1559T_ch04_55-6910/22/0520:19Pa9e69 EQA Solutions to Problems.69 shrinking the volume in space taken up by this substituent.As a better conformation for re Is one in wh ts th on about the arorementone eh away from the axial hydrogens on the same face of the cyclohexane ring: CH CH Diaxial (b)Con rates a ert-b eer.for d by onds to these carbon 109. 5 upon ring-opening.The by lase in drawings.Now that we know what to look for.let's examine conformations of B comparable to those of A drawn above.The diaxial form is just about impossible,there being no available ial H'e ic comp een one o the diequtor nfoer ffromrtobeween methyohi oge occupy the s The molecule escapes this dilemma by adopting a shape based upon the boat shown below,but of the im diate situation is not terribly different from that described in Problem 30.) H 47.(a 48.(d 49.(d) 50.(a)Smallest=most stable (dicquatorial group (such as CH3) causes it to be more sterically demanding than is a comparable molecular fragment contained in a ring. Furthermore, the ring angles in cyclopropanes are only 60°, further shrinking the volume in space taken up by this substituent. As a result, a better conformation for the diequatorial structure is one in which a 120° rotation about the aforementioned bond takes the two methyl groups as far apart as possible and puts the compressed cyclopropane ring on the side of the bond closest to the methyl group at C3. Similar rotation in the diaxial moves the CH3 group away from the axial hydrogens on the same face of the cyclohexane ring: (b) Compound B results from opening of the cyclopropane ring in A. This process generates a 1,1-dimethylethyl (tert-butyl) group in place of the 1-methylcyclopropyl substituent. For purposes of evaluating steric interactions, this new tert-butyl group is effectively much larger, for two reasons. First of all, the rigid OCH2OCH2O fragment of the cyclopropane ring is replaced by two freely spinning CH3 groups. Secondly, the original 60° angle between the bonds to these carbon atoms widens to the full tetrahedral value of 109.5° upon ring-opening. The increase in steric size of the substituent is considerable and not at all communicated adequately by line drawings. Now that we know what to look for, let’s examine conformations of B comparable to those of A drawn above. The diaxial form is just about impossible, there being no available conformation lacking prohibitive steric compression between one of the CH3’s of the tert-butyl group and one or both axial H’s on the same ring face. However, as indicated by the intersecting arcs, the diequatorial conformer suffers from a similar interaction between methyl groups, which is unavoidably present in all staggered conformations of the substituent. Basically, hydrogen atoms on these two CH3’s are being forced to try to occupy the same volume of space: The molecule escapes this dilemma by adopting a shape based upon the boat shown below, but twisted to relieve eclipsing interactions. This unorthodox conformation places the methyl group at C3 in a pseudoaxial position, out of the immediate proximity of the tert-butyl group. (The situation is not terribly different from that described in Problem 30.) 47. (a) 48. (d) 49. (d) 50. (a) Smallest H°comb most stable (diequatorial) CH3 O CH3 CH3 CH3 H H CH3 O O CH3 CH3 CH3 H3C CH3 H H CH3 CH3 Diequatorial Diaxial CH3 O CH3 CH3 Diequatorial CH3 Diaxial O Solutions to Problems • 69 1559T_ch04_55-69 10/22/05 20:19 Page 69