Question

1800 1600 1-00 120 100 TSH) Amm) CH2 HEtylPimay amine ) δ000 16 15 14131211 10 987 1 023 ri ppm)

The pictures above show an nmr of benzylamine in cdcl3 at 300mhz. the bottom picture apparently shows the solvents used (h20 and cdcl3) I believe since water has a peak at 1.56 in cdcl BUT Any answers I see say that The primary amine is actually the cause of this peak! I am very confused and am having troubles Finding what caused each peak, how many H are in each peak and How I can show that. I know there are 9 H in the compound benzylamine but I cannot seem to understand the whole integration thing. What peak should I use to compare the relative number of hydrogen in these integrals?  

Answer 1

Let given compound is benzylamine and its structure is, see your HNMR spectrum you can see there are only three type of proton. Viz methylene (H_2-) 2 mone (-NH_2) and proton. Methyl proton (-CH_2-) is attached ring and due to this methylene protons are de-shielded CH_2 are attached to amine-NH_2 group causes shielding of methylene (-CH_2) proton. Thus methylene protons observed at 1.54 ppm. This is main correction in your consideration. At 8.386 ppm is due to primary amine protons. The 7.33 ppm is due to mono substituted benzene ring. The integrated values are 4(-CH_2-), 2.91(-NH_2) and 7.52 (aromatic protons) dividing by 2 for each value we will set respective protons. 4/2 = 2 therefore two methylene protons present 2.91/2 = 1.45 about 2 thus 2 protons of primary amine. Thus only one -NH_2 group present in molecule \r\n due to low resolution this value should be decreased but this must be due to five aromatic proton and confirmed from