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Finding possible isomers

  1. Oct 14, 2003 #1
    I was given a molecular formula C7H16 and asked to find all possible isomers. I know there are nine because the question said it. But I am just wondering how to find all isomers correctly without repeating myself. Is there a way to do this mathematically or a general rule that makes it easier?
     
  2. jcsd
  3. Oct 14, 2003 #2
    hmm i dont really know if there's a mathematical formula (probably would be a waste of time), but if you think you're repeating yourself just try flipping the structure in your head, then try to match it with any of the structures you have already written down, if none match it's a new isomer :smile:. you could also number the carbons as you would if you were naming the molecule, but if you haven't covered that yet then just stick with trying to visualize it.
     
  4. Oct 14, 2003 #3
    There's a lot more than nine.
     
  5. Oct 14, 2003 #4

    Monique

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    Notice that C7H16 is Cn H2n +2 ?

    That means it can only be a straight chain of atoms (not a cycloalkene) and it may not be unsaturated. The rest is for you to figure out. I don't know of any formulas :)


    Basically start with heptane, and start taking of a C atom from the end and place it somewhere else, making sure you don't make a mirror image. Every time you take one off and finally you should come to 9 different forms.
     
    Last edited: Oct 14, 2003
  6. Oct 14, 2003 #5

    Monique

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    There are only nine!
     
  7. Oct 14, 2003 #6

    Monique

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    Maybe you can figure out a formule from the following data:

    Code (Text):

    Pentane C5H12  3 isomers
    Hexane  C6H14  5 isomers
    Heptane C7H16  9 isomers
    Octane  C8H18  18 isomers
    Nonane  C9H20  35 isomers
    Decane  C10H22 75 isomers
     
  8. Oct 14, 2003 #7
    Whoops. Sorry. Just saw the C7 and not the H16.
     
  9. Oct 14, 2003 #8
    how come they don't count optical isomers? if they did C7H16 would have 11 isomers correct?
     
  10. Oct 15, 2003 #9

    Monique

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    There are no optical isomers, since there are no double bonds in the molecule.. molecules are able to rotate freely around their axes. *edit* I guess I am talking about geometric isomers (cis/trans)

    Well, no optical isomers either (L/D), since you would need different sidechains right? You cannot have a chimeric atom when all groups are identical.. I guess?

    I drew all 9 of them out.. in order for a C atom to be chiral, it needs to have 4 groups attached.. there are a few.. no, no enantiomers.
     
    Last edited: Oct 15, 2003
  11. Oct 15, 2003 #10
    butyl, ethyl, methyl, and hydrogen. Four different groups. You CAN have more than nine isomers, if you count stereoisomers and not just structural isomers.
     
  12. Oct 15, 2003 #11

    Monique

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  13. Oct 15, 2003 #12
    3-methylhexane and 2,3-dimethylpentane each have a chiral carbon.
     
  14. Oct 15, 2003 #13

    Monique

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    You are correct, my err!
     
  15. Oct 22, 2003 #14
    Erm...

    The isomers don't have to be straight chained, (if they did then there would be only one isomer), and since it is in the general formula CnH2n+2 then all of the isomers will be saturated.
    Sorry for being so nit picky.

    Anyway, the best way to spot isomers at a glance is to do many many examples. At least that's what my chemistry teacher made me do. It's also best if you learn to spot them now, that way when you get on to cis/trans and then other more complicated molecules (with benzene rings in, cis/trans and some halides thrown in for fun) you can spot the possibilities a lot easier.
     
  16. Oct 22, 2003 #15
    He means acyclic.
     
  17. Oct 23, 2003 #16
    That makes more sense. I blame the fact that I was tired.
     
  18. Nov 11, 2003 #17
    2-ethylpentane

    Correct me if i´m wrong please, but isn´t there a 2-ethylpentane isomer?




    Rui.
     
  19. Nov 11, 2003 #18
    Re: 2-ethylpentane

    You mean 3-methylhexane?
     
  20. Nov 11, 2003 #19
    No, i mean 2-ethylpentane.



    Rui.
     
  21. Nov 11, 2003 #20
    Code (Text):


      C
      |
      C
      |
    C-C-C-C-C

     
    The above is the molecule that you (RuiMonteiro) named (2-ethylpentane).

    It just so happens that it is also the same as this molecule (3-methylhexane).

    Code (Text):


      C
      |
    C-C-C-C-C-C

     
    Chemists prefer to name the molecule after the longest chaain in the molecule (ie: hexane) and so it's proper name is 3-methylhexane, although 2-ethylpentane is not technically wrong.
     
    Last edited: Nov 11, 2003
  22. Nov 11, 2003 #21
    That´s right, i knew i was missing something cause nobody talked about that isomer lol, thanks for clarifying. :)




    Rui.
     
  23. Nov 12, 2003 #22

    ShawnD

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    The simplest way to find all the isomers is to draw the the possible ways to arrange it as a straight chain, then ways to arrange with a methyl, then ways with several methyls, then an ethyl, then several ethyls and so on. I think 1 ethyl is as high as she goes.
     
  24. Nov 21, 2003 #23
    With any hyrdrocarbon there is a way to narrow down your possible isomers. I always use this formula to start a problem of this kind. This formula is used to find the unsaturation of the molecule which in turn can help you figure out how many rings, double or triple bonds are in the structure.

    (2C+2-H-X+N)/2

    You just multiply the number of Carbons by 2 add 2 subtract the number of hydrogens and halides (X= Cl, I, Br) and add the number of nitrogens. Divide all of that by 2. NOTE: the number of oxygens does not factor into the unsaturation

    Here's an example of how to find unsaturation:

    C6H9Cl

    (2(6) + 2 - 9 - 1)/2 = 2

    How to interpret your results:
    An unsaturation of o indicates no unsaturation and therefore the molecule has no rings, or double or triple bonds
    An unsaturation of 1 indicates that there is either a ring or a double bond
    An unsaturation of 2 indicates there is either a ring and a double bond, 2 double bonds or a triple bond
    etc...

    We now know that the example will have isomers with 1 triple bond, 2 double bonds, and ones with a ring and a double bond.

    This formula is very useful and a great tool for organic chemistry.
     
    Last edited: Nov 21, 2003
  25. Nov 21, 2003 #24
    I'm always wondering how chemists figered out this formula. [?]
     
  26. Nov 21, 2003 #25
    Consider a straight chain hydrocarbon. Each carbon has two hydrogens, except for the ones at the end, which each have an additional hydrogen. So the number of H is 2n+2, where n is the number of carbon atoms, follow? You can than arrange it in anyway you want and it will still work. Halogen atoms will take up the place of a hydrogen atom. So bromopropane has three carbons, and one bromine, so it would have 8 hydrogens just given the carbons but it really has only seven, due to the bromine, so you subract the number of halogens (-X). Oxygen won't affect it at all, it just spaces out the molecule, it's kind of like a filler. Nitrogen is kind of like oxygen, except it's got one more bond for a hydrogen (or something else as the case may be), so you add a hydrogen for every nitrogen. So your final equation is 2n+2-X+N for how many hydrogens you should have. This, minus actual number of hydrogens you actually have, divided by two (each pi bond takes the place of two hydrogens, or the ring forming bond takes the place of those two protons at the end of the chain) gives the degree of unsaturation.
     
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