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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. Saturated Hydrocarbons. C n H 2n+2. GENERAL FORMULA. CH 4. C 2 H 6. C 3 H 8.

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slide1

INDEX OF HYDROGEN DEFICIENCY

and

THE BASIC THEORY OF

INFRARED SPECTROSCOPY

slide2

WHAT CAN YOU LEARN FROM A

MOLECULAR FORMULA ?

YOU CAN DETERMINE THE NUMBER OF

RINGS AND / OR DOUBLE BONDS.

slide3

Saturated Hydrocarbons

CnH2n+2

GENERAL FORMULA

CH4

C2H6

C3H8

C4H10

C5H12

C9H20

branched compounds

also follow the formula

slide4

FORMATION OF RINGS AND DOUBLE BONDS

-2H

-4H

-2H

Formation of each ring or double bond causes the loss of 2H.

slide5

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
  • Subtract the actual formula of your compound
  • The difference in H’s divided by 2 is the

(explained later)

Index of Hydrogen-deficiency

slide6

C5H8

C5H12

= ( CnH2n+2 )

C5H8

H4

Index = 4/2 = 2

Two Unsaturations

double bond and

ring in this example

slide7

Index of Hydrogen Deficiency

CORRECTIONS FOR ATOMS OTHER THAN HYDROGEN

  • 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

C-H

C-O-H

+O

+1

C-H

C-NH2

+N,+H

-1

C-H

C-X

-H,+X

slide8

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

slide9

The index gives the number of

  • double bonds or
  • triple bonds or
  • ringsin 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

slide10

INDEX

C6H14

-C6H8

Index = 6/2 = 3

H6

HYDROGENATION

Pd

C6H8

+

2 H2

C6H12

Hydrogenation shows only two double bonds.

Therefore, there must also be a 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.

slide11

A FEW POSSIBLE ANSWERS

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

slide13

THE ELECTROMAGNETIC SPECTRUM

Frequency (n)

high

low

high

low

Energy

MICRO-

WAVE

X-RAY

ULTRAVIOLET

INFRARED

RADIO

FREQUENCY

Nuclear

magnetic

resonance

Vibrational

infrared

Visible

Ultraviolet

2.5 mm

15 mm

1 m

5 m

200 nm

400 nm

800 nm

BLUE

RED

Wavelength (l)

short

long

slide14

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

slide15

Infrared

Spectrum

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)

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

1000

500

WAVELENGTH (cm-1)

KETONE

4-Methyl-2-pentanoneC-H < 3000, C=O @ 1715 cm-1

AN INFRARED SPECTRUM

slide17

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

c

cm/sec

1

n=

c

=

=

l

l

cm

sec

wavenumbers are directly proportional to frequency

slide18

Molecular vibrations

Two major types :

STRETCHING

C C

C

BENDING

C

C

both of these types are “infrared active”

( excited by infrared radiation )

slide19

BONDING CURVES

AND VIBRATIONS

MORSE CURVES

STRETCHING

slide20

+

+

+

+

+

+

o

o

+

+

BOND VIBRATIONAL ENERGY LEVELS

e

n

e

r

g

y

MORSE CURVE

zero point energy

rmin

rmax

decreasing distance

ravg

(average bond length)

slide21

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)

slide22

2.5

4

5

5.5

6.1

6.5

15.4

4000

2500

2000

1800

1650

1550

650

Typical Infrared Absorption Regions

(stretching vibrations)

WAVELENGTH (mm)

C-Cl

C=O

C=N

O-H

C-H

C N

Very

few

bands

C-O

C=C

N-H

C C

C-N

X=C=Y

C-C

*

N=O N=O

(C,O,N,S)

FREQUENCY (cm-1)

* nitro has

two bands

slide23

HARMONIC OSCILLATOR

MATHEMATICAL DESCRIPTION

OF THE

VIBRATION IN A BOND

…. assumes a bond is like a spring

slide24

HOOKE’S LAW

force

constant

compress

K

stretch

Dx

x0

x1

restoring

force

-F = K(Dx)

=

Molecule

as a

Hooke’s

Law

device

m1

m2

K

slide25

THE MORSE CURVE APPROXIMATES

AN HARMONIC OSCILLATOR

HOOKE’S

LAW

Harmonic Oscillator

ACTUAL

MOLECULE

Morse Curve

(anharmonic)

Using Hooke’s Law and the

Simple Harmonic Oscillator

approximation, the following

equation can be derived to

describe the motion of a bond…..

slide26

1

K

n

=

2pc

m

m1 m2

m

=

m1 + m2

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

>

>

multiple bonds have higher K’s

m= atomic masses

This equation describes the

vibrations of a bond.

m=reduced mass

slide27

1

K

n

=

2pc

m

=

C=C > C=C > C-C

larger K,

higher frequency

larger atom masses,

lower frequency

increasing K

constants

1650

2150

1200

increasing m

C-H > C-C > C-O > C-Cl > C-Br

3000 1200 1100 750 650

slide29

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

slide30

+

+

d+

-

-

d-

oscillating dipoles couple and

energy is transferred

STRONG ABSORBERS

d-

The carbonyl group is one

of the strongest absorbers

d+

Also O-H and C-O bonds

infrared beam

slide31

RAMAN SPECTROSCOPY

Another kind of vibrational spectroscopy that

can detect symmetric bonds.

Infrared spectroscopy and Raman spectroscopy

complement each other.

slide32

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

pp*

absorbs IR

transition

no dipole

symmetric

….. we will not talk further about this technique

slide34

IR TUTOR

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

IR TUTOR ACTUALLY ILLUSTRATES

INFRARED VIBRATIONS AND

THEORY WITH ANIMATIONS