[giants12]

in 3-chloro-2-butanol, there are 4 stereoisomers. They make the conformations of (2R, 3R), (2S, 3S), (2R, 3S), and (2S, 3R).

I know that the 2r3r and the 2s3s should have the same heat of formation and the other two (2s3r and 2r3s) should be equal also. Which set is more stable? I thought that the 2r3r and 2s3s were because they had an anti conformation and that the others were gauche, but many people have disagreed with me.

Can someone please help me out! Tell me what type of conformations they each have (gauche, anti, or eclipsed)

[giants12]

anyone? please! I need help fast!

[djay]

which one is easier to watch?

[giants12]

I couldn't watch anything. It was just a picture...

I just want to know the conformations of them

[giants12]

somebody please help me out  ..i need to figure this out by tonight

[RBF]

I have not had time to analyze the specific conformers but....  I believe the steric requirements of the methyl groups is greater than the Cl or OH.  OH is probably less sterically demanding than Cl.  Therefore, you want to minimize the methyl methyl interactions (the more stable conformers will have them anti to each other, I believe.)  None of the stable conformers should be in eclipsed conformations. 
That's how I would analyze it.  Hope that helps a bit. 

[giants12]

Yes, that does help me understand it a little more. I used ArgusLab to make these molecules and it comes up as the 2r3r and 2s3s having anti for both the OH and Cl and the methyl groups. For the other two stereoisomers it shows a gauche for the OH and Cl and eclipsed for the methyl groups. So that would mean the first two are more stable for both reasons (regarding the Cl and OH and the methyl groups), right?

[xc]

Hello giants12

Maybe I'm arriving late, but I think your reasoning is not completely correct. Just developing what RBF already said (which I think is correct), you shoul draw the three stable conformers of each pair and compare them. I attach the drawing.

All the eclipsed conformations are excluded, because they're always less stable.

Going on with what RBF said, the most sterically demanding substituents are the methyl groups, so in each case the most stable conformation is the one that has them in anti (e.g. the a and a'). They're highlighted in the drawing.

So the problem now is reduced to comparison of a and a' and this is easy:

In both of them the methyls are anti, and in both of them the OH is gauche to a CH3 and the Cl is gauche to the other CH3. But in a OH and Cl are gauche one to each other, while in a' they're anti to each other.

So a' is more stable, which means (R,S) and (S,R) are more stable than (R,R) and (S,S).

I hope you understand better now.

[mylkoa]

I attach the drawing.

What do you use to make such nice drawings?

Thanks,
Andrew

[xc]

Hi

I use chemdraw. There are other programs such as isis draw which I think are free.