Hydroboration and Oxymercuration of Alkynes

by James

in Aldehydes, Alkenes, Alkynes, Ketones, Organic Chemistry 1

Alkyne chemistry bears many resemblances to alkene chemistry, but in these first few posts on the subject, the purpose is to illustrate how one seemingly minor change – an extra π bond – can lead to significant differences in chemical behavior. Previously, we saw that the sp hybridization of alkynes leads to increased acidity, and the second π bond of alkynes leads to the possibility for partial reduction to either cis or trans alkenes. In this post we’ll see again how the addition of that extra π bond has a very important and surprising consequence.

Several posts ago we talked about the hydration of alkenes. This can be done either with aqueous acid, or with mercury and water (“oxymercuration” – more on that later). Looking at the reaction with alkenes, the pattern is fairly straightforward: break a C-C π bond, and form a C-H and C-OH bond. Also recall that the oxygen ends up on the most substituted carbon [“Markovnikov” selectivity].


So what happens when we try this reaction on alkynes? We might expect to observe the same pattern, right? After all, it’s just a simple addition reaction.

Well… here’s what we actually observe. We get… a ketone !?


Now what’s going on here? This seems like the type of thing that drives new organic chemistry students around the bend. Just when you think you understand your surroundings, you pick up the most innocuous looking rock, and underneath it find a poisonous snake!

Don’t panic! It’s a new concept in organic chemistry we’ll be exploring here (tautomerism) – one that gets much more discussion in Org 2 – but it’s not as weird as you initially might think.

Look at the bonds formed and broken. The first set we should understand. Form C-O and form C-H, break C-C π.  It’s that next set of bonds formed/broken that are a big surprise.

If you monitor this reaction closely – one way to do it is in an NMR tube – it’s actually possible to observe the first product of this reaction, which is the one shown below. We call this an “enol“, by the way – kind of like a spork (half spoon half fork) it is part alkene, part alcohol.

2-step 1 alkyne oxymerc

Over time, this enol spontaneously converts into the ketone. Note that the two have the same molecular formula – they are constitutional isomers.  And they are in equilibrium with each other. We call these constitutional isomers which interconvert, “tautomers”. This equilibrium generally favors formation of the ketone due to the strong C-O π bond (compared to C-C π).

3-step 2 alkyne oxymerc

Here’s how the whole process works – arrow by arrow.



Wait – we’re not done! There’s another way to “hydrate” alkynes, just like there was with alkenes. We can also perform the same reaction with mercury, water and strong acid [sulfuric acid, H2SO4 is the usual acid of choice]. For interesting reasons we wont get into at the moment, NaBH4 is not generally needed here; it is sufficient to merely have water and acid present.

5-alkyne oxymerc

There’s also hydroboration. Remember how it performs “anti-Markovnikov” hydration of alkenes?

Likewise, we can use the same reaction to perform “anti-Markovnikov” hydroboration of alkynes. [Note: while BH3 can be used for this, we often use different boranes, such as disiamyl borane or 9-BBN that, due to their greater steric bulk, increase the proportion of addition to the less substituted carbon] .

Just as in the cases above, we initially obtain an enol. However, under the reaction conditions, keto-enol tautomerism results in formation of the aldehyde.

Bottom line here: if we start with a “terminal” alkyne, that is an alkyne where one of the carbons is attached directly to H – then we will obtain ketones with H3O+/H2SO4 or via oxymercuration, and aldehydes via hydroboration.


One final note: if we use an alkyne where both ends are directly attached to carbon, we will obtain a mixture of products. That’s just “Markovnikov’s rule” – remember that if each carbon in the multiple bond is attached to an identical number of hydrogens, then we can’t determine which is the “most substituted” for our purposes. Like in this example.


Next Post: Alkyne Reaction Patterns – The Carbocation Pathway


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{ 7 comments… read them below or add one }


You made this so easy to understand. thanks! :)



What I would really like to know is that WHY THE ANTIMARKONIKOV product is formed in the Hydroboration reactions.Basically I want the simplified mechanism which I can’t find in my books.
And congratulations of becoming a Dad, Mr.James.



It has to do with the fact that hydrogen is more electronegative than boron.



In the mechanism “Step 1 – enol formation” a plus charge is missing in the 3rd structure.


James Ashenhurst

fixed. Thank you!



If an alkyne was on the end of a molecule and being reacted, how would it react with H2SO4 and HgSO4. Would it be an aldehyde or a ketone?


Kumar Neelabh

In the very last example (“One final note”), why don’t we use carbocation stability, the reason for Markovnikov’s Rule?
We could say that the Carbon on the right would form a more stable carbocation, because of hyperconjugation. (As it has 3 alpha-hydrogens while the left carbon has only 2)
So this would mean that the 2nd product would dominate.
Or am I missing something?


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