Pyridine is a weaker base in water than 2,6 lutidine (dimethylpyridine), but the pyridine-BF3 complex is more stable than the comparable lutidine-BF3. What is the reason for this inversion in stabilities?

Question

Pyridine is a weaker base in water than 2,6 lutidine (dimethylpyridine), but the  pyridine-BF3 complex is more stable than the comparable lutidine-BF3. What is the reason for  this inversion in stabilities?

abb0t · Answer

steric effects of the two methyl groups. the addition of methyl groups would actually increase the bascitity of the compound. Can you think of why this might be? :)

anonymous · Answer

They cause it to be more difficult to remove the lone pair from N?

anonymous · Answer

Is this due to resonance?

abb0t · Answer

Yes. That's another reason. Any other reasons?

anonymous · Answer

The methyl would not be considered electronegative so I do not believe it would be due to induction as well. I do not know another reason the basicity would be affected.

abb0t · Answer

You might want to draw the resonance strucutres (if possible) as an answer to explain. Also, keep in mind, methyls groups are electron-donating.

anonymous · Answer

thank you. I have another question in the same subject. It is the last question on my homework. Would you be able to explain this: "Cu(I) and Ag(I) form donor-acceptor complexes with organic -systems. Explain why 
Cu(I) forms a better complex with ethylene than Ag(I).?

anonymous · Answer

I believe it is due to the valence orbital ionization potential. But I do not know how to construct a correct energy graph

abb0t · Answer

Oh wow, I am not that good with organometallic chemistry. I was going to suggest comparing M.O diagrams if you're good in MO theory.

abb0t · Answer

But sure, you could argue that due to the Coppers higher ionization/ (electronegativity in comparison to Ag, affecting it's nucleophilicity.