Question 21.3: Converting Fisher Projections to Cyclic Hemiacetals The open......
Converting Fisher Projections to Cyclic Hemiacetals
The open-chain form of D-altrose, an aldohexose isomer of glucose, has the following structure. Draw D-altrose in its cyclic hemiacetal form:
\begin{matrix} \underset{\mid}{H} \underset{\mid}{H} \underset{\mid}{H} \underset{\mid}{H} \underset{\mid}{O}H \underset{\mid\mid}{O}\\ HO — C — C — C — C — C — C — H\\ \overset{\mid}{H} \overset{\mid}{O}H \overset{\mid}{O}H \overset{\mid}{O}H \overset{\mid}{H} \end{matrix}
D-Altrose
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:
\begin{matrix} \underset{\mid}{H} \underset{\mid}{H} \underset{\mid}{H} \underset{\mid}{H} \underset{\mid}{O}H \underset{\mid\mid}{O}\\ HO — \underset{6\mid}{C} — \underset{5\mid}{C} — \underset{4\mid}{C} — \underset{3\mid}{C} — \underset{2\mid}{C} — \underset{1}{C} — H\\ H OH OH OH H \\ \\ \\ \end{matrix} \begin{matrix}\xrightarrow[]{Coil up}\\ \\ \end{matrix} 
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:
\begin{matrix} \xrightarrow[]{Rotate}\\ \\ \\ \end{matrix} 
Finally, add the —OH group at C5 to the carbonyl C\xlongequal[]{}O to form a hemiacetal ring. The new —OH group formed on C1 can be either up (β) or down (α):
\begin{matrix}\rightleftarrows\\ \\ \\ \\ \end{matrix} 
β α