13-3p545 - 13-3p545.wpd Page 1 of 6 Summary: P545 #1-4;...

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13-3p545.wpd Page 1 of 6 Summary: P545 #1-4; P549 #5; P551 #6-7; P557 #8-14; P560 #15,16; P563 #17,18; P567 #4-8 P545 71 6 1) C H 92 0 2) C H 3) 10 4) 25 P549 5) a) 2-methylbutane b) 2,2-dimethylpropane c) 2,5-dimethyl-3-ethylheptane d) 2,2,4,4-tetramethylhexane e) 4-propyl-2,2,4-trimethyloctane P551 6) a) b) c) d) 7) a) Longest chain is 6 carbons, not 5 2,4-dimethylhexane b) Hexane chain is numbered from the wrong end 2,3-dimethylhexane
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13-3p545.wpd Page 2 of 6 c) Longest chaing is 6 carbons, not 5. 3-ethyl-3-methylhexane P557 8) a) 3-hexene b) 3-propyl-2-heptene c) 2,3-dimethyl-4-ethyl-4-octene 9) a) b) 10) 1-butene 2-butene 2-methyl-1-propene 11) cis-2-pentene trans-2-pentene 12) For a double bond to be able to exist in a cis or trans isomeric form, it must have two different ‘things’ attached to each of its double bond carbons. In the case of carbon #1 of 1-butene, this carbon is attached to two hydrogen atoms, not two different things. It is impossible therefore for it to have a cis or trans isomeric forms.
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This note was uploaded on 04/09/2010 for the course SCIENCE CHEM taught by Professor M during the Spring '09 term at McMaster University.

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13-3p545 - 13-3p545.wpd Page 1 of 6 Summary: P545 #1-4;...

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