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- Thread starter arnav9
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- #2

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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)

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- #4

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its not that tough really, its just an "n choose r" type of problem

- #5

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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.arnav9 said:

The Bob (2004 ©)

- #6

LeonhardEuler

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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:quetzalcoatl9 said:its not that tough really, its just an "n choose r" type of problem

11111111CH3

11111111CH2

111CH3-*C-CH=CH2

11111111H

111111%1CH3

111111%1CH2

1CH=CH2*C-CH3

11111%11H

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