Now that you have identified the more stable chair conformations of compounds A and B, identify which compound is expected to be converted into an epoxide more rapidly upon treatment with NaOH. Compound A reacts more rapidly because an axial nucleophile and an axial leaving group are needed for backside attack. O Compound A reacts more rapidly because an axial leaving group more readily leaves to form a carbocation. O Compound A reacts more rapidly because the axial alcohol can undergo hydrogen bonding with the tert-butyl group. O Compound Breacts more rapidly because an equatorial leaving group more readily leaves to form a carbocation. O Compound Breacts more rapidly because the nucleophile and leaving group are both in the more stable equatorial position. O Compound B reacts more rapidly because an equatorial nucleophile and an equatorial leaving group are needed for backside attack. Consider the following two compounds. When treated with NaOH, one of these compounds forms an epoxide quite rapidly, while the other forms an epoxide very slowly. Identify which compound reacts more rapidly and explain the difference in rate between the two reactions. (Hint: You may find it helpful to review the conformations of substituted cyclohexanes in Section 4.13.) OH OH Br Br Compound A Compound B