IChO 1991 · Theory problem 5

Halogen derivatives of hydrocarbons

Download PDFIChO 1991

After passing 0.25 mol of hydrocarbon A over heated pumice (950 K) in an iron pipe, compound B in yield of 80 % (i. e. 15.4 g) and 2.4 dm\(^3\) of hydrogen, (295 K, 102 kPa) were obtained.

B and hydrogen are the only products. A mixture of halogen derivatives C, D, E, F and G is produced from B by reaction with a halogen in presence of a Lewis acid. Compounds C to G contain each one halogen atom more then the preceding compound. For the compounds C to F only one of the possible isomers is formed. In compound G there is no such preference and its three isomers G₁, G₂ and G₃ are found in the mixture. Compounds C to F racemize easily so that no optical isomerism occurs. However racemization is difficult for G₁, G₂, and especially for G₃. In the mass spectrum of E only three isotopic peaks were observed. Their relative intensities are 1: 1: 0.3.

Information:

  • \(k_{\mathrm{ortho}}>k_{\mathrm{para}}\) in compound B.
  • The effect of the first halogen in the ring: \(k_{\mathrm{para}}>k_{\mathrm{ortho}}\).
  • Compounds D and F in one of their conformations have a center of symmetry.
  • The contribution of carbon and hydrogen isotopes in the mass spectrum of E are negligible.
  • Natural abundance of halogen isotopes:

\(\ce{^{19}F}=100\ \%\);

\(\ce{^{35}Cl}=75.53\ \%\); \(\ce{^{37}Cl}=24.47\ \%\);

\(\ce{^{79}Br}=50.54\ \%\); \(\ce{^{81}Br}=49.46\ \%\);

\(\ce{^{127}I}=100\ \%\).

5.1 Give the structures of A, B, C, D, E, F, G₁, G₂ and G₃.

5.2 Explain your choice for the halogen.

5.3 Draw and label stereo-formulas of the rotational isomers of D for \(\Phi=0,\pi/2,\pi,\) and \(3\pi/2\), where \(\Phi\) denotes the dihedral or torsional angle in radians and \(\vartheta=0\) describes the configuration with maximal energy.

5.4 Draw profiles of the energy changes as a function of the angle of rotation around the \(\ce{C-C}\) bond for compounds C and D.

5.5 Give a sequence of G₁, G₂, G₃ according to their increasing difficulty of racemization. Explain your answer.

5.6 Draw stereo-formulas of the enantiomers of G₃.

5.7 Suggest a chemical reaction or a biological method by which compounds like these can be destroyed.

Official English text of the problem.

Answers

  • 5.1 The structures are given below. Number the two rings of biphenyl from the atoms forming the connecting bond, \(1\) and \(1'\). Define G₁, G₂, G₃ by adding the fifth chlorine to F at positions \(3,4,6\), respectively.
Compound Name SMILES
A Benzene c1ccccc1
B Biphenyl c1(-c2ccccc2)ccccc1
C 2-Chlorobiphenyl c1(-c2ccccc2)c(Cl)cccc1
D 2,2′-Dichlorobiphenyl c1(-c2c(Cl)cccc2)c(Cl)cccc1
E 2,2′,5-Trichlorobiphenyl c1(-c2c(Cl)cccc2)c(Cl)ccc(Cl)c1
F 2,2′,5,5′-Tetrachlorobiphenyl c1(-c2c(Cl)ccc(Cl)c2)c(Cl)ccc(Cl)c1
G₁ 2,2′,3,5,5′-Pentachlorobiphenyl c1(-c2c(Cl)ccc(Cl)c2)c(Cl)c(Cl)cc(Cl)c1
G₂ 2,2′,4,5,5′-Pentachlorobiphenyl c1(-c2c(Cl)ccc(Cl)c2)c(Cl)cc(Cl)c(Cl)c1
G₃ 2,2′,5,5′,6-Pentachlorobiphenyl, equivalently 2,2′,3,5′,6-pentachlorobiphenyl c1(-c2c(Cl)ccc(Cl)c2)c(Cl)ccc(Cl)c1Cl

  • 5.2 The halogen is chlorine. Three chlorine atoms give the isotope-peak ratio \[ 1:0.972:0.315:0.0340. \] The first three peaks match the stated \(1:1:0.3\); a weak fourth peak is predicted.

  • 5.3 For D: \(0\), coplanar with the two ortho chlorines on the same side, highest energy; \(\pi/2\), one twisted enantiomer; \(\pi\), coplanar with the chlorines on opposite sides, possessing an inversion center; \(3\pi/2\), the mirror-image twisted enantiomer.

  • 5.4 C has equivalent coplanar barriers at \(0,\pi,2\pi\). D has higher barriers at \(0,2\pi\) and a lower coplanar barrier at \(\pi\). Both have intervening regions of lower energy with noncoplanar rings.

  • 5.5 With the labels defined above, increasing difficulty of racemization is \[ \boxed{\mathbf{G_2<G_1<G_3}}. \]

  • 5.6 G₃ has two axial enantiomers. Fixing the front ring, perpendicular rear-ring orientations at \(+\pi/2\) and \(-\pi/2\) give an explicit mirror-image pair, described below. Their ordinary connectivity SMILES is the same as that listed for G₃.

  • 5.7 Controlled complete incineration destroys these compounds. For a pentachlorobiphenyl: \[ \boxed{\ce{C12H5Cl5 + 12 O2 -> 12 CO2 + 5 HCl}}. \]

Solution

5.1 — Identification of the hydrocarbons and substitution patterns

The amount of hydrogen, hydrogen gas [H][H], is

\[ n(\ce{H2})=\frac{pV}{RT} =\frac{(102\ \mathrm{kPa})(2.4\ \mathrm{dm^3})} {(8.314\ \mathrm{kPa\,dm^3\,mol^{-1}\,K^{-1}})(295\ \mathrm K)} =0.0998\ \mathrm{mol}\approx0.10\ \mathrm{mol}. \]

At 80 % yield, the amount of A converted into the stated products is

\[ n_{\mathrm{converted}}(\mathbf A) =0.80(0.25\ \mathrm{mol}) =0.20\ \mathrm{mol}. \]

The hydrogen has mass approximately \(0.20\ \mathrm g\). Since B and hydrogen are the only products, mass balance gives

\[ M(\mathbf A) =\frac{15.4\ \mathrm g+0.20\ \mathrm g}{0.20\ \mathrm{mol}} \approx78\ \mathrm{g\,mol^{-1}}. \]

This is consistent with benzene, \(\ce{C6H6}\). The aromatic substitution and rotation about a bond joining two rings identify B as biphenyl, formed by dehydrogenative coupling:

\[ \ce{2 C6H6 ->[950 K,\ pumice,\ iron] C12H10 + H2}. \]

The predicted biphenyl amount and mass are

\[ n(\mathbf B)=\frac{0.80(0.25)}{2}=0.10\ \mathrm{mol}, \]

\[ m(\mathbf B)=(0.10\ \mathrm{mol})(154\ \mathrm{g\,mol^{-1}}) =15.4\ \mathrm g. \]

Thus both the mass of B and the hydrogen volume agree with the proposed reaction.

As established in 5.2, the substituent is chlorine. Chlorine, ClCl, undergoes successive electrophilic aromatic substitutions in the presence of a Lewis acid, producing hydrogen chloride, Cl. The balanced overall equation for each degree of substitution is

\[ \ce{C12H10 + n Cl2 ->[Lewis acid] C12H_{10-n}Cl_n + n HCl}, \qquad n=1,2,3,4,5. \]

The substitution sequence follows from the directing and symmetry information:

  1. C: In biphenyl, the stated \(k_{\mathrm{ortho}}>k_{\mathrm{para}}\) selects position \(2\). Hence C is 2-chlorobiphenyl.

  2. D: An inversion center at the midpoint of the inter-ring bond must interchange identically substituted rings. The second chlorine therefore occupies the equivalent ortho position on the other ring, giving 2,2′-dichlorobiphenyl. Its coplanar, opposite-side conformation has the required inversion center.

  3. E: Each ring now carries one chlorine. The next chlorine enters para to an existing chlorine: position \(5\), which is para to position \(2\). The two rings are equivalent before this substitution, so E is 2,2′,5-trichlorobiphenyl.

  4. F: The inversion-center condition requires matching substitution patterns on the two rings. The fourth chlorine therefore enters position \(5'\), giving 2,2′,5,5′-tetrachlorobiphenyl.

  5. G: In F, the remaining positions on either ring are \(3,4,6\). Interchanging the rings does not generate additional constitutional isomers. Thus there are exactly three possibilities: \[ \begin{aligned} \mathbf{G_1}:&\quad 2,2',3,5,5',\\ \mathbf{G_2}:&\quad 2,2',4,5,5',\\ \mathbf{G_3}:&\quad 2,2',5,5',6. \end{aligned} \] These are the structures and SMILES listed in Answers.

5.2 — Identification of the halogen from the isotope pattern

E contains three halogen atoms. Fluorine and iodine each have only one isotope in the supplied abundance data, so neither can produce the observed halogen-derived isotope cluster.

For chlorine, put

\[ p=0.7553,\qquad q=0.2447. \]

The probabilities of incorporating zero, one, two, or three \(\ce{^{37}Cl}\) atoms are the terms of

\[ (p+q)^3=p^3+3p^2q+3pq^2+q^3. \]

Consequently,

\[ I_M:I_{M+2}:I_{M+4}:I_{M+6} =p^3:3p^2q:3pq^2:q^3. \]

Normalizing to the first peak gives

\[ 1:3\frac qp:3\left(\frac qp\right)^2: \left(\frac qp\right)^3 = \boxed{1:0.9719:0.3149:0.03401}. \]

For E, \(\ce{C12H7Cl3}\), these correspond to nominal molecular-ion masses \(256,258,260,262\).

The first three intensities round to the reported \(1:1:0.3\). The fourth is much smaller, although it is not mathematically zero.

By contrast, three bromine atoms, using the supplied abundances, would give approximately

\[ 1:2.936:2.873:0.937, \]

which does not match. Thus chlorine is the intended identification.

5.3 — Rotational isomers of D

Use the following precise end-on convention:

  • Look along the inter-ring bond from unprimed \(C1\) toward primed \(C1'\).
  • Keep the front ring fixed, with its \(C2\) chlorine at 12 o’clock and its \(C6\) hydrogen at 6 o’clock.
  • Define \(\Phi=0\) when the rear \(C2'\) chlorine is also at 12 o’clock.
  • Increase \(\Phi\) by rotating the rear ring clockwise in this view.

The requested drawings then have these features:

\(\Phi\) Rear \(C2'\) chlorine Rear \(C6'\) hydrogen Conformation and stereochemical label
\(0\) 12 o’clock 6 o’clock Coplanar; chlorines on the same side of the axis; achiral; greatest steric crowding and maximal energy
\(\pi/2\) 3 o’clock 9 o’clock Perpendicular ring planes; chiral conformer I
\(\pi\) 6 o’clock 12 o’clock Coplanar; chlorines on opposite sides; achiral; inversion center at the inter-ring bond midpoint
\(3\pi/2\) 9 o’clock 3 o’clock Perpendicular ring planes; chiral conformer II, the mirror image of I

At \(\Phi=0\), the ortho chlorine atoms approach one another directly. At \(\Phi=\pi\), each ortho chlorine instead encounters an ortho hydrogen. The latter coplanar passage therefore has the lower steric barrier.

The SMILES for D specifies connectivity; these angular descriptions specify its conformations.

5.4 — Torsional-energy profiles of C and D

Plot energy vertically and \(\Phi\), from \(0\) to \(2\pi\), horizontally, using the convention in 5.3.

For C, the unsubstituted ring has equivalent ortho positions. Rotating it by \(\pi\) produces an equivalent arrangement, so

\[ E_C(\Phi+\pi)=E_C(\Phi). \]

The coplanar arrangements at \(0,\pi,2\pi\) have identical steric interactions and therefore equal barrier heights. Draw equivalent lower-energy, noncoplanar regions between successive coplanar barriers.

For D, the two coplanar arrangements differ:

  • At \(0\) and \(2\pi\), an ortho chlorine faces an ortho chlorine: the higher barrier.
  • At \(\pi\), the two close contacts are chlorine–hydrogen contacts: the lower barrier.
  • Between these coplanar arrangements lie two mirror-related, noncoplanar energy wells.

Thus a qualitative steric sketch follows this pattern:

Angle \(0\) Near \(\pi/2\) \(\pi\) Near \(3\pi/2\) \(2\pi\)
C High Low Equally high Equally low High
D Highest Low Lower local maximum Equally low Highest

For D, reflection requires

\[ E_D(\Phi)=E_D(2\pi-\Phi). \]

These are qualitative profiles. Conjugation favors coplanarity whereas steric repulsion favors twisting, so the exact minimum angles need not be precisely \(\pi/2\) and \(3\pi/2\); the supplied information does not determine numerical energies or fine details of the curves.

5.5 — Increasing difficulty of racemization

For these axially chiral biphenyls, interconversion of the enantiomers requires rotation through a coplanar arrangement. The relevant comparison is therefore the steric cost of the easiest coplanar passage.

  • G₂, with the additional chlorine at \(4\), retains only two chlorines ortho to the inter-ring bond. The added chlorine is relatively remote from that bond.
  • G₁, with the additional chlorine at \(3\), also has two ortho chlorines, but the \(3\)-chlorine crowds the adjacent \(2\)-chlorine. This buttressing effect increases the steric obstruction to rotation.
  • G₃, with the additional chlorine at \(6\), has three ortho chlorines. One ring now carries chlorine on both sides of the inter-ring bond. Either coplanar passage therefore brings an ortho chlorine on one ring against an ortho chlorine on the other.

The expected steric order is consequently

\[ \boxed{ \underbrace{\mathbf{G_2}}_{\text{additional }4\text{-Cl}} < \underbrace{\mathbf{G_1}}_{\text{additional }3\text{-Cl}} < \underbrace{\mathbf{G_3}}_{\text{additional }6\text{-Cl}} }. \]

5.6 — Enantiomers of G₃

Retain the common numbering used above: the unprimed ring has chlorines at \(2,5,6\), and the primed ring at \(2',5'\).

The two axial enantiomers of 2,2′,5,5′,6-pentachlorobiphenyl—equivalently 2,2′,3,5′,6-pentachlorobiphenyl—have connectivity SMILES

c1(-c2c(Cl)ccc(Cl)c2)c(Cl)ccc(Cl)c1Cl.

Ordinary SMILES does not distinguish the frozen axial configurations here. The following end-on descriptions specify an explicit enantiomeric pair.

Look from unprimed \(C1\) toward primed \(C1'\). Fix the front ring vertically, with \(C2\) at 12 o’clock and \(C6\) at 6 o’clock. Its chlorine at \(C5\) lies on the \(C6\) side.

Feature Enantiomer I Enantiomer II
Front ring Fixed as described Identical fixed orientation
Rear \(C2'\), bearing chlorine 3 o’clock 9 o’clock
Rear \(C6'\), bearing hydrogen 9 o’clock 3 o’clock
Rear \(C5'\), bearing chlorine On the \(C6'\) side, to the left On the \(C6'\) side, to the right
Ring-plane torsion \(+\pi/2\) \(-\pi/2\)

Reflection in the plane of the front ring converts I into II. They are nonsuperimposable mirror images: the axial \(R_{\mathrm a}/S_{\mathrm a}\) pair.

Although both ortho positions of the front ring bear chlorine, its two paths around the ring are different: the path through \(C6\) encounters chlorine at \(C5\), whereas the path through \(C2\) encounters hydrogen at \(C3\). The primed ring likewise has inequivalent paths through \(C2'\) and \(C6'\). Thus the substitution pattern supports axial chirality.

5.7 — Destruction

A suitable chemical method is controlled high-temperature incineration in excess oxygen, with sufficient residence time for complete oxidation and capture of the hydrogen chloride produced.

For any of the pentachlorobiphenyl isomers:

\[ \ce{C12H5Cl5 + 12 O2 -> 12 CO2 + 5 HCl}. \]

The reactant is a pentachlorobiphenyl listed above; the other molecular species are oxygen O=O, carbon dioxide O=C=O, and hydrogen chloride Cl.

The equation balances all atoms: twelve carbons form twelve carbon dioxide molecules, while the five hydrogens and five chlorines form five hydrogen chloride molecules.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.