IChO 1988 · Theory problem 6

Identification of a bio-accumulated pollutant X from its mass spectrum

Download PDFIChO 1988

Upon analyzing sea mussels a new bio-accumulated pollutant X was found as determined by mass spectroscopy coupled to a gas chromatograph. The mass spectrum is illustrated in figure. Determine the structural formula of X assuming that it is produced out of synthetic rubber used as insulation in electrolysis cells that are used for the production of chlorine. Give the name of the compound X. The isotopic abundances of the pertinent elements are shown in the figure and table below. Intensities of the ions m/e = 196, 233, 268 and 270 are very low and thus omitted. Peaks of the \(\ce{^{13}C}\) containing ions are omitted for simplicity.

Figure 1
Element Mas Norm.abundanc Mass Norm.abundanc Mas Norm.abundanc
H 1 100.0 2 0.015
C 12 100.0 13 1.1
N 14 100.0 15 0.37
O 16 100.0 17 0.04 18 0.20
P 31 100.0
S 32 100.0 33 0.80 34 4.4
Cl 35 100.0 37 32.5
Br 79 100.0 81 98.0

Official English text of the problem.

Answers

  • Problem 6: X is hexachlorobutadiene, systematically 1,1,2,3,4,4-hexachlorobuta-1,3-diene, with molecular formula \(\ce{C4Cl6}\) and structural formula \[ \boxed{\ce{Cl2C=C(Cl)-C(Cl)=CCl2}}. \] SMILES: ClC(Cl)=C(Cl)C(Cl)=C(Cl)Cl.

Solution

Problem 6

The molecular-ion cluster begins at \(m/e=258\), with successive peaks separated by 2. This is characteristic of substitution of \(\ce{^{35}Cl}\) by \(\ce{^{37}Cl}\). The supplied relative abundances give

\[ r=\frac{\text{abundance of }\ce{^{37}Cl}} {\text{abundance of }\ce{^{35}Cl}} =\frac{32.5}{100.0}=0.325. \]

For a species containing \(n\) chlorine atoms, the intensity of the peak containing \(k\) heavy chlorine atoms, relative to the peak containing only light chlorine atoms, is

\[ \frac{I_{M+2k}}{I_M}=\binom{n}{k}r^k. \]

In particular,

\[ \frac{I_{M+2}}{I_M}=nr. \]

The first two molecular-ion peaks have an intensity ratio of approximately \(1:2\). Six chlorine atoms predict \(I_{260}/I_{258}=6(0.325)=1.95\), consistent with the figure. The complete predicted envelope is:

\(m/e\) Intensity relative to \(I_{258}=100\)
258 100
260 195
262 158.4
264 68.7
266 16.7
268 2.18
270 0.118

This also explains why the peaks at 268 and 270 are very weak. Thus, X contains six chlorine atoms.

Using the integer isotope masses supplied, the mass remaining after accounting for six light chlorine atoms is

\[ 258\,\mathrm{u}-6(35\,\mathrm{u})=48\,\mathrm{u}. \]

For the carbon framework derived from the rubber,

\[ N_{\mathrm C}=\frac{48\,\mathrm{u}}{12\,\mathrm{u}}=4. \]

There is no remaining mass for hydrogen, giving the molecular formula

\[ \boxed{\ce{C4Cl6}}. \]

The fragment clusters provide an independent consistency check. Their lowest-mass peaks are

\[ 223=258-35,\qquad 188=258-2(35), \]

consistent with loss of one chlorine atom and net loss of two chlorine atoms, respectively. The corresponding isotope patterns fit five and four remaining chlorine atoms:

Cluster begins at \(m/e\) Chlorine atoms Predicted first-to-second peak ratio Highest-mass member
223 5 \(1:1.625\) \(48+5(37)=233\)
188 4 \(1:1.300\) \(48+4(37)=196\)

The highest-mass members require every remaining chlorine atom to be \(\ce{^{37}Cl}\), explaining the other two weak, omitted peaks.

Counting each halogen atom as a hydrogen atom, the degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(4)+2-6}{2}=2. \]

The synthetic-rubber origin points to a four-carbon diene framework. Buta-1,3-diene, also called 1,3-butadiene, has SMILES C=CC=C. Replacing all six hydrogen atoms of this framework by chlorine gives

\[ \ce{Cl2C=C(Cl)-C(Cl)=CCl2}, \]

which has exactly two double bonds and the required formula and mass spectrum.

Therefore, X is hexachlorobutadiene, or 1,1,2,3,4,4-hexachlorobuta-1,3-diene, SMILES ClC(Cl)=C(Cl)C(Cl)=C(Cl)Cl. Neither double bond has \(E/Z\) stereochemistry because each terminal carbon bears two identical chlorine atoms.

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