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The Baeyer Laboratory, Munich, 1893 (This photograph is in the hallway across from 110 SCL) Adolf von Baeyer (1835-1917); Nobel Prize 1905. (center, seated with derby), who was a student of Kekulé, succeeded Liebig at Munich. In the photograph (second row; third from right) is Henry Lord Wheeler (1867-1914); Yale Faculty 1896-1911. As was the custom in the 19th century, many Americans, such as Wheeler, did advanced study in chemistry in Europe. Karl is the laboratory assistant. (The only person wearing an apron and no tie; upper left.) In 1885, as an addendum to a paper on acetylenic compounds, Baeyer proposed that cyclopentane was the least strained of the cycloalkanes . While he accepted the idea that the carbon atoms in cycloalkanes were tetrahedral, he treated the cycloalkanes as though they were flat. He argued that there is only one cyclohexane carboxylic acid, not two (axial and equatorial) as was predicted by a chair cyclohexane. Equatorial is frequently misspelled.

Sir Derek H. R. Barton (1918-1998) 1969 Nobel Prize with Odd Hassel for their work on conformational analysis For a video of Barton talking about conformational analysis, click here. Cyclohexane in the chair conformation (How to manipulate Jmol structures)

a. Work through How to Draw Cyclohexanes (PowerPoint)

b. The Conformation Module in the Study Aids will give you a good overview of the subject of conformation of cycloalkanes.

1. Compound A (MW=140.22), a 1,4-disubstituted cyclohexane, has the following composition: C, 77.09% and H, 11.50%. (What else is there?) The difference in conformational energy for the two chair conformations of A is 1.3 kcal/mol. Using the A-value data (Energy Differences Between ..... Cyclohexanes), determine the structure of A. Illustrate and explain. What is the conformational energy difference for the stereoisomer of A, ---namely A'. Explain and illustrate. Show the chair comformations of A and A' with the appropriate equilibrium arrows to illustrate the major and minor conformations. Label each conformation with its energy.

Using general chemistry math to determine an empirical formula from percentages (see table on the right), one obtains C9H16O, which is also the molecular formula because MW = 140.22. Oxygen is determined by difference. Carbon (as CO2) and hydrogen (as water) are determined directly. The cyclohexane ring less two 1,4-positions for substituents has a formula C6H10. The composition of the substituents is C9H16O - C6H10 = C3H6O. There are only three groups that contain a single oxygen. CH3O requires a companion group C2H3, which doesn't exist on the list. Secondly, OH has a complement of C3H5, which is not present. CHO (0.6 kcal/mol) requires a complement of C2H5 (1.9 kcal/mol). There difference is the same as shown forA. Therefore, A is cis-4-ethyl cyclohexanecarboxaldehyde. A' is the trans isomer with an energy difference between its chair conformers of 0.6 + 1.9 = 2.5 kcal/mol. Only axial groups increase the energy.[Note: If given the two energies of the equilibria, say A + B = 2.5 and A - B = 1.3. Two equations; two unknowns. Solve for the individual A -values.]

Atoms % Atom Mass Atom Mass/% AtomMass/%/0.71 C 77.09 12.01 6.49 9.1 H 11.50 1.008 11.41 16 O 11.41 16.00 0.71 1

2. Unknown compound (R)-(+)-A (C₁₀H₁₆; [α]_d=124^o) reacts with two molar equivalents of hydrogen to produce two 1,4-disubstituted cyclohexanes, B and C, both of which are optically inactive. Compound B has an energy difference between its two chair conformations of 0.3 kcal/mol, while compound C's difference is 3.9 kcal/mol. Ozonolysis of A provides one mole of formaldehyde and one mole of an optically active diketoaldehyde D. What are the structures of A-D? Show your reasoning. A is has three degrees of unsaturation. Since A absorbs two equivalents of hydrogen, there is one ring, which is a cyclohexane. There could be one triple bond or two double bonds. Because ozonolysis provides four carbonyl groups (formaldehyde, one aldehyde and two ketones), there must be two double bonds. In B and C, which are stereoisomers with formula C₁₀H₂₀, the two groups are C₄H₁₀ (See above solution). The two groups cannot be ethyls because cis-1,4-dimethylcyclohexane would have a 0 kcal/mol energy between chair conformers. One group must be methyl (C1) and the other n-propyl or isopropyl. The A-value table shows methyl = 1.8 kcal/mol and isopropyl = 2.1 kcal/mol. The three carbon group in A must be isopropenyl. It affords formaldehde and a ketone in D on ozonolysis. The remaining double bond in A is trisubstituted giving the remaining aldehyde and ketone in D upon ozonolysis. The structures of A-D are shown below. A is (+)-limonene. For more on this problem, see the end of StudyAids/mechanisms/ozonolysis. A confession: There are two possible answers to this problem !While the set on the left is what I had in mind, the Alternative answer is also acceptable. In the past I had given the structure of D. Not doing so gave two possibilities.

3. Draw the 3-D line-angle structures of the major products in each of the following reactions. Pay attention to stereochemistry, optical activity and racemic mixtures. Explain briefly.

5. Using the Heats of Formation Table list the values for the cycloalkanes in column 2. Which of these cycloalkanes is the least strained? Place its heat of formation in column 3. Compute the hypothetical heats of formation for the remaining alkanes (Remember the magic number in kcal/mol!). Compute the difference in heats of formation of each cycloalkane (pay attention to + and - signs) in column 4 (column 2 minus column 3). What have you calculated in column 4? Find these data in the Thermochemistry Module and compare your answer. Good or bad correlation? Cyclohexane is taken as strain free. To determine the hypothetical heat of formation of the other cycloalkanes, add or subtract 5 kcal/mole/CH₂. Subtracting column 3 from column 2 gives the total strain energy. Compare with this link.