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2. CаН100 100 60 48- 20 3500 2500 1520 1000 6H 1H 1H 2H allle 4.0 3.5 3.0 2.5 1.5 1.0 FPM 0.5 Degree of Unsaturation: Proposed Structure: B0 70 60 ppm 6CCE

2. CаН100 100 60 48- 20 3500 2500 1520 1000 6H 1H 1H 2H allle 4.0 3.5 3.0 2.5 1.5 1.0 FPM 0.5 Degree of Unsaturation: Proposed Structure: B0 70 60 ppm 6CCE

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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2. C4H100 100 80 80- 48- 20 3500 3000 2500 200a 1520 1808 6H 1H 1H 2H 4.0 3.5 3.0 2.5 2.0 1.5 1.0 PPM 0.5 Degree of Unsaturation: Proposed Structure: 60 50 ppm 6ECE
Transcribed Image Text:2. C4H100 100 80 80- 48- 20 3500 3000 2500 200a 1520 1808 6H 1H 1H 2H 4.0 3.5 3.0 2.5 2.0 1.5 1.0 PPM 0.5 Degree of Unsaturation: Proposed Structure: 60 50 ppm 6ECE
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.
Expert Solution
Step 1

Degree of unsaturation = C - (H/2) + (N/2) +1

Where, C is number of carbon atoms, H is number of hydrogen and/or halogen atoms, and N is number of nitrogen atoms

IR: Tells us about the functional group

1H NMR: Tells us more details about the different groups

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