[wildbiologist2]

Hey everyone, real quick question for you all. Writing another lab report (yet again) and my professor asks the following question about lidocaine:

"One of the differences among some of the H on the aromatic is due to their positions relative to the nitrogen attached to the ring. What is the one most significant result of this situation? Your answer should be one sentence that states whether meta or para is more affected, whether the effect is shielding or deshielding, and the proper term for the electronic situation that causes it. Also  include diagram(s) illustrating how this N has this effect on this position of the ring."

Meta is more affected, deshielding effect, but is it because of the dipole between the C-N bond of the ring? Does it relate to dipole induction?

Thanks again.

[Archer]

Draw the dipole for the C-N then the carbon adjacent and so on you should see how the electrons are shielded / desheilded around the ring.

[wildbiologist2]

To reiterate, the dipole of the C-N induces a dipole effect around the ring, with the para being least deshielded, correct?

[Archer]

Which would be the more desheilded proton, one attached to a δ⁺ or a δ^- atom?

[wildbiologist2]

Now I hear some people saying that resonance causes different shielding effects of aromatic H's. Any input?

[wildbiologist2]

Well the more deshielded proton would be the one closer to the δ+. The inductive effect working its way around the ring contributes to electronic deshielding with the ortho position being affected the most and the para position being affected the least.

[wildbiologist2]

Then again, I see these shift values: http://www.biosite.dk/leksikon/lidocain_uk.htm

It appears that the orto positions are deshielded the least, contrary to what I thought. Am I on the right track of thinking?

[Archer]

Then again, I see these shift values: http://www.biosite.dk/leksikon/lidocain_uk.htm

It appears that the orto positions are deshielded the least, contrary to what I thought. Am I on the right track of thinking?

There are no ortho protons in lidocaine.

Well the more deshielded proton would be the one closer to the δ+. The inductive effect working its way around the ring contributes to electronic deshielding with the ortho position being affected the most and the para position being affected the least.

You also have the mesomeric effect to take into account with electron donating groups. If a carbon is δ- would this shield or deshield the attached proton? 

[wildbiologist2]

shield

[Archer]

So if the carbon bonded to the nitrogen is δ+ what would the partial charge on the adjacent (ortho) carbon be?

[wildbiologist2]

δ+, so there is dipole induction and mesomeric effects, with the para hydrogen being the most shielded

[Archer]

δ+, so there is dipole induction and mesomeric effects, with the para hydrogen being the most shielded

Would two adjacent δ+ be likely? Think about the resonance structures

[wildbiologist2]

I am talking about the dipole effect of the C-N bond. Wouldn't the relatively large dipole of the C-N bond cause each C in the aromatic to be partially positive though due to inductive effects of the C-N dipole?

As for mesomeric effects, the ortho and para carbons have a negative character, causing the attached hydrogens to be shielded (none on the ortho position, of course.) The mesomeric effect does not contribute to shielding the meta position...

[Archer]

I am talking about the dipole effect of the C-N bond. Wouldn't the relatively large dipole of the C-N bond cause each C in the aromatic to be partially positive though due to inductive effects of the C-N dipole?

As for mesomeric effects, the ortho and para carbons have a negative character, causing the attached hydrogens to be shielded (none on the ortho position, of course.) The mesomeric effect does not contribute to shielding the meta position...

I have not explained this very well have I

Both the inductive and mesomeric effects have an impact on shielding.

With inductive the whole pi system  is δ+, each atom becoming less so as you get further from the C-N bond. As you get this decrease each carbon is δ- relative to the one before it. Does this make sense?

Mesomeric effect is usually the more influencing factor. Best example is fluorobenzene and iodobenzene. Both display inductive and mesmeric effects but in iodobenzene the inductive effect deshields the ortho protons the most but the inductive effect influences the para position. In fluorobenzene, despite the high electronegativity of fluorine the inductive effect has less influence than the mesomeric effect.

[wildbiologist2]

That was my train of thought but I guess I just had difficulty writing it down. My apologies. Thanks again for your thorough explanation.