Among the following, the most stable isomer is?

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I am aware of the fact that equatorial substituents are more stable than axial substituents but couldn't proceed to apply it here. However the answer key gives the answer as d) in which both substituents are in axial position.

  • 4
    $\begingroup$ Note that there is no space between the substituent prefixes (3-methoxy) and the parent compound (cyclohexan-1-ol). Furthermore, (b) and (c) are diastereomers, not different conformers, of (a) and (d). Anyway, the answer in three words is "intramolecular hydrogen bonding". $\endgroup$ Commented Dec 29, 2016 at 6:41
  • $\begingroup$ @orthocresol There would be a lot of steric hinderence too if both substituents are on the same side. $\endgroup$
    – Pink
    Commented Dec 29, 2016 at 6:51
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    $\begingroup$ Not steric hindrance, but steric repulsions. The word "hindrance" means that something is being blocked, such as an incoming nucleophile in an SN2 reaction. Nothing is attacking anything here, and consequently nothing is being blocked here. So, the word "hindrance" doesn't apply. Anyway, the question really boils down to whether the destabilising steric repulsions are bigger, or the stabilising intramolecular H-bond is bigger. The answer is you probably need a computer to work that out, but in this case the H-bond is more important. $\endgroup$ Commented Dec 29, 2016 at 6:59
  • $\begingroup$ On top of that (b) and (c) are the same isomer/conformer...? $\endgroup$ Commented Dec 29, 2016 at 7:55
  • $\begingroup$ @orthocresol yes b) and c) are the same conformers.it must have been misprinted in the book.Sorry for that. $\endgroup$
    – Pink
    Commented Dec 29, 2016 at 7:59

2 Answers 2


Sometimes, the intuitive rule of thumb that favours conformations with $\ce{OH}$ substituents in equatorial positions miserably fails.

conformers of a geranyl acetate cyclization product

For 1, a cyclization product of geranyl acetate, one might assume that the favoured conformation in solution and in the crystal is 1eq. However, this is not the case! Both NMR spectroscopy and X-ray crystallography indicate that 1ax is the actual conformation.

In the case of 1, the reason most likely isn't a (stabilizing) intramolecular hydrogen bond that would form a seven-membered ring, but the sterical interaction of the methyl groups. The unfavourable interaction of the 1,3-diaxial methyl groups in 1eq is avoided in the conformer with the axial $\ce{OH}$ group.

  • $\begingroup$ So why is this the case? Has it something to do with a 1,3-diaxial hydrogen bonding of the hydroxyl and ester carbonyl group? $\endgroup$ Commented Dec 29, 2016 at 11:17

I would say that the answer to the question depends strongly on the solvent used.

In case anybody still doesn’t see it: in the orientation (d), the compound can form an intramolecular hydrogen bond from the hydroxy group to the methoxy group. This is especially favourable in solvents that cannot participate in hydrogen bonding, e.g. dichloromethane.

Dissolving the same molecule in methanol, however, could change the entire story. The intramolecular hydrogen bond is favoured in the absence of other hydrogen bond donors or acceptors, but in a hydrogen bonding solvent there is absolutely no shortage of donors and acceptors and it can be assumed that all sites that can participate in hydrogen bonding in any way are saturated with hydrogen bonds. At this point, the steric interaction probably becomes more important and I would assume that the molecule preferentially assumes a diequatorial configuration.


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