ORGANIC - EGE 5E CH UV AND INFRARED MASS SPECTROMETRY

Transcription

1 !!

2 CONCEPT: IR SPECTROSCOPY- FREQUENCIES There are specific absorption frequencies in the functional group region that we should be familiar with EXAMPLE: What are the major IR absorptions for each compounds? Page 2

3 PRACTICE: Answer each of the following questions based on the images below. O OH O O O H F 3 C O CF 3 A B C D E a) Which compounds show an intense peak ~ 1700 cm -1? b) Which compound shows an intense, broad peak at ~ 3400 cm -1? c) Which compound has a peak at ~1700 cm -1, but no peaks at 2700 cm -1? d) Which compound has no signal beyond the fingerprint region? Page 3

4 PRACTICE: The following compound contains two carbonyl groups. Identify which carbonyl group will exhibit a signal at a lower wavenumber. O O Page 4

5 CONCEPT: IR SPECTROSCOPY- DRAWING HYDROCARBONS Alkanes: Alkenes: Terminal Alkynes: Page 5

6 CONCEPT: IR SPECTROSCOPY- DRAWING ALCOHOLS AND AMINES Alcohols: 1 Amines: 2 Amines: Page 6

7 CONCEPT: IR SPECTROSCOPY- DRAWING SIMPLE CARBONYLS Ketones: Esters: Page 7

8 CONCEPT: IR SPECTROSCOPY- DRAWING COMPLEX CARBONYLS Aldehydes: Carboxylic Acids: Page 8

9 CONCEPT: IR SPECTROSCOPY- DRAWING CONCEALED FUNCTIONAL GROUPS Alkyl Haldies: Ethers: 3 Amines: Page 9

10 PRACTICE: Based on IR data given determine the structure of the unknown. Unknown compound A has molecular formula C4H11N. It shows a peak at 2900 cm -1 and peaks in the fingerprint region. PRACTICE: Based on IR data given determine the structure of the unknown. Unknown compound B has molecular formula C4H11N. It shows a single peak at approximately 3400 cm -1 as well as peaks at 2900 cm -1 and in the fingerprint region. Compound B also possesses a branched alkyl group. PRACTICE: Based on IR data given determine the structure of the unknown. Unknown compound C has molecular formula C6H10O3. It shows peaks at 2900, 1850, 1740 cm -1 and in the fingerprint region. Page 10

11 PRACTICE: Match the following functional group choices with the supplied infrared spectra data A) Ether B) Ketone C) Alcohol D) Alkene E) Nitrile Page 11

12 PRACTICE: Match the following functional group choices with the supplied infrared spectra data. A) Alkyl Halide B) Alkyne C) Carboxylic Acid D) Alkene E) Ketone Page 12

13 PRACTICE: Match the following functional group choices with the supplied infrared spectra data. A) Aldehyde B) Alkane C) Carboxylic Acid D) Ester E) Ether PRACTICE: Match the following functional group choices with the supplied infrared spectra data. A) Ketone B) Alkyne C) Alkene D) Alkyl Halide E) Amine Page 13

14 CONCEPT: STRUCTURE DETERMINATION MOLECULAR SENTENCES The holy grail of this section is structure determination. You may be asked to produce a structure from scratch given only a MF, NMR Spectrum and IR Spectrum. Our goal is to build a strong molecular sentence by gathering clues, then propose drawings. How to build a molecular sentence: 1. Determine IHD. 2. Analyze NMR, IR and splitting patterns, integrations for major clues (i.e.). NMR = 9.1 ppm IR = 1710 cm -1 Triplet/Quartet 9.1 ppm (2H) 3. Calculate 1 H NMR Signal : Carbon Ratio. Ratio < ½ suggests symmetrical, whereas ratio > ½ suggests asymmetrical Never rule out a structure based on symmetry (you may not be able to visualize it) 4. State the number of 1 H NMR signals needed. --- DRAW POSSIBLE STRUCTURES Use a combination of Shifts, Integrations, and Splitting to confirm which structure is correct. EXAMPLE: Build a strong molecular sentence using the following data. MF: C4H6O2 IR: peak at 2950 cm -1 1 H NMR peak at 2700 cm (doublet, 4H) peak at 1720 cm (triplet, 2H) Page 14

15 PRACTICE: Propose a structure for the following compound that fits the following 1 H NMR data: Formula: C3H8O2 1 H NMR: 3.36 δ (6H, singlet) 4.57 δ (2H, singlet) Page 15

16 PRACTICE: Propose a structure for the following compound that fits the following 1 H NMR data: Formula: C2H4O2 1 H NMR: 2.1 δ (singlet, 1.2 cm) 11.5 δ (0.5 cm, D 2 O exchange) Page 16

17 PRACTICE: Propose a structure for the following compound that fits the following 1 H NMR data: Formula: C10H14 1 H NMR: 1.2 ppm (6H, doublet) 2.3 ppm (3H, singlet) 2.9 ppm (1H, septet) 7.0 ppm (4H, doublet) Page 17

18 PRACTICE: Propose a structure for the following compound, C7H12O2 with the given 13 C NMR spectral data: Broadband decoupled 13 C NMR: 19.1, 28.0, 70.5, 129.0, 129.8, δ DEPT-90: 28.0, δ DEPT-135: 19.1 δ ( ), 28.0 ( ), δ ( ), 70.5 δ ( ) & δ ( ) Page 18

19 PRACTICE: Propose a structure for the following compound, C5H10O with the given 13 C NMR spectral data: Fully Broadband decoupled 13 C NMR and DEPT: δ ( ); 55.0 δ ( ); 21.0 δ ( ) & 11.0 δ ( ). Page 19

20 PRACTICE: Provide the structure of the unknown compound from the given information. Formula: C4H10O IR: cm -1 1 H NMR: 0.9 ppm (6H, doublet) 1.8 ppm (1H, nonatet) 2.4 ppm (1H, singlet) 3.3 ppm (2H, doublet) Page 20

21 PRACTICE: Provide the structure of the unknown compound from the given information. Formula: C4H9N IR: 2950 cm -1, 3400 cm -1 1 H NMR: 1.0 ppm (4H, triplet) 2.1 ppm (4H, triplet) 3.2 ppm (1H, singlet) Page 21

22 CONCEPT: MASS SPECT- INTRODUCTION Mass Spectrometry is usually accomplished through a technique called electron impact ionization (EI) Electrons are beamed at molecules, generating high energy intermediates called radical cations This is known as the molecular ion or as the parent ion. Only fragment cations are deflected by the magnetic field, the smaller ones more than the bigger ones. Detects the mass-to-charge ratio, which means that it detects the MW of cationic fragments Page 22

23 CONCEPT: MASS SPECT- FRAGMENTATION Ionization Potentials: Some electrons require less energy to ionize than others. Simple Fragmentation Mechanisms: The molecular ion will often fragment into smaller, sometimes more stable ion fragments. The stability of the cation fragment usually determines the relative amounts of fragments observed Radicals tend to form on the less stable side of the fragment Common Splitting Fragments: EXAMPLE: Fragmentation of Butane Page 23

24 PRACTICE: Draw the most likely ion fragment for the following molecules a. b. PRACTICE: Propose the molecular ion and likely fragmentation mechanism for the following molecule. What would be the value of the base peak? Page 24

25 CONCEPT: MASS SPECT- COMMON ISOTOPES Isotopes are often visible on a mass spectrum, due to their differing weights. They can be used for structure determination. Understanding the (M + 1) Peak 1.1% of all carbon is found as 13 C, adding a small but distinctive (M + 1) peak proportional in size to the number of carbons. This proportion is fairly consistent, so it gives rise to two helpful equations Understanding the (M + 2) Peak The halogens Cl and Br give distinctive (M + 2) peaks due to their unusual patterns of isotopic abundance 35 Cl = 75.8% and 37 Cl = 24.2%, yielding an approximate 3:1 ratio at (M + 2) 79 Br = 50.7% and 81 Br = 49.3%, yielding an approximate 1:1 ratio at (M + 2) The Nitrogen Rule Unlike carbon, nitrogen forms 3 bonds. We can use this information to determine the number of nitrogens in a molecule. Even or odd molecular weight of parent ions usually indicates and even or odd number of nitrogens present Page 25

26 PRACTICE: Propose the number of carbons for a compound that exhibits the following peak in its mass spectrum: a. (M) + at m/z = 72, relative height = 38.3% of base peak (M+1) + at m/z = 73, relative height = 1.7% of base peak b. Predict the approximate height of the (M + 1) peak for the molecule icosane, molecular formula C20H42. c. Draw the expected isotope pattern that would be observed in the mass spectrum of CH2Br2. In other words, predict the relative heights of the peaks at M, (M + 2), and (M + 4) peaks. Page 26