938 CHAPTER 22 Carbohydrates
Why is there more -D-glucose than -D-glucose in an aqueous solution at equilibri-
um? The OH group bonded to the anomeric carbon is in the equatorial position in -D-
glucose, whereas it is in the axial position in -D-glucose. Therefore, -D-glucose is
more stable than -D-glucose, so -D-glucose predominates at equilibrium in an
aqueous solution.
If you remember that all the OH groups in -D-glucose are in equatorial positions,
it is easy to draw the chair conformation of any other pyranose. For example, if you
want to draw -D-galactose, you would put all the OH groups in equatorial positions,
except the OH groups at C-4 (because galactose is a C-4 epimer of glucose) and at C-1
(because it is the ). You would put these two OH groups in axial positions.
To draw an L-pyranose, draw the D-pyranose first, and then draw its mirror image.
For example, to draw -L-gulose, first draw -D-gulose. (Gulose differs from glucose
at C-3 and C-4, so the OH groups at these positions are in axial positions.) Then draw
the mirror image of -D-gulose to get -L-gulose.
PROBLEM 23
Which OH groups are in the axial position in
a. -D-mannopyranose? c. -D-allopyranose?
b. b-D-idopyranose?
b a
the OH at
C-4 is axial
OH
CH 2 OH HOCH 2
HO OH
HO
HO
OH HO
OH
O O
-D-gulose -L-gulose
the OH at
C-3 is axial
the OH at C-1 is
equatorial ( )
b b
b b
OH
CH 2 OH
HO
HO
OH
O
-D-galactose
the OH at
C-4 is axial
the OH at
C-1 is axial ( )
a-anomer
a
b
a b
a b
b
b a
3-D Molecules:
- D-Galactose;
- D-Gulose;
b-L-Gulose
b
a
OH
CH 2 OH
HO
HO
HO
O
-D-glucose
36%
axial
OH
CH 2 OH
HO
HO
HO
O
CH O
CH 2 OH
HO
HO
HO
OH
-D-glucose
64%
equatorial