- Alcohols and phenols questions
- Alcohol nomenclature
- Properties of alcohols
- Biological oxidation of alcohols
- Oxidation of alcohols
- Oxidation of alcohols (examples)
- Protection of alcohols
- Preparation of mesylates and tosylates
- SN1 and SN2 reactions of alcohols
- Biological redox reactions of alcohols and phenols
- Aromatic stability of benzene
- Aromatic heterocycles
How to convert an alcohol into a mesylate or tosylate, making it a good leaving group. Created by Jay.
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- Do we need to know this detailed mechanism for the MCAT 2015?(13 votes)
- The OChem mechanisms tested on the MCAT are generally not as in depth as these. Know a few basic mechanisms (particularly your substitution and elimination reactions) and for the rest it's just good enough to know what the products would likely be.
If you got a passage on mesylate and tosylate mechanisms, I would be very shocked.(15 votes)
- At2:50and6:25, for preparation of tosylate, the order of the reactions was first nucleophilic attack by alcohol then acid-base reaction. For preparation of mesylate, the order was opposite. Are these orders fixed or random? If they are fixed, what factors determine the particular order of a reaction?(8 votes)
- The order is not random. It is determined by the structure of the sulfonyl chloride.
In methanesulfonyl chloride, the S atom enhances the acidity of the α-hydrogen. So the pyridine can remove it in the first step.
In p-toluenesulfonyl chloride, there is no α hydrogen, so the only useful choice is for the alcohol to attack the S.(12 votes)
- How is OTs a good leaving group? Thanks :)(2 votes)
- TsO⁻ is a good leaving group because it is:
a. an extremely weak base.
b. stabilized by resonance.(13 votes)
- If the objective of this is to substitute the alcohol functional group with a nucleophile, why not just put it into water with your nucleophile? Wouldn't the water protonate the alcohol, making it a better leaving group, allowing you to get a lot of your substituted product?(3 votes)
- Consider the equilibrium protonation of an alcohol by an acid:
ROH + HA ⇌ ROH₂⁺ + A⁻
An acid like water (pK_a = 15.7) is far too weak to protonate the alcohol to any great extent. The equilibrium will lie far to the left.
Methanesulfonic acid (pK_a = -2.6) is over 10¹⁸ times as strong as water. The equilibrium will be far to the right.(3 votes)
- at0:04, mesylate and tosylate are better leaving groups than alcohols are. What factors determine a group better at leaving than other groups?(2 votes)
- Usually, the best leaving groups are very weak bases.
Methanesulfonic acid and p-toluenesulfonic acid are extremely strong acids,
Their conjugate bases are extremely weak, so the conjugate bases are excellent leaving groups.(5 votes)
- couldn't the extra S-O bond push the electrons in the S=O bond to form an oxyanion(which still has an octet) leaving the chlorine as is?(2 votes)
- Hypothetically, yes it could kick those electrons to the oxygen to form TsCl anion except it is less stable than Chlorine bumping off and forming hydrochloric acid with that free floating proton from the alcohol. Hope that helps!(4 votes)
- I'm a little confused. Without outside knowledge, how would you discern the order? For instance, for tosylates the oxygen in the R-O-H attacked first then the base. In mesylates, the base attacked first. What am I suppose to look out for to understand the order?(3 votes)
- at4:44Triethylamine was used to function as a base and then the reaction follows. Is it possible to use the same mechanism as tosylate? The O of the OH attacking the S and then CL leaves. then triethylamine attacks the H which makes the O+ to O. I am getting the same product at the end too. Is this wrong?(3 votes)
You can make mesylates and tosylates from alcohols. And you might want to do this, because mesylates and tosylates are better leaving groups. So if we look at a general reaction to form a tosylate, you would start with an alcohol and you'd add tosyl chlorides and also pyridine, and you would form your tosylate over here on the right. When we look at the mechanism, we start with tosyl chloride. If you focus in on the sulfur here, the sulfur is bonded to two oxygens and a chlorine. And we know that oxygen and chlorine are more electronegative than sulfur, so they're going to withdraw some electron density from that sulfur. And so since sulfur's losing some electron density, the sulfur becomes partially positive, and we have an electrophilic center. So the sulfur wants electrons. It can get electrons from the alcohol, so a lone pair of electrons on the alcohol here can attack the sulfur, so nucleophile attacks the electrophile. And these electrons could kick off onto the chlorine here to form the chloride anion. And let's go ahead and draw what we would make. So we would have an R group. We would have an oxygen. The oxygen would now be bonded to sulfur. We would also have still a hydrogen attached to that oxygen, and we'd still have a lone pair of electrons on that oxygen. So the oxygen gets a plus 1 formal charge. The sulfur is still double bonded to this oxygen and to another oxygen. And then we still have our ring attached to the sulfur. So put in our pi electrons here, and-- oops. That was a bad one. Let me fix that. So we have our pi electrons right here, and then a methyl group like that. So let's go ahead and follow those electrons. So the electrons in magenta right here on the oxygen formed a new bond to the sulfur, so here are those electrons in magenta. And the next step, we're going to take the proton off the oxygen here. And so the pyridine is going to function as a base. The lone pair of electrons on nitrogen is going to take this proton, leaving these electrons behind on the oxygen. So let's go ahead and draw what we would form. We would have an R group. We would have an oxygen, and our oxygen would have now two lone pairs of electrons. And we would have our sulfur double bonded to this oxygen, double bonded to this oxygen. And then we would have our ring like this. So let me just sketch that in really quickly, so we would have electrons here, here, and here. Our methyl group. And let's follow some of those electrons. So the electrons in this bond now end up on the oxygen like that. And we formed our toluenesulfonate ester, so also called a tosylate, so this is the exact same thing. So this compound and this compound are the same, the top way is just a way to abbreviate it. And so we formed our tosylate. So one reason to form a tosylate would be to have a better nucleophilic substitution reaction. So let's look at first forming a tosylate from this alcohol over here on the left. And so we know that this carbon is a chiral center, so that as a chiral center. But if we're forming a tosylate, the tosylate forms at this oxygen here. So let's go ahead and draw the product. We'd still have a wedge here, because again the reaction does not occur at the chirality center, the reaction occurs at the oxygen here. So the oxygen would now be bonded, so we'd form a tosylate group, which is a much better leaving group than this OH over here. So the tosylate's an excellent leaving group for nucleophilic substitution reactions. So if we went ahead and a did a nucleophilic substitution reaction, we could add something like sodium bromide, so Na plus and Br minus. So if this was an SN2 type mechanism, the bromide anion would attack this carbon right here, which is a little bit positive, so we have a partially positive carbon right here. And then we get nucleophilic attack from the bromide anion, so it attacks right here. And then SN2 type mechanism, you're going to get inversion of configuration. So you go ahead and draw your products like that. And so the formation of a tosylate just makes this process much easier. We could talk about formation of another good leaving group and that's a mesylate, so very similar to a tosylate. So if we look at the general reaction, once again we start with an alcohol. This time we add mesyl chloride, and this time triethylamine as the base that we will use to form our mesylate over here on the right. The mechanism is a little bit different from the formation of a tosylate, so let's go ahead and look and see what happens. At first, the triethylamine is going to function as a base and take this proton right here, so these electrons are going to remain behind on this carbon. So triethylamine reacts with-- this is mesyl chloride right here. So if we take a proton off, let's go ahead and draw what we would have. We would have our sulfur double bonded to this oxygen, sulfur double bonded to another oxygen, the chlorine right here. And we would now have a carbon bonded to only two hydrogens and a lone pair of electrons on this carbon, so it's a carbanion, so it's a negative 1 formal charge. So let's show those electrons. So these electrons in this bond were left behind on the carbon to form our carbanion. So in the next step, these electrons in magenta are going to move in here to form a double bond between the sulfur and the carbon, and that would kick these electrons off onto chlorine to form the chloride anion. And let's go ahead and draw what we would make. So now we would have sulfur, sulfur would be double bonded to an oxygen, sulfur is double bonded to another oxygen, and now there's a double bond between sulfur and this carbon. This carbon is bonded to two hydrogens like that. Once again, we can think about sulfur as being electrophilic, because this sulfur right here is bonded to these oxygens, which are more electronegative. They're going to withdraw electron density from that sulfur, leaving that sulfur partially positive. And so our electrophilic center, once again we're going to have our alcohol function as a nucleophile. So a lone pair of electrons on our alcohol are going to go all the way to here, they're going to attack our electrophile, and when they do that they push these electrons back off onto the carbon. So let's get a little more space down here so we can draw what happens. So this is a sulfine right here, so the alcohol attacks the sulfine nucleophile, electrophile. Now let's go ahead and draw what we would make. So now we would have an R group bonded to an oxygen. We would have a hydrogen. And this is the oxygen from the alcohol, which attacked the sulfur so now there's a bond between the oxygen and the sulfur. There's still a lone pair of electrons on that oxygen, giving it a plus 1 formal charge. And the sulfur is double bonded to this oxygen, double bonded to this oxygen, and bonded to this carbon. This carbon has two hydrogens on it. It also has a lone pair of electrons, so that's a negative 1 formal charge. So once again, let's identify those electrons. So these electrons right here moved off onto the carbon to form our carbanion. And we could identify some electrons on our alcohol, too. So let's make it red here. These electrons right here on the oxygen, those are the ones that form the bond between the oxygen and the sulfur, so you could say that those electrons are these electrons in here. And so in the last step of our mechanism, the carbanion is going to function as a base. And this lone pair of electrons here is going to take this proton, leaving these electrons behind on our oxygen. So let's go ahead and draw what we would make. So we would have an R group over here, we would have oxygen, and the oxygen would have two lone pairs of electrons. The oxygen is bonded to a sulfur. The sulfur is double bonded to this oxygen, double bonded to this oxygen. And then we would have a CH3 group over here. So a CH3 group, because this carbon-- I'm going to go ahead and identify it in magenta. This carbon picked up this proton here-- so you could say it's that one if you wanted to-- and we formed our mesylate. So this is the exact same thing. We could just abbreviate it here. We could say it's R, O, and then an Ms here. So that's how to form mesylates and tosylates, which are excellent leaving groups.