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
  • 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, 2017 at 22:36

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.