Chapter 11 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 11 The Chemistry of Ethers, Epoxides, Glycols, and Sulfides Solutions to In-Text Problems 11.1 (b) A thiolate ion, formed by reaction of NaOH with the thiol, is alkylated by allyl chloride to give H 2 C A CH—CH 2 —S—CH 3 (allyl methyl sulfide) + Na + Cl . NaOH is a strong enough base to form the thiolate anion, but would not be strong enough to form the conjugate base of an alcohol. (d) Neopentyl halides and sulfonate esters do not undergo S N 2 reactions at room temperature; furthermore, they do not undergo b -elimination because there are no b - hydrogens. Thus, no reaction occurs. 11.2 (b) The alternative synthesis, reaction of the methanethiolate ion with isopropyl bromide, is less desirable because secondary alkyl halides react more slowly than methyl halides and give some elimination products. 11.5 Because each carbon of the alkene double bond has one alkene substituent, there is no strong preference for the reaction of methanol at either carbon of the resulting mercurinium ion. Consequently, two constitutional isomers of the product ether are formed. (The same result would be obtained regardless of the stereochemistry of the alkene starting material.) 11.6 (b) Isobutylene (2-methylpropene) is subjected to alkoxymercuration in isobutyl alcohol, and the resulting organomercury compound is reduced with NaBH 4 . 11.7 If we start with two primary alcohols, ROH and R´OH, we would expect them to have similar basicities and similar nucleophilicities. Each alcohol could react with each protonated alcohol. Consequently, three possible products would be formed: R—O—R, R´—O—R, and R´—O—R´. None of the alcohols would be formed in very high yield, and separation of the products could be quite laborious. The reason we can let a tertiary alcohol react with a primary alcohol to give an unsymmetrical ether is that the tertiary alcohol forms a carbocation in acidic solution much faster than either it or the primary ether react by the S N 2 mechanism; and, once the carbocation is formed, it is rapidly consumed by its Lewis acid–base association reaction with the large excess of primary alcohol that is present. 11.8 (b) The carbocation formed from the tertiary alcohol reacts with ethanol to give the following ether:
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 11 2 11.9 (b) Use the same approach as in part (a): 11.10 (b) Because this is an ether with one tertiary alkyl group and one methyl group, it can be prepared by dehydration of the tertiary alcohol in the presence of methanol, or by acid-catalyzed addition of methanol to either of two possible alkenes. (d)
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This note was uploaded on 12/30/2011 for the course CHEM CHEM231 taught by Professor Dixon during the Fall '09 term at Maryland.

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Chapter 11 - Instructor Supplemental Solutions to Problems...

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