During SN2 reaction with a silicon compound, a pentacovalent intermediate is formed, while there is no such intermediate formed with carbon compounds (only a pentacovalent transition state). Why is it so? I need an MO explanation (no d orbitals involved).


1 Answer 1


Bonds involving carbon are typically shorter than bonds involving silicon. Therefore a pentacoordinated carbon species would be sterically more crowded around the central carbon atom than its pentacoordinated silicon analogue would be. This crowding would raise the energy of pentacoordinate carbon compared to pentacoordinate silicon; which is consistent with the observation that the SN2 reaction with carbon produces a higher energy pentacoordinate transition state, while in the case of silicon a more stable pentacoordinate intermediate can be observed.

\begin{array}{lcc} \hline \ce{Bond} & d(\ce{X=C})/\pu{pm} & d(\ce{X=Si})/\pu{pm} \\ \hline \ce{X-O} & 143 & 163 \\ \ce{X-Cl} & 177 & 202 \\ \ce{X-Br} & 194 & 215 \\ \ce{X-I} & 214 & 243 \\ \hline \end{array}

Recasting this into MO terms, we would say that both the carbon and silicon pentacoordinate species involve a trigonal bipyramid structure where the two ligands (the "attacking" and "leaving" groups in the carbon case) occupy axial positions in the trigonal bipyramid arrangement.

Structure of the SN2 transition state


The figure above makes it easy to see that we have a hypercoordinated species involving a 3-center 4-electron bond. Since a silicon p-orbital is larger than a carbon p-orbital, the 4 electrons are spread out over a larger volume in the silicon case, reducing electron-electron repulsion. This makes the pentacoordinate silicon case more stable than pentacoordinate carbon.


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