H1 NMR Explanation

Feb 2014
1
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Hello all,

I am working on the spectra from the following link:
http://www.chem.ucla.edu/cgi-bin/webspectra.cgi?Problem=bp2

For the most part, I understand how the spectra available contribute to the answer, but the left (upfield) side of the proton NMR is proving confusing. I see how the methyl group around 4ppm and the lone aldehyde hydrogen around 10.5 ppm produce their respective patterns, but the remaining 4 hydrogens on the left side of the benzene ring mystify me. To be more specific in my confusion:

1. Why does the 7.0 ppm shift have a multiplicity of two? Given that each hydrogen is a different distance from the either and aldehyde, shouldn't each hydrogen have a different peak?

2. What could account for the two-multiplicty 4-peak split between 6.9-7.0 ppm? Since this is a benzene ring, each hydrogen should be adjacent to a maximum of two hydrogens, no? One on each carbon on each side of the bond?

I understand the other two shifts within the 7-8 ppm range, but not how they might fit in with the one above. What am I not seeing?

Thanks for your help.
 
Feb 2014
1
0
From my guess, the doublet between 7.7 and 7.8 refers to the hydrogen ortho to the aldehyde group. The triplet @ around 7.5 refers to the proton para to the aldehyde group(or meta to the ether group) The signal between 6.9 and 7.0 may look like a quartet at first sight. However, it probably is two signals(one doublet and one triplet) overlapping over each other and should be designated multiplet instead of quartet. The two protons happen to have around the same chemical shift and overlap over each other.