Bicyclic molecules such as Camphor and the # of stereoisomer... its confusing :(

[AlwaysLucky]

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I just finished Passage 8 from the Orgo section of TPRH Workbook and got totally stumped on the first question.

The question is: What is the maximum # of stereoisomers of camphor?

I counted two chiral carbons, so I figured it would be 2^n (where n = 2), and thus, 4 stereoisomers ... however, it's wrong!

The explanation says because "camphor is a bicyclic molecule, the bridge carbon stereocenters cannot be inverted." WHHHAATTTT??? "Therefore, camphor only has one enantiomer and no diastereomers." HUH???

It's been 6 years since I've last taken orgo, and it definitely isn't my strength ... could someone explain what "cannot be inverted" means? Do I just take it as a given that whenever I'm given a bicyclic molecule, the # of stereoisomer will always be 2? What if they give me a tricyclic molecule?

lol .. sorry, this question just threw me off completely and I have no idea how I should approach such a question if I see it on the actual MCAT 🙁

Thanks in advance for the help!

 

[mcloaf]

Well in the case of camphor your two stereocenters are both on carbons that are involved in the bicyclic ring system. So in this case you can't just flip/invert for example the methyl group because you'd have to twist and flip the entire molecule in on itself to accommodate this change, which is a contortion that realistically is not going to happen. So here, you have camphor, and its mirror image, giving you two enantiomers.

If you had something like a halogen attached to a carbon that's not part of the bicyclic bridge you'd be free to flip it around with the hydrogen and get multiple stereoisomers.

 

[MedPR]

I just finished Passage 8 from the Orgo section of TPRH Workbook and got totally stumped on the first question.

The question is: What is the maximum # of stereoisomers of camphor?

I counted two chiral carbons, so I figured it would be 2^n (where n = 2), and thus, 4 stereoisomers ... however, it's wrong!

The explanation says because "camphor is a bicyclic molecule, the bridge carbon stereocenters cannot be inverted." WHHHAATTTT??? "Therefore, camphor only has one enantiomer and no diastereomers." HUH???

It's been 6 years since I've last taken orgo, and it definitely isn't my strength ... could someone explain what "cannot be inverted" means? Do I just take it as a given that whenever I'm given a bicyclic molecule, the # of stereoisomer will always be 2? What if they give me a tricyclic molecule?

lol .. sorry, this question just threw me off completely and I have no idea how I should approach such a question if I see it on the actual MCAT 🙁

Thanks in advance for the help!

Does typing stuff like "WHHHAATTTT???" and "HUH???" make you feel better? Just curious.

Anyway, how would you invert the configuration of a carbon that is bound to two different rings? It's like trying to rotate a pi bond, it just doesn't happen. As always with orgo, if you can't visualize it in your head, it is best to make a model and prove it to yourself. You might also want to read up on the 2^n rule and understand it (rather than just memorize it) because if you understand why the rule exists, you won't always be wondering if certain situations obey the rule or not; you'll just be able to figure it out. Knowing that the 2^n rule is the maximum number of stereoisomers is a good start (and key to remember). Another good example of a tricky 2^n based question is meso compounds.

There is no hard and fast rule to how many stereoisomers a bicyclic compound will have because it depends on 1. how many chiral carbons and 2. which carbons (bridgehead or not) are the chiral ones.

 

[AlwaysLucky]

Well in the case of camphor your two stereocenters are both on carbons that are involved in the bicyclic ring system. So in this case you can't just flip/invert for example the methyl group because you'd have to twist and flip the entire molecule in on itself to accommodate this change, which is a contortion that realistically is not going to happen. So here, you have camphor, and its mirror image, giving you two enantiomers.

If you had something like a halogen attached to a carbon that's not part of the bicyclic bridge you'd be free to flip it around with the hydrogen and get multiple stereoisomers.

Thx for the clarification!

So, let's say we have camphor, but have a halogen attached to a carbon that isn't part of the bicyclic bridge (instead of a hydrogen) ... then would we have 4 stereoisomers (because the two bicyclic carbons account for 2 stereoisomers, while the 1 chiral carbon accounts for 2 stereoisomers .. so 2+2 = 4?)

 

[AlwaysLucky]

Does typing stuff like "WHHHAATTTT???" and "HUH???" make you feel better? Just curious.

Anyway, how would you invert the configuration of a carbon that is bound to two different rings? It's like trying to rotate a pi bond, it just doesn't happen. As always with orgo, if you can't visualize it in your head, it is best to make a model and prove it to yourself. You might also want to read up on the 2^n rule and understand it (rather than just memorize it) because if you understand why the rule exists, you won't always be wondering if certain situations obey the rule or not; you'll just be able to figure it out. Knowing that the 2^n rule is the maximum number of stereoisomers is a good start (and key to remember). Another good example of a tricky 2^n based question is meso compounds.

There is no hard and fast rule to how many stereoisomers a bicyclic compound will have because it depends on 1. how many chiral carbons and 2. which carbons (bridgehead or not) are the chiral ones.

Haha, yes it does!

And yes, I can't visualize it in my head 🙁 .. I wish TPRH had a bunch of these 2^n trick questions I can practice; so far, just encountered one meso question in the TPR book & camphor