The O from carbon 5 becomes the O in the ring.
Once you know a few strategies, changing a Fischer projection to a Haworth projection is not too difficult.
If the OH is on the right side of the Fischer for C-2, C-3, and C-4, it will be lower in the Haworth. If it is on the Fischer's left side, it will be higher up in the Haworth.
Draw the C-5 CH₂OH pointing up if the sugar is D.
Draw the C-1 OH pointing downward for the alpha for D-sugars (and up for the beta)
To further understand the stereochemistry on the ring, a bond rotation on the C5 carbon is helpful. A bond rotation is accomplished by switching any three groups on a carbon. We'll therefore do the following three "moves":
H → OH
CH₂OH, OH
CH₂OH→ H
The C5-OH is no longer pointing downward after rotation; it is now to the side. Because of the tetrahedral structure of the carbon, I depicted the bond in the picture below as "dashed." I further dashed the C2-C1 bond.
This gives us the opportunity to depict the C5-OH attacking the carbonyl carbon and rupturing the C1-O pi link to generate a new O-C1 single bond.
We now have one of two rings after proton transfer, which we can represent as Haworth projections.
Hence, The O from carbon 5 becomes the O in the ring.
To learn more about Fischer projection refer- https://brainly.com/question/16906404
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