orgo question about axial orientation

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Meredith92

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I am not sure why the answer implies that its better to have the deuterium in axial rather than equatorial. I thought equatorial was always better due to the decreased steric hindrance

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 the molecules?
(shows a picture of a cyclohexane with three cis deuterium bonds... i guess you would call it: 1,2,3 cis-trideuterium cyclohexane)
A) 0
B) 1
C) 2
D) 3

The answer says because the three deuterium atoms are cis with respect to one another, they cannot all be axial nor all be equatorial. the most stable orientation has as many deterium atoms with axial orientation as possible. however, because the deuterium atoms are all mutually cis to one another, the structure must have at least one deuterium with equatorial orientation. the best choice is C

Why is axial better in this case? is it because the shorter bond decreases steric hindrance? Or is it becauae in the equatorial position you would have gauche interactions?

Thanks!
 
Good question.

Yes, the bigger the atom, the more likely it would take up the equatorial position which is further away from the ring carbons and thus less sterically hindered.

The bond length is of little relevance here imo. The question boils down to: which is bigger, dueterium or hydrogen? They have the same number of electrons but dueterium has more protons. More protons means a strong pull on the electrons and a smaller radius. Thus dueterium is actually smaller (heavier, yes, but also smaller. You can also think of it as more dense.).

So, the bigger hydrogens take up the equatorial positions. and dueterium can deal with the axial spots. So the answer should indeed be 2 axials for Dueterium, or C.
 
Good question.

Yes, the bigger the atom, the more likely it would take up the equatorial position which is further away from the ring carbons and thus less sterically hindered.

The bond length is of little relevance here imo. The question boils down to: which is bigger, dueterium or hydrogen? They have the same number of electrons but dueterium has more protons. More protons means a strong pull on the electrons and a smaller radius. Thus dueterium is actually smaller (heavier, yes, but also smaller. You can also think of it as more dense.).

So, the bigger hydrogens take up the equatorial positions. and dueterium can deal with the axial spots. So the answer should indeed be 2 axials for Dueterium, or C.

Deuterium has an extra neutron which does not effect atomic radius.

Look at bond length. Usually we place bulky substituents on the axial position. The deuterium-carbon bond length is shorter than the hydrogen-carbon bond length. We can therefore look at hydrogen as the the 'bulky' substituent and place it on the equatorial position.
 
Deuterium has an extra neutron which does not effect atomic radius.

Look at bond length. Usually we place bulky substituents on the axial position. The deuterium-carbon bond length is shorter than the hydrogen-carbon bond length. We can therefore look at hydrogen as the the 'bulky' substituent and place it on the equatorial position.

whoops! my mistake, yeah. for some reason i mistook a nuetron for a proton. but dueterium does indeed have a smaller radius than hydrogen and is thus the less bulky of the two.

the only other explanation i can give for why is a bit p-chemy. 1s orbital radius is given by the equation ℏ²/2μe².. μ (reduced mass) is larger for dueterium due to that extra neutron thus giving it a slightly smaller radius than hydrogen.
 
Deuterium has an extra neutron which does not effect atomic radius.

Look at bond length. Usually we place bulky substituents on the axial position. The deuterium-carbon bond length is shorter than the hydrogen-carbon bond length. We can therefore look at hydrogen as the the 'bulky' substituent and place it on the equatorial position.
Equatorial*
whoops! my mistake, yeah. for some reason i mistook a nuetron for a proton. but dueterium does indeed have a smaller radius than hydrogen and is thus the less bulky of the two.

the only other explanation i can give for why is a bit p-chemy. 1s orbital radius is given by the equation ℏ²/2μe².. μ (reduced mass) is larger for dueterium due to that extra neutron thus giving it a slightly smaller radius than hydrogen.
The size of a deuterium atom is irrelevant and the bond length is absolutely what matters here. The question stem tells you that a C-D bond is shorter than a C-H bond, and so we're supposed to recognize that H's are the bulky groups, so we want as many axial D's as possible.

We aren't expected to be able to determine that deuterium has a smaller atomic size than hydrogen, and the question doesn't require it.
 
I just don't see how you make this connection. C-D has shorter bond --> H is bulkier???

The question stem specifically tells you that C-D bonds are shorter. Deuterium substituents are therefore held more closely to the ring, while the longer C-H bonds mean the hydrogen substituents are further out. They're longer "spikes" coming off of the ring, if you will. Bonds are perfectly static, things wiggle and vibrate, and substituents held further out will move further through space and have a greater potential to sterically interact than substituents with shorter bonds given the same angle of bond bend.

Regardless, the only information given in the question stem is bond length, and given the context and appropriate information there's no way to know that deuterium is smaller by atomic radius.
 
The question stem specifically tells you that C-D bonds are shorter. Deuterium substituents are therefore held more closely to the ring, while the longer C-H bonds mean the hydrogen substituents are further out. They're longer "spikes" coming off of the ring, if you will. Bonds are perfectly static, things wiggle and vibrate, and substituents held further out will move further through space and have a greater potential to sterically interact than substituents with shorter bonds given the same angle of bond bend.

Regardless, the only information given in the question stem is bond length, and given the context and appropriate information there's no way to know that deuterium is smaller by atomic radius.

This is seriously the key to so many MCAT questions.
 
The question stem specifically tells you that C-D bonds are shorter. Deuterium substituents are therefore held more closely to the ring, while the longer C-H bonds mean the hydrogen substituents are further out. They're longer "spikes" coming off of the ring, if you will. Bonds are perfectly static, things wiggle and vibrate, and substituents held further out will move further through space and have a greater potential to sterically interact than substituents with shorter bonds given the same angle of bond bend.

Regardless, the only information given in the question stem is bond length, and given the context and appropriate information there's no way to know that deuterium is smaller by atomic radius.
Check out section I, page 8, the last sentence above figure1-5. it says:
"The longer bond is associated with the larger orbitals..."
 
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