How to Quickly Solve Spectrometry Problems

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Kory Snow
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1 How to Quickly Solve Spectrometry Problems You should be looking for: Mass Spectrometry (MS) Chemical Formula DBE Infrared Spectroscopy (IR) Important Functional Groups o Alcohol O-H o Carboxylic Acid O = H o Carbonyl Group C = O o Nitrogen groups: Amine N-H, Amide N-H DBE Analysis o Vinyl Proton Nuclear Magnetic Resonance (H-NMR) Number of Hydrogens Benzene Ring Singlets with 1H Alkyl Chains Chemical Shift Trends Carbon Nuclear Magnetic Resonance (C-NMR) Chemical Shift Trends (Double check) This tutorial is meant to streamline the process by cutting out redundancies and saving time. Do not think of this as an algorithm but as second nature. These strategies are what I noticed when I was completing the practice problems. While this is less useful in a more advanced spectroscopy/ spectrometry course, they certainly will help shave off some time for exam and practice problems for this class s purposes. Chemical Formula MS: Find number of Carbons. MS: Find number of Chlorine, Bromine, Iodine, and Fluorine. H-NMR: Find number of hydrogen. Add up all integrals. If you are unsure whether that is the right number of hydrogen, remember that there is usually only one or two possibilities that it follows the H-Rule. MS: Subtract the total mass of lowest mass isotopes of carbon, hydrogen, and halogens from M m/z to get a leftover m/z number. If it is divisible by 16, then all of the leftovers are oxygen. If it is divisible by 14, then all of the leftovers are nitrogen. If it is divisible by neither 14 nor 16, then it has both nitrogen and oxygen. Then you have to figure it out the old fashioned way. Don t forget the N-rule. (Even M m/z means even number of nitrogen and vice versa) DBE DBE = #C - #H/2 + #N/2 +1 IR Important Functional Groups IR: If you have nitrogen, check amine N-H and amide N-H, and Nitrile C---H IR: Three functional groups that you should be immediately be able to see are Alcohol O-H, Carboxylic Acid O = H and Carbonyl Group C = O. The stretching frequency of the Carbonyl group will tell you which functional group is probably there.

Chemical Shift Trends Carbon Nuclear Magnetic Resonance (C-NMR) Chemical Shift Trends (Double check) This tutorial is meant to streamline the process by cutting out redundancies and saving time.

2 We are saving vinyl for later. Why aren t we looking for other functional groups? Aryl Most benzene groups have them. Alkane Most compounds have them. Alkene DBE s are sufficient. Alkyne DBE s are sufficient Benzene Ring H-NMR s are more decisive and easier to spot. H-NMR If there is a multiplet with ppm greater than 6.5, then there most likely is a benzene. If there is a singlet with integral 1, then there probably is an alcohol (1-5ppm), aldehyde (9.5-11ppm), or carboxylic acid (10-13ppm) group. Recognize alkyl chains Integral 2 (Triplet) CH 2 CH 2 Integral 2 (Pentet) CH 2 CH 2 CH 2 = ~CH 2 CH 2 CH 2 CH 3 Integral 2 (Sextet) CH 2 CH 2 CH 3 Integral 3 (Triplet) CH 2 CH 3 Recognize Chemical Shift Trends Chemical shift decreases in both H-NMR and C-NMR going away from a relatively electronegative atom, such as Oxygen or Nitrogen. Finally, piece together the molecule The greatest weapons that you have to piece together the molecule are the unused carbons (singlets in the C-NMR) and Oxygens. They can be used to connect multiple alkyl chains as ethers or forks. If you have leftover DBE s, check for vinyl in the IR. If there is no evidence for vinyl, there probably is a ring. Use C-NMR to double check your answer. Unless they are equivalent, every carbon should have a signal. Try to label each carbon with a signal and verify if it is consistent with the chemical shift trends we were talking about. Now, let s see these strategies in action!

If there is a singlet with integral 1, then there probably is an alcohol (1-5ppm), aldehyde (9.5-11ppm), or carboxylic acid (10-13ppm) group.

3 Mass spectrum: m/z = 226 (M; 100%), m/z = 227 (13.4%), and m/z = 228 (32.5%). No fluorine or iodine. 1 H-NMR: ppm (multiplet; Integral=5), 4.82 ppm (triplet; Integral=1), 3.56 ppm (triplet; Integral=2), 2.72 ppm (triplet; Integral=2), 1.74 ppm (pentet, Integral=2), 0.90 ppm (triplet; Integral=3) 13 C-NMR: ppm (singlet), ppm (singlet), ppm (doublet), ppm (doublet), ppm (doublet), 99.4 ppm (doublet), 60.8 ppm (triplet), 40.5 ppm (triplet), 31.3 ppm (triplet), and 5.3 ppm (quartet). How many carbons? 13.4/1.1 so approximately 12. How many hydrogens? Sum of Integrals: 15, so multiples of 15. Could there be 30 hydrogens? No, breaks H-Rule Are there any chlorines? Yes, one. How many nitrogen or oxygen? m/z leftover: C 35 1Cl 15 15H = 32, which is divisible by 16, so 2 oxygen and no nitrogen. Chemical Formula? C 12 H 15 O 2 Cl DBE= = 5, which is used up by benzene and carbonyl. IR Functional Groups:

1 ppm (doublet), 128.8 ppm (doublet), 128.6 ppm (doublet), 99.4 ppm (doublet), 60.8 ppm (triplet), 40.5 ppm (triplet), 31.3 ppm (triplet), and 5.3 ppm (quartet). How many carbons? 13.4/1.

4 Can see a carbonyl group at 1687cm -1, but cannot be an amide. So we are unsure which functional group it is. Is there a benzene ring? Yes, ppm (multiplet; Integral=5) Are there any singlets with integral=1? No, so no carboxylic acid, aldehyde, or alcohol. Carbonyl must be a ketone that is conjugated. Build Alkyl Chains 4.82 ppm (triplet; Integral=1) CHCH ppm (triplet; Integral=2) Cl CH 2 CH 2 (3.5ppm usually have halogens attached) 2.72 ppm (triplet; Integral=2) CH 2 CH ppm (pentet, Integral=2) CHCH 2 CH ppm (triplet; Integral=3) CH 2 CH 3 Unused oxygen becomes ether. Put all parts down. ClCH 2 CH 2~ ~CHCH 2 CH 3 There are only two structures consistent with the C-NMR and H-NMR shifts.

Build Alkyl Chains 4.82 ppm (triplet; Integral=1) CHCH 2 3.56 ppm (triplet; Integral=2) Cl CH 2 CH 2 (3.5ppm usually have halogens attached) 2.

5 13 C-NMR: ppm (singlet), ppm (singlet), ppm (doublet), ppm (doublet), ppm (doublet), 99.4 ppm (doublet), 60.8 ppm (triplet), 40.5 ppm (triplet), 31.3 ppm (triplet), and 5.3 ppm (quartet). Benzene Ketone O-CHCH 2 CH 3 O-CH 2 CH 2 -Cl ppm (singlet) ppm (singlet) 99.4 ppm (doublet) 60.8 ppm (triplet) ppm (doublet) 40.5 ppm (triplet) 31.3 ppm (triplet) ppm (doublet) 5.3 ppm (quartet) ppm (doublet) Sample Practice Problem from Chemistry 14C Lecture 2 Spring 2011 Exam 2 by Dr. Steve Hardinger

7 ppm (singlet) 198.1 ppm (singlet) 99.4 ppm (doublet) 60.8 ppm (triplet) 133.1 ppm (doublet) 40.5 ppm (triplet) 31.3 ppm (triplet) 128.

For example: (Example is from page 50 of the Thinkbook)

For example: (Example is from page 50 of the Thinkbook) SOLVING COMBINED SPECTROSCOPY PROBLEMS: Lecture Supplement: page 50-53 in Thinkbook CFQ s and PP s: page 216 241 in Thinkbook Introduction: The structure of an unknown molecule can be determined using