INDEX OF HYDROGEN DEFICIENCY

1
INDEX OF HYDROGEN DEFICIENCY
and
THE BASIC THEORY OF INFRARED SPECTROSCOPY
2
WHAT CAN YOU LEARN FROM A MOLECULAR
FORMULA ?
YOU CAN DETERMINE THE NUMBER OF RINGS AND /
OR DOUBLE BONDS.
3
Saturated Hydrocarbons
CnH2n2
GENERAL FORMULA
CH4
C2H6
C3H8
C4H10
C5H12
C9H20
branched compounds also follow the formula
4
FORMATION OF RINGS AND DOUBLE BONDS
-2H
-4H
-2H
Formation of each ring or double bond causes the
loss of 2H.
5
Index of Hydrogen Deficiency
CALCULATION METHOD

• Determine the expected formula for a noncyclic,
  saturated compound ( CnH2n2 ) with the same
• number of carbon atoms as your compound.
• Correct the formula for heteroatoms
• Subtract the actual formula of your compound
• The difference in Hs divided by 2 is the

(explained later)
Index of Hydrogen-deficiency
6
C5H8
C5H12
( CnH2n2 )
C5H8
H4
Index 4/2 2
Two Unsaturations
double bond and ring in this example
7
Index of Hydrogen Deficiency
CORRECTIONS FOR ATOMS OTHER THAN HYDROGEN

• O or S -- doesnt change H in calculated
  formula
• N or P -- add one H to the calculated formula
• F, Cl, Br, I -- subtract one H from calculated
• formula

0
C-H
C-O-H
O
1
C-H
C-NH2
N,H
-1
C-H
C-X
-H,X
8
C4H5N
C4H10
( CnH2n2 )
C4H11N
add one H for N
C4H5 N
H6
Index 6/2 3
two double bonds and ring in this example
9
The index gives the number of

• double bonds or
• triple bonds or
• rings in a molecule

one ring and the equivalent of three double bonds
gives an index of 4
Benzene
If index 4, or more, expect a
benzene ring
10
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.
11
A FEW POSSIBLE ANSWERS
..... there is still work required to fully solve
the problem
12
INFRARED SPECTROSCOPY
13
THE ELECTROMAGNETIC SPECTRUM
Frequency (n)
low
high
high
low
Energy
MICRO- WAVE
X-RAY
ULTRAVIOLET
INFRARED
RADIO
FREQUENCY
Nuclear magnetic resonance
Vibrational infrared
Ultraviolet
Visible
2.5 mm
15 mm
1 m
5 m
200 nm
400 nm
800 nm
BLUE
RED
Wavelength (l)
short
long
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Types of Energy Transitions in Each Region
of the Electromagnetic Spectrum
REGION
ENERGY TRANSITIONS
X-ray
Bond-breaking
UV/Visible
Electronic
Infrared
Vibrational
Microwave
Rotational
Radio Frequency
Nuclear and
(NMR)
Electronic Spin
15
Simplified Infrared Spectrophotometer
NaCl plates
focusing mirror
Detection Electronics and Computer
Determines Frequencies of Infrared Absorbed
and plots them on a chart
Infrared Source
Sample
intensity of absorption
Absorption peaks
frequency
(decreasing)
16
4-Methyl-2-pentanoneC-H lt 3000, CO _at_ 1715 cm-1
KETONE
AN INFRARED SPECTRUM
17
THE UNIT USED ON AN IR SPECTRUM IS
WAVENUMBERS ( n )
n wavenumbers (cm-1)
1
n
l wavelength (cm)

l
(cm)
c speed of light
n frequency nc
c 3 x 1010 cm/sec
or
( )
1
cm/sec
1
c
n
c


l
l
cm
sec
wavenumbers are directly proportional to frequency
18
Molecular vibrations
Two major types
STRETCHING
C C
C
BENDING
C
C
both of these types are infrared active
( excited by infrared radiation )
19
BONDING CURVES AND VIBRATIONS
MORSE CURVES
STRETCHING
20
BOND VIBRATIONAL ENERGY LEVELS
e n e r g y
MORSE CURVE
zero point energy
rmin
rmax
decreasing distance
ravg
(average bond length)
21
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.
e n e r g y
bond dissociation energy
vibrational energy levels
zero point energy
rmin
rmax
distance
ravg
(average bond length)
22
Typical Infrared Absorption Regions
(stretching vibrations)
WAVELENGTH (mm)
C-Cl
CO
CN
O-H
C-H
C N
Very few bands
C-O
CC
N-H
C C
C-N
XCY
C-C

NO NO
(C,O,N,S)
FREQUENCY (cm-1)
nitro has two bands
23
HARMONIC OSCILLATOR
MATHEMATICAL DESCRIPTION
OF THE VIBRATION IN A BOND
. assumes a bond is like a spring
24
HOOKES LAW
force constant
compress
K
stretch
Dx
x0
x1
restoring force
-F K(Dx)

Molecule as a Hookes Law device
m1
m2
K
25
THE MORSE CURVE APPROXIMATES AN HARMONIC
OSCILLATOR
HOOKES LAW
Harmonic Oscillator
ACTUAL MOLECULE
Morse Curve (anharmonic)
Using Hookes Law and the Simple Harmonic
Oscillator approximation, the following equation
can be derived to describe the motion of a bond..
26
n

THE EQUATION OF A
frequency in cm-1
SIMPLE HARMONIC OSCILLATOR
c velocity of light
( 3 x 1010 cm/sec )
K force constant in dynes/cm
where
gt
gt
multiple bonds have higher Ks
m atomic masses
This equation describes the vibrations of a bond.
m reduced mass
27
larger K, higher frequency
larger atom masses, lower frequency
increasing K
constants
1650
2150
1200
increasing m
C-H gt C-C gt C-O gt C-Cl gt C-Br
3000 1200 1100 750
650
28
DIPOLE MOMENTS
29
DIPOLE MOMENTS
Only bonds which have significant dipole moments
will absorb infrared radiation.
Bonds which do not absorb infrared include

• Symmetrically substituted alkenes and alkynes

• Many types of C-C Bonds

• Symmetric diatomic molecules

H-H Cl-Cl
30
STRONG ABSORBERS
d-
The carbonyl group is one of the strongest
absorbers
d
Also O-H and C-O bonds
infrared beam
31
RAMAN SPECTROSCOPY
Another kind of vibrational spectroscopy that can
detect symmetric bonds.
Infrared spectroscopy and Raman
spectroscopy complement each other.
32
RAMAN SPECTROSCOPY
In this technique the molecule is irradiated with
strong ultraviolet light at the same time that
the infrared spectrum is determined.
Ultraviolet light promotes electrons from
bonding orbitals into antibonding orbitals. This
causes formation of a dipole in groups that were
formerly IR inactive and they will absorb
infrared radiation.
p
.
induced dipole
d
d-
.
hn
..
UV
p p
absorbs IR
transition
no dipole symmetric
.. we will not talk further about this technique
33
SUGGESTED SOFTWARE
34
IR TUTOR

• Select ChemApps folder
• Select Spectroscopy icon
• Select IR Tutor icon

IR TUTOR ACTUALLY ILLUSTRATES INFRARED
VIBRATIONS AND THEORY WITH ANIMATIONS