37 th Middle Atlantic Regional Meeting
MARM 2005
May 25, 2005
Busch Campus, Rutgers University

Abstract
The structural features inherent in acyclic monoterpenes that follow the isoprene rule often lead to unique sets of ozonolysis products from which their structures, excluding stereochemistry, can be determined from molecular formulas only. Three examples show how students may elucidate the structures of these compounds by analysis of the oxidative and reductive workup products. The complete problem set contains 50 problems.

Introduction
The molecular formula of a C
10 hydrocarbon and the molecular formulas of its products of ozonolysis are provided. The hydrocarbon follows the isoprene rule.
From this data, students determine the structure of the hydrocarbon. Some problems are solvable from the reductive workup products and some from the oxidative workup products. Other problems require the products of both workups for their solutions .

Methodology
The reductive workup of an ozonide invariably produces two carbonyl groups from a double bond and two carboxylic acids from a triple bond. These results lead to the following information from molecular formulas.
Reductive Workup (R)
•
The number of degrees of unsaturation
•
The number of triple bonds (tb) in the reactant
• The number of p bonds ( p
) in the reactant
• The number of double bonds (db) in the reactant
• The number of rings (r) in the reactant

•
The presence of CH
2
O proves the presence of a terminal alkene in the reactant.
• The presence of CH
2
O
2 alkyne in the reactant.
proves the presence of a terminal
Oxidative Workup (O)
• Excluding CH
2
O, a one-oxygen atom difference between an oxidative and reductive product’s formula indicates an aldehyde-acid pair (e.g., a C
6
H
12
O aldehyde (R) becomes a
C
6
H
12
O
2 acid (O).
•
Any ketone isolated in the reductive workup is also isolated in the oxidative workup (e.g., C
4
H
8
O remains C
4
H
8
O).

• Any acid isolated in the reductive workup is also isolated in the oxidative workup (e.g., C
9
H
18
O
2 remains C
9
H
18
O
2
).
• A dialdehyde in the reductive workup becomes a diacid in the oxidative workup as evidenced by a two-oxygen atom increase in the diacid (O
2 ketoacid (O
2 to O
3 to O
4
); a ketoaldehyde becomes a
), an aldehyde-acid becomes a diacid (O
3 to O
4
), etc.

Example 1
Structure of a Triene Hydrocarbon (HC) from its Reductive Workup Products
Determine the structure of a C
10
H
16
2 C
3
H
4
O
2 hydrocarbon that follows the isoprene rule and yields CH
2
O, C
3
H
6
O and upon ozonolysis by a reductive workup .

C
10
H
16 p
+ r = 3 tb = 0 p
= 3
1 db = 3
r = 0
2
O
3
(R)
3
CH
2
O + C
3
H
6
O + C
3
H
4
O
2
+ C
3
H
4
O p
= 1 p
= 1 p
= 2 p
= 2
4
5
6
7
9 10 isoprene-rule template
8
8
7
6
5
4
3
2
2
1
3,7-dimethyl-1,3,6-octatriene
• Structural features in red are determined from the molecular formulas.
• Asymmetry in the template makes the C
3
H
6
O compound acetone, which arises from a C2-C3 db.
• The other products must arise from db’s as shown in the template.

Example 2
Structure of a Diene Hydrocarbon (HC) from its Oxidative Workup Product
Determine the structure of a C
10
H
18 hydrocarbon that follows the isoprene rule and yields C
2
H
2
O
2
, C
4
H
8
O and C
4
H
8
O
2 upon ozonolysis by an oxidative workup.

C
10
H
18 p
+ r = 2
db = 2
r = 0 1
O
3
(O)
C
2
H
2
O
4
+ C
4
H
8
O + C
4
H
8
O
2 p
= 2 p
= 1 p
= 1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
9 10 isoprene-rule template 2,6-dimethyl-3,5-octadiene
•
The three products require two multiple bonds, which must both be db’s because p
+ r = 2.
• The db’s must be placed as shown on the template to give a four-carbon acid and a four-carbon ketone.
•
Structural features in red are determined from the molecular formulas.

Example 3
Structure of a Diene Hydrocarbon (HC) from its
Reductive and Oxidative Workup Products
Determine the structure of a C
10
H
16 hydrocarbon that follows the isoprene rule and yields the ozonolysis products shown below.

C
10
H
16 p
+ r = 3 p
= 3
tb = 1
db = 1
r = 0
1
(O)
(R)
2
3
4
CO
2
+ C
4
H
8
O
2
+ C
5
H
8
O
4
CH
2
O + C
4
H
8
O
2
+ C
5
H
8
O
3 p
= 1 p
= 1 p
= 2
5
6
7
8
8
7 6 5 4
3
9 10 isoprene-rule template
2
1
3,7-dimethyl-1-octen-5-yne
• Structural features in red are determined from the molecular formulas.
•
The tb must be placed on the template as shown to account for the four-carbon acid (R).
• The db must be placed on the template as shown to account for the five-carbon diacid (O).

Limitations
• Of 66 compounds, 16 in the series give two solutions.
• Neither E/Z nor R/S stereochemistry can be determined by ozonolysis.

Conclusions
• The structures of monoterpenes (enes, ynes, di- and trienes, and enynes) that follow the isoprene rule can be determined from their molecular formulas and those of their products of ozonolysis.
•
The fifty problems vary in difficulty, affording instructors of beginning organic or qualitative organic analysis courses a set of degradation problems that complement spectral and synthetic problems in engaging students to apply their knowledge and enhance their reasoning skills while solving challenging problems.

Notes
1. The complete problem set will be published as a supplement in J. Chem. Educ ., in press.
2. The author’s email address is rgross@pgcc.edu
.