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Organic chemistry
Hydroboration-oxidation of alkynes
Two-step reaction in which an alkene is converted to an alkene with a hydroxyl substituent (which may tautomerize to a carbonyl). Created by Jay.
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- Why is this reaction anti-Markovnikov?(5 votes)
- Whenever you see the following reagents, it will be an anti-markovnikov addition.
1. BH3/THF
2. H2O2/NaoH/H2O(19 votes)
- At, Sal mentions a video on hydroboration/oxidation of alkenes. Which video is this? 2:00(3 votes)
- What is the purpose of THF and hydrogen peroxide in this reaction? Someone else answered that the boron in BH3 is oxidized to an OH group to make the overall reaction anti Markovinikov and the OH- group is the base in base catalyzed tautomerization but I don't see where the other molecules come in.(3 votes)
- BH3 is violently reactive on its own, so THF is used to stabilize the molecule. The BH3 essentially accepts an electron pair from the THF to complete its octet and form a stable BH3-THF complex. Think of THF as a calming agent. As far as the others are concerned: NaOH serves to remove a proton from H2O2, making it HO2, a better nucleophile (conjugate bases are always stronger nucleophiles). HO2 then attacks the boron to complete the rearrangement reaction, and the O is ultimately protonated with H2O to form a neutral alcohol. Keto-enol tautomerism kicks in here and the (C=O) bond is highly favored.(4 votes)
- In the resonance structures, the first one should be more stable than second [as structure 1 has -ve charge on more electronegative element ie.oxygen & hence conc. of enol >conc. of aldehyde [in products] but it is not so.why?(2 votes)
- We must consider not just the stability of the O atom, but the stability of the system as a whole. A C=O bond is more stable than a C=C bond, so the position of equilibrium favours the aldehyde.(4 votes)
- Are you able to use hydroboration on a symmetrical internal alkyne? If so would it also be a mixture of both ketones, or would BH3/H2O2 force the ketone to create a more stable molecule?(3 votes)
- If you have a symmetrical internal alkyne, you will get a single ketone.(2 votes)
- Why does he bring up resonance structures, and how are you supposed to know when to do a resonance structure?(3 votes)
- The molecule wishes to become more stable, which means to shift the double-bond to the oxygen; from that point, it's much easier to stabilize it.
If you can't see the clues (example: an atom with lone pairs of electrons nearby a double bound) which tell you that there might be a resonance structure, you should go back and re-watch the video about Resonance Structures.(1 vote)
- In my textbook instead of using BH3 it uses Sia2BH, which I believe is a dialkylborane. I want to know if I can use this reagent instead of what was shown in the video?(2 votes)
- Sia₂BH or disiamylborane is bis(1,2-dimethylpropyl)borane. It is a hindered borane, so it is selective for the hydration of terminal alkenes and alkynes. If you have a terminal alkene or alkyne, then you can use Sia₂BH. If not, then you have to use BH₃/THF.(3 votes)
- Would it not be a more efficient way of teaching regiochemistry to have the alkyne have a more subsitued carbon and a less subsitiued carbon? That way you could see the regiochemistry taking place?(1 vote)
- With an alkyne, all you can have is a terminal or an internal triple bond.
The regiochemistry is important only for a terminal alkyne.
You can't make reliable predictions for an internal alkyne, because there is an alkyl group on each end of the triple bond.
Steric factors then become more important than electronic factors.(4 votes)
- can you get a geminal diol, by hydroboration of the enol?
the geminal diol would not be stable right? and could throw out water which would also lead to a keton/aldhyde. or is this reaction unlikely?(2 votes) - What are elaboration reactions?(2 votes)
Video transcript
Here's the
hydroboration-oxidation of alkyne reaction. So we start with our alkyne. And, once again, usually
it's a terminal alkyne. So here's your hydrogen here. And on the other side
of our triple bond, let's say there's some R
group attached to this carbon. So two are alkyne. We're going to add, in the
first step, borane, BH3, in THF, as our solvent. And in the second step, we're
going to add hydrogen peroxide and hydroxide anion OH minus. And we're going to
get addition of water across our double bond. So it's very similar
to the last reaction. The difference here
is the regiochemistry. This regiochemistry is actually
anti-Markovnikov addition. So this is antimark's rule. So anti-Markovnikov,
meaning the OH is going to add to the least
substituted carbon, which, of course, must be the
carbon over here on the left. So we end up adding H2O,
water, across our triple bond. So now there's a
double bond present, and then there's an
OH in our molecule. So just like in the last video,
we form an enol here like that. And the enol, of course, is
not the most stable form. We're going to get
some rearrangement, and we're going to end
up with an aldehyde tier. So we can see we get an
aldehyde functional group from a terminal alkaline
undergoing this reaction. Now, sometimes if you use borane
in this first step over here, sometimes borane is so
reactive that it's actually going to add two molecules
across your triple bond, as your triple bond
consists of two pi bonds. So sometimes you'll see a
different molecule used. Sometimes you'll see
a dialkyl borane. So put two alkyl groups. I'll put R2BH and if you
make them really bulky the steric hindrance
prevents the second addition of the borane. Because in the mechanism,
you want only one addition of the borane. And the borane is going to
add onto the same carbon that the OH does. So check out the video on
hydroboration-oxidation of alkenes for much more
detail of that first mechanism that we discussed. So the net result
of this reaction is to form an aldehyde
from your terminal alkyne. So let's look at
a reaction here. So let's make our triple
bond right here like that, and we'll make it
a terminal alkyne, and we'll make this
side a benzene ring. So we'll make a rather
large R group here. So let me put in some
electrons on my benzene ring. So let's just go like that. All right, so if I
react this alkyne with either a dialkyne borane
or I'll just put borane here, and you could use
a dialkyne borane if you want to-- whatever
your professor wants to use-- and THF. And second step, hydrogen
peroxide and OH minus, like that. So the first thing you
have to think about is, what's going to happen? I'm going to be adding
water across my triple bond. And I need to think about
what side gets the hydrogen and what side gets the OH. So, once again, I
have two choices. I could add the OH onto
the carbon on the right, or I could add the OH to
the carbon on the left. And the regiochemistry
here is anti-Markovnikov, so I'm going to add that OH to
the least substituted carbon, which, of course, is the
carbon on the right side. So I'm going to end
up adding the OH to the carbon on the right side. So let's go ahead and draw
our double bond in here. First, let's go ahead and
draw our benzene rings. So I'll go ahead and
make it like this. And that benzene ring is
connected to a carbon. This carbon right here. There's now a double
bond presence like that. And I added the OH to the
least substituted carbon. I added the OH to the carbon
on the right like that. And I add a hydrogen to
the carbon on the left. So that's adding my H20
across my triple bond. And then there was
still a hydrogen on this carbon on
the right like that. So that's my enol
intermediate like that. So let's take a
look at what's going to happen to that
enol intermediate. It's going to rearrange. And on the last video, we saw an
acid-catalyzed tautomerizaton. In this video, base is present. So we have our hydroxide
anions in the solution. So let's go ahead
and redraw our enol, and show the enol
reacting with base. So let's go ahead and
draw the enol here again. So I'll draw that enol
on the right side. So there's my benzene ring. And I have my double
bond in here like that. And I know I have
an OH like that. And put in my lone
pairs of electrons. And I'm going to react
this with hydroxide anions. So hydroxide anions
are floating around. Hydroxide is negatively charged. So hydroxide will
function as a base. So let me get in my lone pairs
of electrons here on my oxygen. And I'm going to
pick up that proton and kick these electrons
off onto this oxygen. So let's go ahead
and draw the results of that acid-base reaction. So it's an equilibrium. So let me go ahead and draw my
equilibrium arrows like that. So what's going to happen? I have my benzene ring still. And I have my double bond. And let's see, now
I have my oxygen with three lone pairs of
electrons around it like that, so negatively charged. So now I can draw a resonance
structure for this molecule. So let's go ahead and put in
our resonance brackets here like that. So what can I do to spread
that formal charge out throughout the molecule? Well, I could take a lone
pair of electrons right here, and I could move it into
here to form a double bond. And now I give too many
bonds to this carbon. So the pi bond here
is going to break and the two electrons
here are going to move out onto this
carbon like that. So let's go ahead
and draw the results of the movement of
those electrons. I still have my
benzene ring like that. And there's my
carbon skeleton here. Now I have a carbon double
bonded to an oxygen. There used to be three
lone pairs of electrons around this top oxygen. One moved in to form
the double bonds, and now there's only two
lone pairs like that. And the electrons that
used to be in this pi bond have moved onto this
carbon right here. So this carbon
right here is still bonded to another hydrogen. So the two electrons
moving out onto it are actually going to
form a carbon anion here. So this is actually a negatively
charged carbanion like that. And water is now present. OH minus fix of a
proton become H2O. So now water is floating
around like that. So now water is actually
going to function as an acid. It's going to donate a proton. And this lone pair of electrons
are going to take that proton. So they take that proton and
kick these electrons in here off onto the oxygen
atom like that. This is an acid-base reaction. So this is going to
be at equilibrium. I'm going to go ahead and put
in our other resonance bracket like that. So what are we going to get? And we still have our benzene
ring untouched over here. And let's see, now we have
our carbonyl like that. And we just added on a
proton to this carbon right here like that. So now we are done. And you can see
this is an aldehyde. I didn't draw on the
hydrogen over here, so you can leave that
out if you want to, but if you want to
draw on your hydrogen, there's a hydrogen there. There's also a hydrogen
on this carbon right here. I just didn't draw it in to show
some clarity in the mechanism. So the end result is
we form a an aldehyde from a terminal alkyne. And we showed this
tautomerization using base. So let me just go ahead and
write one more thing here. This is an example of a
base-catalyzed tautomerization, where we go from the enol
form to, in this case, an aldehyde as your product,
which is more stable because of the carbon-oxygen
double bond.