[Lo.Lee.Ta.]

I am trying to understand the hydroboration-oxidation reaction in alkynes.

For internal and terminal alkynes, it seems the reaction is a bit different.
In my chem book, they depict internal alkynes only having 1 BH3 being added for every triple bond, but for terminal alkynes there are 2 molecules of BH3 added for every triple bond...

INTERNAL ALKYNE
I wrote out the reaction mechanism for an internal alkyne. On the reaction arrow there are the compounds: 1. BH3, THF and 2. H2O2, H2O, and NaOH. The OH that adds to the overall molecule comes from the H2O2, right?
Is it that the OH in the parentheses dissociates from the hydrogen peroxide: (H-O)-O-H,
and then it replaces the B that attached to the overall molecule? 

But it seems that if this is the case, then what is the point of the H2O and NaOH?

TERMINAL ALKYNE
I am also trying to write out the hydroboration-oxidation reaction mechanism for 1-hexyne. On the reaction arrow is: 1. BH3 2. H2O2, H2O, pH 8
There are 2 BH2 molecules that add to the same terminal carbon... But how does an OH from the H2O2 break off and replace the 2 BH2 molecules? It seems that there would be too many electrons...
And there is again the H2O that is not utilized...

I really have tried to look this up before posting this question! But I can find nowhere that shows a mechanism for how the OH gets added to the overall molecule!

Thank you so much for helping me!!!

[discodermolide]

Hint the boron gets oxidised. Look here; http://en.wikipedia.org/wiki/Hydroboration–oxidation_reaction

[Lo.Lee.Ta.]

From the link you gave me, it says: "The resulting trialkylborane is treated with hydrogen peroxide in the second step. This process replaces the B-C bonds with HO-C bonds."

So I guess I'm right- the H2O2 does supply the OH.

But what are the H2O and NaOH there for? It does not seem that the water or sodium hydroxide participate in the reaction...

[discodermolide]

The product is boric acid after hydrolysis of the B-O-C bond, this is removed as the Sodium salt.

[Lo.Lee.Ta.]

=_= hmmm. For the hydroboration-oxidation of 3-hexyne we have:

(What takes place after the Boron bonds with all 3 carbons simultaneously)
1. There are 3 molecules of H2O2. In each, OH dissociates from the H2O2, leaving us with: OH+ and OH-

2. The 3 OH+ go to replace the B in each B-C bond.

3. Once this happens, we have: (H)(CH3CH2)C=C(CH2CH3)(OH) and 3OH- (from H2O2) and B+ and H2O and NaOH off to the side.

4. Now, 2 electrons from the double bond take the H away from the OH, causing a double bond to form between the O and the C.

5. The molecule now looks like: (H)(CH3CH2)C-C(CH2CH3)(=O) and 3OH- and B+ and H2O and NaOH.

6. Now what about the 3OH- and B+ and H2O and NaOH? I don't see how boric acid is formed...
Boric acid is: (OH)(OH)-B-(OH)

And you say it ends up producing a sodium salt? ...I don't see how it works! Would you please explain how it happens.

It seems in alkyne reaction mechanisms I look up, they just forget about the stuff that dissociated from the overall molecule... But shouldn't we keep track of that too???

Thank you!

[discodermolide]

Boric acid is formed. See picture. The reaction consists of a migration of the R group to an oxygen , which after hydrolysis with NaOH/H₂O gives boric acid as a sodium salt and the alcohol.

[Lo.Lee.Ta.]

I am redrawing my reaction mechanism according to the picture you provided. Thanks!

But I cannot for the life of me figure out where the O-O-H comes from that got added in the 3rd step! 

The H2O is not used until the last step and the H2O2 has already been used (except for the H+), so the other O that adds to the OH- to form O-O-H must be coming from the NaOH when it dissociates into Na+ and OH-...
But this would not make sense because we would have an OH- and an O- adding together, and that would be too many electrons!   

[discodermolide]

The ^-OOH comes from the reaction of NaOH and HOOH.

[Lo.Lee.Ta.]

But the O-O-H that came from the H-O-O-H already got added to the overall molecule...

Now it is: (R)(R)-B-(O-R)

And now they say another O-O-H has to add to the B.

So the OH- that dissociated from the first O-O-H is what reacts with the OH- from the NaOH?

How does an OH- and an OH- bond together? :/ Too many electrons, right?

[discodermolide]

There is an excess of HOOH.
Don't try and make things too complicated.

[Lo.Lee.Ta.]

What? An excess? =_=!

...How am I supposed to know that...? I was trying to work out these mechanisms by taking a log of everything that gets used and what we are left with.

Is that not what you do to solve mechanisms? Do you not worry about how many molecules of each substance on the reaction arrow gets added?

I thought if there was H2O, NaOH, and H2O on the reaction arrow, then only 1 molecule of each gets added. Is that not right?

[discodermolide]

No that is not correct. This tells you the reagents used but says nothing about the molar quantities used.
You need to find an experimental procedure.

[Lo.Lee.Ta.]

Oh, wow! Well, that sure was an important thing to figure out! Fundamental!

Thank you for working with me on this problem! You're the best!

[discodermolide]

I'm sorry if my explanations are not very good. I am not a teacher, but I am trying.

[Lo.Lee.Ta.]

No, you're great! If it weren't for you, my post would still be a lonely question...no answers at all! And I would have figured nothing out!
Thanks!