IChO 2002 · Theory problem 3

Inulin, a New Renewable Raw Material

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Theme II - Chemistry of Industrial Relevance

In our daily life we use many products that are produced on an industrial scale. Mastering the underlying chemistry is at the heart of this business.

PROBLEM 3

Inulin, a New Renewable Raw Material

Inulin, which is produced from chicory roots in Belgium and The Netherlands, is used as a food additive as it has a beneficial effect on the intestinal flora. It is also used as source of fructose which is 1.9 times sweeter than sucrose, and for the production of mannitol which is used in chewing gum. Inulin is a linear polymer of fructose units with a glucose unit at one end; its Haworth projection formula is shown at the left. In this problem inulin has 10 fructose units (\(n = 9\)).

Figure 1

3.1 Inulin may be hydrolyzed under \(\ce{H+}\)-catalysis conditions. Of the four options below (A, B, C and D) indicate which C-O bond cleavage is most likely to occur.

Figure 2

Mark the correct cleavage mechanism for the most efficient hydrolysis.

Hydrolysis with isotopically labelled water can provide information about the mechanism of hydrolysis using modern NMR techniques, which can “see” deuterium (\(\ce{^{2}H}\)) and the oxygen isotope \(\ce{^{17}O}\).

3.2 Indicate which labelled water can best be used for this purpose. Mark the correct answer.

  • ☐ \(\ce{^{2}H2O}\)
  • ☐ \(\ce{H2^{17}O}\)
  • ☐ \(\ce{^{2}H2^{17}O}\)
  • ☐ None of them.

Upon catalytic hydrogenation glucose gives sorbitol (S), whilst fructose (F) gives mannitol (M) and sorbitol (S).

3.3 Draw the Fischer projections of fructose (F), sorbitol (S) and mannitol (M).

Figure 3

1.00 mol of inulin in 2.00 kg of water with added catalysts, is subjected to hydrolysis and hydrogenation at \(95\,^\circ\mathrm{C}\) in a one step process. The selectivity of the hydrogenation of fructose to mannitol / sorbitol is 7 / 3.

3.4 How many moles of mannitol and sorbitol are obtained?

Figure 4

After completion of the reactions the catalysts are removed and the reaction mixture is cooled to \(25\,^\circ\mathrm{C}\). The solubility of M is \(0.40\ \mathrm{mol\,kg^{-1}}\) in water at \(25\,^\circ\mathrm{C}\) and the solubility of S is so high that it will not precipitate.

3.5 Calculate how many moles of M will precipitate.

\(\underline{\hspace{4cm}}\)

Official English text of the problem.

Answers

  • 3.1: B — cleavage of the protonated glycosidic bond at the anomeric carbon, C2, of fructose.
  • 3.2: \(\ce{H2^{17}O}\), oxygen-17-labelled water, [17OH2].
  • 3.3: The required structures are:
  • F: D-fructose, OCC(=O)[C@@H](O)[C@H](O)[C@H](O)CO.
  • S: Sorbitol (D-glucitol), OC[C@H](O)[C@@H](O)[C@H](O)[C@H](O)CO.
  • M: D-mannitol, OC[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)CO.

Their Fischer projections are specified in part 3.3 below. - 3.4: \(\boxed{n(\mathrm M)=7.00\ \mathrm{mol},\qquad n(\mathrm S)=4.00\ \mathrm{mol}}\). - 3.5: \(\boxed{6.27\ \mathrm{mol}}\) of mannitol precipitates.

Solution

3.1 — Acid-catalysed bond cleavage

The fructose residues are connected through glycosidic bonds between anomeric C2 of one residue and the oxygen attached to C1 of another.

In B, the bridging oxygen is protonated, and the bond between this oxygen and anomeric C2 breaks. The curved arrow indicates that the bonding electron pair moves onto the protonated oxygen. This produces an alcohol on the departing fragment and an oxocarbenium centre on the fructose residue. The positive charge is stabilised by electron donation from the adjacent ring oxygen.

Water, O, then attacks the anomeric carbon. Subsequent deprotonation produces the anomeric hydroxyl group and regenerates the acid catalyst, represented by \(\ce{H+}\), [H+].

The alternatives explain the preference:

  • A would generate an unstabilised primary carbocation at C1.
  • C opens the fructofuranose ring without directly breaking the connection between the two residues.
  • D breaks the ring at a non-anomeric carbon, without the favourable oxocarbenium stabilisation obtained in B.

Thus B directly cleaves the polymer chain through the favourable anomeric pathway.

3.2 — Choice of isotopically labelled water

Choose oxygen-17-labelled water, \(\ce{H2^{17}O}\), [17OH2].

In mechanism B, the oxygen of the attacking water becomes the oxygen of the newly formed hydroxyl group at fructose C2. The original bridging oxygen remains attached to C1 of the other fragment. Consequently, the location of the oxygen isotope identifies where water enters during hydrolysis.

Deuterated water, \(\ce{^{2}H2O}\), [2H]O[2H], is less informative because hydroxyl hydrogens readily exchange with the solvent. Deuterium would therefore appear at many hydroxyl groups, independently of glycosidic bond cleavage.

Doubly labelled water, \(\ce{^{2}H2^{17}O}\), [2H][17O][2H], would provide the useful oxygen label, but its deuterium labels add no necessary positional information. The best choice is therefore the oxygen-only label.

3.3 — Fischer projections and hydrogenation

Use a vertical six-carbon chain, numbered C1 at the top to C6 at the bottom, with \(\ce{CH2OH}\) at both ends. The required arrangements are:

Carbon D-fructose, F Sorbitol, S Mannitol, M
C2 Carbonyl, \(\ce{C=O}\) OH on the right OH on the left
C3 OH on the left OH on the left OH on the left
C4 OH on the right OH on the right OH on the right
C5 OH on the right OH on the right OH on the right

At every stereogenic carbon, hydrogen occupies the horizontal position opposite OH. In the fructose projection, draw the carbonyl oxygen to the right of C2 with a double bond. As usual for Fischer projections, horizontal bonds point towards the viewer and vertical bonds point away. The stereochemical SMILES are given in the Answers section.

Hydrogenation of fructose reduces its C2 ketone and creates a new stereogenic centre. The two possible C2 configurations give sorbitol and mannitol, while the configurations at C3, C4 and C5 remain unchanged.

For comparison, D-glucose has open-chain SMILES O=C[C@H](O)[C@@H](O)[C@H](O)[C@H](O)CO. Reducing its C1 aldehyde to \(\ce{CH2OH}\) gives sorbitol without creating a new stereogenic centre.

Molecular hydrogen is \(\ce{H2}\), [H][H]. Each individual hydrogenation—glucose to sorbitol, fructose to sorbitol, or fructose to mannitol—has the balanced stoichiometry

\[ \ce{C6H12O6 + H2 ->[catalyst] C6H14O6}. \]

One mole of fructose therefore yields one mole of total hexitol products.

3.4 — Amounts of mannitol and sorbitol

Each inulin molecule contains one glucose residue and ten fructose residues: 11 monosaccharide residues joined by 10 glycosidic bonds.

The molecular formula follows from condensation of the eleven monosaccharides:

\[ 11(\ce{C6H12O6})-10(\ce{H2O}) =\ce{C66H112O56}. \]

The balanced hydrolysis is

\[ \ce{C66H112O56 + 10H2O ->[H+] C6H12O6 + 10C6H12O6}, \]

where the first product term is glucose and the second is fructose. Thus, \(1.00\ \mathrm{mol}\) of inulin produces

\[ n(\text{glucose})=1.00\ \mathrm{mol}, \qquad n(\text{fructose})=10.0\ \mathrm{mol}. \]

The fructose hydrogenation selectivity is

\[ \mathrm{M:S}=7:3. \]

Hence

\[ n(\mathrm M) =10.0\ \mathrm{mol}\times\frac{7}{10} =\boxed{7.00\ \mathrm{mol}}, \]

and the fructose-derived sorbitol is

\[ n(\mathrm S;\text{from fructose}) =10.0\ \mathrm{mol}\times\frac{3}{10} =3.00\ \mathrm{mol}. \]

The glucose supplies another \(1.00\ \mathrm{mol}\) of sorbitol:

\[ n(\mathrm S)=3.00+1.00 =\boxed{4.00\ \mathrm{mol}}. \]

The combined hydrolysis and hydrogenation balance is

\[ \ce{C66H112O56 + 10H2O + 11H2 ->[catalysts] 11C6H14O6}, \]

with the eleven product moles distributed as seven moles of mannitol and four moles of sorbitol.

3.5 — Precipitation of mannitol

Hydrolysis consumes ten moles of water per mole of inulin:

\[ n(\text{water consumed}) =10(1.00\ \mathrm{mol}) =10.0\ \mathrm{mol}. \]

Using \(M(\ce{H2O})=18.0\ \mathrm{g\,mol^{-1}}\), the remaining water mass is

\[ \begin{aligned} m(\text{water}) &=2.00\ \mathrm{kg} -(10.0\ \mathrm{mol})(18.0\ \mathrm{g\,mol^{-1}}) \left(10^{-3}\ \mathrm{kg\,g^{-1}}\right)\\ &=1.82\ \mathrm{kg}. \end{aligned} \]

Hydrogenation does not change this water balance.

At \(25\,^\circ\mathrm{C}\), the amount of mannitol that remains dissolved is

\[ n(\mathrm M;\text{dissolved}) =(0.40\ \mathrm{mol\,kg^{-1}})(1.82\ \mathrm{kg}) =0.728\ \mathrm{mol}. \]

Retaining these digits until the final subtraction,

\[ \begin{aligned} n(\mathrm M;\text{precipitated}) &=7.00-0.728\\ &=6.272\ \mathrm{mol}\\ &\approx\boxed{6.27\ \mathrm{mol}}. \end{aligned} \]

The solution was checked against the official solution.