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Some substituents of saturated heterocycles prefer to be axial: the anomeric effect
المؤلف:
Jonathan Clayden , Nick Greeves , Stuart Warren
المصدر:
ORGANIC CHEMISTRY
الجزء والصفحة:
ص801-803
2025-07-16
25
Some substituents of saturated heterocycles prefer to be axial: the anomeric effect
Many of the stereoelectronic effects in the list above govern reactivity, but the next section will deal with how stereoelectronic effects affect structure—and in particular conformation. Some of the most important saturated oxygen heterocycles are the sugars. Glucose is a cyclic hemiacetal—a pentasubstituted tetrahydropyran if you like—whose major conformation in solution is shown below. About two-thirds of glucose in solution exists as this stereoisomer, but hemiacetal formation and cleavage is rapid, and this is in equilibrium with a further one-third that carries the hemiacetal hydroxyl group axial (<1% is in the open-chain form).
you will not be surprised that glucose prefers all its substituents to be equatorial. For four of them, of course, there is no choice: they are either all-equatorial or all-axial, and the only way they can get from one to the other is by ring-flipping. But for the fifth substituent, the hydroxyl group next to the ring oxygen (known as the anomeric hydroxyl group), a choice between axial or equatorial is made available by hemiacetal cleavage and re-formation—it can invert its configuration. What is perhaps surprising is that the equatorial preference of this hydroxyl group is so small—only 2:1. Even more surprising is that, for most derivatives of glucose, the anomeric substituents prefer to be axial rather than equatorial.
Move away from glucose and the effect is still there in other substituted tetrahydro pyrans. Listed below are the NMR signals of the chloro compound in the margin. There are now only two possible conformations (no configurational changes are possible because this is not a hemiacetal)—both shown—and from the NMR spectrum you should be able to work out which one this compound has.
The key point is that axial–axial couplings are large (>8 Hz, say), even with adjacent electro negative atoms (which tend to lower coupling constants). So, if H1 were an axial proton, you would expect it to have a large coupling to H2. But it doesn’t—it couples to H2 with J of only 2.0 Hz. (The other coupling is a W-coupling to H3, also of 2.0 Hz: see p. 296.) Similarly, we know that the 12.9 Hz coupling shared by the two H5 protons must be a geminal (2J) coupling. One of H5a or H5b must be axial, yet both couple to H4 with J < 4 Hz, so H4 cannot be axial. With this evidence, we have to conclude that H1 and H4 (and therefore H2 and H3) are equatorial, so the compound must exist mainly in the all-axial conformation. (The 0.6 Hz coupling to H5b is another W-coupling, and shows that H5b is the equatorial proton and H5a therefore the axial one.) This axial preference is called the anomeric effect.
●The anomeric effect In general, any tetrahydropyran bearing an electronegative substituent in the 2-position will prefer that substituent to be axial. This is known as the anomeric effect.
making conformations carrying axial substituents disfavoured. The key again is stereoelectronics, and we can now link up with the message we left you with at the end of the last section: elimination reactions are possible only when the orbitals involved are parallel. An amide is more stable (less reactive) than a ketone because the p orbital of the N and the low-lying C=O π* of the carbonyl can lie parallel—they can overlap and electron density can move from nitrogen into the C=O bond, weakening C=O. (Evidence for this comes from the lower IR stretching frequency of an amide C=O, among other things.) But C–X bonds also have low-lying antibonding orbitals—the C–X σ*—so we would expect a molecule likewise to be stabilized if an adjacent heteroatom could donate electrons into this orbital. Take the generalized tetrahydropyran in the box above, for example, with X=Cl, say. This molecule is most stable if an oxygen lone pair can overlap with C–Cl σ*, as shown in the margin. But it can do this only if the chlorine is axial! Remember what we pointed out earlier: the oxygen’s equatorial lone pairs are parallel with nothing but bonds in the ring, so the oxygen’s axial lone pair is the only one that can help stabilize the molecule, and it can only do this when the Cl is axial. Only the axial conformation benefi ts from the stabilization, and this is the origin of the anomeric effect. How shall we represent the stabilization? Comparing again with the amide stabilization, you might think about how to represent it with curly arrows: this is straightforward with the amide and you have seen it many times. But it looks odd with our heterocycle: electron density moves from O to Cl, and the C–Cl bond is weakened. If the process carried right on, Cl− would leave. This is exactly what did happen in the acetal we presented you with as an example on p. 801: only the axial OAr could leave, however, because of the same requirement for overlap with an oxygen lone pair. In the real structure that we are now looking at, the Cl is still there: the C–Cl bond is weaker, and some of the oxygen’s electron density is delocalized on to Cl. This can be seen in crystal structures: compounds exhibiting an anomeric effect have a longer (and there fore weakened) bond outside the ring and a shorter, stronger C–O bond within the ring.
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