1
$\begingroup$

Are these all valid isomers for the formula $\ce{C_3H_6O}$? I drew the first five; my friend came up with the last three, which have double bonds. Are these all isomers or different? More importantly, are there any more?

Also in general, how can one come up with all the isomers for a molecule? I heard of graph theory, but I think that's a little bit too advanced for me. I guess the only way is to draw out as many as you can and name them to weed out redundant structures?

enter image description here

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ The figures are for $C_3H_6O$, not $C_3H_8O$. Also, the last configuration from your friend is not very stable. $\endgroup$
    – LDC3
    May 18 '14 at 21:28
  • $\begingroup$ Thank you! Corrected in text. Ah, stability. Let me see if I can figure this out. I don't see any Texas carbons; I don't see any non-zero formal charges. So I'm guessing it's less about electronic factors as it is about steric/geometric factors. The rightmost carbon is tetrahedral; the center two carbons are both trigonal planar_ and we have a hydroxyl group hanging off; I'm guessing there's some kind of steric hindrance? $\endgroup$
    – Dissenter
    May 18 '14 at 21:38
  • $\begingroup$ You can find out about enols here. en.wikipedia.org/wiki/Enol $\endgroup$
    – LDC3
    May 18 '14 at 21:49
  • $\begingroup$ Oops, back to electronic effects; the electron withdrawing hydroxyl group makes a carbon more nucleophilic right? By concentrating electron density in the C = C? $\endgroup$
    – Dissenter
    May 18 '14 at 21:53
  • $\begingroup$ You are missing acetone-enol $\endgroup$
    – Lighthart
    Jan 4 '15 at 0:47
1
$\begingroup$

They are all constitutional isomers. If you are counting enols (your friends #3, it is the enol of your #1), then include the enol of acetone as an additional constitutional isomer.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.