Draw all possible stereoisomers for 3,4-Dibromohexane and 2,3-Diamino-2,3-pentandiol

Answer 1

So Lily A.,

First, let's review how you could draw these various molecules. I recommend using the "flattened" view (such as is commonly done with simple sugars), in which the carbon skeleton extends left to right, and the other substituents (H, Br, OH, NH2) are shown branching off vertically (up or down). Then the space-3D structure bends the ends of the chain *away* from you, and lastly, the individual carbon-chain bonds rotate to achieve steric relaxation, as they will -- but that last positioning doesn't affect the chirality of the molecule.

OK, now what? Look at the crucial pair of central-chain carbon atoms. Note whether there are enough different substituents in each of the 4 tetrahedral bonding directions, such that there is the possibility of chirality there. In the first molecule, although there are always a 2C, a 3C, an H and a Br, because both chiral centers are identically positioned with respect to each other in the carbon chain, there are only "+,+" , "-,-" (identical chiralities), and +,- (opposite chirality) to be distinguished. (The other "choice", -,+ is indistinguishable from +,- The carbon chain doesn't "know" which end is which!) In the flattened view, the Br's are variously either both "above" or "below" the carbon chain, or one is "above", and the other "below". There are only those three possibilities.

I'll leave it to you to sort out the second molecfules. Remember that a chiral center is ordered in a specific way for the 4 substituents; there is only one "+" and one "-" ordering possible for each carbon. But here, the two carbons ARE distinguishable in the total carbon chain, so you will have all 4 chiral crossings possible.

-- Cheers, --Mr. d.