Carbanion Stability

[jws95]

I remember reading somewhere in the DAT Destroyer that carbanion stability was Methyl>Primary>Secondary>Tertiary, but this problem says the opposite due to resonance stablization. Does anyone know which one is correct?

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[Mrhyde]

I remember reading somewhere in the DAT Destroyer that carbanion stability was Methyl>Primary>Secondary>Tertiary, but this problem says the opposite due to resonance stablization. Does anyone know which one is correct?

This is my reasoning , i am not 100 percent sure I am right though :

The correct Answer is A right , for number 2? ---> Look at the difference between Answer A and D . They are both Methyl anions so Dat Destroyer is correct. BUT now lets look at the only difference .... A is connected to an aromatic benzene and D is not . Now, if you did not know , Anything that is Aromatic is very stable . So even though both A and D are stable ... A is a lot more stable just because of the added aromatics.

If the Answer A was just a ring and methyl - and not aromatic then D would be the correct answer. So just because choice A is aromatic and methyl it wins.

 

[tommyngu]

You're right in the order of carbanion stability because the more substituted the carbanion with alkyl groups (which are electron donating) the less stable the carbanion is. To make a carbanion more stable, you want to minimize its negative charge by taking electron density away from it. In this case, even though choice A is tertiary with 3 phenyl groups, those phenyl groups allows that negative charge on the carbanion to be delocalized to the rings more. In other words, the phenyl rings are electron withdrawing here.

 

[jws95]

This is my reasoning , i am not 100 percent sure I am right though :

The correct Answer is A right , for number 2? ---> Look at the difference between Answer A and D . They are both Methyl anions so Dat Destroyer is correct. BUT now lets look at the only difference .... A is connected to an aromatic benzene and D is not . Now, if you did not know , Anything that is Aromatic is very stable . So even though both A and D are stable ... A is a lot more stable just because of the added aromatics.

If the Answer A was just a ring and methyl - and not aromatic then D would be the correct answer. So just because choice A is aromatic and methyl it wins.

You're right in the order of carbanion stability because the more substituted the carbanion with alkyl groups (which are electron donating) the less stable the carbanion is. To make a carbanion more stable, you want to minimize its negative charge by taking electron density away from it. In this case, even though choice A is tertiary with 3 phenyl groups, those phenyl groups allows that negative charge on the carbanion to be delocalized to the rings more. In other words, the phenyl rings are electron withdrawing here.

Thanks!!! So essentially any EWG attached to a carbanion stabilizes it more than a typical methyl carbanion?

 

[Mrhyde]

Thanks!!! So essentially any EWG attached to a carbanion stabilizes it more than a typical methyl carbanion?

whats EWG stand for ?

 

[tommyngu]

Thanks!!! So essentially any EWG attached to a carbanion stabilizes it more than a typical methyl carbanion?

Pretty much. But make sure to look at resonance effects, too. They can be more stabilizing than an EWG (electron withdrawing group)

 

[jws95]

whats EWG stand for ?

Electron-Withdrawing Group

 

[jws95]

Pretty much. But make sure to look at resonance effects, too. They can be more stabilizing than an EWG (electron withdrawing group)

That makes sense, thanks!!

 

[Mrhyde]

That makes sense, thanks!!

Also realize the differences of stability when comparing a + charge and a - charge on a carbon when you have induction present from halides

 

[jws95]

Also realize the differences of stability when comparing a + charge and a - charge on a carbon when you have induction present from halides

Wouldn't a halide destabilize a carbocation? Because it is drawing electron density away from a + charge

 

[Mrhyde]

Wouldn't a halide destabilize a carbocation? Because it is drawing electron density away from a + charge

Example 1) If you had a '+' carbon next to 3 F groups VS the same group without the 3 F groups -- then the group without the 3 F groups will be more stable. Since the F groups are trying to take electrons that a carbon with a + obviously does not have to share.

Example 2 ) If you had a '-' carbon next to 3 F groups VS the same group without the 3 F groups -- then the group with the 3 F groups will be more stable. Since the F groups are trying to take electrons that a carbon with a '-' obviously has excess electrons so will be happy to share.

 

[jws95]

Example 1) If you had a '+' carbon next to 3 F groups VS the same group without the 3 F groups -- then the group without the 3 F groups will be more stable. Since the F groups are trying to take electrons that a carbon with a + obviously does not have to share.

Example 2 ) If you had a '-' carbon next to 3 F groups VS the same group without the 3 F groups -- then the group with the 3 F groups will be more stable. Since the F groups are trying to take electrons that a carbon with a '-' obviously has excess electrons so will be happy to share.

This example helps a lot!

 

[Mrhyde]

This example helps a lot!

Of course Glad I could help 🙂 Just a small side note, remember that distance has an effect on induction so an F group 5 carbons away from a charge will have a different effect on an F group 1 carbon away from a charge.