Some Cycloalkanes Have Angle Strain

Convert the following Newman projections to skeletal structures and name them. 

Using Newman projections, draw the most stable conformer for each of the following  

In addition to the angle strain of the C—C bonds, all the adjacent C—H bonds in cyclopropane are eclipsed rather than staggered, making it even more unstable. 

If cyclobutane were planar, the bond angles would have to be compressed from 109.5° to 90°. Planar cyclobutane would therefore have less angle strain than cyclopropane because the bond angles in cyclobutane would be only 19.5° (not 49.5°) less than the ideal bond angle. It would, however, have eight pairs of eclipsed hydrogens, compared with six pairs in cyclopropane. Because of the eclipsed hydrogens, we will see that cyclobutane is not planar. 

The bond angles in a regular polygon with n sides are equal to 

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If a cycloalkane requires bond angles other than 109.5°, the orbitals of its carbon-carbon bonds cannot achieve optimum overlap, and the cycloalkane must have some angle strain (sometimes called Baeyer strain) associated with it. Figure 3-14 shows that a planar cyclobutane, with 90° bond angles, is expected to have significant angle strain. 

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