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Why this Chapter? Finding structures of new molecules synthesized is critical To get a good idea of the range of structural techniques available and how.

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Presentation on theme: "Why this Chapter? Finding structures of new molecules synthesized is critical To get a good idea of the range of structural techniques available and how."— Presentation transcript:

1 Chapter 12 Structure Determination: Mass Spectrometry and Infrared Spectroscopy

2 Why this Chapter? Finding structures of new molecules synthesized is critical To get a good idea of the range of structural techniques available and how they should be used

3 12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments
Measures molecular weight Sample vaporized and subjected to bombardment by electrons that remove an electron Creates a cation radical Bonds in cation radicals begin to break (fragment) Charge to mass ratio is measured

4 The Mass Spectrum Plot mass of ions (m/z) (x-axis) versus the intensity of the signal (roughly corresponding to the number of ions) (y-axis) Tallest peak is base peak (100%) Other peaks listed as the % of that peak Peak that corresponds to the unfragmented radical cation is parent peak or molecular ion (M+)

5 Other Mass Spectral Features
If parent ion not present due to electron bombardment causing breakdown, “softer” methods such as chemical ionization are used Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample (M+1) from 13C that is randomly present

6 Interpreting Mass-Spectral Fragmentation Patterns
The way molecular ions break down can produce characteristic fragments that help in identification Serves as a “fingerprint” for comparison with known materials in analysis (used in forensics) Positive charge goes to fragments that best can stabilize it

7 Mass Spectral Fragmentation of Hexane
Hexane (m/z = 86 for parent) has peaks at m/z = 71, 57, 43, 29

8 12.3 Mass Spectrometry of Some Common Functional Groups
Alcohols: Alcohols undergo -cleavage (at the bond next to the C-OH) as well as loss of H-OH to give C=C

9 Mass Spectral Cleavage of Amines
Amines undergo -cleavage, generating radicals

10 Fragmentation of Carbonyl Compounds
A C-H that is three atoms away leads to an internal transfer of a proton to the C=O, called the McLafferty rearrangement Carbonyl compounds can also undergo  cleavage

11 12.5 Spectroscopy and the Electromagnetic Spectrum
Radiant energy is proportional to its frequency (cycles/s = Hz) as a wave (Amplitude is its height) Different types are classified by frequency or wavelength ranges

12 Absorption Spectra An organic compound exposed to electromagnetic radiation can absorb energy of only certain wavelengths (unit of energy) Transmits energy of other wavelengths. Changing wavelengths to determine which are absorbed and which are transmitted produces an absorption spectrum

13 12.6 Infrared Spectroscopy
IR region lower energy than visible light (below red – produces heating as with a heat lamp) IR energy in a spectrum is usually measured as wavenumber (cm-1), the inverse of wavelength and proportional to frequency Specific IR absorbed by an organic molecule is related to its structure

14 Infrared Energy Modes IR energy absorption corresponds to specific modes, corresponding to combinations of atomic movements, such as bending and stretching of bonds between groups of atoms called “normal modes” Corresponds to vibrations and rotations

15 12.7 Interpreting Infrared Spectra
Most functional groups absorb at about the same energy and intensity independent of the molecule they are in IR spectrum has lower energy region characteristic of molecule as a whole (“fingerprint” region)

16 Figure 12.14

17 Regions of the Infrared Spectrum
cm-1 N-H, C-H, O-H (stretching) N-H, O-H 3000 C-H cm-1 CºC and C º N (stretching) cm-1 double bonds (stretching) C=O C=C cm-1 Below 1500 cm-1 “fingerprint” region

18 Differences in Infrared Absorptions
Bond stretching dominates higher energy modes Light objects connected to heavy objects vibrate fastest: C–H, N–H, O–H For two heavy atoms, stronger bond requires more energy: C º C, C º N > C=C, C=O, C=N > C–C, C–O, C–N, C–halogen

19 12.8 Infrared Spectra of Some Common Functional Groups
Alkanes, Alkenes, Alkynes C-H, C-C, C=C, C º C have characteristic peaks absence helps rule out C=C or C º C

20 IR: Aromatic Compounds
Weak C–H stretch at 3030 cm1 Weak absorptions cm1 range Medium-intensity absorptions 1450 to 1600 cm1 See spectrum of phenylacetylene, Figure 12.15

21 IR: Alcohols and Amines
O–H 3400 to 3650 cm1 Usually broad and intense N–H 3300 to 3500 cm1 Sharper and less intense than an O–H

22 IR: Carbonyl Compounds
Strong, sharp C=O peak 1670 to 1780 cm1 Exact absorption characteristic of type of carbonyl compound 1730 cm1 in saturated aldehydes 1705 cm1 in aldehydes next to double bond or aromatic ring

23 C=O in Ketones C=O in Esters 1735 cm1 in saturated esters
1715 cm1 in six-membered ring and acyclic ketones 1750 cm1 in 5-membered ring ketones 1690 cm1 in ketones next to a double bond or an aromatic ring C=O in Esters 1735 cm1 in saturated esters 1715 cm1 in esters next to aromatic ring or a double bond

24 Let’s Work a Problem Propose structures for a compound that fits the following data: It is an alcohol with M+ = 88 and fragments at m/z = 73, m/z = 70, and m/z = 59

25 Answer Answer: We must first decide on the the formula of an alcohol that could undergo this type of fragmentation via mass spectrometry. We know that an alcohol possesses an O atom (MW=16), so that leads us to the formula C5H12O for an alcohol with M+ = 88, with a structure of: One fragmentation peak at 70 is due to the loss of water, and alpha cleavage can result in m/z of 73 and 59.


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