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Chapter 15

CHM 2211 Chapter Notes - Chapter 15: Ketone, Unpaired Electron, Alkyne


Department
Chemistry
Course Code
CHM 2211
Professor
Mohammed Daoudi
Chapter
15

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INFRARED SPECTROSCOPY AND
MASS SPECTROMETRY
1. INTRODUCTION TO SPECTROSCOPY
Spectroscopy: used for structure determination
A. The Nature of Light
i. Wavelength (lambda): describes the distance between adjacent peaks of an
oscillating field
ii. Frequency(v): describes the number of wavelengths that pass a particular point
in space per unit time
iii. Wavelength and frequency are inversely proportional (v=c/lambda (c= constant
speed of light)
iv. Photons: packets of energy; directly proportional to its frequency (E=hv
(h=planck’s constant=6.66 x ^-34 Jxs)
v. Electromagnetic spectrum: range of all frequencies
2. IR SPECTROSCOPY
A. Vibrational Excitation
i. IR radiation causes vibrational excitation of the bonds in a molecule; different
types due to the different ways bonds store vibrational energy
1. Stretching and beinding vibrations
B. Identification of Functional Groups with IR Spectroscopy
i. Energy gap for a CH bond is much larger than the energy gap for a CO bond
1. CH will absorb a higher energy in photons
C. The General Shape of an IR Absorbance Spectrum
i. Absorption spectrum: a plot that is created when an IR spectrometer measures
the percent transmittance as a function of frequency
1. All signals point down
2. Location of the signal on the spectrum can be specified by corresponding
wavelengths or frequency of radation that was absorbed
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ii. Wave number: frequency of light/speed of light (a constant); measured in
inverse centimeters
iii. Every signal in an IR spectrum has three characteristics: wavenumber, intensity,
and shape
3. SIGNAL CHARACTERISTICS: WAVENUMBER
A. Hooke’s Law
i. Wavenumber of absorption is associated with bond stretching is dependent on
bond strength and the masses of the atoms sharing the bond
ii. Hooke’s law:enables us to approximate the frequency of vibration for a bond
between two atoms of mas m1 and m2 (see book for equation)
1. From the equation we see that smaller atoms give bonds that vibrate at
higher frequencies, creating a higher wavenumber
a. Example: CH (3000 cm^-1) > CD (2200 cm^-1) > CO (1100cm^-
1) > CCl (700 cm ^-1)
2. Stronger bonds will vibrate at higher frequencies, creating a higher
wavenumber of absorption
a. Example: C (triple bond) N (2200 cm ^-1) > C (double bond) N (1600
cm ^-1) > CN (1100 cm ^-1)
iii. View figure 15.6
iv. There are two main regions of the IR spectra
1. Diagnostic region: fewer peaks and provides the clearest information;
contains all signals from double, triple, and xH bonds
2. Fingerprint region: contains signals from vibrational excitation of most
single bonds; more difficult to analyze
3. View figure 15.7
B. Effect of Hybridization States on Wavenumber of Absorption
i. Wavenumber of carbon is dependent on hypbridization states
1. Example: CH carbon
a. Sp^3 (2900 cm^-1) < sp^2 (3100 cm^-1) < sp (3300 cm ^-1)
ii. To compare the spectra of an alkane, alkene, and alkyne:
1. Draw a line at 3000 cm ^-1
2. View figure 15.10
C. Effect of Resonance on Wavenumber of Absorption
i. An unsaturated, conjugated ketone produces a signal with a lower wavenumber
compared to a carbonyl group of a saturated ketone
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