IChO 1996 · Theory problem 7

Stereochemistry of 5-norbornene-2,3-dicarboxylic acids

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Stereochemistry of organic compounds can sometimes be determined by studying their chemical behaviour. The stereochemical configuration of one of the isomers of 5-norbornene-2,3-dicarboxylic acids (compound X)

Figure 1

was established by the following experiments.

On heating this substance decomposes producing water and a new compound Y. Compound Y slowly dissolves in excess of aqueous \(\ce{NaOH}\) with the formation of product \(X_1\) same to that is formed in the reaction of X with \(\ce{NaOH}\). The resulting solution of \(X_1\) is treated by \(\ce{I2}\) to give compounds containing iodine. Acidification of the solution leads to a mixture of two isomeric compounds, A and B in the 3 : 1 ratio. The titration of 0.3913 g of compound A by 0.1000 M aqueous solution of \(\ce{NaOH}\) in the presence of phenolphthalein takes 12.70 \(\mathrm{cm^3}\) of alkali. The same amount of 0.1000 M solution of \(\ce{NaOH}\) is required for the titration of 0.3913 g of compound B. On heating compound A slowly transforms into compound C, which contains no iodine and is able to react with water. Under the same conditions compound B does not undergo this transformation, but on heating with hydrochloric acid slowly transforms into A.

All reactions must be written as balance equations. No mechanisms are required.

7.1 Mark by asterisks (*) the asymmetric carbon atoms in the structure of 5-norbornene-2,3-dicarboxylic acids.

7.2 Draw the stereochemical formulas of all stereoisomers of compound X, and the structures of products of their dehydration in those cases when it is possible.

7.3 Write the reactions of \(\ce{NaOH}\) with a stereoisomer of X and a stereoisomer of Y.

7.4 Calculate the molar mass of compound A. Write the reactions leading from \(X_1\) to A.

7.5 Write the reaction of the formation of C from A and the reaction of C with water.

7.6 Draw the stereochemical formula of compound X which satisfies all of the data given in the problem.

7.7 Write the reactions leading from B to A.

7.8 Are the compounds A and B diastereomers?

Official English text of the problem.

Answers

  • 7.1: Mark C1, C2, C3 and C4, including both bridgehead carbon atoms.
  • 7.2: There are four stereoisomers: the meso cis-endo,endo and cis-exo,exo forms, and a pair of trans enantiomers. Each cis form gives its corresponding five-membered cyclic anhydride; neither trans form can do so without changing configuration. Their stereochemical SMILES are given below.
  • 7.3: Both the diacid and its anhydride consume two equivalents of sodium hydroxide to give the same disodium dicarboxylate, \(X_1\).
  • 7.4: \(\boxed{M(A)=308.1\ \mathrm{g\,mol^{-1}}}\). A is a five-membered iodolactone with one remaining carboxylic acid group.
  • 7.5: A loses hydrogen iodide to form the 2,6:3,5-dilactone C. Complete hydrolysis of C consumes two water molecules and gives the corresponding endo,endo-dihydroxy diacid.
  • 7.6: X is cis-endo,endo-5-norbornene-2,3-dicarboxylic acid, O=C(O)[C@@H]1[C@H](C(=O)O)[C@@H]2C=C[C@H]1C2.
  • 7.7: Acid-catalysed hydrolysis opens the six-membered lactone of B; subsequent esterification using the other carboxyl group produces the five-membered lactone A.
  • 7.8: No. A and B are constitutional isomers, differing in which carboxyl group forms the lactone.

Solution

Throughout, use the original norbornene numbering: the bridgeheads are C1 and C4, and the three bridges are

\[ \mathrm{C1-C2-C3-C4},\qquad \mathrm{C1-C6=C5-C4},\qquad \mathrm{C1-C7-C4}. \]

The carboxyl groups are attached to C2 and C3. An endo carboxyl group points toward the other two-carbon bridge of the cage; an exo group points away from it.

A 2,6-lactone, for example, means that the carboxyl group attached to C2 forms an ester linkage to an oxygen attached to C6. The carboxyl carbon is an additional atom, not C2 itself.

For A, B, their sodium salts and the iodohydroxy intermediate below, each SMILES specifies one enantiomer. The reactions starting from achiral X give the corresponding racemic products.

The small molecules and salts used in the equations are: water O; sodium hydroxide [Na+].[OH-]; iodine II; sodium iodide [Na+].[I-]; hydrochloric acid, represented in aqueous ionic form, [H+].[Cl-]; sodium chloride [Na+].[Cl-]; and hydrogen iodide I.

7.1 — Asymmetric carbon atoms

The parent compound is 5-norbornene-2,3-dicarboxylic acid, with unspecified stereochemistry:

O=C(O)C1C(C(=O)O)C2C=CC1C2

Place asterisks on

\[ \boxed{\mathrm{C1^{*},\ C2^{*},\ C3^{*},\ C4^{*}}}. \]

C2 and C3 each carry hydrogen, a carboxyl group and two different paths through the cage. Each bridgehead, C1 and C4, carries hydrogen and three distinguishable carbon paths.

C5 and C6 are trigonal alkene carbons; C7 bears two hydrogen atoms.

7.2 — All stereoisomers and their dehydration products

The rigid bridge system constrains the bridgehead configurations; the four asymmetric carbons do not give \(2^4\) independently accessible configurations.

Stereoisomer of 5-norbornene-2,3-dicarboxylic acid Stereochemical description SMILES
cis-endo,endo Both carboxyl groups endo; meso O=C(O)[C@@H]1[C@H](C(=O)O)[C@@H]2C=C[C@H]1C2
cis-exo,exo Both carboxyl groups exo; meso O=C(O)[C@@H]1[C@H](C(=O)O)[C@H]2C=C[C@@H]1C2
trans-endo,exo C2 carboxyl endo and C3 carboxyl exo in the fixed numbered cage O=C(O)[C@H]1[C@H](C(=O)O)[C@H]2C=C[C@@H]1C2
trans-exo,endo Mirror image of the preceding trans isomer O=C(O)[C@@H]1[C@@H](C(=O)O)[C@H]2C=C[C@@H]1C2

Thus there are two achiral cis isomers and two enantiomeric trans isomers.

In either cis isomer, the neighbouring carboxyl groups can form a five-membered cyclic anhydride. The original configurations at C2 and C3 are retained.

Starting acid Dehydration product SMILES
cis-endo,endo cis-endo,endo-5-norbornene-2,3-dicarboxylic anhydride, \(Y_{\mathrm{endo}}\) O=C1OC(=O)[C@H]2[C@@H]1[C@H]1C=C[C@@H]2C1
cis-exo,exo cis-exo,exo-5-norbornene-2,3-dicarboxylic anhydride, \(Y_{\mathrm{exo}}\) O=C1OC(=O)[C@H]2[C@@H]1[C@@H]1C=C[C@H]2C1

For each cis isomer, the balanced dehydration equation is

\[ \boxed{\ce{C9H10O4 ->[\Delta] C9H8O3 + H2O}}. \]

To sketch either product, connect the two carboxyl carbonyl carbons through a single anhydride oxygen, giving the ring sequence

\[ \mathrm{C2-C3-C(=O)-O-C(=O)-C2}. \]

The trans carboxyl groups lie on opposite faces and cannot form this small anhydride ring without a configurational change.

7.3 — Reactions with sodium hydroxide

Choose the cis-endo,endo acid and its anhydride. Both produce disodium cis-endo,endo-5-norbornene-2,3-dicarboxylate, \(X_1\):

O=C([O-])[C@@H]1[C@H](C(=O)[O-])[C@@H]2C=C[C@H]1C2.[Na+].[Na+]

Neutralization of the diacid gives

\[ \boxed{\ce{C9H10O4 + 2NaOH -> C9H8Na2O4 + 2H2O}}. \]

Hydrolysis and neutralization of the anhydride give

\[ \boxed{\ce{C9H8O3 + 2NaOH -> C9H8Na2O4 + H2O}}. \]

The corresponding exo acid and exo anhydride likewise give the same exo dicarboxylate as one another.

7.4 — Molar mass of A and its formation from \(X_1\)

Iodination permits intramolecular lactone formation: one carboxyl group becomes an ester, while the other remains available for acid–base titration. Thus A and B are monobasic iodolactones.

The amount of sodium hydroxide used is

\[ n(\ce{NaOH}) = (0.1000\ \mathrm{mol\,L^{-1}}) (12.70\times10^{-3}\ \mathrm L) = 1.270\times10^{-3}\ \mathrm{mol}. \]

Since one mole of A consumes one mole of sodium hydroxide at the acid–base endpoint,

\[ M(A) = \frac{0.3913\ \mathrm g}{1.270\times10^{-3}\ \mathrm{mol}} = \boxed{308.1\ \mathrm{g\,mol^{-1}}}. \]

Strictly, the titration first gives the acid equivalent mass; its equality to the molar mass follows from the presence of one free carboxyl group.

The formula is \(\ce{C9H9IO4}\), consistent with the original diacid incorporating one iodine atom and losing one carboxyl hydrogen during iodolactonization.

The structures established by the subsequent heating and hydrolysis tests are:

Compound Name and precise connectivity SMILES
A 2,6-lactone of exo-5-iodo-endo-6-hydroxynorbornane-endo-2,endo-3-dicarboxylic acid. A five-membered \(\gamma\)-lactone; the C3 carboxyl group remains free. O=C1O[C@@H]2[C@@H](I)[C@H]3C[C@@H]2[C@@H]1[C@H]3C(=O)O
B 3,6-lactone of exo-5-iodo-endo-6-hydroxynorbornane-endo-2,endo-3-dicarboxylic acid. A six-membered \(\delta\)-lactone; the C2 carboxyl group remains free. O=C(O)[C@@H]1[C@H]2C[C@@H]3[C@H](I)[C@H]2OC(=O)[C@@H]31
\(A_{\mathrm{Na}}\) Sodium carboxylate of A O=C1O[C@@H]2[C@@H](I)[C@H]3C[C@@H]2[C@@H]1[C@H]3C(=O)[O-].[Na+]
\(B_{\mathrm{Na}}\) Sodium carboxylate of B O=C([O-])[C@@H]1[C@H]2C[C@@H]3[C@H](I)[C@H]2OC(=O)[C@@H]31.[Na+]

For A, the lactone ring contains C2, C1, C6, the ester oxygen and the C2 carboxyl carbon: five atoms. For B, the corresponding ring through C3, C2, C1 and C6 contains six atoms.

Formation of the sodium iodolactone from \(X_1\) is balanced as

\[ \boxed{\ce{C9H8Na2O4 + I2 -> C9H8INaO4 + NaI}}. \]

The organic product on the right is \(A_{\mathrm{Na}}\) or \(B_{\mathrm{Na}}\), according to which carboxyl group closes the ring.

Using hydrochloric acid to express the acidification explicitly,

\[ \boxed{\ce{C9H8INaO4 + HCl -> C9H9IO4 + NaCl}}. \]

The resulting A:B ratio is the stated 3:1.

For either isolated iodolactone, the acid–base titration is

\[ \ce{C9H9IO4 + NaOH -> C9H8INaO4 + H2O}. \]

Both therefore have the same molar mass and consume the same amount of alkali for equal masses.

7.5 — Formation of C and its reaction with water

In A, the remaining endo carboxyl group at C3 is suitably placed to form a second five-membered lactone involving C5. The exo iodine at C5 is replaced by an endo ester oxygen, with loss of hydrogen iodide.

The product C is the 2,6:3,5-dilactone of endo-5,endo-6-dihydroxynorbornane-endo-2,endo-3-dicarboxylic acid:

O=C1O[C@H]2[C@@H]3C[C@@H]4[C@H]2OC(=O)[C@@H]4[C@H]13

It has formula \(\ce{C9H8O4}\). Its two lactone connections are:

  • the C2 carboxyl group to the endo oxygen at C6;
  • the C3 carboxyl group to the endo oxygen at C5.

The balanced thermal reaction is

\[ \boxed{\ce{C9H9IO4 ->[\Delta] C9H8O4 + HI}}. \]

Complete hydrolysis opens both ester rings, producing endo-5,endo-6-dihydroxynorbornane-endo-2,endo-3-dicarboxylic acid, designated E:

O=C(O)[C@@H]1[C@H]2C[C@H]([C@@H](O)[C@H]2O)[C@@H]1C(=O)O

The balanced hydrolysis equation is

\[ \boxed{\ce{C9H8O4 + 2H2O <=> C9H12O6}}. \]

In a drawing of E, both hydroxyl groups and both carboxyl substituents are endo.

7.6 — Identification of X

Anhydride formation establishes that the two carboxyl groups are cis, leaving the endo,endo and exo,exo possibilities.

Formation of the lactone connections to C5 and C6 requires the carboxyl groups to face inward toward that bridge. The formation of the second lactone in C establishes the necessary orientation of the remaining carboxyl group as well.

Consequently,

\[ \boxed{X=\text{cis-endo,endo-5-norbornene-2,3-dicarboxylic acid}.} \]

Its stereochemical SMILES is

O=C(O)[C@@H]1[C@H](C(=O)O)[C@@H]2C=C[C@H]1C2

To draw X, retain the original norbornene skeleton and its C5=C6 double bond, place both carboxyl groups endo, and place the hydrogens on C2 and C3 exo. Its dehydration product is \(Y_{\mathrm{endo}}\) from part 7.2.

7.7 — Conversion of B into A

B already contains the 3,6-lactone connection. Its free C2 carboxyl group is not positioned to give the second five-membered lactone formed from A, explaining the different thermal behaviour.

In hot aqueous hydrochloric acid, B can first undergo lactone hydrolysis. This gives exo-5-iodo-endo-6-hydroxynorbornane-endo-2,endo-3-dicarboxylic acid, designated D:

O=C(O)[C@@H]1[C@H]2C[C@H]([C@H](I)[C@H]2O)[C@@H]1C(=O)O

The balanced opening of B is

\[ \boxed{\ce{C9H9IO4 + H2O <=>[HCl,\ \Delta] C9H11IO5}} \qquad\text{(B to D)}. \]

The C2 carboxyl group of D can then esterify the endo hydroxyl group at C6, forming the five-membered 2,6-lactone A:

\[ \boxed{\ce{C9H11IO5 <=>[HCl,\ \Delta] C9H9IO4 + H2O}} \qquad\text{(D to A)}. \]

Water cancels from the overall conversion, and hydrochloric acid acts catalytically. The iodine position and the configurations of the cage carbon atoms are retained; the carboxyl group participating in the ester linkage changes.

7.8 — Are A and B diastereomers?

No. Diastereomers must have the same atom connectivity.

In A, the carboxyl group at C2 forms the ester bond to the oxygen at C6. In B, the carboxyl group at C3 forms that bond. Their lactone rings consequently contain five and six atoms, respectively.

Thus A and B are constitutional isomers, specifically regioisomeric iodolactones.

The solution was checked against the official solution.