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9. C10H140 108- 82 40- 28- 3000 2520 2008 1500 1800 9H 2H 2H 1H Í PPM Degree of Unsaturation: Proposed Structure: 200 180 160 140 120 100 80 60 40 20 CDS-00-331 ppm S esac

9. C10H140 108- 82 40- 28- 3000 2520 2008 1500 1800 9H 2H 2H 1H Í PPM Degree of Unsaturation: Proposed Structure: 200 180 160 140 120 100 80 60 40 20 CDS-00-331 ppm S esac

Chemistry
Chemistry
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ISBN:9781305957404
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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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9. C10H140 108- 82 40- 28- 400a 35ag 2520 2000 1808 9H 2H 2H 1H 6. Í PPM Degree of Unsaturation: Proposed Structure: 200 180 160 140 120 100 80 60 40 20 CDS-00-331 ppm 1450
Transcribed Image Text:9. C10H140 108- 82 40- 28- 400a 35ag 2520 2000 1808 9H 2H 2H 1H 6. Í PPM Degree of Unsaturation: Proposed Structure: 200 180 160 140 120 100 80 60 40 20 CDS-00-331 ppm 1450
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

Given: 

Molecular formula = C10H14O 

From HNMR spectra: 

Types of Hydrogens= 4 types 

From CNMR : 

Types of carbons= 6 types 

From IR spectra: 

3258 cm-1 : broad peak 

2983 cm-1 , 1514 cm-1 strong peaks 

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