Chapter 10 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 10 The Chemistry of Alcohols and Thiols Solutions to In-Text Problems 10.1 (b) 10.2 The OH group of the alcohol is protonated in a Brønsted acid–base reaction to form the conjugate acid of the alcohol. This loses water to form a carbocation in a Lewis acid–base dissociation reaction. Finally, in a Brønsted acid–base reaction, water acts as a Brønsted base to remove a b -proton from the carbocation, which acts as a Brønsted acid, to give the alkene. The formation of product A by removal of proton (a) is shown here; the formation of products B and C occurs in an analogous manner by removal of b -protons (b) and (c) , respectively. 10.5 (b) Both 3-methyl-3-pentanol and 3-methyl-2-pentanol should give 3-methyl-2-pentene as the major product. The tertiary alcohol 3-methyl-3-pentanol should dehydrate more rapidly. 10.6 (a) Protonation of the OH group and loss of water as shown in several of the previous solutions, as well as in Eqs. 10.3a–b on p. 437 of the text, gives a secondary carbocation. As the text discussion of Eq. 10.6 suggests,
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 10 2 the mechanism involves a rearrangement of the initially formed secondary carbocation to a more stable tertiary carbocation. Loss of the two possible b -protons gives the two alkene products. (b) Rearrangement occurs because a more strained secondary carbocation is converted into a less strained, and therefore more stable, secondary carbocation. 10.8 (b) 10.9 This reaction involves a carbocation rearrangement. We use HBr as the acid, although, because water is generated as a product, H 3 O + could also be used. The product in part (c) results from a carbocation rearrangement. 10.10 (b) The product is I—CH 2 CH 2 CH 2 —I. (d) The compound, neopentyl alcohol, is a primary alkyl halide and cannot react by the S N 1 mechanism; and it has too many b -substituents to react by the S N 2 mechanism. Consequently, there is no reaction.
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 10 3 10.11 (b) (d) 10.12 (b) 10.13 (b) Although a polar aprotic solvent would accelerate the last step, it would probably work in an alcohol solvent. The nucleophile, CH 3 S , can be generated by allowing the thiol CH 3 —SH to react with one equivalent of sodium ethoxide in ethanol. 10.14 (a) Cyanide ion displaces the tosylate ester formed in the first step. 10.15 (b) 10.16 (b) (c)
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 10 4 10.18 (b) Eq. 10.26 of the text shows that the nucleophilic reaction of the bromide ion on the reactive intermediate occurs by a concerted (S N 2) substitution reaction. Therefore, the reaction should occur with inversion of stereochemistry, and the product would then be ( S )-2-bromopentane. The S N 2 reaction occurs at an acceptable rate on a secondary carbon in the absence of b substituents. In addition, the reaction is fast because the leaving group is a
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Chapter 10 - Instructor Supplemental Solutions to Problems...

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