IChO 1975 · Theory problem 7

Oxidation products of a saturated hydrocarbon

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A certain saturated hydrocarbon A is transformed by a catalytic oxidation partly into a secondary alcohol B, and partly into a ketone C. The ketone when oxidised with nitric acid in the presence of catalyst yields a compound D with a formula of \(\ce{C6H10O4}\). From compound D when heated in the presence of acetic anhydride, a ketone E is formed, its formation being accompanied by evolution of \(\ce{CO2}\) and \(\ce{H2O}\). Compounds E and C have similar structures but compound E contains one methylene group less than ketone C. Compound D is one of the starting materials in the production of an important polycondensation synthetic fibre.

Problem:

7.1 Write the structural formulas of compounds A, B, C, D, and E.

Official English text of the problem.

Answers

  • 7.1: The compounds are:
Compound Name Structural description SMILES
A Cyclohexane Six-membered ring of methylene groups C1CCCCC1
B Cyclohexanol Six-membered carbon ring with one \(\ce{CH(OH)}\) group and five methylene groups OC1CCCCC1
C Cyclohexanone Six-membered ring containing one carbonyl carbon and five methylene groups O=C1CCCCC1
D Adipic acid (hexanedioic acid) \(\ce{HOOC(CH2)4COOH}\) O=C(O)CCCCC(=O)O
E Cyclopentanone Five-membered ring containing one carbonyl carbon and four methylene groups O=C1CCCC1

Solution

7.1

The fibre-production clue identifies D as adipic acid, a monomer used to manufacture nylon-6,6. Its structure, \(\ce{HOOC(CH2)4COOH}\), has exactly the required molecular formula \(\ce{C6H10O4}\). Its two degrees of unsaturation,

\[ \mathrm{DBE}=\frac{2(6)+2-10}{2}=2, \]

are accounted for by the two carboxyl carbonyl groups.

On heating in the presence of acetic anhydride, CC(=O)OC(C)=O, adipic acid undergoes intramolecular ketonic decarboxylation. The four methylene carbons and one carboxyl carbon form a five-membered ketone ring; the remaining carboxyl carbon is lost as carbon dioxide, O=C=O. The balanced net reaction also produces water, O:

\[ \ce{HOOC(CH2)4COOH ->[acetic\ anhydride][\Delta] C5H8O + CO2 + H2O}. \]

Thus E is cyclopentanone.

Since C has a similar structure but contains one additional methylene group, C must be cyclohexanone. Its six-membered ring contains five methylene groups and one carbonyl carbon, whereas cyclopentanone contains four methylene groups and one carbonyl carbon.

Oxidation of cyclohexanone with nitric acid, O=[N+]([O-])O, cleaves a carbon–carbon bond adjacent to the carbonyl group and opens the ring to give adipic acid:

\[ \ce{C6H10O + 3[O] ->[HNO3,\ catalyst] HOOC(CH2)4COOH}. \]

Here \(\ce{[O]}\) represents a formal oxygen-atom equivalent supplied by the oxidant. This balanced oxidation equation leaves the nitrogen-containing reduction products unspecified, as in the problem.

The corresponding secondary alcohol B is therefore cyclohexanol. The carbon bearing its hydroxyl group is bonded to two other carbon atoms, making it a secondary alcohol. Oxidation of this group gives the ketone:

\[ \ce{C6H12O + [O] -> C6H10O + H2O}. \]

Finally, the saturated hydrocarbon retaining this six-membered carbon skeleton is cyclohexane, A. Its catalytic oxidation with oxygen, O=O, gives cyclohexanol and cyclohexanone through the following balanced overall reactions:

\[ \ce{2C6H12 + O2 ->[catalyst] 2C6H12O}, \]

\[ \ce{C6H12 + O2 ->[catalyst] C6H10O + H2O}. \]

All five structures therefore satisfy the oxidation sequence, the loss of one methylene group between C and E, and the synthetic-fibre clue.

The solution was checked against the official solution.