The most stable conformation of optically inactive butane-2,3-diol is:

(A)enter image description here  (B)enter image description here  (C)enter image description here  (D)enter image description here

I chose C as the correct answer but, in the answer key it is given to be B. C is the staggered form and the most stable one. It can also have hydrogen bonding between -OH and -H groups. B can also have hydrogen bonding but there would be repulsion between the two Me groups. So, which one is correct?


3 Answers 3


Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's A value, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. According to the Wikipedia article on hydrogen boning, the typical $\ce{O-H-OH}$ hydrogen bond has an energy stabilization of −21 kJ/mol. Since these groups are not quite properly aligned, a safe assumption is that the hydrogen bond interaction would be about half of that ideal value.

However, in a nonpolar solvent, there might be a benefit to conformation C, which has a net-zero molecular dipole moment. The entropic repulsion between the nonpolar solvent and B would have to be more serious than the two Me-OH gauche interactions (probably about +1.8 kJ/mol each). However, without that information, conformation B should be correct.

The correct answer cannot be A or D since they are representing a chiral diastereromer.

  • $\begingroup$ Is the gauche effect applicable here? Hyperconjugation stabilises 2,3-difuoroethane. So does it increase the stability of butane-2,3-diol as well? $\endgroup$
    – Aditya Dev
    Commented Jan 2, 2016 at 14:51
  • $\begingroup$ well what if the condition of optical inactivity is removed? then wouldn't it be A $\endgroup$
    – Prakhar
    Commented Jan 2, 2016 at 17:48
  • $\begingroup$ how is A chiral by the way $\endgroup$
    – Prakhar
    Commented Jan 2, 2016 at 18:06
  • $\begingroup$ If the condition of optical activity were removed, it would be D (fewest gauche). If you are having trouble determining that A is chiral, try drawing it in a different style (perhaps line-bond with perspective wedges). Or, rotate it into an eclipsed conformation and note that while one pair of substituents line up in mirror-symmetry, the other two do not. A is the R,R enantiomer, B and C are the meso compound, and D is the S,S enantiomer. $\endgroup$
    – Ben Norris
    Commented Jan 2, 2016 at 19:08
  • $\begingroup$ @PrakharSankrityayan - It sounds as if you are on the verge of asking another question, and the comments are not the place for it. $\endgroup$
    – Ben Norris
    Commented Jan 2, 2016 at 19:09

The argument made by Ben can be supported by computational chemistry. I calculated three conformers of the compound.
Conformation B1 includes an intramolecular hydrogen bond, while in conformation B2 the proton was rotated away from the second hydroxyl group. Conformation C is the staggered one. The displayed structures were optimised in the gas phase.
three conformations of meso-butandiol

Computations were performed with Gaussian09 rev. D at the DF-BP86/def2-SVP level of theory. Solvent effects were estimated with the polarised continuum model, taking the dielectric constants as $\epsilon=78.3553$ for water and $\epsilon=2.0165$ for cyclohexane. Energies in $\mathrm{kJ\,mol^{-1}}$. The energies of each row are relative to conformer C. \begin{array}{lrrr}\hline \phantom{\hspace{3cm}} & \hspace{2cm}\mathbf{B1} & \hspace{2cm}\mathbf{B2} & \hspace{2cm}\mathbf{C}\\\hline \text{gas phase} & -7.2 & 10.2 & 0.0 \\ \ce{H2O} & -7.0 & 5.3 & 0.0 \\ \ce{C6H12} & -7.3 & 8.8 & 0.0 \\\hline \end{array}

As you can see, the intramolecular hydrogen bond accounts for much more than the repulsion between hydroxyl and methyl moieties. While the dihedral angle in B2 is $\angle(\ce{OCCO},\mathbf{B2})=-75.2^\circ$ - larger than expected -, it is smaller than expected in B1 with $\angle(\ce{OCCO},\mathbf{B1})=-57.0^\circ$.

We can assume that the hydrogen bond supplies about $-12$ to $-16~\mathrm{kJ\,mol^{-1}}$ depending on the solvent, which is in quite good agreement of what is known in the literature.

It is remarkable, that the energy difference between B1 and C in the various solvation states is negligible, but in the open form it has quite a significant impact, i.e. a polar solvent stabilises the open form more than an non-polar solvent. This is somewhat a little oversimplification because the used solvent model is quite crude. To obtain a concise picture one would, of course, have to increase the level of theory and treat the solvent explicitly.

Take home message: If you have two hydroxyl groups in proximity, don't forget that there might be an internal hydrogen bond stabilisation.

  • 3
    $\begingroup$ Huge "Thank yous" to pH13 for creating this awesome picture! $\endgroup$ Commented Sep 16, 2015 at 10:17
  • $\begingroup$ Didn't quite get the home message. I am just a newbie in opticak isometism so I wanted to confirm. Are you saying that hydtogen bonding internally is decreasing energy of the molecule more than the staggered configuration $\endgroup$
    – Aladdin
    Commented Sep 19, 2018 at 3:52

Stabilisation due to H-bonding(intramolecular) is far more stabilising factor in this case than the destabilisation between interactions of the adjacent methyl groups. Again this depends on solvent (in aprotic solvents 'C' could possibly be more accurate). If this was from a competitive exam book , say one for JEE or AIPMT (seems very familiar, hence the reference) , then the solvent is assumed to be water, in which case 'B' is correct.


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