IChO 2021 · Theory problem 5

Mysterious Silicon

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Figure 1

Although silicon is also a group 14 element like carbon, their properties differ significantly.

PART A

Unlike the carbon–carbon triple bond, the silicon–silicon triple bond in a compound formulated as \(\ce{R^{1}-Si#Si-R^{1}}\) (R: organic substituent) is extremely reactive. For example, it reacts with ethylene to form a cyclic product that contains a four-membered ring.

Figure 2

When \(\ce{R^{1}-Si#Si-R^{1}}\) is treated with an alkyne (\(\ce{R^{2}-C#C-R^{2}}\)), the four-membered-ring compound A is formed as an initial intermediate. Further reaction of another molecule of \(\ce{R^{2}-C#C-R^{2}}\) with A affords isomers B and C, both of which have benzene-like cyclic conjugated structures, so-called ‘disilabenzenes’ that contain a six-membered ring and can be formulated as \(\ce{(R^{1}-Si)2(R^{2}-C)4}\).

Figure 3

The \(\ce{^{13}C}\) NMR analysis of the corresponding six-membered ring skeletons \(\ce{Si2C4}\) shows two signals for B and one signal for C.

5.1 Draw the structural formulae of A, B, and C using \(\ce{R^{1}}\), \(\ce{R^{2}}\), \(\ce{Si}\), and \(\ce{C}\), with one of the possible resonance structures.

5.2 Calculate the aromatic stabilization energy (ASE) for benzene and C (in the case of \(\ce{R^{1}}=\ce{R^{2}}=\ce{H}\)) as positive values, considering the enthalpy change in some hydrogenation reactions of unsaturated systems shown below (Fig. 1).

Figure 4

Fig. 1

When a xylene solution of C is heated, it undergoes isomerization to give an equilibrium mixture of compounds D and E. The molar ratio is D : E = 1 : 40.0 at 50.0 °C and D : E = 1 : 20.0 at 120.0 °C.

5.3 Calculate \(\Delta H\) for the transformation of D to E. Assume that \(\Delta H\) does not depend on temperature.

The isomerization from C to D and to E proceeds via transformations of \(\pi\)-bonds into \(\sigma\)-bonds without breaking any \(\sigma\)-bonds. A \(\ce{^{13}C}\) NMR analysis revealed one signal for the \(\ce{Si2C4}\) skeleton of D and two signals for that of E. The skeleton of D does not contain any three-membered rings, while E has two three membered rings that share an edge.

5.4 Draw the structural formulae of D and E using \(\ce{R^{1}}\), \(\ce{R^{2}}\), \(\ce{Si}\), and \(\ce{C}\)

PART B

Silicon is able to form highly coordinated compounds (> four substituents) with electronegative elements such as fluorine. As metal fluorides are often used as fluorination reagents, highly coordinated silicon fluorides also act as fluorination reagents.

The fluorination reaction of \(\ce{CCl4}\) using \(\ce{Na2SiF6}\) was carried out as follows.

Standardization of \(\ce{Na2SiF6}\) solution:

  • Preparation

Aqueous solution F: 0.855 g of \(\ce{Na2SiF6}\) (188.053 g mol\(^{-1}\)) dissolved in water (total volume: 200 mL).

Aqueous solution G: 6.86 g of \(\ce{Ce2(SO4)3}\) (568.424 g mol\(^{-1}\)) dissolved in water (total volume: 200 mL).

  • Procedure

Precipitation titration of a solution F (50.0 mL) by dropwise adding solution G in the presence of xylenol orange, which coordinates to \(\ce{Ce^{3+}}\), as an indicator. After adding 18.8 mL of solution G, the color of the solution changes from yellow to magenta. The generated precipitate is a binary compound that contains \(\ce{Ce^{3+}}\), and the only resulting silicon compound is \(\ce{Si(OH)4}\).

5.5 Write the balanced equation for the reaction of \(\ce{Na2SiF6}\) with \(\ce{Ce2(SO4)3}\).

  • Reaction of \(\ce{CCl4}\) with \(\ce{Na2SiF6}\):

(Substance losses by e.g. evaporation are negligible during the following operations.)

\(\ce{Na2SiF6}\)(\(x\) [g]) was added to \(\ce{CCl4}\) (500.0 g) and heated to 300°C in a sealed pressure-resistant reaction vessel. The unreacted \(\ce{Na2SiF6}\) and generated \(\ce{NaCl}\) were removed by filtration. The filtrate was diluted to a total volume of 1.00 L with \(\ce{CCl4}\) (solution H). The \(\ce{^{29}Si}\) and \(\ce{^{19}F}\) NMR spectra of solution H showed \(\ce{SiF4}\) as the only silicon compound. In the \(\ce{^{19}F}\) NMR spectrum, in addition to \(\ce{SiF4}\), signals corresponding to \(\ce{CFCl3}\), \(\ce{CF2Cl2}\), \(\ce{CF3Cl}\), and \(\ce{CF4}\) were observed (cf. Table 1). The integration ratios in the \(\ce{^{19}F}\) NMR spectrum are proportional to the number of fluorine nuclei.

Table 1

\(\ce{^{19}F}\) NMR data \(\ce{CFCl3}\) \(\ce{CF2Cl2}\) \(\ce{CF3Cl}\) \(\ce{CF4}\)
Integration ratio 45.0 65.0 18.0 2.0

\(\ce{SiF4}\) is hydrolyzed to form \(\ce{H2SiF6}\) according to the following eq. 8:

Figure 5

Solution H (10 mL) was added to an excess amount of water, which resulted in the complete hydrolysis of \(\ce{SiF4}\). After separation, the \(\ce{H2SiF6}\) generated from the hydrolysis in the aqueous solution was neutralized and completely converted to \(\ce{Na2SiF6}\) (aqueous solution J).

The precipitate of unreacted \(\ce{Na2SiF6}\) and \(\ce{NaCl}\), which was removed by filtration in the initial step (underlined), was completely dissolved in water to give an aqueous solution (solution K; 10.0 L).

Then, additional precipitation titrations using solution G were carried out, and the endpoints of the titrations with G were as follows:

  • For solution J (entire amount): 61.6 mL.
  • For 100 mL of solution K: 44.4 mL.

It should be noted here that the coexistence of \(\ce{NaCl}\) or \(\ce{SiO2}\) has no effect on the precipitation titration.

5.6 Calculate the mass of the \(\ce{NaCl}\) produced in the reaction vessel (information underlined), and calculate the mass (\(x\) [g]) of the \(\ce{Na2SiF6}\) used as a starting material.

5.7 77.8% of the \(\ce{CCl4}\) used as a starting material was unreacted. Calculate the mass of \(\ce{CF3Cl}\) generated.

Official English text of the problem.

Answers

In the generic SMILES below, each [*:1] represents an \(\ce{R^{1}}\) substituent and each [*:2] an \(\ce{R^{2}}\) substituent. The names identify the parent skeletons. Stereochemical marks in D and E specify their relative cage arrangements; they do not specify stereochemistry within the R groups.

  • 5.1
  • A: substituted 1,2-disilacyclobuta-1,3-diene, [*:1][Si]1=[Si]([*:1])C([*:2])=C1[*:2].
  • B: substituted 1,2-disilabenzene, [*:1][Si]1=[Si]([*:1])C([*:2])=C([*:2])C([*:2])=C1[*:2].

  • C: substituted 1,4-disilabenzene, [*:1][Si]1=C([*:2])C([*:2])=[Si]([*:1])C([*:2])=C1[*:2].

  • 5.2 Benzene, c1ccccc1: \(\boxed{\mathrm{ASE}=232\ \mathrm{kJ\,mol^{-1}}}\).
    C, with \(\ce{R^{1}}=\ce{R^{2}}=\ce{H}\): \(\boxed{\mathrm{ASE}=172\ \mathrm{kJ\,mol^{-1}}}\).

  • 5.3 \(\boxed{\Delta H_{\mathrm{D\rightarrow E}}=-10.5\ \mathrm{kJ\,mol^{-1}}}\).

  • 5.4

  • D: substituted 1,4-disilabicyclo[2.2.0]hexa-2,5-diene, the disila-Dewar benzene structure, [*:1][Si@]12C([*:2])=C([*:2])[Si@@]1([*:1])C([*:2])=C2[*:2].

  • E: substituted 2,5-disilabenzvalene, [*:1][Si@@]12[C@@]3([*:2])[C@]1([*:2])[Si@@]3([*:1])C([*:2])=C2[*:2].

  • 5.5 \[ \boxed{\ce{Na2SiF6 + Ce2(SO4)3 + 4H2O -> 2CeF3(s) + Si(OH)4 + Na2SO4 + 2H2SO4}} \]

  • 5.6 Sodium chloride, [Na+].[Cl-]: \(\boxed{65.2\ \mathrm{g}}\).
    Starting sodium hexafluorosilicate, [Na+].[Na+].F[Si-2](F)(F)(F)(F)F: \(\boxed{x=155\ \mathrm{g}}\).

  • 5.7 Chlorotrifluoromethane, FC(F)(F)Cl: \(\boxed{5.38\ \mathrm{g}}\).

Solution

5.1 — Structures of A, B, and C

The reaction with ethylene shows that cycloaddition consumes one of the two \(\pi\)-bonds of the silicon–silicon triple bond. With an alkyne, one \(\pi\)-bond from each triple bond is consumed, leaving a four-membered ring with opposite \(\ce{Si=Si}\) and \(\ce{C=C}\) bonds.

Thus, to draw A, place two adjacent silicon atoms along one edge of a square and two carbon atoms along the opposite edge. Draw both these edges as double bonds and the other two edges as single bonds. Each silicon carries \(\ce{R^{1}}\), and each carbon carries \(\ce{R^{2}}\).

For the six-membered conjugated products, the possible relative silicon positions give the following carbon environments:

Silicon positions Equivalent carbon positions Number of \(\ce{^{13}C}\) signals
1,2 \(3\equiv6;\ 4\equiv5\) 2
1,3 \(2;\ 5;\ 4\equiv6\) 3
1,4 \(2\equiv3\equiv5\equiv6\) 1

The observed two signals therefore identify B as the 1,2-disilabenzene, while one signal identifies C as the 1,4-disilabenzene.

Precise drawings corresponding to the SMILES in the answers are:

  • B: a six-membered ring numbered consecutively, with silicon at positions 1 and 2 and carbon at positions 3–6. Draw double bonds \(1=2\), \(3=4\), and \(5=6\).
  • C: a six-membered ring with silicon at positions 1 and 4. Draw double bonds \(1=2\), \(3=4\), and \(5=6\).

Attach \(\ce{R^{1}}\) to every silicon and \(\ce{R^{2}}\) to every carbon. These are individual Kekulé contributors; the NMR equivalence concerns the conjugated molecules.

5.2 — Aromatic stabilization energies

The isolated-double-bond references are ethene, C=C, for \(\ce{C=C}\), and silene, C=[SiH2], for \(\ce{Si=C}\).

The positive stabilization energy is the difference between the actual hydrogenation enthalpy and the sum expected for isolated double bonds:

\[ \mathrm{ASE} =\Delta H_{\mathrm{hyd,actual}} -\sum \Delta H_{\mathrm{hyd,isolated}}. \]

For benzene, the isolated-bond reference contains three \(\ce{C=C}\) bonds:

\[ \Delta H_{\mathrm{hyd,reference}} =3(-135)=-405\ \mathrm{kJ\,mol^{-1}}. \]

Using reaction (4),

\[ \boxed{\mathrm{ASE}_{\mathrm{benzene}} =-173-(-405)=232\ \mathrm{kJ\,mol^{-1}}}. \]

For C with hydrogen substituents, the molecule is 1,4-disilabenzene, [SiH]1=CC=[SiH]C=C1. A Kekulé structure contains two \(\ce{Si=C}\) bonds and one \(\ce{C=C}\) bond:

\[ \Delta H_{\mathrm{hyd,reference}} =2(-213)+(-135) =-561\ \mathrm{kJ\,mol^{-1}}. \]

The opposite silicon positions identify its hydrogenation as reaction (7), with \(\Delta H=-389\ \mathrm{kJ\,mol^{-1}}\). Hence,

\[ \boxed{\mathrm{ASE}_{\mathrm C} =-389-(-561)=172\ \mathrm{kJ\,mol^{-1}}}. \]

5.3 — Enthalpy change for D to E

For the equilibrium \(\mathrm D\rightleftharpoons\mathrm E\),

\[ K=\frac{[\mathrm E]}{[\mathrm D]}. \]

Thus,

\[ K_1=40.0,\quad T_1=323.15\ \mathrm K, \qquad K_2=20.0,\quad T_2=393.15\ \mathrm K. \]

The integrated van ’t Hoff equation is

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

Using \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\),

\[ \Delta H =-\frac{8.314\ln(20.0/40.0)} {1/393.15-1/323.15} =-1.046\times10^4\ \mathrm{J\,mol^{-1}}. \]

Therefore,

\[ \boxed{\Delta H_{\mathrm{D\rightarrow E}} =-10.5\ \mathrm{kJ\,mol^{-1}}}. \]

The exothermic sign agrees with the greater proportion of E at the lower temperature.

5.4 — Structures of D and E

Retain the numbering of C:

\[ \ce{Si}_{1}-\ce{C}_{2}-\ce{C}_{3} -\ce{Si}_{4}-\ce{C}_{5}-\ce{C}_{6} -\ce{Si}_{1}. \]

Every edge in this original six-membered ring must remain as a \(\sigma\)-bond. Silicon atoms retain their \(\ce{R^{1}}\) substituents and carbon atoms their \(\ce{R^{2}}\) substituents.

For D, add a single bond between silicon atoms 1 and 4. The remaining double bonds are \(2=3\) and \(5=6\).

This gives two four-membered rings:

\[ (1,2,3,4),\qquad (1,4,5,6), \]

sharing the \(\ce{Si-Si}\) edge. There are no three-membered rings. Symmetry makes all four carbon atoms equivalent, giving the required one carbon signal.

Thus D is the disila-Dewar benzene structure. Draw the two four-membered rings folded about their shared silicon–silicon bond; the molecular framework is not planar.

For E, add single bonds \(1-3\) and \(2-4\), and retain only the double bond \(5=6\).

The resulting three-membered rings are

\[ (1,2,3),\qquad (2,3,4), \]

which share the carbon–carbon edge \(2-3\). The carbon environments are:

  • the two equivalent saturated carbons, 2 and 3;
  • the two equivalent alkene carbons, 5 and 6.

This gives the required two carbon signals. E is therefore 2,5-disilabenzvalene. Draw the two \(\ce{SiC2}\) triangles sharing their carbon–carbon edge, with the two silicon vertices also connected through the \(\ce{C=C}\)-containing path \(1-6=5-4\). This is a folded cage structure.

Both constructions preserve all six original ring \(\sigma\)-bonds and satisfy the valence of every skeletal atom.

5.5 — Precipitation reaction

The amount of sodium hexafluorosilicate in the titrated portion of F is

\[ n_F =\frac{0.855\ \mathrm g} {188.053\ \mathrm{g\,mol^{-1}}} \frac{50.0}{200} =1.13665\times10^{-3}\ \mathrm{mol}. \]

The amount of cerium(III) supplied at the endpoint is

\[ n_{\mathrm{Ce}} =2\frac{6.86\ \mathrm g} {568.424\ \mathrm{g\,mol^{-1}}} \frac{18.8}{200} =2.26887\times10^{-3}\ \mathrm{mol}. \]

Consequently,

\[ \frac{n_{\mathrm{Ce}}}{n_F}=1.996\approx2. \]

Two cerium(III) ions therefore react per hexafluorosilicate ion. The binary precipitate is cerium(III) fluoride, \(\ce{CeF3}\), containing the six fluorine atoms as two formula units of precipitate.

The net ionic equation is

\[ \ce{SiF6^{2-} + 2Ce^{3+} + 4H2O -> 2CeF3(s) + Si(OH)4 + 4H+}. \]

Restoring the sodium and sulfate ions gives

\[ \boxed{\ce{Na2SiF6 + Ce2(SO4)3 + 4H2O -> 2CeF3(s) + Si(OH)4 + Na2SO4 + 2H2SO4}}. \]

The species used here have the following names and SMILES:

Name SMILES
Cerium(III) sulfate [Ce+3].[Ce+3].[O-]S(=O)(=O)[O-].[O-]S(=O)(=O)[O-].[O-]S(=O)(=O)[O-]
Cerium(III) fluoride [Ce+3].[F-].[F-].[F-]
Orthosilicic acid O[Si](O)(O)O
Sodium sulfate [Na+].[Na+].[O-]S(=O)(=O)[O-]
Sulfuric acid OS(=O)(=O)O
Water O
Hexafluorosilicate(2−) ion F[Si-2](F)(F)(F)(F)F
Cerium(III) ion [Ce+3]
Hydrogen ion, or proton [H+]

The disconnected salt SMILES express composition and charges rather than crystal structures. The hexafluorosilicate ion has octahedral coordination, which these SMILES do not explicitly encode.

5.6 — Sodium chloride produced and starting sodium hexafluorosilicate

From its stated preparation, the formula-unit concentration of G is

\[ c_G =\frac{6.86/568.424}{0.200} =0.0603423\ \mathrm{mol\,L^{-1}}. \]

The balanced titration equation establishes a 1:1 molar ratio between \(\ce{Ce2(SO4)3}\) and \(\ce{Na2SiF6}\).

The hydrolysis species are silicon tetrafluoride, F[Si](F)(F)F, and hexafluorosilicic acid, formally represented in ionic form by [H+].[H+].F[Si-2](F)(F)(F)(F)F. Silicon dioxide is an extended network solid and has no discrete molecular SMILES.

For the entire solution J,

\[ n_J(\ce{Na2SiF6}) =c_G(0.0616\ \mathrm L) =3.71708\times10^{-3}\ \mathrm{mol}. \]

From the given hydrolysis equation,

\[ \ce{3SiF4 + 2H2O -> SiO2 + 2H2SiF6}, \]

the 10 mL aliquot of H originally contained

\[ n_{\text{10 mL}}(\ce{SiF4}) =\frac32 n_J =5.57563\times10^{-3}\ \mathrm{mol}. \]

Therefore the complete 1.00 L solution contained

\[ n_{\text{total}}(\ce{SiF4}) =5.57563\times10^{-3}\frac{1000}{10} =0.557563\ \mathrm{mol}. \]

Every consumed formula unit of \(\ce{Na2SiF6}\) supplies one silicon atom to \(\ce{SiF4}\). Thus,

\[ n_{\mathrm{reacted}}(\ce{Na2SiF6}) =0.557563\ \mathrm{mol}. \]

Its two sodium atoms appear in the sodium chloride removed by filtration:

\[ n(\ce{NaCl})=2(0.557563)=1.11513\ \mathrm{mol}. \]

Using \(M(\ce{NaCl})=58.44\ \mathrm{g\,mol^{-1}}\),

\[ \boxed{m(\ce{NaCl}) =1.11513(58.44)=65.2\ \mathrm g}. \]

The 100 mL aliquot of K contains

\[ n_{\text{100 mL}}(\ce{Na2SiF6}) =c_G(0.0444\ \mathrm L) =2.67920\times10^{-3}\ \mathrm{mol}. \]

Since K has a total volume of 10.0 L,

\[ n_{\mathrm{unreacted}}(\ce{Na2SiF6}) =2.67920\times10^{-3}\frac{10.0}{0.100} =0.267920\ \mathrm{mol}. \]

The initial amount was the sum of reacted and unreacted material:

\[ n_{\mathrm{initial}}=0.557563+0.267920 =0.825482\ \mathrm{mol}. \]

Hence,

\[ \boxed{x =0.825482(188.053) =155.234\ \mathrm g \approx155\ \mathrm g}. \]

5.7 — Mass of chlorotrifluoromethane

The starting reagent is tetrachloromethane, ClC(Cl)(Cl)Cl.

Dividing each NMR integral by the number of fluorine atoms per molecule gives the molecular mole ratios:

Product Name and SMILES Relative amount
\(\ce{CFCl3}\) Trichlorofluoromethane, FC(Cl)(Cl)Cl \(45.0/1=45.0\)
\(\ce{CF2Cl2}\) Dichlorodifluoromethane, FC(F)(Cl)Cl \(65.0/2=32.5\)
\(\ce{CF3Cl}\) Chlorotrifluoromethane, FC(F)(F)Cl \(18.0/3=6.0\)
\(\ce{CF4}\) Tetrafluoromethane, FC(F)(F)F \(2.0/4=0.5\)

Thus, the mole fraction of chlorotrifluoromethane among the fluorinated carbon products is

\[ y(\ce{CF3Cl}) =\frac{6.0}{45.0+32.5+6.0+0.5} =\frac1{14}. \]

The mass of starting tetrachloromethane that reacted was

\[ m_{\mathrm{reacted}}(\ce{CCl4}) =500.0(1-0.778)=111.0\ \mathrm g. \]

Using \(M(\ce{CCl4})=153.81\ \mathrm{g\,mol^{-1}}\),

\[ n_{\mathrm{reacted}}(\ce{CCl4}) =\frac{111.0}{153.81} =0.721670\ \mathrm{mol}. \]

Each fluorinated carbon product contains one carbon atom, so their combined amount equals the amount of tetrachloromethane consumed:

\[ n(\ce{CF3Cl}) =\frac{0.721670}{14} =0.0515478\ \mathrm{mol}. \]

With \(M(\ce{CF3Cl})=104.46\ \mathrm{g\,mol^{-1}}\),

\[ \boxed{m(\ce{CF3Cl}) =0.0515478(104.46) =5.38\ \mathrm g}. \]

The solution was checked against the official solution.