Synthesis of alcohols using Grignard reagents II | Organic chemistry | Khan Academy

By Brian Lemay 13 comments

In the last video, we saw
how to make Grignard reagents and how to use Grignard reagents
to make primary, secondary, or tertiary alcohols. In this video, we’ll
take a look at how to retrosynthesize alcohols, how
to think backwards and figure out what Grignard
reagent you would need to make a certain alcohol. And then, we’ll take
a look at the reaction of Grignard reagents
with esters as well. So here is our alcohol. And we’re going to
think backwards. We’re going to think
using retrosynthesis and try to figure out how
to make this alcohol using a Grignard reagent. And there are actually several
different ways to do it. So if you understand the
mechanism really well, you know that you
add an alkyl group onto what used to be
your carbonyl carbon. And your carbonyl carbon was the
one connected to your oxygen. So if you’re thinking backwards,
you can think to yourself– this must be my carbonyl carbon. All right. This must be the one. And I added an alkyl group
onto that carbonyl carbon. So if you’re taking backwards,
you can disconnect a bond. So I could say– oh,
I’m going to make my alkyl group this
group right here. So I’m going to say
it’s that methyl group. So I’m going to
disconnect that bond. And that will allow me to figure
out my starting materials. So if that’s the bond that
isn’t going to exist over here when I’m drawing my
starting materials– I have my benzene
ring right here. And the benzene ring is
directly attached to what was my carbonyl carbon. So I’m going to go ahead
and draw my carbonyl carbon in here. Like that. And I’m saying that this ethyl
group over here in the right is still there. And what I added on in the
mechanism was a methyl group. So the methyl group must have
come from my Grignard reagent. So if I’m going to go ahead
and write my Grignard reagent, it must be a methyl group. And so I’ll make methyl
magnesium bromide– which is what we used
in the previous video for our examples. So these two things in the right
would be my starting materials to make the alcohol on the left. And if you’re
unsure of that, just think about working backwards. So you could think to
yourself– well, once again, from the last video, we
know that this methyl group is actually negatively
charged and is a carbanion. And it’s going to attack
our carbonyl carbon. Like that. And then, those
electrons are going to kick off onto the oxygen. So when we add in our
reagents– in the first step, when you add your Grignard
reagent and your ketone up here– you would need to
use ether as your solvent. And the second step– you need
to protonate your alkoxides to form your alcohol
over here on the left. So you need H3O+. So that’s how to think
backwards and come up with your starting
materials used to synthesize the
alcohol on the left. This isn’t the
only way to do it. Right? Let’s redraw this alcohol,
and let’s look at another way that we can synthesize
the same alcohol here. So let’s go ahead and redraw
the alcohol on the left. And think about how
else could I make it. OK. So here’s my alcohol. And before, we first
identified our carbonyl carbon. It’s going to be the one
attached to the oxygen right here. And I have all these alkyl
groups attached to it. So I’m just going to
disconnect another alkyl group. This time I’ll disconnect
the alkyl group on the right. And so I’m thinking
retrosynthesis. I’m thinking about
working backwards. All right. So if I’m going to go ahead and
draw the carbonyl compounds, that has my benzene
ring untouched. And the carbon in red is going
to become my carbonyl carbon. Like that. And what’s attached
to that carbonyl? It’s going to be a methyl group. Right? And this time, the
methyl group was left. So when I’m thinking about what
should be my alkyl group here, it would be the
methyl group that I disconnected on the right. So my Grignard
reagent is going to be the source of my ethyl group. So I’m going to go ahead and
draw an ethyl group attached to the magnesium. So we have ethyl
magnesium, a bromide. Like that. So this would give us
the alcohol on the left. Again, same concept. Same mechanism. This carbon right here would
end up attacking my carbonyl. Kick these electrons off. And once again,
we need to specify that the first step
of this synthesis is done in diethyl ether. And the second step– we
protonate to form our alcohol. Like that. All right. So let’s look at yet another
way to form this alcohol. We’re kind of ignoring
stereochemistry here. So if some of you were
thinking, that carbon is a chirality center. We’re just trying
to think about how to make these
alcohols here and not worry about what we’re getting
in terms of stereochemistry for our products. So if I start here
with my benzene ring, and once again, we’re going to
make the exact same alcohol. And we’re going to
think backwards again. And once again, we know this is
going to be my carbonyl carbon. This time, we’re going
to disconnect this bond. So that’s the alkyl
group that we’re going to add on as
our Grignard reagent. So there’s a third
way of doing it. So now, this time, I’ve
disconnected my benzene ring. And this is going to be my
Grignard agent this time. So this carbon on
my benzene ring is going to be ionically
bonded to magnesium. So MgBr. Like that. And so now it’s called
a “phenyl group.” So you would term this Grignard
reagent “phenylmagnesium bromide.” And I can see the
skeleton for the ketone that I’m going to use
on the right there. So I have a carbonyl. And let’s see. There’s a methyl
group on the left side and an ethyl group
on the right side. So once again, if you mix
those two together in ether and then protonate
it, you will end up with the alcohol on the left. So three different ways to
synthesize the same alcohol. And it’s just thinking
about what’s my alkyl group and how can I add it on? All right. Let’s look at the general
reaction for a Grignard reagent with esters. OK. So there is my Grignard
reagent, and I’m going to react it with
an ester this time. So I’ll make this R prime. And I’ll make this one R
double prime to distinguish it from the R group on
our Grignard reagent. And the second step,
once again, we add H3O+. In this situation,
the Grignard reagent is going to add on twice
to our carbonyl carbon. So the carbon attached to
the OH for our product, this is another way to
synthesize alcohols. It’s going to have the R prime
group still attached to it. And to the R group from
the Grignard reagent is going to add on twice. So if you look at
this, you can see that the R prime group and this
carbon came from our ester. So did this oxygen right here. So this R group here
from our Grignard reagent is going to add on
twice in the mechanism. So let’s take a look at what
happens in our mechanism. We start with our ester. All right. So R prime, C double bond
O, O, R double prime. And we’ll go ahead and put in
our lone pairs of electrons. We saw– in the last
video– that that carbonyl is a polarized bond. Right? The oxygen ends up being
partially negative, and the carbon ends up
being partially positive. So the carbon is
partially positive, making it an electrophile. The source of the electrons
come from the Grignard reagent because we know that in
that Grignard reagent we’re treating that R
group as a carbanion. All right. So this R group up here is
actually negatively charged. So we saw that in
the last video. So it is R with 2 electrons
around it giving it a negative 1 formal
charge, making it a carbanion, which is
a good nucleophile. So the nucleophile’s going
to attack our electrophile. Opposite charges attract. All right. So nucleophilic attack. We’ll kick these electrons
off onto your oxygen in the first step
of the mechanism. So now we have R prime,
carbon with an oxygen up here, Three lone pairs
of electrons around it– giving it a negative
1 formal charge. And over here in the right, we
have oxygen with an R prime. And then, our lone
pairs of electrons. Like that. Our double prime, I should say. And then, down here, we
attached our R group. OK. So in the next step
of the mechanism, a lone pair electrons
are going to move in here to reform our pi bond, to
reform our double bond. And that’s going to kick the
electrons in this bond off onto the oxygen, since carbon
cannot have five bonds. All right. So let’s go ahead and
draw the product of that. So we’re going to
reform our carbonyl. So we still have an R prime
group attached to our carbonyl. Like that. We’ve just added an R
group onto our carbonyl. And the leaving group will
be this alkoxide over here. So R double prime with a
negative 1 formal charge on the oxygen. So now we have a ketone. And if you have a molar excess
of your Grignard reagent, the Grignard reagent is
going to attack your ketone. So once again, another
molar equivalent of our carbanion from
our Grignard reagent is going to attack our carbonyl
carbon– just like before. Kick these electrons
off onto the oxygen. So let’s get some space
here to show what happens. So our next
intermediate is going to have our R prime
attached to our carbon. This top oxygen here
has three lone pairs of electrons and a
negative 1 formal charge. We already added on one R group. We’re going to add
on another R group from our Grignard reagent. And in the last step, it’s
just acid-base chemistry. Right? We’re going to add on H3O+. And this will allow us to
protonate our alkoxide. Right? So a lone pair of
electrons on here. Pick up a proton. Like that. Kick those electrons off. And then, we are done
with our product. R prime. Carbon bonded to an oxygen
bonded to a hydrogen. 2 lone pairs like that. And then, two R groups,
which came from our Grignard reagent– like that–
to form our alcohol. OK. So let’s look at a reaction. So let’s react an ester
with excess Grignard reagent and see what our
product will be. So we’ll do a nice simple
ester over here on the left. So here is our ester. And in the first step, we
add excess methyl magnesium bromides. And in the second
step, we add our H3O+. So we need to think to ourself–
what’s going to happen? Well, I know that
in my mechanism, this negatively charged
carbanion methyl group is going to add on to this carbon. This is going to leave. It’s going to be
our leaving group. And then, the reaction
happens again. All right. And the reaction happens again
adding on another methyl group, and then, finally, protonated
to form an alcohol. So when we draw the
product, we know that this portion is going
to come from our ester. We know we’re going
to form an alcohol. And we know that we’re going to
add our methyl group on twice. So all we have to do is put
two methyl groups on there. Like that. So that, of course, would
form a tertiary alcohol. So this is tertiary right here. This carbon bonded to
three other carbons. So that does it for Grignard
reagents and different ways to make alcohols.



Jun 6, 2013, 1:17 pm Reply

How can I thank you?

Maha Tamimi

Feb 2, 2014, 10:45 pm Reply

You are the best. Thank You 


May 5, 2014, 11:43 pm Reply

very cool

Jenn Lee

Aug 8, 2014, 5:55 am Reply


Demise Slapper

Apr 4, 2015, 5:46 pm Reply

thank you organic god

Sarah Swager [ARC Email]

Dec 12, 2015, 4:17 am Reply

Khan Academy should make tee shirts to raise funds.

N Park

Dec 12, 2015, 4:47 am Reply

why is the reagent for the first problem 1.Et2O instead of CH3MgBr? im confused lol

Collin Crowson

Feb 2, 2016, 7:59 pm Reply

how do you know if there is excess or not?

ghassen bettaieb

Mar 3, 2016, 9:58 pm Reply

thanks a lot


Jan 1, 2017, 1:12 pm Reply

How we know which group disconnection is more favourable..

Hussien Jaafar

Feb 2, 2017, 4:42 pm Reply

so i understand the mechanism of how to make a tertiary alcohol out of adding grignard to ester but what about the OR'' that detached from the R'COOR'', it has -ve charge on the oxygen so cant it be protonated with H+ from H3O+ to form a primary alcohol ?, and what if we added grignard reagent to an ester of formic acid? wouldn't that give us a secondary alcohol? thank you, very helpfull video but i hope u reply

The Odessa Cubbage Foundation

Aug 8, 2017, 9:05 pm Reply

Would this be considered an addition reaction?

Ryan Ally

Aug 8, 2018, 10:07 am Reply

Beautifully explained!!!

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