IChO 2025 · Theory problem 3

Lanterns and the Burj Khalifa

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7.5% of the total

3.1 3.2 3.3 3.4 3.5 3.6 3.7 Total
6 4 4 5 40 20 32 111
Figure 1

At 829.8 m, Dubai’s Burj Khalifa is the world's tallest structure. To simplify construction, many floors were made from similar pieces. Some artificial "molecular skyscrapers" can be made through coordination chemistry of just two pieces: metal ions and organic ligands.

In these 3D structures, every vertex contains a square planar \(\ce{Pd}\)(II) ion coordinated to four pyridine ligands. Along each edge is a bent organic ligand consisting of two pyridines linked together.

Figure 2

The angle between the two \(\ce{N -> Pd}\) bonds from the same ligand controls which structure forms. Larger angles give larger structures.

Figure 3

Where free rotation around a bond is possible, the ligand adopts a conformation to form a structure with the smallest possible angle.

3.1 Tick the smallest angle (approximately) between the \(\ce{N -> Pd}\) bonds which could be achieved by ligands A, B, and C.

Figure 4

3.2 Using “<”, rank ligands D, E, and F, from smallest angle to largest angle between the \(\ce{N -> Pd}\) bonds that they could form.

Figure 5

One series of structures can be modelled as polyhedra with tetracoordinated \(\ce{Pd}\)(II) ions at the vertices. All structures have eight triangular faces (\(F_3\)), with various numbers of square faces (\(F_4\)).

The triangular and square faces can be placed onto a square grid, where the relative position of two nearest triangles is described with integer coordinates \(h\) and \(k\). \(Q\) is then given by the equation:

\[ Q=h^2+k^2 \]

Part of the grid for a structure with \(h=2\) and \(k=4\) is shown below.

Figure 6

The smallest possible structure in this series is an octahedron, where \(Q=1\), and \((h,k)=(0,1)\) or \((1,0)\). It has six vertices, eight triangular faces, and 12 edges. Nearest triangles are a move of one face in one direction apart.

Figure 7

The second smallest structure is a cuboctahedron, where \(Q=2\). Nearest triangles are a move of one face in each direction apart.

Figure 8

3.3 Complete the column in the table for \(Q=2\), describing the number of vertices, \(V\), number of edges \(E\), the total number of faces, \(F\), and the number of square faces, \(F_4\).

\(Q\) 1 2
\((h,k)\) \((0,1)\) or \((1,0)\) \((1,1)\)
\(V\) 6
\(E\) 12
\(F\) 8
\(F_3\) 8 8
\(F_4\) 0

All structures obey the following rules:

\[ V-E+F=2 \]

\[ V=6Q \]

\[ F-V=2 \]

3.4 Complete the table for the third smallest structure, where \(Q=4\).

\(Q\) 4
\((h,k)\)
\(V\)
\(E\)
\(F\)
\(F_3\) 8
\(F_4\)
Figure 9

In contrast to these large structures, the smallest structures contain two metals and four ligands and are called lanterns. They are similar in shape to the traditional lanterns seen at the Dubai light festival. Lanterns are formed by ligands such as L, which make two parallel \(\ce{N -> Pd}\) bonds. Ligand L has two planes of symmetry and can adopt a planar conformation with the angles shown.

Figure 10

Hint:

  • The transformation from H to I could also be achieved by Lewis acid catalysis.

3.5 Draw the structures of compounds H, I, J, K, and ligand L.

Unlike symmetrical ligand L, chiral ligand M lacks symmetry and exists as a pair of \(R\) and \(S\) enantiomers.

Figure 11

Multiple lantern isomers can be formed from ligand M. We can group these isomers in two ways. Firstly we categorise by the number of each type of ligand in the lantern: (i) all the same (\(RRRR\) or \(SSSS\)), (ii) one different (\(RRRS\) or \(SSSR\)), (iii) cis-\(RRSS\), and (iv) trans-\(RSRS\).

Figure 12

Secondly, the two ends of ligand M are not the same; the ligand has a head (H) and a tail (T). There are four categories of structure considering head-to-tail orientation: (1) all up (HHHH), (2) three up (HHHT), (3) cis-HHTT, and (4) trans-HTHT.

Figure 13

The 16 possible combinations of these categories are represented by empty cells in the table. Some combinations have only one possible isomer, whereas others have multiple possible isomers. In any one combination, all possible isomers are either chiral or achiral.

3.6 Complete the table by writing "C" for chiral and "A" for achiral. Note, in this part you will be awarded points for each correct answer but deducted points for each incorrect answer down to a minimum of zero. There is no penalty for empty cells.

Figure 14
Figure 15
(1) (2) (3) (4)
(i)
(ii)
(iii)
(iv)

3.7 Complete the table with the number of possible isomers of each combination.

Figure 16
Figure 17
(1) (2) (3) (4)
(i) 2
(ii) 4
(iii)
(iv)

Official English text of the problem.

Answers

  • 3.1: \(\boxed{A:180^\circ,\quad B:120^\circ,\quad C:0^\circ}\).

  • 3.2: \(\boxed{F<D<E}\).

  • 3.3: \(\boxed{V=12,\ E=24,\ F=14,\ F_4=6}\).

  • 3.4: \(\boxed{(h,k)=(0,2)\text{ or }(2,0),\ V=24,\ E=48,\ F=26,\ F_3=8,\ F_4=18}\).

  • 3.5: The following structures specify the required drawings.

H — Tetramethyl (2Z,2′Z)-2,2′-[(2-fluorobenzene-1,3-diyl)bis(azanediyl)]bis(but-2-enedioate), the double aza-Michael adduct:

COC(=O)/C=C(\Nc1cccc(N/C(=C\C(=O)OC)C(=O)OC)c1F)C(=O)OC

I — Dimethyl 9-fluoro-4,5-dihydroxy-1,8-diazaanthracene-2,7-dicarboxylate, shown in its hydroxy tautomer:

COC(=O)c1cc(O)c2cc3c(O)cc(C(=O)OC)nc3c(F)c2n1

J — Dimethyl 9-fluoro-4,5-bis(2-methylpropoxy)-1,8-diazaanthracene-2,7-dicarboxylate:

COC(=O)c1cc(OCC(C)C)c2cc3c(OCC(C)C)cc(C(=O)OC)nc3c(F)c2n1

K — 9-Fluoro-4,5-bis(2-methylpropoxy)-1,8-diazaanthracene-2,7-dicarboxylic acid, after acid work-up:

O=C(O)c1cc(OCC(C)C)c2cc3c(OCC(C)C)cc(C(=O)O)nc3c(F)c2n1

L — 9-Fluoro-4,5-bis(2-methylpropoxy)-\(N^2,N^7\)-bis(pyridin-3-yl)-1,8-diazaanthracene-2,7-dicarboxamide:

O=C(Nc1cccnc1)c1cc(OCC(C)C)c2cc3c(OCC(C)C)cc(C(=O)Nc4cccnc4)nc3c(F)c2n1

  • 3.6: Columns and rows follow the problem’s numbering.
Ligand configurations (1) HHHH (2) HHHT (3) cis-HHTT (4) trans-HTHT
(i) RRRR or SSSS C C C C
(ii) RRRS or SSSR C C C C
(iii) cis-RRSS A C A C
(iv) trans-RSRS A C A A
  • 3.7: Enantiomers are counted separately.
Ligand configurations (1) HHHH (2) HHHT (3) cis-HHTT (4) trans-HTHT
(i) RRRR or SSSS 2 2 2 2
(ii) RRRS or SSSR 2 8 4 2
(iii) cis-RRSS 1 4 3 2
(iv) trans-RSRS 1 2 1 1

There are 39 stereoisomers: 7 achiral structures and 16 enantiomeric pairs.

Solution

3.1 — Minimum angles

The donor direction of a 4-pyridyl group lies along the line joining its nitrogen and its para substituent. Rotating that ring about its attachment bond therefore does not change the donor direction.

  • A, 1,2-bis(pyridin-4-yl)ethyne, n1ccc(C#Cc2ccncc2)cc1, has a linear ethynyl connection. Its two donor vectors point in opposite directions: \[ \boxed{\theta_A=180^\circ}. \]

  • B, 1,3-bis(pyridin-4-yl)benzene, c1cc(-c2ccncc2)cc(-c2ccncc2)c1, has two outward substituent directions separated by approximately \(120^\circ\) at the central meta-substituted benzene: \[ \boxed{\theta_B=120^\circ}. \]

  • C, 1,3-bis[3-(pyridin-4-yl)phenyl]benzene, c1cc(-c2cccc(-c3ccncc3)c2)cc(-c2cccc(-c3ccncc3)c2)c1, contains three meta-phenylene bends. Rotation about the inter-ring single bonds permits a folded conformation in which the successive \(60^\circ\) changes of chain direction all have the same sense. The terminal donor vectors then point in the same direction: \[ \boxed{\theta_C=0^\circ}. \]

For the required drawing of C’s minimum-angle conformation, fold the aromatic chain into a U-shaped arrangement with the two terminal \(\ce{N -> Pd}\) arrows parallel and pointing the same way.

3.2 — Heterocycle-dependent angles

The ligands are:

Ligand Name SMILES
D 2,5-Bis(pyridin-4-yl)thiophene s1c(-c2ccncc2)ccc1-c1ccncc1
E 2,5-Bis(pyridin-4-yl)selenophene [se]1c(-c2ccncc2)ccc1-c1ccncc1
F 2,5-Bis(pyridin-4-yl)furan o1c(-c2ccncc2)ccc1-c1ccncc1

The internal \(\ce{C-X-C}\) angle at the heteroatom decreases in the order oxygen, sulfur, selenium. The outward directions of the two pyridyl substituents consequently spread farther apart.

In an idealized trigonal-carbon construction, their angle is approximately

\[ \theta_{\text{donors}}\simeq240^\circ-\angle\ce{C-X-C}. \]

Thus,

\[ \boxed{F<D<E}. \]

3.3 — Cuboctahedron

The metal is palladium(II), [Pd+2]; its four donors are pyridine nitrogens, with pyridine represented by n1ccccc1. The local coordination geometry is square planar.

Each vertex has degree four. Counting edge ends gives

\[ 4V=2E,\qquad E=2V. \]

Euler’s equation then gives

\[ F=2-V+E=V+2. \]

Since eight faces are triangular,

\[ F_4=F-8=V-6. \]

For \(Q=2\),

\[ V=6Q=12,\qquad E=24,\qquad F=14,\qquad F_4=6. \]

The completed column is therefore:

Quantity \(Q=2\)
\((h,k)\) \((1,1)\)
\(V\) 12
\(E\) 24
\(F\) 14
\(F_3\) 8
\(F_4\) 6

3.4 — Third-smallest structure

For nonnegative grid coordinates,

\[ h^2+k^2=4 \]

has the solutions \((0,2)\) and \((2,0)\). Reversing coordinate directions gives equivalent signed choices.

Using the same relations,

\[ V=6(4)=24,\qquad E=2(24)=48, \]

\[ F=24+2=26,\qquad F_4=26-8=18. \]

These are the counts of a rhombicuboctahedron.

3.5 — Synthesis and structural identification

The SMILES in Answers specify the connectivity of each product.

Formation of H. The starting materials are 2-fluorobenzene-1,3-diamine, Nc1cccc(N)c1F, and dimethyl acetylenedicarboxylate, COC(=O)C#CC(=O)OC.

Each amino group adds across one electron-poor alkyne. Nitrogen attacks an alkyne carbon, and proton transfer gives an enamine arm:

\[ \ce{Ar-NH-C(CO2Me)=CH-CO2Me}. \]

Thus two equivalents of the alkyne are incorporated without elimination:

\[ \ce{C6H7FN2 + 2 C6H6O4 -> C18H19FN2O8}. \]

This reproduces H’s supplied molecular formula. Draw two identical enamine arms on the original amino nitrogens. In the cyclization-competent arrangement, each nitrogen and the distal ester are on the same side of the alkene.

Formation of I. Each enamine arm cyclizes at the unoccupied ortho position of its attached benzene ring. Fluorine occupies the other ortho position.

The aromatic carbon forms a bond to the distal ester carbonyl carbon, closing a six-membered ring. Acyl substitution and proton transfer eliminate methanol, CO, and produce a quinolin-4-one unit. Its hydroxy tautomer is a 4-hydroxyquinoline unit.

Both sides undergo this reaction:

\[ \ce{C18H19FN2O8 ->[\Delta] C16H11FN2O6 + 2 CH3OH}. \]

The formula difference is exactly two methanol molecules. The two new rings share opposite edges of the original benzene, giving the linear 1,8-diazaanthracene framework. Lewis acid activation of the ester carbonyl provides the alternative cyclization route mentioned in the hint.

For the drawing, retain fluorine on the central ring; place ester groups at positions 2 and 7 and hydroxy groups at positions 4 and 5. The corresponding doubly lactamized tautomer represents the same tautomeric system.

Formation of J. I assign this as Mitsunobu etherification of both hydroxy groups with 2-methylpropan-1-ol, CC(C)CO.

The reagents are triphenylphosphine, P(c1ccccc1)(c1ccccc1)c1ccccc1, and diisopropyl azodicarboxylate, CC(C)OC(=O)N=NC(=O)OC(C)C.

Each hydroxy oxygen attacks the activated primary alcohol carbon, installing

\[ \ce{-O-CH2-CH(CH3)2}. \]

Replacing two hydroxy hydrogens with two isobutyl groups adds \(\ce{C8H16}\):

\[ \ce{C16H11FN2O6 -> C24H27FN2O6}. \]

The complete Mitsunobu material balance is

\[ \begin{aligned} &\ce{C16H11FN2O6 + 2 C4H10O + 2 C18H15P + 2 C8H14N2O4}\\ &\qquad\ce{-> C24H27FN2O6 + 2 C18H15OP + 2 C8H16N2O4}. \end{aligned} \]

The additional products are triphenylphosphine oxide, O=P(c1ccccc1)(c1ccccc1)c1ccccc1, and diisopropyl hydrazine-1,2-dicarboxylate, CC(C)OC(=O)NNC(=O)OC(C)C.

Formation of K. Hydroxide, [OH-], saponifies the two methyl esters:

\[ \ce{C24H27FN2O6 + 2 OH- -> C22H21FN2O6^{2-} + 2 CH3OH}. \]

The resulting dicarboxylate is

O=C([O-])c1cc(OCC(C)C)c2cc3c(OCC(C)C)cc(C(=O)[O-])nc3c(F)c2n1

Acidification with protons, [H+], gives the diacid K:

\[ \ce{C22H21FN2O6^{2-} + 2 H+ -> C22H23FN2O6}. \]

The alkoxy substituents remain intact.

Formation of L. Oxalyl chloride, O=C(Cl)C(=O)Cl, first converts both carboxylic acids to acid chlorides:

\[ \ce{C22H23FN2O6 + 2 (COCl)2 -> C22H21Cl2FN2O4 + 2 CO + 2 CO2 + 2 HCl}. \]

The intermediate is K’s dicarbonyl dichloride:

O=C(Cl)c1cc(OCC(C)C)c2cc3c(OCC(C)C)cc(C(=O)Cl)nc3c(F)c2n1

The byproducts are carbon monoxide, [C-]#[O+], carbon dioxide, O=C=O, and hydrogen chloride, Cl.

Reaction with two equivalents of pyridin-3-amine, Nc1cccnc1, forms the two amides. Triethylamine, CCN(CC)CC, captures the hydrogen chloride:

\[ \ce{C22H21Cl2FN2O4 + 2 C5H6N2 + 2 NEt3 -> C32H31FN6O4 + 2 [HNEt3]Cl}. \]

Triethylammonium chloride is CC[NH+](CC)CC.[Cl-].

Each terminal connection in L must therefore be

\[ \ce{core-C(=O)-NH-(pyridin-3-yl)}. \]

For the requested ligand drawing, place the two amide–pyridyl arms as mirror-related arms on the fused core. Use the indicated \(120^\circ\) relationship between the core’s exit directions and the indicated \(180^\circ\) arrangements in the arms. Orient the terminal pyridine rings so that their two \(\ce{N -> Pd}\) vectors are parallel. These conformational choices are not encoded by ordinary connectivity SMILES.

3.6 — Chirality of the lanterns

The supplied ligand M is a cyclopentane-fused dipyridyl dihydropyridazine. Its enantiomers can be represented as:

  • \(R\)-M: C1(c2cccnc2)=C2CCC[C@H]2C(c2cccnc2)=NN1
  • \(S\)-M: C1(c2cccnc2)=C2CCC[C@@H]2C(c2cccnc2)=NN1

Number the four ligand positions \(0,1,2,3\) around the metal–metal axis. At each position record two attributes: \(R/S\), and whether the head or tail lies at the upper metal.

There are eight proper rotations that can identify arrangements:

  1. Four rotations about the metal–metal axis, giving cyclic permutations.
  2. Four half-turns about perpendicular axes. These reverse cyclic order and interchange upper and lower ends, so they also interchange H and T.

Proper rotations never interchange \(R\) and \(S\).

To test chirality, reflect the cage through its equatorial plane. This interchanges \(R\leftrightarrow S\) and H\(\leftrightarrow\)T at every position. An arrangement is achiral precisely when a proper rotation superposes it on this reflected arrangement.

For rows (i) and (ii), reflection changes the numbers of \(R\) and \(S\) ligands. A rotation cannot undo that change. Every entry in these rows is therefore C.

For the two balanced rows:

  • HHHH: Both cis-RRSS and trans-RSRS permit a vertical reflection exchanging the appropriate \(R\) and \(S\) positions. Both are A.
  • HHHT: The unique reversed ligand prevents a symmetry operation from exchanging all \(R\) and \(S\) positions while preserving the head pattern. Both are C.
  • cis-HHTT: With cis-RRSS, inversion exchanges both pairs of attributes correctly. With trans-RSRS, a suitable vertical reflection does so. Both are A.
  • trans-HTHT: The cis-RRSS combination lacks the required improper symmetry and is C. The trans-RSRS combination has an improper fourfold rotation, \(S_4\), and is A.

The last case illustrates why absence of an obvious mirror plane alone does not establish chirality: an \(S_4\) operation also makes a structure achiral.

3.7 — Number of isomers

We identify arrangements only under the eight proper rotations described above. Mirror-related structures remain separate unless superposable.

Before rotational identification, the numbers of head patterns in columns (1)–(4) are

\[ \mathbf n=(2,8,4,2). \]

These count HHHH/TTTT, the eight three-to-one patterns, the four adjacent two-to-two patterns, and the two alternating patterns.

Row (i): all configurations the same.

For fixed RRRR, all head patterns within each column are equivalent under rotations. Thus there is one arrangement per column. SSSS supplies its distinct enantiomer:

\[ \boxed{(2,2,2,2)}. \]

Row (ii): one configuration different.

Fix the position of the unique \(S\) ligand in RRRS. Two proper rotations preserve this \(R/S\) pattern: the identity and a perpendicular half-turn through the unique ligand and the opposite ligand.

The half-turn complements the head pattern and fixes none of the labeled head assignments. There are consequently \(\mathbf n/2\) arrangements for RRRS. The SSSR enantiomeric composition doubles this:

\[ \boxed{(2,8,4,2)}. \]

Row (iii): cis-RRSS.

Fix the cyclic pattern RRSS. Its preserving rotation group has two elements. The nonidentity element exchanges the two \(R\) positions, exchanges the two \(S\) positions, and complements H/T.

Its numbers of fixed head assignments are

\[ \mathbf f=(0,0,2,2). \]

Burnside’s counting formula gives

\[ \frac{\mathbf n+\mathbf f}{2} = \frac{(2,8,4,2)+(0,0,2,2)}2 = \boxed{(1,4,3,2)}. \]

Row (iv): trans-RSRS.

The alternating \(R/S\) pattern has four preserving proper rotations. Apart from the identity, the axial half-turn fixes

\[ \mathbf g=(2,0,0,2) \]

head assignments. Neither perpendicular half-turn fixes a head assignment, because each would have to complement H/T at a position it leaves fixed.

Thus,

\[ \frac{\mathbf n+\mathbf g}{4} = \frac{(2,8,4,2)+(2,0,0,2)}4 = \boxed{(1,2,1,1)}. \]

As an independent check, there are initially \(2^4\times2^4=256\) labeled assignments. Burnside’s formula over all eight proper rotations gives

\[ N=\frac{256+2(4)+16+2(0)+2(16)}8 =\boxed{39}. \]

Summing the entries marked A gives seven achiral structures. The remaining \(39-7=32\) structures form sixteen enantiomeric pairs.

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