instructor_supplement_17

instructor_supplement_17 - Instructor Supplemental...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 17 Allylic and Benzylic Reactivity Solutions to In-Text Problems 17.1 (b) The allylic carbons are indicated with an asterisk (*). 17.2 (b) The benzylic carbons are indicated with an asterisk (*). 17.3 (b) The reactivity order is (2) < (3) < (1). The S N 1 reaction of compound (2) is slowest because the polar effect of the meta-chloro substituent destabilizes the intermediate carbocation. The reaction of compound (3) is faster because the resonance effect of the para -chloro group partially offsets its polar effect. Compound (1) reacts most rapidly because the carbocation intermediate is not destabilized by the deactivating polar effect of a chloro substituent, which outweighs its resonance effect. 17.5 The carbocation formed when trityl chloride ionizes, the trityl cation (Ph 3 C + ), is stabilized by delocalization of electrons from all three phenyl rings. This carbocation has more resonance structures than the carbocations formed from the other alkyl halides in the table, and is thus so stable that the transition state leading to its formation also has very low energy; consequently, it is formed very rapidly. 17.6 The number of products depends on (1) whether all of benzylic or allylic positions are equivalent, and (2) whether the resonance structures of the free-radical intermediate are identical.
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 17 2 (a) All allylic positions of cyclohexene are chemically equivalent, and the two resonance structures are identical. Hence, only one allylic bromination product is possible. (c) Trans- 2-pentene has two distinguishable allylic positions. In each of the allylic radicals that results from hydrogen abstraction at the two positions, the unpaired electron is delocalized to two different carbons. (See Study Problem 17.1.) The abstraction of H a leads to two products, but abstraction of H b leads to one product, because the resonance structures of the intermediate free radical are identical (assuming that the double bond retains its trans stereochemistry).
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3 (e) A benzylic hydrogen is abstracted from the isopropyl group rather than a hydrogen of the two methyl groups because a more stable benzylic free-radical intermediate is obtained. 17.7 (b) Because the two Grignard reagents in rapid equilibrium are identical, only one product is obtained: 17.8 (b) The benzylic proton is abstracted; b -elimination gives a vinylic ether. 17.10
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instructor_supplement_17 - Instructor Supplemental...

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