IChO 1985 · Theory problem 1

Determination of aluminium with 8-hydroxyquinoline and bromate

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A solution was formed from 0.5284 g of a sample of an alloy containing aluminium. The aluminium was then precipitated as aluminium 8-hydroxyquinolate. The precipitate was separated, dissolved in hydrochloric acid and the 8-hydroxyquinoline formed was titrated with a standard solution of potassium bromate containing potassium bromide. The concentration of the standard potassium bromate solution was 0.0200 M and 17.40 cm³ of it were required. The resultant product is a dibromo derivative of 8-hydroxyquinoline.

The structural formula of 8-hydroxiquinoline is:

Figure 1

The relative atomic mass of aluminium is 26.98.

Problems:

1.1 Write the balanced equation for the reaction of the aluminium (III) ion with 8-hydroxyquinoline, showing clearly the structure of the products.

1.2 Give the name of the type of compound which is formed during the precipitation.

1.3 Write the balanced equation for the reaction in which bromine is produced.

1.4 Write the balanced equation for the reaction of bromine with 8-hydroxyquinoline.

1.5 Calculate the molar ratio of aluminium ions to bromate ions.

1.6 Calculate the percentage by weight of aluminium in the alloy.

Official English text of the problem.

Answers

  • 1.1 \[ \ce{Al^3+ + 3 C9H7NO -> Al(C9H6NO)3(s) + 3 H+} \] The precipitate is aluminium tris(quinolin-8-olate), with three bidentate ligands coordinated through oxygen and nitrogen; its structure and SMILES are specified below.

  • 1.2 A chelate complex, specifically a neutral complex with three five-membered chelate rings.

  • 1.3 \[ \ce{BrO3^- + 5 Br^- + 6 H+ -> 3 Br2 + 3 H2O} \]

  • 1.4 \[ \ce{C9H7NO + 2 Br2 -> C9H5Br2NO + 2 HBr} \] The organic product is 5,7-dibromoquinolin-8-ol, SMILES Oc1c(Br)cc(Br)c2cccnc12.

  • 1.5 \[ \boxed{n(\ce{Al^3+}):n(\ce{BrO3^-})=1:2} \]

  • 1.6 \[ \boxed{\text{Aluminium content}=0.888\%\text{ by mass}} \]

Solution

1.1 — Precipitation and structure

8-Hydroxyquinoline, also called quinolin-8-ol or oxine, has formula \(\ce{C9H7NO}\) and SMILES Oc1cccc2cccnc12. Loss of its hydroxyl proton gives the quinolin-8-olate ion, \(\ce{C9H6NO^-}\), SMILES [O-]c1cccc2cccnc12.

Each ligand binds aluminium through two donor atoms:

  • the negatively charged phenoxide oxygen;
  • the quinoline nitrogen, using its lone pair.

Three singly charged ligands neutralize one aluminium(III) ion, \(\ce{Al^3+}\), SMILES [Al+3]. Therefore, \[ \boxed{\ce{Al^3+(aq) + 3 C9H7NO -> Al(C9H6NO)3(s) + 3 H+(aq)}} \] where the hydrogen ion, or proton, has SMILES [H+].

The product is aluminium tris(quinolin-8-olate), also called aluminium 8-hydroxyquinolate. An extended SMILES explicitly representing all six donor-to-metal bonds is:

[Al+3]123(<-[O-]c4cccc5cccn->1c45)(<-[O-]c4cccc5cccn->2c45)<-[O-]c4cccc5cccn->3c45

Here the arrows denote coordinate bonds directed from the donor atom towards aluminium.

To draw the product, place aluminium at the centre and surround it with three quinolin-8-olate ligands. Join both the oxygen and nitrogen of each ligand to the same aluminium centre. Each ligand forms a five-membered ring following the sequence aluminium–oxygen–carbon-8–carbon-8a–nitrogen. The six donor atoms give a distorted octahedral coordination geometry. No particular geometric or optical isomer is specified.

1.2 — Type of compound

The product is a chelate complex. A chelating ligand binds the same metal ion through at least two donor atoms, forming a ring that includes the metal. Here each quinolin-8-olate ligand is bidentate, and three such ligands produce three chelate rings.

1.3 — Production of bromine

In acidic solution, bromate oxidizes bromide to bromine. The species are bromate, SMILES [O-][Br](=O)=O; bromide, [Br-]; bromine, BrBr; and water, O.

Bromine in bromate has oxidation state \(+5\), so its reduction to oxidation state zero requires five electrons. Five bromide ions, each initially at oxidation state \(-1\), supply these electrons. Balancing oxygen with water and hydrogen with protons gives \[ \boxed{\ce{BrO3^- + 5 Br^- + 6 H+ -> 3 Br2 + 3 H2O}}. \]

Thus, one mole of bromate produces three moles of bromine.

1.4 — Bromination of 8-hydroxyquinoline

The hydroxyl group activates the benzene portion of the quinoline system towards electrophilic substitution. Bromination occurs at positions 7 and 5, respectively ortho and para to the hydroxyl group.

The product is 5,7-dibromo-8-hydroxyquinoline, or 5,7-dibromoquinolin-8-ol, SMILES Oc1c(Br)cc(Br)c2cccnc12. Its structure retains the original quinoline skeleton and hydroxyl group at position 8, with bromine atoms replacing the hydrogens at positions 5 and 7.

Each substitution consumes one bromine molecule and produces one hydrogen bromide molecule, SMILES [H]Br. Consequently, \[ \boxed{\ce{C9H7NO + 2 Br2 -> C9H5Br2NO + 2 HBr}}. \]

One mole of 8-hydroxyquinoline therefore consumes two moles of bromine.

1.5 — Aluminium-to-bromate molar ratio

Acid dissolution of the precipitate releases three molecules of 8-hydroxyquinoline per aluminium ion. Representing the liberated ligand in its neutral form, \[ \ce{Al(C9H6NO)3(s) + 3 H+ -> Al^3+ + 3 C9H7NO}. \]

Each of these three ligand molecules consumes two bromine molecules. Hence, \[ n(\ce{Br2})=2\,n(\ce{C9H7NO})=6\,n(\ce{Al^3+}). \]

From part 1.3, \[ n(\ce{Br2})=3\,n(\ce{BrO3^-}). \]

Equating these expressions gives \[ 6\,n(\ce{Al^3+})=3\,n(\ce{BrO3^-}), \] so \[ \boxed{n(\ce{Al^3+}):n(\ce{BrO3^-})=1:2}. \]

1.6 — Aluminium mass percentage

The volume of bromate solution is \[ V=17.40\ \mathrm{cm^3}=0.01740\ \mathrm{L}. \]

The amount of bromate used is \[ n(\ce{BrO3^-}) =cV =(0.0200\ \mathrm{mol\,L^{-1}})(0.01740\ \mathrm{L}) =3.48\times10^{-4}\ \mathrm{mol}. \]

Using the \(1:2\) aluminium-to-bromate ratio, \[ n(\ce{Al}) =\frac{3.48\times10^{-4}}{2} =1.74\times10^{-4}\ \mathrm{mol}. \]

The given relative atomic mass corresponds to a molar mass of \(26.98\ \mathrm{g\,mol^{-1}}\). Retaining guard digits, \[ m(\ce{Al}) =(1.74\times10^{-4}\ \mathrm{mol})(26.98\ \mathrm{g\,mol^{-1}}) =0.00469452\ \mathrm{g}. \]

Therefore, \[ \text{Aluminium mass percentage} =\frac{0.00469452\ \mathrm{g}}{0.5284\ \mathrm{g}}\times100\% =0.8884406\%. \]

The bromate concentration has three significant figures, so the final result is \[ \boxed{0.888\%\text{ aluminium by mass}}. \]

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.