D-Glucose ( acetic anhydride )/( anhydrous ) D-Glucopyranose...

D-Glucose $\frac{\text { acetic anhydride }}{\text { anhydrous }}$ D-Glucopyranose pentaacetate, sodium acetate $$\begin{array}{c}\text { acetic anhydride } \\\text { cat. HA }\end{array} \begin{array}{c}\text { D-Glucopyranose pentaacetate, } \\\text { anomer } \mathbf{V I}\end{array}$$ The 'H NMR data for the two anomers included very comparable peaks in the $\delta 2.0-5.6$ region but differed in that, as their highest $\delta$ peaks, anomer $\mathbf{V}$ had a doublet at $\delta 5.8(1 \mathrm{H}, J=12 \mathrm{Hz})$ while anomer $\mathbf{V I}$ had a doublet at $\delta 6.3(1 \mathrm{H}, J=4 \mathrm{Hz})$ (a) Which proton in these anomers would be expected to have these highest $\delta$ values? (b) Why do the signals for these protons appear as doublets? (c) The relationship between the magnitude of the observed coupling constant and the dihedral angle (when measured using a Newman projection) between $\mathrm{C}-\mathrm{H}$ bonds on the adjaccht carbons of a $\mathrm{C}-\mathrm{C}$ bond is given by the Karplus cquation. It indicates that an axialaxial relationship results in a coupling constant of about $9 \mathrm{Hz}$ (observed range is $8-14 \mathrm{Hz}$ ) and an cquatorial-axial relationship results in a coupling constant of about $2 \mathrm{Hz}$ (observed range is $1-7 \mathrm{Hz}$ ). Which of $\mathrm{V}$ and $\mathrm{VI}$ is the $\alpha$ anomer and which is the $\beta$ anomer? (d) Draw the most stable conformer for each of $\mathbf{V}$ and $\mathbf{V}$.

D-Glucose $\frac{\text { acetic anhydride }}{\text { anhydrous }}$ D-Glucopyranose pentaacetate, sodium acetate $$\begin{array}{c}\text { acetic anhydride } \\\text { cat. HA }\end{array} \begin{array}{c}\text { D-Glucopyranose pentaacetate, } \\\text { anomer } \mathbf{V I}\end{array}$$
The 'H NMR data for the two anomers included very comparable peaks in the $\delta 2.0-5.6$ region but differed in that, as their highest $\delta$ peaks, anomer $\mathbf{V}$ had a doublet at $\delta 5.8(1 \mathrm{H}, J=12 \mathrm{Hz})$ while anomer $\mathbf{V I}$ had a doublet at $\delta 6.3(1 \mathrm{H}, J=4 \mathrm{Hz})$
(a) Which proton in these anomers would be expected to have these highest $\delta$ values?
(b) Why do the signals for these protons appear as doublets?
(c) The relationship between the magnitude of the observed coupling constant and the dihedral angle (when measured using a Newman projection) between $\mathrm{C}-\mathrm{H}$ bonds on the adjaccht carbons of a $\mathrm{C}-\mathrm{C}$ bond is given by the Karplus cquation. It indicates that an axialaxial relationship results in a coupling constant of about $9 \mathrm{Hz}$ (observed range is $8-14 \mathrm{Hz}$ ) and an cquatorial-axial relationship results in a coupling constant of about $2 \mathrm{Hz}$ (observed range is $1-7 \mathrm{Hz}$ ). Which of $\mathrm{V}$ and $\mathrm{VI}$ is the $\alpha$ anomer and which is the $\beta$ anomer?
(d) Draw the most stable conformer for each of $\mathbf{V}$ and $\mathbf{V}$.

Key Concepts

Spin–Spin Coupling in NMR

Spin–spin coupling occurs when a proton's magnetic environment is split due to interactions with neighboring nonequivalent protons. In these pyranose derivatives, the anomeric proton couples with the adjacent proton on C-2, resulting in a doublet. The magnitude of the coupling constant (J value) reflects the dihedral angle between the coupled C–H bonds and provides insight into the spatial arrangement of those protons.

Stereochemistry of Pyranoses

The ? and ? anomers of pyranoses differ in the configuration at the anomeric carbon. In the ? anomer, the substituent attached to C-1 (such as an acetyl group) is oriented axially, which places the anomeric proton in an equatorial position, leading to smaller coupling with the adjacent C-2 proton. In contrast, the ? anomer has the substituent equatorial and the anomeric proton axial, resulting in a larger axial–axial coupling constant. The observed differences in the NMR coupling constants thus allow the identification of the respective anomers.

Anomeric Proton

The anomeric proton in a carbohydrate is the hydrogen attached to the anomeric carbon (C-1 in pyranoses). This proton is generally found at the downfield region in the 1H NMR spectrum, due to the electron-withdrawing effects of the adjacent acyl groups (in this case, acetyl groups) and the involvement in the hemiacetal (or acetal) system. Its chemical shift and coupling behavior are key indicators of the anomer’s configuration.

Karplus Equation

The Karplus equation relates the magnitude of the vicinal proton–proton coupling constant to the dihedral angle between them. In the context of carbohydrate chemistry, an axial–axial relationship (with a dihedral angle near 180° or 0°) yields large coupling constants, typically around 8–14 Hz, while an equatorial–axial relationship (with a dihedral angle around 60°) results in much smaller J values (approximately 1–7 Hz). This relationship is utilized to deduce the relative stereochemistry of protons in sugar rings.

Transcript

00:04 Let's examine the differences and similarities of the enol and keto forms of acetyl acetone.

00:14 So first off, is the enol form and the keto form resonance? and what is meant by that is really what's changing first.

00:25 We start with something that kind of looks like this group, and we end with this group.

00:31 So what happened? well, what happened was this hydrogen got taken by this lone pair, or the electrons from this double bond took that hydrogen, and this hydrogen gave those electrons to make a double bond, as shown right here.

00:58 Is that resonance? as a matter of fact, it's not exactly resonance.

01:05 It works similar to resonance, but it's not the movement of solar.

01:09 Electrons.

01:10 It's also the movement of atoms and electrons.

01:15 And this is something that is specifically called tautomerism.

01:23 And this works with enols and anemines.

01:27 So like instead of an oxygen, if it was a nitrogen involved, it works in those situations where you have a double bond adjacent to an alcohol or an amine.

01:40 Really electro -negative atoms next to an alkenes is not a very stable form, and it would want nothing more than to get rid of that form.

01:50 So we never really see an enol form in nature.

01:54 We only see a ketone form because of that.

01:58 Now let's talk about hybridization.

02:00 What would the hybridization of each atom be? well, this carbon forms four bonds, so that's sp3.

02:08 This carbon forms three bonds.

02:10 That's sp2.

02:11 This oxygen has two lone pairs and two bonds.

02:16 That is sp3.

02:18 This carbon makes three bonds.

02:21 Sp2.

02:22 This carbon makes three bonds.

02:23 Sp2.

02:24 This oxygen makes one bond and two lone pairs.

02:29 So sp2 and this carbon is sp2.

02:35 And our keto form, well, this carbon is the exact same as the pre -20.

02:40 Example so it's sp3 this carbon still makes those three bonds sp2 this carbon makes three or four bonds so now it's sp3 the oxygen in the carbonyl is sp2 this carbon as p2 this oxygen would be sp2 and this carbon is sp3 oh my apologies this carbon is also sp3 back there so that is going to be the hybridization of each atom.

03:18 And if you notice what changes in the hybridization is, well, this atom goes from sp2 to sp3, and this oxygen goes from sp3 to sp2.

03:36 And those are the two differences of there, which is also why it's not resonance, because resonance is three adjacent sp2 atoms and we're starting with two sp2s and an sp3 going to two sp2s and an sp3.

03:52 So it's not resonance because those electrons are not in the p orbital entirely.

04:04 If we had to draw resonance structures for the ion, what we would have is, well, if it reacts with an ion or an an ion or an oh h group it would react to form this compound which looks very similar to what we had before except now it looks like there's just some deprotonation somewhere right and the way we can know this is well we have a ch3 group bonded to a c bonded to an o bonded to another c so either this is a double bond which is probably the case and we bonded to a c which is bonded to an h which is bonded to and there's bonded to another c o c h 3 so really we have a negative here which is not really a good thing to have and if we had a negative charge in a carbonyl group what would happen is well, resonance, because now that is an sp2 hybridized orbital, because negative charges always exist in the p -orbital.

05:53 So now we can have resonance.

05:55 So one possible structure would be donating those electrons to this alken, this carbon donates its electrons, or i'm sorry, this negative charge would donate its electrons to this car, carbon, so it would make it carbonyl, and this oxygen would go up, and we'd get one resin structure, which looks something like, i'm going to draw the bond line structures, or i'll just draw those write them out.

06:37 So we'd have one resonant structure as this molecule, and alternatively, well, it could have gone the opposite direction, which is the same way...

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