Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a $\ce{Cl}$ group? In the same vein, isn't a trialkylammonium a much weaker acid than $\ce{HCl}$?

$\ce{R-NR3+ -> Nuc-R + NR3}$ (Stronger leaving group)

$\ce{R-Cl -> Nuc-R + Cl-}$ (Weaker Leaving group)

Compared to:

$\ce{H-NR3+ -> Nuc-H + NR3}$ (Weaker acid, NR3 thus weaker leaving group)

$\ce{H-Cl -> Nuc-H + Cl-}$ (Stronger acid, Cl- thus stronger leaving group)

  • $\begingroup$ The trialkylamine is likely a lot less nucleophilic than a chloride ion. This reduces the possibility of reverse reactions. $\endgroup$ – Dissenter Dec 29 '14 at 2:22
  • $\begingroup$ Wouldn't the same hold true with acids? If Cl is more nucleiphillic, by extension, it should also be more basic, no? $\endgroup$ – user32134 Dec 29 '14 at 2:42
  • $\begingroup$ no. Iodide ion (I-) is way way more nucleophilic than fluoride ion (F-) in a polar protic solvent. Yet F- is clearly way way more basic than I-. Nucleophilicity is a kinetic property; basicity and acidity are both thermodynamic properties. $\endgroup$ – Dissenter Dec 29 '14 at 2:48

Leaving group ability refers to how well a leaving group can stay off whatever it came from.

Consider the following prototypical nucleophilic displacement reaction:

$\ce{R-LG + Nuc- <=> R-Nuc + LG-}$

There are two issues at play here:

1) Thermodynamics. How stable is the reactant with the leaving group, $\ce{R-LG}$, and the nucleophile, $\ce{Nuc-}$? How stable are the products, $\ce{R-Nuc}$ and $\ce{LG-}$? And when examining stability, we must look at not only the reactants and products but also their environments - i.e. the solvent(s).

2) Kinetics. What's the physical ability of the nucleophile to actually displace the leaving group? What's the physical ability of the leaving group to displace the nucleophile again?

Usually we're taught that leaving group ability has to do with acidity and basicity. The less basic the leaving group is, the better its ability to leave. This isn't a perfect rule of thumb, but it's acceptable.

That's the thermodynamic part of the question. If $\ce{LG-}$ weren't thermodynamically stable, then it would be inclined to attack $\ce{R-Nuc}$ and return to the status quo. (Or just grab a proton).

However, how capable is $\ce{LG-}$ of carrying out such an attack? Here we come to the kinetic part of the question. The bulkier the leaving group, the more likely it is to be sterically hindered and/or strongly solvated (trapped by solvent molecules). Trialkylamine is a very genetic term, but when I think of amines with alkyl groups I think of quaternary ammonium compounds, which have alkyl groups that can contain upwards of 18 carbons (!). Such bulk affects the ability of the leaving group to line up properly with the substrate to perform nucleophilic attack. Nucleophilic attack often has specific steric demands; $\ce{SN2}$ requires backside attack; the approach of attacking nucleophiles with respect to flat carbonyl groups has an angle of 104.5 degrees, etc.

Here is a good summary of sterics and amine nucleophilicity.


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