[jkulier]

Hello,

I have been given the compound (1S,2R,3S)-2-(1-ethylpropyl)-3-iodo-cyclohexan-1-ol, and I cannot figure out how to make the second Carbon an R! According to priority, it would be: 1st - Iodine, 2nd - Ethylpropyl, 3rd - OH (I think), but that's as far as I know how to go.

For reference, I've attached a picture of what I think the compound looks like...
any help would be great,
thanks! 

[sjb]

I think you have the structure right, but would disagree that the ethylpropyl branch is higher priority than the hydroxyl. Can you draw out the structure with carbons 1 and 3 in stereo?

[jkulier]

You're right, the Ethylpropyl is of lesser priority than the -OH group. Thanks for correcting me on that! In that case then, then I do not have to do anything with the structure, as the 2nd Carbon is already an R?

Oh also, with regards to making the 1st and 3rd Carbons S's, I believe I can do that 

Thanks again.

[Squirmy]

Right now the stereochemistry is ambiguous...to show whether each chiral center is R/S, you need to use dashed/hatched bonds. See http://www.cem.msu.edu/~reusch/VirtualText/intro3.htm#strc4

For your cyclohexane, it's easiest to leave all of the carbons of the ring in the same plane. On each carbon, there will be one substituent that's a hatched bond (going away from you), and one that's a wedged bond (coming out at you).

One of the first rules for assigning R and S is that the lowest priority group (#4) must be a hatched bond. Otherwise you'll get the wrong answer. The good news is that with only R and S as possibilities, knowing the wrong answer also gives you the right one.

For the problem at hand, you actually haven't identified the lowest priority group, yet. Can you figure out what it is?

If the order of 1 --> 2 --> 3 already goes in a clockwise fashion, which group coming off the ring must be hatched bond and which must be a wedged bond?

[jkulier]

I am quite sure that for all 3 Carbons, Hydrogen would take the lowest priority (which also means that H is always the hatched wedge).

For the second Carbon, if going from I -> OH -> Ethylpropyl is already in a clockwise fashion, then should Ethylpropyl be on the solid wedge and H be on the hatched wedge? (I made the model, and this seems to make sense  )

Also, with the model, I realized I don't quite have the hang of the "S" on the first Carbon. I put H on the hatched wedge and the OH group on a solid wedge, but this seemed to give an R configuration. Going from OH -> Ethylpropyl -> 6-C was clockwise. What am I doing wrong?

Sorry for the trouble,
thanks in advance.

[Squirmy]

You're right about the H being the lowest priority group in each case and from what I read, sounds like you have the 2 and 3 carbons correct.

Also, with the model, I realized I don't quite have the hang of the "S" on the first Carbon. I put H on the hatched wedge and the OH group on a solid wedge, but this seemed to give an R configuration. Going from OH -> Ethylpropyl -> 6-C was clockwise. What am I doing wrong?

Exactly...if the OH were a wedge, it would be R. So try something else 

Here's an analogy for you...say you set up a rule for figuring out which of your hands was left and right by saying "if pinky -> thumb is clockwise, it's my right hand". This rule only works if you look at the hand with the palm facing you. If the palm's facing away, clockwise pinky -> thumb is actually the left hand.

The same goes for R/S. The rule that the lowest priority group must be pointing away (a hatched bond) ensures that you're looking at the right face/side of the molecule. If the lowest priority group is a wedge, you get the wrong assignment for R/S. However, with only 2 possible answers, knowing the wrong answer is pretty useful.

[orgopete]

Let's not mix up CIP nomenclature rules with drawing this structure. The assignment rules do specify the lowest priority group should be to the back.

In drawing a structure or examining a model, the absolute stereochemistry will remain fixed when viewed from all angles. Because three groups are being asked for assignment, you must perform the reverse of doing a CIP assignment. You already know the first carbon is S. The easiest method for this problem is to make the cyclohexane planar. Therefore the H and OH will be hashed or wedged. You just need to assign them properly. Since OH is 1, CHR  is 2, CH2 is 3, and H is 4. If H is to the back, if the provided structure is used, that will give 1R. In order to get 1S, you must either use the mirror image (equivalent to flipping the cyclohexane over) or reverse the H and OH. That is, make H wedged and OH hashed.

I think that will make the hydrogens at C2 and C3 hashed and hashed.

[jkulier]

Thank you so much for all the input, everyone! I now understand this =) Thx again