Lecture Notes on Structure Determination



PROBLEM: Compound A has a molecular formula C₇H₁₂. Reaction of Compound A with cold aqueous KMnO₄ gives Compound B (C₇H₁₄O₂). When Compound A is treated with peroxybenzoic acid in chloroform, Compound C (C₇H₁₂O) is formed. Compound C reacts with dilute aqueous sulfuric acid to give Compound D (C₇H₁₄O₂) which is isomeric with Compound B. Ozonolysis of Compound A gives 2,6-heptanedione. Assign structures to Compounds A, B, C, and D.

ANALYSIS: Look at the molecular formula for Compound A. C₇H₁₂ Fully saturated formula would be C₇H₁₆.	Two degrees of unsaturation so Compound A may have: two C=C two rings one C=C & one ring
Reaction of Compound A to form Compound B is the reaction in the Baeyer Test.	Compound A must have at least one C=C.
Get the structure of Compound A from the ozonolysis data.
The organic product in the Baeyer Test is a vicinal diol. This addition is a syn addition (This is new information for you).
Compound C is an epoxide.
Figuring out the structure of Compound D is a bit of a challenge since you have not yet studied any reactions of epoxides. However, you can use what you have learned so far this term to come up with a solution. Look at the reagents in the reaction: Compound C has an oxygen that can act as a Lewis base. H₂SO₄ is an acid which can protonate a Lewis base The protonated epoxide like a protonated alcohol will be susceptible to attack by a nucleophile. Water can serve as a nucleophile. Compound D is an isomer of Compound B.