isomers of C6H10 and C6H12


by arnav9
Tags: c6h10, c6h12, isomers
arnav9
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#1
Aug4-05, 07:39 PM
P: 3
could someone please tell me how many isomers there are for C6H10 and C6H12, even better could u tell me what they are, I was trying to get them but i keep getting too many repeats
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gravenewworld
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#2
Aug5-05, 06:56 AM
P: 1,389
For n number of carbons you need 2n+2 hydrogens for a carbon skeleton with no double or triple bonds anywhere. so 6 carbons means you need 14 hydrogens, you have 10 which means either a) you have 1 triple bond in there somewhere or b) you must have 2 double bonds. The easiest way to do this is by doing all the compounds with triple bond first and then with the 2 double bond. Right off the top of my head, I do not know the amount of isomers possible. There are ways to find out how many isomers there are for a given formula for a organic compound with just carbons and hydrogens but it is an extremely tough math problem involving graph theory and combinatorics. (c6h12 means you have only 1 double bond of course)
Artermis
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#3
Aug5-05, 11:33 AM
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P: 59
Yes the math for it is complicated I believe, but there is some complex method of figuring out how many isomers there are; best way is just to draw them out :P Think "primary", "secondary" and "tertiary"... It wouldn't help you if we just did it for you because then you wouldn't learn much :P

quetzalcoatl9
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#4
Aug6-05, 01:49 PM
P: 701

isomers of C6H10 and C6H12


its not that tough really, its just an "n choose r" type of problem
The Bob
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#5
Aug6-05, 02:43 PM
P: 1,116
Quote Quote by arnav9
could someone please tell me how many isomers there are for C6H10 and C6H12, even better could u tell me what they are, I was trying to get them but i keep getting too many repeats
I would suggest you list the ones you do have here so that we can see what is missing and then help you to complete your list of isomers.

The Bob (2004 )
LeonhardEuler
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#6
Aug6-05, 03:00 PM
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P: 864
Quote Quote by quetzalcoatl9
its not that tough really, its just an "n choose r" type of problem
It actually is a quite complicated problem in general. You have to take into account symetry to eliminate repititions and you also have to consider the steriochemistry of the situation to find all of the enantiomers. In this case the simplest solution is to list all of the isomers rather than finding a general formula. Even this relatively small hydrocarbon will have one pair of enantiomers:

11111111CH3
11111111CH2
111CH3-*C-CH=CH2
11111111H

111111%1CH3
111111%1CH2
1CH=CH2*C-CH3
11111%11H
so-crates
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#7
Aug7-05, 03:08 PM
P: 309
The way I did it for hydrocarbons was to consider how many isomers there were given a certain number of primary, secondary, tertiary and quaternary carbons. This helped me to limit the confusion somewhat when the number of carbons gets large. And as Euler said you have to account for steriochiometry as well for each isomer with a stereogenic carbon.


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