IChO 1977 · Theory problem 3

Isomeric cyclobutanedicarboxylic acids

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The letters A, B, C, D, and E represent isomeric cyclobutane dicarboxylic acid, one of them being a racemic form.

It was shown that:

a) only compound C forms a cyclic anhydride easily,

b) B yields a cyclic anhydride only at higher temperatures,

c) of all the acids under investigation, only A releases carbon dioxide when heated,

d) D and E do not change at higher temperatures,

e) 2 moles of diethyl ester of malonic acid when reacted with sodium ethoxide (\(\ce{EtONa}\)), are transformed to sodium salt which by reaction with methylene iodide (\(\ce{CH2I2}\)) yields a tetraester \(\ce{C15H24O8}\). This new ester gives a tetraester \(\ce{C16H24O8}\) in the reaction with 2 moles of sodium ethoxide and 1 mole of methylene iodide. The last mentioned tetraester is transformed by alkaline hydrolysis and subsequent acidification to a tetracarboxylic acid which when heated, gives a mixture of B and E.

Problems:

3.1 Give formulas corresponding to the letters A, B, C, D, and E. Mark the group which is above the plane of the cycle

Figure 1

, that under the plane with

Figure 2

For example:

Figure 3

3.2 Express processes a), b), and c) by chemical equations.

Official English text of the problem.

Answers

  • 3.1 The assignments are:
Letter Identity SMILES
A Cyclobutane-1,1-dicarboxylic acid O=C(O)C1(C(=O)O)CCC1
B cis-Cyclobutane-1,3-dicarboxylic acid O=C(O)[C@H]1C[C@@H](C(=O)O)C1
C cis-Cyclobutane-1,2-dicarboxylic acid, the meso form O=C(O)[C@H]1[C@@H](C(=O)O)CC1
D trans-Cyclobutane-1,2-dicarboxylic acid, a 1:1 racemate of the \((1R,2R)\) and \((1S,2S)\) enantiomers O=C(O)[C@H]1[C@H](C(=O)O)CC1 and O=C(O)[C@@H]1[C@@H](C(=O)O)CC1
E trans-Cyclobutane-1,3-dicarboxylic acid O=C(O)[C@H]1C[C@H](C(=O)O)C1

For the requested drawings, number an idealized square cyclobutane ring clockwise:

Letter Positions of the two carboxyl groups Wedge/dash representation
A Both at carbon 1 One solid wedge and one hashed wedge
B Carbons 1 and 3 Both solid wedges
C Carbons 1 and 2 Both solid wedges
D Carbons 1 and 2 One solid wedge and one hashed wedge; show its mirror image alongside it, in equal proportion
E Carbons 1 and 3 One solid wedge and one hashed wedge
  • 3.2 The balanced reactions are:

\[ \begin{aligned} \text{a) C:}\quad& \ce{C6H8O4 ->[\text{mild heating}] C6H6O3 + H2O}\\ \text{b) B:}\quad& \ce{C6H8O4 ->[\text{higher temperature}] C6H6O3 + H2O}\\ \text{c) A:}\quad& \ce{C6H8O4 ->[\Delta] C5H8O2 + CO2} \end{aligned} \]

The organic products, respectively, are:

Process Product SMILES
a) Cyclic anhydride of cis-cyclobutane-1,2-dicarboxylic acid O=C1OC(=O)[C@H]2CC[C@@H]12
b) Cyclic anhydride of cis-cyclobutane-1,3-dicarboxylic acid O=C1OC(=O)[C@H]2C[C@@H]1C2
c) Cyclobutanecarboxylic acid O=C(O)C1CCC1

The small-molecule products are water, O, and carbon dioxide, O=C=O.

Solution

3.1 — Identification of the five acids

There are three possible arrangements of two carboxyl groups on a cyclobutane ring: 1,1, 1,2, and 1,3. The 1,2- and 1,3-arrangements each have cis and trans forms. Counting an enantiomeric pair as one racemic form gives the five substances in the question.

Identification of A.
The 1,1-isomer contains two carboxyl groups attached to the same carbon, the characteristic arrangement of a substituted malonic acid. It can lose one carboxyl group as carbon dioxide on heating, replacing that group by hydrogen. Consequently,

\[ \boxed{\text{A is cyclobutane-1,1-dicarboxylic acid.}} \]

The synthesis in observation e) determines which positional isomers B and E are.

The starting reagents are diethyl malonate, CCOC(=O)CC(=O)OCC; sodium ethoxide, [Na+].CC[O-]; and diiodomethane (methylene iodide), C(I)I. Ethanol, CCO, and sodium iodide, [Na+].[I-], are the coproducts of the alkylation steps.

Ethoxide removes an acidic methylene proton from each of two diethyl malonate molecules. Their carbon atoms then replace the two iodides of one diiodomethane molecule, giving the linkage

\[ \ce{(EtO2C)2CH-CH2-CH(CO2Et)2}. \]

Here \(\ce{Et}\) denotes an ethyl group. The product is tetraethyl propane-1,1,3,3-tetracarboxylate, CCOC(=O)C(CC(C(=O)OCC)C(=O)OCC)C(=O)OCC. The balanced net reaction is

\[ \ce{2 C7H12O4 + 2 NaOC2H5 + CH2I2 -> C15H24O8 + 2 C2H5OH + 2 NaI}. \]

Its molecular formula agrees with the stated \(\ce{C15H24O8}\).

Each terminal malonate carbon still bears one hydrogen. Removing these two hydrogens with two equivalents of ethoxide allows a second diiodomethane molecule to link the same two carbons through another methylene group. This closes a four-membered ring:

  • Ring carbons 1 and 3 are the original malonate carbons, each bearing two ester groups.
  • Ring carbons 2 and 4 are the two methylene groups introduced from diiodomethane.

The product is therefore tetraethyl cyclobutane-1,1,3,3-tetracarboxylate, CCOC(=O)C1(C(=O)OCC)CC(C(=O)OCC)(C(=O)OCC)C1:

\[ \ce{C15H24O8 + 2 NaOC2H5 + CH2I2 -> C16H24O8 + 2 C2H5OH + 2 NaI}. \]

Hydrolysis and acidification replace the four ethoxy groups by hydroxyl groups, producing cyclobutane-1,1,3,3-tetracarboxylic acid, O=C(O)C1(C(=O)O)CC(C(=O)O)(C(=O)O)C1.

The overall ester-to-acid conversion is

\[ \ce{C16H24O8 + 4 H2O ->[\text{alkaline hydrolysis}][\text{then acidification}] C8H8O8 + 4 C2H5OH}. \]

Heating removes one carboxyl group from each geminal pair:

\[ \ce{C8H8O8 ->[\Delta] C6H8O4 + 2 CO2}. \]

The remaining carboxyl groups are necessarily at positions 1 and 3. Their relative orientations can be cis or trans; the stated product mixture therefore establishes that B and E are the two cyclobutane-1,3-dicarboxylic acids. No mixture ratio is specified.

Distinguishing B and E.
In the cis-1,3-isomer, both carboxyl groups lie on the same face of the ring and can form an oxygen bridge by dehydration. The constrained bridged anhydride requires higher temperature. The trans-1,3-isomer has the groups on opposite faces and remains unchanged under the stated conditions. Thus,

\[ \boxed{\text{B is cis-1,3; E is trans-1,3.}} \]

Distinguishing C and D.
The remaining substances are the cis- and trans-1,2-isomers. Adjacent carboxyl groups on the same face can readily form a five-membered anhydride ring, so

\[ \boxed{\text{C is cis-1,2; D is trans-1,2.}} \]

The cis-1,2-isomer has opposite configurations at its two stereogenic carbons and is meso. The trans-1,2-isomer has the enantiomeric configurations \((1R,2R)\) and \((1S,2S)\), making D the racemic form.

The square drawings in the Answers section specify relative configurations. Actual cyclobutane rings are puckered; the wedges and hashes indicate the two ring faces, rather than asserting a perfectly planar conformation.

3.2 — Equations and explanation of the thermal reactions

a) Easy anhydride formation by C

\[ \ce{C6H8O4 ->[\text{mild heating}] C6H6O3 + H2O}. \]

The reactant is cis-cyclobutane-1,2-dicarboxylic acid; the product is its cyclic anhydride, whose structure is given above. The two adjacent carboxyl groups lie on the same face. Dehydration joins their carbonyl carbons through an oxygen atom, forming a five-membered anhydride ring fused to the cyclobutane ring.

b) Anhydride formation by B at higher temperature

\[ \ce{C6H8O4 ->[\text{higher temperature}] C6H6O3 + H2O}. \]

The reactant is cis-cyclobutane-1,3-dicarboxylic acid; the product is its cyclic anhydride. Here the bridge spans opposite ring carbons. An anhydride-containing circuit is six-membered, but it is constrained within a bridged bicyclic framework. This less favorable geometry accounts for the higher temperature required.

Although the two anhydrides have the same molecular formula, their connectivities differ, as their SMILES show.

c) Decarboxylation of A

\[ \ce{C6H8O4 ->[\Delta] C5H8O2 + CO2}. \]

Cyclobutane-1,1-dicarboxylic acid undergoes malonic-acid-type decarboxylation to give cyclobutanecarboxylic acid. One carboxyl group is replaced by hydrogen, while the cyclobutane ring remains intact. One mole of A therefore releases one mole of carbon dioxide.

The solution was checked against the official solution.