It's hard to draw structures here, but I'll try naming a few. First thing you have to remember is that it's got two degrees of unsaturation, which could be rings or pi-bonds.
As a straight chain, it could be 1-pentyne, 2-pentyne, 1,2-pentadiene, 1,3-pentadiene, 1,4-pentadiene, or 2,3-pentadiene.
As a branched chain, it could be 3-methyl-1-butyne or 2-methyl-1,3-butadiene.
As a ring, it could be cyclopentene. Smaller rings are really too unstable, especially with a double-bond.
2006-12-10 03:14:58
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answer #1
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answered by Amy F 5
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here are a few
structure 1
Arrange 5 carbons in a pentagon. Make a C=C bond between 2 carbons and place one H on each of those carbons. Add 2 H's to each of the 3 remaining carbons.
put H's attached to carbons on formulas below making sure there are 4 bonds to each carbon and H only forms one single bond.
structure 2
H₂C=CH-CH=CH-CH₃
structure 3
H₂C=CH-CH₂-CH=CH₂
structure 4
HC≡C-CH₂-CH₂-CH₃
structure 5
H₃C-C≡C-CH₂-CH₃
and any other isomers of above
2006-12-10 04:09:48
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answer #2
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answered by rm 3
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