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. $\endgroup$ – Immortal Player Mar 19 '14 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? $\endgroup$ – scienceauror Mar 19 '14 at 6:32

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 $\ce{CO2}$ attack 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.

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  • $\begingroup$ 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 $\endgroup$ – scienceauror Mar 19 '14 at 6:36
  • $\begingroup$ @scienceauror I added some lines to explain the possible reasons. $\endgroup$ – Klaus-Dieter Warzecha Mar 19 '14 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. $\endgroup$ – scienceauror Mar 19 '14 at 7:55
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    $\begingroup$ 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. $\endgroup$ – Klaus-Dieter Warzecha Mar 19 '14 at 8:03
  • $\begingroup$ @KlausWarzecha, $$\text {Perfect Answer.}$$ ...+1 $\endgroup$ – Apurv Mar 19 '14 at 13:00

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