[Toonsummoner]

Hey guys, I have an exam next Wednesday and I'm getting a head start as I normally do. I'm hitting a wall, however, in reviewing the past exams that my professor has placed online. One of the biggest questions is to draw a fischer projection (easy enough, I think) of 1,2,3-pentanetriol. I understand how to draw most projections- it's just that this one has a substituent on the 1st carbon. Also, their are 4 possible stereoisomers I believe, which makes me even more confused, because I only got three (I believe one of them is a meso compound). I'm really sorry if this is a repost- I did a search and a few things came up, but the links in the replies to the threads were expired. Thanks for the help in advance!

[discodermolide]

If you have two chiral centers then there are 4 possible stereoisomers, (S)(S), (R)(R), (S)(R) and (R)(S).
See if this points you in the right direction.

[Dan]

I don't quite see what the issue with the Fischer projections is, you draw it in the normal way. E.g. take the Fischer projection of a sugar pentitol  and remove the appropriate OH groups.

If you have two chiral centers then there are 4 possible stereoisomers, (S)(S), (R)(R), (S)(R) and (R)(S).
See if this points you in the right direction.

Also, in the case of the (R,S)/(S,R), the central carbon is pseudoasymmetric. <- Edit: This is not true for 1,2,3-pentanetriol, I misread the original post as 2,3,4-pentanetriol

[sjb]

Also, in the case of the (R,S)/(S,R), the central carbon is pseudoasymmetric.

Not for 1,2,3-pentan-triol, surely? (Agreed for glycerol)

[Dan]

Also, in the case of the (R,S)/(S,R), the central carbon is pseudoasymmetric.

Not for 1,2,3-pentan-triol, surely? (Agreed for glycerol)

Whoops, sorry I read it as 2,3,4-pentanetriol, my mistake.