C h a p t e r F o u r t e e n: Structure Determination: Mass Spectrometry and Infrared Spectroscopy

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1 C h a p t e r F o u r t e e n: Structure Determination: Mass Spectrometry and Infrared Spectroscopy Cl OH Cl An electron ionization mass spectrum of 2,5-dichlorophenol

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3 CHM 323: Summary of Important Concepts YConcepts for Chapter 14: Structure Determination: Mass Spectrometry and Infrared Spectroscopy I. Different from other types of spectrometry we will study: Mass spectrometry (MS) A. MS uses particles rather than photons to add lots of energy to molecules. This causes then to fall apart. MS measures the masses of the fragments. 1. Usually, particles are fast-moving electrons a. In this case, the type of MS is called electron ionization mass spectrometry (EI MS) 2. These electrons cause the molecule to loose an electron while giving it excess energy, turning it into an energetic cation called the molecular ion, M +C 3. Fate of M +C : a. It's mass can be measured, providing the molecular weight of the compound. If done with great precision, a molecular formula can be determined. This is what MS is most commonly used for in organic chemistry b. If it fragments, it generates daughter ions whose masses can be determined. How an M +C fragments is characteristic of its structure; therefore, analysis of the fragmentation pattern of M +C provides information as to its structure. This requires more knowledge of gas-phase reaction mechanisms than many organic chemists know and so the fragmentation information is not extensively exploited by organic chemists B. Remember, MS weighs individual molecules. So, it requires only tiny amounts of a compound. Also, it is able to determine if isotopes are present in the molecules it weighs II. Infrared Spectroscopy (IR) A. When IR energy is absorbed by a molecule, its chemical bonds vibrate as if they were springs. Vibrations can be 1. Stretches (the length of the bond varies with time) a. symmetric b. asymmetric 2. Bends (the linear bond is deformed into an arc over time) a. also can be described as symmetric or asymmetric 3. The exact energy of an absorption is related to a. the strength of the bond b. the atoms connected through the bond B. Generally, IR spectroscopy is functional group spectroscopy. It tells us what functional groups may be present or absent in a molecule 71

4 1. Three IR regions: a cm -1 is the Functional Group Region i. most functional groups have characteristic absorptions in this region b cm -1 is the Substitution Region i. if double bonds or an aromatic ring is present, absorptions in this region indicate mono, di, tri, tetra, cis, trans, o, m, or p substitution patterns c cm -1 is the Fingerprint Region i. this region is very complex and usually is not examined unless the other two regions are not informative C. Refer to table 21.1 of text or to accompanying table of characteristic IR absorptions when attempting to determine the structure of molecules using IR. You also may find the section Approach to the Analysis of an IR Spectrum on page 95 of the Organic Laboratory Manual useful 72

5 BASIC IR VIBRATIONS Frequency Functional Group and Type of Vibration (cm -1 ) Utility a Intensity b C H Alkanes (stretch) l s CH 3 (bend) 1450 & 1375 l & g m Alkenes (stretch) l m Aromatics (stretch) l s Alkynes (stretch) 3300 g s Aldehydes g w g w cis-alkenes (bend) g m-s trans-alkenes (bend) g m-s R 2 C4CH 2 (bend) 900 g m-s R 2 C4CHR (bend) g m-s C 6 H 5 R (bend) & g m-s R C 6 H 4 R ortho- (bend) g m-s meta- (bend) , g m-s & para- (bend) g m-s C4C Alkenes l m-w Aromatics 1600 & 1475 l m-w C/C Alkynes g m-w C4O Aldehydes g s Ketones g s Carboxylic Acids g s Esters g s Amides g s Anhydrides 1810 & 1760 g s Acid Chlorides 10 g s C O Alcohols, Ethers, Esters, g s Carboxylic Acids, & Anhydrides O H Alcohols & Phenols g m Carboxylic Acids g m N H Amines & Amides (stretch) g m (bend) g m-s C N Amines l m-s 73

6 BASIC IR VIBRATIONS, CONTINUED Frequency Functional Group and Type of Vibration (cm -1 ) Utility a Intensity b C4N Imines & Oximes g w-s C/N Nitriles g m X4C4Y Allenes, Ketenes, Isocyanates, g m-s & Isothiocyanates N4O Nitro groups 1550 &1350 g s S4O Sulfoxides, Sulfones, Sulfonyl Chlorides, Sulfates, & Sulfonamides & g s C X Fluorides l s Chlorides l s Bromides & Iodides <667 l s a l = of little use; generally not reliable in determining the presence of a functional group. g = good or useful; generally very reliable in determining the presence of a functional group. b w = weak; m = medium; s = strong 74

7 1. The most prominent cationic fragment generated in the mass spectra of nearly all alkylbenzenes occurs at m/z 91, as illustrated below for 1-phenyloctane. Propose a structure for this cation. How does it form and why is it so abundant? Mass spectrum of 1-phenyloctane 75

8 2. Using the IR spectra below, provide structures for compounds A and B: a. Compound A: C 8 H %Transmittance Wavenumbers (cm-1)

9 b. Comound B: C 6 H %Transmittance Wavenumbers (cm-1)

10 Four of compounds A - F have the IR spectra shown below. Match the correct compound with its IR spectrum. %Transmittance CH A Cl Wavenumbers (cm-1) B 70 %Transmittance Cl C Wavenumbers (cm-1) O D %Transmittance O Wavenumbers (cm-1) E %Transmittance H 3 C CN F Wavenumbers (cm-1)

11 4. 1. Using the IR spectra below, provide structures for compounds C and D: a. Compound C: C 7 H 7 Cl %Transmittance Wavenumbers (cm-1)

12 b. Compound D: C 8 H %Transmittance Wavenumbers (cm-1)

13 5. Consider the following typical E1 reaction: Cl H 2 O CF 3 CH 2 OH ) It is simple to follow the progress of this reaction by IR spectroscopy. The IR spectra of the starting material and the product are shown below. Label each as starting material or product, and briefly explain the reasoning behind your choice. Briefly but clearly, the reason for your choice: 81

14 6. (16 points) Four of compounds A - G have the IR spectra shown below. Match the correct compound with its IR spectrum by placing the compound s letter in each spectrum s blank. OH A B H O C OCH 3 O D N H E O F OH G OCH 3 82

15 7. Decalin, C 10 H 18, has MW = 138. Its mass spectrum is shown at right, and clearly shows the molecular ion with a mass of 138 and a relative intensity of 100. Look closely, though, and you will see that there is an ion appearing at mass 139 with a relative intensity of 10. Briefly but clearly, explain (a) what is responsible for the presence of this mass 139 ion, and (b) why its relative intensity is about 10% of that of the mass of the molecular ion at

16 8. Solvolysis of trans-1-bromo-2-methylcyclopentane in water at 50 EC for 12 hours gives two products, A and B. The infrared and mass speecta of product A are shown below: IR spectrum A Mass spectrum A a. Clearly label two features on the IR spectrum indicating the functional group class of product A. b. Cearly label the molecular ion (M + A) of the mass spectrum. c. The fragment appearing at mass 67 suggests a loss of what functional group from the molecular ion? iv. Provide a structural formula for product A. v. Product B has molecular formula C 6 H 12 O. Provide a structural formula for product B. 84