I came across a question where we have to find out the stability of species that are in chair conformations. The species are as follows:

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Comparing the 'A' values tells me that compound I, III and VI should be most stable, but the answer given is: I,IV and V.

Can anybody tell me where did I go wrong?

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    $\begingroup$ You can't use A-values directly except for the first row. The others have oxygen atoms in the ring. The oxygen atoms lack flaghole hydrogens, so the steric repulsion is less. Off the top of my head, I don't remember ever seeing numbers for this, so I don't have intuition on how much it changes the A-value. However, for V/VI, I still think VI. And the repulsion between the t-Bu and the lone pairs in IV might still strongly prefer III, but I'm just guessing. $\endgroup$ – Zhe Jun 12 '20 at 14:28
  • $\begingroup$ Why is V according to you is correct? $\endgroup$ – Sanu_012 Jun 12 '20 at 14:30
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    $\begingroup$ Sorry, I meant to write VI! $\endgroup$ – Zhe Jun 12 '20 at 14:31
  • $\begingroup$ Wow, there were a lot of typos in that original comment... :( $\endgroup$ – Zhe Jun 12 '20 at 14:32
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    $\begingroup$ @mathewmahindaratne Good call on hydrogen bonding. I missed that. $\endgroup$ – Zhe Jun 13 '20 at 21:29

According to the Ref.1, 5-hydroxy-1,3-dioxane (the compound of structures V and VI) both in the gaseous state and in dilute $\ce{CCl4}$ solution exists as a chair conformer, the hydroxy group in an axial position with an intramolecular hydrogen bond of the $\ce{O—H⋯O}$ type:


Accordingly, the axial conformer (V) is more stable by $\Delta G^\circ = \pu{-1.2 kcal mol−1}$). Microwave spectroscopy and the $^3J_\ce{H(5eq),OH}$ coupling constant suggest that the $\ce{OH}$ group lies in the plane of symmetry $(\ce{C_{(2)}-C_{(5)}-O}\text{-plane})$ and is a part of a bifurcated hydrogen bond to the two ring oxygen atoms (Ref.2). Therefore, it is safe to say the conformer V is more stable than conformer VI between two possible conformers of 5-hydroxy-1,3-dioxane (marked with $\color{green}{\text{green}}$ oval).

It is well known that the most stable conformer of tert-butylcyclohexane is tert-butyl in equatorial position. In a 1,4-substituted version of 1-tert-butyl-4-methylcyclohexane, since methyl group is so small compared to tert-butyl group that I is almost exclusive in equilibrium (marked with $\color{green}{\text{green}}$ oval).

However, I wasn't so sure about stability between III and IV. There is clearly no H-bonding opportunities except theoretical chemist may argue hyper conjugation with ring oxygen by MO calculations. I leave it open for any computational chemist to prove that possibility, yet my best guess is III more stable between them (marked with $\color{orange}{\text{orange}}$ oval with question mark).


  1. J. C. Jochims, Y. Kobayashi, “Bifurcated hydrogen bonds in z-2-phenyl-1,3-dioxan-5-ols,” Tetrahedron Letters 1976, 17(24), 2065-2068 (https://doi.org/10.1016/S0040-4039(00)93819-X).
  2. Jose L. Alonso, E. Bright Wilson, “Study of an intramolecular, bifurcated hydrogen bond in 1,3-dioxan-5-ol by microwave spectroscopy,” J. Am. Chem. Soc. 1980, 102(4), 1248–1251 (https://doi.org/10.1021/ja00524a005).
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    $\begingroup$ Okay, this is understandable. But in para 2, I think you meant to write VI instead of IV. Isn't it? $\endgroup$ – Sanu_012 Jun 14 '20 at 1:55
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    $\begingroup$ @Sanu_012: Good catch! Yes, I meant VI. I corrected it.:-) $\endgroup$ – Mathew Mahindaratne Jun 14 '20 at 5:32
  • $\begingroup$ That means, according to you the answer should be I, III and V and I can assume that the answer I, IV and V provided in the the answer key is wrong? Isn't it? $\endgroup$ – Sanu_012 Jun 14 '20 at 10:18
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    $\begingroup$ @Sanu_012: To my knowledge, yes. But, they might have a reason. Let's wait and see how others have responded. $\endgroup$ – Mathew Mahindaratne Jun 14 '20 at 15:59
  • $\begingroup$ Okay. Let's wait some days :) $\endgroup$ – Sanu_012 Jun 14 '20 at 16:13

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