2. WHAT IS SPECTROSCOPY ?
• Atoms and molecules interact with electromagnetic radiation (EMR) in a
wide variety of ways.
• Atoms and molecules may absorb and/or emit EMR.
• Absorption of EMR stimulates different types of motion in atoms and/or
molecules.
• The patterns of absorption (wavelengths absorbed and to what extent)
and/or emission (wavelengths emitted and their respective intensities) are
called ‘spectra’.
• spectroscopy is the interaction of EMR with matters to get specta ,which
gives information like, bond length, bond angle, geometry and molecular
structure.
3.
4. The entire electromagnetic spectrum is used by chemists:
UVX-rays IR RadioMicrowave
Visible
nuclear
excitation
(PET)
core
electron
excitation
(X-ray
cryst.)
electronic
excitation
(p to p*)
molecular
vibration
molecular
rotation
Nuclear Magnetic
Resonance NMR
(MRI)
g-rays
5. Electromagnetic radiations, are a form of energy, displays the properties of both
,particles and waves. The particle component is called a photon
The term “photon” is implied to mean a small, massless particle that contains a small
wave-packet of EM radiation/light –
. The important parameters associated with electromagnetic radiation are:
Energy (E):
Frequency (n) E = hn
Wavelength (l)
l= distance of one wave
n = frequency: waves per unit time (sec-1, Hz)
c = speed of light (3.0 x 108 m • sec-1)
h = Plank’s constant (6.63 x 10-34 J • sec)
5
: Energy is directly proportional to frequency, and inversely
proportion to wavelength, as indicated by the equation below
l
6. UV-Vis: valance electron transitions
- gives information about p-bonds and conjugated systems
Infrared: molecular vibrations (stretches, bends)
- identify functional groups
Radiowaves: nuclear spin in a magnetic field (NMR)
- gives a map of the H and C framework
organic
molecule
organic
molecule
(excited state)
organic
molecule(gro
und state)
relaxationlight
hn
Principles of molecular spectroscopy
7. Ultra-Violet Spectroscopy
• It is a branch of spectroscopy in which transition occur due to the excitation
of electrons from one energy level to higher one by the interaction of
molecules with ultraviolet and visible lights.
• Absorption of photon results in electronic transition of a molecule, and
electrons are promoted from ground state to higher electronic states.
• Since it involves an electron excitation phenomenon, so, also called as
Electronic Spectroscopy.
HOMO
LUMO
hvE
9. Absorption in UV Spectroscopy
Absorption of light in U V region is governed by Lambert’s-Ber’s law.
When a abeam of monochromatic light is passed through the homogenous
medium of substance, a fraction of light is absorbed. The decreased in the intensity of
light with the concentration of medium depends on the thickness and concentration
of the medium and the phenomenon is given by the equation:
A = - logT = ebc
A = absorbance ( optical density ) ; and A = log I0 / I,
T= Transmittance
C= concentration of solutios
b = length of the sample tube
= molar extinction coefficient
I0 = Intensity of incident light
I = Intensity of transmitted light
Absorption in UV Spectroscopy
10. There are three types of electronic transition which can be considered;
1. Transitions involving charge-transfer electrons
2. Transitions involving d and f electrons
3. Transitions involving p, s, and n electrons
Electronic transitions
.
Absorption of ultraviolet and visible radiation in organic molecules is
restricted to certain functional groups (chromophores) that contain valence
electrons of low excitation energy.
12. Selection Rules
1. Not all transitions that are possible are observed. Only those
transitios which follow certain rule ,are called as allowed
transitions.
2. For an electron to transition, certain quantum mechanical
constraints apply – these are called “selection rules”
3. For example, an electron cannot change its spin quantum number
during a transition – these are “forbidden”
Other examples include:
• the number of electrons that can be excited at one
time
• symmetry properties of the molecule
• symmetry of the electronic states
13. Transition Probability
( Allowed and Forbidden Transitions )
The molar extinction coefficient, ξ depends upon:
ξ max = 0.87 x 1020 x P x a
Where,
P = transition Probability varies from 0 - 1
a = target area of absorbing system(Chromophores)
Depending upon the value of ξ , and symmetry of orbital , the transition may be
allowed or forbidden
If ,
ξ > 104 Strong / high intensity peak ( allowed transition)
ξ ~ 103 - 104 Medium / Moderate intensity peak( allowed Trans.)
ξ < 103
; weak/low intensity peak ( forbidden transition)
14. Transitions
s->s*
– UV photon required, high energy, shorter wavelength
• Methane at 125 nm ( All saturated comp )
• Ethane at 135 nm
n-> s*
– Saturated compounds with unshared e-
N, O, S and Halogens,
• Absorption between 150 nm to 250 nm
• e between 100 and 3000 L cm-1 mol-1
• Shifts to shorter wavelengths with polar solvents
• R-X, R-OH R-NH2
15. Transitions
• n->p*and p->p*
– Unsaturated Organic compounds, containing non-bonded
electrons (O, N , S) and X
– wavelengths 200 to 700 nm
• n->p*low e (10 to 100),
• Shorter wavelengths 200 - 400
– Shorter wavelengths
• p->p*higher e (1000 to 10000) , 200- 260
Longer wavelength as compared to n->p*
17. TERMINOLOGIES
Chromophores and Auxochromes
1. A functional group capable of having characteristic electronic transitions is
called a chromophore (color loving)
C=C ; C=N, C=S , C=C; N=N etc
2. Structural or electronic changes in the chromophore can be quantified
and used to predict shifts in the observed electronic transitions
3. The attachment of substituent groups (other than H) can shift the energy
of the transition
4. Substituent's ,that could not absorbed light in UV region, but,increase the
intensity and often wavelength of an absorption are called auxochromes
5. Common auxochromes include alkyl, hydroxyl, alkoxy and amino
groups and the halogens
18. 200 nm 700 nm
Hyperchromic
Hypochromic
BathochromicHypsochromic
Substituents may have any of four effects on a chromophore
I. Bathochromic shift (red shift) – a shift to longer l; lower energy
II. Hypsochromic shift (blue shift) – shift to shorter l; higher energy
III. Hyperchromic effect – an increase in intensity
IV. Hypochromic effect – a decrease in intensity
20. Conjugation And Ethylene Chromophores
Ethylene absorbs at 175 nm. Presence of either chromophores or
auxochromes gives bathochromic shift.
lmax nm
175 15,000
217 21,000
258 35,000
465 125,000
Lmax = 2 17 253 220 227 227 256 263 nm
21.
22. EXPLANATION OF HIGHER WAVE LENHTH FOR
CONJUGATION
When we consider butadiene, we are now mixing 4 p orbitals giving 4 MOs of an
as compared to ethylene which has only 2 M Os
Y2*
pY1
Y1
Y2
Y3
*
Y4
*
DE for the HOMO LUMO transition is reduced So, lmax is increased .
HOMO
HOMO
LUMO
LUMO
217
nm
23. Extending this effect to longer conjugated systems the
energy gap becomes progressively smaller:
Energy
ethylene
butadiene
hexatriene
octatetraene
Lower energy =
Longer wavelengths
24. Empirical Approach To Structure
Determination
WOODWARD-FEISER RULE
*Woodward ( 1914 ) : gave certain rules for correlate with molecular
structures with lmax .
*Scott-Feiser(1959):modified rule with more experimental data , the
modified rule is known as Woodward-Feiser rule.
*A more modern interpretation was compiled by Rao in 1975 – (C.N.R.
Rao, Ultraviolet and Visible Spectroscopy, 3rd Ed., Butterworths, London,
1975)
It is used to correlate the lmax of for a given structure by
relating position and degree of substitution of chromophores.
25. Woodward-Fieser Rules for Conjugated acyclic Dienes
Woodward had predicted an empirical rule for calculating lmax of conjugated acyclic and cyclic
dienes based on base value and contribution of different substituent. The equation is:
lmax = Base value + ∑ Substituent's contribution + ∑ Other contribution
Base value : Take a base value of 214/217 for any conjugated. diene or triene
Incrementals: Add the following to the base value
Group Increment
Extended conjugation +30
Each exo-cyclic C=C
Phenyl group
Auxochromes:
+5
+60
Alkyl +5
-OCOCH3 +0
-OR +6
-SR +30
-Cl, -Br +5
-NR2
+60
27. Woodward-Fieser Rules – Cyclic Dienes
There are two major types of cyclic dienes, with two different base values
Heteroannular (transoid): Homoannular (cisoid):
base lmax = 214 base lmax = 253 base lmax = 253
Groups Increment
Extended conjugation
Pheny group
+30
+60
Each exo-cyclic C=C +5
Alkyl / ring residue +5
-OCOCH3 +0
-OR +6
-SR +30
-Cl, -Br +5
-NR2 +60
Additional homoannular +39
31. R
This compound has three exocyclic
double bonds; the indicated bond is
exocyclic to two rings
This is not a heteroannular diene; you would
use the base value for an acyclic diene
Likewise, this is not a homooannular diene;
you would use the base value for an acyclic
diene
Be careful with your assignments – three common
errors:
32. Base - Transoid/Heteroannular + 215 nm
Substituents- 5 Ring Residues
1 Double bond extending conjugation
5 x 5 = + 25 nm
+ 30 nm
Other Effects- 3 Exocyclic Double Bond + 15 nm
Calculated λmax 285 nm
Observed λmax 283 nm
35. Limitations of Woodwards rule for dienes
The Woodward rule is applicable to conjugated acyclic dienes with a
maximum no. of 04 double bonds , it fails beyond that.
Conjugated dienes more than 04 double bonds ; Fischer and Khun rule
is applicable.
λmax = 114 + 5M + n (48.0 – 1.7 n) – 16.5 Rendo – 10 Rexo
εmax = (1.74 x 104) n
M = the number of alkyl substituents and ring residues
n = the number of conjugated double bonds
Rendo = the number of rings with endocyclic double bonds
Rexo = the number of rings with exocyclic double bonds
36. Name of Compound all-trans-lycophene
Base Value 114 nm
M (number of alkyl & RR substituents) 8
n (number of conjugated double bonds) 11
Rendo (number of endocyclic double bonds) 0
Rexo (number of exocyclic double bonds) 0
Substituting in equation
λmax = 114 + 5M + n (48.0 – 1.7 n) – 16.5
Rendo – 10 Rexo
= 114 + 5(8) + 11 (48.0-1.7(11)) – 16.5 (0)
– 10 (0)
= 114 + 40 + 11 (29.3) – 0 – 0
= 114 + 40 + 322.3 – 0
Calc. λmax = 476.30 nm
λmax observed practically 474 nm
Calculate εmax using equation:
εmax = (1.74 x 104) n
= (1.74 x 104) 11
Calc. εmax= 19.14 x 104
Practically observed εmax 18.6 x 104
37. Name of Compound β-Carotene
Base Value 114 nm
M (number of alkyl & RR substituents) 10
n (number of conjugated double bonds) 11
Rendo (number of endocyclic double bonds) 2
Rexo (number of exocyclic double bonds) 0
Substituting in equation
λmax = 114 + 5M + n (48.0 – 1.7 n) – 16.5 Rendo –
10 Rexo
= 114 + 5(10) + 11 (48.0-1.7(11)) – 16.5 (2) – 10
(0)
= 114 + 50 + 11 (29.3) – 33 – 0
= 114 + 50 + 322.3 – 33
Calc. λmax = 453.30 nm
λmax observed practically 452nm
Calculate εmax using equation:
εmax = (1.74 x 104) n
= (1.74 x 104) 11
Calc. εmax= 19.14 x 104
Practically observed εmax 15.2 x 104
38. p
p*
n
Enones
Enone, called as alpha beta unsaturated carbonyl compound shows two
types of electronic transitions.
The, p p* and n p* ;
Both show longer λmax than isolated enone
Conjugation of a double bond with a
carbonyl group leads to
π→π*, at 200 ~ 250 nm absorption (ε =
8,000 to 20,000), Predictable
And
n→π*, at 210 ~ 330 nm, much less intense
(ε= 50 to 100), not predictable
39. The effects of conjugation are apparent from the MO diagram for
an enone:
p
Y1
Y2
Y3
*
Y4
*
p*
n
p
p*
n
OO
E
40. Woodward-Fieser Rules - Enones
Group Increment
6-membered ring or acyclic enone Base 215 nm
5-membered ring parent enone Base 202 nm
Acyclic dienone Base 245 nm
Aldehyde,Ester & Acid
Substituents:
Base 208 nm
Double bond extending conjugation
Phenyl group b
30
60
Alkyl group or ring residue a,b,gand higher 10, 12, 18
-OH a,b,gand higher 35, 30, 18
-OR a,b,g,d 35, 30, 17, 31
-O(C=O)R a,b,d 6
-Cl a,b 15, 12
-Br a,b 25, 30
-NR2 b 95
Exocyclic double bond 5
Homocyclic diene component 39
C C Cb
b a
C C CC
b a
C
gd
d
O O
41. SOLVENT CORRECTION
Unlike conjugated alkenes, solvent does have an effect on lmax
These effects are also described by the Woodward-Fieser rules
Solvent correction Increment
Water +8
Ethanol, methanol 0
Chloroform -1
Dioxane -5
Ether -7
Hydrocarbon -11
Common solvent
cuttoff:
acetonitrile 190
chloroform 240
cyclohexane 195
1,4-dioxane 215
95% ethanol 205
n-hexane 201
methanol 205
isooctane 195
water 190
42.
43.
44. Examples
keep in mind these are more complex than dienes
cyclic enone = 215 nm
2 x b- alkyl subs. (2 x 12) +24 nm
239 nm
Experimental value 238 nm
cyclic enone = 215 nm
extended conj. +30 nm
b-ring residue +12 nm
d-ring residue +18 nm
exocyclic double bond + 5 nm
280 nm
Experimental 280 nm
O
R
O
45. Base - cyclopentenone + 202 nm
Substituents at α-position 0
Substituents at β-position-
2 Ring Residue
2 x 12= + 24 nm
Other Effects- 1 Exocyclic Double
Bond
+ 5 nm
Calculated λmax 231 nm
Observed λmax 229 nm
46. Base - cyclohexenone + 215 nm
Substituents at α-position: 0
Substituents at β-position: 1 Ring Residue + 12 nm
Substituents at γ-position
0
Substituents at δ-position: 0
Substituents at ε-position: 1 Ring Residue + 18 nm
Substituents at ζ-position: 2 Ring Residue 2 x 18 = + 36 nm
Other Effects: 2 Double bonds extending
conjugation
Homoannular Diene system in ring B
1 Exocyclic double bond
2 x 30 = + 60 nm
+ 35 nm
+5
Calculated λmax 381 nm
Observed λmax 388 nm
47. Problem – can these two isomers be discerned by UV-spectroscopy ?
O
O
Eremophilone allo-Eremophilone
48. Aromatic Compounds
Benzene gives three bands in UV absorption.One would expect there to be four
possible HOMO-LUMO p p* transitions at observable wavelengths
(conjugation)
Due to symmetry concerns and selection rules, the actual transition energy
states of benzene are illustrated at the right:
p4* p5*
p6*
p2
p1
p3
260 nm
(forbidden)
200 nm
(forbidden)
180 nm
(allowed)
49. Substituent Effects Aromatic Compound
No matter what electronic influence a group exerts, EWG or EDG, their
presence shift l towards longer side.
Substituent lmax
H 203.5
-CH3 207
-Cl 210
-Br 210
-OH 211
-OCH3 217
-NH2 230
-CN 224
C(O)OH 230
-C(O)H 250
-C(O)CH3 224
-NO2 269
50. 50
Polynuclear aromatics
• When the number of fused aromatic rings increases, the lmax
for the primary and secondary bands also increased.
• For heteroaromatic systems spectra ,become complex
51. Woodward Rule for Aromatic Ketones
RO
G
Substituent increment
G o m p
Alkyl or ring residue 3 3 10
-O-Alkyl, -OH, -O-Ring 7 7 25
-O- 11 20 78
-Cl 0 0 10
-Br 2 2 15
-NH2 13 13 58
-NHC(O)CH3 20 20 45
-NHCH3 73
-N(CH3)2 20 20 85
Parent Chromophore lmax
R = alkyl or ring
residue
246
R = H 250
R = OH or O-Alkyl 230
52. Base value = 246 nm
Ring residue in o- position = 1 x 3 = 3 nm
Polar group -OCH3 in p- position = 25 nm
λmax = 274 nm
Observed = 274 nm
53. PROBLEMS
1. The enol acetate (ester) of the following compounds has λmax at 238nm, suggest
the structure of the ester.
2. The following triene on practical hydrogenation gives three products, separated by
GLC. How UV spectra anWoodwards rule will help to identify the products.
54. 3.. Pentene-1 absorbs at λmax 178nm. Three isomeric dienes A,B and C of molecular
composition C5H8 absorbs at:
λmax , for A=179nm
for B=217nm
for C=219nm
Find out A, B and C
4. Four isomeric ketones A, B , C and D having molecular formula C6H8O The
UV spectra gives absorption at 228 nm. Which is probable structure of the
ketone.
55.
56. Fluorescence is a process that involves a singlet to singlet transition.
Phosphorescence is a process that involves a triplet to singlet transition
• Fluorescence: absorption of radiation to
an excited state, followed by emission of
radiation to a lower state of the same
multiplicity
• Phosphorescence: absorption of
radiation to an excited state, followed by
emission of radiation to a lower state of
different multiplicity
• Singlet state: spins are paired, no net
angular momentum (and no net magnetic
field)
• Triplet state: spins are unpaired, net
angular momentum (and net magnetic
field)
More Complex Electronic Processes