Reimer-Tiemann reaction is known to give ortho major product when NaOH is used as a base. However there seems to be some confusion regarding the major product in case KOH is used as the base.

It is known that when KOH is used as a base in Kolbe-Schmitt reaction para product is favoured whereas ortho product is favoured in case when NaOH is used for the same.
Now the accepted answer in this post states that the same thing will happen even in reimer-tiemann i.e. para will be the major product when KOH is used. However it does not states any credible sources which back it up and hence this creates confusion since the Wikipedia page on Reimer Tiemann displays the reaction with KOH as its base and it prefers the ortho product. But again I don't see any specific talk about KOH on this page or the books I have.

Hence to clear the confusion, I'm posting this question once again. If anyone has any citable/reliable source which clearly and specifically examines the case of KOH in reimer-tiemann reaction please answer this question. Also reasons as to why it should or why it should not are welcome.


1 Answer 1


From here:

Reimer and Tiemann (Ber., 1876, 9, 824) state that, contrary to experience in the Kolbe-Schmidt reaction, in their synthesis of hydroxy-aldehydes the proportion of $ortho-$ and $para-$ derivatives is unaffected by changing from sodium hydroxide to potassium hydroxide

The improvement in technique afforded by the determination of aldehydes by dinitro phenylhydrazine has enabled us to show the effect of change of alkali in the Reimer-Tiemann reaction where the original authors failed (loc. cit.). In 15N-alkali the ortho-para-ratio was $\ce{NaOH} \ 2.08$, $\ce{KOH}\ 1.24$, $\ce{CsOH}\ 0.98$, and $\ce{NMeEt3.OH}\ 0.52$. In 2N-alkali the difference almost disappeared, the ratio being $\ce{NaOH}$ $0.90$, $\ce{KOH} \ 0.91$, and $\ce{NMeEt3.OH}\ 0.86$. In 15N-lithium hydroxide the ratio was $0.9$ but, as much lithium phenoxide crystallised out, the concentration in solution was low.

When the Reimer-Tiemann reaction is carried out in less concentrated solutions, e.g., in 2N-alkali, the ortho-para ratio, when tetra-alkylammonium hydroxide is used, corresponds to the substitution of the phenoxide ion plus that of free phenol formed by hydrolysis. When 2N-sodium hydroxide is used, only a small amount of co-ordination compound is formed, so that the ortho-para ratio is not appreciably altered.

When a concentrated $\ce{NaOH}$ or $\ce{KOH}$ solution is taken, clearly the ortho product is major (o/p ratio $2.08$ and $1.24$ respectively). In the 2N-alkali solution, the para product is major and the o/p ratio for $\ce{NaOH}$ and $\ce{KOH}$ are nearly the same ($0.90$, $0.91$); this can be explained by saying that the concentration of the $\ce{Na+}$ and $\ce{K+}$ cations in the solution is low enough that they do not effect the reaction, hence the para-product is major and the o/p ratio is same for both $\ce{NaOH}$ and $\ce{KOH}$.

In the Kolbe-Schmidt reaction, the formation of the para-product with $\ce{KOH}$ is due to the inability of the $\ce{K+}$ cation to sufficiently stabilize the transition state due to it's larger size when compared to $\ce{Na+}$. The same effect can be seen with the Reimer-Tiemann reaction, where the o/p ratio for $\ce{KOH}$ ($1.24$) is significantly lower than the ratio for $\ce{NaOH}$ ($2.08$). Nevertheless, the ortho product is still the major product with $\ce{KOH}$, and the linked answer is wrong.

Reference: O.L. Brady and J. Jakobovits, J. Chem. Soc., 1950, 767-777

  • $\begingroup$ +1, interesting indeed. Had a doubt, what is the plausible reason for para product being major at low concentrations, is it due to steric reasons? $\endgroup$
    – Ashish
    Commented Jul 22, 2021 at 8:42
  • 1
    $\begingroup$ @Ashish Yes, you can explain it using both steric reasons and the negative inductive effect of the oxygen atom. $\endgroup$ Commented Jul 22, 2021 at 8:49
  • $\begingroup$ Ortho product is also stabilized by intramolecular h bonding $\endgroup$ Commented Mar 25, 2022 at 18:57

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