Chapter 22 Even Problem Solutions

Chapter 22 Even Problem Solutions - CHAPTER TWENTY-TWO...

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CHAPTER TWENTY-TWO ORGANIC CHEMISTRY Hydrocarbons 2. There is only one consecutive chain of C-atoms in the molecule. They are not all in a true straight line since the bond angles at each carbon atom are the tetrahedral angles of 109.5°. 4. i. CH 3 – CH 2 – CH 2 – CH 2 – CH 2 – CH 3 hexane ii. 2- methylpentane iii. 3-methylpentane iv. 2,2 – dimethylbutane v. 2,3-dimethylbutane All other possibilities are identical to one of these five compounds.
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6. 8. a. isopropylcyclobutane; C 7 H 14 b. 1-tert-butyl-3-methylcyclopentane; C 10 H 22 c. 1,3-dimethyl-2-propylcyclohexane; C 11 H 22 10. a. CH 3 – CH 2 – CH ═ CH – CH 2 – CH 3 b.CH 3 – CH ═ CH – CH CH – CH 2 CH 3 12. a. b. c.
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d. Isomerism 14. resonance: All atoms are in the same position. Only the position of π electrons are different. isomerism: Atoms are in different locations in space. Isomers are distinctly different substances. Resonance is the use of more than one Lewis structure to describe the bonding in a single compound. Resonance structures are not isomers. 16. CH 2 Cl – CH 2 Cl, 1-2-dichloroethane; There is free rotation about the C– C single bond which doesn't lead to different compounds. CHCl CHCl, 1,2dichloroethene; There is no rotation about the C C double bond. This creates the cis and trans isomers which are different compounds. 18. In Exercise 22.10, 3-hexene,2,4-heptadiene,and 2-methyl-3-octene meet the requirements for cis-trans isomerism as outlined in Exercise 22.17. Only 4-methyl-1-pentyne does not exhibit cis-trans isomerism. Because of the triple bond in alkynes, the carbons with the restricted rotation only have one group bonded to them (not two groups as is a necessity for geometric isomerism). See the structure below. 4-methyl-1-pentyne 20.
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22.
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Note: 1-bromo-1-chlorocyclopropane, cis-1-bromo-2-chlorocyclopropane and trans-1- bromo-2chlorocyclopropane are the ring structures that are isomers of bromochloropropene. We did not include the ring structures in the answer since their base name is not bromochloropropene. 26. a. There are three different types of hydrogens in n- pentane (see asterisks). Thus there are three mono-chloro isomers of n-pentane (1- chloropentane, 2-chloropentane and 3- chloropentane). b. There are four different types of hydrogens in 2-methylbutane, so four monochloro isomers of 2-methylbutane are possible. c. There are three different types of hydrogens, so three monochloro isomers are possible. d. There are three trichlorobenzenes (1,2,3-trichlorobenzene, 1,2,4-trichlorobenzene and 1,3,5- trichlorobenzene).
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28. There are many possibilities for isomers. Any structure with four chlorines in any four of the numbered positions would be an isomer, i.e., 1, 2, 3, 4-tetrachloro-dibenzo-p-dioxin is a possible isomer. Functional Groups
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This note was uploaded on 04/07/2008 for the course CHEM 027 taught by Professor Mccallum during the Spring '08 term at Pacific.

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Chapter 22 Even Problem Solutions - CHAPTER TWENTY-TWO...

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