I've read that in Kolbe's reaction of phenol, i.e. reaction of phenol $(\ce {C6H5OH})$ with $\ce{NaOH}$ forming sodium phenoxide $(\ce {C6H5O^{-}Na^{+}})$, followed by reaction with $\ce{CO_2}$ and acidification; salicylic acid (2-hydroxybenzoic acid ) is formed, but when $\ce{KOH}$ is used instead of $\ce{NaOH}$, 4-hydroxybenzoic acid is formed.

I don't understand why two different products are formed in the same reaction using very similar bases. Moreover, there is electrophilic aromatic substitution on the phenoxide ion in both reactions and $\ce{O^{-}}$ being an ortho-para directing group must form both products in both the reactions.

  • $\begingroup$ At temperatures above 410 K, 4-hydroxybenzoic acid is also formed if you use NaOH. $\endgroup$
    – Sensebe
    Mar 19, 2014 at 6:26
  • $\begingroup$ Why the temperature difference? In fact there should be steric repulsion between bulky COO- and O- hence the other product should be easy to form right? $\endgroup$ Mar 19, 2014 at 6:32
  • $\begingroup$ Intramolecular hydrogen bonding maybe?! $\endgroup$ Dec 7, 2021 at 10:47

2 Answers 2


I don't understand why two different products are formed in the same reaction using very similar bases.

  • If the bases make a a difference, there must be a difference between them.
  • If the anion is the same, the cation must make the difference.

The difference between $\ce{Na+}$ and $\ce{K+}$ lies in the ionic radii of 0.95 and 1.33 $\mathring{A}$, respectively.

It is believed that $\ce{Na+}$ serves as an anchor and guides $\ce{CO2}$ into the ortho position during the formation of the Wheland intermediate. In the latter, $\ce{Na+}$ is chelated between the oxygen atoms of carboxylate and the phenolate. Apparently, this is very effective.

This does not mean that the phenolate doesn't attack $\ce{CO2}$ via C-4 at all. However, the formation of a Wheland intermediate (= the rate-determining step) is normally (and definitely in the case of the para attack) reversible.

It is open to question if full reversibility can be claimed for the $\ce{Na+}$ directed ortho attack under modest reaction conditions.

It is however likely that the back reaction (= cleavage Wheland intermediate) will stronger come into play at elevated temperatures. Consequently, some 4-hydroxybenzoic acid will be formed then.

With the bigger $\ce{K+}$, the anchoring and chelating effect is seemingly not possible, $\ce{K+}$ doesn't fit and the para direction effect of the phenolate prevails.

  • $\begingroup$ That's a nice explanation.can you explain why 4-hydroxybenzoic acid isn't formed in the NaOH reaction?It should be easy, given that steric factors don't come into the way,right? I assume that steric factors also aid in difficulty of formation of salicylic aid in the reaction with KOH $\endgroup$ Mar 19, 2014 at 6:36
  • $\begingroup$ @scienceauror I added some lines to explain the possible reasons. $\endgroup$ Mar 19, 2014 at 7:28
  • $\begingroup$ Okay so that means in each reaction the other product can form in difficult-to-reach conditions like high temperature, and are easily reversible; due to the reasons you specified. $\endgroup$ Mar 19, 2014 at 7:55
  • 5
    $\begingroup$ In reactions, in which the first step is reversible and the following one is not, the product formed first (fastest) is not necessarily the most stable one. At low temperature, the faster one makes the race. At higher temp, the system equilibrates and the more stable product is eventually formed. $\endgroup$ Mar 19, 2014 at 8:03
  • $\begingroup$ @Klaus-DieterWarzecha But using the same analogy as you told, it suggests that in reimer tiemann reaction, the para salicylaldehyde shall be major product if KOH is used. However on wikipedia, ortho product is formed with NaOH as well as KOH in reimer tiemann reaction. Can you please explain why this don't work in reimer tiemann reaction? $\endgroup$ Aug 24, 2022 at 13:59

The question: Why isn't 4-hydroxybenzoic acid formed along with salicylic acid in Kolbe's reaction?

There was a good answer provided to OP's question elsewhere using the cation sizes. However, it didn't provide the clear answer to the main question about why isn't 4-hydroxybenzoic acid formed when $\ce{NaOH}$ is used as a base. And there is also another question within the text: Why isn't salicylic acid formed along with 4-hydroxybenzoic acid when $\ce{KOH}$ is used as a base. Yet, OP has not provided any reference to these claims. Therefore, we can conclude that there are no scientific evidence to support them.

The Kolbe-Schmitt reaction is a carboxylation reaction of alkali metal phenoxides with carbon dioxide where hydroxybenzoic acids are formed. This reaction has been used for more than a century for industrial production of aromatic hydroxy acids such as salicylic acid (2-hydroxybenzoic acid). The accepted general mechanism of the formation of salicylic acid is presented in the bottom of following image:

Reaction mechanism of the Kolbe–Schmitt reaction

The Kolbe-Schmitt reaction is a classical example of a reaction in which the nature of the reaction products is dependent on the alkali metal cation (Ref.1). For example, Marković et al. have investigated the mechanism of the Kolbe-Schmitt reaction (using phenolates of lithium, potassium, rubidium, and cesium phenoxides for the carboxylation) by means of computational analysis. It is shown that the reactions of all alkali metal phenoxides with carbon dioxide occur via very similar reaction mechanisms and the reactions can proceed in the ortho- and para-positions. The exception is lithium phenoxide, which yields only salicylic acid in the Kolbe−Schmitt reaction. It is found that the yield of the para substituted product increases with increasing the ionic radius of the alkali metal used.

Experimentally, under comparable conditions, lithium and sodium phenoxides have yielded salicylic acid as the major product (Ref.2). In contrast, potassium, rubidium, and cesium phenoxides have yielded the mixtures of salicylic acid and 4-hydroxybenzoic acid in different ratios (Ref.2 & 3).It is worth mentioning here that before the work in Ref.3 (1954), it was generally accepted that carbonation of sodium phenoxide leads to the formation of salicylic acid, while potassium phenoxide, under similar conditions, converts primarily into 4-hydroxybenzoic acid. However, Ref.3 has shown that this is not always true.

Because of these new insights the mechanism of the Kolbe−Schmitt reaction has been the focal point of both chemical and computational studies. A mechanism involving an intermediate chelate complex between carbon dioxide and metal phenoxide or 2-naphthoxide ($\ce{MOPh–CO2}$ and $\ce{MONaph–CO2}$ where $\ce{M}$ stands for a metal) is a postulation of early researchers. The presence of the intermediate $\ce{PhONa–CO2}$ complex was confirmed on the basis of the IR absorption spectra (Ref.4). Based on these results and others indicated that direct carboxylation is not an operative mechanism of the Kolbe–Schmitt reaction.

Mean time, a standpoint of initial formation of the $\ce{MOPh–CO2}$ or $\ce{MONaph–CO2}$ complex has been supported by a number of theoretical works (Ref.5). It was shown that the $\ce{CO2}$ moiety of the $\ce{MOPh–CO2}$ complex performs an electrophilic attack at the benzene ring in the ortho- and para-positions, leading to the new intermediates and final products (similar to pathway 1 in the image; e.g., Ref.5&6):

Suggested mechanism

This mechanism clearly shows that the ortho-product is possible whenever $\ce{KOH}$ is used as a base. When 2-naphthol is the substrate, Ref.4 listed following results (Reference reaction is at the top of ist image):

$$ \begin{array}{c|ccc} \hline \ce{MOH}^a & \text{Rxn Temp.} & \text{Rxn Time} & \text{% yield ($\bf{1}$}) & \text{% yield ($\bf{2}$}) & \text{% yield ($\bf{3}$}) & \text{% SM recovery} \\ \hline \ce{KOH} & \pu{293 K} & \pu{1 h} & 52.0 & 0.0 & 0.0 & 47.0\\ \ce{KOH} & \pu{503 K} & \pu{10 h} & 4.0 & 72.0 & 0.0 & 24.0\\ \ce{KOH} & \pu{523 K} & \pu{10 h} & 2.0 & 87.0 & 0.0 & 10.0\\ \ce{NaOH} & \pu{503 K} & \pu{10 h} & 14.5 & 71.5 & 2.5 & 11.0\\ \ce{NaOH} & \pu{523 K} & \pu{10 h} & \text{Trace} & 84.5 & 2.0 & 13.0\\ \hline \end{array}\\ ^a: \text{Carboxylation of pottasium and sodium 2-naphthoxides with $\ce{CO2}$ at $\pu{5 MPa}$ pressure.} $$

The above Table shows there is not clear preference for pottasium and sodium 2-naphthoxides. The recent green approach shows adjusting conditions can make different isomers (Ref.7):

Green Approach


  1. Zoran Marković, Svetlana Marković, Nebojša Begović, "Influence of Alkali Metal Cations upon the Kolbe−Schmitt Reaction Mechanism," J. Chem. Inf. Model. 2006, 46(5), 1957–1964 (https://doi.org/10.1021/ci0600556).
  2. Alan S. Lindsey, Harold Jeskey, "The Kolbe-Schmitt Reaction," Chem. Rev. 1957, 57(4), 583–620 (https://doi.org/10.1021/cr50016a001).
  3. Ogden Baine, G. F. Adamson, J. W. Barton, J. L. Fitch, D. R. Swayampati, Harold Jeskey, "A study of the Kolbe-Schmitt reaction. II. The carbonation of phenols," J. Org. Chem. 1954, 19(4), 510–514 (https://doi.org/10.1021/jo01369a006).
  4. J. L. Hales, J. Idris Jones, A. S. Lindsey, "Mechanism of the Kolbe–Schmitt reaction. Part I. Infra-red studies," J. Chem. Soc. 1954, 3145-3151 (https://doi.org/10.1039/JR9540003145).
  5. Svetlana Marković, Igor Đurović, Zoran Marković, "Revisiting the Kolbe–Schmitt reaction of sodium 2‑naphthoxide," Theoretical Chemistry Accounts 2015, 134, Article # 45 (9 pages) (DOI 10.1007/s00214-015-1648-0).
  6. Zoran Marković, Svetlana Marković, Nedeljko Manojlović, Jasmina Predojević-Simović, "Mechanism of the Kolbe–Schmitt reaction. Structure of the intermediate potassium phenoxide-$\ce{CO2}$ complex," J. Chem. Inf. Model. 2007, 47(4), 1520–1525 (https://doi.org/10.1021/ci700068b).
  7. Khakim A. Suerbaev, Mayliby K. Aldabergenov, Nurbolat Zh. Kudaibergenov, "Carboxylation of hydroxyarens with metal alkyl carbonates," Green Processing and Synthesis 2015, 4(2), 91-96 (https://doi.org/10.1515/gps-2014-0098).
  • $\begingroup$ Decarboxylation of 4-hydroxy-3,5-diisopropyl benzoic acid with NaOH/ethylcellosolve at 130 deg. C gives 2,6-diisopropyl phenol, propofol. [pubs.acs.org/doi/10.1021/op400300t]. Table I, pg. 588, in ref. 2 (vide supra) gives data on the effect of temperature of the consumption of p-hydroxybenzoic acid with potassium counter ion in the K-S-M reaction. $\endgroup$
    – user55119
    Jul 27, 2020 at 20:41

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