IChO 1974 · Theory problem 1

From electrolytic hydrogen to methyl orange

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By electrochemical decomposition of water, there are in an electric circuit a voltmeter, platinum electrodes and a battery containing ten galvanic cells connected in series, each of it having the voltage of 1.5 V and internal resistance of 0.4 Ω. The resistance of the voltmeter is 0.5 Ω and the polarisation voltage of the battery is 1.5 V. Electric current flows for 8 hours, 56 minutes and 7 seconds through the electrolyte. Hydrogen obtained in this way was used for a synthesis with another substance, thus forming a gaseous substance A which can be converted by oxidation with oxygen via oxide to substance B.

By means of substance B it is possible to prepare substance C from which after reduction by hydrogen substance D can be obtained. Substance D reacts at 180 °C with a concentration solution of sulphuric acid to produce sulphanilic acid. By diazotization and successive copulation with p-N,N-dimethylaniline, an azo dye, methyl orange is formed.

Problems:

  1. Write chemical equations for all the above mentioned reactions.
  2. Calculate the mass of product D.
  3. Give the exact chemical name for the indicator methyl orange. Show by means of structural formulas what changes take place in dependence on concentration of \(\ce{H3O+}\) ions in the solution.

Relative atomic masses: \(A_\mathrm{r}(\ce{N}) = 14;\quad A_\mathrm{r}(\ce{O}) = 16;\quad A_\mathrm{r}(\ce{C}) = 12;\quad A_\mathrm{r}(\ce{H}) = 1.\)

Official English text of the problem.

Answers

  1. The substances are:
Label Identity SMILES
A Ammonia, \(\ce{NH3}\) N
B Nitric acid, \(\ce{HNO3}\) O=[N+]([O-])O
C Nitrobenzene, \(\ce{C6H5NO2}\) O=[N+]([O-])c1ccccc1
D Aniline, \(\ce{C6H5NH2}\) Nc1ccccc1

Aniline is sulfonated to sulfanilic acid, which is diazotized and coupled with N,N-dimethylaniline to give methyl orange. All balanced equations appear in Solution 1.

  1. \(\boxed{m(\mathbf D)\approx31\ \mathrm{g}}\), assuming quantitative conversion and an additional hydrogen supply for the reduction of C.

  2. Methyl orange is sodium 4-{[4-(dimethylamino)phenyl]diazenyl}benzenesulfonate, also called sodium 4-[(4-dimethylaminophenyl)azo]benzenesulfonate. Its usual \(E\) form has SMILES
    CN(C)c1ccc(/N=N/c2ccc(S(=O)(=O)[O-])cc2)cc1.[Na+].

Its azo anion is yellow; increasing the hydronium concentration produces the red protonated form, whose resonance contributors are shown in Solution 3.

Solution

1. Identification and balanced reaction equations

Working backwards, sulfanilic acid is an aniline derivative with a sulfonic acid group para to the amino group. Therefore D is aniline. Hydrogenation of nitrobenzene gives aniline, identifying C. Nitrobenzene can be prepared by nitrating benzene with nitric acid, identifying B. Finally, ammonia is synthesized from hydrogen and nitrogen and can be oxidized through nitrogen oxides to nitric acid, identifying A.

Electrolysis of water. Water (O) produces hydrogen ([H][H]) and oxygen (O=O). Using the conventional aqueous proton notation, \(\ce{H+}\) ([H+]), the electrode reactions are

\[ \begin{aligned} \text{Cathode:}\qquad &\ce{2H+ + 2e- -> H2},\\ \text{Anode:}\qquad &\ce{2H2O -> O2 + 4H+ + 4e-}. \end{aligned} \]

Thus,

\[ \boxed{\ce{2H2O -> 2H2 + O2}}. \]

The proton notation suppresses its hydration in water.

Synthesis of ammonia, A. Hydrogen reacts with nitrogen (N#N), using an iron catalyst at elevated temperature and pressure:

\[ \boxed{\ce{N2 + 3H2 <=> 2NH3}}. \]

Conversion of A into nitric acid, B. Catalytic oxidation first gives nitrogen monoxide ([N]=O), which is oxidized further to nitrogen dioxide (O=[N+][O-]):

\[ \ce{4NH3 + 5O2 ->[Pt] 4NO + 6H2O}, \]

\[ \ce{2NO + O2 -> 2NO2}. \]

Absorption in water gives nitric acid and regenerates nitrogen monoxide:

\[ \ce{3NO2 + H2O -> 2HNO3 + NO}. \]

The regenerated nitrogen monoxide is reoxidized. Consequently, absorption with sufficient oxygen has the overall equation

\[ \ce{4NO2 + O2 + 2H2O -> 4HNO3}. \]

The net conversion of ammonia into nitric acid is therefore

\[ \boxed{\ce{NH3 + 2O2 -> HNO3 + H2O}}. \]

Preparation of nitrobenzene, C. Benzene (c1ccccc1) is nitrated with nitric acid in the presence of concentrated sulfuric acid (OS(=O)(=O)O):

\[ \boxed{\ce{C6H6 + HNO3 ->[H2SO4] C6H5NO2 + H2O}}. \]

Reduction to aniline, D. Catalytic hydrogenation, for example over nickel, gives

\[ \boxed{\ce{C6H5NO2 + 3H2 ->[Ni] C6H5NH2 + 2H2O}}. \]

Sulfonation of aniline. Heating with concentrated sulfuric acid at \(180\,^\circ\mathrm C\) produces sulfanilic acid, or 4-aminobenzenesulfonic acid: Nc1ccc(S(=O)(=O)O)cc1.

\[ \boxed{\ce{C6H5NH2 + H2SO4 -> H2N-C6H4-SO3H + H2O}}. \]

The amino and sulfonic acid groups occupy positions 1 and 4. Sulfanilic acid commonly exists as the zwitterion [NH3+]c1ccc(S(=O)(=O)[O-])cc1; the equation above uses its neutral structural representation.

Diazotization. Nitrous acid (O=NO) converts the amino group into a diazonium group in acidic solution at approximately \(0\text{–}5\,^\circ\mathrm C\). The product is 4-diazoniobenzenesulfonate, an inner salt: N#[N+]c1ccc(S(=O)(=O)[O-])cc1.

Its balanced net formation is

\[ \boxed{ \ce{H2N-C6H4-SO3H + HNO2 -> ^{-}O3S-C6H4-N2^{+} + 2H2O} }. \]

The diazonium and sulfonate groups remain para to each other.

Azo coupling and formation of methyl orange. N,N-Dimethylaniline (CN(C)c1ccccc1) couples at the carbon para to its dimethylamino group. Coupling followed by neutralization with sodium hydroxide ([Na+].[OH-]) has the combined equation

\[ \boxed{ \begin{aligned} &\ce{^{-}O3S-C6H4-N2^{+} + C6H5-N(CH3)2 + NaOH}\\ &\qquad\ce{-> NaO3S-C6H4-N=N-C6H4-N(CH3)2 + H2O}. \end{aligned} } \]

Both benzene rings in the product are para-disubstituted. The product is the sodium salt methyl orange identified in Answer 3.

2. Mass of aniline

Treating the stated resistances as series resistances, the total electromotive force and resistance are

\[ \mathcal E=10(1.5\ \mathrm V)=15\ \mathrm V, \]

\[ R=10(0.4\ \Omega)+0.5\ \Omega=4.5\ \Omega. \]

The polarization voltage opposes the applied electromotive force. Hence

\[ I=\frac{\mathcal E-E_{\mathrm{pol}}}{R} =\frac{15-1.5}{4.5} =3.0\ \mathrm A. \]

The electrolysis time is

\[ t=8(3600)+56(60)+7=32167\ \mathrm s. \]

Therefore the charge passed is

\[ Q=It=(3.0\ \mathrm A)(32167\ \mathrm s) =96501\ \mathrm C. \]

Using the standard Faraday constant \(F\approx96500\ \mathrm{C\,mol^{-1}}\), and assuming complete current efficiency, two moles of electrons produce one mole of hydrogen:

\[ n(\ce{H2})=\frac{Q}{2F} =\frac{96501}{2(96500)} \approx0.500\ \mathrm{mol}. \]

Ammonia synthesis gives

\[ n(\ce{NH3})=\frac23n(\ce{H2}) \approx\frac13\ \mathrm{mol}. \]

The subsequent stoichiometric relationships are

\[ n(\ce{NH3}) =n(\ce{HNO3}) =n(\ce{C6H5NO2}) =n(\ce{C6H5NH2}), \]

assuming quantitative conversion, adequate other reagents, and additional hydrogen for the nitrobenzene reduction.

Aniline has molecular formula \(\ce{C6H7N}\), so

\[ M(\ce{C6H7N}) =6(12)+7(1)+14 =93\ \mathrm{g\,mol^{-1}}. \]

Thus,

\[ \boxed{ m(\mathbf D) =\left(\frac13\ \mathrm{mol}\right) \left(93\ \mathrm{g\,mol^{-1}}\right) \approx31\ \mathrm g }. \]

3. Methyl orange: name, structures and indicator equilibrium

The exact chemical name is sodium 4-{[4-(dimethylamino)phenyl]diazenyl}benzenesulfonate, with the usual \(E\) azo configuration:

CN(C)c1ccc(/N=N/c2ccc(S(=O)(=O)[O-])cc2)cc1.[Na+].

The sodium ion, \(\ce{Na+}\) ([Na+]), is a spectator. Omitting it, the yellow methyl-orange anion is

\[ \boxed{\ce{^{-}O3S-C6H4-N=N-C6H4-N(CH3)2}} \]

with SMILES CN(C)c1ccc(/N=N/c2ccc(S(=O)(=O)[O-])cc2)cc1.

Hydronium, \(\ce{H3O+}\) ([OH3+]), protonates the azo nitrogen attached to the sulfonated ring:

\[ \begin{aligned} &\underbrace{\ce{^{-}O3S-C6H4-N=N-C6H4-N(CH3)2}}_{\text{yellow}} +\ce{H3O+}\\ &\qquad\rightleftharpoons \underbrace{\ce{^{-}O3S-C6H4-NH^{+}=N-C6H4-N(CH3)2}}_{\text{red protonated form}} +\ce{H2O}. \end{aligned} \]

This protonated methyl-orange zwitterion has the azo-form SMILES

CN(C)c1ccc(/N=[NH+]/c2ccc(S(=O)(=O)[O-])cc2)cc1.

Its positive charge is delocalized. An important alternative resonance contributor is the para-quinoid form:

\[ \ce{^{-}O3S-C6H4-NH^{+}=N-C6H4-N(CH3)2} \;\longleftrightarrow\; \ce{^{-}O3S-C6H4-NH-N=C6H4=N^{+}(CH3)2}. \]

The para-quinoid resonance contributor of protonated methyl orange has SMILES

C[N+](C)=C1C=CC(=NNc2ccc(S(=O)(=O)[O-])cc2)C=C1.

To draw this contributor precisely, keep the sulfonated ring aromatic. In the other ring, place the azo-derived nitrogen at carbon 1 and the dimethylamino nitrogen at carbon 4. Draw exocyclic double bonds from these two carbons to their respective nitrogens, and internal ring double bonds between carbons 2–3 and 5–6. The dimethylamino nitrogen carries the positive charge; the proton remains on the nitrogen attached to the sulfonated ring.

These are resonance contributors of the same protonated species. SMILES records each contributor separately; it does not itself show the delocalization of electrons between them.

Consequently:

  • Increasing \([\ce{H3O+}]\) favors the protonated form and gives a red solution.
  • Decreasing \([\ce{H3O+}]\) favors the azo anion and gives a yellow solution.
  • Intermediate proportions produce the orange transition color.

The solution was checked against the official solution.