Until NMR became the main method for the determination of structures of organic molecules, the ozono — Organic Chemistry Chemistry Question
Essential ozone
Until NMR became the main method for the determination of structures of organic molecules, the ozonolysis reaction, disclosed by Schönbein in 1840, was intensively used for the ascertainment of the unsaturated bond(s) position(s) in molecules. Imagine that you are in a similar situation (but with modern reagents). You have found that some hydrocarbon C10H16 participates in the transformations given in Scheme 1.
[VISUAL]
Determine the structural formulae of the hydrocarbon C10H16 and the molecules A – D accounting for the fact that compounds C and D are isomers of the initial hydrocarbon; the ozonolysis of C followed by the treatment of the reaction mixture with alkaline H2O2 produces a single product while the same transformations of D afford two compounds.
Model Answer
Treatment of compound A with base produces the unsaturated bicyclic ketone (C10H14O) which has the same number of carbon atoms as initial hydrocarbon C10H16. It allows for concluding that: a) the initial hydrocarbon has endocyclic unsaturated bond; b) compound A contains two carbonyls groups and its molecular formula is C10H16O2; c) the unsaturated bicyclic ketone is a product of the intramolecular aldol condensation. Therefore, A is cyclodecane-1,6-dione and initial hydrocarbon is octahydronaphthalene. The ozonolysis of octahydronaphthalene followed by reduction of ozonide with NaBH4 affords cyclodecane-1,6-diol dehydration of which gives two cyclodecadienes C and D. Formation of a single product during the ozonolysis of compound C demonstrates clearly that C is a symmetric product, i.e. it is cyclodeca-1,6-diene. So, D is cyclodeca-1,5-diene.
Some other hydrocarbon E (wC = 90.6 %) under ozonolysis (1. O3, CH2Cl2, –78 oC; 2. Me2S) forms three carbonyl compounds – F (C2H2O2), G (C3H4O2), and H (C4H6O2) in a ratio of 3:2:1. Initial hydrocarbon E doesn’t decolorize bromine water. Write down the structural formulae of hydrocarbon E and products of its ozonolysis F – H.
Model Answer
From the known composition of compound E we can determine its molecular formula as (C4H5)x. At the same time compound E doesn’t decolorize bromine water. It allows one to conclude that E is an aromatic compound. If x = 2, E is C8H10. The possible alternatives are ethylbenzene and isomeric dimethylbenzenes. The ozonolysis of substituted aromatic compounds leads to two sets of products as there are two Lewis structures for a given aromatic molecule. From molecular formulae of products it is seen that these are glyoxal (F) and two its substituted derivatives: monomethyl (2-oxopropanal, G) and dimethyl (butan-2,3-dione, or biacetyl, H). Therefore, E is o-xylene (1,2-dimethylbenzene).
Hydrocarbon I having center of symmetry was used as an initial material in the total synthesis of pentalenene (Scheme 2):
[VISUAL]
The ozonolysis of hydrocarbon I furnishes a single compound P or Q depending on the treatment of the ozonolysis product. Under treatment with I2 and NaOH, compound Q forms a yellow precipitate containing 96.7% of iodine. Under basic conditions compound Q is transformed into compound R containing 4 types of hydrogen atoms (4 signals in 1H NMR spectrum with integral intensity of signals 1 : 1 : 2 : 2). Molecular formula of R is C5H6O. Molecule of compound N has bicyclic framework containing R as a fragment. Molecule of O consists of three rings. Descript the scheme of the synthesis of pentalenene.
Model Answer
Сompound L contains 13 carbon atoms. During transformation of I to L 3 carbon atoms are introduced into molecule. Therefore, hydrocarbon I has 10 carbon atoms. The ozonolysis of the hydrocarbon I furnishes a single compound P (after oxidative treatment of ozonide) or Q (after reductive treatment). Accounting for molecular formulae of P and Q, it is possible to deduce that P is ketoacid and Q is ketoaldehyde. The positive iodoform test (formation of yellow precipitate of CHI3 under treatment with I2 and NaOH) indicates the presence of CH3CO- fragment in the molecule of Q. So, Q is 4-oxopentanal and P is 4-oxopentanoic acid (levulinic acid). Thus, compound I is 1,5-dimethylcycloocta-1,5-diene or 1,6-dimethylcycloocta-1,5-diene. However, only the first compound has a center of symmetry.
During K-to-L transformation hydrogen atom is substituted by the allyl fragment CH2=CHCH2–. Therefore, the molecular formula of K is C10H16O, i.e. formula of K differs from formula of I by 1 oxygen atom. The I-to-J is anti-Markovnikov hydration of C=C bond; next step is alcohol-to-ketone oxidation. From formula of K it is possible to conclude that only one C=C bond was hydrated during the first step. LiN(SiMe3)2 is a strong bulky base which selectively deprotonates ketone K at the more accessible CH2 group. Steric effects prevent deprotonation of methane CH-fragment. Alkylation of enolate with allyl bromide accomplishes the synthesis of L which is formed as a mixture of cis- and trans-isomers.
Analysis of the final part of synthesis: transformation of O (C13H18O) into pentalenene (C15H24). O has the same tricyclic framework as pentalenene but instead of two methyl groups compound O has oxygen atom. Molecule N is bicyclic and contains cyclopentenone fragment. The second ring in N is the 8-membered carbocycle. The transformation of N into O is an acid-induced transannular cyclization. The cyclopentenone ring in N is formed by aldol condensation. This leads to the conclusion that L-to-M transformation is the oxidation of C=C double bond producing methyl ketone (Waker process). Oxidation of the second C=C bond cannot produce pentalenene.