Question

Which of the following is most stable carbocation and why?

Answer 1

I would think it is `(2)`. Adjacent `pi` bonds are a good indication, if it was a neutral molecule, but these are cations.

`(1)` spreads out the electrons over three atoms, but `(2)` spreads out the electrons only over two atoms. Normally, if it was a neutral molecule, it would mean that `(1)` is more stable.

But since we are spreading out positive charge, I could see how `(2)` could probably be more stable. Positive charge in a cation is what makes it unstable, so if the positive charge is less spread out, it makes the cation more stable.

What if the positive charge in `(4)` was on one carbon to the left?

---

Before I make any conclusions for these structural questions, I draw resonance structures. Which one stabilizes (through delocalization) the positive charge best? (Or does it?)

• In this case we find that `bb((2))` delocalizes the positive charge over more atoms than `(1)`, and the more spread out the positive charge it is, the less stable the molecular cation becomes.

The partial positive lands on the oxygen, which is an electronegative atom... it prefers to be partially negative. However, this only spreads out positive charge over two atoms, so that is OK.

• Structures `(3)` and `(4)` have no resonance structures that do anything for the positive charge.

Answer 2

Here is my contribution to the discussion.

Explanation:

I would start my argument at the other end.

4. is the least stable because it is a primary carbocation with no resonance stabilization.

3. is a little more stable because it is a secondary carbocation, but it has no resonance stabilization.

We must choose between 1. and 2. Both of these are stabilized by resonance.

Which is more stable: an allylic carbocation or a carbocation with an adjacent alkoxy group?

The only difference I see between the two carbocations is this:

`bb((1))color(white)(m)"H"_2"C=CH-"stackrelcolor(blue)("+")("C")"H"_2 ⟷ "H"_2stackrelcolor(blue)("+")("C")"-CH=CH"_2`

`bb((2))color(white)(m)"H"_3"C-O-"stackrelcolor(blue)("+")("C")"H"_2 ⟷ underbrace("H"_3"C-"stackrelcolor(blue)("+")("O")"=CH"_2)_color(red)("complete octet")`

Resonance participation of the `"O"` atom gives a structure in which all atoms have a complete octet. All the other contributors have incomplete octets.

This may be the reason why structure 2. is more stable.

Confirmatory evidence

We can compare the substituent effects on `"S"_text(N)1` solvolysis rates of `"G-CH"_2"-Cl"`.

`ulbb("Substrate"color(white)(mmmmmmm)"G"color(white)(mmmml)k_text(rel)`
`"Methyl chloride"color(white)(mmmmml)"H-"color(white)(mmmmml)1`
`"Allyl chloride"color(white)(mmmmmml)"H"_2"C=CH-"color(white)(m)58`
`"Methoxymethyl chloride"color(white)(m)"H"_3"CO-"color(white)(mmll)10^14`

An adjacent `"O"` atom is far more effective than an adjacent `"C=C"` double bond in stabilizing a carbocation.