IChO 2010 · Theory problem 9

α-Cyclodextrin

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α-Cyclodextrin (αCyD), which is a cyclic oligosaccharide of six α(1→4) linked α-D-glucopyranoside units, can be topologically represented as toroids (Figure 1). α-D-glucopyranoside units in αCyD are usually in the most stable chair conformation.

Figure 1
Figure 2

Figure 1. Space filling model of αCyD.
Left: view through the hole.
Right: side view.

9.1 Give the absolute configuration (\(R\) or \(S\)) at stereogenic carbons C-2 and C-5 of D-glucose. Also, draw a stereostructure of the open chain form of D-glucose.

9.2 Choose the most stable conformation from the following four incomplete α-D-glucopyranose formulas. Also, add four \(\ce{OH}\) groups and four \(\ce{H}\) atoms to complete the α-D-glucopyranose formula.

Figure 3

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.

Figure 4

where \(k_1\) and \(k_{-1}\) are the rate constant for the forward and backward reaction, respectively. The complexation of a guest to αCyD causes a chemical shift change in \(\ce{^1H}\) NMR spectra.

Figure 2 shows a part of \(\ce{^1H}\) NMR spectra (signals from H–1 of αCyD) showing the chemical shift change in the presence of varying amounts of BTAD (1,10-bis(trimethylammonium)decane diiodide). The doublet peak at 5.06 ppm is from H–1 of free αCyD, while the doublet at 5.14 ppm is from H–1 of αCyD complexed with BTAD. (Note that the spectra given in Figure 2 were measured in the complexation equilibrium state.)

Figure 5
Figure 6

Figure 2.
Expanded \(\ce{^1H}\) NMR spectra (signals from H–1 of αCyD) of solutions with concentration of αCyD equal to \(5.0\cdot10^{-3}\ \mathrm{mol\,dm^{-3}}\) and that of BTAD equal from 0 to \(3.0\cdot10^{-2}\ \mathrm{mol\,dm^{-3}}\).

9.3 Consider a solution in which concentrations of αCyD as well as BTAD are equal to \(5.0\cdot10^{-3}\ \mathrm{mol\,dm^{-3}}\) and the relative peak areas of the doublets at 5.06 and 5.14 ppm are 0.41 and 0.59, respectively. Calculate (to 2 significant figures) the concentration equilibrium constant, \(K\), for the inclusion complexation of αCyD / BTAD.

Complexation of αCyD with hexyltrimethylammonium bromide (HTAB) appears in NMR spectra in a way different from the αCyD/BTAD complexation. Figure 3 shows a part of \(\ce{^1H}\) NMR spectra (H-6 signal of HTAB) in αCyD/HTAB solutions. The signal appears as one triplet (not two triplets), which shifts depending on the concentration of αCyD from the position of free HTAB to the position of αCyD/HTAB in proportion to the fraction of the complex in the solution. The H-6 signals from free HTAB and HTAB complexed with αCyD are triplets at 0.740 ppm and 0.860 ppm, respectively.

Figure 7
Figure 8

Figure 3.
Expanded \(\ce{^1H}\) NMR spectra (H-6 signal of HTAB) of solutions with concentration of HTAB equal to \(1.0\cdot10^{-2}\ \mathrm{mol\,dm^{-3}}\) and that of αCyD equal from 0 to \(3.0\cdot10^{-2}\ \mathrm{mol\,dm^{-3}}\).

9.4 The signal of HTAB in αCyD/HTAB solutions appears as one triplet that shifts depending on the concentration of αCyD. Choose the rational interpretation(s) just from these spectra.

Hint: When a guest molecule move in and out of αCyD rapidly and repeatedly, only one signal of the guest is observed at the weighted average of the chemical shifts of the free guest and the shift of the guest included in αCyD.

a) \(k_1\) of αCyD/HTAB \(>\) \(k_1\) of αCyD/BTAD

b) \(k_1\) of αCyD/HTAB \(<\) \(k_1\) of αCyD/BTAD

c) \(K\) of αCyD/HTAB \(>\) \(K\) of αCyD/BTAD

d) \(K\) of αCyD/HTAB \(<\) \(K\) of αCyD/BTAD

9.5 The signals of HTAB in the solution with concentrations of αCyD as well as HTAB equal to \(1.0\cdot10^{-2}\ \mathrm{mol\,dm^{-3}}\) are positioned at 0.815 ppm. Calculate (to 2 significant figures) the value of \(K\) for the complexation of αCyD/HTAB.

9.6 At 40.0 °C and 60.0 °C, \(K\) for the complexation of αCyD / HTAB are \(3.12\cdot10^2\) and \(2.09\cdot10^2\) respectively. Calculate (to 2 significant figures) the enthalpy change, \(\Delta H^\circ\) [\(\mathrm{kJ\,mol^{-1}}\)] and the entropy change, \(\Delta S^\circ\) [\(\mathrm{J\,K^{-1}\,mol^{-1}}\)]. (Ignore the temperature dependence of \(\Delta H^\circ\) and \(\Delta S^\circ\).)

Official English text of the problem.

Answers

  • 9.1: C-2: \(R\); C-5: \(R\). Open-chain D-glucose: O=C[C@H](O)[C@@H](O)[C@H](O)[C@H](O)CO. Its Fischer projection has \(\ce{CHO}\) at the top, \(\ce{CH2OH}\) at the bottom, and the \(\ce{OH}\) groups at C-2 through C-5 right, left, right, right.
  • 9.2: Choose the lower-right structure, the \(^{4}C_1\) chair of α-D-glucopyranose: OC[C@H]1O[C@H](O)[C@H](O)[C@@H](O)[C@@H]1O. The C-1 \(\ce{OH}\) is axial down; the C-2, C-3 and C-4 \(\ce{OH}\) groups and C-5 \(\ce{CH2OH}\) group are equatorial. Complete the labels as specified below.
  • 9.3: \(\boxed{K_{\mathrm{BTAD}}=7.0\times10^2\ \mathrm{dm^3\,mol^{-1}}}\).
  • 9.4: a) is the intended qualitative interpretation of the exchange pattern. The quantitative population data also establish d); see the qualification below.
  • 9.5: \(\boxed{K_{\mathrm{HTAB}}=4.4\times10^2\ \mathrm{dm^3\,mol^{-1}}}\).
  • 9.6: \(\boxed{\Delta H^\circ=-17\ \mathrm{kJ\,mol^{-1}}}\), \(\boxed{\Delta S^\circ=-7.7\ \mathrm{J\,K^{-1}\,mol^{-1}}}\).

Solution

9.1

The required Fischer projection of open-chain D-glucose has the carbon chain vertical, with the aldehyde group \(\ce{CHO}\) uppermost and \(\ce{CH2OH}\) at the bottom.

Stereogenic carbon Left-hand substituent Right-hand substituent
C-2 \(\ce{H}\) \(\ce{OH}\)
C-3 \(\ce{OH}\) \(\ce{H}\)
C-4 \(\ce{H}\) \(\ce{OH}\)
C-5 \(\ce{H}\) \(\ce{OH}\)

In a Fischer projection, horizontal bonds point towards the viewer and vertical bonds point away.

At C-2, the CIP priorities are

\[ \ce{OH} > \text{C-1 side} > \text{C-3 side} > \ce{H}. \]

The aldehyde carbon is compared as attached to \((\ce{O},\ce{O},\ce{H})\), whereas C-3 is attached to \((\ce{O},\ce{C},\ce{H})\). Therefore the aldehyde side has higher priority. Accounting for the horizontal hydrogen pointing towards the viewer gives \(R\).

At C-5, the priorities are

\[ \ce{OH} > \text{C-4 side} > \ce{CH2OH} > \ce{H}. \]

The C-4 side has attached atoms \((\ce{O},\ce{C},\ce{H})\), whereas C-6 has \((\ce{O},\ce{H},\ce{H})\). The same Fischer-projection analysis gives \(R\).

Thus

\[ \boxed{\text{C-2: }R,\qquad \text{C-5: }R}. \]

The complete open-chain configuration is \((2R,3S,4R,5R)\).

9.2

The lower-right drawing is the stable \(^{4}C_1\) chair. Complete it according to this table; “up” and “down” refer to the two faces of the ring.

Carbon Oxygen-containing substituent Hydrogen
C-1 \(\ce{OH}\): axial, down Equatorial, up
C-2 \(\ce{OH}\): equatorial, down Axial, up
C-3 \(\ce{OH}\): equatorial, up Axial, down
C-4 \(\ce{OH}\): equatorial, down Axial, up
C-5 \(\ce{CH2OH}\): equatorial, up Axial, down

Specifically, add the four missing \(\ce{OH}\) labels at C-2, C-3, C-4, and the end of the C-6 side chain. Add the four missing \(\ce{H}\) labels at C-2, C-3, C-4, and C-5.

This chair places the three non-anomeric ring hydroxyl groups and the hydroxymethyl group equatorial, minimizing unfavorable axial interactions. For the α anomer, the anomeric hydroxyl group is on the opposite face from the C-5 hydroxymethyl group, so it is axial down in this chair.

The SMILES given above specifies the stereoisomer; the chair conformation and axial/equatorial placements are specified by the table.

9.3

Let H denote the host, G the guest, and HG their 1:1 inclusion complex:

\[ \ce{H + G <=> HG}. \]

The concentration equilibrium constant is

\[ K=\frac{[\mathrm{HG}]}{[\mathrm H][\mathrm G]}. \]

Each free or complexed host contributes the same number of H–1 protons. Consequently, the integrated peak areas give the fractions of host in the two states:

\[ f_{\mathrm{free}}=0.41,\qquad f_{\mathrm{complexed}}=0.59. \]

With total host and guest concentrations both equal to \(5.0\times10^{-3}\ \mathrm{mol\,dm^{-3}}\),

\[ [\mathrm{HG}] =0.59(5.0\times10^{-3}) =2.95\times10^{-3}\ \mathrm{mol\,dm^{-3}}, \]

\[ [\mathrm H] =0.41(5.0\times10^{-3}) =2.05\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

The 1:1 stoichiometry and equal total concentrations give

\[ [\mathrm G] =5.0\times10^{-3}-2.95\times10^{-3} =2.05\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

Therefore,

\[ K_{\mathrm{BTAD}} =\frac{2.95\times10^{-3}} {(2.05\times10^{-3})^2} =701.96\ \mathrm{dm^3\,mol^{-1}}, \]

or, to two significant figures,

\[ \boxed{K_{\mathrm{BTAD}}=7.0\times10^2\ \mathrm{dm^3\,mol^{-1}}}. \]

9.4

For HTAB, the single triplet at an intermediate chemical shift indicates rapid exchange between free and bound states on the NMR timescale. Its observed shift is

\[ \delta_{\mathrm{obs}} =(1-f)\delta_{\mathrm{free}}+f\delta_{\mathrm{bound}}, \]

where \(f\) is the fraction of guest bound.

For BTAD, separate free-host and complexed-host doublets remain visible, indicating slow exchange on the NMR timescale.

The intended qualitative kinetic interpretation is therefore a): faster association/exchange for HTAB.

An averaged signal does not itself establish stronger or weaker equilibrium binding. At equilibrium,

\[ k_1[\mathrm H][\mathrm G]=k_{-1}[\mathrm{HG}], \qquad K=\frac{k_1}{k_{-1}}. \]

Both forward and reverse exchange can become faster without increasing their ratio. Thus neither c) nor d) follows merely from the distinction between averaged and separate signals.

Using the quantitative populations, however, parts 9.3 and 9.5 establish

\[ K_{\mathrm{HTAB}}<K_{\mathrm{BTAD}}, \]

so d) is also supported when the numerical information in the spectra is included.

9.5

Let \(f\) be the fraction of HTAB present in the complex. The weighted-average shift gives

\[ 0.815=(1-f)(0.740)+f(0.860). \]

Hence

\[ f=\frac{0.815-0.740}{0.860-0.740} =\frac{0.075}{0.120} =0.625. \]

The total guest concentration is \(1.0\times10^{-2}\ \mathrm{mol\,dm^{-3}}\), so

\[ [\mathrm{HG}] =0.625(1.0\times10^{-2}) =6.25\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

Because the total host and guest concentrations are equal,

\[ [\mathrm H]=[\mathrm G] =1.0\times10^{-2}-6.25\times10^{-3} =3.75\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

Therefore,

\[ K_{\mathrm{HTAB}} =\frac{6.25\times10^{-3}} {(3.75\times10^{-3})^2} =444.44\ \mathrm{dm^3\,mol^{-1}}, \]

giving

\[ \boxed{K_{\mathrm{HTAB}}=4.4\times10^2\ \mathrm{dm^3\,mol^{-1}}}. \]

9.6

Convert the temperatures to kelvin:

\[ T_1=313.15\ \mathrm K,\qquad T_2=333.15\ \mathrm K. \]

For thermodynamic logarithms, use the dimensionless constant \(K^\circ=Kc^\circ\), with \(c^\circ=1\ \mathrm{mol\,dm^{-3}}\). Its numerical values are therefore 312 and 209.

Assuming temperature-independent \(\Delta H^\circ\), the integrated van ’t Hoff equation is

\[ \ln\left(\frac{K_2^\circ}{K_1^\circ}\right) =-\frac{\Delta H^\circ}{R} \left(\frac1{T_2}-\frac1{T_1}\right). \]

Using \(R=8.31446\ \mathrm{J\,K^{-1}\,mol^{-1}}\),

\[ \Delta H^\circ = -\frac{R\ln(209/312)} {1/333.15-1/313.15} =-1.73773\times10^4\ \mathrm{J\,mol^{-1}}. \]

Thus

\[ \boxed{\Delta H^\circ=-17\ \mathrm{kJ\,mol^{-1}}}. \]

The negative sign agrees with the decrease of \(K\) as temperature increases.

Next,

\[ \Delta G^\circ=-RT\ln K^\circ =\Delta H^\circ-T\Delta S^\circ, \]

so

\[ \Delta S^\circ=R\ln K^\circ+\frac{\Delta H^\circ}{T}. \]

Using \(T_1\) and retaining the unrounded enthalpy,

\[ \Delta S^\circ =(8.31446)\ln312 +\frac{-17377.3}{313.15} =-7.7419\ \mathrm{J\,K^{-1}\,mol^{-1}}. \]

Therefore,

\[ \boxed{\Delta S^\circ=-7.7\ \mathrm{J\,K^{-1}\,mol^{-1}}}. \]

The solution was checked against the official solution.