TBR Orgo Ch. 2, Passage 1, Q. 2

[vanillabear55]

Can someone explain to me how the answer is C? I thought the more equatorial deuterium the better, but the explanation says "The most stable orientation has as many deuterium atoms with axial orientation possible."

Lost.

Thanks!

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[popopopop]

You want to maximize the C-H bonds in the most stable (equitorial position) since it has the most steric hindrance, and have the deuterium in the axial position. I drew it out and you only have two combos which is either 2 equ. deuterium + 1 axial deuterium, or 2 axial deuterium + 1 equ. deuterium.

 

[kraskadva]

They tell you that the C-D bond is shorter than the C-H bond, which means that D would cause less steric hinderance than H.
The most stable conformation is always the one with the most bulky groups in the equatorial position, so typically we think of H in axial and the other things in equatorial. But here, H is your 'bulky' group, so you want it equatorial and D axial.
Then, as @popopopop said, there are only 2 conformations with all 3 D up, and the most you can have axial is 2 D.

 

[vanillabear55]

Yeah I was thinking the opposite- this makes sense now thanks everyone!