SCH 306 (307)

SPECTROSCOPYAS APPLIED TO

ORGANIC CHEMISTRY

• Lecturer : Abiy Yenesew(Associate Professor of Chemistry)

Office: G6 (Chemistry Department)Consultation: My office is open even when I

am NOT in.abiyenesew@yahoo.com orayenesew@uonbi.ac.ke

Monday Wedensday Friday8.00-10.00 8.00-10.00

Venue; Room 122 (Chemistry Department)

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LECTURES• 1. UV-VIS SPECTROSCOPY (6 HOURS)• 2. IR SPECTROSCOPY (6 HOURS)• C.A.T. 1 (10 MARKS)• 3. NMR SPECTROSCOPY (12 HOURS)• 4. MASS SPECTROSCOPY (6 HOURS)• C.A.T. 2 (10 MARKS)• 5. PRACTICE IN STRUCTURE

DETERMINATION (4 HOURS)• ASSIGNMETS AND TUTORIALS (10 MARKS)• FINAL EXAM (70 MARKS)

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RECOMMENDED BOOKS • 1) Spectrometric Identification of organic

compounds (by Silverstein et al., John Wiley & Sons, INC.).

• 2. Spectroscopic Methods in Organic Chemistry (by Williams and Fleming, McGraw-Hill Company).

• 3. Spectroscopic Methods in Organic Chemistry (by Hesse et al., Thieme).

• 4. Organic Structural Spectroscopy (by Lamberiet al., Prentice Hall).

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Unit Objectives• At the end of this unit, you should be able

to:• 1. Explain the principles behind different

spectroscopic techniques.• 2. Propose the expected spectroscopic

features of organic molecules.• Determine the structure of organic

compounds using different spectroscopic methods.

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Organic Spectroscopy is the study of the interaction of electromagnetic radiation with organic molecules.

• Purpose: structure determination of organic molecules.

• Electromagnetic radiation has both the properties of a wave and a particle.

• It can be described as a wave occurring simultaneously in electrical and magnetic fields.

• And it can also be described as if it is consisted of particles called quanta or photons.

Electromagnetic Radiation

INTRODUCTION

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Speed = 3X1010 cm/sec

Electromagenetic Radiation

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Electromagnetic spectrum

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Wave length (l) can be used to describe a wave;

•which is the distance between consecutive crests (or troughs) or • the distance for a complete cycle of a wave.

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• Electromagnetic radiation can also be characterized by its frequency (n),

• which is defined as the number of complete cycles per second (cps) passing a point; it is also called in modern times as Hertz (Hz);

• Or number of complete cycles per centimetre called wave number (cm-1)

one complete cycle time1 cycle/sec = 1 Hzfreqency (n)

distance1cycle/cm = 1cm-1

wave number 10

• The energy of the quantum (photon) of electromagentic radiation is directlyproportional to frequency (n)

• E = hn,• h (Plank’s constant) = 6.63X10-34Js

n = c/l• where c is the speed of electromagnetic

radiation (c = 3X1010 cm/sec)• E = hn = hc/l• Energy is inversely proportional to the

wavelength (l).11

• Thus the higher the frequency of radiation the greater will be its energy.

• X-rays are more energetic than rays of visible light.

• Radiation of long wavelength has low energyand that of short wavelength has high energy.

• i.e., a photon of ultraviolet light has more energy than photon of visible light.

E = hn = hc/l= hc

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Electromagnetic spectrum

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Symbol Definition

n Frequency in Hz (cycles per second)

l Wavelength (mm, nm or Å)

mm Micrometer (10-6m)

nm Nanometer (10-9m)

Å Angstron (10-10m)

Wave number (cm-1)

E = hn = hc/l= hc

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Spectroscopic Methods and Quantized Transitions

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ULTRAVIOLET (UV) AND VISIBLE (VIS) SPECTROSCOPY

(l = 190-780 nm)UV (190-400 nm)VIS (400-780 nm)

INTRODUCTIONMany molecules absorb ultraviolet or visible light.The type of light (qualitative) and the amount of

light (quantitative) they absorb varies.Quantitative:Beer's Law states that: Absorbance is directly

proportional to the path length, b, and the concentration, c, of the absorbing species.

A = ebc, where e is a proportionality constant , called the molar absorbtivity or

molar extinction coefficient.

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• Different molecules with different structural groups absorb radiation of different wavelengths.

• An absorption spectrum will show a number of absorption bands corresponding to structural groups within the molecule.

• For example, the UV absorption of carbonyl group in acetone is at the same wavelength as carbonyl group in diethyl ketone.

CH3 CH3

O OCH3CH3

acetone diethyl ketone

Qualitative:

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Electronic Transitions• Absorption of UV and VIS radiation in organic

molecules is restricted to certain functional groups (chromophores),

• that contain valence electrons of low excitation energy.

• The types of electronic transition which are considered here are:

1) Transitions involving s (s, p) electrons,2) Nonbonding (n) electrons and3) p electrons.

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The absorption of UV or visible radiation correspondsto the excitation of outer electrons.• When an atom or a molecule absorbs energy,• electrons are promoted from their ground state

to an excited state.

ground state (E0)

excited state (E*)

DE = E* - E0 = hc/l

OH

H

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In a molecule, the atoms can rotate and vibratewith respect to each other.

However continuous absorption bands are observedReason:

These vibrations and rotations also have discrete energy levels; these are packed on top of each electronic level.

A

l

A

l

If monochromic light (of l wavelength) is absorbed,then absorption lines are expected.

expected observed

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ground state (E0)

excited state (E*)

DE = E* - E0 = hc/l

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The spectrum of a molecule containing chromophores is complex.

• This is because the superposition of rotational and vibrational transitions on the electronic transitions gives a combination of overlapping lines.

There is also solvent-solute interaction, which smoothens the spectrum.

• Thus UV-VIS spectra in solutions appear as a continuous absorption band.

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NNN

[1,2,4]Triazine

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Possible Electronic Transitions

Which of these are important transitions?25

PROBABILITY OF TRANSITIONA = ebc or elcwhere e is the molar absorbtivity ormolar extinction coefficient.b (or l ) is cell length, c is concentration

The magnitude of e determines the probability of transition

Determined by the symmetry of the orbitals involvedin electronic transitions.

Different symmetry Same symmetry(e < 1,000) (e > 1,000)

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sp3 orbital (often non-bonding)

Transitions between the same symmetry, allowed

Transitions between different symmetry, forbidden

sorbital s*orbital

porbital p*orbital

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√

√

√ = ‘allowed’

X

X X

X

X = ‘forbidden’

28Important transitions

s s* Transitions• ‘Allowed’ transition.• An electron in a bonding s orbital is excited to the

corresponding antibonding orbital. • The energy required is large. • Eg. methane (which has only C-H bonds) • can only undergo s s* transitions • shows an absorbance maximum at 125 nm. • Absorption maxima due to s s* transitions are

not seen in typical UV-VIS spectra (190-780 nm).

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p p* Transitions

• ‘Allowed’ transition.One of the most common absorption spectroscopy of organic compounds is based on transitions of p electrons to the p*excited state.

• Because the absorption peaks for these transitions fall in the region 190 - 780 nm.

• This transition needs an unsaturated groupin the molecule to provide the p electrons.

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Absorption spectra of alkenes (p p )H

H H

H

lmax 165 nm

R

H H

H

lmax 177 nm

R

H R

H

lmax 180 nm

R

R H

R

lmax 187 nm

R

R R

R

lmax 200 nm

Alkyl substituents to sp2-carbon result in smallshift to longer wave length (bathochromic shift).Due to hyperconjugation.

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Cyclic alkenes (p p )Cyclohexenelmax 190 nm (e 7,250)

A B

C D

lmax 193 nm (e 10,700)

A B

C D

lmax 206 nm (e 11,200)

The more substituted the double bond the greater theobserved shift to longer wave length.

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CONJUGATED DIENES

H

H

H

H

1,5-hexadiene

lmax 217 (e 21,000) lmax 185 (e 20,000)

Why the difference?

Conjugated diene Non-conjugated diene

Conjugation:- the mixing of two or more isolated p-bonds to form new bonding molecular orbitals and associated anti-bonding orbitals.

H

HH

H1,3-Butadiene

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DE

p

p*

p

p*

DEDE’

ψ1

ψ2

ψ3

ψ4

Isolated p system Isolated p system

conjugated systemΨ2 Highest Occupied Molecular Orbital (HOMO)Ψ3 Lowest Unoccupied Molecular Orbital (LUMO)

Conjugation brings stability34

DE’ < DE l’ > l

The energy separation between HOMO and LUMOIn conjugated system is less than that of isolated p-bond (un-conjugated diene).absorbtion observed at longer wavelength in

conjugated system than in un-conjugated system

H

HH

H

H

HH H

Which of the two is more stable?

Orbitals are co-planarMaximum overlapping

Due to steric effectOrbitals are not co-planarLevel of overlappingdecreased

ls-trans >ls-cisS-trans-1,3-butadiene

S-cis-1,3-butadiene

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more stable

Since isoprene is colourless, it does not absorb in the visible part of the spectrum (above 400 nm)and this region is not displayed on the graph. Absorbance usually ranges from 0 (no absorption) to 2 (99% absorption), and is precisely defined in context with spectrometer operation.

c = 4 X 10-5

moles per litre, in EtOH

A = ebc

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(3-methyl-1,3-butadiene)

• Because the absorbance of a sample will be proportional to the number of absorbing molecules,

• it is necessary to correct the absorbance value.

• The corrected absorption value is called "molar absorptivity"

• This is useful when comparing the spectra of different compounds and

• determining the relative strength of light absorbing functions (chromophores).

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Molar Absorptivity (ε)Molar Absorptivity, ε = A ⁄ clwhere A = absorbance, c = sample concentration in moles/litre,l = length of light path through the sample in cmFor isoprene:c = 4 X 10-5 moles per litre, l = 1 cm (1 cm sample

cuvette used), A = 0.8

ε = A ⁄ cl = 0.8 ⁄ 4X10-5X1 = 20,000 at the maximum absorption wavelength

Isoprene lmax (e) = 222 nm (20,000)lmax (log e) = 222 nm (4.3) 38

Woodward-Fieser Rule• On the basis of observations, Woodward-Fieser

noted that:• The position of lmax of conjugated dienes varies

in a regular manner with pattern of substitutionabout the chromophore.

• These observations have resulted in a set of empirical rules.

(S-trans diene)

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Core Chromophore

Transoid diene(S-trans diene)

215 nm

Substituent and Influence(auxochroms)

R- (Alkyl Group) .... +5 nmRO- (Alkoxy Group) +6 X- (Cl- or Br-) ........ +10 RCO2- (Acyl Group) .. 0 RS- (Sulfide Group) +30R2N- (Amino Group) +60

Further π -ConjugationC=C (Double Bond) ... +30C6H5 (Phenyl Group) ... +60

Cyclohexadiene (cisoid diene)253 nm 40

Each exocyclic double bond adds 5 nm. In the example, there are two exocyclic double bondcomponents: one to ring A and the other to ring B.

Example

Abiatic acid

A B

C

Parent chromophore 215 nm

Ring residue 5X3=15Alkyl substituent 5Exocyclic double bond 5Total 240 nmlobserved 241 nm

Difference of less than ± 5nm, good correlation 41

OHA B

C D1

4 6

8

11 13

10

15

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Ergosterol

Parent chromophore 253 nm

lobserved 282 nm

Ring residue 5X4=20

Exocyclic double bonds 5X2=10Total 283 nm

good correlation 42

1,2-Dimethylenecyclohexane

Parent chromophore

253 nm

Ring residue 5X2=10

Exocyclic double bonds

5X2=10

Total 273 nm

The chromophore isDistorted from co-planarity Loss of conjugation;hence absorb at shorter l

lobserved 220 nm

bad correlationDifference 53 nm43

1,2-Dimethlenecyclopentane

Parent chromophore

253 nm

Ring residue 5X2=10

Exocyclic double bonds

5X2=10

Total 273 nmThe chromophore is lessdistorted from co-planarityresulting better conjugation.Hence the observed value is closer to the calculate value

lobserved 248 nm

Puckered conformation

Difference 25 nm44

The predictions are for the most intense band,And does not mean that this is the only band

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conjugation of double and triple bonds also shifts the absorption maximum to longer wavelengths.

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Each additional double bond inthe conjugated p-electron system shifts the absorption maximum about 30 nm in the same direction. the molar absorptivity (ε)roughly doubles with each newconjugated double bond.

Thus, extending conjugation results in bathochromic andhyperchromic shifts in absorption.

EXTENDED CONJUGATION

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The appearance of several absorption peaks or shoulders (sh) for a given chromophore is common for highly conjugated systems.

This fine structure reflects:different conformations such

systems may assume as the result of rotation at single bonds

(s-cis and s-trans conformations).but also electronic transitions

between the different vibrational energy and rotational energy levels possible for each electronic state.

s-cis, s-cis

s-trans, s-trans

s-cis, s-trans 49

Terminology for Absorption ShiftsNature of Shift Descriptive Term

To Longer Wavelength Bathochromic (red shift)

To Shorter Wavelength Hypsochromic (blue shift)

To Greater Absorbance Hyperchromic shift

To Lower Absorbance Hypochromic shift

Auxochrome Functional group which causes a shift 50

b-carotene (pigment responsible for the orangecolour of carrots).11 conjugated double bondslmax 452 nmWhy is it orange?Does it mean it absorbs orange light?

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Red 620 - 780 nmOrange 585 - 620 nmYellow 570 - 585 nmGreen 490 - 570 nm

Indigo 420 - 440 nmViolet 400 - 420 nm

Blue 440 - 490 nm

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When white light passes through or is reflected by a colouredsubstance, a portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary colour.Complementary colours are diametrically opposite each other. Absorption of 420-430 nm light renders a substance yellow, Absorption of 500-520 nm light makes it red. Green is created by absorption of light close to 400 nm as well as absorption near 800 nm.

Colour Wheel

b-carotene

lmax 452 nm

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AROMATIC COMPOUNDS•180 nm (ε > 65,000) •200 nm (ε = 8,000)

•group of weaker bands (fine structure)•at 254 nm (ε = 204) - called B-Band.

lmax 254 (e 204)

CH3

lmax 261 (e 300)

CH3

CH3lmax 274 (e 460)54

CH3

CH3

CH3

CH3

lmax 279 (e 460)

lmax 279 (e 820)

Alkylation results in shifts towardslonger wavelength and in generalintensification;

due to hyperconjugation which reduces the energy gap (DE)

Sterric effect forces two methyl groups out of plane, NO further hyperconjugation.

-

CH2

HCH2

CH2H+ H+

-

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OH

lmax 270 (e 1,470)

O

lmax 287 (e 2,600)Red shift (due toincreased conjugation)

lmax 254 (e 204)

KOH

O CH3

O

O+

CH3

O

lmax 261 (e 300), blue shift, (due to reduced conjugation)

O

-

O

-

N

O CH3

O

CH3

O

Auxochromes

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The added conjugation in naphthalene, anthracene and tetracene causes bathochromic shifts of these absorption bands.naphthalene & anthracene are colourless, but tetracene is orange.

Effect of Further Conjugation

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lmax 254 (B-band)

HH

H H

lmax 270 (B-and)

Conjugation is greatest when the interacting functionsallow maximum overlap of their p-orbitals system.

In biphenyls this would be achieved if the two benzene rings are coplanar.

Due to sterric reasons the two benzene rings arenot coplanar, the average angle between the two benzene rings is 45º

Biphenyls

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CH3

CH3 CH3

lmax 266 (B-band) CH3

CH3CH3

CH3

CH3 CH3

lmax 266 (B-band)

Due to sterric reasons the two benzene rings areorthogonal, no overlapping between the two rings.

Bulky groups at ortho-positions of biphenyl molecules,serves to increase the angle.

Effective conjugation is further reduced.

Substituted Biphenyls

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a,b-Unsaturated Carbonyl Compounds

p p

n p

O

RR

R

Conjugation of double bondwith carbonyl group,

lowers energy gap betweenLUMO and HOMO

lmax 220-260 nm(e 10,000-15,000)

lmax 310-330 nm(e 50-100)

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DE

p

p*

p

p*

DE

DE’

ψ1

ψ2

ψ4

ψ5

Isolated p system incarbonyl compound

Isolated p systemin alkeneconjugated system

Ψ3 Highest Occupied Molecular Orbital (HOMO)Ψ4 Lowest Unoccupied Molecular Orbital (LUMO)

DE’ψ3n

DEp p

n p

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Woodward-Fieser Rules for Calculating Conjugated Carbonyl Compounds

λmax (calculated) = Base + Substituent Contributions and Corrections

p p ψ2 ψ4The same as

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Core Chromophore Substituent and Influence

R = Alkyl , 215 nmR = H, 207 nm

α- SubstituentR- (Alkyl Group) +10 nmCl- (Chloro Group) +15Br- (Bromo Group) +25HO- (Hydroxyl Group) +35RO- (Alkoxyl Group) +35RCO2- (Acyl Group) +6

β- SubstituentR- (Alkyl Group) +12 nmCl- (Chloro Group) +12Br- (Bromo Group) +30HO- (Hydroxyl Group) +30RO- (Alkoxyl Group) +30RCO2- (Acyl Group) +6RS- (Sulfide Group) +85R2N- (Amino Group) +95

Ra

b

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g & δ- SubstituentsR- (Alkyl Group) +18 nm (both g & δ)HO- (Hydroxyl Group) +50 nm (g)RO- (Alkoxyl Group) +30 nm (g)

Ra

b

g

Further π -Conjugation• C=C (Double Bond) ... +30

C6H5 (Phenyl Group) ... +60

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• (i) Each exocyclic double bond adds 5 nm.

(iii) Solvent Correction: water = –8; methanol/ethanol = 0; ether = +7; hexane/cyclohexane = +11

• (ii) Homoannular cyclohexadiene component adds +39 nm

O

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Parent chromophore 215 nm

O

1-(6,6-Dimethylcyclohex-1-enyl)-ethanone

Examples

a-ring residue 10

b-ring residue 12Total 237 nmlobserved 232 nm66

Parent chromophore 215 nm

Examples

a-ring residue 0

b-ring residue 2 x 12exocyclic double bond 5Total 244 nmlobserved 241 nm

OH

OTestosterone

A B

C D

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Parent chromophore 215 nm

a-ring residue 10

Double bond extending conjugation

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homodiene 39

Total 317 nmlobserved 314 nm

exocyclic double bond 5

d-ring residue 18

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O

O

O

OH

tautomerism

lmax 260 nm (e 8,500), keto-form enol-form

OCH2lmax 284 nm (e 10,000)

Not conjugated in a formal senseBut interaction does occur to account for the lmax

C=C and C=O bonds are in close proximity and co-plannar.

CH3 CH3

O

CH2 CH3

OHtautomerism

keto-form enol-form

enol-form is favoured due to conjugationcalculated (215 + 30+12) =257 nm

keto-form is favoured(has more bond energies)

p p

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• This is the second most important absorption transitions of organic compounds.

• Molar absorbtivities are relatively low, e10 to100 L mol-1 cm-1 .

‘forbidden’ transition

n p* Transitions

OCH3

CH3

lmax (e) 190 nm (1,800), p plmax (e) 280 nm (13), n p

In hexane

DE’

p

n

p*

DE

Example

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lmax (e) 270 nm (16), n p (in ethanol)

The position of absorption of carbonyl compounds is solvent dependent.

DE’

n

p*

DE

In hexane

CH3

CH3

O HO CH2CH3

In ethanol

d- d+DE’

n

p*

DE

H-bondingLowers the n-orbitalDE’ increased(Blue shift)

In ethanolp p

lmax (e) 280 nm (13), n p (in hexane)

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O

CH3

CH3

CH3

CH3

CH3CH3

di-tert-butyl ketone

lmax (e) 295 nm (13), n p

In ethanol

Up on increasing a-substitution, the position ofn p shifts to longer wave length due to:i) The energy of the n-orbital is raised and/orii) Energy gap reduced through hyperconjugation

(energy of p reduced)

Effect of a-Substitution

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n s Transitions• Saturated compounds containing atoms with

lone pairs (non-bonding electrons) are capable of n s* transitions.

• They can be initiated by light whose wavelength is in the range of 150 - 250 nm (e < 100).

• The number of organic functional groups with• n s* peaks in the UV region is small.• -OH, -OR, -NH2, -SH, halogens• When in conjugation with chromophors, these

groups are called auxochromes;• they modify the absorption positions of

chromophors73

• When sample molecules are exposed to light• having an energy that matches a possible

electronic transition within the molecule, • some of the light energy will be absorbed as the

electron is promoted to a higher energy orbital. • An optical spectrometer records the wavelengths at

which absorption occurs, together with the degree of absorption at each wavelength.

• The resulting spectrum is presented as a graph of absorbance (A) versus wavelength (l), 180-780 nm

• 180-380 nm corresponds to UV light• 380-780 nm corresponds to Vis light

SUMMARY

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• only molecular moieties likely to absorb light in the 180 to 780 nm region are:

• p-electron functions and • hetero atoms having non-bonding valence-

shell electron pairs. • Such light absorbing groups are referred

to as chromophores.

• Tutorial • 5) Why is the sky blue?

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Of the six transitions outlined, only the two lowest energy ones (left-most, coloured blue) are achieved by the energies available in the 200 to 780 nmspectrum.

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Chromophore Example Excitation λmax, nm ε Solvent

C=C Ethene π __> π* 171 15,000 hexane

C≡C 1-Hexyne π __> π* 180 10,000 hexane

C=O Ethanal n __> π*π __> π*

290180

1510,000

hexanehexane

N=O Nitromethane n __> π*π __> π*

275200

175,000

ethanolethanol

C-X X=BrX=I

Methyl bromide

Methyl Iodide

n __> σ*n __> σ*

205255

200360

hexanehexane

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