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Chapter 38 Suggested solutions for Chapter 38 343 Problem 12 These related spirocyclic compounds give different naphthalenes when treated with sodium borohydride or with 5M HCI. Each reaction starts with a different fragmentation. Give mechanisms for the reactions and explain why the fragmentations are different. Treatment of the starting ketone with instead of NaBH₄ gives the alcohol below without fragmentation. Comment on the difference between the two reagents and the stereo- chemistry of the alcohol. 0 Ph NMe₂ Ph Me N NaBH₄ Ho Me Ph N Ph N Me 5M HCI OH Purpose of the problem A subtle effect of stereochemistry and reagent combined on two fragmentation pathways for the same compound. Suggested solution Sodium borohydride is a weak reducing agent and fragmentation (push by N and pull by C=O) is faster. The borohydride reduces the iminium salt to a methyl group and, at the other end of the molecule, the product is the 'keto' form of a phenol. 0 X H Ph OH Ph Ph Me Me N Me N N H CH₂ CH₃ N BH₃ The second reaction starts with the rapid reduction of the ketone by the more powerful reducing agent Hydride is added from the less hindered face of the ketone opposite the phenyl group. Then, on treatment with acid, the alcohol does fragment but by a different pathway. The amine can no longer participate as it is protonated under the reaction conditions. It may look awkward to do the fragmentation with two separate cations as intermediates but both are stable and the stereochemistry is all wrong for anything concerted. OH Ph Ph Me Me N Me N N Ph H H