1H-NMR Correlation

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 H-NMR Correlation_New

Explanations:

δ9.50 (br s, 1H) corresponds to Hf because it is a singlet and Hf does not have any 3-bond hydrogen neighbors. Also, it is a broad peak. According to what we learned in class, hydrogens bonded to nitrogen and oxygen often have broad peaks (Molecule. Spectrum).

δ8.73 corresponds to Hq because it is very close to an electronegative atom. It couples with Hp and Hf. Since it is spatially close to Hf, it is possible for Hq to couple with Hf (Molecule. Spectrum).

δ8.56 corresponds to Hp because it is very close to an electronegative atom. It couples with Hq and Hf. Since it is spatially close to Hf, it is possible for Hp to couple with Hf (Molecule. Spectrum).

δ8.09 corresponds to Hm. It is a doublet of doublet so it should have 2 3-bond hydrogen neighbors. It is also close to nitrogen so it has a high chemical shift (Molecule. Spectrum).

δ7.48–7.36 corresponds to Hl, Hn, and Ho. They are in the aromatic rings, so they have relatively high chemical shifts, especially there is one nitrogen in the ring. However, because of the resonance, they can bear negative charges, making them more shielded (Molecule. Spectrum).

δ6.81 corresponds to Hk. Although it seems that Hk should be a singlet, it is close to many hydrogens, so it is possible for Hk to weakly couple with another hydrogen (the coupling constant is 1.8 Hz, which is small). It possibly couples with Hj, which has a coupling constant of 1.9 Hz (Molecule. Spectrum).

δ6.77 corresponds to Hj. It couples with Hi and is close to the oxygens, so it should have a high chemical shifts. In addition, it also couples with another hydrogen with a coupling constant of 1.8 Hz. Given the similar position of Hk and Hj, Hj is assigned at δ6.77. It possibly couples with Hk, which has a coupling constant of 1.8 Hz (Molecule. Spectrum).

δ6.55 corresponds to Hi. It couples with Hj with a coupling constant of 8.0 Hz (Molecule. Spectrum).

δ5.72 corresponds to Hh. The integration shows that 2 hydrogens belong to this peak. Only Hh has 2 identical hydrogens (Molecule. Spectrum).

δ4.29 corresponds to Hd. Although it is bonded to 3 different hydrogens. Jcd (9.7) is very close to Jce (9.8). It is possible that the NMR spectrometer does not have a high enough resonance frequency to detect this difference, so it shows a triplet instead of doublet of doublet. It also couples with Ha with a coupling constant 2.2 Hz (Molecule. Spectrum).

δ3.84 and δ3.67 corresponds to Hc and He respectively. They have the same coupling constants, so these two peaks should represent two hydrogens which have identical neighbors. Both Hc and He have Ha, Hb, and Hd as its neighbors (Molecule. Spectrum).

δ3.45 – 3.33 corresponds Hb and Hg. This region has a strong peak representing 3 hydrogens. This peak should be resulted from Hg. Hb is assigned in this region because of its low chemical shift. Also, the splitting pattern is very similar to that of Ha (Molecule. Spectrum).

δ2.54 corresponds to Ha. It couples with Hc and He with a coupling constant of 8.7 Hz. It also couples with Hb with a large coupling constant of 12.2 Hz. It weakly couples with Hd with a coupling constant of 2.4 Hz (Molecule. Spectrum).