HM 233 b 1: NMR DRY LAB Directions: In each of these problems you are given vari

Question

HM 233 b 1: NMR DRY LAB Directions: In each of these problems you are given various spectra that may include IR, 'H NMR, 13C NMR, and molecular formula. The best approach for spectroscopy problems is the following steps: Using this information, your task is to determine the structure of the compound. 1. From the molecular formula, calculate the degree of unsaturation to limit the number of possible structures. Remember, each degree of unsaturation is a ring or pi bond (likely an alkene or carbonyl). An alkyne has two degrees of unsaturation (2 pi bonds), and an aromatic ring has four (3 pi bonds plus a ring.) Although there's no guarantee, if your structure has more than four degrees of unsaturation it's quite likely to have an aromatic ring 0 2. If applicable, look at the IR absorption bands at wavenumbers above 1500 cm-1 to determine what functional groups are likely in the compound. Remember that these functional groups must be consistent with the degree of unsaturation 8 2 3. Look at the NMR spectra to determine the connectivity of the compound. If you can't figure out the entire structure at once, it helps to come up with fragments of the molecule that you can stick together into larger and larger groups until you have the entire structure. When applicable, specific peaks of the spectra have been zoomed into in order to help distinguish different peaks. This lab activity as a whole is worth 20 points. Each problem is worth up to four points. 3 points for showing work done on interpreting the spectra, and 1 point for having the correct structure. You are able to utililize tables from your lecture and lab textbooks to help with the interpretation of the spectra. There are two bonus problems at the end of the activity that can be done for up to 8 extra points. Approach this dry lab as a puzzle it can be fun!

Explanation / Answer

1) Degree of unsaturation formula-

DOU=(2C+2+N-H-X)/2=(2*4+2+0-7-1)/2=1

C=number of carbons, N=number of nitrogen atoms,H=N=number of hydrogen atoms,X=N=number of halogen atoms

2) Analysis of 1H NMR:

2.85-2.9 ppm ,tripet

--(this proton group/proton has n=2 neighboring protons as it is split into n+1=2+1=3 peaks)

-- CH2-CH2-CO-

3.5-3.6ppm ,triplet ((this proton group/proton has n=2 neighboring protons as it is split into n+1=2+1=3 peaks)

-CH2-CH2-Br

3.7 ppm, singlet

-R-COOCH3 [ester protons]

-((this proton group/proton has n=0 neighboring protons as has 0+1=1 peak only)

Analysis of 13C NMR:

chemical shift value: [20-30ppm shift for alkyl group carbon R2CH2- or secondary alkyl]

35-40ppm -CH2-X [where X is halogen ,Br]

50-55ppm CH3CO- [shift for carbon in bold]

75-80ppm (triplet) [attached to two protons as it is split into 2+1=n+1 =3 peaks due to carbon and proton coupling]

171 ppm (C=O in esters or acids)

IR absorption band @ 1500 cm-1

C=O group present

Structure of the compound:

Br-CH2-CH2-COOCH3

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