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UV Spectra Interpretation Lecture #2
UV Spectra Interpretation
Lecture #2
Dapatkan kedua senyawa berikut dibedakan dengan menggunakan spektrum UV?
Senyawa A:Sistem induk 217 nmSubstituen alkil 20 nmIkatan rangkap exo 5 nm maks(heksana) 242 nm
Senyawa B:Sistem induk 217 nmSubstituen alkil 15 nmIkatan rangkap exo 0 nm maks(heksana) 232 nm
Hitung maks(heksana) untuk senyawa berikut:
HO
CH3
CH3C8H17
Senyawa :Sistem induk 217 nmDiena homoanular 36 nmSubstituen alkil 20 nmIkatan rangkap exo 0 nm maks(heksana) 273 nm
Hitung maks(heksana) kedua senyawa berikut dengan menggunakan spektrum UV.
H3CCO
CH3
CH3 C9H19
OH3CCO
CH3
CH3 C9H19
O
Senyawa :Sistem induk 217 nmDiena homoanular 36 nmSubstituen alkil 30 nmIkatan rangkap exo 15 nmPerpanjangan sistem 60 nm maks(heksana) 358 nm
Senyawa :Sistem induk 217 nmDiena homoanular 72 nmSubstituen alkil 30 nmIkatan rangkap exo 5 nmPerpanjangan sistem 60 nm maks(heksana) 384 nm
Similar for Enones
O
x
bb O O
X=H 207
X=R 215
X=OH 193
X=OR 193
215 202 227 239
Base Values, add these increments…
Extnd C=C +30Add exocyclic C=C +5
Homoannular diene +39
alkyl +10 +12 +18 +18
OH +35 +30 +50
OAcyl +6 +6 +6 +6
O-alkyl +35 +30 +17 +31
NR2
S-alkyl
Cl/Br +15/+25 +12/+30
b g d,+
With solvent correction of…..
Water +8
EtOH 0
CHCl3 -1
Dioxane -5
Et2O -7
Hydrcrbn -11
O
b
g
d
Senyawa :Sistem induk 215 nmSubstituen beta (1x12) 12 nmSubstituen delta (1x18) 18 nmExo.DB 5 nmPerpanjangan sistem 1 x30 30 nm maks(etanol) 280 nm
Contoh
Hitung maks(etanol) senyawa-senyawa berikut:
O
O
O
O
O
C8H17CH3CH3
C8H17
CH3CH3
C8H17CH3CH3
C8H17
CH3CH3
O
O
A
B
C
D E
F
G
Absorpsi Maksimum Lingkar Benzena
C Z
O
Z = alkil atau sisa lingkar 246 nmZ= H 250 nmZ=OH atau O alkil 230 nmTambahan substituenR=alkil atau sisa lingkar o-,m- 3 nm
p- 10 nmR=OH, Ome, O-alkil o-, m- 7 nm
p- 25 nmR=O o- 11 nm
m- 20 nmp- 78 nm
R=Cl o-, m- 0 nmp- 10 nm
R=Br o-, m- 2 nmp- 15 nm
R=NH2 o-, m- 13 nmp- 58 nm
R=NHAc o-, m- 20 nmp- 45 nm
R=NHMe p- 73 nmR=NMe2 o-, m- 20 nm
p- 85 nm
Contoh.COCH3
OCH3
COOH
H2N
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
From Crewes, Rodriguez, Jaspars, Organic Structure Analysis
UV of Benzene in heptane
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)
-CCH 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)
-CN 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 IIIsosbestic points
Single clear point, can exclude intermediate state, exclude light scattering and Beer’s law applies
Binding of a lanthanide complex to an oligonucleotide
