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1H NMR Spectra

MermaidWonders

Active member
Feb 20, 2018
113
Screen Shot 2019-03-03 at 16.25.08.png

For the colourfully-annotated compound above, why aren't the splitting patterns and integrations for the red and blue set of protons "pentet, 6 H" and "octet, 4 H", respectively?
 

MermaidWonders

Active member
Feb 20, 2018
113
For the colourfully-annotated compound above, why aren't the splitting patterns and integrations for the red and blue set of protons "pentet, 6 H" and "octet, 4 H", respectively?
Never mind. I see why.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123

MermaidWonders

Active member
Feb 20, 2018
113
I'm curious. Would you be willing to share your solution?

-Dan
Hi,

Sorry for the super late reply. Just saw this.

When I first came across that question, I was very confused as to why my professor had put "6 H, t" and "4 H, quintet" for the splitting patterns and peak integrations for the red and blue set of protons, respectively (as in the screenshot), so I tried to look at that question in a "different" way. If we start by looking at the red protons branching out from either of the 2 carbons, we see that there are 2 protons coming off of the adjacent carbon (whether you are looking at the top or bottom adjacent carbon). Because all 6 of the red protons are chemically-equivalent, it makes sense for the splitting pattern to be a triplet according to the n + 1 rule, where n = 2 for the adjacent C. Next, if we look at the blue set of protons, we are in a similar situation. All 4 of the blue protons are chemically-equivalent, and a quintet splitting pattern comes from the fact that the "top" OR "bottom" adjacent C (the C's with red protons branching off of it) gives n = 3 AND the other adjacent C (the one with a green proton branching off of it) gives n = 1, so together, n + 1 = 4 + 1 = 5 --> quintet.

Hope what I said sort of makes sense. I'm really no expert in organic chemistry myself, but at least this is the process I had to go through to understand that solution! :(