TBR O-chem, Section I, Passage II, Question 2

[cysky]

The question is: A carbon-deuterium bond is shorter than a carbon-hydrogen bond. Using this idea, how many deuterium atoms assume axial orientation in the most stable conformation of 1,2,6 tri-deuterium cyclohexane (the 3 deuterium atoms are cis with respect to each other)?

So 2 conformations can be assumed
1). 2 axial deuteriums and 1 equatorial deuterium
2). 1 axial deuterium and 2 equatorial deuteriums

I thought "1 axial deuterium and 2 equatorial deuteriums" would give the most stable conformation. The answer said the most stable orientation has as many deuterium atoms with axial orientation possible, so it should be "2 axial deuteriums and 1 equatorial deuterium".

Can someone please explain this? Thanks!

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[PiBond]

The question is: A carbon-deuterium bond is shorter than a carbon-hydrogen bond. Using this idea, how many deuterium atoms assume axial orientation in the most stable conformation of 1,2,6 tri-deuterium cyclohexane (the 3 deuterium atoms are cis with respect to each other)?

So 2 conformations can be assumed
1). 2 axial deuteriums and 1 equatorial deuterium
2). 1 axial deuterium and 2 equatorial deuteriums

I thought "1 axial deuterium and 2 equatorial deuteriums" would give the most stable conformation. The answer said the most stable orientation has as many deuterium atoms with axial orientation possible, so it should be "2 axial deuteriums and 1 equatorial deuterium".

Can someone please explain this? Thanks!

edit: nvm

 

[Akarat]

The question is: A carbon-deuterium bond is shorter than a carbon-hydrogen bond. Using this idea, how many deuterium atoms assume axial orientation in the most stable conformation of 1,2,6 tri-deuterium cyclohexane (the 3 deuterium atoms are cis with respect to each other)?

So 2 conformations can be assumed
1). 2 axial deuteriums and 1 equatorial deuterium
2). 1 axial deuterium and 2 equatorial deuteriums

I thought "1 axial deuterium and 2 equatorial deuteriums" would give the most stable conformation. The answer said the most stable orientation has as many deuterium atoms with axial orientation possible, so it should be "2 axial deuteriums and 1 equatorial deuterium".

Can someone please explain this? Thanks!

When considering the stability of the chair conformation, steric hindrance is the primary focus. Given that C-D bonds are shorter than C-H bonds, a C-D bond in the axial position will produce less steric hindrance than a C-H bond. The longer the bond, the more of a chance that the atom coming off of the C has a chance to hit another atom coming off of a nearby C.

And it follows that since C-H bonds produce more steric hindrance than C-D bonds, there should be as many C-H bonds in the equatorial position as possible.

 

[srivastk]

I just had the exact same question and a google search sent me right back to SDN hahaha. Thanks.

 

[cpufreak3]

Normally you maximize the substituents so that they are equitorial.

In the case of deuterium, hydrogen is considered the substituent, since it is has a longer bond.

As a result, you maximize the # of H's in the equitorial position. Hence, more deuterium in axial positions.