### A Structural Formula Puzzle

Two isomeric C8H10 hydrocarbons (A & B) both give the same saturated hydrocarbon product, C, on exhaustive catalytic hydrogenation. Both A & B, but not C, show strong absorption of UV light, λmax = 245 nm. Addition of one equivalent of HBr to either A or B produces the same mixture of C8H11Br isomers (D & E).
 A or B } excess H2Pt catalyst C =

 1. How many double bonds must be present in compound A? (enter a number from 0 to 9 in the answer box) In compound B? (enter a number from 0 to 9 in the answer box)
 2. What must be the relationship of these functional groups? (enter a word or a short phrase in the answer space)

Spectroscopic evidence concerning the structures of these compounds:

Compound A:   five 13C nmr signals, two from sp2 carbons & three from sp3 carbons. A complex 1H nmr signal at δ 5.6 ppm, 1:4 ratio with respect to higher-field signals.
Compound B:   four 13C nmr signals, two from sp2 carbons & two from sp3 carbons. A complex 1H nmr signal at δ 5.6 ppm, 1:4 ratio with respect to higher-field signals.
Compound D:   A complex 1H nmr signal at δ 5.5 ppm, 1:9 ratio with respect to higher-field signals.
Compound E:   No 1H nmr signals in the region below (lower field than) δ 5.0 ppm.

3. From the information given above you should be able to write structural formulas for compounds A, B, D & E. When you have done so continue the problem by clicking here.

From the information given at the beginning of this problem you should have concluded that compounds A & B have two double bonds each. The UV absorption indicates that these double bonds must be conjugated. Given the common bicyclic skeleton of these isomers (hydrogenation to C), only five conjugated diene structures can be written: 1 through 5 below.

We can now examine these isomeric dienes and predict how their different structures would be reflected in differences in their 13C and 1H nmr spectra.
You must enter numbers for both carbon and proton nmr before checking your answer.

CompoundNumber of expected
13C nmr signals
Number of H atoms giving
1H nmr signals at δ > 5.0 ppm
1
2
3
4
5

With this analysis accomplished, the spectroscopic data from A & B can be used to make structural assignments:
 4. Which compound (1 through 5) is A? { Five 13C nmr signals, two from sp2 carbons & three from sp3 carbons. A complex 1H nmr signal at δ 5.6 ppm, 1:4 ratio with respect to higher-field signals. 5. Which compound (1 through 5) is B? { Four 13C nmr signals, two from sp2 carbons & two from sp3 carbons. A complex 1H nmr signal at δ 5.6 ppm, 1:4 ratio with respect to higher-field signals.

6. Finally, you should be able to write structural formulas for the HBr addition products, compounds D & E.
Remember, D shows a complex 1H nmr signal at δ 5.5 ppm, 1:9 ratio with respect to higher-field signals; whereas, E has no 1H nmr signals in the region below (lower field than) δ 5.0 ppm. To draw a structural formula, use the drawing window opened by the Draw Formulas button. When you are finished, Submit your formula and then check it by pressing the appropriate Check Compound button. A help screen is available for the drawing window.

This script written by William Reusch, Dept. of Chemistry, Michigan State University. Please send comments and corrections to whreusch@pilot.msu.edu.