[GoldShadow]

The following question was part of a homework assignment:

a) Which is the stronger acid: (CH₃)₃NH⁺ or (CH₃)₃PH⁺
b) Which is the stronger base: (CH₃)₃N: or (CH₃)₃P:


For part a), I said (CH₃)₃PH⁺ is the stronger acid, because the P-H bond is weaker than the other molecule's N-H bond and as a result, it will be more likely to be ionized.

But I'm a little confused on part b.  I would expect that the stronger acid have the weaker base.  But I'm having trouble understanding how (CH₃)₃N: could be a stronger base than (CH₃)₃P:
After all, N is more electronegative right? So it would be less likely to donate its unshared electron pair to an H⁺.  By that logic I would say (CH₃)₃P: is the stronger base, since P is less electronegative than N and as a result, more likely to donate its unshared electron pair to an H⁺.


I'd appreciate it if someone could help with the reasoning behind this question.

[Yggdrasil]

The answer has to do with the atomic radius.  Because it is in the third period, phosphorus has a much larger atomic radius than nitrogen, which lies in the second period.  As a result, the valence shell of phosphorus has a much larger area than the valence shell of nitrogen.  Thus, the lone pair electrons on the phosphorus are spread over a much larger area than the lone pair electrons in the nitrogen, resulting in a lower electron density for the HOMO of phosphine.  This greater density of electrons in the HOMO of ammonia makes it a better base.

Electronegativity also does affect base strength.  However, chemists have found that the effects of atomic radius outweigh the effects of electronegativity when comparing atoms in different rows.  Electronegativity becomes an important factor only when examining elements from the same row of the periodic table because changes in atomic radius are small.

[lizu19]

hello everyone... i am really sorry this doesn't relate to this subject well actually i had a question about conjugated bases... but i am new to this web cite, therefore i don't know how to post my own questions i was wondering if someone can help me with that i would really appreciate it!! thank you SO much!!! 

[lavoisier]

The answer has to do with the atomic radius.  Because it is in the third period, phosphorus has a much larger atomic radius than nitrogen, which lies in the second period.  As a result, the valence shell of phosphorus has a much larger area than the valence shell of nitrogen.  Thus, the lone pair electrons on the phosphorus are spread over a much larger area than the lone pair electrons in the nitrogen, resulting in a lower electron density for the HOMO of phosphine.  This greater density of electrons in the HOMO of ammonia makes it a better base.

Electronegativity also does affect base strength.  However, chemists have found that the effects of atomic radius outweigh the effects of electronegativity when comparing atoms in different rows.  Electronegativity becomes an important factor only when examining elements from the same row of the periodic table because changes in atomic radius are small.

I am puzzled here, because the explanation makes perfect sense, and probably it was exactly what they taught me, but then why are tertiary phosphines better nucleophiles than tertiary amines? Or are they, actually?

I remember that once I prepared a phosphonium salt by just stirring Ph3P with an allylic-type iodide in acetone, but on the other hand you can benzylate many nucleophiles by stirring them in dichlorometane with TEA as the base, without the tertiary amine being quaternarized. Why is that?