I have studied the Reimer–Tiemann reaction of phenol with chloroform in presence of aqueous strong alkali. It takes place with intermediate dichlorocarbene.

I have also studied the Reimer–Tiemann "carboxylation" reaction, which requires carbon tetrachloride, and proceeds with intermediate carbocation.

I am however unfamiliar with what the following question requires from me:

$$\ce{Catechol + CH2I2 + NaOH -> ?}$$

Can anyone please state the step-wise mechanisms through which this reaction occurs? Thanks!

  • 1
    $\begingroup$ Methylene iodide appears to be at the wrong oxidation to give a Reimer-Tiemann product. The more likely product is benzo[d][1,3]dioxole aka methylenedioxy benzene. The catechol dianion does a double SN2 displacement on CH2I2. The second displacement is intramolecular. $\endgroup$
    – user55119
    Commented Dec 8, 2017 at 1:00
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    $\begingroup$ @user55119 I've understood what you're saying in the last 2 sentences. And yes, this is exactly the product given in my textbook! But, the book does not mention any mechanism. Why is the mechanism SN2 here? I mean, in the above two reactions, there was Riemer Tienman reaction. But, in this case, it is not. You gave the reason in your first sentence but sorry I couldn't understand it. $\endgroup$ Commented Dec 8, 2017 at 1:28
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    $\begingroup$ In methylene iodide the carbon has a charge of zero. In chloroform carbon is +2. The carbon in CCl4 is +4. Thus, CCl4 gives a carboxyl group as you have noted. Chloroform leads to an aldehyde group. But CH2I2 undergoes displacement faster than a carbenoid species can arise from CH2I2. One might imagine the carbene "CHI" forming. If if it reacts with the negative oxygen, it is the same intermediate as the first SN2 displacement. If the carbene reacts with a ring carbon, the best you can hope for is a benzylic iodide and then a benzylic alcohol in the presence of base. $\endgroup$
    – user55119
    Commented Dec 8, 2017 at 3:34
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    $\begingroup$ @user55119 Thanks! You're mentioning that "Methylene iodide appears to be at the wrong oxidation to give a Reimer-Tiemann product." So, from what I can infer from all your comments, is it a fact that only certain oxidation numbers can give certain products? Is there a rule through which I can predict which oxidation numbers lead to which intermediates? $\endgroup$ Commented Dec 8, 2017 at 11:23
  • $\begingroup$ Go by the oxidation number of carbon as a general rule. $\endgroup$
    – user55119
    Commented Dec 8, 2017 at 14:06

2 Answers 2


This is not really an answer to your question, but I do not think that it is possible to answer it. From one of the many comments we learn that the reference is actually the "preparation of piperonal". It uses 3-4-dihydroxy-benzenecarbaldehyde instead of catechol, but the reaction is same, with $\ce{CH2I2}$ and $\ce{NaOH}$.

I was able to find this reaction described in Chemistry of Natural Products by Sujata V. Bhat, B.A. Nagasampagi, Meenakshi Sivakumar on pages 307f.

4.22.B. Synthesis of Piperine

Piperine was synthesized (Ladenburg, 1894) by the reaction of the piperic acid chloride with piperidine, which confirmed the structure of the molecule (Figure 4.57). The synthesis of piperic acid was achieved starting from piperonal 4.288, which was obtained from catechol using Reimer-Tiemann reaction followed by the condensation with diiodomethane in the presence of a base. Piperonal was condensed with acetaldehyde in the presence of sodium hydroxide and the product obtained was then heated with acetic anhydride and sodium acetate to yield piperic acid.

chemistry of natural products p 307 and 308, Fig. 4.57

I find this a lot more in accord from what you would expect. I cannot imagine any way in which a Reimer-Tiemann (RT) like reaction would occur given the starting reagents. Instead the first step from catechol to 2,4-dihydroxy-benzenecarbaldehyde occurs via an RT reaction. This is very well discussed by H. Wynberg (Chem. Rev. 1960, 60 (2), pp 169–184, DOI: 10.1021/cr60204a003). The mechanism can also be found on Wikipedia.

The second step then involves diiodomethane and forms the acetal. I suppose after deprotonation of the phenolic hydrogen, it can attack at the diiodomethane in nucleophilic substitution fashion, further following another deprotonation and an intramolecular SN step to close the ring.

2,4-dihydroxy-benzenecarbaldehyde + diiodomethane = acetal formation

Unfortunately I could not find anything to further back up my suspicion, since the original synthesis only starts at 4.288. See here (in German): A. Ladenburg, and M. Scholtz, Ber. Dtsch. Chem. Ges. 1894, 27, 2958–2960. DOI: 10.1002/cber.18940270356.

However, since there is proof that the reaction with 2,4-dihydroxybenzenecarbaldehyde forms the acetal, it would be highly suspicious if it were not doing a similar reaction with catechol.


Catechol does react with diiodomathane in presence of $\ce{NaOH}$ as the base. But it does not perform Reimer-Tiemann reaction, but instead it forms methylenedioxybenzene (2H-1,3-Benzodioxole is the preferred IUPAC name according to Wikipedia). As Martin - マーチン notes in his answer, this reaction is most commonly found in the synthesis of piperonal, but the same reaction can also be used for catechol.

The mechanism is similar and instead of diiodomethane, you can also use dichloromethane, dibromomethane, methylene sulphate and methylene chlorobromide. As this paper explains:

The ionizing base for catechols has always been an alkali hydroxide, alkoxide, or carbonate; the methylenating agents have variously been methylene iodide in alcohol or acetone, methylene bromide, 'methylene sulphate', methylene chlorobromide, and methylene chloride. The best yield (69%) of benzo-1,3-dioxole was obtained with methylene iodide, methanolic potassium hydroxide, catechol, and copper bronze, heated and shaken together at $\pu{100-110°C}$ for $\pu{18 hr}$ in a sealed tube. The yield from methylene chloride with these conditions was 53%.

From the same paper:

The formation of benzo-1,3-dioxole probably proceeds in two steps:

  1. A bimolecular nucleophilic displacement of halide ion from the methylene halide by catechoxide dianion (when the ionizing base is strong enough to form the dianion); yielding the haloguaiacol anion (III).

  2. An intramolecular displacement of a second halide ion from (III)


Source: W. Bonthrone and J. W. Cornforth, "The Methylenation of Catechols," J. Chem. Soc. (C) 1969, 1202-1204 (DOI: 10.1039/J39690001202)

Another paper also mentions the reaction between catechol and diiodomethane with methanolic $\ce{KOH}$.

The reaction mechanism is $\mathrm{S_N2}$ so the mechanism provided by Martin - マーチン in his answer should be correct. In case his answer is deleted or modified in future, I am providing a simple mechanism with catechol and diiodomethane only:mechanism

The final product is methylenedioxybenzene (1,3-Benzodioxole), i.e. methylenated catechol.


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