TPR stereochemistry question

[circulus vitios]

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Isn't the product an achiral molecule?

 

[milski]

Interesting. I am not sure why they drew the double bond out of the plane of the other two bonds which are part of the ring (atom C-4). Normally, I would imagine this to be a flat structure and the compound will be achiral. In the case of cyclohexane the bond angle is close to 109 and that probably makes the double bond pop in one direction.

 

[gettheleadout]

Interesting. I am not sure why they drew the double bond out of the plane of the other two bonds which are part of the ring (atom C-4). Normally, I would imagine this to be a flat structure and the compound will be achiral. In the case of cyclohexane the bond angle is close to 109 and that probably makes the double bond pop in one direction.

That is odd.

As a note though, if this question were on the MCAT, you could reason that A and D are wrong, and B is wrong because without any chiral centers in the reactant there's no difference in steric hindrance of the reaction with NaBH4 on either side of the carbonyl.

Though considering that they're saying the reactant would have the alkene substituent out of the ring plane, the reaction wouldn't actually form a racemic mixture would it? It'd be a major/minor mix.

In any case, this is cyclohexanone:

Looks in the plane to me.

 

[milski]

Yes, it's strange, I don't remember seeing anything like that. With my minimal exposure to organic, that does not mean much.

Why do you think it will be major/minor? The two enantiomers that they have draw seem equally stable/likely to happen to me?

 

[gettheleadout]

Major/minor only because the face of the ring with the alkene substituents will be sterically hindered compared to the opposite face of the ring. It doesn't have to be a huge effect, but that group is rather large anyway relative to the rest of the molecule, and as a rule anytime you have asymmetric reactants you won't get a racemic mixture. Typically that goes for chiral reactants specifically, but when a bent-out substituent like that it might as well be a chiral center (imagine a hydrogen extending out into the other face of the ring on the same carbon as the alkene).

 

[milski]

Major/minor only because the face of the ring with the alkene substituents will be sterically hindered compared to the opposite face of the ring. It doesn't have to be a huge effect, but that group is rather large anyway relative to the rest of the molecule, and as a rule anytime you have asymmetric reactants you won't get a racemic mixture. Typically that goes for chiral reactants specifically, but when a bent-out substituent like that it might as well be a chiral center (imagine a hydrogen extending out into the other face of the ring on the same carbon as the alkene).

While I was trying write a better explanation of why I don't see the sterical hinderance, I realized something else. Even if the double bond stays in the same plane, it cannot rotate and the presence of Cl/Et gives you a specific orientation. At that point, reducing the =O to OH/H will lead to enantiomers with the OH group being above/below the ring. The ring can re-conform so that the OH stays in the less hindered position and you end up with two equally likely enantiomers.

Were you trying to say that during the reduction the O (which becomes OH) is more likely to stay on the same side to which it was bent initially? That would make sense to me but see above paragraph about the possibility of it not being bent.

 

[cheesier]

I think it's out of the plane because they chose for the double bond to oxygen be equatorial, so the other group has to be axial since it is para. You could still get it purely by process of elimination though. B is wrong since racemic mixtures are by definition optically inactive. If you know that NaBH4 is a mild reducing agent A and D are out as well, leaving you with C.

 

[circulus vitios]

Maybe I'm misunderstanding something but for a racemic mixture the product must be chiral, and a molecule must be sp3 hybridized with 4 different substituents for it to be chiral. The site of the reaction (carbon 1 & the attached oxygen) is not chiral because there are not 4 different substituents. The double bond cannot be chiral because it's sp2 hybridized. There are absolutely no chiral centers, so there cannot be a racemic mixture? Am I supposed to choose the "least wrong" answer if all of the answers are wrong?

edit:

Or must a molecule simply have in-plane/out-of-plane wedges to be optically active?

 

[cheesier]

Maybe I'm misunderstanding something but for a racemic mixture the product must be chiral, and a molecule must be sp3 hybridized with 4 different substituents for it to be chiral. The site of the reaction (carbon 1 & the attached oxygen) is not chiral because there are not 4 different substituents. The double bond cannot be chiral because it's sp2 hybridized. There are absolutely no chiral centers, so there cannot be a racemic mixture? Am I supposed to choose the "least wrong" answer if all of the answers are wrong?

edit:

Or must a molecule simply have in-plane/out-of-plane wedges to be optically active?

Carbon 1 is chiral. It has a hydroxyl group, a hydrogen, and carbons on either side that are different, so it is sp3 with four different substituents. That's the only chiral center on the molecule because the double bond between carbons 4 and 5 doesn't react since NaBH4 is a very weak reducing agent.

 

[circulus vitios]

Carbon 1 is chiral. It has a hydroxyl group, a hydrogen, and carbons on either side that are different, so it is sp3 with four different substituents. That's the only chiral center on the molecule because the double bond between carbons 4 and 5 doesn't react since NaBH4 is a very weak reducing agent.

So we go around the ring and across the double bond to find the "point of difference" (the point of difference being the chlorine and ethyl groups) that gives carbon 1 four different substituents?

 

[milski]

So we go around the ring and across the double bond to find the "point of difference" (the point of difference being the chlorine and ethyl groups) that gives carbon 1 four different substituents?

Yes. And it is important that the ring and the double bond are holding the Cl/Et in place to make the two paths to them different. If you replace the double bond with a single one, the molecule will become achiral.

 

[Trayshawn]

Im pretty sure the structure they have for the product is wrong. The fact that one of the ring carbons is involved in a double bond an dis thus sp2 hybridized removes the possibility of existing as a chair. Chair conformations have each carbon sp3 hybridized (thats the whole point of a chair, its the only configuration that gives 109.5 degree bond distances to each carbon as all sp3 carbons want).

In any case, if you draw a mirror image of the flat molecule, you'll still notice that they are not superimposable thanks to the vinyl chloride and ethyl groups.

Lastly, MCAT strategy would say eliminate A and D because you know NaBH4 reacts with carbonyls and there is a carbonyl present at carbon 1. Then eliminate B because a racemic mixture is, by definition, optically inactive. Only choice C is left. Thats probably the best way to do this.

 

[gettheleadout]

While I was trying write a better explanation of why I don't see the sterical hinderance, I realized something else. Even if the double bond stays in the same plane, it cannot rotate and the presence of Cl/Et gives you a specific orientation. At that point, reducing the =O to OH/H will lead to enantiomers with the OH group being above/below the ring. The ring can re-conform so that the OH stays in the less hindered position and you end up with two equally likely enantiomers.

Were you trying to say that during the reduction the O (which becomes OH) is more likely to stay on the same side to which it was bent initially? That would make sense to me but see above paragraph about the possibility of it not being bent.

That's a good point. What I was trying to say is that the configuration of the chiral C1 in the product wouldn't be racemic because steric hindrace from the ethyl group on one face of the ring vs lower steric hindrance from the chlorine atom on the other side would differently affect the approach of NaBH4 to the molecule.

I think it's out of the plane because they chose for the double bond to oxygen be equatorial, so the other group has to be axial since it is para. You could still get it purely by process of elimination though. B is wrong since racemic mixtures are by definition optically inactive. If you know that NaBH4 is a mild reducing agent A and D are out as well, leaving you with C.

"Choosing" the double bond to oxygen to be equatorial doesn't make any sense though, the carbons in the ring where double bonds exist will be sp2 and trigonal planar, they won't bend out from the right and be equatorial or axial. Also, I think you mean trans, not para.

I have a meeting with an organic professor on Thursday, I might bring him this and ask just to be sure.

 

[cheesier]

"Choosing" the double bond to oxygen to be equatorial doesn't make any sense though, the carbons in the ring where double bonds exist will be sp2 and trigonal planar, they won't bend out from the right and be equatorial or axial. Also, I think you mean trans, not para.

I have a meeting with an organic professor on Thursday, I might bring him this and ask just to be sure.

I don't know if double bonds can orient out of the plane so update when you find out.

As for whether it's racemic or not, it's a small group and it's para to the reactant site, so I don't think there's a really strong case for why one product would be preferred over the other. At least in my organic class we were generally taught that para groups don't really affect regiochemistry. I'm sure that's a huge overgeneralization but my guess is it holds true here.

 

[milski]

That's a good point. What I was trying to say is that the configuration of the chiral C1 in the product wouldn't be racemic because steric hindrace from the ethyl group on one face of the ring vs lower steric hindrance from the chlorine atom on the other side would differently affect the approach of NaBH4 to the molecule.


"Choosing" the double bond to oxygen to be equatorial doesn't make any sense though, the carbons in the ring where double bonds exist will be sp2 and trigonal planar, they won't bend out from the right and be equatorial or axial. Also, I think you mean trans, not para.

I have a meeting with an organic professor on Thursday, I might bring him this and ask just to be sure.

Are not the Cl/Et pretty far from the =O group to matter? Although you can always make the argument that it matters, even just a tiny bit. Anyway, I understand what you mean. It will be interesting to hear what your o-chem prof will say.

 

[gettheleadout]

Are not the Cl/Et are pretty far from the =O group to matter? Although you can always make the argument that it matters, even just a tiny bit. Anyway, I understand what you mean. It will be interesting to hear what your o-chem prof will say.

That's the thing is I've been under the impression that it always matters, even if its slight.

 

[milski]

That's the thing is I've been under the impression that it always matters, even if its slight.

Got it. Feels like one of those situations where you can be wrong no matter what you answer.

 

[Docility]

Lurker here. 😀

I have a meeting with an organic professor on Thursday, I might bring him this and ask just to be sure.

Did you have a chance to ask him about this?

I've been dying for some clarification. 😳

 

[gettheleadout]

I asked!

The TPR representation of the product is wrong; the double bond is in the plane of the ring (even though it's technically a chair, but let's consider it flat). Showing the double bond pointing down in the product is just incorrect.

Also, the Et and Cl substituents are in plane with the ring as well, so there would not be an effect on steric hindrance of the NaBH4 reaching the carbonyl.

 

[Docility]

I asked!

The TPR representation of the product is wrong; the double bond is in the plane of the ring (even though it's technically a chair, but let's consider it flat). Showing the double bond pointing down in the product is just incorrect.

Also, the Et and Cl substituents are in plane with the ring as well, so there would not be an effect on steric hindrance of the NaBH4 reaching the carbonyl.

Ah, thanks for sharing! 🙂

 

[milski]

I asked!

The TPR representation of the product is wrong; the double bond is in the plane of the ring (even though it's technically a chair, but let's consider it flat). Showing the double bond pointing down in the product is just incorrect.

Also, the Et and Cl substituents are in plane with the ring as well, so there would not be an effect on steric hindrance of the NaBH4 reaching the carbonyl.

Cool, I'll sleep well tonight! 😀 Thanks for taking the time to do that! 👍