CHEM 211-2009
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Experiment 7
Part A
How Can We
Determine the
Structure of an
Organic Compound?
A. Pre-lab Preparation Assignment:
1. Read: This handout
2. Read: Lab Manual Part V. Appendix D1.
Padías pp. 98-102
3. Read the Mass Spectrometry Background Introduction to Mass Spectrometry links on the
Experiment 7A page of the course website.
4. Work Through: The Mass Spectrometry and Structural Analysis activity on the next two
pages and be ready to respond to the questions at the bottom of the last page.
5. Complete the Prelab Questions linked to the Experiment 7A page on the course website by
7:00 PM, Monday, November 2.
6. Consider possible claims for the QOW (See below)
7. In your lab notebook: (See Lab Manual pp. 19-21 for format.)
Enter the title of the Experiment, group mate names and the Question of the Week.
8. Bring your notebook and claims to the Tuesday AM discussion section on November 3.
B. Objectives:
1. To gain experience with collecting mass spectra of solid compounds.
2. To gain expertise with analyzing IR and mass spectral data to determine the structure of
organic compounds.
C. Question of the Day:
How Do Organic Compounds Interact with High Energy Electrons and What
Structural Information Can We Obtain from the Results?
D. Key Terms/concepts/techniques:
 Mass spectrometry
 Base Peak
 Molecular ion
E. Procedure:
1. You will work as a team to obtain the mass spectrum of your group unknown. Individual
instructions will be provided.
2. You will work with your team to analyze the IR and mass spectra of your unknown. (In-Lab
activities will be distributed).
CHEM 211 Experiment 7A
2
Mass Spectrometry
Mass Spectrometry and Structural Analysis
1. The Mass Selective Detector (MSD).
When a neutral substance enters that mass spectrometer, it is bombarded with an electron, which
causes the substance to ionize as illustrated in Equation 1.
[M] +. + 2 ee + M
Equation 1: Electron Impact Ionization
The species M is both a cation (has a positive charge) and a free radical (has an unpaired
electron) and is referred to as a radical cation. The mass of this first formed radical cation is
equal to that of the original molecule entering the mass spectrometer and is referred to as the
molecular ion.
The molecular ion, because it retains considerable energy from the initial collision, will tend to
break down (fragment) into smaller fragments. These fragmentations usually result from
breakage of one bond dividing the original structure into two pieces one having a positive charge
(a cation) and the other containing the unpaired electron (a free radical). The mass spectrometer
detects only the positive ions and not the free radicals.
2. Characteristics of Mass Spectra.
For our purposes in this activity it is important to know the following:
• The mass spectrometer detects only positive ions and separates them on the according to their
mass to charge ratio, m/z.
• Most ions produced on ionization in the electron beam are uni-positive so numerically,
m/z = mass.
• The mass of the molecular ion can usually be taken as the molecular weight of the substance.
• The mass spectrometer can separate ions that have different isotopes of their atoms.
• The most intense peak in the spectrum is called the base peak and represents the ion of
highest abundance and therefore the highest intensity.
• Knowledge of the isotope composition of the more common elements is essential to the
interpretation of mass spectral data. See Figure 1
Figure 1: Table of Isotope Composition of Some Common Elements
Mass Spectrometry
3
CHEM 211 Experiment 7A
Detection of Halogens in Compounds from Molecular Ions:
•• The presence of bromine in a compound leads to two molecular ions (with 35Br79 or with 35Br81)
differing by 2 mass units. For a molecule with one bromine atom, what should be the relative
intensities of these two molecular ions?
•• The presence of chlorine in a compound leads to two molecular ions (with 17Cl35 or with 17Cl37)
differing by 2 mass units. For a molecule with one chlorine atom, what should be the relative
intensities of these two molecular ions?
Additional Molecular Ion Considerations.
As we learned in developing our methods for calculating double bond equivalents:
•• The general formula for a hydrocarbon is CnH2n+2. Considering the mass of carbon and
hydrogen, predict if the molecular ion of a hydrocarbon should be an even or odd number.
•• A molecule with one oxygen atom adds a total mass of 16 to the mass of a corresponding the
hydrocarbon. That makes the molecular ion of a compound containing an O an (even or odd)
number (choose one).
•• A molecule with one nitrogen atom also has an additional hydrogen atom compared to a
corresponding the hydrocarbon, a total mass of 15 (14 + 1). That makes the molecular ion of a
compound containing an N an (even or odd) number (choose one).