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Organic chemistry
Synthesis of alcohols using Grignard reagents I
Synthesis of primary, secondary, and tertiary alcohols from aldehydes and ketones using Grignard reagents. Created by Jay.
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- So, my question is, what exactly happens to the magnesium-halogen compound? Does it simply leave or does it form a compound with something in the air/the ether.
Thanks a lot for the video, it truly explains a lot, really well!(5 votes)- When the ketone or aldehyde (R = O) reacts with CH3MgBr, it is formed CH3 - R - O - MgBr. Then, when you add H2O (H - OH), you will get CH3 - R - OH (formed with the H from water) and HO - MgBr (formed with the OH from the water).(9 votes)
- At, a carbon "anion radical" is formed, and it looks like there are 9 electrons around carbon. Since carbon's electrons can't expand into d and f orbitals, how is this possible? 2:40(6 votes)
- The Octet rule does not always hold! What you have described is a classical example of a violation of the rule.......(2 votes)
- What's the solvent for?(3 votes)
- Almost every reaction must occur in a solution. Because the grignard reagent acts as a carbanion (have two lone pair of electrons on a carbon), it is highly reactive and a strong nucleophile. Because of this, if we were to use water as our solvent, which contain hydrogens that are easy to steal by the carbanion, we would lose our grignard reagent and can no longer complete the reaction. By using ether (CH3-O-CH3) as our solvent, it is polar APROTIC, meaning that there aren't hydrogens for the carbanion to readily steal, and so we can keep our grignard reagent.(5 votes)
- atJay did the single electron addition and that halogen leaving in two different steps, so the intermediate is a carbon that has 9 electron and so breaking the octet rule, what is the reason that this could happen? 3:00(4 votes)
- I may be wrong, but I thought it was one step. The halogen's electrons are not really around the Carbon (electronegativity differences), so the single electron does have some room to come in(1 vote)
- can you explain to us how to read a chemistry given without get confused with all the information(2 votes)
- Just take it step by step and you can do it!(2 votes)
- Hi. Would you mind explaining to me why the CH3 anion chooses to bind to the carbonyl carbon instead of maintaining it ionic bond to the charged bromide? I understand that carbon is more electronegative than magnesium, however the magnesium is carrying a charge, and I thought a charged particle would always be more attractive than a non-charged one. Thanks(2 votes)
- Interesting perspective. However, it is faulty.
Remember that the methyl anion (Grignard) is extremely strong as a nucleophile and base.
You need to forget about charge and consider what atoms and molecular intermediates can form covalent bonds. Mg cannot ever do that. Therefore, the carbanion must be the nucleophile (MgX is just a spectator).(2 votes)
- With the cyclohexanone molecule, jay draws partial -ve charge on the O and partial +ve charge on the C.
My question is:
The carbonyl carbon is connected to three other carbons, isn't the inductive effect of the three carbons enough to counteract the polarisation of the -C=O ??(2 votes)- The carbonyl carbon is connected to two other carbons. The inductive effect is real, but it is not enough to completely counteract the polarity of the C=O bond.(2 votes)
- In the first example, around, we see that Grignard's reagent is our starting material (left of arrow) instead of the carbonyl containing compound. Is there a reason for this? Would it be wrong to write the carbonyl on the left side of the arrow and Grignard's reagent as being added/reacted in the first step? 05:46(2 votes)
- It makes not one iota of difference. You can switch the starting material and the reagent at any time. There is no etiquette/SOP (std of procedure) for doing this. No order is implied when doing either. It's the exact same thing.(2 votes)
- At, what is driving the magnesium to donate its electrons to the carbon atom? The carbon atom already seems very stable in its octet so why would it want another electron? 2:04(2 votes)
- As Lithium and Magnesium shows diagonal relationship, why not Li is used to form Grignard reagent?(2 votes)
Video transcript
In this video we'll see
how to synthesize alcohols using the Grignard reagents. So first, we have to learn how
to make a Grignard reagent. So you start with an
alkyl halide, so over here on the left. And you add a magnesium metal. And you need to add
something like diethyl ether as your solvent. You can't have any water
present because water will react with the
Grignard reagent. And so this is what you
make, over here on the right. You end up with a carbon
atom bonded to a metal. Right? So carbon is bonded
to magnesium. This is called an
organometallic bond. And you can do this
with other metals. You can do this with
lithium, for example. But Grignard reagents are one
of those things that's always talked about in undergraduate
organic chemistry classes. And you can see that these two
electrons here, these red ones, the ones in red. I've drawn it like
a covalent bond. The bond between
carbon and magnesium. But in reality, it's
more ionic than covalent. So it's equivalent to the
second structure down here. Now, in terms of
electronegativities, carbon is actually more
electronegative than magnesium. So the two electrons
in red are actually going to be closer to
the carbon atom, itself, giving the carbon
a negative charge, and forming a carbanion. And so this is a
carbanion that is formed. And this is unique because
this carbanion can now act as a nucleophile in your
mechanism to make alcohols. So this is the preparation
of a Grignard reagent, it's proved to be a
very, very useful thing in organic synthesis, so much
so that Victor Grignard won the Nobel Prize for his
research into this chemistry. Let's take a look
at the mechanism to form a Grignard reagent. So I'm going to start
with my alkyl halide. And this time I'll draw in all
of my lone pairs on my halogen, like that. And we're going to add
magnesium, which we know, being in Group 2, magnesium
has two valence electrons. I'm going to draw
magnesium's two valence electrons like that. In the first step
of the mechanism, magnesium is going to
donate one of its electrons. All right? So we're going to show
the movement of one of its electrons over
here to this carbon. We're going to use a
half-headed arrow like that. So, we're going to
make a new anion here, because this carbon actually
picks up an electron. So it actually picks
up a negative charge. And we call this an
anion radical, OK? So this intermediate here
is called an anion radical. So I'll go ahead and write that. So it's an anion radical. It's an anion because it
picked up an electron, giving it a negative charge. And it's radical because
that electron is unpaired. So this anion
radical is unstable and it's going to fragment. So these two
electrons right here are going to come
off, onto the halogen. Right? So let's go ahead and draw
what that would give us. We now have this carbon
with one electron around it on the right side, like that. And now, our halogen over
here, it had three lone pairs. It just picked up another. So it is now negatively
charged, like that. And the magnesium
that we started with donated an electron. Right? So this magnesium has one
electron left around it. One valence electron. And it donated an electron,
which gives it a plus 1 charge. Right? So this magnesium is
now positively charged because it donated an
electron in the first step. And in the next step
of the mechanism, magnesium can donate its
second valence electron. And its stable for it to
do so, because then it'll have an electron configuration
like a noble gas. So magnesium is going
to go ahead and donate its second electron over here
to the carbon, like that. And let's go ahead and
draw what would result. So we're going to
now have our carbon, right here, with three bonds. With now, with now two
electrons around it. And that's what gives
it its negative 1 formal charge to
make a carbanion. Magnesium has donated
both of its electrons. So this magnesium is now
a plus 2 charged cation. So Mg2 plus, like that. And then the halogen is
going to form an ionic bond with the magnesium on
the right side here. So we have our halogen,
which is negatively charged. Right? So we have a magnesium
with two positive charges. And then, we have two
things with negative charges around it, forming
ionic bonds, right? So this is our carbanion. And, so I could
redraw this stuff on the left with
this carbon here. I can make it an R group with
the lone pair of electrons, a negative 1 formal charge. And all the stuff
on the right, I could just write it like this. There are several ways that
you'll see this written. I could just say this is
MgX with a plus 1 charge. So this just allows us to
focus in on this carbanion here with a negative 1 charge. Or I could just pretend
like everything's covalent and just save myself
some time, right? I could go like that. And organic chemists
understand what this organometallic
compound means, that the R group is negatively
charged as a carbanion. So, you'll see
several different ways to write a Grignard
reagent, just as long as you understand
what's going on. That's the most important thing. So let's now take the Grignard
reagent we just formed, and let's make an
alcohol with it. OK, so I'll go ahead and
write it the last way I did. It doesn't really
matter how you do it. All right, so this is
our generic reaction. We're going to introduce
a carbonyl compound. So for this generic reaction
I'm just going to say, it's some generic carbonyl. So I'm not going to
show what's attached to either side of my
carbonyl carbon here. And diethyl ether once
again is our solvent. You have to exclude water
from this reaction again, because the Grignard
reagent will react with it. So in the first step, you want
it to react with your carbonyl. And the second step, once it's
reacted with the carbonyl, it's OK to add water in
the form of H3O plus. And this is going
to form our alcohol. So we're going to form an
alcohol as our product. And this carbon here
came from our carbonyl, and this R group is going to
attach to that carbonyl carbon. That comes from our
Grignard reagent. So it's a very useful reaction
because it's a carbon, carbon, bond forming reaction. Right? So this R group had
a carbon on the end. And then this carbon right here. So forming carbon, carbon
bonds is very important when you're trying to build
large, organic molecules using synthesis. So this is a very useful way
to form either a primary, secondary, or a
tertiary alcohol. It all depends on what
sort of carbonyl compound that you're starting with. So let's show the mechanism
for what's happening. All right. So I'm going to say
the Grignard reagent is a source of carbanions, right? So I'm just going to draw
my carbanion here like that. And then my carbonyl, right? I have carbon double
bonded to an oxygen. And I'm going to
make this, you know, anything could be attached
the carbon for right now. Once again, when you're
looking at carbonyl chemistry, all you have to
do is think about electronegativity differences
between carbon and oxygen, right? So go back and watch the
electronegativity video. We know that oxygen, being
more electronegative, will draw these electrons in
the double bond closer to it, giving it a partial
negative charge, leaving our carbon
partially positive. Right, so carbon, being
partially positive, carbon wants electrons, right? Carbon is an electrophile. And from our Grignard reagent,
we have a nucleophile. The carbanion is going
to act as a nucleophile. The negative charge is attracted
to the positive charge. So the carbanion attacks
the carbonyl carbon like that, which would
kick these electrons off onto our oxygen. And I'll go ahead and draw
the intermediate here. So we now have, we now
have our R group directly attached to our, what used
to be our carbonyl carbon. And now, our oxygen
has three lone pairs of electrons around it, which
give our oxygen a negative 1 formal charge, like that. So that's the first
step of our reaction. In the second step, we have
hydronium ions floating around, right? So H3O plus. So I'll go ahead and draw
H3O plus here, like that. And the second step, of course,
will be acid based chemistry. So a lone pair of
electrons on our oxygen takes a proton from H3O plus,
leaving these electrons behind to form water. And that is how we
get our alcohol. OK? So that's going to give us
our alcohol as our product. So, after our acid
based reaction. All right, let's look at
three different examples of synthesis of alcohols. OK, and let's show
the different types of alcohols that
can be produced. So we'll start with
formaldehyde here. So we'll start with a very
simple molecule like that. And we'll go ahead and already
make our Grignard reagent. And we're going to make
methyl magnesium bromide, so methyl magnesium
bromide we're going to add in our first step. We're going to use
her as our solvent. And in our second step,
we're going to add H3O plus. Now, when you're analyzing
a Grignard reagent, you pretty much have to
think, where's my carbanion? Right? So this carbon right here
is negatively charged. That's the carbon that's
going to attack my carbonyl. So if this carbon attacks
my carbonyl, right? And the electrons kick
off onto here, right? As our intermediate,
we would have hydrogens on either
side of our carbon. A negatively charged
oxygen up here. And the R group, this time,
is a methyl group like that. So after you protonate it,
right, in the second step. So after these,
so we'll just say, these lone pair of
electrons are going to pick up a proton
from H3O plus, right? We would form this carbon
with two hydrogens. We're going to protonate our
alc oxide to form an alcohol up here for our product. And if you look at
that molecule closely, you'll notice it is ethanol. Right? So you make a primary alcohols
if you use formaldihide. So this is a primary alcohol. Its primary because the
carbon attached to the OH is attached to one other carbon. Let's see how we can
make secondary alcohols. OK, so this time you need
to start with an aldehide. So instead of two hydrogens
on either side of your carbon, as we did before,
this time you have to have an R group on one side. So this will be our
aldehide, like that. All right, and let's go ahead
and use the same Grignard reagent. We use methyl magnesium
bromide again. Like that, and second
step, H3O plus. So once again, think about
what is your nucleophile. Right? So the negatively charged carbon
is going to be my nucleophile. Is going to attack my carbonyl,
kick these electrons off. So once again, when we
draw the intermediate, all right, up at the top here. We have our alc oxide
anion, negatively charged. The hydrogen is still there. And what we did was, we
added a methyl group on. So this CH3 at the bottom
of our intermediate came from our Grignard reagent. And once again, acid
based chemistry, to protonate the
alc oxide, we'll form our secondary
alcohol like that. So that would be the
secondary alcohol that is produced
from this reaction. Once again, this carbon is
attached to two other carbons, making this a secondary alcohol. All right, so one
more example here. This time we will react our
Grignard reagent with a ketone. So we'll start with our
ketone over here, on the left. So here's our ketone. So it's cyclohexanone. And once again, let's stick
with methyl magnesium bromide. And so we have methyl
magnesium bromide that we add. And again, our solvent is
ether, excluding water. And second step, we're going
to add a source of proton. So H3O plus. So once again, the exact same
mechanism, exact same thinking involved. Right? Our negatively charged carbanion
attacks, our carbonyl carbon, kicking these electrons
off onto our oxygen. All right, so we form
our intermediate. Right, so this top
oxygen here now has three lone pairs of
electrons, negatively charged. And actually, let me
go ahead and take that off there so we can better
show the atoms attached to that carbonyl carbon, right? So, if I'm showing that methyl
group attacking that carbonyl, I'm going to push that alc
oxide over here a little bit to the left. And that's going to
get my negative charge. And that way, that just
gives me some space to put my methyl group
right here, like that. And that just looks
a little bit closer to what my final
product will look like. Right? So this lone pair
of electrons, right? As usual, this lone pair,
one of these one pairs is going to pick up
a proton right here. And that's going to
form our product. Right, so we now protonate our
alc oxide to form our alcohol, like that. So our alcohol is going
to form right here. And then, our methyl
group adds on right here. And, if you look
closely, you can see this is a tertiary
alcohol that we just made. All right, so this
carbon is connected to one, two, three
other carbons. So, Grignard reagents are very
useful for making alcohols. And you can make either primary,
secondary, or tertiary alcohols from them. So it's a very versatile
reagent to use. In the next video,
we'll take a look at more about Grignard reagents,
and we'll talk a little bit about how to work backwards and
think about synthesis problems.