INDEX OF HYDROGEN DEFICIENCYINDEX OF HYDROGEN DEFICIENCY

THE BASIC THEORY OF THE BASIC THEORY OF INFRARED SPECTROSCOPYINFRARED SPECTROSCOPY

and

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.

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

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.

• 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

C5H8

C5H12

C5H8

H4 Index = 4/2 = 2

double bond andring in this example

= ( CnH2n+2 )

Two Unsaturations

• 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

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

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

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.

CH3

CH3

CH3

CH3

H3C

CH2CH3

A FEW POSSIBLE ANSWERSA FEW POSSIBLE ANSWERS

..... there is still work required to fully solve the problem

INFRARED SPECTROSCOPYINFRARED SPECTROSCOPY

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

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)

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

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

( )

= 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

Two major types :

STRETCHING

BENDING

C C

C

C

C

Molecular vibrationsMolecular vibrations

both of these types are “infrared active”( excited by infrared radiation )

BONDING CURVES BONDING CURVES AND VIBRATIONSAND VIBRATIONS

MORSE CURVES

STRETCHING

ravg

decreasing distance

energy

rmin rmax

zero point energy

+ ++ +

++

++

(average bond length)

BOND VIBRATIONAL ENERGY LEVELSBOND VIBRATIONAL ENERGY LEVELS

MORSE CURVE

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.

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

MATHEMATICAL DESCRIPTION MATHEMATICAL DESCRIPTION OF THEOF THE VIBRATION IN A BONDVIBRATION IN A BOND

…. assumes a bond is like a spring

HARMONIC OSCILLATOR

HOOKE’S LAWHOOKE’S LAW

x0 x1

x

K

-F = K(x)

m1 m2

K

Moleculeas aHooke’sLawdevice

restoringforce =

stretch

compress

forceconstant

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…..

=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

=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

DIPOLE MOMENTSDIPOLE MOMENTS

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

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

RAMAN SPECTROSCOPYRAMAN SPECTROSCOPY

Another kind of vibrational spectroscopy thatcan detect symmetric bonds.

Infrared spectroscopy and Raman spectroscopycomplement each other.

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

SUGGESTED SOFTWARE

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

IR TUTORIR TUTOR

IR TUTOR ACTUALLY ILLUSTRATES INFRARED VIBRATIONS AND THEORY WITH ANIMATIONS