INDEX OF HYDROGEN DEFICIENCY THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPY and.
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Transcript of INDEX OF HYDROGEN DEFICIENCY THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPY and.
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INDEX OF HYDROGEN DEFICIENCYINDEX OF HYDROGEN DEFICIENCY
THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPYINFRARED SPECTROSCOPY
and
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WHAT CAN YOU LEARN FROM A WHAT CAN YOU LEARN FROM A MOLECULAR FORMULA ?MOLECULAR FORMULA ?
YOU CAN DETERMINE THE NUMBER OF RINGS AND / OR DOUBLE BONDS.
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Saturated HydrocarbonsSaturated Hydrocarbons
CH4
CH3CH3
CH3CH2CH3
CH3CH2CH2CH3
CH3CH2CH2CH2CH3
CH4
C5H12
C3H8
C4H10
C2H6
CH3 CH
CH3
CH
CH3
C CH3
CH3
CH3
C9H20
CCnnHH2n+22n+2 GENERAL FORMULA
branched compoundsalso follow the formula
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C C
H H
C C
C CC C
H H
HH
CH2H2C
H2CCH2
CH2 H
HCH2
CH2H2C
H2CCH2
CH2
CH2
-2H
-4H
-2H
FORMATION OF RINGS AND DOUBLE BONDSFORMATION OF RINGS AND DOUBLE BONDS
Formation of each ring or double bond causes the loss of 2H.
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• 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
• Subtract the actual formula of your compound
• The difference in H’s divided by 2 is the
Index of Hydrogen DeficiencyIndex of Hydrogen DeficiencyCALCULATION METHOD
(explained later)
Index of Hydrogen-deficiency
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C5H8
C5H12
C5H8
H4 Index = 4/2 = 2
double bond andring in this example
= ( CnH2n+2 )
Two Unsaturations
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• O or S -- doesn’t 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
+1
-1
Index of Hydrogen Deficiency
C-H C-X
C-H C-O-H
C-H C-NH2+N,+H
-H,+X
+O
CORRECTIONS FOR ATOMS OTHER THAN HYDROGEN
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C4H5N
C4H10
C4H11N
H6 Index = 6/2 = 3
two double bonds andring in this example
= ( CnH2n+2 )
N
H
add one H for N
C4H5 N
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The index gives the number of
one ring and theequivalent of threedouble bonds givesan index of 4
If index = 4, or more, expect a benzene ring
• double bonds or • triple bonds or • rings in a molecule
Benzene
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PROBLEM A hydrocarbon has a molecularformula of C6H8. It will react with hydrogen and a palladium catalystto give a compound of formulaC6H12. Give a possible structure.
INDEX C6H14
-C6H8
H6Index = 6/2 = 3
C6H8 + 2 H2 C6H12
HYDROGENATIONPd
Hydrogenation shows only two double bonds. Therefore, there must also be a ring.
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CH3
CH3
CH3
CH3
H3C
CH2CH3
A FEW POSSIBLE ANSWERSA FEW POSSIBLE ANSWERS
..... there is still work required to fully solve the problem
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INFRARED SPECTROSCOPYINFRARED SPECTROSCOPY
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lowhigh Frequency ()
Energy
X-RAY ULTRAVIOLET INFRARED MICRO- WAVE
RADIO FREQUENCY
Ultraviolet VisibleVibrationalinfrared
Nuclear magneticresonance
200 nm 400 nm 800 nm
2.5 m 15 m 1 m 5 m
short longWavelength ()
high low
THE ELECTROMAGNETIC SPECTRUMTHE ELECTROMAGNETIC SPECTRUM
BLUE RED
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X-ray
UV/Visible
Infrared
Microwave
Radio Frequency
Bond-breaking
Electronic
Vibrational
Rotational
Nuclear and Electronic Spin
REGION ENERGY TRANSITIONS
Types of Energy Transitions in Each Region Types of Energy Transitions in Each Region of the Electromagnetic Spectrumof the Electromagnetic Spectrum
(NMR)
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Detection Electronics and Computer
InfraredSource
Determines Frequenciesof Infrared Absorbed andplots them on a chart
Sample
Simplified Infrared SpectrophotometerSimplified Infrared SpectrophotometerNaClplates
Absorption “peaks”
Infrared Spectrum
frequency
intensity ofabsorption
(decreasing)
focusingmirror
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4-Methyl-2-pentanoneC-H < 3000, C=O @ 1715 cm-1
KETONE
100
80
60
40
20
0CH3 CH CH2 C CH3
OCH3
3500 3000 2500 2000 1500 1000 500
100
80
60
40
20
0
WAVELENGTH (cm-1)
%
TRANSMITTANCE
AN INFRARED SPECTRUM
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( )
= wavenumbers (cm-1)
= 1
(cm) = wavelength (cm)
THE UNIT USED ON AN IR SPECTRUM ISTHE UNIT USED ON AN IR SPECTRUM IS
= frequency = cc = speed of light
WAVENUMBERS ( WAVENUMBERS ( ))
c = 3 x 1010 cm/sec
wavenumbers are directly proportional to frequency
= =
or
c1 cm/sec
cm=
sec1
c
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Two major types :
STRETCHING
BENDING
C C
C
C
C
Molecular vibrationsMolecular vibrations
both of these types are “infrared active”( excited by infrared radiation )
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BONDING CURVES BONDING CURVES AND VIBRATIONSAND VIBRATIONS
MORSE CURVES
STRETCHING
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ravg
decreasing distance
energy
rmin rmax
zero point energy
+ ++ +
++
++
(average bond length)
BOND VIBRATIONAL ENERGY LEVELSBOND VIBRATIONAL ENERGY LEVELS
MORSE CURVE
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bond dissociation energy
ravg
distance
energy
rmin rmax
zero point energy
vibrationalenergy levels
(average bond length)
BOND VIBRATIONAL ENERGY LEVELSBOND VIBRATIONAL ENERGY LEVELSBonds 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.
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2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
Typical Infrared Absorption Regions Typical Infrared Absorption Regions (stretching vibrations)(stretching vibrations)
N=O N=O*
* nitro has two bands
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MATHEMATICAL DESCRIPTION MATHEMATICAL DESCRIPTION OF THEOF THE VIBRATION IN A BONDVIBRATION IN A BOND
…. assumes a bond is like a spring
HARMONIC OSCILLATOR
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HOOKE’S LAWHOOKE’S LAW
x0 x1
x
K
-F = K(x)
m1 m2
K
Moleculeas aHooke’sLawdevice
restoringforce =
stretch
compress
forceconstant
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Harmonic Oscillator
Morse Curve(anharmonic)
THE MORSE CURVE APPROXIMATES THE MORSE CURVE APPROXIMATES AN HARMONIC OSCILLATORAN HARMONIC OSCILLATOR
HOOKE’SLAW
ACTUALMOLECULE
Using Hooke’s Law and theSimple Harmonic Oscillatorapproximation, the followingequation can be derived todescribe the motion of a bond…..
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=1
2c K
=m1 m2
m1 + m2
= frequencyin cm-1
c = velocity of light
K = force constant in dynes/cm
m = atomic masses
SIMPLE HARMONIC OSCILLATOR
C CC CC C > >multiple bonds have higher K’s
=reduced mass
( 3 x 1010 cm/sec )
THE EQUATION OF A
This equation describes the vibrations of a bond.
where
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=1
2c K
larger K,higher frequency
larger atom masses,lower frequency
constants
2150 1650 1200
C=C > C=C > C-C=
C-H > C-C > C-O > C-Cl > C-Br3000 1200 1100 750 650
increasing K
increasing
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DIPOLE MOMENTSDIPOLE MOMENTS
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DIPOLE MOMENTSDIPOLE MOMENTS
Only bonds which have significant dipole moments will absorb infrared radiation.
Bonds which do not absorb infrared include:
• Symmetrically substituted alkenes and alkynes
C C RR
R
R R
R
• Many types of C-C Bonds
• Symmetric diatomic molecules
H-H Cl-Cl
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C
OThe carbonyl group is oneof the strongest absorbers
O H C OAlso O-H and C-O bonds
+
-
STRONG ABSORBERSSTRONG ABSORBERS
+ +
- -
CO
C
O
+
-
oscillating dipoles couple andenergy is transferred
infrared beam
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RAMAN SPECTROSCOPYRAMAN SPECTROSCOPY
Another kind of vibrational spectroscopy thatcan detect symmetric bonds.
Infrared spectroscopy and Raman spectroscopycomplement each other.
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RAMAN SPECTROSCOPYRAMAN SPECTROSCOPYIn this technique the molecule is irradiated with strong ultraviolet light at the same time that theinfrared spectrum is determined.
Ultraviolet light promotes electrons from bondingorbitals into antibonding orbitals. This causes formation of a dipole in groups that were formerlyIR inactive and they will absorb infrared radiation.
C C RR C C RR ..+ -h
UV *
*
..
transitionabsorbs IR
induceddipole
no dipolesymmetric
….. we will not talk further about this technique
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SUGGESTED SOFTWARE
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• Select ChemApps folder• Select Spectroscopy icon• Select IR Tutor icon
IR TUTORIR TUTOR
IR TUTOR ACTUALLY ILLUSTRATES INFRARED VIBRATIONS AND THEORY WITH ANIMATIONS