Can NH2 with no formal charge on it be protonated?

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Discussion Overview

The discussion centers on the protonation of an NH2 group in the context of a specific molecular structure featuring a benzene ring, an NH2 group, and a methoxy group. Participants explore the reasoning behind the protonation preference of the NH2 group over the methoxy oxygen, considering factors such as basicity and the influence of the aromatic ring.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how an NH2 group, which already has two hydrogen bonds and a lone pair, can form another bond upon protonation.
  • Another participant asserts that amines, including NH2, can be protonated similarly to ammonia, implying a general principle of amine protonation.
  • A participant reflects on the implications of protonation not violating the octet rule for nitrogen when considering H+ addition.
  • It is noted that nitrogen is typically a stronger base than oxygen, suggesting a preference for protonation of the NH2 group over the methoxy oxygen.
  • Examples are provided comparing the basicity of ammonia and water, highlighting the pKa values of NH4+ and H3O+ to illustrate the relative strengths of bases.
  • Participants discuss the effect of the aromatic ring on the basicity of aromatic amines, indicating that the ring withdraws electron density, making the amine a weaker base.
  • A question is raised about how an ether group across the aromatic ring might influence the pKa of an aromatic amine.

Areas of Agreement / Disagreement

Participants express differing views on the factors influencing protonation, particularly regarding the roles of nitrogen and oxygen in the context of basicity. There is no consensus on how the methoxy group affects the protonation preference or the overall basicity of the structure.

Contextual Notes

The discussion includes assumptions about the basicity of functional groups and the influence of the aromatic system, which may not be fully resolved. The participants do not clarify the specific conditions under which the protonation occurs.

Who May Find This Useful

This discussion may be of interest to those studying organic chemistry, particularly in the areas of acid-base chemistry and the behavior of functional groups in aromatic compounds.

ngu9997
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So I have this question that asks which is the strongest base. I have one structure with a benzene ring with an NH2 group on one side and a methoxy group on another side. Automatically I knew that the NH2 group would be what is protonated in the situation that this structure acts as a base and becomes protonated. But the more I think about it, the less it makes sense. N already has 2 H bonds on it and is attached to the benzene ring and has a lone pair giving it a formal charge of 0. How could it form another H bond in the case it becomes protonated (I assume there has to be some logic as to why the N is what I'm looking at to become protonated rather than the oxygen on the methoxy group).
 
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ngu9997 said:
How could it form another H bond in the case it becomes protonated
How does any amine (or ammonia) get protonated? Just the same.
 
Oh wow , I forgot that it was H+ so it wouldn't violate any rules of having more than 8 electrons in N's octet. That makes sense, but how would we determine that the NH2 group gains the hydrogen in protonation rather than the oxygen on the methoxy group?
 
ngu9997 said:
Oh wow , I forgot that it was H+ so it wouldn't violate any rules of having more than 8 electrons in N's octet. That makes sense, but how would we determine that the NH2 group gains the hydrogen in protonation rather than the oxygen on the methoxy group?

Because N is usually a stronger base than O.

If you have NH3 and H2O in ammonia water solution and add acid, then you can protonate either water - forming H3O+ - or ammonia - forming NH4+.
But ammonia is much stronger base. H3O+ has pKa of -1,5, while NH4+ has pKa of +9,25.
Now, in aromatic amines, the aromatic ring actually withdraws some of the free electron pair and makes the amine a weaker base. Aniline, C6H5NH3+ has pKa of +4,6.

But it is still much more basic than water, or presumably phenylether.
How does an ether group across the aromatic ring affect pKa of an aromatic amine?
 

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