TBR cyclohexane question

[LoLCareerGoals]

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Page 147 question 2:
A C-D bond is shorter than a C-H bond. Using this idea, how many D
atoms assume axial orientation in the most stable
conformation of the following molecule?

Picture of 1,2,3 cis-triDeuterium cyclohexane.

Frankly I don't know how to use info that C-D bond is shorter. So it is stronger...so what?
My level of understanding (so it can be corrected/augmented):
"axial bad. Substituent on axial bad. Bulky substituent on axial very bad. D has extra neutron, so bulkier than H. D on axial bad. For 3 in a row at least 1 has to be axial though.
Answer: 1.
Book's answer is 2. 🙁

 

[BerkReviewTeach]

Page 147 question 2:
A C-D bond is shorter than a C-H bond. Using this idea, how many D
atoms assume axial orientation in the most stable
conformation of the following molecule?

Picture of 1,2,3 cis-triDeuterium cyclohexane.

Frankly I don't know how to use info that C-D bond is shorter. So it is stronger...so what?
My level of understanding (so it can be corrected/augmented):
"axial bad. Substituent on axial bad. Bulky substituent on axial very bad. D has extra neutron, so bulkier than H. D on axial bad. For 3 in a row at least 1 has to be axial though.
Answer: 1.
Book's answer is 2. 🙁

You have all of the understanding you need in your "axial bad" quote. Axial is bad because bulky groups can collide with one another in that orientation (1,3-diaxial interactions). The most stable structure is the one that minimizes steric hindrance (as long as their isn't some extreme attractive force like H-bonding). In this question, they tell us that C-D is shorter than C-H, so "C-H is bad". You want to put as many C-H bonds equatorial as possible and in doing so C-D gets stuck in axial. In an ideal situation, all three C-D bonds would be pushed to axial, but because the structure is 1,2,3-cis, at least one of them has to be equatorial. So the two possible chair confirmations have (1) 2 axial C-D bonds and 1 equatorial C-D bond and (2) 2 equatorial C-D bonds and 1 axial C-D bond. The better option is (1), which leads to the best answer.

 

[LoLCareerGoals]

You have all of the understanding you need in your "axial bad" quote. Axial is bad because bulky groups can collide with one another in that orientation (1,3-diaxial interactions). The most stable structure is the one that minimizes steric hindrance (as long as their isn't some extreme attractive force like H-bonding). In this question, they tell us that C-D is shorter than C-H, so "C-H is bad". You want to put as many C-H bonds equatorial as possible and in doing so C-D gets stuck in axial. In an ideal situation, all three C-D bonds would be pushed to axial, but because the structure is 1,2,3-cis, at least one of them has to be equatorial. So the two possible chair confirmations have (1) 2 axial C-D bonds and 1 equatorial C-D bond and (2) 2 equatorial C-D bonds and 1 axial C-D bond. The better option is (1), which leads to the best answer.

I don't understand why if C-D is shorter then C-H, so C-H is bad part. Shorter means bulkier now? D is bulkier than H, so C-D is bad! What's wrong with logic of the previous sentence.

 

[BerkReviewTeach]

I don't understand why if C-D is shorter then C-H, so C-H is bad part. Shorter means bulkier now? D is bulkier than H, so C-D is bad! What's wrong with logic of the previous sentence.

Think of someone swing a large stick around versus a smaller stick. Which would clear space in a crowd? The longer stick sweeps over larger area, so it offers more crowd repulsion if you will. This why we consider an ethyl group to prefer equatorial over a methyl group. The larger group gets relegated to the equatorial position. The C-H is longer than C-D, so it bumps into more things than a C-D, making it the larger group.