IChO 1995 · Theory problem 3

Glyceraldehyde and its cyclic dimer

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1,3-Dihydroxyacetone can be converted to glyceraldehyde. On standing this glyceraldehyde changes spontaneously into a six member cyclic dimer \(\ce{C6H12O6}\). The infrared spectrum of the dimer shows no absorption peak between 1600 – 1800 \(\mathrm{cm^{-1}}\) and the dipole moment of the dimer is determined to be zero.

3.1 Write the Fischer projection structural formula(e) for the resulting glyceraldehyde and indicate configuration using D(+) and/or L(-).

3.2 Write the structural formula for the reaction intermediate of the conversion of 1,3-dihydroxyacetone to glyceraldehyde.

3.3 Write the structural formula for the dimer.

3.4 Using Haworth projection formula represent the possible stereoisomers which fit the dipole moment data.

3.5 Denote each chiral carbon atom in the above formulae with \(R\) or \(S\).

Official English text of the problem.

Answers

  • 3.1: A racemic mixture of D(+)-glyceraldehyde, (2R)-2,3-dihydroxypropanal, O=C[C@H](O)CO, and L(-)-glyceraldehyde, (2S)-2,3-dihydroxypropanal, O=C[C@@H](O)CO. In their Fischer projections, place \(\ce{CHO}\) above and \(\ce{CH2OH}\) below; \(\ce{OH}\) is on the right for D(+) and on the left for L(-).

  • 3.2: The 1,2-enediol, prop-1-ene-1,2,3-triol, \(\ce{HO-CH=C(OH)-CH2OH}\), OC=C(O)CO.

  • 3.3: The cyclic bis-hemiacetal 3,6-bis(hydroxymethyl)-1,4-dioxane-2,5-diol, O1C(O)C(CO)OC(O)C1CO. Its ring sequence is O1–C2–C3–O4–C5–C6; C2 and C5 bear \(\ce{OH}\), while C3 and C6 bear \(\ce{CH2OH}\).

  • 3.4: Two centrosymmetric, meso stereoisomers satisfy the dipole-moment condition:

I: (2R,3R,5S,6S)-3,6-bis(hydroxymethyl)-1,4-dioxane-2,5-diol, O1[C@@H](O)[C@@H](CO)O[C@H](O)[C@@H]1CO. Each adjacent hydroxyl/hydroxymethyl pair is cis.

II: (2S,3R,5R,6S)-3,6-bis(hydroxymethyl)-1,4-dioxane-2,5-diol, O1[C@H](O)[C@@H](CO)O[C@@H](O)[C@@H]1CO. Each adjacent hydroxyl/hydroxymethyl pair is trans.

  • 3.5: D(+)-glyceraldehyde is \(R\); L(-)-glyceraldehyde is \(S\). For ring positions C2, C3, C5, C6, respectively, dimer I is \(R,R,S,S\), and dimer II is \(S,R,R,S\). The enediol has no chiral carbon atoms.

Solution

3.1 — Glyceraldehyde formation and Fischer projections

The starting compound, 1,3-dihydroxyacetone (1,3-dihydroxypropan-2-one), OCC(=O)CO, is achiral. Conversion of its ketone group into a terminal aldehyde creates a stereogenic centre at C2. With no chiral reagent or catalyst specified, the two enantiomers form in equal amounts.

Draw the Fischer projections with the carbon chain vertical:

Position D(+)-glyceraldehyde L(-)-glyceraldehyde
Top \(\ce{CHO}\) \(\ce{CHO}\)
Left of central carbon \(\ce{H}\) \(\ce{OH}\)
Right of central carbon \(\ce{OH}\) \(\ce{H}\)
Bottom \(\ce{CH2OH}\) \(\ce{CH2OH}\)

As usual for Fischer projections, horizontal bonds point toward the viewer and vertical bonds point away. The mixture is racemic and therefore has zero net optical rotation.

3.2 — The reaction intermediate

The ketose–aldose conversion proceeds through the enediol:

\[ \ce{HOCH2-C(=O)-CH2OH <=> HO-CH=C(OH)-CH2OH <=> O=CH-CH(OH)-CH2OH}. \]

Each species has molecular formula \(\ce{C3H6O3}\), so both steps are balanced tautomerizations.

In the first step, proton transfers and redistribution of the carbonyl \(\pi\) electrons generate the C1–C2 double bond and the second hydroxyl group. In the second step, the terminal hydroxyl oxygen forms the aldehyde carbonyl while C2 receives a proton.

C2 is planar in the enediol. Protonation from its two faces produces the two glyceraldehyde configurations equally. The intermediate’s E/Z geometry need not be specified for this conclusion.

3.3 — Connectivity of the cyclic dimer

The missing carbonyl absorption indicates that the aldehyde groups have been consumed. Each glyceraldehyde molecule supplies:

  • Its aldehyde carbon as a hemiacetal carbon.
  • Its C2 hydroxyl oxygen as an ether oxygen in the ring.

The C2 hydroxyl oxygen of one molecule attacks the aldehyde carbon of the other; the carbonyl \(\pi\) electrons move onto oxygen, and proton transfer gives a hemiacetal. The corresponding reaction between the remaining hydroxyl and aldehyde groups closes the ring.

Each molecule contributes two carbons and one oxygen to the ring, giving six ring atoms. The connectivity, with the final carbon bonded back to the initial oxygen, is

\[ \text{cyclic }\left[ \ce{O-CH(OH)-CH(CH2OH)-O-CH(OH)-CH(CH2OH)} \right]. \]

Thus the product is the cyclic bis-hemiacetal identified in Answer 3.3. Its two former carbonyl oxygens are now hydroxyl oxygens.

The balanced dimerization is

\[ \ce{2 O=CH-CH(OH)-CH2OH <=> C6H12O6}. \]

No water is eliminated. Formation of the two hemiacetal linkages preserves all atoms of the two monomers.

3.4 — Haworth projections and the zero dipole moment

Use the following precise Haworth convention:

  • Viewed from above the ring, the atoms run clockwise as O1–C2–C3–O4–C5–C6.
  • Draw O1 at the upper right, C2 at the right, C3 at the lower right, O4 at the lower left, C5 at the left, and C6 at the upper left.
  • “Up” and “down” mean above and below the ring plane. At each substituted carbon, its hydrogen points to the opposite face.

An inversion centre at the ring centre exchanges O1 with O4, C2 with C5, and C3 with C6. Therefore, corresponding substituents must occupy opposite faces:

\[ \ce{OH}\text{ at C2 opposite to }\ce{OH}\text{ at C5}, \] \[ \ce{CH2OH}\text{ at C3 opposite to }\ce{CH2OH}\text{ at C6}. \]

The two distinct Haworth projections are:

Dimer C2–\(\ce{OH}\) C3–\(\ce{CH2OH}\) C5–\(\ce{OH}\) C6–\(\ce{CH2OH}\)
I Up Up Down Down
II Down Up Up Down

In I, each adjacent hydroxyl/hydroxymethyl pair lies on the same face; in II, each pair lies on opposite faces. Reversing all four up/down entries does not give additional stereoisomers: the resulting configurations are equivalent after exchanging the two identical halves.

These configurations permit centrosymmetric chair conformations, with corresponding side chains also related by inversion. Since inversion reverses a dipole vector while leaving such a molecule unchanged,

\[ \boldsymbol{\mu}=-\boldsymbol{\mu} \quad\Longrightarrow\quad \boldsymbol{\mu}=0. \]

Both structures are meso and contain one D-glyceraldehyde residue and one L-glyceraldehyde residue. Haworth projections and stereochemical SMILES specify their configurations; they do not imply that the actual ring is flat.

3.5 — Absolute configurations

For glyceraldehyde, the priorities at C2 are

\[ \ce{OH}>\ce{CHO}>\ce{CH2OH}>\ce{H}. \]

The aldehyde carbon outranks the hydroxymethyl carbon because the carbonyl double bond is treated as attachment to two oxygens in the CIP comparison. Applying these priorities to the Fischer projections gives D(+): \(R\) and L(-): \(S\).

For the dimer:

  • At C2 and C5: ring oxygen \(>\) hydroxyl oxygen \(>\) neighbouring ring carbon \(>\) hydrogen. The ring oxygen outranks the hydroxyl oxygen because it is attached to carbon rather than hydrogen.
  • At C3 and C6: ring oxygen \(>\) neighbouring hemiacetal carbon \(>\) hydroxymethyl carbon \(>\) hydrogen. The hemiacetal carbon has two oxygen neighbours, whereas the hydroxymethyl carbon has only one.

With the clockwise Haworth convention defined above, an up hydroxyl at C2/C5 gives \(R\), and an up hydroxymethyl group at C3/C6 also gives \(R\). Down gives \(S\).

Structure C2 C3 C5 C6
Dimer I \(R\) \(R\) \(S\) \(S\)
Dimer II \(S\) \(R\) \(R\) \(S\)

The original glyceraldehyde stereocentres become C3 and C6 without changing configuration. The two additional stereocentres, C2 and C5, arise during hemiacetal formation.

The solution was checked against the official solution.