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Protection of alcohols

Use of protecting groups to "hide" alcohols while other groups in a molecule react. Created by Jay.

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  • blobby green style avatar for user Anshul Ramachandran
    At , why do those electrons kick back onto the oxygen when the silicon is fine with five bonds? Also, if a bond does have to be removed, why the Si-O bond, why not the bond to the tertbutyl for example?
    (6 votes)
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    • leaf green style avatar for user Courtney Smith
      The electrons kicked back to the oxygen because when you protonated the molecule, the oxygen picked up a hydrogen, giving it a +1 charge. While silicon is okay with five bonds, oxygen is very electronegative, so it wants its electrons back. That's also why it wouldn't be the tertbutyl bond that leaves, because the bonds to silicon aren't the issue, the bonds to oxygen are.
      (3 votes)
  • blobby green style avatar for user EvaS
    At , why does the carbanion attack the C attached to the Br, and not the C attached to the O? Isn't the C attached to the O more electrophilic? Would the O-TBDMS not be a good leaving group, like Br-?
    (2 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      Silyl ethers are still ethers. R-OSiR₃ is analogous to R-OCR₃.
      An alkoxy group is a notoriously poor leaving group. But if you first protonate the ether to form an alkyloxonium ion, you have a much better leaving group.
      This is why silyl ethers make good protecting groups. They are exceptionally stable in base solution. In acid solution, they are easily removed to regenerate the original alcohol. You can even avoid strong acid, because F⁻ ions also react with the silane and regenerate the alcohol.
      The acetylide ion is a powerful base/nucleophile. It can either take part in an SN2 displacement of Br or E2 elimination of HBr. Since it is completely unhindered, the SN2 displacement is the major reaction.
      (6 votes)
  • blobby green style avatar for user Kristina Kerolus
    what other protection groups can be used?
    and what removes them?
    (2 votes)
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  • blobby green style avatar for user markdavison513
    At you mentioned that Si can make 5 bonds because it has d orbitals. Being in the 3rd period and above the d block, how does Si have d orbitals?
    (1 vote)
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    • female robot grace style avatar for user Anna
      It is because in the 3rd shell there is 3s2 3p6 3d10 in a full shell and even before the transition metals there are d orbitals since while the 3s and 3p orbitals fill the 3d orbitals form from the electrons having more places to go.

      This makes all elements past neon hypervalent and thus Iron is able to have a +7 oxidation state instead of just +3 like you would expect from an oxide.
      (2 votes)
  • leaf green style avatar for user studentn
    At , how do we lose HCl in the process? Don't we just lose a Cl? Where did the H come from?
    (1 vote)
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  • blobby green style avatar for user Nicoleta Ese
    Hi! What is the name of the reaction?
    (1 vote)
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  • duskpin ultimate style avatar for user kitty
    Are Silyl Ethers better for non stericly hindered molecules? what if we had tertiary alcohol? would it still be a good protecting group?
    (1 vote)
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  • aqualine ultimate style avatar for user Moises Jerez Schachtler
    So F- attacks Si because it has longer bonds than C-C and C-O bonds. Since Si can hold negative charges because of its size, couldn't F- attack the C-O bond making TBDMSCl a good leaving group?
    (1 vote)
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  • blobby green style avatar for user James OConnor
    The book uses H2SO4 + Isopropene--> tertiary carbocation as a protector. I am still trying to understand why it's a protector. I assume steric hinderence?
    (1 vote)
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  • blobby green style avatar for user guy elgar
    why does the oxygen take the excess electrons rather than fluorine given its decreased electronegativity relative to fluorine?
    (1 vote)
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Video transcript

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.