chap11a

force constant
HOOKE'S LAW
compress K stretch x
restoring = force
-F = K(x)
x0
x1
m1 K
m2
Molecule as a Hooke's Law device
THE MORSE CURVE APPROXIMATES AN HARMONIC OSCILLATOR
CH2 CH2 CH2 CH2
Formation of each ring or double bond causes the loss of 2H.
Index of Hydrogen Deficiency
CALCULATION METHOD
Determine the expected formula for a noncyclic, saturated compound ( CnH2n+2 ) with the same number of carbon atoms as your compound. Correct the formula for heteroatoms
short
Wavelength (λ)
long
Types of Energy Transitions in Each Region of the Electromagnetic Spectrum
REGION ENERGY TRANSITIONS
X-ray UV/Visible Infrared Microwave Radio Frequency
CH3 CH3
H3C
CH3
CH3
CH2CH3
..... there is still work required to fully solve the problem
INFRARED SPECTROSCOPY
THE ELECTROMAGNETIC SPECTRUM
high high Frequency (ν) Energy low low
(stretching vibrations)
WAVELENGTH (m)
2.5 4 5 5.5 6.1 6.5 15.4
O-H N-H
C-H
C N C C X=C=Y (C,O,N,S)
C=O Very few bands
C-Cl C-O C=C C-N C-C N=O N=O * C=N 650
(NMR)
Bond-breaking Electronic Vibrational Rotational Nuclear and Electronic Spin
Simplified Infrared Spectrophotometer
focusing mirror NaCl plates Detection Electronics and Computer
m1 m2 = m1 + m2
C C > C C >C C
multiple bonds have higher K's
m = atomic masses
This equation describes the vibrations of a bond.
= reduced mass
larger K, higher frequency
C C
C
C
both of these types are "infrared active" ( excited by infrared radiation )
BONDING CURVES AND VIBRATIONS
MORSE CURVES STRETCHING
BOND VIBRATIONAL ENERGY LEVELS
C-H C-X -H,+X
-1
H N
C4H5N
C4H10 = ( CnH2n+2 ) C4H11N add one H for N C4H5 N H6 Index = 6/2 = 3
two double bonds and ring in this example
The index gives the number of double bonds or triple bonds or rings in a molecule
= ( CnH2n+2 )
Index = 4/2 = 2
Two Unsaturations
double bond and ring in this example
Index of Hydrogen Deficiency
CORRECTIONS FOR ATOMS OTHER THAN HYDROGEN
e n e r g y
vibrational energy levels
bond dissociation energy
zero point energy
rmin ravg
rmax
distance
(average bond length)
Typical Infrared Absorption Regions
CH3
O CH3
500
CH3 CH CH2 C
1000
AN INFRARED SPECTRUM
THE UNIT USED ON AN IR SPECTRUM IS WAVENUMBERS ( ν )
ν= ν=
ν ν
wavenumbers (cm-1) 1 λ(cm)
λ νc
= wavelength (cm) c = speed of light c = 3 x 1010 cm/sec
++ ++
e n e r g y
+ +
+
οο
+
MORSE CURVE
zero point energy
rmin ravg
rmax
decreasing distance
(average bond length)
BOND VIBRATIONAL ENERGY LEVELS
Bonds do not have a fixed distance. They vibrate continually even at 0oK (absolute). The frequency for a given bond is a constant. Vibrations are quantized as levels. The lowest level is called the zero point energy.
(explained later)
Subtract the actual formula of your compound The difference in H's divided by 2 is the Index of Hydrogen-deficiency
C5H8
C5H12 C5H8 H4
ν
1 = 2 πc
K
larger atom masses, lower frequency
X-RAY
ULTRAVIOLET
INFRARED MICROWAVE
RADIO
FREQUENCY
Ultraviolet
Visible
Vibrational infrared 2.5 m 15 m
Hale Waihona Puke Nuclear magnetic resonance 1m 5m
200 nm
400 nm BLUE
800 nm RED
Saturated Hydrocarbons CnH2n+2
CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 CH3CH2CH2CH2CH3
CH3 CH3 CH3 CH CH C CH3 CH3 CH3
GENERAL FORMULA
CH4 C2H6 C3H8 C4H10 C5H12
INDEX OF HYDROGEN DEFICIENCY
and
THE BASIC THEORY OF INFRARED SPECTROSCOPY
WHAT CAN YOU LEARN FROM A MOLECULAR FORMULA ?
YOU CAN DETERMINE THE NUMBER OF RINGS AND / OR DOUBLE BONDS.
C9H20
branched compounds also follow the formula
FORMATION OF RINGS AND DOUBLE BONDS
C C H H H H C C H H H2C H2C CH2 CH2 H CH2 CH2 H
-2H
C C
-4H
C C
-2H
H2C H2C
(decreasing)
KETONE
4-Methyl-2-pentanone
C-H < 3000, C=O @ 1715 cm-1
100 100 % T R A N S M I T T A N C E 80 80 60 60 40 40 20 20 0 0 3500 3000 2500 2000 1500 WAVELENGTH (cm-1)
Benzene one ring and the equivalent of three double bonds gives an index of 4
If index = 4, or more, expect a benzene ring
PROBLEM
A hydrocarbon has a molecular formula of C6H8. It will react with hydrogen and a palladium catalyst to give a compound of formula C6H12. Give a possible structure.