider17

Original post by Mango88

What value did everyone get for the percentage yield?

I got around 23% I think (it was denitely in the 20s), I thought it was too low, but the next question did say "the yield is low," so I think it was nearly right. If not, only one mark lost becuase of a slip somewhere.

Reply 38

Formless

i accidentally put down butanal wasn't looking at the splitting correctly. I talked about the ppm and its correspondence to groups, the splitting was my downfall. methyl propanal is correct. How many marks do you think i will lose, so pissed at myself.