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The above is a representation of magic acid. The left part is the acid; the right part is the solvent. Why would this combination make for a particularly good acid?

The left part looks like a sulfuric acid derivative, except one of the oxygens has been replaced by an electronegative fluorine. Why can't we replace all the oxygens with fluorines? Wouldn't that make for an even stronger acid through inductive stabilization of negative charge after proton loss?

Also, does the solvent play a role in magic acid's acidity? Does the highly partially positive Sb central atom play a role in stabilizing negative charge on the solute after the solute has been deprotonated?

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3 Answers 3

up vote 14 down vote accepted

The explanation is that sufficiently strong Lewis acids like $\ce{SbF5}$ will form adducts even with very weak, traditionally non-coordinating Lewis bases like $\ce{SO3F-}$. The resulting anionic complexes are extremely stable, with the negative charge distributed over a large, highly polarizable molecule and inductively stabilized by the fluorine atoms. The formation of this complex, in turn, promotes the autoprotolysis equilibrium:

$$\ce{HSO3F + HSO3F <=> SO3F- + H2SO3F+}$$

The protonated species above is the most active molecule in sufficiently concentrated solutions. (In reality, the composition of these mixtures may be much more complex, with dimers of the form $\ce{Sb2F10(SO3F)-}$ and possibly more complex oligomeric structures. See, for example, this book.)

While there are some more recent and exotic examples of such Bronsted/Lewis acid mixtures, the idea is actually a fairly old one. Perhaps the most common example: sulfur trioxide in sulfuric acid gives a solution known as oleum, which contains a proportion of disulfuric acid molecules. The resulting mixture is considerably more acidic than pure sulfuric acid alone and has various uses in industrial applications as well as laboratory-scale synthetic organic chemistry.

As for why the oxygens cannot be subsituted with fluorines to yield an even stronger acid system, I can only give a speculative answer. It seems to me that a hypothetical molecule such as, e.g., $\ce{SF3OH}$ (if this is what you're imagining) would probably be very liable to decompose:

$$\ce{SF3OH -> SOF2 + HF}$$

In any case, I can find absolutely no relevant literature, which in and of itself says something about the stability of such a molecule (or perhaps my ineptitude, as the case may be).

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Isn't sulfuric acid such a strong acid because of the delocalization effect? Replacing oxygen with fluorine would then only decrease the acidity. –  Jori Aug 20 at 7:25
1  
@Jori, I was addressing Dissenter's secondary question as to why sulfuric acid derivatives with all but one $\ce{-OH}$ replaced aren't seen (and probably don't exist). Yes, the structure with oxygens has an advantage in polarizability over fluorines, but (to the best of my knowledge) inductive through-bond and through-space effects are chiefly influenced by electronegativity, similarly for any negative hyperconjugation that may exist, and fluorine is dominant there. The molecules in question are purely hypothetical, so I really have no definitive answer. –  Greg E. Aug 20 at 7:58

Why is magic acid a superacid?

Because it meets the definition of a superacid: A medium having a high acidity, generally greater than that of 100 wt.% sulfuric acid

Why would this combination make for a particularly good acid?

I'm simplifying a bit, but let's examine the following equilibrium (it is one of the early equilibria involved in the formation of Magic acid) $$\ce{HSO_{3}F + SbF5 <=> H^+ + [SbF_{5}-O-SO_{2}F]^-}$$ The extreme stability of the hexacoordinate antimony anion is what drives the equilibrium to the right.

Why can't we replace all the oxygens with fluorines?

First, the oxygen(s) is used to coordinate to the $\ce{SbF_5}$ structures. In some structures there is coordination to multiple $\ce{SbF5}$ units through multiple oxygens. In other structures the oxygens have been removed and replaced with fluorine. Also, you need at least one oxygen so you can have a proton. The equilibria that occur in magic acid are very complex and not entirely understood (for example, see p 52)

does the solvent play a role in magic acid's acidity?

There is no solvent.

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I thought the antimony complex was the solvent –  Dissenter Aug 19 at 16:06
    
Also why would the complex be stable ? Induction? –  Dissenter Aug 19 at 16:09
    
"I thought the antimony complex was the solvent" I misunderstood, I thought you meant some traditional solvent. "why would the complex be stable" With all of the oxygens and fluorines inductive effects must play a role. I also suspect that the Sb is playing a role. –  ron Aug 19 at 16:19

In addition to the answers of ron and Greg E. I just want to add another concept, that was not mentioned so far: non-coordinating anions. You can stop reading here or continue and see a short summary.

As a second remark, it is important to know, that fluorosulfuric acid is already a superacid.


Acidity is strongly dependent on the strength of the proton-conjugated-base bond. Basically stating, the weaker the bond, the higher is the acidity.

All currently known superacids operate on this principle and "magic acid" is one of the strongest. The reason for the extraordinary acidity is the high potency of antimony pentafluoride as a lewis acid. It will coordinate to almost anything with electrons, especially itself in liquid phase, forming chain oligo-/polymers $\ce{[SbF5]_{n}}$ with a length between $\ce{n=5-10}$.

If there is something more potent than this, it will go for it. Antimony pentafluoride is a very dangerous substance. This is exactly what is happening in magic acid, forming weakly or non-coordinating anions as conjugated base. $$\ce{SbF5 + HSO3F <=> H+ + [SbF5SO3F]^{-}}.$$ Now this is a fairly large anion with a highly delocalised (due to symmetry constraints) charge. With excess of antimony pentafluoride, even larger anions can be formed, which is the case when used as a solvent. $$\ce{[SbF5SO3F]^{-} + SbF5 <=> [(SbF5)2SO3F]^{-}}$$ As a consequence, the bond to the proton is extremely weak. In other words, there is almost no coordination of this proton, leaving it basically naked.

There is another superacid, using the same concept, which is even stronger: Fluoroantimonic acid. It is in principle that, what you already suggested, substituting the oxygens by fluorines. Instead of $\ce{HSO3F}$, hydrofluoric acid $\ce{HF}$ is used, giving rise to the (to date) most powerful acid. It also makes use of the high affinity of $\ce{SbF5}$ towards fluoride anions. $$\ce{2SbF5 + HF <=>> H+ + [Sb2F11]- }$$ With these naked protons it is possible to protonate aliphatic hydrocarbons to form carbocations.

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