# Why do self-replicating molecules dissociate?

I am new to chemistry (have only done a GCSE) so I apologize for my lack of knowledge. But I am confused about why self-replicating molecules dissociate.

It is my understanding that a self-complimentary molecule can gather the parts needed through bonding to create another copy of itself on its recognition surface. Once the new molecule is constructed I am unsure why it would then depart from the original one during this dissociation phase. Why would all of these bonds suddenly break apart? What exact mechanisms are involved in this process?

Well, allow me to answer since I worked on self-replication during my master’s thesis.

The thing about all these associationen is that there is always an equilibrium to be considered. That is also true for chemical reactions: Technically, every reaction is at an equilibrium of some kind.

$$\ce{A + B <=> C + D}$$

or in the special case of associations:

$$\ce{A + B + AB <=> A-B-AB}$$

For nomal bond-creating and bond-breaking reactions, the energy involved in the whole process is relatively large — breaking a carbon-carbon bond requires approximately $400\,\mathrm{\frac{kJ}{mol}}$! However, if we are merely talking about association as in complexation, the energies involved (i.e. gained or lost) are rather small. This es especially also true for the activation energy, meaning that forming and dissociation of such complexes are rather fast and at equilibrium the whole time. It is important to realise that the bonds formed for the self-replicating complex are all intermolecular and only the bond between the fragments is a covalent, intramolecular bond.

Let’s take a look at an exemplary self-replicating complex:[1]

On the right half of this picture we have the two fragments, on the left we see the template. All solid lines are actual (intramolecular) bonds while all dashed lines are intermolecular associations — in this case hydrogen bonds. The intermolecular dashed interactions build up and break down rapidly with low energy barriers. The new intramolecular bond is formed between carbon and nitrogen on the aromatic rings (not formed on the right, formed on the left) and represents the energy liberated in the reaction. This one is only formed, not broken.

To form the bond, the two fragments need to linger on the template long enough. That is generally no problem, despite rapid association and dissociation.

In fact, form too stable of a complex (i.e. one that won’t dissociate quickly) and you have effectively eliminated the complex’ ability to self-replicate.

[1] A. Terford, G. von Kiedrowski, Angew. Chem. 1992, 104, 626. DOI: 10.1002/ange.19921040523
In English: Angew. Chem. Int. Ed. 1992, 31, 654. DOI: 10.1002/anie.199206541

• Great! So the constituent fragments of a self-complimentary molecule are bonded together inTRAmolecularly as they are gathered on the recognition surface (presumably before fragments have a chance to dissociate individually). These bonds require lots of energy to break (and so won't at equilibrium). The gathering process is done easily/quickly through association. Then at some point the formed molecule will then dissociate, which like association requires little energy, is based on intERmolecular forces, and can happen at equilibrium. Is that right? – Jonathan Sep 30 '15 at 21:47
• @Jonathan I added an image and a reference; maybe that can clear the confusion that arose with intramolecular and intermolecular bonds. – Jan Oct 1 '15 at 11:58
• Jan, I respectfully disagree with your statement: "Technically, every reaction is at an equilibrium of some kind." Many reactions, such as the burning of hydrogen in air, are not in equilibrium--there is no "reverse burning" going on that dissociates water back into hydrogen and oxygen. I do agree that in considering any reaction, it is worth asking whether there is an equilibrium involved. – iad22agp Oct 1 '15 at 17:08
• @iad22agp I respectfully disagree with your respectful disagreement. Reversablility of all reactions is a fundamental thermodynamic property. Only that in the case of combustion (no matter which one) the equilibrium is 99.999999… % towards the product side. – Jan Oct 1 '15 at 17:11
• @Jan, I think you are confusing the term equilibrium with the term reversible. Take a simple system, where two chemicals A and B can interconvert, but have a high kinetic barrier (activation energy) for interconversion. At a sufficiently high temperature, they do interconvert, and an equilibrium is established, where the A:B ratio reflects their relative stabilities. However, at a low temperature, they do not interconvert, so you can add more A or more B and obtain a non-equilibrium mixture that remains stable at low temperature. (contd.) – iad22agp Oct 1 '15 at 19:26