I need help in making and analyzing both 1H and 13C NMR (in purple color) to be

Question

I need help in making and analyzing both 1H and 13C NMR (in purple color) to be the same as the attached report.

[ the main three things that need to be included for each one is to identify if it is singlet, dublet, triplet, etc + J value + number of Hydrogen].

The first photo is the sample of the report, the second photo is the 1H NMR, the third photo is the 13C NMR

Example of NMR Report Butanoic acid: H NMR (CDCI) 11.6 (bs, 1H), 2.32 (t, J=7.4 Hz, 2H), 1.65 (m, 1-1.4 Hz, 2H), 096 (t,J=7.45 Hz, 3H); 13C NMR (CDCl) 180.05, 35.89, 18.13, 13.57 H NMR 11.6 2.32 1.65 0.96 OH 13C NMR 180.05 35.89 18.13 13.57

Explanation / Answer

J value shall be calculated as follows.

For givn splitted peak, (Upper chemical shift value-lower chemical shift)/Instrument frequency.....for the given spectra it is 500 MHz for Proton NMR

5.28-5.25(dd, may be -CH=CH-, J = 1)

5.79-5.75(dd, may be -CH=CH-, J= 1)

6.77-6.72 (m, May be aromatic)

7.29-7.26 (m, May be aromatic)

7.44-7.34 (m, May be aromatic)

13CNMR:

137.5 (aromatic C),

128.5 ( aromatic 3c's, intensitywise)

127.7 ( aromatic 2c's, intensitywise)

126.2 ( aromatic 3c's, intensitywise)

113.77 ( aromatic 2c's, intensitywise)

77 (triplet for CDCl3)

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