With the provided 1 H NMR and IR. I have come up with the structure but am sure

Question

With the provided 1 H NMR and IR. I have come up with the structure but am sure if it correct. Need help analyzing both spectrum to see if my structure is correct.

Spin Works 3 no title 8.2 8.0 7.8 7.6 7.4 7.2 7.0 6.8 6.6 6.4 6.2 6.0 PPM 8.6 8.4 freq, of 0 ppm: 300 1300os MHz file: nd Settings WNMR Files WThanhuowd exper processed size: 32768 complex points transmitter freq.: 300.131853 MHz LB: 0.300 GF: 0,0000 Hz/cm: 958 ppm/am: 0.11648 time domain size: 65536 points 0.094893 Hz/pt width: 6218.91 Hz 20.7206 ppm of scans: 16

Explanation / Answer

1H NMR: The proposed structure indeed tallies with the available spectrum. Normally, unsymmetric para-disubstituted benzene products give two doublet of doublets (in high resolution) or doublets in low resolution. But, fluorine also involves in the splitting of neighbouring protons (a known phenomenon). It just splits the neighbouring protons as if it is a proton. Therefore, in 4-fluorobenzoic acid, protons present on 3,5 positions appear as a triplet integrating to two protons (7.15 ppm in this case) and protons a 2,6 positions give a weak doublet of doublet (8.15 ppm) integrating to two protons. Thus, the splitting pattern and the integration support the proposed structure.

IR: Broad peak (unlabelled) between 3200- 3000 are indicative of a O-H str. 1674 is due to C=O str. 1508, 1603 bands are due to aromatic C=C stretch. 1292 is due to C-O str. 1312, 1328 are due to C-F aromatic stretch

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