Chemistry: Concepts and Applications

Chapter 18: Organic Chemistry

Problem of the Week

 
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Teachers Notes:
Characterizing a hydrocarbon

Problem 1:
a. C3H8 → C3H6 + H2
<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_5.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (7.0K)</a><a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_6.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (7.0K)</a> + H-H

b. C7H16 → C3H8 + C4H8

<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_7.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (11.0K)</a><a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_5.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (7.0K)</a> + <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_8.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (8.0K)</a>

c. C10H22 → C5H12 + C5H10

<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_9.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (14.0K)</a><a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_10.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (9.0K)</a> + <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_11.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (9.0K)</a>

Problem 2:
1. Location of the double bond is one possible isomer. These could include 1-dodecene, 2-dodecene, 3-dodecene, 4-dodecene, 5-dodecene, and 6-dodecene.

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(Hydrogen omitted to simplify the drawing)

2. Branched compounds are a second type of isomer.

3,4 dimethyl, 2-decene       <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_14.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (6.0K)</a>

3. Geometric isomers include attached groups on the carbons the double bond.

3,4 diethyl-3-octene (trans)       <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_15.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (7.0K)</a>

3,4 diethyl-3-octene (cis)          <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=jpg::::/sites/dl/free/0078617987/179001/ch22_16.jpg','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (8.0K)</a>>

or trans-6-dodecene and cis-6-dodecene

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