Chirality in TBR

[powersellingmom]

So there is this question:

Which of the following CANNOT form optical isomers?
A . A four carbon gem diol
B. A four carbon vicinal diol
C. A fourcarbon secondary amine
D. A four carbon secondarv alcohol

I didn't remembered what vicinal and geminal meant, but in retrospect it's easy to see that A is the most obvious answer. However, when I was actually trying to guess an answer, I thought it would be C, because a secondary amine can either have two ethyl groups attached, or one methyl and one propyl group attached. Either way, none of the carbons are stereogenic. Does this mean that the amine is the stereocenter, and if so, does that mean that the lone pair of an atom acts a different substituent?

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[Testosterone]

The second scenario you described is certainly chiral. A methyl, propyl, hydrogen and amine groups are all different functional groups.

 

[sciencebooks]

So there is this question:

Which of the following CANNOT form optical isomers?
A . A four carbon gem diol
B. A four carbon vicinal diol
C. A fourcarbon secondary amine
D. A four carbon secondarv alcohol

I didn't remembered what vicinal and geminal meant, but in retrospect it's easy to see that A is the most obvious answer. However, when I was actually trying to guess an answer, I thought it would be C, because a secondary amine can either have two ethyl groups attached, or one methyl and one propyl group attached. Either way, none of the carbons are stereogenic. Does this mean that the amine is the stereocenter, and if so, does that mean that the lone pair of an atom acts a different substituent?

It's not how I solved it/envisioned things, but for future reference, chirality actually isn't limited to tetrahedral carbons.

 

[powersellingmom]

The second scenario you described is certainly chiral. A methyl, propyl, hydrogen and amine groups are all different functional groups.

In the diethyl amine, all of the carbons have either two or three hydrogens. same in the methyl/propyl amine. If it's methyl/isopropyl then one of the carbons has two CH3 groups.

@Sciencebooks, so you're saying that any atom with four non-equivalent groups attached can be chiral, and the a lone pair counts as a substituent? Doesn't that make most secondary amines chiral centers then?

 

[sciencebooks]

In the diethyl amine, all of the carbons have either two or three hydrogens. same in the methyl/propyl amine. If it's methyl/isopropyl then one of the carbons has two CH3 groups.

@Sciencebooks, so you're saying that any atom with four non-equivalent groups attached can be chiral, and the a lone pair counts as a substituent? Doesn't that make most secondary amines chiral centers then?

I literally asked about this to a teacher last week and he answered that it was possible but not expected on the MCAT. I just Googled for some affirmation:

Other elements can produce chiral centers besides carbon, although their importance in industry and in
health studies is not as great. The sulfur atom can produce chiral centers for example the sulfoxides,
sulfoximides, sulphonates and the sulfonium ion.

http://sci.kufauniv.com/teaching/akeel/Chiral Chromatography - Thomas E. Beesley.pdf

Other elements in addition to carbon can be stereocenters. The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter.

http://chemwiki.ucdavis.edu/Organic...erism_–_chirality,_stereocenters,_enantiomers

I hadn't asked about lone pairs but this site seems to confirm it:

Any atom which has four different groups bonded to it is a chirality center. The more common of these atoms, with which an organic chemist should be familiar, are Si, N and P (note: a lone pair is included as one of the four different groups)

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-3.html

 

[stevenz]

From what I understand, chiral centers with lone pairs can spontaneously flip along the axis of the lone pair, creating the enantiomer. That's why secondary amine solutions are not optically active; they spontaneously racemize.

On the other hand, it is impossible for A to even have a stereocenter.

 

[sciencebooks]

From what I understand, chiral centers with lone pairs can spontaneously flip along the axis of the lone pair, creating the enantiomer. That's why secondary amine solutions are not optically active; they spontaneously racemize.

On the other hand, it is impossible for A to even have a stereocenter.

That makes absolute sense as well. Thanks for adding on, .

 

[TactFR44]

So there is this question:

Which of the following CANNOT form optical isomers?
A . A four carbon gem diol
B. A four carbon vicinal diol
C. A fourcarbon secondary amine
D. A four carbon secondarv alcohol

I didn't remembered what vicinal and geminal meant, but in retrospect it's easy to see that A is the most obvious answer. However, when I was actually trying to guess an answer, I thought it would be C, because a secondary amine can either have two ethyl groups attached, or one methyl and one propyl group attached. Either way, none of the carbons are stereogenic. Does this mean that the amine is the stereocenter, and if so, does that mean that the lone pair of an atom acts a different substituent?

I'm brining this up because I'm having the same issue and glad someone else had it in the past. The explanation states that a secondary amine with four carbons can have a CARBON that is chiral...uhh I can't see this. I understand how the Nitrogen can be chiral but the CARBON?

Nitrogen in a secondary amine has to be bonded to two other carbons. So the only way I'm coming up with this is Nitrogen (with one H and lone pair) is bonded to the carbon chain in a variety of ways with two of the carbons being bonded directly to nitrogen. There are no chiral carbons! Am I wrong?

 

[AwayFromReality]

I'm brining this up because I'm having the same issue and glad someone else had it in the past. The explanation states that a secondary amine with four carbons can have a CARBON that is chiral...uhh I can't see this. I understand how the Nitrogen can be chiral but the CARBON?

Nitrogen in a secondary amine has to be bonded to two other carbons. So the only way I'm coming up with this is Nitrogen (with one H and lone pair) is bonded to the carbon chain in a variety of ways with two of the carbons being bonded directly to nitrogen. There are no chiral carbons! Am I wrong?

The only way I can picture a 4 carbon secondary amine is if one of the carbons is bonded to another element. So we could potentially have this amine:

CH2OH
|
HC----NH-----CH3
|
CH3

Notice how the carbon that is bonded to the left of the nitrogen have four unique substituents making it chiral. I think this question is one of those annoying "A is more correct than C" types. A is obviously not chiral whereas there could be some doubt as to how C can be chiral

PS: I made up that molecule, I'm not sure if it even exist in nature or lab. The point is, it is possible to have a chiral carbon

 

[knv2u]

Oops! I'm wrong - it was sloppiness on my part.

 

[AwayFromReality]

OP, try to visualize the structure or use a modeling kit to actually see what it would like like in three dimensions. For instance, with the figure above, free rotation about the HC-NH sigma bond is possible and will produce an identical molecule, which would mean that it is not optically active. Put another way, flipping the CH2OH and CH3 in the figure above doesn't create an enantiomer. I don't think a four carbon secondary amine can be chiral.

Well the original question in the OP is asking "which of the following cannot form optical isomers?" The answer explanation states a four carbon secondary amine can be chiral so there must be some structure that makes a four carbon secondary amine chiral.

The structure that I posted above seems chiral to me, based on what I learned from TBR orgo (a carbon with four unique substituent is chiral). Sure if we have both of the enantiomers together there would be no optical rotation due to racemization. But according to you we would not get any enantionmer? Could you explain that a bit more, it seems I might be missing something. For a carbon to be chiral it has to have only a sigma bond (since double and triple bonds would mean it does not have all unique substituent) therefore there is always free rotation. Am I interpreting this wrong?

EDIT: are you saying that because the CH2OH and CH3 groups are drawn in the plane of the paper, the molecule is not chiral? If CH2OH was on wedge and CH3 on a dash then flipping the two will definitely create an enantiomer. I drew both of those in the plane of the paper because there was no way for me to draw wedges and dashes using sdn's text editor

 

[Cawolf]

I agree @AwayFromReality, the structure you posted is chiral.

 

[knv2u]

Well the original question in the OP is asking "which of the following cannot form optical isomers?" The answer explanation states a four carbon secondary amine can be chiral so there must be some structure that makes a four carbon secondary amine chiral.

The structure that I posted above seems chiral to me, based on what I learned from TBR orgo (a carbon with four unique substituent is chiral). Sure if we have both of the enantiomers together there would be no optical rotation due to racemization. But according to you we would not get any enantionmer? Could you explain that a bit more, it seems I might be missing something. For a carbon to be chiral it has to have only a sigma bond (since double and triple bonds would mean it does not have all unique substituent) therefore there is always free rotation. Am I interpreting this wrong?

EDIT: are you saying that because the CH2OH and CH3 groups are drawn in the plane of the paper, the molecule is not chiral? If CH2OH was on wedge and CH3 on a dash then flipping the two will definitely create an enantiomer. I drew both of those in the plane of the paper because there was no way for me to draw wedges and dashes using sdn's text editor

I was being sloppy. It is my error, not yours.

 

[knv2u]

I agree @AwayFromReality, the structure you posted is chiral.

Yes, you two are right - the carbon to the left clearly has four different substituents. Sorry for any confusion.