Chem 310
Host Guest Chemistry I
β-Cyclodextrin Binding Assay
Fall 2009
In this experiment you will determine the stability constant KS of the inclusion complex (H⊂G)
formed between β-cyclodextrin (H) and 2-methylcyclohexanone (G) and verify the complex
stoichiometry via a Job plot.
Background
β-Cyclodextrin is a cyclic oligosaccharide composed of seven glucose units and shaped like a
cone with a hydrophobic cavity. In water, small organic molecules (guests, G) of appropriate size
and shape, such as 2-methylcyclohexanone, may bind inside the cavity to form an inclusion
complex (H⊂G). The symbol ‘⊂’ means that the guest G is included inside the host H (βcyclodextrin). The host-guest binding equilibrium in Figure 1C describes this binding process:
Figure 1
HO
O
HO
O
O
HO
O
O
O
OH
6.0-6.5 A
O
HO
OH
O
OH
O
OH
OH
O
OH
O
HO
HO
OH
OH
OH
HO
OHO
HO
OHO
O OH
OH
OH
O
OH
O
OH
OH
OOH
OHOOH O
OH
O
~8A
Cavity Volume ~ 262
A3
HO
(Α) β-cyclodextrin
O
O
+
CH3
H
G
Host
Guest
KS
CH3
HCG
(B) Space-filling models of β-cyclodextrin
and of the inclusion complex with 2methylcyclohexanone. red: O; white: H;
black: C
Inclusion Complex
(C) Host guest binding equilibrium
The stability constant KS of the inclusion complex is defined by equation 1:
[H⊂G]
KS = 
[H] x [G]
(1)
[H⊂G], [H], and [G] are the concentrations of H⊂G, H and G at equilibrium. KS depends on the
hydrophobicity of the guest, the size and shape match between host cavity and guest,
temperature and the solvent. Since complexation is driven by hydrophobic interactions, binding
is only observed in water.
You will study this binding process and will use 1H NMR spectroscopy to detect complexation
of 2-methylcyclohexanone, to quantify the stability of the inclusion complex KS and to verify the
complex stoichiometry (1:1, 2:1 or 1:2 Host:Guest ratio is complex).
You will prepare an NMR sample solution by mixing a solution of β-cyclodextrin in D2O of
known concentration [H]stock with a solution of 2-methylcyclohexanone in D2O of the
concentration [G]stock at the volume ratio r, that is assigned to you. The total host and guest
concentration in your NMR sample solution will be [H]0 and [G]0, respectively. You will record
a 1H NMR spectrum of your sample at 25oC and measure the chemical shift of the guest’s methyl
group δexp. Your spectrum will be referenced to the HDO signal at δ 4.768 ppm.
For this experiment, the following stock solutions will be provided:
Stock Solution 1: β-Cyclodextrin in D2O; [H]stock = 15.41 mM
Stock Solution 2: 2-Methylcyclohexanone in D2O; [G]stock = 15.41 mM
1. Detection of complexation by 1H NMR spectroscopy and determination of the fraction of
complexed and free guest at equilibrium
Binding of 2-methylcyclohexanone inside the β-cyclodextrin leads to a small shift of the guest’s
1
H NMR signals. One refers to that shift as the complexation induced shift (CID). Most
characteristic is the CID for the CH3 protons of the guest. The protons of the free guest resonate
at δfree and those of complexed guest at δcompl. The maximal CISmax is:
δfree = 0.9606 ppm
(2)
δcompl = 1.085 ppm
(3)
CISmax = δcompl - δfree = 0.1245 ppm
(4)
δcompl is the measured chemical shift of the methyl protons if at very high β-cyclodextrin
concentration and low guest concentration. At lower cyclodextrin concentration, fraction of the
guest will be complexed (fcompl) and the remaining will be free (ffree).
fcompl + ffree = 1
(5)
fcompl and ffree are the mole fraction of complexed and free guest. At equilibrium, complexed guest
exchanges with free guest. This exchange is rapid on the 1H NMR time scale (µs – ms time
scale). Thus, if you measure the 1H NMR spectrum of a solution containing host and guest, the
measured chemical shift of the guest’s methyl protons (δexp) will be the weighted average of
δcompl and δfree:
δexp = fcompl x δcompl + ffree x δfree
(6)
Calculate fcompl and ffree from δexp and equations 2-6 and calculate the concentrations of host
[H], guest [G] and inclusion complex [H⊂G] at equilibrium.
2. Determination of the stoichiometry of the host-guest inclusion complex via a Job Plot
The Job plot method allows determination of the stoichiometry of a binding event and was
introduced by P. Job in 1928 (Job, P. Ann. Chim. (1928) 9, 113-203). In this methodology, a
series of binding experiments are carried, in which the total concentration of host and guest are
held constant, but their mole fractions are varied. In each binding experiment, the concentration
of complex is determined and is plotted against the mole fractions of host and guest. The
maximum of a Job plot corresponds to the stoichiometry of the complex.
Construct a Job plot by plotting [[H⊂G] against the mole fraction of host and guest in the
NMR sample. The spectrum that you record at your assigned stock solutions ratio provides one
data point. For a full analysis, you will need 1H NMR spectra at additional ratio such that you
cover the full range between 0 and 1. A good set of ratio is 1:9; 2:8; 3:7; 4:6; 5:5; 6:4; 7:3; 8:2;
and 9:1. You will receive spectra of all ratios other than your assigned ratio for your analysis.
3. Complex Stability Constant KS
Once you know the complex stoichiometry (it should be 1:1), you can calculate KS for each ratio
of Host and Guest stock solution. Calculate the average value for KS.
4. Instructions
Using the 1 mL syringe provided to remove your assigned amount of stock solution 1 and place
it in your drum vial. Then remove the corresponding amount of stock solution 2 and place it into
the drum vial.
IMPORTANT: Don’t contaminate the syringe for solution 1 with solution 2 and vise versa!
Close the vial with the screw cap and mix the solution. Then transfer about 700 microliter into a
clean and dry NMR tube and cap it immediately.
IMPORTANT: D2O solutions are hydroscopic. Thus, expose your D2O solutions as little as
possible to the atmosphere. Moisture adsorbed from the atmosphere will increase the HDO
signal and may reduce the quality of your NMR spectrum.
Once the sample is prepared, record a 1H NMR spectrum at 25oC using the standard proton
parameters. (Remember to select D2O as solvent after inserting the sample)
Reference your sample to the HDO signal at 4.768 ppm
Determine the chemical shift of the methyl group of 2-methylcyclohexanone
Record the chemical shift in the corresponding column in the Table below.
For each experiment, calculate fcompl (guest), ffree (guest), [H⊂G], [G], [H] and KS for a 1:1
binding model.
Experiment #
Sol 1/µL
Sol 2/µL
[G]0/mmol/L
[H]0/mmol/L
δexp
fcompl (guest)
ffree (guest)
[H]/mmol/L
[G]/mmol/L
[H⊂G]/mmol/L
KS/M-1
1
100
900
2
200
800
3
300
700
4
400
600
5
500
500
6
600
400
7
700
300
8
800
200
9
900
100
5. Discussion
In your discussion of the results include the following points:
- How good is the Job Plot? A typical Job plot for a 1:1 complex should have roughly the
shape of an inverted parabola centered at 0.5. True? Are data points jumping out? How
well does your own data point fit? Can you say with confidence that it is a 1:1 complex?
- How large is spread in the binding constant for the individual experiments?
- How well does your own experimental data point fit? What is the deviation to the average
KS in comparison to that of the data points calculated from the provided 1H NMR spectra.
- In your discussion of the quality of your and of the data points extracted from the provided
spectra, you should consider possible experimental errors in the measurements:
o How precise is your measured chemical shift? Since CISmax is quite small, a small
error in δexp could have a large effect on the calculated mole fractions of free and
complexed guest. Possible errors of δexp could result from broad lines as a
consequence of bad shimming. Another error may result from the temperature
dependence of the chemical shift of the HDO reference signal: δHDO = 5.060 –
0.0122 x T + (2.1 x 10-5) x T2. The guest’s methyl signal shows very little
temperature dependence.
o Another, perhaps larger source of error may be the total concentration of host and
guest in the samples. To prepare the solutions for the spectra provided, an
Eppendorf microliter pipette was used that has an error of < 2-3%. The syringe, that
you used has probably an error of +/- 20-30 µL if you measure 1 mL and perhaps a
very similar absolute error if you measure 0.2 mL.
- How does your measured stability constant compare with those of other similar sized
guests (see lecture handout ‘Host-Guest Chemistry’)?
-