0
$\begingroup$

Are these structures equivalent?

Image 1:

http://i.minus.com/jnojtCx6wYvqD.png

Image 2:

enter image description here

$\endgroup$
2
$\begingroup$

Yes, the two structures you've drawn are the same; both substituents are in axial positions. The two structures shown in the link you provided are not the same; they are conformational isomers. The one on the left is the same as the one you've drawn with both substituents in the axial position. The one on the right has both substituents in the equatorial position. These two conformational isomers exist in equilibrium with the (e,e) isomer favored. The two conformers interconvert by a cyclohexane ring flip.

$\endgroup$
  • $\begingroup$ Thank you. I was trying to make a comparison between image 1 and image 2; are image 1 and image 2 the same? Because in image 1, the picture has the chlorine with dashes and the ethyl group with a wedge. I however have it switched. So would both be valid representations of 1-chloro-2-ethylcyclohexane? $\endgroup$ – Dissenter May 9 '14 at 14:35
  • $\begingroup$ Image 1 and Image 2 are not the same. Image 1 shows two conformers (the a,a and the e,e conformers) of 1-chloro-2-ethylcyclohexane that interconvert with one another via a cyclohexane ring flip. Image 2 shows just the a,a conformer. For completeness I'll mention that there are more isomers of 1-chloro-2-ethylcyclohexane, namely the a,e and e,a conformers which again interconvert via a cyclohexane ring flip. $\endgroup$ – ron May 9 '14 at 14:51
  • $\begingroup$ Okay! I just wanted to make sure that both images were images of 1-chloro-2-ethylcyclohexane. $\endgroup$ – Dissenter May 9 '14 at 14:52
  • $\begingroup$ Also when you say (e,e) isomer is that referring to the conformational isomer with the equatorial bonds because you say "favored" in equilibrium - and as I understand, placing the substituents in the equatorial positions helps minimize the diaxial interactions/steric strain? Also I see what you mean by how they're non-equivalent; one is favored at equilibrium. $\endgroup$ – Dissenter May 9 '14 at 14:53
  • $\begingroup$ Yes, (e,e) means both substituents are in the equatorial position; (a,a) would have both in the axial position. In equilibria, conformers with substituents in the equatorial position are generally favored over conformers with axial substituents. If you look at a model of cyclohexane you'll see that an axial substituent at position 1 has an unfavorable steric interaction with hydrogens (or substituents) at the axial 3 and 5 positions. $\endgroup$ – ron May 9 '14 at 15:02

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.