IChO 1978 · Theory problem 5

Liquid compound X from coal tar

Download PDFIChO 1978

A certain liquid organic compound X (being present in coal tar) with a mass of 1.06 g was burned to produce 0.90 g of water and 3.52 g of carbon dioxide. Its vapours were 3.79 times as dense as nitrogen.

The compound X was oxidised by a hot mixture of \(\ce{CrO3 + H2SO4}\). A colourless crystalline substance A was isolated from the reaction mixture. It was soluble in an aqueous solution of \(\ce{NaOH}\) or \(\ce{NaHCO3}\).

Compound A when heated loses water and converts to compound B. Condensation of compound B with phenol in the presence of \(\ce{H2SO4}\) or \(\ce{ZnCl2}\) yields a substance Y which is very often used as an acid-base indicator.

Both compound A and compound B when heated with an access of 1-butanol (some drops of a concentrated \(\ce{H2SO4}\) solution are added) gives the same liquid compound C.

If accepted that a carbon atom shows a tendency to form four bonds in organic compounds, it is possible to write formally two different formulas for the compound X. The formulas written in this way do not correspond, however, to the chemical structure of the molecule X because up to date nobody has succeeded in the preparation of the two hypothetical isomers.

Haayman and Witbaut carried out in 1941 an ozonisation of the compound X in a \(\ce{CH3Cl}\) solution. After hydrolysis of ozonides, it was found that the water layer contains three different organic compounds in a molar ratio of D : E : F = 3 : 2 : 1. Only two of them formed new compounds G and H by a mild oxidation, the third one remains unchanged under these conditions but the effect of stronger oxidising agents as \(\ce{H2O2}\) for example, results in forming a well known liquid compound I with a characteristic sharp smell.

A sample of a pure anhydrous compound G was dissolved in an aqueous 1-molar solution of \(\ce{H2SO4}\) and the resulting solution was titrated with a volumetric 0.05-molar \(\ce{KMnO4}\) solution. An amount of 0.288 g of substance G required 25.6 \(\mathrm{cm^3}\) of the \(\ce{KMnO4}\) solution.

Problems:

5.1 Write the summary formula for the compound X.

5.2 Based on the information and data in the task, write chemical equations for the reactions by which products A, B, and C are formed.

5.3 Give the name for compound Y and write the equation of its synthesis. Write its structural formula and colour in both acidic and basic solutions.

5.4 Write two formal structural formulas for the compound X as well as a more correct structural formula according to the latest findings.

5.5 Write the chemical equation for the ozonolysis of compound X by which the fact can be explained why compounds D, E, and F are after hydrolysis of ozonides present in water layer in a molar ration of 3 : 2 : 1.

5.6 Write chemical equations for the reactions of formation of compounds G, H, and I.

5.7 Write both formal structural formulas used before, and the more correct modern structural formula of another liquid organic compound if you know that the compound is also present in coal tar and it is a derivative of compound X. What name of a known chemist is connected in the history with the formula of this basic compound? What are the products obtained by its ozonolysis?

5.8 Write the names of the substances X, Y, A – I under the corresponding compounds in the equations.

Note:

Use in your calculation:

\(M(\ce{C}) = 12\ \mathrm{g\,mol^{-1}};\quad M(\ce{O}) = 16\ \mathrm{g\,mol^{-1}};\quad M(\ce{H}) = 1\ \mathrm{g\,mol^{-1}};\quad M(\ce{KMnO4}) = 158\ \mathrm{g\,mol^{-1}}.\)

Official English text of the problem.

Answers

  • 5.1 \(\boxed{\ce{C8H10}}\). X is \(o\)-xylene, or 1,2-dimethylbenzene.
  • 5.2 Oxidation gives A, phthalic acid; dehydration gives B, phthalic anhydride. Esterification of either with excess 1-butanol gives C, di-\(n\)-butyl phthalate.
  • 5.3 Y is phenolphthalein, formed from one molecule of B and two molecules of phenol, with elimination of one molecule of water. Its neutral lactone is colourless; its quinoid dianion is pink/magenta in the usual basic indicator range.
  • 5.4 With methyl groups at ring positions 1 and 2, the two formal Kekulé structures have double bonds at \(1=2,\,3=4,\,5=6\) and \(2=3,\,4=5,\,6=1\). The actual molecule has a delocalized aromatic ring.
  • 5.5 D = glyoxal, E = methylglyoxal, F = diacetyl. The two formal cleavage patterns give \(D+2E\) and \(2D+F\); equal contributions give \(D:E:F=3:2:1\).
  • 5.6 G = oxalic acid, H = pyruvic acid, I = acetic acid. Mild oxidation converts D into G and E into H; peroxide cleavage of F produces two molecules of I. The titration confirms anhydrous oxalic acid, \(M=90\ \mathrm{g\,mol^{-1}}\).
  • 5.7 The intended parent compound is benzene, c1ccccc1, associated historically with August Kekulé. Its carbonyl-forming ozonolysis gives three molecules of glyoxal per molecule of benzene. The wording about the derivative relationship is reversed.
  • 5.8 The complete identification key is:
Label Name SMILES
X \(o\)-Xylene; 1,2-dimethylbenzene Cc1ccccc1C
Y Phenolphthalein, neutral lactone O=C1OC(c2ccc(O)cc2)(c2ccc(O)cc2)c2ccccc21
A Phthalic acid; benzene-1,2-dicarboxylic acid O=C(O)c1ccccc1C(=O)O
B Phthalic anhydride O=C1OC(=O)c2ccccc21
C Di-\(n\)-butyl phthalate CCCCOC(=O)c1ccccc1C(=O)OCCCC
D Glyoxal; ethanedial O=CC=O
E Methylglyoxal; 2-oxopropanal CC(=O)C=O
F Diacetyl; butane-2,3-dione CC(=O)C(=O)C
G Oxalic acid; ethanedioic acid O=C(O)C(=O)O
H Pyruvic acid; 2-oxopropanoic acid CC(=O)C(=O)O
I Acetic acid; ethanoic acid CC(=O)O

Solution

5.1 — Molecular formula of X

The combustion products give the amounts of carbon and hydrogen:

\[ n(\ce{C})=n(\ce{CO2}) =\frac{3.52\ \mathrm{g}}{44\ \mathrm{g\,mol^{-1}}} =0.0800\ \mathrm{mol}, \]

\[ n(\ce{H})=2n(\ce{H2O}) =2\frac{0.90\ \mathrm{g}}{18\ \mathrm{g\,mol^{-1}}} =0.100\ \mathrm{mol}. \]

The corresponding masses, retaining intermediate digits, are

\[ m(\ce{C})=0.960\ \mathrm{g}, \qquad m(\ce{H})=0.100\ \mathrm{g}. \]

Their sum equals the sample mass, \(1.060\ \mathrm{g}\). Thus the analysis accounts for the sample entirely as carbon and hydrogen.

The atomic ratio is

\[ n(\ce{C}):n(\ce{H})=0.0800:0.100=4:5, \]

so the empirical formula is \(\ce{C4H5}\), with empirical formula mass

\[ 4(12)+5(1)=53\ \mathrm{g\,mol^{-1}}. \]

At equal temperature and pressure, gas densities are proportional to molar masses. Taking \(M(\ce{N2})=28\ \mathrm{g\,mol^{-1}}\),

\[ M(X)=3.79(28)=106.12\ \mathrm{g\,mol^{-1}} \approx106\ \mathrm{g\,mol^{-1}}. \]

Consequently,

\[ \frac{M(X)}{M(\ce{C4H5})}\approx\frac{106}{53}=2, \qquad \boxed{X=\ce{C8H10}}. \]

Its degree of unsaturation is

\[ \frac{2(8)+2-10}{2}=4, \]

consistent with a benzene ring.

5.2 — Identification and formation of A, B, and C

Solubility in aqueous bicarbonate indicates an acidic compound, here a carboxylic acid. Formation of an anhydride on heating, followed by condensation with phenol to form an indicator, identifies the phthalic acid–phthalic anhydride pair.

Phthalic acid has adjacent carboxyl groups and readily forms a five-membered cyclic anhydride. Therefore X must have adjacent methyl groups: it is \(o\)-xylene. Hot chromium(VI) oxidizes each methyl group to a carboxyl group.

A fully balanced equation in acidic solution is

\[ \underbrace{\ce{C6H4(CH3)2}}_{\text{X: ortho-xylene}} +\ce{4CrO3 + 12H+} \xrightarrow{\Delta} \underbrace{\ce{C6H4(COOH)2}}_{\text{A: phthalic acid}} +\ce{4Cr^3+ + 8H2O}. \]

The two substituents in these condensed aromatic formulas occupy adjacent ring positions.

Heating A eliminates water:

\[ \underbrace{\ce{C6H4(COOH)2}}_{\text{A: phthalic acid}} \xrightarrow{\Delta} \underbrace{\ce{C6H4(CO)2O}}_{\text{B: phthalic anhydride}} +\ce{H2O}. \]

In B, the oxygen between the two acyl groups closes the five-membered anhydride ring fused to the benzene ring.

Excess butan-1-ol, CCCCO, esterifies both carboxyl groups of A:

\[ \begin{aligned} &\underbrace{\ce{C6H4(COOH)2}}_{\text{A: phthalic acid}} +\ce{2HO(CH2)3CH3}\\ &\quad\xrightarrow{\ce{H2SO4},\,\Delta} \underbrace{\ce{C6H4[COO(CH2)3CH3]2}}_{\text{C: di-n-butyl phthalate}} +\ce{2H2O}. \end{aligned} \]

Starting with B gives the same diester, but only one net molecule of water is formed:

\[ \begin{aligned} &\underbrace{\ce{C6H4(CO)2O}}_{\text{B: phthalic anhydride}} +\ce{2HO(CH2)3CH3}\\ &\quad\xrightarrow{\ce{H2SO4},\,\Delta} \underbrace{\ce{C6H4[COO(CH2)3CH3]2}}_{\text{C: di-n-butyl phthalate}} +\ce{H2O}. \end{aligned} \]

5.3 — Phenolphthalein: synthesis, structures, and colours

Phthalic anhydride condenses with two molecules of phenol, Oc1ccccc1, to form phenolphthalein:

\[ \underbrace{\ce{C6H4(CO)2O}}_{\text{B: phthalic anhydride}} +\underbrace{\ce{2C6H5OH}}_{\text{phenol}} \xrightarrow[\Delta]{\ce{H2SO4}\ \text{or}\ \ce{ZnCl2}} \underbrace{\ce{C20H14O4}}_{\text{Y: phenolphthalein}} +\ce{H2O}. \]

The relevant structures are:

Medium Form, name, and colour SMILES
Acidic Neutral phenolphthalein lactone; colourless O=C1OC(c2ccc(O)cc2)(c2ccc(O)cc2)c2ccccc21
Basic, usual indicator range Phenolphthalein quinoid dianion; pink/magenta O=C([O-])c1ccccc1C(c1ccc([O-])cc1)=C1C=CC(=O)C=C1

To draw the acidic structure, draw a phthalide ring whose tetrahedral carbon bears two para-hydroxyphenyl groups.

For the basic structure, open the lactone ring to give an ortho-carboxylate group. In the displayed resonance contributor, the central carbon is double-bonded to a quinonoid ring and single-bonded to a para-phenoxide ring. The charge and conjugation are delocalized; the single SMILES represents one contributor.

The overall acid–base conversion is

\[ \underbrace{\ce{C20H14O4}}_{\text{colourless lactone}} +\ce{2OH-} \rightleftharpoons \underbrace{\ce{C20H12O4^2-}}_{\text{pink quinoid dianion}} +\ce{2H2O}. \]

Ring opening and the resulting extended conjugation account for the visible colour. The stated basic colour refers to the normal indicator range; very strongly alkaline solutions can undergo further changes.

5.4 — Two formal structures and the aromatic structure of X

Number the ring carbons consecutively from 1 to 6, with methyl groups attached to positions 1 and 2.

Representation of \(o\)-xylene Ring double bonds SMILES
Kekulé contributor \(K_1\) \(1=2,\ 3=4,\ 5=6\) CC1=C(C)C=CC=C1
Kekulé contributor \(K_2\) \(2=3,\ 4=5,\ 6=1\) CC1C(C)=CC=CC=1
Delocalized aromatic structure No fixed alternating localization Cc1ccccc1C

In \(K_1\), the two methyl-bearing ring carbons share a formal double bond; in \(K_2\), they share a formal single bond.

For the modern drawing, draw a hexagon with an internal circle and methyl groups at two adjacent vertices. The aromatic ring is approximately planar, with six delocalized \(\pi\) electrons.

The Kekulé drawings are resonance contributors, not separately existing isomers or molecules in equilibrium. Their SMILES may therefore be converted to the same aromatic structure by chemical software.

5.5 — Ozonolysis and the \(3:2:1\) ratio

Formal ozonolytic cleavage converts each carbon of a localized double bond into a carbonyl carbon:

  • A ring carbon bearing hydrogen becomes an aldehyde carbonyl.
  • A ring carbon bearing a methyl group becomes a ketone carbonyl.

Cleaving the three formal double bonds leaves three two-carbon fragments, corresponding to the three remaining ring bonds.

Formal pattern Remaining carbon pairs Organic products
\(K_1\): cleave \(1=2,\ 3=4,\ 5=6\) \(2–3,\ 4–5,\ 6–1\) One glyoxal D and two methylglyoxal E
\(K_2\): cleave \(2=3,\ 4=5,\ 6=1\) \(1–2,\ 3–4,\ 5–6\) One diacetyl F and two glyoxal D

Thus one occurrence of each formal cleavage pattern gives

\[ (D+2E)+(2D+F)=3D+2E+F. \]

This reproduces the observed ratio:

\[ \boxed{D:E:F=3:2:1}. \]

Using the conventional formal balance for ozonide hydrolysis, with hydrogen peroxide as coproduct, the combined equation is

\[ \begin{aligned} &2\underbrace{\ce{C8H10}}_{\text{X: ortho-xylene}} +\ce{6O3 + 6H2O}\\ &\quad\longrightarrow 3\underbrace{\ce{OHC-CHO}}_{\text{D: glyoxal}} +2\underbrace{\ce{CH3COCHO}}_{\text{E: methylglyoxal}} +\underbrace{\ce{CH3COCOCH3}}_{\text{F: diacetyl}} +\ce{6H2O2}. \end{aligned} \]

The solvent is chloromethane, CCl. Carbonyl products are written in their conventional unhydrated forms.

This is a formal cleavage account of the stipulated initial product mixture. It does not mean that X contains two independently existing Kekulé isomers, nor does resonance alone establish quantitative reaction branching.

5.6 — Formation of G, H, and I; verification by titration

Here \(\ce{[O]}\) denotes one oxygen equivalent supplied by an oxidizing reagent, rather than a free oxygen-atom intermediate.

Mild oxidation converts both aldehyde groups of glyoxal into carboxyl groups:

\[ \underbrace{\ce{OHC-CHO}}_{\text{D: glyoxal}} +\ce{2[O]} \longrightarrow \underbrace{\ce{HOOC-COOH}}_{\text{G: oxalic acid}}. \]

Methylglyoxal has one aldehyde group and one ketone group. Oxidation of the aldehyde group gives pyruvic acid:

\[ \underbrace{\ce{CH3COCHO}}_{\text{E: methylglyoxal}} +\ce{[O]} \longrightarrow \underbrace{\ce{CH3COCOOH}}_{\text{H: pyruvic acid}}. \]

Diacetyl has two ketone groups and no aldehyde group, explaining its resistance to the stipulated mild oxidation. Hydrogen peroxide cleaves the bond between its carbonyl carbons:

\[ \underbrace{\ce{CH3COCOCH3}}_{\text{F: diacetyl}} +\ce{H2O2} \longrightarrow 2\underbrace{\ce{CH3COOH}}_{\text{I: acetic acid}}. \]

Acetic acid is the familiar liquid with a sharp, vinegar-like smell.

The titration independently checks the assignment of G. The relevant half-reactions are

\[ \ce{H2C2O4 -> 2CO2 + 2H+ + 2e-}, \]

\[ \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}. \]

Combining them gives

\[ 5\underbrace{\ce{H2C2O4}}_{\text{G: oxalic acid}} +\ce{2MnO4- + 6H+} \longrightarrow \ce{10CO2 + 2Mn^2+ + 8H2O}. \]

The amount of permanganate is

\[ n(\ce{MnO4-}) = 0.05\ \mathrm{mol\,L^{-1}} \times25.6\times10^{-3}\ \mathrm{L} = 1.28\times10^{-3}\ \mathrm{mol}. \]

From the \(5:2\) stoichiometry,

\[ n(G)=\frac52\,n(\ce{MnO4-}) =3.20\times10^{-3}\ \mathrm{mol}. \]

Hence the implied molar mass is

\[ M(G)= \frac{0.288\ \mathrm{g}} {3.20\times10^{-3}\ \mathrm{mol}} = 90\ \mathrm{g\,mol^{-1}}. \]

For anhydrous oxalic acid,

\[ M(\ce{H2C2O4})=2(1)+2(12)+4(16) =90\ \mathrm{g\,mol^{-1}}, \]

exactly matching the titration data.

5.7 — Benzene, Kekulé structures, and ozonolysis

The intended basic or parent compound is benzene. More precisely, \(o\)-xylene is a dimethyl derivative of benzene.

Using fixed numbering around a six-membered ring:

Benzene representation Ring double bonds SMILES
First Kekulé contributor \(1=2,\ 3=4,\ 5=6\) C1=CC=CC=C1
Second Kekulé contributor \(2=3,\ 4=5,\ 6=1\) C1C=CC=CC=1
Aromatic structure Six delocalized \(\pi\) electrons c1ccccc1

The corresponding drawings are two hexagons with alternating double bonds shifted by one edge, and a modern hexagon containing a circle. Every ring carbon bears one hydrogen. Benzene is planar and all six carbon–carbon bonds are equivalent.

The historical chemist is August Kekulé.

Because every ring carbon bears hydrogen, either formal cleavage pattern produces only glyoxal as the organic carbonyl product. The conventional hydrolysis balance is

\[ \underbrace{\ce{C6H6}}_{\text{benzene}} +\ce{3O3 + 3H2O} \longrightarrow 3\underbrace{\ce{OHC-CHO}}_{\text{glyoxal}} +\ce{3H2O2}. \]

Thus one molecule of benzene gives three molecules of glyoxal under the carbonyl-forming work-up considered here.

5.8 — Names beneath the formulas and remaining SMILES

The names of X, Y, and A–I appear beneath their formulas in the equations above; their complete name–SMILES key is given in Answers 5.8.

For completeness, the other molecular and ionic species used in the solution are:

Species SMILES
Water O
Carbon dioxide O=C=O
Dinitrogen N#N
Ozone [O-][O+]=O
Hydrogen peroxide OO
Sulfuric acid OS(=O)(=O)O
Chromium(VI) oxide O=[Cr](=O)=O
Chromium(III) ion [Cr+3]
Hydrogen ion [H+]
Hydroxide ion [OH-]
Sodium hydroxide [Na+].[OH-]
Sodium hydrogen carbonate [Na+].OC(=O)[O-]
Zinc chloride [Zn+2].[Cl-].[Cl-]
Potassium permanganate [K+].[O-][Mn](=O)(=O)=O
Permanganate ion [O-][Mn](=O)(=O)=O
Manganese(II) ion [Mn+2]

The disconnected salt SMILES specify ionic composition; they do not specify solvation or solid-state coordination.

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.