Protection of alcohols | Alcohols, ethers, epoxides, sulfides | Organic chemistry | Khan Academy

Protection of alcohols | Alcohols, ethers, epoxides, sulfides | Organic chemistry | Khan Academy


Sometimes when you’re trying
to synthesize a molecule, you have to use a
protecting group. And in this video,
we’re going to talk about how to protect alcohols
using a trialkylsilyl group. And so let’s say, our goal
is to make this target compound over here on the right. And we need to start with this
compound over here on the left. And so you might think that this
organolithium compound right here would function
as a nucleophile. So we have a negative 1
formal charge in this carbon. So this lone pair of electrons
is going to be our nucleophile and attack this carbon, which is
a little bit partially positive right here. And so these electrons will
kick off onto your bromine, and you would end up adding
this carbon and this carbon on to form your target molecule. So like that. Unfortunately, this is not
the reaction that occurs. Because not only can
organolithium compounds be strong nucleophiles, they
can also be strong bases. And so what would
actually happen is this lone pair of electrons
here would function as a base, take this proton, leaving these
electrons behind on the oxygen to form an alkoxide. So really, you would form
this product over here. So we would have an oxygen. The oxygen would have now
three lone pairs of electrons around it, giving it a
negative 1 formal charge and it would have lithium plus. So we would form an
alkoxide product instead. So the point of a
protecting group is we need to protect this
hydroxyl group to prevent it from reacting. So if we could somehow
protect this group, we can allow our
reaction to occur at this portion of the molecule. And then we could remove
our protecting group to form our target compound. And so that’s the
idea behind it. So let’s go ahead and show how
we can use a protecting group. So over here on the
right, we have– this would be
t-butyldimethylsilyl chloride. So there’s a tert-butyl
group attached to a silicon, and then there’s two methyl
groups attached to the silicon and also a chlorine. So this would be t-butyl,
or tert-butyldimethylsilyl chloride. So TBDMSCl. And if you think
about this silicon, this silicon is
bonded to a carbon here, a carbon here, a
carbon here, and a chlorine. And all those, the
carbon and the chlorine, are more electronegative
than the silicons, so they’re going to
withdraw some electron density from the silicon making
the silicon partially positive. And so the silicon can
function as an electrophile. It’s electrophilic. And we can get some
electrons from the oxygen. So the alcohol
over here is going to function as a nucleophile,
and the lone pair of electrons is going to attack the silicon. And then these electrons will
kick off onto your chlorine. And so we would lose
HCl in the process. And the imidazole, one of
the things the imidazole does is help to remove the HCl. And the mechanism is a
little more complicated than what I’ve shown. But this is just a simple
way of thinking about it. So you have a nucleophile
and electrophile, and you’re going to put
your protecting group onto your alcohol. So let’s go ahead and draw
the product of this reaction. We would now have our oxygen
bonded to the silicon. And our oxygen would
have two lone pairs of electrons around it. And the silicon is bonded to
two methyl groups, and also a tert-butyl group, like that. So we put on our
protecting group. And sometimes you might
see instead of drawing out all that stuff around silicon,
you might just see an oxygen, and then you might see TBDMS
for our protecting group, which is our t-butyldimethylsilyl
protecting group, like that. So this would be another
way of representing that portion of the molecule. And so now that we’ve
added our protecting group, we can go ahead and react with
our organolithium compounds. Let me go ahead and draw in our
organolithium compound again. So we had a
carbanion here, which now can function
as a nucleophile. So this lone pair of electrons
could attack this carbon right here. And these electrons would
kick off onto the bromine. And so we can go ahead and draw
what we would get from that. So now we would add on our
triple bond right here. So once again, let’s
highlight some carbons. This carbon would
have added on to here. This carbon is right here. And then we have
these electrons, formed this bond
right here, like that. And so we still have
our protecting group. So let’s go ahead
and draw that, too. We have our oxygen and
bonded to our oxygen, we have our silicon
with our methyl groups, and also our tert-butyl
group, like that. And so now that we have
done the desired reaction, now we can take off
our protecting group. So we can remove it to
form our target compound. And so we need to have something
that reacts selectively with the silicon here. And so we’re going to use
tetrabutylammonium fluoride, which is really just a good
source of fluoride anions. So I’m going to go ahead
and draw in a fluoride anion here, which is normally an
extremely poor nucleophile. So it’s actually
selective for silicon. So if the fluoride
functions as a nucleophile, it’s going to attack
the silicon here. And it could do this
for a couple of reasons. So let’s talk about
those reasons here. So first of all, the silicon
is bonded to some carbons. And silicon is
bigger than carbon, if you look at where it
is in the periodic table. And so the silicon
carbon bonds are longer than we’re
used to seeing. And that means that there’s
decreased steric hindrance. So the silicon is a
little bit more exposed, and that allows
the fluoride anion to attack it a little more. So another factor that
allows this is silicon is in the third period
on the periodic table. So it has vacant d orbitals. And so we can go ahead
and show a bond forming between the fluorine
and the silicon. So let me go ahead
and draw what we would get after the fluoride
attacks the silicon. So we would have this
portion of the molecule. And we would have
our oxygen bonded to our silicon in
this intermediate. And now we could show the
fluorine bonded to the silicon, like that. And the silicon is still
bonded to two methyl groups and also a tert-butyl
group, like that. This will give the silicon
a negative 1 formal charge. And it looks a little
bit weird, because we see silicon has
five bonds to it. But that’s, again,
it’s OK because of where silicon is
on the periodic table. It has those has
those d orbitals, and so forming five bonds
for an intermediate is OK. It’s OK for it to have
an expanded octet. Another reason why fluoride can
attack the silicon very well is because the bond that forms
between fluorine and silicon happens to be very strong. So it’s a very strong
single bond here. And we can finish up by
kicking these electrons back onto the oxygen and
protonating and forming our target compound. So we would go ahead and form
our target compound here. So we would get back
our alcohol, like that. And we also successfully added
on this portion of the molecule on the right. And then we would also
form– we would now have the fluorine bonded
to the silicon, like that. So we selectively removed
our protecting group and we formed our
target compound. And so that’s the idea
of a protecting group. It allows you to protect
one area of the molecule and react with another
area of the molecule. And it’s also nice
to have it easily removed to get back
your target molecule.

2 Replies to “Protection of alcohols | Alcohols, ethers, epoxides, sulfides | Organic chemistry | Khan Academy”

  1. +Khan Academy Organic Chemistry Wouldn't it be easier to protect the alcohol group with (CH3)3COH and then take it off using H2O?

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