I am stuck on the last question of the January 2010 Unit 4 Chemistry Paper (EDEXCEL) It is question 26 showing two NMR Spectra and I am being asked to identify the structures of X and Y. I would really appreciate any help! I attempted the question and understood the first bit but the splitting patterns that I got don't match with the MS :/ Does it matter what order it is in?
I am stuck on the last question of the January 2010 Unit 4 Chemistry Paper (EDEXCEL) It is question 26 showing two NMR Spectra and I am being asked to identify the structures of X and Y. I would really appreciate any help! I attempted the question and understood the first bit but the splitting patterns that I got don't match with the MS :/ Does it matter what order it is in?
Hiya! I moved this to the chemistry forum for you - you're more likely to get a good answer here
I am stuck on the last question of the January 2010 Unit 4 Chemistry Paper (EDEXCEL) It is question 26 showing two NMR Spectra and I am being asked to identify the structures of X and Y. I would really appreciate any help! I attempted the question and understood the first bit but the splitting patterns that I got don't match with the MS :/ Does it matter what order it is in?
It would help if you posted your working and the question, if possible. Do you know how to link to an image or attach one?
This is the link, it's the very last question (Q26). Hope this helps
This question is about compounds X, C4H10O, and Y, C4H8O. (a) Compound X, C4H10O, can be oxidized to compound Y, C4H8O.
The IR of X suggests hydrogen bonded OH group, i.e. we are dealing with an alcohol.
NMR X:
multiplet 3.7 singlet 2.4 multiplet 1.5 doublet 1.2 triplet 0.9
five different environments primary alcohol CH3CH2CH2CH2OH = five environments
secondary alcohol CH3CHOHCH2CH3 = five environments
tertiary alcohol can't be oxidised
So we look at the splitting patterns: primary alcohol CH3CH2CH2CH2OH
In order (left to right) CH3 gives a triplet (adjacent to CH2) CH2 gives a triplet of quartets (nasty multiplet) CH2 gives a triplet of triplets (nasty multiplet) CH2 gives a triplet (not split by OH) OH gives a singlet (not split by CH2)
secondary alcohol CH3CHOHCH2CH3
CH3 gives a doublet (adjacent to CH) CH gives a triplet of quartets (nasty multiplet) CH2 gives a doublet of quartets (nasty multiplet) CH3 gives a triplet OH gives a singlet (not split by CH)
Hence spectrum given is the secondary alcohol.
On oxidation the secondary alcohol would produce a ketone, butan-2-one
CH3COCH2CH3
NMR of this would expect: CH3 gives a singlet CH2 gives quartet (adjacent to CH3) CH3 gives a triplet
Which matches perfectly the spectrum given.
The last stage is to show that the chemical shift of each signal corresponds to its data book value.
This question is about compounds X, C4H10O, and Y, C4H8O. (a) Compound X, C4H10O, can be oxidized to compound Y, C4H8O.
The IR of X suggests hydrogen bonded OH group, i.e. we are dealing with an alcohol.
NMR X:
multiplet 3.7 singlet 2.4 multiplet 1.5 doublet 1.2 triplet 0.9
five different environments primary alcohol CH3CH2CH2CH2OH = five environments
secondary alcohol CH3CHOHCH2CH3 = five environments
tertiary alcohol can't be oxidised
So we look at the splitting patterns: primary alcohol CH3CH2CH2CH2OH
In order (left to right) CH3 gives a triplet (adjacent to CH2) CH2 gives a triplet of quartets (nasty multiplet) CH2 gives a triplet of triplets (nasty multiplet) CH2 gives a triplet (not split by OH) OH gives a singlet (not split by CH2)
secondary alcohol CH3CHOHCH2CH3
CH3 gives a doublet (adjacent to CH) CH gives a triplet of quartets (nasty multiplet) CH2 gives a doublet of quartets (nasty multiplet) CH3 gives a triplet OH gives a singlet (not split by CH)
Hence spectrum given is the secondary alcohol.
On oxidation the secondary alcohol would produce a ketone, butan-2-one
CH3COCH2CH3
NMR of this would expect: CH3 gives a singlet CH2 gives quartet (adjacent to CH3) CH3 gives a triplet
Which matches perfectly the spectrum given.
The last stage is to show that the chemical shift of each signal corresponds to its data book value.
Thank you so much for explaining that Such a great detailed answer, I'm starting to have hope with these questions hehe