The Student Room Group

Please help! Quick proton NMR confusion question OCR F324

I cannot hack this NMR stuff! :L
The specific question I'm stuck on is in the june 2011 paper, question 5c). http://www.ocr.org.uk/Images/62264-question-paper-unit-f324-rings-polymers-and-analysis.pdf

Firstly, what the heck is that little peak at 0? Can I just ignore it? Also in the mark scheme it says there's at peak at 3.4 which is due to HC-C=O but in the data sheet (http://www.ocr.org.uk/Images/74947-datasheet.pdf) I got this as HC-O?? Is this just an updated data sheet or something?

Thanks for your time :colondollar:
Reply 1
Peak @ 0 = TMS
The question states that the peak @ 3.4 should be @ about 2.4
Reply 2
Original post by Pigster
Peak @ 0 = TMS
The question states that the peak @ 3.4 should be @ about 2.4


*facepalm*
I'm such an idiot. Thank you so much :colondollar:
Reply 3
Original post by Pigster
Peak @ 0 = TMS
The question states that the peak @ 3.4 should be @ about 2.4


Sorry to ask again, but please can you help me with another question? It's from Jan 2012 question 4. The mark scheme shows the correct structure and I've included a picture of the spectrum, but I don't understand why there isn't a peak which has been split into 10 in there?? Since there are 9 protons on one carbon, and so for the CH2-C=O next to it in the structure, the peak that's for the CH2-C=O should be split into 10 shouldn't it? :L Instead there are two singlets but I thought there should be a 'declet' (don't know if that's the correct term aha) and a triplet (since for the (CH3)3C it's attached to the CH2 which have two protons on)??

Thanks for your time :smile:
Reply 4
Too far. Tertiary carbon has no protons that could split those at -CH2-
Reply 5
The CH2 left of the ester link isn't adjacent to the 3xCH3 on the left (there is a C in between).
Reply 6
Original post by Borek
Too far. Tertiary carbon has no protons that could split those at -CH2-



Original post by Pigster
The CH2 left of the ester link isn't adjacent to the 3xCH3 on the left (there is a C in between).


Ahh thank you both! It seemed to contradict another question that I had done so I was stressing but yeah on that previous question there isn't a carbon in between. I have a real problem with noticing obvious things -.-' Thanks again for your time :biggrin:

Quick Reply

Latest