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Why is proton transfer always kinetically favored? In other words, why are Brønsted acid-base reactions so quick?

Is it because protons are generally unhindered, sterically? This seems plausible; consider an alcohol. The proton isn't blocked by any bulky groups. It's on the end of a oxygen atom.

What else could contribute to the kinetic favorability of a proton transfer? Reasoning backwards - given that protons transfers are very fast - this implies that the activation energy of a proton transfer reaction is small. And as outlined above - the activation energy is small in part because most of the time, protons are unhindered. What else might contribute to the small activation energy of a proton transfer/kinetic favorability of a proton transfer?

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    $\begingroup$ Quantum tunneling. $\endgroup$ – Marko Sep 22 '14 at 15:13
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Due to their low mass, protons can tunnel through barriers, they don't always have to go over the barrier in an activated process. Proton tunneling becomes particularly important at low temperatures where the available thermal energy may be inadequate for an activated process (e.g. over the barrier) to occur. A nice, brief and not overly-mathematical introduction to tunneling can be found here.

dashed line - thermally activated process

solid line - tunneling process

enter image description here

A second reason why proton transfer is so facile relates to hydrogen bonding. If hydrogen bonding is present, then the proton only needs to elongate one bond (the initial bond) while simultaneously shortening the hydrogen bond in order to effect transfer from one molecule to the next. This energetically compensating process reduces the activation energy required for proton transfer. In solvents with hydrogen bonded clusters or chains (water would be one such example), this process can be repeated within the cluster thus allowing the proton to cover large distances (from one side of the cluster to the other) on a very short timescale. In other words, only minor mass movement is required; by simply shortening some bonds a bit, and slightly elongating others, a proton at the left side of a group of molecules (top row in the following figure) can "reappear" on the right side of the chain (bottom row) extremely rapidly. This process is referred to as the Grotthuss mechanism after its discoverer. Some of the finer points of the mechanism still constitute an active areas of research.

enter image description here

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  • $\begingroup$ Hydrogen bonding is just an elongated covalent bond? $\endgroup$ – Dissenter Sep 22 '14 at 15:30
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    $\begingroup$ Yes, an elongated, weaker, polar covalent bond. $\endgroup$ – ron Sep 22 '14 at 15:51
  • $\begingroup$ What about the Steric factor I mentioned? @ron $\endgroup$ – Dissenter Sep 22 '14 at 16:05
  • $\begingroup$ Yes, steric considerations also play a role. $\endgroup$ – ron Sep 22 '14 at 17:23
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    $\begingroup$ Proton transfer in many ways exists in its own unique domain of chemistry since the proton does not have any inner shells beneath its valence electron. Recommended classic book: "The Proton in Chemistry" by Bell. $\endgroup$ – iad22agp Sep 22 '14 at 18:48

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