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Why does racemisation occur?

My doubts are:

  1. Is it due to entropy?

  2. If it is due to entropy, how is the reduction of optical activity by racemisation a consequence of increased thermodynamic stability?

  3. Will the entropy increase or decrease? I have learnt in thermodynamics that it is increase in entropy that favours thermodynamic stability.

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    $\begingroup$ No the racemisation occurs because of thermodynamic instability. In other words there is enough thermal energy to allow racemisation. $\endgroup$ – MaxW Nov 5 '18 at 18:57
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    $\begingroup$ Yes, there is, and it’s called entropy. @MaxW I’m not sure what you’re trying to say. If something occurs because of instability then it follows that the products are more stable than the reactants. $\endgroup$ – orthocresol Nov 5 '18 at 19:39
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    $\begingroup$ @orthocresol My only hesitation in saying yes to the question is just to make clear that it is not the reduction in optical activity causing thermodynamic stability, but that the reduction is a consequence of it being more thermodynamically stable (due to entropy) to have a mixture, which happens to be optically inactive. $\endgroup$ – Tyberius Nov 6 '18 at 20:53
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    $\begingroup$ Having thought about it a little (unfortunately, I don't always have time to write answers on the spot), I'm finding it surprisingly difficult to formalise why a racemate has a larger entropy than an enantiomerically pure compound. It is intuitive, to me at least - the best parallel I can draw is to the entropy of mixing of two ideal gases - but I'm not sure how to do the maths. Your second question is basically "why do thermodynamically favoured processes occur?", to which I cannot give any real answer apart from the Second Law of Thermodynamics. And your third question answers itself. $\endgroup$ – orthocresol Nov 8 '18 at 0:30
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    $\begingroup$ (@orthocresol, here's my attempt at formalization, if you're interested.) $\endgroup$ – a-cyclohexane-molecule Nov 11 '18 at 1:21
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Thermodynamics is not usually helpful in understanding racemisation: think mechanisms and kinetics instead

The thing about enantiomers is that, from a thermodynamic point of view, they are the same so any process is not being driven by differences in the energy between the molecules.

What matters, if racemisation is to occur, is that there is some accessible pathway that allows interconversion. For example, in acidic or basic ethanol solution, (3​R)-3-phenyl-2-butanone will racemise via the (achiral) enol form of the molecule. There is a small amount of the enol present which interconverts to the (chiral) molecule but without remembering which chiral molecule the enol was formed from. This will, ultimately, give a racemic mixture. If no such mechanism existed the molecule would not interconvert. What matters is that some such pathway exists.

The notorious medicine thalidomide is an interesting example. One enantiomer is a useful therapeutic, the other a dangerous teratogen. But making a pure enantiomer doesn't help as the two forms interconvert in the body (by a complicated mechanism).

These conversions are not driven by the different stabilities of the end products but by the inability of non-chiral intermediate molecules in an equilibrium with the chiral starting materials to remember the chirality of their progenitor. This, in effect, increases entropy. But that entropic force is only possible if there is an available reaction mechanism that passes through a non-chiral form allowing the increased disorder that results from a chiral starting material turning into a racemic mixture.

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  • $\begingroup$ Thanks a lot. While I understood most of your answer, I think I don't understand how inability to remember chirality of progenitor molecule the reason for increasing entropy. Please say more on that $\endgroup$ – Anubhab Das Nov 7 '18 at 19:13
  • $\begingroup$ @AnubhabDas Imagine the non-chiral compound A interacting with B to regenerate the original chiral molecule C. Which enantiomer you get depends on which orientation (or which "side") of A the second molecule B approaches from. But that will be random in most situations meaning that a random mixture of the enantiomers of C will result. $\endgroup$ – matt_black Nov 7 '18 at 20:40
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Racemization can indeed be understood from an entropic perspective. While an activation energy barrier exists for spontaneous racemization (hence the importance of kinetics), it is ultimately the increase in entropy that drives racemization.

A simple model: consider a series of $N$ coins, and label heads $H$ and tails $T$. A solution of enantiomers corresponds to a sequence $$\underbrace{HHH\cdots HHH}_{N\;\text{occurrences}},$$ with $N \sim 10^{24}$. Racemization corresponds to picking up a random coin in your sequence and flipping it over. Kinetics determines how quickly you can pick up and flip coins. Eventually, if the kinetics are fast relative to laboratory timescales, we expect some equilibrated sequence like $$\underbrace{HTT\cdots HTH}_{N\;\text{occurrences}},$$ which corresponds to a racemate. The original enantiomeric solution corresponds to just one sequence of coin flips, and the equilibrated racemate to $2^N$ possible sequences. The increase in entropy is then $$\Delta S = k_\text{B}\ln\frac{2^N}{1} = Nk_\text{B}\ln 2.$$ (We could instead have modeled this scenario as doubling the volume of an ideal gas, with the same results. I find coin flips more intuitive.)

The optical activity of your solution corresponds to the excess number of heads over tails in the average sequence. In the enantiomeric solution all our coins are heads, so we have $N$ excess heads. In the racemate we have $0$ excess heads on average.


There are also a few statements from matt_black's answer that I'd like to clarify.

  • While both an enantiomeric solution and a racemate have the same internal energy $U$, they do not have the same free energy $G$ (or $F$), because the free energy depends on entropy.
  • The entropic force for racemization is always present regardless of its rate. Kinetics controls this rate, but cannot affect the thermodynamics. I dislike calling entropy a force, because it's really statistical in origin, but that's just me.
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  • $\begingroup$ This answer surely completes my question. I really wish I could accept more than one answer. Well, if I understood correctly, you said that the racemate has equal number of heads and tails. Will that not mean than any N/2 flips will give heads and rest tails? That will lead to the number of possible sequences as N choose N/2 because all we have to do is choose which flips are head and rest will be tails. $\endgroup$ – Anubhab Das Nov 11 '18 at 7:29
  • $\begingroup$ @AnubhabDas, the racemate will have equal numbers of heads and tails on average. It is not a constraint: each coin can independently be heads or tails. $\endgroup$ – a-cyclohexane-molecule Nov 11 '18 at 12:24
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In addition to @matt_black 's answer, I think it's prudent to add some mechanisms that have partial racemisation instead of complete racemisation, which would have happened in case entropy completely determined the reaction products.

In an $S_N1$ reaction, we have a possibility for the nucleophile to attack the carbocation from either side. However, there are multiple cases where the carbocation doesn't form before the attack of the nucleophile. In these cases, there's a marked difference between the two chiral forms.

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