What is the mechanism for the hydroxymercuration-demercuration of n-butyne?

1 Answer
Jun 9, 2016

Wow, what a throwback question; haven't seen this reaction in forever! For reference, my textbook calls this "Mercuric-ion-catalyzed hydration".

You should form 2-butanone.

MERCURIC-ION-CATALYZED HYDRATION: REACTION

Mercuric-ion-catalyzed hydration of an alkyne is a type of hydration reaction, adding water across the triple bond in a Markovnikov fashion.

Unlike for the oxymercuration of alkenes, it actually uses significantly different reactants due to the lesser comparative reactivity. Here's what's up...

The main differences are:

• For alkynes, we are in acidic conditions. For alkenes, we are not.
• For alkynes, we do not require two separate steps. For alkenes, we do.
• For alkynes, we have keto-enol tautomerization (not at the end, but in the MIDDLE of the mechanism. Weird!). For alkenes, we do not.

MERCURIC-ION-CATALYZED HYDRATION: MECHANISM

The mechanism goes as follows. For your reaction, $R$ is ${\text{CH"_3"CH}}_{2}$.

1. We form a mercury cyclopropene analog, known as a "$\pi$ complex"; similar to a bromonium cation in alkene bromination.

Mercury donates its $5 d$ electron pair in this scenario (as ${\text{Hg}}^{2 +}$, its configuration is $\left[X e\right] 4 {f}^{14} 5 {d}^{10}$).

2. Water attacks the more-substituted carbon, since the transition state stabilizes the more-substituted vinylic carbocation via hyperconjugation.

This is the step that makes this reaction a Markovnikov addition.

3. Another water molecule abstracts a proton from the protonated enol, which is a strong acid. This forms a mercuric enol.

4. The oxygen conjugates its $\sigma$ electrons down and the $\text{C"="C}$ $\pi$ electrons obtain a proton from the previously-generated hydronium.

(If this didn't happen in one step, the electronegative oxygen would have been adjacent to a carbocation, a thermodynamically unstable scenario. It would have poorly attempted to draw electron density away from an electropositive atom.)

5. Mercury donates away its bonding electron pair (which it got in step 1), forming a $\pi$ bond, while oxygen withdraws its $\pi$ electrons and breaking its $\pi$ bond.

6. A hydronium approaches, and we basically repeat step 4, except without the mercury ion.
7. A water molecule grabs a proton to regenerate the acid catalyst.