Chiral Question

[deleted92121]

Maybe I am missing something here but the question asks how many stereoisomers are possible from 2-bromo-1,3-pentadiene?

My observations are as follows:

1. No chiral center
2. No plane of symmetry
3. Mirror image is possible, not superimposable

Based on these observations and after drawing out the molecule, I say there are two possible stereoisomers, each of which would be an enantiomer of one another. Also said it was a chiral molecule.

Not looking for an answer but any tips or advice in wrapping my mind around this would be great.

Thanks,
DU

---

Members don't see this ad.

 

[MedPR]

Maybe I am missing something here but the question asks how many stereoisomers are possible from 2-bromo-1,3-pentadiene?

My observations are as follows:

1. No chiral center
2. No plane of symmetry
3. Mirror image is possible, not superimposable

Based on these observations and after drawing out the molecule, I say there are two possible stereoisomers, each of which would be an enantiomer of one another. Also said it was a chiral molecule.

Not looking for an answer but any tips or advice in wrapping my mind around this would be great.

Thanks,
DU

Not sure which part you are having trouble with. The stereoisomerism comes from the E,Z isomers at carbon 3. You are correct that there is no chiral center, and no plane of symmetry.

 

[deleted92121]

Not sure which part you are having trouble with. The stereoisomerism comes from the E,Z isomers at carbon 3. You are correct that there is no chiral center, and no plane of symmetry.

So there would be four possible stereoisomers, correct?

There are two enantiomers and then if you flip what is attached to Carbon 3 so you have one E and one Z. Does this sound right?

Thanks for your help.

 

[milski]

Two stereoisomers - the E,Z based on the third carbon.

There are no enantiomers. There is a plane of symmetry - the plane in which all the carbons are. That means that you can get from the left to right mirror image by just 'flipping' the molecule like a pancake.

 

[MedPR]

So there would be four possible stereoisomers, correct?

There are two enantiomers and then if you flip what is attached to Carbon 3 so you have one E and one Z. Does this sound right?

Thanks for your help.

Are you trying to apply the 2^n rule? Is there how you are getting 4 possible stereoisomers? If yes, then let me remind you that 2^n possible stereoisomers only applies for tetrahedral stereocenters, for instance, chiral carbons. There aren't any tetrahedral stereocenters here, so the 2^n is out. There are, however, two double bonds, which are potentially stereogenic. Only carbon 3, however, is stereogenic, since carbon 1 is bound to two identical groups (hydrogens). If an sp2 carbon is bound to two identical groups, there cannot be E,Z isomerism at that double bond.

There's only 2 stereoisomers, 1 is the E isomer, the other is the Z isomer. You can't "flip" what is attached to carbon 3 because there is a double bond there. Double bonds cannot rotate like single bonds can due to the pi bond.

Two stereoisomers - the E,Z based on the third carbon.

There are no enantiomers. There is a plane of symmetry - the plane in which all the carbons are. That means that you can get from the left to right mirror image by just 'flipping' the molecule like a pancake.

How is there a plane of symmetry when there's only 1 Br?

 

[milski]

How is there a plane of symmetry when there's only 1 Br?

If you draw the molecule, the page itself is a plane of symmetry. All atoms lie in the that plane except two of the hydrogens on C-5 but they're still symmetrical to each other. What's below the page is symmetrical to what's above it.

 

[MedPR]

If you draw the molecule, the page itself is a plane of symmetry. All atoms lie in the that plane except two of the hydrogens on C-5 but they're still symmetrical to each other. What's below the page is symmetrical to what's above it.

Right, thank you. I forgot that the sp2 substituents were coplanar.

 

[deleted92121]

Essentially, all I can do is swap the positions of the methyl group and the hydrogen atom on carbon 5. So based on this, there is no plane of symmetry and I cannot find a mirror image that is superimposable, ergo this is a chiral molecule with a total of 4 possible stereoisomers for 2-bromo-1,3- pentadiene. 2 of the stereoisomers are enantiomers.

If you are wondering how I got 4 possible stereoisomers, I mirrored the (E) and (Z) configurations. Am I on the right track?

 

[milski]

Essentially, all I can do is swap the positions of the methyl group and the hydrogen atom on carbon 5. So based on this, there is no plane of symmetry and I cannot find a mirror image that is superimposable, ergo this is a chiral molecule with a total of 4 possible stereoisomers for 2-bromo-1,3- pentadiene. 2 of the stereoisomers are enantiomers.

If you are wondering how I got 4 possible stereoisomers, I mirrored the (E) and (Z) configurations. Am I on the right track?

The mirror images are superimposable. The plane defined by all the C atoms is the plane of symmetry.

As a generalization, you cannot have enantiomers without a chiral center or a chiral axis. The opposite is not true - you can have chiral centers and no enantiomers, but that's not relevant here.

 

[MedPR]

Essentially, all I can do is swap the positions of the methyl group and the hydrogen atom on carbon 5. So based on this, there is no plane of symmetry and I cannot find a mirror image that is superimposable, ergo this is a chiral molecule with a total of 4 possible stereoisomers for 2-bromo-1,3- pentadiene. 2 of the stereoisomers are enantiomers.

If you are wondering how I got 4 possible stereoisomers, I mirrored the (E) and (Z) configurations. Am I on the right track?

Mirror images of E,Z isomers are superimposable. Also, there is no methyl group on Carbon 5. If there was, the compound would be named 2-bromo-1,3-hexadiene. Carbon 5 has 3 hydrogens on it.

Make models of it to prove to yourself that the mirror images are superimposable.