1
$\begingroup$

Suppose we have 2-bromo butane. The carbon connecting to the bromine will have the following valence electron configuration:

$$\underset{2s}{[\uparrow]}\underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]} $$

This hybridizes to an sp3 configuration which yields a trigonal pyramidal configuration.

Suppose in an SN1 reaction the bromine comes off. Then the connecting carbon will have a valence electron configuration of

$$\underset{2s}{[\uparrow]}\underset{2p}{[\uparrow \vert \uparrow \vert\;]} $$

which yields an sp2 or trigonal planar configuration.

Suppose we instead have an SN2 reaction such as with NaOH. Then the intermediate would be a carbanion that has a carbon that has 5 bonds.

$$ \underset{2s}{[\uparrow ]} \underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]} \underset{3s}{[\uparrow]} $$

would be the wrong configuration but I'm having trouble understanding what the right one would be.

A trigonal bipyramidal molecular geometry seems to me like it'd fit the shape of what the carbanion would be but the numbers don't seem to add up to me.

In a trigonal bipyramidal geometry there needs to be two $sp$ orbitals two point up and down.

I learned from How are the hybrid orbitals of sulfur hexafluoride shaped? that higher orbitals are usually only slightly involved so the resulting bonds would only have a small d characteristic. For example, instead of a $sp^2d$ hybrid something like a $sp^{2.75}d^{0.25}$ characteristic is more likely.

Would the rest of the bonds have something like a $sp^{2.67}d^{0.33}$ characteristic?

According to molecular orbital theory what would the shape of that carbanion be and how would the groups attached to it be arranged?

$\endgroup$
  • 1
    $\begingroup$ Molecular orbitals theory works fine, but it is incompatible with hybrid orbitals. The transition states involves a 3-center-4-electron bond. $\endgroup$ – Zhe Mar 23 '17 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.