IChO 2019 · Theory problem 5

Azobenzene–β-cyclodextrin complexes for the formation of nanomachines

Nanomachines are molecular assemblies that enable the transformation of an energy source into a nano-movement for applications such as drug delivery. Numerous nanomachines make use of the isomerization of azo compounds (\(\ce{R-N=N-R'}\)) upon irradiation.

5.1 Draw the stereoisomers of azobenzene (\(\ce{H5C6-N=N-C6H5}\)) and draw a line between the two carbon atoms that are the furthest apart. Compare these two distances (\(d_{\mathrm{trans}}\) and \(d_{\mathrm{cis}}\)).

Figure 1

Fig. 1. Possible reactants for the synthesis of M.

5.2 M can be synthesized in two steps from simple reactants (Fig 1). Choose among the suggested reactants (N to Q) the ones that can provide M with very high regioselectivity. Sodium nitrite (\(\ce{NaNO2}\)) in cold aqueous hydrochloric acid is used as reagent for the first step of the synthesis.

Determination of the association constant \(K_t\)

β-cyclodextrin (C, Fig. 2) is a cyclic heptamer of glucose, which can form inclusion complexes with azo compounds. In tasks 3 to 6, we will determine by spectroscopy the association constant \(K_t\), corresponding to the formation of the inclusion complex \(\mathbf{CM}_{\mathrm{trans}}\) as depicted in Fig. 2.

Figure 2

Fig. 2 – Formation of the \(\mathbf{CM}_{\mathrm{trans}}\) inclusion complex.

Several solutions are prepared by mixing C and \(\mathbf{M}_{\mathrm{trans}}\) in different proportions to reach initial concentrations \([\mathbf C]_0\) and \([\mathbf M_{\mathrm{trans}}]_0\). While \([\mathbf M_{\mathrm{trans}}]_0\) is identical for all solutions, \([\mathbf C]_0\) varies. We follow, at a fixed wavelength, the evolution of the difference in absorbance \(\Delta A\) between the absorbance of each solution and the pure \(\mathbf M_{\mathrm{trans}}\) solution. We note the molar absorption coefficients of \(\mathbf{CM}_{\mathrm{trans}}\) and \(\mathbf M_{\mathrm{trans}}\), \(\varepsilon_{\mathrm{CM}_{\mathrm{trans}}}\) and \(\varepsilon_{\mathrm M_{\mathrm{trans}}}\), respectively. \(L\) is the path length of the beam through the sample. The absorbance of C (\(\varepsilon_{\mathrm C}\)) is negligible.

5.3 Demonstrate that \(\Delta A=\alpha[\mathbf{CM}_{\mathrm{trans}}]\) and express \(\alpha\) in terms of known constant(s).

5.4 Demonstrate that in case when C is in large excess with respect to \(\mathbf M_{\mathrm{trans}}\) (i.e. \([\mathbf C]_0\gg[\mathbf M_{\mathrm{trans}}]_0\)), the concentration of C may be considered as constant, \([\mathbf C]\simeq[\mathbf C]_0\).

5.5 Demonstrate that in case when C is in a large excess with respect to \(\mathbf M_{\mathrm{trans}}\), (i.e. \([\mathbf C]_0\gg[\mathbf M_{\mathrm{trans}}]_0\)),

\[ \Delta A=\alpha\frac{\beta[\mathbf C]_0}{1+K_t[\mathbf C]_0} \]

and express \(\beta\) in terms of constant(s) and initial concentration(s).

5.6 Determine \(K_t\) using the following experimental curve (Fig. 3).

Figure 3

Fig. 3 – Evolution of \(1/\Delta A\) as a function of \(1/[\mathbf C]_0\).

Determination of the association constant \(K_c\)

In tasks 7 to 9, we will determine by kinetic studies the association constant \(K_c\), corresponding to the formation of the inclusion complex with \(\mathbf M_{\mathrm{cis}}\), \(\mathbf{CM}_{\mathrm{cis}}\). A sample containing only \(\mathbf M_{\mathrm{trans}}\) is irradiated, thus producing a known amount of \(\mathbf M_{\mathrm{cis}}\), \([\mathbf M_{\mathrm{cis}}]_0\). \(\mathbf M_{\mathrm{cis}}\) (free or within the inclusion complex) then thermally isomerizes into \(\mathbf M_{\mathrm{trans}}\). In the absence of C, the isomerization follows a first order kinetics with a rate constant \(k_1\). All complexation equilibria are faster than the isomerization processes. The kinetic scheme corresponding to this experiment is provided in Fig. 4.

Figure 4

Fig. 4 – Kinetic scheme for the isomerization of \(\mathbf M_{\mathrm{cis}}\) in the presence of C.

The rate of disappearance \(r\) for the total amount of \(\mathbf M_{\mathrm{cis}}\) (free and complexed) is defined as

\[ r=k_1[\mathbf M_{\mathrm{cis}}]+k_2[\mathbf{CM}_{\mathrm{cis}}]. \]

Experimentally, \(r\) follows an apparent first order kinetic law with an apparent rate constant \(k_{\mathrm{obs}}\):

\[ r=k_{\mathrm{obs}}\bigl([\mathbf M_{\mathrm{cis}}]+[\mathbf{CM}_{\mathrm{cis}}]\bigr). \]

5.7 Demonstrate that

\[ k_{\mathrm{obs}}= \frac{\gamma+\delta k_2[\mathbf C]}{1+K_c[\mathbf C]} \]

and express \(\gamma\) and \(\delta\) in terms of known constant(s).

5.8 Choose in which condition(s) the half-life \(t_{1/2}\) corresponding to \(k_{\mathrm{obs}}\) can be expressed as \(t_{1/2}=\ln(2/\gamma)(1+K_c[\mathbf C]_0)\) given that \([\mathbf C]_0\gg[\mathbf M_{\mathrm{cis}}]_0\). Mathematically justify your answer.

5.9 Assuming the condition(s) in task 8 satisfied, determine \(K_c\) by a linear regression using the data below. You may use a calculator or plot a graph.

\([\mathbf C]_0\) (\(\mathrm{mol\,dm^{-3}}\)) \(t_{1/2}\) (s) \([\mathbf C]_0\) (\(\mathrm{mol\,dm^{-3}}\)) \(t_{1/2}\) (s)
0 3.0 \(3.0\cdot10^{-3}\) 5.9
\(1.0\cdot10^{-4}\) 3.2 \(5.0\cdot10^{-3}\) 7.7
\(5.0\cdot10^{-4}\) 3.6 \(7.5\cdot10^{-3}\) 9.9
\(1.0\cdot10^{-3}\) 4.1 \(1.0\cdot10^{-2}\) 12.6

Formation of nanomachines

Figure 5

Fig. 5 – Cleavage of an azobenzene–cyclodextrin inclusion complex induced by a light-triggered isomerization, which allows delivery of a dye (grey circles).

Another azobenzene compound (for which \(K_c\ll K_t\)), initially in the trans form, is covalently grafted on silica (Fig. 5). The silica pores are filled with a dye (rhodamine B, grey circles in Fig. 5). Upon addition of C, an inclusion complex is formed, which blocks the pores and prevents the release of the dye.

5.10 Choose the most appropriate condition (one choice only) so that the pores are initially blocked in the presence of C, and the dye can be released upon irradiation.

This azobenzene-silica powder loaded with a dye is placed in the corner of a cuvette (Fig. 6) so that the powder cannot move into solution. The powder is irradiated at a wavelength \(\lambda_1\) to trigger the release of the dye from the pores (Fig. 5). To monitor this release by absorbance spectroscopy we measure the absorbance of the solution at wavelength \(\lambda_2\).

Figure 6

Fig. 6 – Left: experimental setup used to monitor the release of the dye; right: absorption spectra of trans-azobenzene (full line), cis-azobenzene (dotted line) and rhodamine B (dashed line).

5.11 Determine \(\lambda_1\).

5.12 Determine \(\lambda_2\).

Official English text of the problem.

Answers

  • 5.1: trans-\((E)\)-azobenzene, c1ccc(/N=N/c2ccccc2)cc1, and cis-\((Z)\)-azobenzene, c1ccc(/N=N\c2ccccc2)cc1. Their maximum carbon-to-carbon distances satisfy \(\boxed{d_{\mathrm{trans}}>d_{\mathrm{cis}}}\).
  • 5.2: Choose N, anthranilic acid, Nc1ccccc1C(=O)O, and P, \(N,N\)-dimethylaniline, CN(C)c1ccccc1. Diazotize N, then couple with P at its para position.
  • 5.3: \(\boxed{\alpha=L(\varepsilon_{\mathrm{CM}_{\mathrm{trans}}}-\varepsilon_{\mathrm M_{\mathrm{trans}}})}\).
  • 5.4: \(0\le[\mathbf C]_0-[\mathbf C]\le[\mathbf M_{\mathrm{trans}}]_0\ll[\mathbf C]_0\), hence \(\boxed{[\mathbf C]\simeq[\mathbf C]_0}\).
  • 5.5: \(\boxed{\beta=K_t[\mathbf M_{\mathrm{trans}}]_0}\).
  • 5.6: \(\boxed{K_t\approx1.9\times10^2\ \mathrm{L\,mol^{-1}}}\).
  • 5.7: \(\boxed{\gamma=k_1,\qquad\delta=K_c}\).
  • 5.8: With constant free host concentration, require \(\boxed{k_2K_c[\mathbf C]_0\ll k_1}\), including the limiting case \(k_2=0\). The corrected expression is \(\boxed{t_{1/2}\simeq\frac{\ln2}{\gamma}(1+K_c[\mathbf C]_0)}\).
  • 5.9: The regression gives \(t_{1/2}=(3.0804\ \mathrm s)+(935.675\ \mathrm{s\,L\,mol^{-1}})[\mathbf C]_0\), hence \(\boxed{K_c\approx3.0\times10^2\ \mathrm{L\,mol^{-1}}}\).
  • 5.10: \(\boxed{K_c[\mathbf C]\ll1\ll K_t[\mathbf C]}\), equivalently \(\boxed{1/K_t\ll[\mathbf C]\ll1/K_c}\). Under host excess, use \([\mathbf C]\simeq[\mathbf C]_0\).
  • 5.11: \(\boxed{\lambda_1=330\ \mathrm{nm}}\).
  • 5.12: \(\boxed{\lambda_2=550\ \mathrm{nm}}\).

Solution

5.1 — Azobenzene stereoisomers and distances

Azobenzene has two geometrical isomers about its \(\ce{N=N}\) double bond:

Isomer Name Isomeric SMILES
trans \((E)\)-Azobenzene, or \((E)\)-diphenyldiazene c1ccc(/N=N/c2ccccc2)cc1
cis \((Z)\)-Azobenzene, or \((Z)\)-diphenyldiazene c1ccc(/N=N\c2ccccc2)cc1

For the requested drawings, place the \(\ce{N=N}\) bond horizontally:

  • In the trans drawing, put one phenyl ring above-left and the other below-right. Join the two distal para carbon atoms, one on each ring, and label the segment \(d_{\mathrm{trans}}\).
  • In the cis drawing, put both phenyl rings on the same side, above-left and above-right. For the usual planar schematic with approximately \(120^\circ\) bond angles at nitrogen, join the leftmost carbon of the left ring to the rightmost carbon of the right ring—the outward-facing meta positions—and label the segment \(d_{\mathrm{cis}}\).

The trans form is extended, whereas the cis form is folded:

\[ \boxed{d_{\mathrm{trans}}>d_{\mathrm{cis}}}. \]

These are schematic drawings. Actual cis-azobenzene is nonplanar; the SMILES specifies the double-bond stereochemistry but does not specify phenyl-ring torsions or numerical interatomic distances.

5.2 — Synthesis of M

The proposed reactants are:

Label Name SMILES
N Anthranilic acid, or 2-aminobenzoic acid Nc1ccccc1C(=O)O
O 4-Amino-\(N,N\)-dimethylaniline CN(C)c1ccc(N)cc1
P \(N,N\)-Dimethylaniline CN(C)c1ccccc1
Q Benzoic acid O=C(O)c1ccccc1

Choose N and P.

First, diazotize the primary aromatic amino group of N using cold sodium nitrite, [Na+].[O-]N=O, in aqueous hydrochloric acid. This gives the 2-carboxybenzenediazonium ion, O=C(O)c1ccccc1[N+]#N.

The balanced net ionic equation is

\[ \ce{H2N-C6H4-COOH + NO2- + 2H+ -> HOOC-C6H4-N2+ + 2H2O}. \]

The amino/diazonium group and carboxyl group occupy adjacent positions throughout. Here nitrite is [O-]N=O, the hydrogen ion is [H+], and water is O; sodium ion, [Na+], and chloride ion, [Cl-], are spectators.

Next, carry out azo coupling with P:

\[ \ce{HOOC-C6H4-N2+ + C6H5-N(CH3)2 -> HOOC-C6H4-N=N-C6H4-N(CH3)2 + H+}. \]

The resulting product is M, methyl red, shown as \((E)\)-2-[(4-(dimethylamino)phenyl)diazenyl]benzoic acid:

CN(C)c1ccc(/N=N/c2ccccc2C(=O)O)cc1

The regioselectivity follows from two features:

  1. N already fixes the carboxyl group ortho to the future azo linkage.
  2. P’s dimethylamino group strongly activates its aromatic ring toward electrophilic azo coupling. Its less hindered para position is preferred over its ortho positions.

Mechanistically, the activated aromatic ring of P attacks the terminal nitrogen of the diazonium group; loss of the proton from the attacked carbon restores aromaticity.

The alternative O/Q pairing is unsuitable: Q’s carboxyl group deactivates its ring and directs electrophilic substitution meta, whereas M would require coupling ortho to that group.

5.3 — Absorbance change

Let

\[ m_0=[\mathbf M_{\mathrm{trans}}]_0, \qquad x=[\mathbf{CM}_{\mathrm{trans}}]. \]

Because complexation is 1:1,

\[ [\mathbf M_{\mathrm{trans}}]=m_0-x. \]

By the Beer–Lambert law, neglecting the host’s absorption,

\[ A=L\left[ \varepsilon_{\mathrm M_{\mathrm{trans}}}(m_0-x) +\varepsilon_{\mathrm{CM}_{\mathrm{trans}}}x \right]. \]

For the reference solution containing the same total concentration of uncomplexed guest,

\[ A_0=L\varepsilon_{\mathrm M_{\mathrm{trans}}}m_0. \]

Consequently,

\[ \Delta A=A-A_0 =L\left( \varepsilon_{\mathrm{CM}_{\mathrm{trans}}} -\varepsilon_{\mathrm M_{\mathrm{trans}}} \right)x. \]

Thus

\[ \boxed{ \Delta A=\alpha[\mathbf{CM}_{\mathrm{trans}}], \qquad \alpha=L\left( \varepsilon_{\mathrm{CM}_{\mathrm{trans}}} -\varepsilon_{\mathrm M_{\mathrm{trans}}} \right) }. \]

With concentrations in \(\mathrm{mol\,L^{-1}}\), \(\alpha\) has units \(\mathrm{L\,mol^{-1}}\).

5.4 — Host concentration in large excess

The host balance is

\[ [\mathbf C]_0=[\mathbf C]+[\mathbf{CM}_{\mathrm{trans}}]. \]

At most all the guest can be complexed, so

\[ 0\le[\mathbf{CM}_{\mathrm{trans}}] \le[\mathbf M_{\mathrm{trans}}]_0. \]

Therefore,

\[ 0\le \frac{[\mathbf C]_0-[\mathbf C]}{[\mathbf C]_0} \le \frac{[\mathbf M_{\mathrm{trans}}]_0}{[\mathbf C]_0} \ll1. \]

The fractional depletion of free host is negligible:

\[ \boxed{[\mathbf C]\simeq[\mathbf C]_0}. \]

5.5 — Dependence of absorbance on host concentration

The association constant is

\[ K_t= \frac{[\mathbf{CM}_{\mathrm{trans}}]} {[\mathbf C][\mathbf M_{\mathrm{trans}}]}. \]

Writing \(c_0=[\mathbf C]_0\), and using the approximation established above,

\[ K_t=\frac{x}{c_0(m_0-x)}. \]

Rearranging,

\[ x=K_tc_0(m_0-x), \]

\[ x(1+K_tc_0)=K_tc_0m_0, \]

\[ x=\frac{K_tm_0c_0}{1+K_tc_0}. \]

Substitution into \(\Delta A=\alpha x\) gives

\[ \boxed{ \Delta A= \alpha\frac{K_t[\mathbf M_{\mathrm{trans}}]_0[\mathbf C]_0} {1+K_t[\mathbf C]_0} }. \]

Comparison with the requested expression yields

\[ \boxed{\beta=K_t[\mathbf M_{\mathrm{trans}}]_0}. \]

Since \(K_t\) has units of inverse concentration, \(\beta\) is dimensionless.

5.6 — Determination of \(K_t\)

Taking the reciprocal of the expression in 5.5,

\[ \frac1{\Delta A} = \frac1{\alpha\beta}\frac1{[\mathbf C]_0} +\frac{K_t}{\alpha\beta}. \]

Thus, for a plot of \(y=1/\Delta A\) against \(X=1/[\mathbf C]_0\),

\[ \text{slope }s=\frac1{\alpha\beta}, \qquad \text{intercept }b=\frac{K_t}{\alpha\beta}, \]

and therefore

\[ \boxed{K_t=\frac bs}. \]

Using the labelled points \(Y=(100,4.2)\) and \(Z=(1000,17.2)\),

\[ s= \frac{17.2-4.2} {(1000-100)\ \mathrm{L\,mol^{-1}}} = 0.014444\ \mathrm{mol\,L^{-1}}. \]

The intercept is

\[ b=4.2- (0.014444\ \mathrm{mol\,L^{-1}}) (100\ \mathrm{L\,mol^{-1}}) =2.7556. \]

Hence

\[ K_t= \frac{2.7556}{0.014444\ \mathrm{mol\,L^{-1}}} =190.77\ \mathrm{L\,mol^{-1}}, \]

or, to sensible precision,

\[ \boxed{K_t\approx1.9\times10^2\ \mathrm{L\,mol^{-1}}}. \]

5.7 — Apparent isomerization rate constant

The cis guest is \((Z)\)-methyl red, or \((Z)\)-2-[(4-(dimethylamino)phenyl)diazenyl]benzoic acid:

CN(C)c1ccc(/N=N\c2ccccc2C(=O)O)cc1

Its inclusion complex contains this guest noncovalently inside β-cyclodextrin. The guest’s SMILES does not encode its placement within the host cavity.

Fast complexation equilibrium gives

\[ K_c= \frac{[\mathbf{CM}_{\mathrm{cis}}]} {[\mathbf C][\mathbf M_{\mathrm{cis}}]}, \]

so

\[ [\mathbf{CM}_{\mathrm{cis}}] =K_c[\mathbf C][\mathbf M_{\mathrm{cis}}]. \]

The disappearance rate is therefore

\[ r= \left(k_1+k_2K_c[\mathbf C]\right) [\mathbf M_{\mathrm{cis}}]. \]

The total concentration of cis guest is

\[ [\mathbf M_{\mathrm{cis}}]+[\mathbf{CM}_{\mathrm{cis}}] = \left(1+K_c[\mathbf C]\right)[\mathbf M_{\mathrm{cis}}]. \]

Dividing rate by total cis concentration,

\[ \boxed{ k_{\mathrm{obs}} = \frac{k_1+k_2K_c[\mathbf C]} {1+K_c[\mathbf C]} }. \]

Comparison with the stated expression gives

\[ \boxed{\gamma=k_1,\qquad\delta=K_c}. \]

The dimensions are consistent: \(\gamma\) is in \(\mathrm{s^{-1}}\), while \(\delta\) is in \(\mathrm{L\,mol^{-1}}\).

5.8 — Conditions for the half-life expression

First, the printed factor \(\ln(2/\gamma)\) must be corrected to \((\ln2)/\gamma\). A logarithm cannot have a dimensional argument, and \(\gamma=k_1\) has units \(\mathrm{s^{-1}}\).

Let the total cis concentration be

\[ u=[\mathbf M_{\mathrm{cis}}]+[\mathbf{CM}_{\mathrm{cis}}]. \]

When free host concentration remains constant,

\[ -\frac{du}{dt}=k_{\mathrm{obs}}u, \]

and integration gives

\[ u(t)=u(0)e^{-k_{\mathrm{obs}}t}. \]

Thus

\[ t_{1/2}=\frac{\ln2}{k_{\mathrm{obs}}}. \]

With \([\mathbf C]\simeq[\mathbf C]_0\), the expression from 5.7 becomes

\[ t_{1/2} = \frac{\ln2\left(1+K_c[\mathbf C]_0\right)} {k_1+k_2K_c[\mathbf C]_0}. \]

To obtain the requested linear dependence on host concentration, the second term in the denominator must be negligible:

\[ \boxed{k_2K_c[\mathbf C]_0\ll k_1}. \]

Equivalently, isomerization through the complexed population must contribute negligibly to the total rate:

\[ k_2[\mathbf{CM}_{\mathrm{cis}}] \ll k_1[\mathbf M_{\mathrm{cis}}]. \]

Then

\[ \boxed{ t_{1/2}\simeq \frac{\ln2}{k_1}\left(1+K_c[\mathbf C]_0\right) = \frac{\ln2}{\gamma}\left(1+K_c[\mathbf C]_0\right) }. \]

The expression is exact within the constant-host model when \(k_2=0\). Merely requiring \(k_2\ll k_1\) is insufficient if \(K_c[\mathbf C]_0\) is very large.

The host-excess approximation must also cover binding by the trans product. If appreciable unconverted trans guest remains after irradiation, host excess must apply to the total binding guest, including that residual trans population.

5.9 — Linear regression and \(K_c\)

Under the conditions above,

\[ t_{1/2} = \frac{\ln2}{k_1} + \frac{\ln2}{k_1}K_c[\mathbf C]_0. \]

Writing

\[ t_{1/2}=a+b[\mathbf C]_0, \]

we identify

\[ a=\frac{\ln2}{k_1}, \qquad b=aK_c, \]

so

\[ \boxed{K_c=\frac ba}. \]

For the eight tabulated points, let \(x_i=[\mathbf C]_{0,i}\) and \(y_i=t_{1/2,i}\). Ordinary least squares gives

\[ b= \frac{n\sum x_iy_i-(\sum x_i)(\sum y_i)} {n\sum x_i^2-(\sum x_i)^2}, \qquad a=\frac{\sum y_i-b\sum x_i}{n}. \]

Using the supplied data,

\[ a=3.0804004\ \mathrm s, \qquad b=935.67515\ \mathrm{s\,L\,mol^{-1}}. \]

Thus the required plot is an increasing straight line of \(t_{1/2}\) against \([\mathbf C]_0\), with this intercept and slope. Its \(R^2\) is approximately \(0.9990\).

The association constant is

\[ K_c= \frac{935.67515\ \mathrm{s\,L\,mol^{-1}}} {3.0804004\ \mathrm s} = 303.75\ \mathrm{L\,mol^{-1}}. \]

Therefore,

\[ \boxed{K_c\approx3.0\times10^2\ \mathrm{L\,mol^{-1}}}. \]

Using the graph’s rounded coefficients, \(a=3.1\ \mathrm s\) and \(b=936\ \mathrm{s\,L\,mol^{-1}}\), gives \(302\ \mathrm{L\,mol^{-1}}\), consistent with this result.

5.10 — Conditions for light-triggered pore opening

For a guest with association constant \(K\), the fraction complexed at free host concentration \([\mathbf C]\) is

\[ f_{\mathrm{bound}}=\frac{K[\mathbf C]}{1+K[\mathbf C]}. \]

To block the pores initially, the trans guest should be predominantly complexed:

\[ f_{\mathrm{trans}}\simeq1 \quad\Longrightarrow\quad K_t[\mathbf C]\gg1. \]

After irradiation, the cis guest should be predominantly uncomplexed:

\[ f_{\mathrm{cis}}\simeq0 \quad\Longrightarrow\quad K_c[\mathbf C]\ll1. \]

The appropriate condition is therefore

\[ \boxed{K_c[\mathbf C]\ll1\ll K_t[\mathbf C]}, \]

or equivalently,

\[ \boxed{\frac1{K_t}\ll[\mathbf C]\ll\frac1{K_c}}. \]

If host depletion is negligible, this can be expressed using \([\mathbf C]_0\).

This part explicitly concerns another azobenzene compound, with \(K_c\ll K_t\). The numerical constants determined for M in 5.6 and 5.9 do not apply to this grafted compound.

5.11 — Irradiation wavelength

The solid absorption curve for trans-azobenzene has its strong maximum at \(330\ \mathrm{nm}\). At that wavelength, cis-azobenzene absorbs less strongly.

Therefore, preferential excitation of the initially trans guest is achieved with

\[ \boxed{\lambda_1=330\ \mathrm{nm}}. \]

The resulting trans-to-cis isomerization weakens host binding, removes the cyclodextrin caps, and allows dye release.

5.12 — Monitoring wavelength

Rhodamine B has its strong absorption maximum at \(550\ \mathrm{nm}\), where both azobenzene isomers have negligible absorption.

Thus

\[ \boxed{\lambda_2=550\ \mathrm{nm}}. \]

At this wavelength, the solution absorbance tracks the concentration of released dye through the Beer–Lambert law:

\[ A_{550} = \varepsilon_{\mathrm{RhB},550}\,L\, [\mathrm{RhB}]_{\mathrm{solution}}. \]

The solution was checked against the official solution.