Exception to "CARDIO" for acidity? (Berkley Review)

[manohman]

So the rule for acidity, going from most important to least important is...CARDIO (Charge, Atom, Resonance, Inductive Effect, Orbitals).

But in the question below, which asks for the compound with the LOWEST pkB (most basic), and thus the least acidic, the order of importance between Resonance and Orbitals is switched.

Lone pairs/Electron delocalization DECREASE Acidity.
hybridization Increases Acidity in the following order: sp>sp2>sp3 from most acidic to least acidic.

So Berkley Review says choice A is more basic than Choice D. But if we compare which is less acidic (and thus more basic), we see that
choice D has a resonance because of the elctrons in the double bond that can delocalize onto the nitrogen itself. So it has resonance which should DECRESE Acidity. However it is also, due to its double bond sp2 hybridized, which INCREASES acidity compared to choice A.

So choice D has Decreased acidity from Resonance and Increased acidity from its hybridization compared to Choice A. And since resonance trumps hybridiation in considerations of acidity, shouldnt choice D be the less acidic, and thus MORE Basic compound? But berkley only mentions hybridization? Do you just ignore electron delocalizatoin if you have a double bond?

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[quicheduty]

Hmm I'm not following what you mean by resonance/delocalization in D. The only compound with the e- pair delocalized into a pi bond system is B. You can imagine the p orbital on the N lining up with the p orbitals of the carbons in the double bond. However in both C and D the lone pairs on the N don't really "delocalize" into the double bond. They occupy a different set of p orbitals that by definition won't align.

So just based on hybridization, A is the most basic.

 

[Thoroughbred_Med]

Why go through all that trouble?

Just pick the best base. The N on A. has a lone pair ready for action without any other complicating factors on the compound.

 

[manohman]

But so does the Nitrogen in choice D?

By delocalizatoin i mean the double bond pi electrons. They could delocalize couldnt they?

 

[quicheduty]

But so does the Nitrogen in choice D?

By delocalizatoin i mean the double bond pi electrons. They could delocalize couldnt they?

No, they can't. The lone pairs can only delocalize if they're in p orbitals that are aligned with the pi bond. In both C and D your nitrogen is already in a pi bond where the p orbitals on N are aligned with the p orbitals on the neighboring C. The lone pair occupies an sp2 orbital that is basically perpendicular to the p orbitals (in the plane of the molecule).

This is not the best pictoral representation but maybe it'll help to distinguish between the p orbitals & sp2 orbitals. For delocalization you'd have to have that lone pair in a p orbital, usually only possible if the N is not the same N involved in the double bond:

 

[manohman]

No, they can't. The lone pairs can only delocalize if they're in p orbitals that are aligned with the pi bond. In both C and D your nitrogen is already in a pi bond where the p orbitals on N are aligned with the p orbitals on the neighboring C. The lone pair occupies an sp2 orbital that is basically perpendicular to the p orbitals (in the plane of the molecule).

This is not the best pictoral representation but maybe it'll help to distinguish between the p orbitals & sp2 orbitals. For delocalization you'd have to have that lone pair in a p orbital, usually only possible if the N is not the same N involved in the double bond:

I see okay that makes a lot more sense now. So electrons in a double bond cannot result in resonance structures?

the scenario that i am envisioning is like what happens here

but that may be only becaus all of the p orbitals are lined up, yes? so lone pair delocalization can happen any time, but for double bond delocalization, there must be a p ortibal available and it must be in the same plane.

 

[quicheduty]

I see okay that makes a lot more sense now. So electrons in a double bond cannot result in resonance structures?

the scenario that i am envisioning is like what happens here

View attachment 187929

but that may be only becaus all of the p orbitals are lined up, yes? so lone pair delocalization can happen any time, but for double bond delocalization, there must be a p ortibal available and it must be in the same plane.

Oh no, yes they can (double bonds are part of the reason why resonance exists!). In your example, that CH2 carbon is actually sp2 hybridized, since there are only 3 ligands. It being a cation means there's a free p orbital that can align with the conjugated pi bonds.

However, if you look at C and D, all of the neighboring carbons are sp3 and do not align. Therefore there are no resonance structures and no delocalization.

In B, the N is actually sp2 because its p orbital can align with the pi bond.

 

[manohman]

Oh no, yes they can (double bonds are part of the reason why resonance exists!). In your example, that CH2 carbon is actually sp2 hybridized, since there are only 3 ligands. It being a cation means there's a free p orbital that can align with the conjugated pi bonds.

However, if you look at C and D, all of the neighboring carbons are sp3 and do not align. Therefore there are no resonance structures and no delocalization.

In B, the N is actually sp2 because its p orbital can align with the pi bond.

Ah i see! Thank you thats so much clearer now. Cheers man. Happy Monday!

 

[quicheduty]

Ah i see! Thank you thats so much clearer now. Cheers man. Happy Monday!

You too dude! No prob