Question

How do I interprete this data?

thank you

lood 4" 1 1 C Brukeroppin).2.6\data asa-galan PROTONA COC ( C e ker Topspin.2.6) 1 -a 4 SC- 3096 12 500 CO2- -7.1670 LELES- -3.8401 -2.8171 COOL 6198 (ppm)

Answer 1

ANSWER:

• Protons between 7 - 7.6 ppm are aromatic protons. There is three type of aromatic protons and apparently there are 8 aromatic protons => this means that the compound must have two benzene rings
• Proton at 5.37 ppm is an olefinic proton and there is only one proton of this type => the compound has a C=C bond where one sp2 C has one proton and the other sp2 carbon is a quternary carbon => -HC=C<
• Protons between 3.63 and 4.16 ppm are protons placed over a carbon with one heteroatom (O, N, halogen)
  • Signals at 3.63 and 3.84 ppm appears to be protons of -CH2-OH group (each signal integrates for two protons)
  • Signal at 3.98 ppm appears to be a >CH-OH (the signal integrates for one proton)
  • Signal at 4.16 ppm appears to be an impurity or >CH-OH (the signal integrates for 0.13 protons).
• Signal at 2.82 ppm appears to be a proton attached to C with a phenyl group as subtituent => >CH-Ph (the signal at 2.82 integrates for 1 proton)
• Signal at 2.67 ppm is similar to previous signal. However it also could be an impurity because the signal at 3.67 ppm integrates for 0.33 protons
• Protons at 1.95 ppm are usually protons near to a double bond C=C. In this case it is not clear if this signals are impurities or not (the signals integrates for 0.1 H)
• Signals at 1.60, 1.50 and 1.25 ppm are protons in CH2 or CH groups in an aliphatic chain. Integration does not allows determine if these signas are impurities or not
• Signal at 1.05 ppm is a -CH3 because the signal integrates for 3H

For a deeper analysis it is needed additional information as molecular formula, IR or 13C-NMR spectra, etc.

NOTE: the integration for each signal are the red numbers at the bottom of the spectrum