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What is the order of basicity between $\ce{Me2NH}$, $\ce{Me3N}$, $\ce{MeNH2}$ in a protic solvent.

It has something to do with field effect which I could not understand and could not find anywhere.

Can someone please explain?

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  • $\begingroup$ Try searching for the inductive effect. $\endgroup$
    – Dissenter
    Commented Jun 4, 2014 at 4:04
  • $\begingroup$ Here's the IUPAC definition on the field effect: goldbook.iupac.org/F02358.html $\endgroup$
    – Dissenter
    Commented Jun 4, 2014 at 4:04

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The basicity of amines has to do not only with the degree of substitution with electron-releasing alkyl groups but also factors such as steric hindrance and degree of solvation.

The electronegative nitrogen is able to isolate electron density from the alkyl groups. So having more electron-releasing alkyl groups to stabilize the positive formal charge created by the abstraction of a proton from the protic solvent might seem like a good thing.

On the other hand steric hindrance due to bulky alkyl groups reduces basicity.

Also, having more alkyl groups reduces solvation due to hydrogen "bonding." Consider ammonia, $\ce{NH_3}$. This molecule can form four hydrogen bonds. On the other hand, a tertiary amine that has been protonated can form only "good" hydrogen bond; one might argue that the hydrogens on the alkyl groups can form hydrogen "bonds", but these are obviously going to be weaker "bonds" compared to the ones among ammonia molecules.

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    $\begingroup$ So according to you what will be the order of basicity? $\endgroup$
    – AJ_
    Commented Jun 4, 2014 at 4:40
  • $\begingroup$ I don't know for sure. $\endgroup$
    – Dissenter
    Commented Jun 4, 2014 at 16:00
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In case of Ethyl in protic solvent the basicity is: $$\ce{Et2NH \gt Et3N \gt EtNH2}$$ This is a result of two opposing factors, inductive Effect and Solvation. In case of other alkyl groups than Ethyl: $$\ce{R2NH \gt RNH2 \gt R3N}$$ And it is as the following when inductive effect becomes dominant in gaseous state: $$\ce{R3N \gt R2NH \gt RNH2}$$

Field Effect (or I may say inductive effect) as stated on Wikipedia

The 'Inductive Effect' is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule.
The permanent dipole induced in one bond by another is called inductive effect. The electron cloud in a $\sigma$-bond between two unlike atoms is not uniform and is slightly displaced towards the more electronegative of the two atoms. This causes a permanent state of bond polarization, where the more electronegative atom has a slight negative charge ($\delta^{–}$) and the other atom has a slight positive charge ($\delta^{+}$).

If the electronegative atom is then joined to a chain of atoms, usually carbon, the positive charge is relayed to the other atoms in the chain. This is the electron-withdrawing inductive effect, also known as the $-I$ effect.

Some groups, such as the alkyl group, are less electron-withdrawing than hydrogen and are therefore considered as electron-releasing. This is electron releasing character and is indicated by the $+I$ effect. In short, alkyl groups tend to give electrons, leading to induction effect.


The $\ce{R}$ groups are $+I$ that give $\ce{N}$ a more electron density and hence $\ce{N}$ becomes more basic, and hence, we can say that $\ce{R3N}$ would be most basic as it has the most alkyl groups, and thus the basicity in gaseous state can be predicted. But in solution, the water molecule tend to solvate amines and decrease its basicity, the solvation depends, here, mainly on size as difference between $\ce{-R}$ and $\ce{-H}$ groups and thus the two orders of basicity arise for alkyl amines.

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  • $\begingroup$ also theoretically, it is a permanent effect and not a temporary one and thus can be seen experimenally with electrostatic potential models $\endgroup$
    – RE60K
    Commented Jun 5, 2014 at 4:25
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    $\begingroup$ Wikipedia is not a solid source. The goldbook however provides a similar definition and can be referred to as a standard. Please provide sources in your answers whenever possible. $\endgroup$ Commented Jun 5, 2014 at 5:04

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