Ch9 - J 78 S olutions Chapter 8 HaJoalkanes-Halogenation...

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_Chapter 9: Nucleophilic Substitution and ~-Eliminatjon Solutions 179 CHAPTER 9 Solutions to the Problems ErQblem 9.1 CQmplete t.he fQIlQwing nucleophilic substitution reactiQns. 'In each reaction, shQW all electron pairs Qn both the· nucleQphile and the leavmg grQup, o o eY C1: II .. :0: OCCH 3 II •• _ Q-13C-~: Na+. ethanol (a) + .+ NaCI 'J • (b)k - acetone + ~.~:Na+ +. Nat Br: (c) ~ + + LiBr Problem 9.2 The reaction Qf methyl bromide with azide ion (N3-) in methanQI is a typical SN2 reactiQn. What happens t() the rate of the reactiQn if [N3-] is dQubled? An SN2 reaction is bimolecular, meaning that the rate of the reaction depends on th~ con~entration of both the nucleophile and alkyl halide. Therefore, when the [N3-] is doubled, the rate of the reactIOn also doubles. . Problem 9.3 C~mplete these SN2 reactions, shQwing the configuratiQn Qf each product. In both cases, the stereochemistry at the site of reaction is due to the nucieophile's backside attack that occurs during an S N 2 reaction. - Br - (a) H3~ + Na+N 3 - ~ C--J:;:::::T- ~ + NaBr ~/ if Br + NaBr (b) Na+ I I + CH 3 S - ()~ # # ' Problem 9.4 Write an additiQnal resonance cQntributing structure -for each carbQcation, and state which of the tWQ makes the greater cQntribution to the reSQnance hybrid. A more highly substituted carbocation is more stable, s? the contributing structu~e that has the more highly substituted carbocation will make the greater contributIOn to the resonance hybrId. {a)~cH, _.. ~~- ~CH2 a--. ..... -- +0-- Greater contribution (2° carbocation) (3° carbocation)
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180 Solutions Chapter 9: Nucleophilic Substitution and ~-Elj"mination Problem 9.5 .Write the expected substitution product(s) fot each reaction and predict the mechanism by which each product is formed. + - aceton~ (a) ~r + Na SH The HS- is a go~d nUcleoph~le a~d; since ~he re~ction involves a secondary haloalkane with a good lea\'ing group, the reactIOn mechamsm IS SN2 s~ mverslOn of configuration is observed. . . CI ~ (b) + I+-C-OH '/lv (R)- 2-chloro-butane The haloalkane is se~ondary and chlorid~ is ~ good leaving group. Formic acid is an excellent ionizing solvent and a'poor nucleophIle. Therefore, substItutIOn takes place by an SNI mechanism and leads to a mixture of enantlOmers. o /I .' H o ~H CH 3 ,- ,}'- II S 1 C·,\\\H H-C:-OH ~ CH /' ~ r~CH2CH3 + 3 CH CH formation of carbocation 2 3 HCO II followed by reaction o with formic acid I' Problem 9.6 Predict the f:l-elimination product(s) fon:ned when each chloroalkane is treated with sodium efhoxide in ethanol. If two or more products might be formed, predict which is the major product. . When there is a choice, the mor.e highly substituted alkene will be the major product as predicted by Zaitsev's rule.' , CH3 (a) 'cf a + Major .Product (b) ~CI rY CI (c) ~ '. Problem? 7 1-S:hloro-4-isopropylcyclohexane exists as two stereoisomers: one cis and one trans. Treatmentof either Isomer with sodIUm ethoxide in ethanol gives 4-isopropylcyclohexene by an E2 reaction. .
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This note was uploaded on 07/09/2011 for the course CHE 301 taught by Professor Diver during the Spring '08 term at SUNY Buffalo.

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Ch9 - J 78 S olutions Chapter 8 HaJoalkanes-Halogenation...

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