Chapter24ISM - CHAPTER 24 ORGANIC CHEMISTRY Problem...

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CHAPTER 24 ORGANIC CHEMISTRY Problem Categories Conceptual : 24.11, 24.12, 24.13, 24.14, 24.16, 24.18, 24.19, 24.35, 24.40, 24.43, 24.46, 24.53, 24.57, 24.65 Descriptiv e: 24.15, 24.17, 24.20, 24.21, 24.23, 24.37, 24.38, 24.41, 24.45, 24.55, 24.61, 24.62, 24.63, 24.67, 24.68, 24.74. Difficulty Level Easy : 24.16, 24.18, 24.19, 24.21, 24.22, 24.23, 24.24, 24.25, 24.31, 24.36, 24.38, 24.42, 24.44, 24.45, 24.46, 24.50, 24.51, 24.56, 24.58, 24.59, 24.60, 24.66. Medium : 24.12, 24.13, 24.14, 24.15, 24.17, 24.20, 24.26, 24.27, 24.28, 24.32, 24.37, 24.39, 24.41, 24.43, 24.47, 24.48, 24.49, 24.55, 24.57, 24.61, 24.62, 24.64, 24.66, 24.69. Difficult : 24.11, 24.35, 24.40, 24.52, 24.53, 24.54, 24.63, 24.65, 24.67, 24.68. 24.11 The structures are as follows: 24.12 Strategy: For small hydrocarbon molecules (eight or fewer carbons), it is relatively easy to determine the number of structural isomers by trial and error. Solution: We are starting with n -pentane, so we do not need to worry about any branched chain structures. In the chlorination reaction, a Cl atom replaces one H atom. There are three different carbons on which the Cl atom can be placed. Hence, three structural isomers of chloropentane can be derived from n pentane: CH 3 CH 2 CH 2 CH 2 CH 2 Cl CH 3 CH 2 CH 2 CHClCH 3 CH 3 CH 2 CHClCH 2 CH 3 CH 3 2 2 3 2 2 2 3 2 3 3 3 2 3 2 3 3 2 2 3 2 3 3 2 3 2 3 2 C 3 3 3 2 3 2 C 3 3 2 3 3 3 3 3 3 C 3 3 3 3 2 2 3 3 2
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CHAPTER 24: ORGANIC CHEMISTRY 670 24.13 The molecular formula shows the compound is either an alkene or a cycloalkane. (Why?) You can't tell which from the formula. The possible isomers are: The structure in the middle (2 butene) can exist as cis or trans isomers. There are two more isomers. Can you find and draw them? Can you have an isomer with a double bond and a ring? What would the molecular formula be like in that case? 24.14 Both alkenes and cycloalkanes have the general formula C n H 2 n . Let’s start with C 3 H 6 . It could be an alkene or a cycloalkane. Now, let’s replace one H with a Br atom to form C 3 H 5 Br. Four isomers are possible. There is only one isomer for the cycloalkane. Note that all three carbons are equivalent in this structure. 24.15 The straight chain molecules have the highest boiling points and therefore the strongest intermolecular attractions. Theses chains can pack together more closely and efficiently than highly branched, cluster structures. This allows intermolecular forces to operate more effectively and cause stronger attractions. 24.16 (a) This compound could be an alkene or a cycloalkane ; both have the general formula, C n H 2 n . (b) This could be an alkyne with general formula, C n H 2 n 2 . It could also be a hydrocarbon with two double bonds (a diene). It could be a cyclic hydrocarbon with one double bond (a cycloalkene). (c) This must be an alkane ; the formula is of the C n H 2 n + 2 type. (d) This compound could be an alkene or a cycloalkane ; both have the general formula, C n H 2 n . (e) This compound could be an alkyne with one triple bond, or it could be a cyclic alkene (unlikely because of ring strain). 24.17 The two isomers are: H C H H H H H HH C CC H H C H H H H H H C H H H H H H C C C C C H H C CH 3 H C C C H H H H H H C H Br C 3 H C H C 3 H C H H C 2 H C C C H H H H H HC H 3 H C C H 3 C trans H 3 H C C H 3 C cis
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CHAPTER 24: ORGANIC CHEMISTRY 671 A simplified method of presenting the structures is: The cis structure is more crowded and a little less stable. As a result, slightly more heat (energy) would be
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This note was uploaded on 09/24/2011 for the course CHEM 369 taught by Professor Wang during the Spring '11 term at University of Houston.

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Chapter24ISM - CHAPTER 24 ORGANIC CHEMISTRY Problem...

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