TBR O-chem chapter 2 passage 1

[typicalindian]

Question #2 asks:

I put choice C.1 -because I always was taught that the most stable chair conformation would have the most equatorial atoms, but the explanation states that the most stable conformation would be the one with the most deuterium's in axial positions.

edit: also what is the purpose of the first sentence of the problem, what was that sentence supposed to lead me to?

---

Members don't see this ad.

 

[Ssina]

here we discussed that specific question a bit:

http://forums.studentdoctor.net/showthread.php?t=879287&highlight=axial+chair+conformation
http://forums.studentdoctor.net/showthread.php?t=877812&highlight=axial+chair+conformation

hope that helps

 

[MedPR]

This question pops up a lot, lol.

The fact that the D bond is shorter than the H bond means that H contributes more steric hindrance than D, thus you want as many H's equatorial as possible, meaning as many D's axial as possible.

 

[typicalindian]

Thank you both! So the longer the bond, the more it contributes to steric hindrance?

 

[ljc]

For two identical (or nearly identical) groups, yes. But you also have to consider what is bonded to those groups. In this case, nothing is bonded to H or D so it is irrelevant, but if these were carbons instead, you could have a completely different scenario.

 

[MedPR]

Thank you both! So the longer the bond, the more it contributes to steric hindrance?

In this case you assume that the longer bond is to the substituent that contributes more to steric hindrance because it is trying to get as far away from the ring.

Remember, the goal is to spread out as much as possible (unless there is H-bonding).