Figure 1. Space filling model of αCyD. Left: view through the hole. Right: side view. αCyD in water — Analytical Chemistry Chemistry Question
Cyclodextrins
Figure 1. Space filling model of αCyD. Left: view through the hole. Right: side view.
αCyD in water is able to host hydrophobic molecules. When the host : guest (H : G) stoichiometry is 1 : 1, the inclusion complexation can be given by the following equilibrium.
BTAD. (Note that the spectra given in Figure 2 were measured in the complexation equilibrium state.)
Figure 2. Expanded 1H NMR spectra (signals from H–1 of αCyD) of solutions with concentration of αCyD equal to 5.0 · 10-3 mol dm-3 and that of BTAD equal from 0 to 3.0 · 10-2 mol dm-3.
glucopyranose formulas. Also, add four OH groups and four H atoms to complete the α-D-glucopyranose formula.
Model Answer
Absolute configuration at C-2: R
Absolute configuration at C-5: R
Chain form:
Draw the structure of the complex.
Consider a solution in which concentrations of αCyD as well as BTAD are equal to 5.0 · 10–3 mol dm-3 and the relative peak areas of the doublets at 5.06 and 5.14 ppm... equal to 1.0 · 10–2 mol dm-3 are positioned at 0.815 ppm. Calculate (to 2 significant figures) the value of K for the complexation of αCyD/HTAB.
Model Answer
a5.06: relative area of the peak at 5.06 ppm = mole fraction of free αCyD
a5.14: relative area of the peak at 5.14 ppm = mole fraction of αCyD complexed with BTAD
Identify the correct option.
Model Answer
The correct answer: a
Determine the equilibrium properties of the αCyD/HTAB system.
Model Answer
(M = mol dm–3)
In 1.0 · 10–2 M / 1.0 · 10–2 M αCyD/HTAB
10/10 free 10/10
complex free
0.815 0.740 = = = 0.625
0.860 0.740 s s
f s s
− − − −
sfree : chemical shift of HTAB in free, and complexed state
scomplex: chemical shift of HTAB in a complexed state
s10/10: chemical shift of HTAB in 10.0 mM / 10.0 mM αCyD/HTAB
f10/10: mole fraction of complexed HTAB in 10.0 mM / 10.0 mM αCyD/HTAB
K = 4.4 · 102
At 40.0 ºC and 60.0 ºC, K for the complexation of αCyD / HTAB are 3.12 · 102 and 2.09 · 102 respectively. Calculate (to 2 significant figures) the enthalpy change, ∆Hº and the entropy change, ∆Sº [J K–1 mol–1]. (Ignore the temperature dependence of ∆Hº and ∆Sº.)
Model Answer
From ∆Go = – RT ln K
∆Go(40.0 °C) = – 8.314 × 313. 2 × ln (3.12 · 102) = –14.94 · 103 J mol–1
∆Go(60.0 °C) = – 8.314 × 333. 2 × ln (2.09 · 102) = –14.79 · 103 J mol–1
From ∆Go = ∆Ho – T∆So
–14.94 · 103 = ∆Ho – 313. 2 × ∆So
–14.79 · 103 = ∆Ho – 333. 2 × ∆So
∆So = –7.5 J K–1mol–1; ∆Ho = –17 kJ mol–1