IChO 1973 · Theory problem 1

Nitration product of a hydroxybenzene and its electroreduction

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In nitrating a hydroxy derivative of benzene a compound is formed which contains 49.0 % by mass of oxygen. A charge of 4350 C is required for a total electroreduction of 0.458 g of the compound, efficiency being 80 %.

Problem:

1.1 Determine the stoichiometric as well as structural formulas of the compound if the product of the electrochemical reduction is an aromatic hydroxy amino derivative.

F (Faraday's charge) = \(96\,500\ \mathrm{C\,mol^{-1}}\)

Official English text of the problem.

Answers

  • 1.1 The stoichiometric formula is \(\boxed{\ce{C6H3N3O7}}\). The expected nitration product is 2,4,6-trinitrophenol (picric acid), Oc1c([N+](=O)[O-])cc([N+](=O)[O-])cc1[N+](=O)[O-]. Its structural formula is \(\ce{HO-C6H2(NO2)3}\), with the hydroxyl group at position 1 and nitro groups at positions 2, 4 and 6.

Solution

1.1

Let the nitration product contain \(h\) hydroxyl groups and \(n\) nitro groups on a benzene ring. It therefore contains six carbon atoms, \(6-n\) hydrogen atoms, \(n\) nitrogen atoms and \(h+2n\) oxygen atoms. Using the usual rounded atomic masses, its molar mass is

\[ M=\left[6(12)+(6-n)+14n+16(h+2n)\right]\mathrm{g\,mol^{-1}} =(78+16h+45n)\mathrm{g\,mol^{-1}}. \]

Reduction of a nitro group to an amino group changes the nitrogen oxidation state from \(+3\) to \(-3\), requiring six electrons per nitro group.

Only 80 % of the supplied charge effects the desired reduction:

\[ Q_{\mathrm{effective}}=0.80(4350\ \mathrm C)=3480\ \mathrm C. \]

Thus the amount of electrons used is

\[ n(\ce{e-}) =\frac{3480\ \mathrm C}{96\,500\ \mathrm{C\,mol^{-1}}} =0.0360622\ \mathrm{mol}. \]

Each mole of compound requires \(6n\) moles of electrons, so

\[ \frac{0.458\ \mathrm g}{M}(6n)=0.0360622\ \mathrm{mol}, \]

giving

\[ \boxed{\frac{M}{n}=76.20\ \mathrm{g\,mol^{-1}}}. \]

The oxygen mass fraction supplies a second relation:

\[ 0.490=\frac{(16.0\ \mathrm{g\,mol^{-1}})(h+2n)}{M}. \]

Consequently,

\[ \frac{h}{n} =\frac{0.490(76.2017)}{16.0}-2 =0.33368. \]

Combining this result with the molar-mass expression, with masses expressed numerically in \(\mathrm{g\,mol^{-1}}\),

\[ 76.2017=\frac{78}{n}+16(0.33368)+45, \]

which yields

\[ n\simeq3.016,\qquad h\simeq1.006. \]

The required integer group counts are therefore

\[ \boxed{n=3,\qquad h=1}. \]

The compound is a trinitrophenol, with molecular—and also empirical—formula

\[ \boxed{\ce{C6H3N3O7}},\qquad M=229\ \mathrm{g\,mol^{-1}}. \]

The corresponding checks are

\[ w_{\mathrm O}=\frac{7(16)}{229}\times100\%=48.91\%, \]

and

\[ Q=\frac{(0.458/229)\,\mathrm{mol}\times18\times 96\,500\ \mathrm{C\,mol^{-1}}}{0.80} =4342.5\ \mathrm C, \]

both close to the supplied values.

The starting hydroxybenzene is phenol, Oc1ccccc1. Its hydroxyl group directs nitration to the two ortho positions and the para position. Introducing three nitro groups therefore gives the expected product 2,4,6-trinitrophenol, identified above. To draw its structure, place hydroxyl at carbon 1 of a benzene ring, nitro groups at carbons 2, 4 and 6, and hydrogen atoms at carbons 3 and 5.

The balanced nitration equation is

\[ \ce{C6H5OH + 3HNO3 -> C6H2(NO2)3OH + 3H2O}. \]

Here the reagent is nitric acid, O=[N+]([O-])O, and the coproduct is water, O.

Complete electroreduction gives 2,4,6-triaminophenol, Oc1c(N)cc(N)cc1N, with formula \(\ce{C6H9N3O}\). The balanced reduction half-equation is

\[ \boxed{\ce{C6H2(NO2)3OH + 18H+ + 18e- -> C6H2(NH2)3OH + 6H2O}}. \]

The hydrogen ions in this acidic-medium half-equation are protons, [H+].

The solution was checked against the official solution.