# Understanding the mechanism of Blanc chloromethylation

I have seen this question: Mechanism for chloromethylation of benzene.

The answer there uses a picture from wikipedia which confuses me. It gives $$\ce{H2C=O <=>C[HCl] H2C+-OH <-> H2C=O+H}$$ and the benzene attacking the $\ce{C}$ from $\ce{H2C=O+H}$, not the $\ce{C}$ from $\ce{H2C+-OH}$ as I would have expected.

Is it only a quirk in the picture or do I have some fundamental misconceptions?

• I found the question this is a duplicate of. chemistry.stackexchange.com/questions/37870/… Different reaction, same question, although it is a poor answer. Maybe mark that as a dupe of this – orthocresol Jan 7 '16 at 16:14
• @orthocresol I agree, but my resonance structure is easier, as it has only two parts ;-) – Gyro Gearloose Jan 7 '16 at 16:18

Resonance is a concept which expresses a more realistic bonding situation through a set of conformations. Therefore the following can only be treated as one structure: $$\Bigg[~\ce{H2\overset{+}{C}-OH <-> H2C=\overset{+}{O}H}~\Bigg]$$ It is therefore irrelevant in which structure you indicate the nucleophilic attack since it is the same carbon. Usually you would rather draw it the way that the arrow points to the structure where the carbon carries the positive charge, since you would expect this structure to have a higher contribution to the ground state.

• So my misunderstanding is that resonance is not some equilibrium state but rather a quantum mechanic "dead and alive" at the same time, like Schroedingers Cat? – Gyro Gearloose Jan 7 '16 at 16:13
• @GyroGearloose yes that is pretty much the case. There would be a couple of more resonance structures, but they are not too important, so most of the times you neglect them. Whenever you see resonance structures think of them as one state. – Martin - マーチン Jan 7 '16 at 16:59

The result is the same no matter which resonance form you use. Resonance forms are only formal representations, extreme forms of the true structure which is somewhere in between.

Note that to get from the blue resonance form to the red resonance form, you'd need to push an arrow:

So there is no real difference which resonance form you use - in the end you would draw exactly the same arrows. In the blue reaction, you draw two arrows. In the red reaction, you only draw one arrow. But, to get from the blue resonance form to the red resonance form, you have to draw one arrow anyway.

Don't read too much into the number of arrows or whatever - the only point I am trying to make is that from a mechanism-drawing perspective, it does not matter which resonance form you use, because all roads lead to the same product anyway. All those arrows are just formal bookkeeping.

• I could (naively?) interpret $\ce{H2C+-OH}$ as Lewis acid, but not so $\ce{H2C=O+H}$. – Gyro Gearloose Jan 7 '16 at 16:28
• @GyroGearloose No, since those are the same thing, if one is a Lewis acid, the other must be as well. – jerepierre Jan 7 '16 at 16:59
• Bookkeeping... You like that word, don't you. Nice pictures! +1 – Martin - マーチン Jan 7 '16 at 17:01