Splitting pattern nmr

  • Thread starter katiecool
  • Start date
  • #1
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I am struggling to understand how to determine splitting patterns for molecules. For example, (see attached) I was trying to determine the splitting pattern for CH2 in the molecule. On the right i believe n is equal to 1 and on the right it is equal to 3, but i am not sure if this is correct. A doublet quartet doesn't make any sense to me and that is what i seem to conclude. The addition of the OH is really throwing me off. Also, on top of that how would i begin to determine which is farthest down field! Any help would be great!
 

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Answers and Replies

  • #2
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I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
 
  • #3
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I think that i am miss understanding how to determine which are equivalent hydrogen's. Equivalent hydrogen's act as a group. On both CH3's all the hydrogen's are equivalent. so "n" for both CH3's are equal to one. Thus i think it may be n=4 and n+1=5 so the splitting pattern would be a pentet? Can confirm that i am right?
 
  • #4
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You were more right the first time. It may be that the J values are similar so it looks like a quintet, but this is just a resolution thing. As to which is more downfield, what kind of groups have downfield resonances in general?
 
  • #5
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I agree, but what i wrote the first time is something i learned outside of my notes. A pentet was something in my notes. So a pentet is possible? and i think that it would be the the part of the molecule with the fewest hydrogens.....so CH i think would be the farthest downfield?
 
  • #6
445
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A quintet would be more possible if you were only coupling to a nucleus with spin I=2, it's certainly not impossible.

I agree that the CH would be the most down field, but not because it has the fewest hydrogens. The CH in terminal alkynes for instance resonates at about δ3-4 from memory. Look into the electronic environment of the protons. I can see a similar thread on chemicalforums.com; it may be worth looking there too.
 

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