How can the formation of an alkene be explained during the hydrolysis of an ester when the alkoxy moiety is tertiary?

Question

$$ \ce{RCOOR' + H2O -> RCOOH + R'OH} $$ But when $\ce{R'}=3^\circ$ it may result in the formation of alkene. But by the normal mechanism given in books and in various sites I am not able to explain its formation. I think that somewhere $\ce{R'^{+}}$ must be formed for alkene to be the product.

Answer 1

By the normal mechanism given in books and in various sites I am not able to explain its formation

That's right, and it means that there must be a different mechanism predominating.

This is the "standard" mechanism for ester hydrolysis in acid:

If you look closely at the colour-coding, you will notice that it is the blue bond in the ester which connects the $\ce{C=O}$ to the $\ce{OR'}$ that is broken. This is called the acyl bond. The rate-determining step involves two molecules - it is a bimolecular reaction. Therefore, this ester hydrolysis mechanism is called $\mathrm{A_{AC}2}$: $\mathrm{A}$ for acid, $_\mathrm{AC}$ for acyl bond, and $2$ for bimolecular.

Let's use $\ce{R'} = \ce{C(CH3)3}$, the prototypical example of a tertiary alkyl group. The alternative mechanism goes as follows:

Notice here it is the green bond, between the $\ce{O}$ and the $\ce{R'}$, that is broken. This is called the alkyl bond. The slow step is unimolecular. So, following the same logic as before, this hydrolysis is called an $\mathrm{A_{AL}1}$ mechanism. (It doesn't matter which of the red and blue bonds ends up as the carbonyl group and which one ends up as the $\ce{OH}$ group - they are interchangeable via a simple proton transfer.)

As you correctly suspected, the formation of the $\ce{C(CH3)3}$ carbocation occurs. Depending on the reaction conditions, it could be attacked by a nucleophile (e.g. if you use ethanol as the solvent, you might get some ether byproduct), or it could just simply lose a proton to give isobutene.

What factors may work to favour the $\mathrm{A_{AL}1}$ pathway? Well, the most obvious thing is the formation of the relatively stable tertiary carbocation. Another possibility is steric hindrance arising from the bulky $\ce{R'}$ group, which disfavours nucleophilic attack of water in the "normal" pathway. In fact, t-butyl esters like the one shown above are so resistant to nucleophilic attack (under basic conditions) that they are sometimes used as protecting groups for carboxylic acids!