In Concise Inorganic Chemistry by J. D. Lee (adapted by Sudarsan Guha), in the chapter "Chemical Bonding", under the topic "Back Bonding with Nitrogen as a donor atom" the following reaction is given:

$\ce{(SiH3)3N +4HCl \to NH4Cl + 3SiH3Cl}$ (bond cleavage through $\mathrm{S_N2}$ mechanism)

I have learnt about $\mathrm{S_N2}$ in organic chemistry. I know its mechanism there and tried to do the same here (there is no reaction mechanism given for this in my book).

As suggested by this comment, $\ce{(SiH3)3N}$ is still basic so the following set of reactions takes place:

$$\ce{\color{orange}{(SiH3)3N} + \color{blue}{HCl} \to (SiH3)3NH+ +Cl-}\tag{Protonation}$$ $$\ce{(SiH3)3NH+ +Cl- \to (SiH3)2NH + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$ $$\ce{(SiH3)2NH + \color{blue}{HCl} \to (SiH3)2NH2+ + Cl-}\tag{Protonation}$$ $$\ce{(SiH3)2NH2+ +Cl- \to (SiH3)NH2 + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$ $$\ce{(SiH3)NH2 + \color{blue}{HCl} \to (SiH3)NH3+ + Cl-}\tag{Protonation}$$ $$\ce{(SiH3)NH3+ + Cl- \to NH3 + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$ $$\ce{NH3 + \color{blue}{HCl} \to \color{green}{NH4+Cl-}}\tag{Protonation}$$

Summing up all the above reactions we get the overall reaction as follows:

$$\ce{\color{orange}{(SiH3)3N} + 4\color{blue}{HCl} \to \color{green}{NH4+Cl-} + 3\color{red}{SiH3Cl} }$$

I got the mechanism I was looking for! But the main problem is the first step - protonation. In my book, it's given that in $\ce{(SiH3)3N}$, due to the presence of back bonding, the lone pair on nitrogen atom is not available for donation. Or in other words, $\ce{(SiH3)3N}$ cannot act as a lewis base (electron-pair donor). Thus the first step of protonation (acid-base reaction) is ruled out.

So, how does the above reaction take place? Do we have any other reaction mechanism for the above reaction? If yes, could you please specify the mechanism? Or is the above one is itself correct?

I am unable to find relevant information regarding this on the internet.

  • $\begingroup$ chemistry.stackexchange.com/questions/69568/… $\endgroup$
    – Mithoron
    Nov 6, 2019 at 15:15
  • $\begingroup$ Hmm, I have a feeling your textbook may be wrong after all, and while what you wrote is just what I thought about, it still may be not exactly SN2, but more like addition-elimination, because Si has no problem with being pentacoordinated. $\endgroup$
    – Mithoron
    Nov 6, 2019 at 15:20

1 Answer 1


I am not sure what back bonding you are referring to. It may be related to the outdated concept of silicon using its vacant, high-energy d orbitals in any meaningful way which is not actually the case.

There is a second effect which consists of overlap of nitrogen’s p orbitals with the σ* orbitals of the $\ce{Si-H}$ bonds (which are silicon-centred due to the low electronegativity of silicon). This indeed reduces nitrogen’s basicity slightly compared to other nitrogen compounds but the effect is not that important overall. If you check page 2 of the Evans $\mathrm pK_\mathrm a$ table, you can see that the difference between $\ce{(Me3Si)2NH}$ and $\ce{(iPr)2NH}$ is about 10 logarithmic units which is noticeable (i.e. the HMDS salts are milder bases than LDA) but it does nothing to counteract their overall behaviour. (I am aware that I am arguing with the wrong $\mathrm pK_\mathrm a$ values but sadly the table does not contain data for $\ce{(Me3Si)2NH2+}$ so this is the closest equivalent.)

It is also important to point out that you are using a very strong acid $(\mathrm pK_\mathrm a \approx 8)$ which will protonate practically any nitrogen lone pair that it can come across.


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