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178 Chapter 9 FURTHER REACTIONS OF ALCOHOLS AND THE CHEMISTRY OF ETHERS (a) CH₃CH₂CHCH₂CH₃ The substrate is a primary haloalkane: SN2 with any but tertiary alkoxides (this one is secondary). This reaction is a :0: good Williamson ether synthesis. (b) Reverse the roles and the substrate is now a secondary haloalkane: SN2 fails with alkoxides H because they are strong bases. E2 prevails, and the desired ether is no longer the major product. (c) CH₃ Yes, I see that the alkoxide is tertiary and would give E2 with almost 0:- any haloalkane. Except methyl! Iodomethane has only one carbon. It cannot eliminate (where would you put the double bond!?) and CH₃ instead simply follows the SN2 pathway, its only option. (d) CH₃CHCH₃ (e) H Secondary haloalkane + alkoxide; H 0: you know the drill. E2, same as (b). :0: H H Essentially the same as (a). CH₃ (f) C Yes, the substrate is primary. But now the alkoxide is tertiary. E2 results (except when structurally impossible, as was the case in (c) above). 46. (b) Reaction is given in Problem 44(a). (e) Use conditions: chlorocyclohexane in neutral cyclohexanol (solvolysis, without a basic nucleophile). (f) Use another solvolysis, this time with the 3° haloalkane. CH₃ CH₃ C CH₂CH₃ 47. (a) An oxacyclopropane forms, via an internal displacement of bromide by the alkoxide: OH 0:- O Br Br