Question

Expt 9: 'H NMR Data The following is a 'H NMR of a student's sample from Experiment 9. This is a crude reaction mixture, which means it has solvent, starting materials, and product all mixed in together. Therefore, we do NOT expect you to interpret all of the peaks or tabulate the data. We will only look at signature peaks that can help us determine what has happened. Note that this sample may not exactly match your reaction, but it may help you understand what you see in your GC. 674

Answer 1

1. The large multiplet of peaks at 7-8 ppm is due to the presence of aromatic H of Phenyl ring present in the molecule which has 3 diff type of Protons present at its orhto meta and para position of Phenyl ring. Both ortho adjacent to CH2OH group will so same peak due to same environment and have Doublet peak beacause it has one more proton near to its (meta H). So due to the fact that spilliting is (n+1) for n no of neighbouring Protons. Same way the meta proton have 2 neighbouring H present at ortho and para position. hence spilitting will be (n+1)= (2+1)=3. And for para also it is the same as for meta one.

2. The peak at 10.15 ppm is due the proton of OH group which is very very less and is also variable due to the fact that this proton gets interchanged with the solvent during the analysis. While the peak at 4.80 ppm is due the H of -CH2- of CH2OH group present at Phenyl ring. The electron cloud of CH2 proton is withdrawan by the present of two electron withdrawing group Phenyl and OH hence this is deshielded at 4.80 ppm.

3. The integral value tells everything about starting material and we can assume that the product is almost not formed. Beacuase the NMR data tells that the starting material taken is as it is in the reaction.