Glucose: Molecular Modeling with CAChe
Mark R. Ams
Department of Chemistry, University of Pittsburgh
June 2006
Update: February 2007
Carbohydrates comprise the most abundant group of natural products. They are produced in billions of tons every year by photosynthesis in plants and microorganisms. The widespread distribution of cellulose and starch in the enormous numbers and kinds of plants make glucose the most abundant organic compounds on earth! In aqueous solution, glucose exists as three forms in dynamic equilibrium:
HO HO
H
OH
H
OH
H
O
H
H
OH
HO
OH
OH OH
OH H
O
H
OH
H
OH
H
O
H
OH
H
OH OH

-glucose open form 
-glucose
The two cyclic forms of glucose differ only in the orientation of the OH group at C-1, upward in

-glucose and downward in

-glucose. At equilibrium, the proportions in solution are 36%

-glucose, 0.02% open-chain glucose, and 64%

-glucose.
PART 1.
Just like cyclohexane, the 6-membered ring of monosaccharides (such as glucose) also exists in two isomeric chair conformations. (Detailed procedures for using CAChe are included in the Supplementary material at the back of this document. When reporting numbers, keep in mind that the differences are much more reliable than the absolute valutes . Do not report any values to greater than an accuracy of 0.1 kcal/mol.)
1.
Carefully draw by hand the two possible chair conformations of β-glucose.
All Equatorial Chair
Δ f
H =
All Axial Chair
Δ f
H =
1

Although these two conformations are both possible, only the chair with all equatorial substituents is observable in H
2
O by NMR spectroscopy. Would CAChe have predicted this observation?
2. To answer this, use CAChe to calculate the energies (Δ f
H ) of the two chairs and report them above. (Note: attempt the axial calculation several times, redrawing it differently for each attempt – report only the global minimum that you find. It is wise to look carefully at each optimized structure you generate.)
You will see that the Δ f
H values for each chair are very close in energy and that
CAChe falsely predicts the all axial conformer to slightly predominate. Why? If you look at the geometry of the all axial conformer that CAChe generates from its calculations, you will see that the –OH substituent at C3 is H-bonding to its 1,3-diaxial –
OH neighbor:
HO
O
H
O
O
H
H H
H
OH OH
As chemists, we realize that this 1,3-diaxial interaction is not likely to occur in H
2
O. Hbonding with H
2
O is much more likely. This lesson teaches us that computational calculations are not always correct. We must use our own chemical intuition to convince ourselves that the computational data fits our chemical theory.
This 1,3-diaxial interaction issue can be eliminated by changing the OH group at C3 into
OCH
3
as shown below.
3. Use CAChe to calculate the Δ f
H’s for each chair . Which chair predominates now?
H
OH HO
OH
HO
CH
3
O
H
H
O
OH
OH
H
OCH
3
O
H
H
H
H
OH OH
Δ f
H =
Chair 1
Δ f
H =
Chair 2
2

PART 2.
Use CAChe to calculate the lowest energy minima (Δ f
H in kcal/mol) of the alpha and beta forms of glucose below (in H
2
O).
HO HO
H
OH
H
OH
H
O
H
H
OH
H
OH
H
OH
H
O
H
OH
H
OH OH

-glucose
(either chair conformation)

-glucose
(use the all equatorial chair
value determined in question 2)
Δ f
H =
Δ f
H =
4. Which would you expect to see most abundant in an aqueous mixture? Justify your answer with data from your calculation.
PART 3.
In the first exercise, your Δ f
H values calculated by CAChe should lead you to the conclusion that β-glucose (not α-glucose) is the expected form in aqueous solution.
However, in aqueous solution, the α- and β- anomers are actually observed in a ratio of
1:2. If β-glucose is the more energetically favored anomer, how is this possible? There must be another operating effect that competes successfully with the aforementioned destabilizing one. This is the anomeric effect .
In the anomeric effect , the nonbonding electrons of the endocyclic oxygen atom have a p-orbital that is axially oriented and is synperiplanar to the antibonding orbital of the anomeric substituent when it’s in an α-position. As a result, the two orbitals can mix, establishing a n→σ* interaction, which results in favorable energy stabilization. orbital overlap - anomeric effect no orbital overlap - no anomeric effect
O
O
X X = F, Cl, Br, N, O
X
 anomer


-anomer

3

5. Calculate the Δ f
H values for modified α/β-glucose (in the indicated chair conformations in H
2
O) that we should expect due to the anomeric effect when X is changed to a more electronegative atom:
OH
O
HO
HO
OH
X
 anomer
 f
H
OH
HO
HO

-anomer
 f
H
O
OH
X
Energy
Difference
Chair
Favored
Cl
Br
 f
H
O
6. According to your calculated values, how does the anomeric effect affect the relative stabilities of the modified α/β-glucose?
PART 4.
A commonplace table ingredient in every American diner, table sugar (sucrose, and carbohydrates in general) are often thought of being water soluble (hydrophilic) due to the hydroxyl group substituents attached to the ring backbone. However, they are capable of generating apolar surfaces depending on the monomer ring conformation.
7. Carefully redraw by hand the two chair conformations of β-glucose and calculate their energies in hexane solvent.
Chair 1 Chair 2
4

Δ f
H =
Δ f
H =
8. Which chair is preferred in hexane? Which chair is preferred in water? Use your drawings to explain your answers.
PART 5.
A disaccharide is a compound that is formed from two monosaccharide units (such as glucose) by the elimination of one molecule of water. The bond between the two components of the acetal is called the glycoside bond. The molecule formed when many saccharide units are linked by glycoside bonds is called a polysaccharide .
The 3-D properties of polysaccharides are of particular importance to molecular recognition processes in biochemistry. When investigating the shapes of polysaccharides, their energy minimum conformations must be addressed, i.e. is the polysaccharide a flexible or rigid structure?
The flexibility of a polysaccharide chain depends on the ease of rotation around the glycoside link:
H
H
OH
OH
 
HO O
HO
O OH
O HO
OH
OH
 linked glucose disaccharide
Bond rotation changes the energy of the structure and this can be visualized by CAChe as a potential energy map (shown below).
9. Using CAChe, create this potential energy map and determine the torsion angles phi (

H
, H
1
C
1
OC
4
) and psi (

H
, C
1
OC
4
H
4
) for the maximum and minimum energy
5

conformations of β-(1,4)-linked glucose (in H
2
O). Carefully draw by hand the 3-D structures for each.
(Note – the calculation takes 75 minutes for CAChe to complete.)
Maximum Energy Confirmation
Δ f
H =

=

=
Minimum Energy Confirmation
Δ f
H


=
=
=
6

Supplement for
Glucose: Molecular Modeling with CAChe
Mark R. Ams
Department of Chemistry, University of Pittsburgh
June 2006
Directions for Conducting Drawing and CAChe calculations
(Based on CAChe Worksystem Pro Version 6.1.10)
Part 1 and 2. Drawing glucose chair conformations and calculating Δ f
H in H
2
0:
To Draw β-Glucose chair 1:
1. Click drawing pencil tool . Choose “C/sp3/blank/single” in the drop down windows in the above left corner of the blue workspace window.
2. Click in the workspace. A black C atom appears.
3. Click on the C atom and drag away. Release. A new C atom appears bonded to your original C atom.
4. Continue to do this until you have made a 6-member carbon ring.
5. Click the icon and use the mouse to highlight the 6-member ring. Click the
“Beautify” icon in menu, then “comprehensive”. C
6
H
12
appears.
6. Use the rotate tool to rotate the molecule so that you can view it “side on”. Click on the icon. Then click 1 of the carbon atoms that you want to change into oxygen.
Once its highlighted, choose “O/sp3/blank/single” in the drop down windows. Highlight the entire molecule and click “beautify/comprehensive” again. C
5
H
10
O appears.
7. Continue with this process to add all the appropriate glucose ring substituents in their proper chirality. “Beautify” when finished.
8. Save as “β-glucose chair 1”
To Draw β-Glucose chair 2:
1. Highlight the entire “β-glucose chair 1” molecule.
2. Click on the rotate icon
. Using your mouse, rotate “glucose chair 1” 180° (along it’s imaginary x-axis) so that it’s shape resembles chair conformation 2:
X
O
O
O
Chair 1 Chair 2
Next highlight the oxygen atom and change it to a C, then convert the appropriate C into an O. Delete the substituents on the new O atom.
7

3. The ring substituents must now be corrected. To do this, highlight the atom substituent/s you want to change (or delete), then choose “atom/sp3/blank/single” in the drop down windows to change them.
3. “Beautify” it, convince your self it looks correct and save as “β-glucose chair 2”.
To Calculate Δ f
H for β-glucose chairs in H
2
O:
1. Once your glucose chair has been constructed (chair 1 or 2), look at the pull-down menu at the top of the screen and click on “Experiment/New”.
2. In the window that appears, choose “ Property of : chemical sample, Property : optimized geometry, Using
: PM3 geometry in water”.
3. Now click “Start” in the “Experiment1” window to begin the calculation. The computer will calculate for 10-20 seconds. Once it’s finished, the molecule’s Δ f
H value will be displayed at the end of the calculation.
4. To get Δ f
H for β-glucose chair 2, upload the drawing that you have saved for it, and then repeat steps 1-3.
To Draw α-glucose chair and Calculate Δ f
H in H
2
O:
1. Upload the more stable β-glucose chair determined in Part 1. The only difference between α and β-glucose is the chirality of 1 carbon atom. It’s chirality must be flipped to make α-glucose.
2. Flip it’s chirality by clicking on the O atom from –OH substituent and “shift”-click on the carbon atom it’s bonded to and then it’s geminal H. On the pull-down menu click
“edit/invert chirality”.
3. “Beautify”,
Save As “α-glucose,” and calculate Δ f
H just like you did above for βglucose.
To Draw β-glucose chairs (Methylated –OH at C-3)
1. If it hasn’t been uploaded already, upload the β-glucose molecule you want to methylate. Using the tool, highlight the H atom of the –OH group at C-3. Choose
“C/sp3/blank/single” in the drop down window. This converts it into a C atom.
2. “Beautify” the entire molecule.
To Calculate Δ f
H for β-glucose chairs (methylated –OH at C-3) in H
2
O:
1. With your molecule in the workspace window, look at the pull-down menu at the top of the screen and click on “Experiment/New”.
2. In the window that appears, choose “ Property of : chemical sample, Property : optimized geometry, Using : PM3 geometry in water”.
3. Now click “Start” in the “Experiment1” window to begin the calculation. The computer will calculate for 10-20 seconds. Once it’s finished, the molecule’s Δ f
H value will be displayed at the end of the calculation.
Part 3. Calculating Δ f
H due to the anomeric affect in H
2
O.
To Calculate the α-anomer:
8

1. Upload the α-glucose molecule you constructed in Part 1. Click on the icon and then highlight the anomeric –OH substituent. Convert that substituent to –Br (or –Cl) by choosing “Br/sp3/blank/single” in the drop down windows.
2. “Beautify”, Save As “α-glucose bromine chair” and then calculate Δ f
H as described above.
To Calculate the β-anomer:
1. Upload the β-glucose you constructed in Part 1. Repeat the same steps you did for the
α-anomer.
Part 4. Calculating Δ f
H in Hexane:
1. Upload the β-glucose chair 1 molecule you constructed in Part 1.
2. Click on “Experiment/New” from the pull-down menu at the top of the screen.
2. In the window that appears, choose “
Property of : chemical sample, Property : optimized geometry, Using
: PM3 geometry”.
3. Click “Edit”. Then click on “Call MOPAC-PM3_Geo using Sample1” twice to open it. Then click “Edit” in the “Procedure to Call” window. Then click on “Run Molecular
Orbital Package using Sample1” twice to open it. At the bottom of the “MOPAC
Settings” window, check the “Simulate Solvents” box. Type “hexane” in the name box and 1.88 for the dielectric constant.
4. Click “ok” at the bottom of the “MOPAC Settings” window and close the remaining
“edit” windows you have opened. Click “yes” to “Save changes to Procedure MOPAC-
PM3_Geo”.
5. Now click “Start” in the “Experiment1” window to begin the calculation. The computer will calculate for 10-20 seconds. Once it’s finished, the molecule’s Δ f
H value will be displayed at the end of the calculation.
4. To get Δ f
H for β-glucose chair 2, open the drawing that you saved earlier, and then repeat step 5. The calculation will still be performed in hexane solvent because you saved your changes.
Part 5. Calculating φ and Ψ for β-linked glucose disaccharide in H
2
O :
1. To construct the β-(1,4)-linked glucose disaccharide molecule shown in the worksheet:
A. Upload your β-glucose chair 1 molecule, then highlight it and “copy/paste” it in the workspace.
B. The 2 molecules are temporarily superimposed. Hold the “control” key down and click on the molecule and use the mouse to drag it away from the 1st. This is how you move 1 molecule in the presence of another. To couple them together at the β position, position the 2 molecules close together at the β position. Draw the proper bond coupling glucose 1 to 2:
9

coupling
OH
OH
HO
HO
O
O
HO
O
OH
OH OH
2. “Beautify” the molecule. CAChe will automatically orient the disaccharide as shown in the worksheet.
3. Use the rotate tool to orient the molecule into the position shown in the worksheet. WARNING! Convince yourself that the structure you drew is correct before proceeding to the next step.
4. The dihedral angles φ and Ψ must be defined. Highlight the 4 atoms involved for φ
(see drawing) by clicking on the first atom and then shift-clicking on the other 3 in the order in which they’re bonded to each other so that they’re all highlighted.

HO
HO
OH
O
OH
H
O
HO
H OH
O
OH
OH

5. On the drop-down menu, click on “adjust, dihedral angle”. Check the “define geometry label” box and then check the “search” box and click “ok”.
6. Repeat steps 4 and 5 to define Ψ.
7. Click on “Experiment/New” from the pull-down menu at the top of the screen.
In the window that appears, choose “
Property of : chemical sample confirmations
(CAChe 5.0 experiments), Property : optimized map, Using : PM3 energies in water (2 labels)”.
8. The calculation takes 75 minutes to complete. Its progress (in %) can be monitored by reading the “Experiment Status” window. When the calculation is done, 2 windows will appear. On the left is a potential energy map showing the regions of energy change as a function of φ and Ψ (CAChe calls them “dihedral angle 1” and “dihedral angle 2”).
The map is 3-dimentional, so use the rotate tool to rotate the map and visualize all 3 axes. Use your mouse/arrow keys to move the little grey ball around the map. As you do this, watch the disaccharide in the right window change shape.
9. Use the pull down window located at the lower left corner of the energy diagram window (scroll all the way to the bottom) to display the energy values of any specified dihedral angle. Report φ and Ψ for the highest and lowest energy conformations.
10