Question 22.3: The open-chain form of D-altrose, an aldohexose isomer of gl...
The open-chain form of D-altrose, an aldohexose isomer of glucose, has the following structure. Draw D-altrose in its cyclic hemiacetal form:
\underset{D-Altrose}{\begin{array}{r c} H \quad H \quad \; H \quad H \quad \; OH \; O \qquad \\ | \quad \; \; | \quad \; \; \; | \quad \; \; | \quad \; \; \; | \quad \; \; || \qquad \; \\ HO-C-C-C-C-C-C-H \\ | \quad \; \; | \quad \; \; \; | \quad \; \; \> | \quad \; \; \> | \qquad \qquad \; \\ H \quad OH \; \> OH \> OH \; H \qquad \qquad \end{array}}First, coil D-altrose into a circular shape by mentally grasping the end farthest from the carbonyl group and bending it backward into the plane of the paper:
Next, rotate the bottom of the structure around the single bond between C4 and C5 so that the – CH_2OH group at the end of the chain points up and the — OH group on C5 points toward the aldehyde carbonyl group on the right:
Finally, add the — OH group at C5 to the carbonyl C = O to form a hemiacetal ring. The new — OH group formed on C1 can be either up (β) or down (α):
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