SPECTROSCOPY DRY LAB General Formula to calculate the degree of unsaturation: #C atoms - (#H atoms/2) + 1 Example: Benzene (C6H6) 6 - (6/2) + 1 = 4 Rules for calculating unsaturation with organic compounds that contain heteroatoms (halogen, O, S, N) Rule 1) Replace any halogens with hydrogens Example: CH3Br change to CH4 and then calculate 1- (4/2) + 1 = 0 Rule 2) Ignore any O or S atoms in the compound Example: Phenol (C6H5OH) calculate for C6H6 6- (6/2) + 1 = 4 Rule 3) Subtract a H for every Nitrogen present Example: Aniline (C6H5NH2) calculate for C6H6 6-(6/2) + 1 = 4 For the following problems given spectral data and the chemical formula: 1) Determine the degree of unsatuaration using the unsaturation rules above and list it in the appropriate box: 2) Label important absorption bands directly on the IR spectra 3) Identify and label all signals directly on the 1H NMR and or 13C NMR spectra (make sure to use Ha, Hb, Ca, Cb, etc... Remember TMS shows up as a singlet at 0 ppm for both the 1H and the 13C NMR CDCI3 (solvent) sometimes shows up as a triplet around 77pm in a 1°C NMR 4) Propose a structure for the compound that matches 1-3 above (If your structure has more than four degrees of unsaturation it's quite likely to have an aromatic ring). Draw the SKELETAL structure in the proposed structure box.

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SPECTROSCOPY DRY LAB
General Formula to calculate the degree of unsaturation:
#C atoms - (#H atoms/2) + 1
Example: Benzene (C6H6) 6 - (6/2) + 1 = 4
Rules for calculating unsaturation with organic compounds that contain heteroatoms
(halogen, O, S, N)
Rule 1) Replace any halogens with hydrogens
Example: CH3B change to CH4 and then calculate 1- (4/2) + 1 = 0
Rule 2) Ignore any O or S atoms in the compound
Example: Phenol (C6H5OH) calculate for C6H6 6- (6/2) + 1 = 4
Rule 3) Subtract a H for every Nitrogen present
Example: Aniline (C6H5NH2) calculate for C6H6 6- (6/2) + 1 = 4
For the following problems given spectral data and the chemical formula:
1) Determine the degree of unsatuaration using the unsaturation rules above and list it in the
appropriate box:
2) Label important absorption bands directly on the IR spectra
3) Identify and label all signals directly on the 1H NMR and or 13C NMR spectra (make sure to
use Ha, Hb, Ca, Cb, etc...
Remember TMS shows up as a singlet at 0 ppm for both the 1H and the 13C NMR
CDCI3 (solvent) sometimes shows up as a triplet around 77pm in a 1°C NMR
4) Propose a structure for the compound that matches 1-3 above (If your structure has more
than four degrees of unsaturation it's quite likely to have an aromatic ring). Draw the
SKELETAL structure in the proposed structure box.
Transcribed Image Text:SPECTROSCOPY DRY LAB General Formula to calculate the degree of unsaturation: #C atoms - (#H atoms/2) + 1 Example: Benzene (C6H6) 6 - (6/2) + 1 = 4 Rules for calculating unsaturation with organic compounds that contain heteroatoms (halogen, O, S, N) Rule 1) Replace any halogens with hydrogens Example: CH3B change to CH4 and then calculate 1- (4/2) + 1 = 0 Rule 2) Ignore any O or S atoms in the compound Example: Phenol (C6H5OH) calculate for C6H6 6- (6/2) + 1 = 4 Rule 3) Subtract a H for every Nitrogen present Example: Aniline (C6H5NH2) calculate for C6H6 6- (6/2) + 1 = 4 For the following problems given spectral data and the chemical formula: 1) Determine the degree of unsatuaration using the unsaturation rules above and list it in the appropriate box: 2) Label important absorption bands directly on the IR spectra 3) Identify and label all signals directly on the 1H NMR and or 13C NMR spectra (make sure to use Ha, Hb, Ca, Cb, etc... Remember TMS shows up as a singlet at 0 ppm for both the 1H and the 13C NMR CDCI3 (solvent) sometimes shows up as a triplet around 77pm in a 1°C NMR 4) Propose a structure for the compound that matches 1-3 above (If your structure has more than four degrees of unsaturation it's quite likely to have an aromatic ring). Draw the SKELETAL structure in the proposed structure box.
3. С,HoN
100
80-
60-
40-
20
4000
3500
3000
2500
2200
1500
1eee
3H
Hint: There is a singlet
4H
that integrates to 1
proton buried in here
2H
2 PPM
Degree of Unsaturation:
Proposed Structure:
200
150
100
50
ő (ppm)
- SLL
2291
2918
S EETE
Transcribed Image Text:3. С,HoN 100 80- 60- 40- 20 4000 3500 3000 2500 2200 1500 1eee 3H Hint: There is a singlet 4H that integrates to 1 proton buried in here 2H 2 PPM Degree of Unsaturation: Proposed Structure: 200 150 100 50 ő (ppm) - SLL 2291 2918 S EETE
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