The Anomeric Effect

In a post several months ago, I was talking about sugars and mentioned:

Note that in the cyclic isomer of glucose – β-D-glucopyranose (left) – all 5 substituents on the pyran ring are in the low-energy equatorial position (actually, the lowest-energy conformation of glucose is α-D-glucopyranose, where one of the -OH substituents is in the axial position. It is stabilized by what is known as the anomeric effect)

To which Mitch commented:

In regards to the anomeric effect, no one else finds it strange that when there is no good stereoelectronic effect explanation all of a sudden hyperconjugation is the key?

First, a mea culpa.  For unsubstituted glucopyranose, the β isomer is the lowest-energy isomer, and the α isomer is disfavored at a ratio of about 64:36.  When the hydroxyl group at the acetal position is changed to a methoxy group, then the α isomer is the lowest-energy isomer at a ratio of about 67:33 – the selectivity reverses. (click images for larger throughout) (update: figure labels fixed)


Second, my PhD research relies heavily on the anomeric effect, and I often get this ‘I don’t understand the anomeric effect’ response from people.  They assume it’s all handwaving.  I’d like to explain the anomeric effect and hopefully clear up some of the confusion surrounding it.  Read more below the jump.

We remember from undergrad organic classes, that substituents on a cyclohexane ring prefer the equatorial position to relieve steric strain.  There is a corresponding ring-flipped chair in which the substituent is axial, and the energy barrier associated with this ring flip can be quantified.  For cyclohexanol, the equatorial position is favored at a ratio of about 9:1.

When an endocyclic heteroatom (typically oxygen, nitrogen, or sulfur) is introduced at the position adjacent to the hydroxy group, for the example of methoxypyran, the hydroxy substituent now prefers the axial orientation – in spite of the associated steric strain – at a ratio of about 4:1.  This trend is not limited to oxygen substituents.  Any electronegative element will prefer an axial position when a heteroatom is on the adjacent endocylic position.  For the fluoro-xylose derivative in the image, the fluorine prefers the axial position EVEN THOUGH all of the other substituents are also axial.


This phenomenon is known for acylic systems, too.  If you were asked to draw a Newman projection for pentane and dimethoxymethane, for pentane you would drawn the Newman projection where the two bulky substituents are antiperiplanar to each other to minimize steric interactions.  And you’d be right.  But for dimethoxymethane, the two bulky substituents are gauche to each other in the lowest-energy Newman projection.  As a hint to what’s going on, this puts a lone pair on oxygen antiperiplanar to the electronegative substituent.


So why is this happening?  There are two main explanations given, and they both work together to explain why the seemingly more sterically hindered conformation is the most stable.  On one hand, when the electronegative substituent is in an equatorial orientation, the local dipole moment of the substituent and the local dipole of the endocyclic heteroatom are pointing in relatively the same direction.  This alignment of dipoles leads to a large net dipole for the molecule.  However, when the electronegative substituent is axial, the local dipoles are more or less pointing away from each other.  This relief of a net dipole is stabilizing for the molecule.  This is a fine explanation, but I don’t think it fully accounts for all of the stabilization seen in these systems.

The other main explanation is a molecular orbital argument.  In short, whenever you can lower the energy of a system, the system is net more stable.  When the electronegative substituent is in an axial orientation, the C-X sigma* antibonding orbital is directly lined up with the axial lone pair of electrons on oxygen.  This allows for some delocalization of the lone pair of electrons into the sigma* antibonding orbital.  The electrons from the lone pair interact with the sigma* antibonding orbital in a stabilizing manner.  The elctron density is now spread over two atoms.  The delocalization of the electrons results in a stabilization of those electrons, and leads to a net stabilization of the molecule.


But why don’t all groups next to an endocyclic heteroatom prefer the axial orientation, you might ask?  Why wouldn’t a regular methyl group be stabilized in the same way?  There is a trend amond endocyclic and exocylic groups which rates the ability to donate or accept electrons.  A carbanion is among the best electron donors, and heteroatom lone pairs are pretty good donors, too.  Things like C-H sigma bonding electrons aren’t such good donors, but do delocalize to some extent (this is why teritary carbocations are more stable than methyl carbocations).  There is a similar trend for electron accepting groups.  An empty p orbital is among the best acceptors, and C-X sigma* antibonding orbitals are pretty good, too.  The reason for this is the relative energy of the donating and accepting orbitals.  Carbanions and heteroatom lone pairs are relatively higher in energy than C-H sigma bonding electrons, and p orbitals and C-X sigma* antibonding orbitals are relatively lower in energy than C-C sigma* antibonding orbtials.  When the donor and acceptor orbitals are closer in energy, the stabilization is more favorable.

I talked about this general phenomenon as it relates to hyperconjugation over on the forums, too, if you want to read more.

So to summarize, for a cyclic system, when an endocyclic atom (Y) has a lone pair of electrons, a neighboring electronegative substituent (X) prefers to reside in an axial orientation.  More generally for cyclic or acylic systems, when a heteroatom (Y) has a lone pair of electrons, an neighboring electronegative group prefers a gauche orientation – in spite of what sterics might dictate.  This allows for maximum overlap of the lone pair of electrons and the neighboring C-X sigma* antibonding orbital.


Update: Forgot to list my sources.  The two sources from which I pulled most of my analysis are:

  • Juaristi, E.; Cuevas, G. The Anomeric Effect, CRC Press: Boca Raton, 1995.


  1. Apparently we are both similar in that we like to re-analyze old discussions.

    I was wrong in implying the anomeric effect is not an electronic effect, it obviously is very much an electronic effect. Question though, in the first figure you have an equilibrium arrow for your two set of compounds, but how is that an equilibrium? I’m not a sugar guru and can’t see the process.

  2. I’ve been sitting on this post for a while.

    As for the mech: (hemi)acetal opens, recloses to other anomer. It’s just an equilibrium between the open-chain sugar and the closed pyranose.

  3. So why are the OH groups not as affected by the anomeric effect as the OMe groups in your examples? From an electronic argument then, since the antibonding O-H sigma orbital is closer in energy to the nonbonding O electrons than the antibonding C-H sigma orbital, shouldn’t the effect be most prominent for the OH systems?

    • Good question. The sources from which I got the deltaG values don’t list the solvent in which the deltaG was measured – and I think it’s the solvent that’s key.

      Solvent actually plays a big role in the magnitude of the anomeric stabilization. The larger the dielectric constant of the solvent (the more polar the solvent) the smaller the anomeric stabilization. For methoxypyran (for example), in carbon tetrachloride, the axial orientation is favored 83:17. But in water, the axial orientation is only favored 52:48.

      This is probably not an MO argument, but a dipole argument. The effect of the aligned dipoles is mitigated in polar solvents which accept polar molecules better – the steric argument has more merit in polar solvents and the anomerically-stabilized axial orientation is less favored. In nonpolar solvents, the minimization of dipoles is more important, and the anomeric stabilization is stronger.

      So my answer is probably that the deltaG’s were measured in different solvents. Absent that, I don’t have a good answer. Maybe it has something to do with the O-H bond dipole compared to the O-CH3 bond dipole and how that modulates the local dipoles and consequently the net dipole of the molecule?

  4. I am confused, it seems that in your first figure showing the equilibrium that the alpha and beta are labeling the wrong anomers. Isn’t beta the anomer with the equatorial hydroxyl?

  5. note about last post:

    Isn’t beta the anomer with the equatorial anomeric hydroxyl?

    • Thank you for your comment. Turns out the text is ok, and I mislabeled the structures in the figure (the text said beta was favored for the top case, but equilibrium pointed to alpha).

      Figure has been fixed, thanks for the notice. Good eye.

      For the record, here’s the official IUPAC rules for alpha/beta anomers:

      the anomeric reference atom is the highest-numbered atom of the group of chiral centres next to the anomeric centre that is involved in the heterocyclic ring and specified by a single configurational prefix. In the α anomer, the exocyclic oxygen atom at the anomeric centre is formally cis, in the Fischer projection, to the oxygen attached to the anomeric reference atom; in the β anomer these oxygen atoms are formally trans.

      If that was too confusing, Wikipedia opines:

      Anomers are identified as “α” or “β” based on the relation between the stereochemistry of the anomeric carbon and the furthest chiral centre in the ring.[2] The α anomer is the one in which these two positions have the same configuration; they are opposite in the β anomer.[3] Thus the structure of α-D-glucopyranose has the same stereochemistry at both C1 and C5 whereas β-D-glucopyranose has opposite stereochemistry at C1 compared to C5. This relationship is strictly based on the absolute stereochemistry, not cis/trans relationships or axial/equatorial positioning in the ring conformation.

      So it’s not quite as simple as beta equals equatorial.

  6. One other thing I might add, since we’re bringing the topic back up.

    In the comments above, Mitch and I mused over why the hydroxy version of glucose is still favored for the equatorial position.

    Here’s my new thought – while the absolute values of the equilibrium unquestioningly favor the equatorial position, compare those values to the equilibrium values for cyclohexanol (also cited in the post).

    Cyclohexanol favors the equatorial position at a ratio of 9:1.

    Hydroxy glucose favors the equatorial position at a ratio of 2:1.

    So even though the axial position isn’t absolutely favored, it is still favored to a much greater extent than it is when compared to cyclohexanol.

    More food for thought.

  7. wow, thanks.(: this explanation is indeed convincing (compared to my lecturer’s approach).

  8. Great Explanation. Have never read so clear but elaborate explanation of “Anomeric Effect”