Which option best depicts trimethylcyclohexane shown in the Haworth projection below? Finding which Haworth projection best depicts trimethylcyclohexane

I thought B because

  • All the methyl groups are equatorial and equatorial groups are more stable.
  • All the methyl groups look as spread out as much as possible, reducing steric repulsion.

But the provided solution states that-

Although A,B and D could all be said to represent the chair form, the best depiction will show the methyl groups repelling one another as far as possible. The arrangement in A with two equatorial and one axial methyl groups makes A the best depiction.

Also, when converting Haworth to Chair, do we have to match the upward facing groups on the Haworth to upward facing positions on the chair?

i.e. the Haworth has 2 upward methyl groups and 1 downward facing methyl. But in A, the left equatorial one is facing upward (as axial in that position is downward), the middle axial methyl is upward, and the right equatorial methyl is downward (as axial in that position is upward). This doesn't match the Haworth.


While you are correct that B should be the most stable conformer of trimethylcylcohexane, the question asked which one best represents the one shown in the Haworth projection. Answer choice B has R stereochemistry on carbon B (see below), while carbon B has S stereochemistry in the original projection. As such, the initially drawn compound is different from the answer choice B. On the other hand, answer choice A has the correct stereochemistry, and also has the maximum number of possible methyl groups in equatorial positions, causing it to be the best depiction of the Haworth projection.

Note: Instead of finding the absolute stereochemistry by using R/S notation, you could also notice that the Haworth projection and answer choice B have different relative stereo chemistry. In the original molecule, carbons B and C are cis, while in answer choice B, these carbons are trans.

Question with labeled carbons

  • 3
    $\begingroup$ As far as this question is concerned, one doesn't even need to look at the absolute stereochemistry; B and C have the methyl groups on entirely different carbons, and between A and D, one only needs to consider the relative stereochemistry (cis or trans) between the two vicinal methyl groups. You're absolutely right, though. Although you might want to clarify which carbons you're talking about with a numbered diagram. $\endgroup$ – orthocresol Nov 8 '16 at 20:02

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.