IChO 2017 · Theory problem 7

Chemical Structure Puzzles

Titanium complexes have been investigated for their antitumor activity. Many factors including isomerism and sizes have shown to affect the potency of the complexes. This question deals with the synthesis and characterization of some titanium complexes.

7.1 A reaction of 2 equivalents of 2-tert-butylphenol, 2 equivalents of formaldehyde, and \(N,N'\)-dimethylethylene-1,2-diamine under acidic conditions at \(75\,^\circ\mathrm{C}\) affords three major products with the same chemical formula of \(\ce{C26H40N2O2}\), as shown in the equation below. Draw the structure of each product.

Figure 1

7.2 If 2,4-di-tert-butylphenol is used as a substrate instead of 2-tert-butylphenol using the same stoichiometry as that in 7.1, only one product X was obtained. Draw the structure of X.

A reaction between X from 7.2 and \(\ce{Ti(O^{i}Pr)4}\) [\(^{i}\mathrm{Pr}\) = isopropyl] in diethyl ether under an inert atmosphere resulted in the six-coordinate Ti complex Y, as a yellow crystalline solid and isopropanol at room temperature.

Figure 2

UV-VIS spectra of X, \(\ce{Ti(O^{i}Pr)4}\), and Y reveal that only the product Y has an absorption at \(\lambda=370\ \mathrm{nm}\). By varying the volumes of X and \(\ce{Ti(O^{i}Pr)}\), each with the concentration of \(0.50\ \mathrm{mol\,dm^{-3}}\), and using benzene as the solvent, the absorbance data at \(\lambda=370\ \mathrm{nm}\) are given below:

Volume of X (\(\mathrm{cm^3}\)) Volume of \(\ce{Ti(O^{i}Pr)4}\) (\(\mathrm{cm^3}\)) Volume of benzene (\(\mathrm{cm^3}\)) Absorbance
0 1.20 1.80 0.05
0.20 1.00 1.80 0.25
0.30 0.90 1.80 0.38
0.50 0.70 1.80 0.59
0.78 0.42 1.80 0.48
0.90 0.30 1.80 0.38
1.10 0.10 1.80 0.17
1.20 0 1.80 0.02

7.3 Fill in appropriate values in the table provided below.

\(\displaystyle\frac{\text{mol of X}}{\text{mol of X}+\text{mol of }\ce{Ti(O^{i}Pr)4}}\) Absorbance
0.05
0.25
0.38
0.59
0.48
0.38
0.17
0.02

Next figure: Plot a graph showing a relationship between

\[ \frac{\text{mole of X}} {\text{mole of X}+\text{mole of }\ce{Ti(O^{i}Pr)4}} \]

and absorbance in the space provided below.

The value of

\[ \frac{\text{mole of X}} {\text{mole of X}+\text{mole of }\ce{Ti(O^{i}Pr)4}} \]

which maximizes the amount of the product Y represents the stoichiometry of X in the chemical formula of Y. Based on the graph above, show what is the molar ratio Ti : X in the complex Y.

7.4 The titanium complex Y is six-coordinated. The IR spectrum of Y does not contain a broad absorption band in the range of \(3200–3600\ \mathrm{cm^{-1}}\). Y exists as three diastereomers. Ignoring stereochemistry at N atoms, draw clearly the structures of all three diastereomers.

Note that you do not need to draw the complete structure of the ligand. Only identify donor atoms that involve in coordination with titanium and the ligand framework between the donor atoms can be drawn as follows:

For example:

Figure 3

(2,2'-bipyridine)

can be drawn as:

Figure 4

**If you did not get a structure of X from 7.2, use the following ligand symbol to represent X (A and Z are donor atoms):

Figure 5

7.5 Under certain conditions, the reaction shown in equation 1 affords only one diastereomer of Y. Given that structures of Y are "fixed" (no intramolecular movement), the \(\ce{^1H}\) NMR spectrum of Y in \(\ce{CDCl3}\) shows four singlet resonances at \(\delta\) 1.25, 1.30, 1.66, and 1.72 corresponding to the tert-butyl groups. Draw a structure of the only possible diastereomer of Y.

(You do not need to draw the complete structure of the ligand. Only identify donor atoms that involve in coordination and the ligand framework between the donor atoms can be drawn as shown in 7.4).

Official English text of the problem.

Answers

  • 7.1: The three products are the ortho/ortho, ortho/para, and para/para bis-Mannich products. Their names and complete SMILES are given in Solution 7.1.
  • 7.2: X is \(N,N'\)-bis(3,5-di-tert-butyl-2-hydroxybenzyl)-\(N,N'\)-dimethylethane-1,2-diamine, \(\ce{C34H56N2O2}\):
    CN(Cc1c(O)c(C(C)(C)C)cc(C(C)(C)C)c1)CCN(C)Cc1c(O)c(C(C)(C)C)cc(C(C)(C)C)c1
  • 7.3: The mole fractions, in order, are \(0,\ 0.167,\ 0.250,\ 0.417,\ 0.650,\ 0.750,\ 0.917,\ 1.000\). The graph peaks near \(0.50\), giving Ti : X = 1 : 1.
  • 7.4: Writing \(\ce{X = H2L}\), Y is the titanium(IV) bis(isopropoxide) complex \(\ce{[Ti(L)(O^{i}Pr)2]}\), with formula \(\ce{C40H68N2O4Ti}\). Its three geometric diastereomers have: I, trans isopropoxide donors and cis phenolate donors; II, cis isopropoxide donors and trans phenolate donors; III, both pairs cis. Exact structures and stereochemical SMILES appear in Solution 7.4. Equation 1 has \(a=b=d=1,\ c=2\).
  • 7.5: Isomer III, with both the isopropoxide pair and the phenolate pair cis, is selected. Its two aromatic rings are inequivalent, allowing four distinct tert-butyl resonances.

Solution

7.1 — The three Mannich products

The starting structures are:

  • 2-tert-Butylphenol: Oc1c(C(C)(C)C)cccc1
  • Formaldehyde: C=O
  • \(N,N'\)-Dimethylethane-1,2-diamine: CNCCNC

The reaction is a Mannich aminomethylation. Each secondary amine group reacts with one formaldehyde molecule and one phenol ring, installing an aromatic \(\ce{-CH2-N}\) linkage and eliminating water, O.

Number each starting phenol ring with the hydroxyl group at C1 and tert-butyl at C2. The hydroxyl group activates the ortho and para positions. C2 is occupied, so the available positions are C6, ortho, and C4, para.

The common product framework is

\[ \mathrm{Ar^1{-}CH_2{-}N(CH_3){-}CH_2CH_2{-}N(CH_3){-}CH_2{-}Ar^2}. \]

Choosing ortho or para attachment independently on the two rings gives three distinct combinations; exchanging the two ends makes ortho/para and para/ortho identical.

Ortho/ortho product, \(P_{oo}\):
\(N,N'\)-Bis(3-tert-butyl-2-hydroxybenzyl)-\(N,N'\)-dimethylethane-1,2-diamine.

CN(Cc1c(O)c(C(C)(C)C)ccc1)CCN(C)Cc1c(O)c(C(C)(C)C)ccc1

For the drawing, attach both aminomethyl groups at C6 of their respective starting phenol rings.

Ortho/para product, \(P_{op}\):
\(N\)-(3-tert-Butyl-2-hydroxybenzyl)-\(N'\)-(3-tert-butyl-4-hydroxybenzyl)-\(N,N'\)-dimethylethane-1,2-diamine.

CN(Cc1c(O)c(C(C)(C)C)ccc1)CCN(C)Cc1cc(C(C)(C)C)c(O)cc1

Attach one aminomethyl group at C6 and the other at C4.

Para/para product, \(P_{pp}\):
\(N,N'\)-Bis(3-tert-butyl-4-hydroxybenzyl)-\(N,N'\)-dimethylethane-1,2-diamine.

CN(Cc1cc(C(C)(C)C)c(O)cc1)CCN(C)Cc1cc(C(C)(C)C)c(O)cc1

Attach both aminomethyl groups at C4.

For each product, the balanced overall equation under the stated acidic conditions is

\[ \ce{2 C10H14O + 2 CH2O + C4H12N2 -> C26H40N2O2 + 2 H2O}. \]

Thus all three structures have the required molecular formula.

7.2 — Identification of X

The new starting phenol is 2,4-di-tert-butylphenol:

Oc1c(C(C)(C)C)cc(C(C)(C)C)cc1

Its C2 ortho position and C4 para position are both occupied. Aminomethylation therefore occurs at the remaining ortho position, C6, on each ring.

Consequently, X is the ortho/ortho product named and specified in the Answers section. To draw it, use the same diamine bridge as in \(P_{oo}\), retaining tert-butyl substituents at C2 and C4 of each phenol ring.

The balanced synthesis is

\[ \ce{2 C14H22O + 2 CH2O + C4H12N2 -> C34H56N2O2 + 2 H2O}. \]

X contains two phenolic hydroxyl groups and two tertiary amine nitrogen atoms. These supply the eventual four-donor sequence

\[ \mathrm{O_1\;{\sim}\;N_1\;{\sim}\;N_2\;{\sim}\;O_2}, \]

where each curved connection represents the ligand framework between consecutive donors.

7.3 — Continuous-variation plot and stoichiometry

Let \(x_X\) denote the mole fraction requested in the table. Since both stock solutions have the same concentration,

\[ x_X =\frac{cV_X}{cV_X+cV_{\mathrm{Ti}}} =\frac{V_X}{V_X+V_{\mathrm{Ti}}} =\frac{V_X}{1.20\ \mathrm{cm^3}}. \]

The combined amount of the two reagents is constant:

\[ n_X+n_{\mathrm{Ti}} =(0.50\ \mathrm{mol\,dm^{-3}}) (1.20\times10^{-3}\ \mathrm{dm^3}) =6.0\times10^{-4}\ \mathrm{mol}. \]

Every mixture also has the same total volume, \(3.00\ \mathrm{cm^3}\). Thus the absorbances can be compared directly in a Job plot.

The completed table is:

\(x_X\) Absorbance at \(370\ \mathrm{nm}\)
0.000 0.05
0.167 0.25
0.250 0.38
0.417 0.59
0.650 0.48
0.750 0.38
0.917 0.17
1.000 0.02

For the requested graph, place \(x_X\) on the horizontal axis from 0 to 1 and absorbance on the vertical axis. Plot these eight points. The rising and falling branches are approximately straight, and their intersection lies near \(x_X=0.50\).

For a numerical check, least-squares lines through the first four and last four points are

\[ A_{\mathrm{left}}\approx0.0454+1.3062x_X, \qquad A_{\mathrm{right}}\approx1.3406-1.3003x_X. \]

Their intersection is

\[ x_{X,\max} =\frac{1.3406-0.0454}{1.3062+1.3003} \approx0.497, \qquad A_{\max}\approx0.69. \]

The peak falls between the sampled compositions. For a complex containing \(a\) ligands X per \(b\) titanium atoms,

\[ x_{X,\max}=\frac{a}{a+b}. \]

Hence

\[ \mathrm{Ti:X} =(1-x_{X,\max}):x_{X,\max} \approx0.503:0.497 \approx\boxed{1:1}. \]

7.4 — Composition and three octahedral diastereomers

Write the neutral ligand as \(\ce{H2L}\). The absence of the broad hydroxyl absorption indicates that both phenolic groups have lost their protons.

The resulting ligand is the bis(phenolate) dianion of X, \(\ce{L^{2-}}\), with formula \(\ce{C34H54N2O2^{2-}}\):

CN(Cc1c([O-])c(C(C)(C)C)cc(C(C)(C)C)c1)CCN(C)Cc1c([O-])c(C(C)(C)C)cc(C(C)(C)C)c1

Its two phenolate oxygen atoms and two tertiary amine nitrogen atoms bind titanium. Two isopropoxide oxygen donors complete the six-coordinate, octahedral complex.

The reagent is titanium(IV) isopropoxide:

CC(C)O[Ti](OC(C)C)(OC(C)C)OC(C)C

The eliminated alcohol is propan-2-ol (isopropanol):

CC(O)C

The balanced complexation reaction is

\[ \boxed{\ce{H2L + Ti(O^{i}Pr)4 -> [Ti(L)(O^{i}Pr)2] + 2 ^{i}PrOH}}. \]

Therefore,

\[ \boxed{a=b=d=1,\qquad c=2}. \]

Charge balance is consistent with a neutral complex:

\[ (+4)_{\mathrm{Ti}}+(-2)_{\mathrm{L}}+2(-1)_{\mathrm{isopropoxide}}=0. \]

The molecular formula is consequently \(\ce{C40H68N2O4Ti}\).

For the geometry, label the phenolate donors \(\mathrm O_1,\mathrm O_2\), their respective amine donors \(\mathrm N_1,\mathrm N_2\), and the two isopropoxide oxygen donors \(\mathrm O_{a1},\mathrm O_{a2}\). The ligand connections are always

\[ \mathrm O_1{\sim}\mathrm N_1{\sim}\mathrm N_2{\sim}\mathrm O_2. \]

Each consecutive donor pair must occupy cis sites because it forms a chelate ring. There are three resulting geometric classes:

Isomer Isopropoxide oxygen pair Phenolate oxygen pair
I trans cis
II cis trans
III cis cis

The following table specifies an exact drawing of one representative of each class. Put Ti at the origin, with \(+x\) right, \(-x\) left, \(+y\) up, \(-y\) down, \(+z\) toward the reader, and \(-z\) away. Draw the \(+z\) bond as a solid wedge and the \(-z\) bond as a dashed wedge, then add the three ligand-framework connections.

Isomer \(+x\) \(-x\) \(+y\) \(-y\) \(+z\) \(-z\)
I \(\mathrm N_1\) \(\mathrm O_2\) \(\mathrm N_2\) \(\mathrm O_1\) \(\mathrm O_{a1}\) \(\mathrm O_{a2}\)
II \(\mathrm N_1\) \(\mathrm O_{a1}\) \(\mathrm N_2\) \(\mathrm O_{a2}\) \(\mathrm O_1\) \(\mathrm O_2\)
III \(\mathrm N_1\) \(\mathrm O_{a1}\) \(\mathrm N_2\) \(\mathrm O_1\) \(\mathrm O_2\) \(\mathrm O_{a2}\)

Complete stereochemical SMILES for these representatives follow. They use the extended @OH notation for octahedral stereochemistry and -> for nitrogen-to-titanium dative bonds. Nitrogen configurations are intentionally unspecified, as requested.

I — trans-Isopropoxide isomer of the titanium(IV) complex:

[Ti@OH23]123(OC(C)C)(OC(C)C)Oc4c(C(C)(C)C)cc(C(C)(C)C)cc4CN->1(C)CCN->2(C)Cc1cc(C(C)(C)C)cc(C(C)(C)C)c1O3

II — cis-Isopropoxide/trans-phenolate isomer of the titanium(IV) complex:

[Ti@OH16]123(OC(C)C)(OC(C)C)Oc4c(C(C)(C)C)cc(C(C)(C)C)cc4CN->1(C)CCN->2(C)Cc1cc(C(C)(C)C)cc(C(C)(C)C)c1O3

III — cis-Isopropoxide/cis-phenolate isomer of the titanium(IV) complex:

[Ti@OH21]123(OC(C)C)(OC(C)C)Oc4c(C(C)(C)C)cc(C(C)(C)C)cc4CN->1(C)CCN->2(C)Cc1cc(C(C)(C)C)cc(C(C)(C)C)c1O3

At the donor-framework level, II and III each have mirror-related forms. Drawing either enantiomer represents the corresponding diastereomeric class.

7.5 — Selection from the tert-butyl NMR signals

There are four tert-butyl groups: one ortho and one para to the phenolate oxygen on each aromatic ring.

In I and II, symmetry interchanges the two ligand arms. Consequently, the two ortho tert-butyl groups are equivalent to each other, and the two para tert-butyl groups are equivalent to each other. These arrangements predict two tert-butyl resonances, each accounting for 18 H.

In III, the two phenolate donors have different trans partners:

  • \(\mathrm O_1\) is trans to the opposite arm’s amine donor, \(\mathrm N_2\).
  • \(\mathrm O_2\) is trans to an isopropoxide oxygen, \(\mathrm O_{a2}\).

No symmetry operation interchanges the two aromatic rings. All four tert-butyl groups can therefore be chemically inequivalent, giving four singlets, each accounting for 9 H.

Thus the required drawing is precisely isomer III in the coordinate table:

\[ \boxed{\text{cis isopropoxide donors; cis phenolate donors; inequivalent ligand arms}}. \]

Its name and stereochemical SMILES are given under III above. Either enantiomer, or a racemic mixture, gives the same spectrum in achiral \(\ce{CDCl3}\); these data do not assign absolute configuration.

The answers were checked against the official answers published by the organisers.