Supplemental Lecture (96/12/28 update) by Stephen T. Abedon (abedon.1@osu.edu)

  1. Chapter title: Glucose Model
    1. A list of vocabulary words is found toward the end of this document
    2. The following is a walk through making a model of the glucose molecule made from balls and sticks (or, more sophisticated, an organic chemistry modeling kit consisting, basically, of balls and sticks). As you go through this, keep in mind how the structure of glucose determines its molecular stability. This stability probably explains, at least in part, why glucose is the dominant monosaccharide employed by life.
  2. Model #1
    1. This model shows (below) a linear form of glucose and galactose with minimal detail, i.e., most carbons and many hydrogens are missing (i.e., not shown). Note that, from the top, the second, third, fourth, and fifth carbons (all but the ends) are not rotatable. That is, when an -OH is to the right of one of these carbons, moving it to the left can be done only by the breaking and rejoining of bonds. You will understand this better as we move through models of increasing sophistication.
  3. Illustration, glucose and galactose, open chain form
  4. Illustration, glucose and galactose, open chain form
    1. The following is simply the above model with all carbons and hydrogens shown:

  5. Model #2
    1. This model connects carbons together with "curved" bonds (below). In nature, the single bonds between carbons are, of course, not curved. However, due to the tetrahedral nature of the electron pairs in carbon's outer shell, drawing glucose with straight bonds between carbons will produce a structure that bends back on itself (a ring, no surprise there, as we will see). However, playing around with the structure of glucose in this manner (i.e., employing "curved" bonds), will give a good idea of where the above representations (model #1) come from, and why the middle carbons don't freely rotate while the end carbons can).
    2. Note that, in the figure below, the -OHs and -Hs should be thought of as angling out of the plane made by the carbon atoms (toward you). Note also that carbon to carbon bonds enter each carbon from behind the plane. Rotating a carbon will literally rotate one end of the carbon chains out of the plane (i.e., thus making two planes). The only way an -OH on one side of a middle carbon can be exchanged for an -H on the other side of a middle carbon, while still retaining all of the carbon atoms in a single plane, is through the breaking of bonds (i.e., removing the -OH and -H on either side, switching sides, then reforming the bonds)
    3. This effect should at this point be fairly obvious. If the concept has not yet snapped into place, then perhaps stare at the figure until it does.
  6. Illustration, glucose and galactose, open chain form
  7. Model #3
    1. Replacing the curved bonds between carbons with straight ones results in a much more complicated looking molecule. Continuing to hold carbon atoms within a single plane gives a figure that looks like the following. Note that -OHs have been drawn as being either above the plane or below the plane (to increase clarity, -Hs have not been shown; note that they would be either above or below the plane, whichever orientation the opposite -OH is not).
  8. Illustration, glucose and galactose, open chain form
  9. Model #4 [beta form of glucose]
    1. The carbon bonds in model 3 may be rotated such that instead of forming a zig-zagging structure within a plane, they form a ring. This model presents glucose in a ring-like structure (prior to ring formation) and a ringed structure (after ring formation) consisting of five of the six carbon atoms and one of the -OH groups of glucose (below). Note that the far right carbon is that double-bonded to an oxygen atom--that is, the carbon found at the top of the above models (e.g., model 1). In this model the thick lines signify which portion of the molecule that is closer to you. That is, all the carbons are in a plane and the thick part of the plane is closer to you than the thin portion (the thin part, by the way, contains the oxygen atom in the ring).
    2. Note where the bond that completes the ring comes from in the below figure, as well as where the -H on the -OH attached to the carbon on the far right comes from. Note that the formation of the ring locks the -OH of that far right carbon into a position it cannot rotate out of (i.e., see arguments associated with model 2). This -OH up isomer is called the beta-form of glucose.
    3. Notice, starting with the far right carbon, beta-glucose has one -OH up, the next -OH down, the next -OH up, the next -OH down, and then, finally, the CH2OH group up (i.e., below the ring and above it, as shown). This staggering of -OH (and CH2OH) groups above and below the plane keeps these groups out of each other's way. As a consequence, the beta-form of glucose is not only the most stable form of glucose, but the most stable of the hexoses (six carbon sugars).
  10. Illustration, glucose, beta ring form
  11. Model #4 [alpha form of glucose]
    1. Also inter-convertible with the linear form of glucose (and therefore, ultimately, with the beta-form as well) is the alpha-form of glucose (below; however, though these bonds are inter-convertible in monomeric glucose, they are not inter-convertible once a bond has formed between the far right carbon and another atom, e.g., see maltose, cellulose, etc.). Note that the alpha-form has the far right -OH below the plane of the molecule. Alpha-glucose consequently lacks the optimal staggering displayed by beta-glucose and therefore is not quite as stable.
  12. Illustration, glucose, alpha ring form
  13. Galactose is less stable than glucose
    1. The following is galactose shown also in a ring-like form. Notice that regardless of how the far right -OH should be arranged following ring formation, galactose is inherently less stable than the equivalent form of glucose since the -OHs attached to the two front carbons are inherently not staggered.
  14. Illustration, galactose, open chain form suggesting ring form
  15. Chair and boat forms
    1. For the sake of countering arguments that the above is absurdly difficult to comprehend, allow me to point out that the models presented above are themselves not quite the whole story. Indeed, note that because the angles between the carbon atoms are actually 109.5ø, the ring of ringed glucose is actually bent into three separate planes. For example, at the bottom of the text figure is what is known as the chair conformation of glucose. In the boat form the tipped down triangle on the right is tipped up (like this \_/ rather than this \/\).
  16. Making glucose models in 3-D
    1. The following is how to make 3-D representations of models 2, 3 and 4:
      1. Take out 6 carbons (black), 6 oxygens (red), 12 hydrogens (balls), 7 bent bonds, 12 short bonds, and 12 long bonds.
      2. Connect all six carbons with curved bonds.
      3. Arrange chain such that each curved bond is concave up (i.e., as in model 3).
      4. Put all hydrogens on short bonds.
      5. Put hydrogens on five of six oxygens.
      6. Attach long bonds to all five -OHs.
      7. Place two curved bonds on remaining oxygen.
      8. Place the double bonded oxygen created in step 7 on the carbon at one end of the chain made in step 2.
      9. Arrange this carbon attached to the double bonded carbon oriented up.
      10. Starting with the second carbon from the top, attach -OHs to the carbons on the right, left, right, and right (i.e., #2, #3, #4, and #5 carbon respectively).
      11. Place the remaining -OH on the last (sixth) carbon.
      12. Place hydrogens on all the remaining carbon slots.
      13. Note how this molecule as formed resembles glucose in model 1 and, especially, model 2. Note that groups cannot be rotated around middle carbons.
      14. Remove -OH and -H on second carbon and switch these groups to form galactose. Reconstitute glucose.
      15. Replace curved bonds between carbons with long bonds.
      16. Arrange as shown for model 3.
      17. Interconvert glucose and galactose, then reconstitute glucose.
      18. Arrange as shown for model 4.
      19. Remove -H from fifth carbon -OH.
      20. Remove double bonded oxygen.
      21. Replace double bonds on freed oxygen (step 20) with one long bond and -H removed in step 19.
      22. Place -OH created in step 21 on number 1 carbon.
      23. Place long bond on oxygen in position created with removal of -H in step 19.
      24. Connect other end of long bond added in step 23 with number 1 carbon such that the ring closes in the á-form.
      25. Note that in the beta-form all -OH groups are found either above or below the plane made by the ring, and that these groups, along with the sixth carbon, stagger one up, one down, etc. Note that groups cannot be rotated around any of the carbons (except carbon 6).
      26. Reverse step 24.
      27. Rotate carbon 1 such that reformation of the bond broken in the step 26 forms the alpha-form.
      28. Note how the alpha-form isn't quite as well staggered as the beta-form.
      29. Again from carbon 2, remove the -OH and -H and switch their positions to form galactose.
  17. Vocabulary
    1. Alpha form of glucose
    2. Beta form of glucose
    3. Boat conformation of glucose
    4. Chair conformation of glucose
    5. Galactose is less stable than glucose
    6. Glucose and galactose, open chain form, illustration
    7. Glucose, beta ring form, illustration
    8. Making glucose models in 3D
    9. "Model #1"
    10. "Model #2"
    11. "Model #3"
    12. "Model #4"
  18. Practice questions
    1. No entry.
  19. Practice question answers
    1. No entry.
  20. References
    1. No entry.