UV Spectra Interpretation
Lecture #2
Dapatkan kedua senyawa berikut dibedakan
dengan menggunakan spektrum UV?
Senyawa A:
Sistem induk
217 nm
Substituen alkil
20 nm
Ikatan rangkap exo
5 nm
 maks(heksana) 242 nm
Senyawa B:
Sistem induk
217 nm
Substituen alkil
15 nm
Ikatan rangkap exo
0 nm
 maks(heksana) 232 nm
Hitung  maks(heksana) untuk senyawa berikut:
CH 3
C8 H17
CH3
HO
Senyawa :
Sistem induk
217 nm
Diena homoanular 36 nm
Substituen alkil
20 nm
Ikatan rangkap exo
0 nm
 maks(heksana) 273 nm
Hitung  maks(heksana) kedua senyawa berikut
dengan menggunakan spektrum UV.
CH 3
C9 H 19
CH 3
CH3
CH3
O
H3 CCO
O
H3 CCO
Senyawa :
Sistem induk
217 nm
Diena homoanular 36 nm
Substituen alkil
30 nm
Ikatan rangkap exo 15 nm
Perpanjangan sistem 60 nm
 maks(heksana) 358 nm
Senyawa :
Sistem induk
217 nm
Diena homoanular 72 nm
Substituen alkil
30 nm
Ikatan rangkap exo
5 nm
Perpanjangan sistem 60 nm
 maks(heksana) 384 nm
C9 H 19
Similar for Enones
b
b
O
O
O
227
202
215
Base Values, add these increments…

X=H 207
x
b

239
g
d,+
Extnd C=C
Add exocyclic C=C
+30
+5
Homoannular diene
+39
alkyl
+10
+12
With solvent correction
of…..
OH
+35
+30
Water
+8
OAcyl
+6
+6
+6
+6
EtOH
0
O-alkyl
+35
+30
+17
+31
CHCl3
-1
+15/+25
+12/+30
X=R 215
X=OH 193
X=OR 193
Dioxane
-5
Et2O
-7
Hydrcrbn -11
NR2
S-alkyl
Cl/Br
+18
+18
+50
d
b
O

g
Senyawa :
Sistem induk
Substituen beta (1x12)
Substituen delta (1x18)
Exo.DB
Perpanjangan sistem 1 x30
 maks(etanol)
215 nm
12 nm
18 nm
5 nm
30 nm
280 nm
Hitung  maks(etanol) senyawa-senyawa berikut:
O
CH3
CH3
C 8H 17
C 8H 17
CH 3
CH 3
O
O
O
CH3
CH3
C 8H 17
CH 3
CH 3
O
O
O
C 8H 17
Absorpsi
Maksimum
Lingkar
Benzena
R=OH, Ome, O-alkil
R=O
R=Cl
O
C
Z = alkil atau sisa lingkar
Z= H
Z=OH atau O alkil
Tambahan substituen
R=alkil atau sisa lingkar
Z
R=Br
R=NH2
R=NHAc
R=NHMe
R=NMe2
246 nm
250 nm
230 nm
o-,mpo-, mpompo-, mpo-, mpo-, mpo-, mppo-, mp-
3 nm
10 nm
7 nm
25 nm
11 nm
20 nm
78 nm
0 nm
10 nm
2 nm
15 nm
13 nm
58 nm
20 nm
45 nm
73 nm
20 nm
85 nm
Contoh.
COCH 3
COOH
OCH 3
H 2N
Generally, extending conjugation leads
to red shift
“particle in a box” QM theory; bigger box
Substituents attached to a chromophore that cause a red shift are called
“auxochromes”
Strain has an effect…
max
253
239
256
248
Interpretation of UV-Visible Spectra
•
•
•
Transition metal complexes;
d, f electrons.
Lanthanide complexes –
sharp lines caused by
“screening” of the f
electrons by other orbitals
One advantage of this is the
use of holmium oxide filters
(sharp lines) for wavelength
calibration of UV
spectrometers.
See Shriver et al. Inorganic Chemistry, 2nd Ed. Ch. 14
Benzenoid
aromatics
UV of Benzene in
heptane
From Crewes, Rodriguez, Jaspars, Organic Structure Analysis
Group
K band ()
B band()
R band
Alkyl
208(7800)
260(220)
--
-OH
211(6200)
270(1450)
-O-
236(9400)
287(2600)
-OCH3
217(6400)
269(1500)
NH2
230(8600)
280(1400)
-F
204(6200)
254(900)
-Cl
210(7500)
257(170)
-Br
210(7500)
257(170)
-I
207(7000)
258/285(610/180)
-NH3+
203(7500)
254(160)
-C=CH2
248(15000)
282(740)
-CCH
248(17000)
278(6500
-C6H6
250(14000)
-C(=O)H
242(14000)
280(1400)
328(55)
-C(=O)R
238(13000)
276(800)
320(40)
-CO2H
226(9800)
272(850)
-CO2-
224(8700)
268(800)
-CN
224(13000)
271(1000)
-NO2
252(10000)
280(1000)
330(140)
Substituent effects don’t really add up
Can’t tell any thing about substitution geometry
Exception to this is when adjacent substituents can interact, e.g
hydrogen bonding.
E.g the secondary benzene band at 254 shifts to 303 in salicylic
acid
In p-hydroxybenzoic acid, it is at the phenol or benzoic acid
frequency
Heterocycles
Nitrogen heterocycles are pretty similar to the benzenoid anaologs that are
isoelectronic.
Can study protonation, complex formation (charge transfer bands)
Quantitative
analysis
Great for non-aqueous
titrations
Example here gives detn
of endpoint for
bromcresol green
Binding studies
Form I to form II
Isosbestic points
Single clear point, can exclude
intermediate state, exclude light
scattering and Beer’s law applies
Binding of a lanthanide complex to
an oligonucleotide