7
$\begingroup$

Excuse me if this a dumb question but it has haunted me for years. Back in my old school days I mixed naphthalene, formaldehyde and sulfuric acid in a mad attempt to make $\mbox{2-Methoxynaphthalene}$. I can't remember if I heated the mixture or not.

It turned black instantly and my teacher at school couldn't tell me what had happened, even less why. Did I produce nano-carbon? Just soot? Or something else?

Edit: way back then, I did try the reaction without naphthalene, and also without formaldehyde, with no (noticeable) result. My teacher also repeated the experiment with the considerably more pure chemicals available at the school lab.

Edit2: as of the comment by Loong, what I did was using Marquis reagent on naphthalene. What I found on the net about this was very little and poor. I couldn't even find anything on the color of $\mbox{polymethylnaphthene}$. (Instinct tells me it should be colorless.)

$\endgroup$
5
  • $\begingroup$ Welcome to chemistry.SE! If you had any questions about the policies of our community, please ‎visit the help center. $\endgroup$ – M.A.R. Dec 25 '15 at 15:27
  • $\begingroup$ @Ϻ.Λ.Ʀ. this has nothing to do with my question here, but could you please give a newcomer like me some hint on how to do all those beautiful formulas? I saw your \mbox and that is quite familiar, but most formulas I have looked at by now are pictures, mostly *.png and I think they are not hand-made with some primitive drawing-tool. $\endgroup$ – Gyro Gearloose Dec 25 '15 at 16:54
  • $\begingroup$ Please visit this page, this page and this ‎one on how to format your posts better using the Mathjax syntax our site uses. Regarding the structural formulas, there are certain molecular editors (some online), that you can use for this purpose. I suggest just Googling for "molecular editor" and choose the one you're most comfy with. $\endgroup$ – M.A.R. Dec 25 '15 at 16:56
  • 2
    $\begingroup$ @GyroGearloose Your description reads as if you made a Marquis reaction. $\endgroup$ – user7951 Dec 26 '15 at 20:20
  • $\begingroup$ @Loong Thank you, so I produced polymethylnaphthalene? $\endgroup$ – Gyro Gearloose Dec 27 '15 at 11:20
3
$\begingroup$

Marquis Reagent (1) is a spot test for alkaloids, first reported in 1896. The original agent was a mixture of 2 drops of 40% formaldehyde and 3 milliliters of concentrated sulfuric acid and was used to detect certain (e.g., opium) alkaloids and distinguish between them. The signature of the alkaloid is both the initial color produced and the sequence of color changes with time. Morphine reacts with Marquis' Reagent to give a purple to violet color. Two molecules of morphine and two of formaldehyde are proposed to condense in the presence of sulfuric acid to the dimeric product which is protonated to the oxoniumcarbenium salt.

The key reaction is protonation of formaldehyde, which attacks an aromatic ring, ortho to the activating phenol group, producing a hydroxymethyl derivative which is immediately protonated and dehydroxylated to give a benzylic cation which attacks another aromatic ring. And it happens again at ortho positions to close a new ring.

enter image description here

Now the ingredients in your old school days reaction (naphthalene, sulfuric acid and formaldehyde) are used to make what was a common concrete admixture. Kao Soap (2), who made Mighty 150 (“Naphthalenesulonic acid-formaldehyde condensate sodium salt”) offered it to my company. We were making a similar material for paint dispersion and promoted our material for concrete strengthening by water reduction.

The condensate is a polymer and is made by first sulfonating napthalene in the beta position with concentrated sulfuric acid. (The alpha sulfonate is the first product, but quickly rearranges to the more stable beta isomer.) Then formalin, 40%, is added, slowly. The formaldehyde protonates, attacks the alpha position of the (still activated) unsulfonated ring (the sulfonated ring is deactivated), forms a hydroxy methyl aryl group, is immediately protonated, dehydrated and attacks another naphthalene sulfonate in the alpha position of the unsulfonated ring.

$H_2SO_4$ + $H_2CO$ --> $H_2COH^+$ +$HSO_4^-$

$H_2COH^+$ + $HO_3SNap$ --> $HO_3SNapCH_2OH$ + $H^+$ --> $HO_3SNapCH_2^+$

$HO_3SNapCH_2^+$ + $HO_3SNap$ --> $HO_3SNapCH_2NapSO_3H$

This methylene group is the first link of a polymer chain. The other alpha positions (one on each unsulfonated ring) are still open and will be attacked if there is enough formaldehyde. Controlling the amount of formaldehyde added is a method of controlling ultimate molecular weight of the polymer, which affects its dispersing power because of the ability of longer polymer chains to adsorb more strongly and repel other particles without deabsorption of the polymer.

Controlling the rate of formaldehyde addition is key to preventing runaway polymerization. The addition of formaldehyde at near room temperature lowers the temperature of the 500-1000 gallon batch; however, the batch temperature must be raised to cause polymerization, which is exothermic - sulfuric acid generates heat as it hydrates - and the batch viscosity is already very high, so the heat cannot be quickly removed.

There is a possibility of a runaway reaction in your mixture: the naphthalene is not even sulfonated, so both rings are activated (relative to benzene). If you did get 2-methoxynaphthalene as an intermediate, the sulfuric acid would protonate the methoxy oxygen, generating a benzylic cation, and water and heat of hydration, and a polymer with chains coming out of up to 4 alpha positions per naphthalene. Now the typical naphthalene sulfonate condensates (the acids and the sodium salts) are brown (light to dark), but I could imagine a polymer coming out with a black color.

This approach to 2-methoxynaphthalene seems destined to fail, because the protonation of 2-methoxynaphthalene is easier than protonation of formaldehyde. But organic chemists frequently generate tars.

Ref.1 https://en.wikipedia.org/wiki/Marquis_reagent

Ref.2 https://chemical.kao.com/en/products/B0012344_en/?region=gl

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.