Ch8 - 162 S olutions Chapter 7 Alkynes siahBH THF OC K2 C03...

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163 162 Solutions Chapter 7: Alkynes (siahBH, THF, O°C K 2 C0 3 , H 2 0 2 , H 2 0 (b) H (c) Hz . o Lindlar Catalyst CH 3 cis alkene Q.. o t in product o ~ o Chapter 8: Haloalkanes, Halogenation, and Radical Reactions. Solutions CHAPTER 8 Solutions to the Problems Problem 8.1 Write the IUPAC name, !lnd where possible, the common name of each compound. Show stereochemistry where relevant. (a) ~CI (b) ~I (C)~CI l-Chloro-2-methylpropane (Z)-2-Chloro-2-butene ChlorocycIohexane (Isobutyl chloride) N CI (d) 2-Chloro-l,3-butadiene Problem 8.2 Name and draw structural formulas for all monochlorination products formed by treatment of 2- methylpropane with C1 . Predict the major product based on the regioselectivity of the reaction of CI 2 with alkanes. 2 CH 3 + CI heat HCI 1 -----'. . - monochloroalkanes + 2 CH CHCH 3 or light 3 CH 3 I " .CH 3 CCH 3 I CI 2-Chloro-2.methY,lpropane l-Chloro-2-methylpropane There is 1 tertiary hydrogen atoms and 9 primary hydrogen atoms on the molecule. The ratio of reactivity for 3°: 1° chlorination is 5: 1. Therefore, the predominant product wiU be the l-chloro-2-methylpropane, formed in approximately: . 9xl x100 = (64%] (9 x 1) + (1 x 5) Problem 8 3 Using tables of bond dissociation enthalpies in Appendix 3, calculate tJ.Ho for bromination of propane to give l-bromopropane' and hydrogen bromide. . .' HBr Mf o equals the difference between the bond dissociation enthalpies of bonds made vs. bonds broken in the reaction. One C-H bond [+ 422(101),kj(kcal)/mol)] and one Br-Br bond [+ 192(46) kj(kcal)/mol)] were broken while one C-Br [-301(-72) kj(kcal)/mol)] bond and one H-Br bond [-368(-88) kj(kcal)/mol)] were made in this reaction. Thus, for the complete reactiop MfO = 422(101) + 192(46) - 301(-72)·368(-88) =- 55(-13) kHkcaJ)/moJ. Problem 8.4 Write a pair of chain propagation steps for the radical bromination of propane to give I-bromopropane, and calculate MfO for each propagation step and for the overall reaction. Following is a pair of chain propagation steps for this reaction. Of these steps, the first involving hydrogen abstraction, has the higher activation energy. Manta ray Manta birostJjs Bora Bora, French Polynesia
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164 Solutions Chapter 8: HaJoalkanes, Halogenation, and Radical Reactions MlO kJ/mol " (kcal/mol) C~-C~-CH3 +"Br- CH3-:-C~-CH2 + H-Br +422 ·368 +S4 (+101) (·88) (+13) Br . I CH 3 -CH 2 ---':CH 2 + Br2 - CH 3 2 2 + " Br +1?2 -301 -109 (+46) P2) (-26) fr r=ssl CH. ..2. ...C~-CH + Br2. ~ CH 3 2 2 H-Br 3. 3 + ~ Problem 8 5 Given the solution to Example 8.5, predict the structure of the product(s) formed when'3-hexene is treated with NBS? .. +" " CH 3 CH=CHCHCH 2 CH 3 + HBr Br I. I. CH 3 CHCH=CHCH 2 CH 3 CH 3 2 CH 3 Both ?f the above products have a chiral center,so they will each be produced as a racemic mixture·of enantlOmers. . . Problem 8.6 Show the products of the following reaction and indicate the major one. The reaction involves ~he initial forr~ation of a .resonance-stabilized allylic radical intermediate which reacts with O 2 The compound WIth tl1e more hIghly substItuted double bond will be the major product.
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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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Ch8 - 162 S olutions Chapter 7 Alkynes siahBH THF OC K2 C03...

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