Protokoll E6b PDF

Preview

Summary

Experiment E6b:
Es handelt sich um ein Beispielprotokoll für das organisch-chemische Praktikum 2 (Literaturpraktikum) für das Experiment E6b. In diesem Protokoll geht es um die Job Plot Analyse der Bindungsstöchiometrie von supramolekularen Host-Guest-Komplexen am Beispiel des Cyclodextrins. ...

Description

Name: Michele Cuciuffo,

Room: F2.044, Place 4

Datum:

Experiment E6b: Job Plot Analysis of the Binding Stoichiometry of Supramolecular Host-Guest complexes: The Example Cyclodextrin 1. Reaction Equation

2. Experimental Procedure Methyl orange and β-Cyclodextrin were dissolved in D 2O to prepare a dilution series. They were made from 8.8 mM stock solutions. The series was contained 10 different concentrations, which were 1:9, 2:8, 3:7, 4:6, 5:5, 6:4 7:3, 8:2, 9:1 (Metyhl orange: β-Cyclodextrin) and pure methyl orange. The solutions were analyzed for their molar ratio [G]0/([H0+G0) via 1H-NMR spectroscopy. 3. Analytics: 3.1 1 H-NMR spectrum

1

H-NMR (200 MHz, D 2O): δ (ppm) = 7.85 (d, 3J = 8.4 Hz, 2 H, C-H(2)), 7.62 (d, 3J = 8.7 Hz, 4 H, CH(3) and C-H(6)), 6.68 (d, 3J = 9.1 Hz, 2 H, C-H(7)), 2.91 (s, 6 H, C-H(9)).

Literature value: 1

H-NMR (200 MHz, D 2O): δ (ppm) = 7.65 (m, 6 H, C-H(2), C-H(3) and C-H(6)), 6.65 (d, 2 H, CH(7)), 2.80 (s, 6 H, C-H(9)).[1] The measured analytical data are largely consistent with the literature values. Minor deviations presumably refer to differences in the preparation of the samples. The signal at δ = 4.79 ppm is due to the used solvent D2O. The signal at δ = 2.21 ppm comes from acetone.

page 1

Name: Michele Cuciuffo,

Room: F2.044, Place 4

Datum:

3.2 Calculation Discussion The change of the chemical shift of the guest ∆δG was calculated by the difference between the measured shift in the mixture δobsG and the chemical shift of the pure guest solution δG0 (see equation (1)). ∆δG = δobsG–δG0.

(1)

For analysis the molar ratio [G] 0/([H0+G0) was plotted against the product of chemical shift difference and the starting concentration of the substrate ∆δG·[S]0, [S]0 = 8.8 mmol L -1. The results of the calculations are depict in table 1. The plot is shown in figure 2. table 1: calculated values. [S]0 = [G]0/([H0+G0)

δobsG [ppm]

∆δG [ppm]

∆δG·[S]0

0.0

/

/

/

0.2

8.007

0.138

1.211

0.3

8.014

0.144

1.268

0.4

7.991

0.121

1.067

0.5

7.999

0.129

1.137

0.6

7.978

0.108

0.949

0.7

7.951

0.081

0.714

0.8

7.934

0.064

0.564

0.9

7.903

0.034

0.296

1.000

7.870

/

/

3.3 Discussion [S]0 = 0.3 ist the concentration with the highest molar ratio. Equation (2) was used to calculate the stoichiometric factor n with the requirement m = 1 (see equation) G H

=

n (m+ n)

0.3 =

n (1+ n)

0.3 + 0.3⋅n = n , n = 0.43 The resulting stoichiometric ratrio is

n m

(2)

(3) (4)

= 0.43.

5. Waste Disposal Solvents (Cyclodextrin and methyl orange) were disposed in the canister for organic solutions.

page 2

Name: Michele Cuciuffo,

Room: F2.044, Place 4

6. Literature [1] L. Fielding, Tetrahedron 2000, 56, 6151—6170.

page 3

Datum:...