I can't figure out why the dichlorocarbene molecule produced in the Reimer-Tiemann reaction gets attached to the ortho position of phenol. The ortho position is sterically slightly more hindered due to the oxygen, so why dosen't the carbene attach to the para position?

Mechanism of carbene generation in Reimer-Tiemann reaction

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    $\begingroup$ Does this answer your question? In Reimer-Tiemann reaction why does phenol attack the carbene from ortho position? $\endgroup$ Jun 27, 2020 at 14:23
  • $\begingroup$ No, because this answer was a bit more simple and was straightforward $\endgroup$
    – user95393
    Jun 27, 2020 at 14:53
  • $\begingroup$ @EVO Answers can be simple and straightforward. If they answers your question, then yours is a duplicate of that. $\endgroup$ Jun 29, 2020 at 9:02
  • $\begingroup$ I don't think this is a duplicate; the linked duplicate asks about C2 attack vs O attack, not C2 vs C4. $\endgroup$ Jun 29, 2020 at 9:37

2 Answers 2


The Reimer–Tiemann reaction will give a mixture of products unless the formation of one product is highly disfavoured. The Kürti-Csakó notes in its introductory paragraph on the Reimer–Tiemann:[1]

  1. the regioselectivity is not high, but ortho-formyl products tend to predominate;

The statement already hints at a product mixture being obtained. To further show that both ortho and para attacks are possible, the 1960 review published by Wynberg contains the following scheme as its very first scheme in the introduction:[2]

General scheme of a Reimer–Tiemann reaction

So while both products are obtained, a back of the envelope calculation already shows that the distribution is not a perfect $2:1$ ratio. This can apparently be explaned by the effect of positive counterions, as Hine and van der Veen report:[3]

The ratio of ortho to para product was found to be 2.21 [under high base concentrations], showing that the tendency towards o-substitution is indeed increased under conditions where ion-pair formation is encouraged. One factor that would certainly be expected to be present and that would tend to favor o-substitution is an electrostatic effect. When a dichloromethylene molecule attacks the o-position of a sodium phenoxide ion-pair to yield the probable initial product, there is less separation of unlike charges than when the analogous para product is formed.

So while the steric hindrance of a single oxygen atom is not much more than that of a hydrogen atom and thus does not play a huge role, the presence of positive counterions – although increasing steric congestion – may serve to enhance the formation of the ortho-product through favourable electronic effects.


[1]: L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 378.

[2]: H. Wynberg, Chem. Rev. 1960, 60, 169–184. DOI: 10.1021/cr60204a003.

[3]: J. Hine, J. M. van der Veen, J. Am. Chem. Soc. 1959, 81, 6446–6449. DOI: 10.1021/ja01533a028.


In Reimer-Tiemann Reaction, a mixture of ortho and para isomers is obtained in which the ortho isomer predominates (it is not the sole product). If one of the ortho positions is occupied, the para-isomer is the major product. The two isomers can be separated by fractional distillation, in which the unreacted phenol and the ortho-isomer distil over leaving behind the para-isomer. Ortho product is major mainly due to 2 reasons-

  1. Probability factor ( there are 2 ortho positions available Vs only 1 para position )
  2. H- bonding in the final salisaldehyde.(There is a formation of 6 membered chelated ring which increases the stability of this product.)
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    $\begingroup$ Is the Reimer-Tiemann reaction thermodynamically controlled, such that the product stability determines the outcome? To me, it seems more likely that it's kinetically controlled, where transition state stability is the key. $\endgroup$ Jun 29, 2020 at 8:02
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    $\begingroup$ Although it is run at refluxing chloroform conditions, I want to agree with orthocresol on this one and think that thermodynamic control is slightly unlikely. $\endgroup$
    – Jan
    Jun 29, 2020 at 8:55