3
$\begingroup$

For a carbocation to form, it needs to lose an electron. In the example below the leaving group is Br and the products are a methyl carbocation and a bromide ion.

enter image description here

What I am unsure about is why the carbocation needs an empty p orbital. For a carbon atom with 6 valence electrons, the molecular orbital diagram looks like the one below. So when an electron is lost in heterolysis the $\ce{2p}$ orbital will be empty and I assume that is why the p orbital is empty. However, why do you include the empty p orbital in the shape. Does it serve a purpose?

enter image description here

$\endgroup$
1
$\begingroup$

You might want to read up on hyperconjugation.

Basically, the R groups (if NOT a hydrogen) can donate some electron density into the carbocation’s empty p orbital, stabilising the positive charge and increasing the ion’s stability. This is the reason for Markovnikov’s Rule when it comes to Sn1 mechanisms; the tertiary carbocation is most stable, followed by a secondary one, while a primary/methyl carbocation hardly exists at all.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.