The two hydroxyl groups on the cyclopentyl ring are cis because they both point behind the plane of the page (both dashed bonds). The carbon chains have a trans relationship (one is dashed and the other is wedged).

All hydrogen atoms on the same side of the cyclopropane ring are eclipsed by neighboring hydrogens. If we draw each hydrogen–hydrogen interaction, we count six eclipsing interactions, three on each side of the ring. Since each of these interactions costs 4.0 kJ/mol, all six cost 24.0 kJ/mol. 24 kJ/mol ÷115 kJ/mol × 100% = 21% of the total strain energy of cyclopropane is due to torsional strain.

The added energy cost of eclipsing interactions causes cis-1,2-dimethylcyclopropane to be of higher energy and to be less stable than the trans isomer.

If cyclopentane were planar, it would have ten hydrogen–hydrogen interactions with a total energy cost of 40 kJ/mol (10 × 4.0 kJ/mol). The measured total strain energy of 26 kJ/mol indicates that 14 kJ/mol of eclipsing strain in cyclopentane (35%) has been relieved by puckering.

The methyl groups are farther apart in the more stable conformation of cis-1,3- dimethylcyclobutane.

The conformation with –OH in the equatorial position is more stable.

Note: The starred ring carbons lie in the plane of the paper.

2 [H ↔ CH3] = 7.6 kJ/mol                           2 [H ↔ CH3] = 7.6 kJ/mol 2 [H ↔ Cl]= 2.0 kJ/mol   = 9.6 kJ/mol

The illustrated conformation is the less stable chair form.

Trans-decalin is more stable than cis-decalin because of three 1,3-diaxial interactions present in the cis isomer. You can recognize these interactions if you think of the circled parts of cis-decalin as similar to axial methyl groups. The gauche interactions that occur with axial methyl groups also occur in cis-decalin.

The conformation shown (the left structure) is the less stable conformation.

The only difference between α-glucose and β-glucose is in the orientation of the –OH group at carbon 1: the –OH group is axial in α-glucose, and it is equatorial in β-glucose. You would expect β-glucose to be more stable because all of its substituents are in the equatorial position.

Two cis–trans isomers of 1,3,5-trimethylcyclohexane are possible. In one isomer (A), all methyl groups are cis; in B, one methyl group is trans to the other two.

After the ring-flip, the relationship of the two substituents is still axial–equatorial. No two adjacent cis substituents can be converted to being both axial or both equatorial by a ring- flip. Don’t forget that there are only two chair conformations of any given cyclohexane.

A cis-1,3-disubstituted isomer exists almost exclusively in the diequatorial conformation, which has no 1,3-diaxial interactions. The trans isomer must have one group axial, leading to 1,3-diaxial interactions. Thus, the trans isomer is less stable than the cis isomer. When a molecule has two conformations available, the molecule exists mainly in the lower energy conformation.

The trans-1,4-isomer is more stable because both substituents can be equatorial.

(1) Find the energy cost of a 1,3-diaxial interaction by using Table 4.1.

(2) Convert this energy difference into a percent by using Figure 4.13.

Use Table 4.1 to find the values of 1,3-diaxial interactions. For the first conformation, the steric strain is 2 × 1.0 kJ/mol = 2.0 kJ/mol. The steric strain in the second conformation is 2 × 3.8 kJ/mol, or 7.6 kJ/mol. The first conformation is more stable than the second conformation by 5.6 kJ/mol.

Use Table 4.1 to find the values of 1,3-diaxial interactions. The first conformation is more stable than the second conformation by a maximum of 9.6 kJ/mol. (A gauche interaction between the two substituents in the diequatorial conformation reduces the value of the energy difference, but its value can’t be determined with the given data.)

In this conformation, all substituents, except for one hydroxyl group, are equatorial.

Conformation A of cis-1-chloro-3-methylcyclohexane has no 1,3-diaxial interactions and is the more stable conformation. Steric strain in B is due to one CH₃ ↔ H interaction (3.8 kJ/mol), one Cl ↔ H interaction (1.0 kJ/mol) and one CH₃ ↔ Cl interaction. Since the total-strain energy of B is 15.5 kJ/mol, 15.5 kJ/mol – 3.8 kJ/mol – 1.0 kJ/mol = 10.7 kJ/mol of strain is caused by a CH₃ ↔ Cl interaction.

The rings of cis-decalin are joined by an axial bond and an equatorial bond. After a ring- flip, the rings are still linked by an equatorial and an axial bond. No additional strain or interaction is introduced by a ring-flip of cis-decalin.

Draw both rings and add –OH groups having the indicated relationships. Perform a ring- flip on the structure you have drawn to arrive at the other conformation.

There are eight cis–trans stereoisomers of myo-inositol. The first isomer is the most stable because all hydroxyl groups can assume an equatorial conformation.

If you build a model of 1-norbornene, you will find that it is almost impossible to form the bridgehead double bond. sp²-Hybridization at the double bond requires all carbons bonded to the starred carbons to lie in a common plane in order for the p orbitals to overlap to form the π bond. The bicyclic ring system forces these atoms out of plane, and the bridgehead double bond can’t form.

Substituents on the bottom side of the ring system also alternate axial and equatorial positions. A substituent at C7 (b) is axial, and one at C11 (c) is equatorial

The two trans-1,2-dimethylcyclopentanes are mirror images.

All four conformations of the two isomers are illustrated. The second conformation of each pair has a high degree of steric strain, and thus each isomer adopts the first conformation. Since only the cis isomer has the hydroxyl group in the necessary axial position, it oxidizes faster than the trans isomer.