Benzene H-NMR

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happyfellow

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After looking through some TBR organic I have a couple basic questions...

If I had fluoro-benzene how many H-NMR peaks would there be? Would it just be 1 or would I have 3?

If I had 1,4-difluoro-benzene would I also just have 1 peak?

Thanks in advance

-Happy

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After looking through some TBR organic I have a couple basic questions...

If I had fluoro-benzene how many H-NMR peaks would there be? Would it just be 1 or would I have 3?

If I had 1,4-difluoro-benzene would I also just have 1 peak?

Thanks in advance

-Happy

I believe you would have 3 on the fluoro benzene and only 1 peak on the difluoro benzene
 
Fluorobenzene has around 30 peaks arranged into 3 multiplets (2 of which are smooshed together). See here. The multiplet that's furthest downfield corresponds to the protons at meta. The ortho protons are the furthest upfield but their signals are so close to the para proton's signals that they tangle.

1.4-difluorobenzene produces a quartet but the middle peaks are so similar that they are difficult to distinguish from one another. Exhibit (the wimpy singlet at d=7.26 corresponds to the solvent (deuterated chloroform)).

There's an explanation for the complex splitting seen from aromatic and allylic protons but I can't grasp it. I'd appreciate input from anyone who has insight on this.
 
As some previous posters said, fluorobenzene has 3 distinct H's so in theory should have 3 distinct peaks. In 1,4-diflurobenzene, all of the H's are equivalent, so there will be one peak.

That's really all you would need to know for the MCAT. But the spectra are actually more complex than just 3 and 1 peak spectra (as the above poster stated), due to spin-spin coupling. For example, the ortho H's in fluorobenzene will be coupled to the meta H's, the F, and weakly to the para H. This results in the splitting of the ortho H peak into the multiplet observed.
 
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