- Singlet at 2.3 ppm corresponds to 3 protons of -CH3
- Singlet at 3.8 ppm corresponds to 3 protons of -OCH3
- There are four doublets at 8.1, 7.8, 7.2 and 7.0 and a multiplet at 7.5-7.6
- The first doublet at 8.1 for two protons corresponds to H1 and H4, as they are near to ketone group which is electron withdrawing and hence the signal goes downfield (J = 8.8 Hz)
- The next doublet at 7.8 for 1 proton corresponds to H5 (J = 15.6 Hz)
- The two doublets at 7.2 (J = 7.6 Hz) and 7.0 (J = 8.4 Hz) for two protons each corresponds to H1' - H3' and H2 '- H4'
- The multiplet at 7.5 is actually a mixture of two doublets, one doublet with chemical shifts at 7.552 and 7.530 for two protons (J = 8.8 Hz) corresponds to H2 and H3, the other doublet with shift value of 7.530 and 7.489 (J = 16 Hz) for one proton corresponds to H6.
- Here, the J value corresponding to H5 and H6 is in the range of 15-16 Hz, which indicates that the alkene exists as E isomer, because in Z form the J value exists around 7 Hz.
Note: Because the NMR operating frequency was not provided, the value was assumed as 400 MHz and accordingly the J value was calculated (J = difference in chemical shift of a doublet * 400)
Can you correspond each of the peaks to the protons in one of the following molecules?...
Can
you correspond each of the peaks to the protons in
4-methoxy-4-methyl chalcone?
4.0 3.00 -7.530 8.051 28.029 7.352 OCUL ΤΙΣ' 800'L- ---7.769 8.2 8.0 7.8 7.6 7.4 7.2 7.0 6.8 6.6 6.4 6.2 ppm 1.BR 2.47 0.67 0.92 1.93
Correspond each of the peaks to the H's in the structure. Is it the
Z or E isomer? Show how you know.
1H NMR (CDCI, 300 MHz) 9-(2-phenylethenyl)anthracene 7.9424 - 7.8874 - 6.9760 6.9207 1H 7.0 8.6 8.4 8.2 8.0 7.8 7.6 7.4 7.2 6.8 6.6
please provide the chemical shift, splitting, how many protons
in each peak, and calculate the J coupling constant for above
spectrum.
The structure is
the 1H NMR was 400 MHz and solvent CDCl3.
II 8.6 8.4 8.2 8.0 7.8 7.6 7.4 7.2 ppm 8 6 5 4 3 2 1 ppm 8.540 8.540 8.528 8.528 7.828 00L 000 T 7.819 CO 90' 90 86S L 7.586 7.476 7.463 1.08 092L 7.182 7.172 7.160 7.828 1.061 7.819 86S 7.586 1.056 7.476...
In the proton NMR for benzil, there are clearly some impurity
peaks below 8 ppm. Although we cannot assign these for certain, is
it possible that these peaks come from unreacted benzoin? Why or
why not?
8.0 8.3 7.5 8.2 NMR nuclei observed: H Solvent: CDCl3 NMR field strength: 400MHz 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 (ppm) f1 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 f1 (ppm) 11 10 12 13 14 2.0 1.5...
Identify products peaks A-D in Diels-Alder reaction where we
used Anthracene and Maleic anhydride to produce Diels-Alder Product
7.41 7.40 7.26 odds 7.34 1-7.34 7.33 1.7.22 7.21 -7.21 17.20 17.20 17.18 6621 17.39 SE2 44.83 14.83 SRB 3.60 3.58 3.56 3.50 3.48 3.46 C (dd) 7.45 7.40 735 3.54 3.52 11 (ppm) 730 725 11 ppm) 720 7.15 7.10 B (da) A (dt) 7.397.20 E dd) 3.53 1.98 1,93 4.00 2.01 6.6 6.4 6.2 6.0 5.8 5.6 5.4 5.2 5.0...
Questions Using the spectrum attached to this procedure, assign the peaks in the 'H NMR spectrum for 3.4-methylenedioxynitrostyrene. Using coupling constants, describe the geometry of the double bond. You should tear out the labeled proton spectrum and turn it in with your notebook pages. Account for the unusual acidity of nitromethane whose pK, is 10. 1) CH3NO2, NaOH 2) HCI, H2O piperonal Mol. Wt.: 150.13 3,4-methylenedioxynitrostyrene Mol. Wt.: 193.16 —3177.4 - 3163.9 -2995.7 -2982.2 -2904.9 2838.0 2836.4 2030.0 2828.3 2801.0...
An aldol condensation reaction was performed where benzaldehyde
and cyclopentanone were used. The product is 2,5
dibenzylidenecyclopentanone or just 2-benzylidenecyclopentanone.
The spectra correspond with the aldehyde being benzaldehyde but the
ketone used could either be acetone, cyclopentanone or cyclohexane.
I need to find the NMR spectra that correspond to cyclopentanone
being the ketone in the aldol condensation. Which NMR spectra
correspond to the correct product?
I think the second set is with cyclohexanone and not
cyclopentanone but I could be...
interpret both h and ch13 NMR, label the structure DMTPP and
TPCPD protons and carbons and locate the signal for them. use chem
shift and integral ratio to help find the peaks. explain how u got
to the peak assignments
We were unable to transcribe this image10 1 /opt/topspin sbachan 20110726uglab CARBON DIMETHYLTETRAPHENYLPHTHALATE 13C NMR: mNNNNN 2 10 [ppm 60 A0 100 120 140 160 168.7169 143.2799 52. 2812 sbachan 20110726uglab 11 1 /opt/topspin 1H NMR: TETRAPHENYLCYCLOPENTADIENONE NO A777* 20...
Match the peaks labeled A,B,C,D,E to the Diels-Adler
product and what part of the products structure creates each peak
on the NMR graph.
3.60 3.58 3.56 3.54 3.52 f1 (ppm) 3.50 3.48 3.46 C (dd) 7.34 7.45 7.40 7.35 7.30 7.25 i (ppm) 7,20 7.15 7.10 B (dd) A (dt) 7.39 7.20 D (t) 4.83 E (dd) 3.53 T T 1.98 1.93 4.00 2.09 2.01 .8 7.6 7.4 7.2 7.0 6.8 6.6 6.4 6.2 6.0 5.8 5.6 5.4 f1 (ppm)...
Write down problem number.
Provide formula, HDI value, and a complete, chemically correct
structure for each problem.
Assign signals in the 1H and 13C spectra to specific atoms,
identify structural units and/or functional groups.
Example:
Problem # 3
Formula: C8H8O2
HDI: 5
1H NMR: p-substituted Ar at 7.27-7.97 ppm, benzylic CH3 group at
2.42 ppm, COOH proton at 12.9 ppm
13C NMR: carbonyl C at 167.8 ppm (ester or acid), benzylic CH3
at 21.40 ppm, 4 remaining signals are Ar...