When converting a Haworth Projection with a six-membered ring to a chair conformation, how do you know which groups are axial and which are equatorial?
A general rule of thumb is to draw the conformation with the most substituents equatorial
What you have drawn there is unfortunately alpha-D-allopyranose.
First thing I will say is that:
For Haworths: In the D series the alpha-anomer projects the anomeric OH down, and the beta-anomer up.In the L sereis it's the other way round
, so for L sugars the alpha-anomer goes up and the beta-anomer goes down.
Secondly, watch your stereochemistry at the C before the ring-O (C5 in this example). As you have to rotate the chain to get the OH into the ring it can get a bit tricky to see, but your CH2OH substituent should be down for L-talopyranose.
and also into beta-L-talofuranose
For this one you've drawn beta-D-gulofuranose
First off, same thing about anomeric position. This is an L sugar so the anomeric OH projects down in the Haworth for a beta-anomer.
Again, you've got mixed up at the C before the ring O (C4 in this example).
You've also inverted at C5 (which you shouldn't have).
Have another shot at it. I'll post some diagrams in a bit.