Can pthalic acid or 1,2-Benzenedimethanol be possible products of Cannizzaro reaction of phthalaldehyde?

Question

I know the main product that is ortho-hydroxymethyl benzoate ion but I think that phthalic acid and catechol must be formed also in very small amounts but I am unable to get any confirmation from the internet. Please help me with the products.

Answer 1

The Cannizzaro reaction is the disproportionation of an aldehyde, which do not have an $\alpha$-hydrogen, to an equimolar mixture of corresponding primary alcohol and carboxylic acid salt. For example, original Cannizzaro's experiment:

$$\ce{2C6H5-CHO ->[aq NaOH] C6H5-CH2OH + C6H5-COO^-Na+}$$

Since there is no $\alpha$-hydrogen, the mixture does not undergo an aldol condensation, which is the characteristic of this reaction. The Cannizzaro reaction is an example of a typical redox reaction, which is considered to be one of the most important synthetic reactions of organic chemistry prior to the discovery of $\ce{LiAlH4}$ in 1946 (Ref.1).

The proposed reaction mechanism is as follows (Ref.1):

According to Ref.1, the hydride transfer must be from one molecule of benzaldehyde to the other molecule (when the reaction is performed in $\ce{D2O}$, no deuterated $\ce{CD2-O}$ is found in benzyl alcohol). The formation of benzyl benzoate was suggested and supported (e.g., Ref.2), evidence in Ref.1 doubts this proposal.

The Cannizzaro reaction of phthalaldehyde is studied in aqueous base ($\ce{NaOH}$) to obtain sodium o-hydroxymethylbenzoate as the product. The studies include the kinetics of the reaction, which have been measured over a range of $\ce{NaOH}$ concentration at $\pu{40 ^\circ C}$. This kinetic study proposed the hydration of phthalaldehyde to hemi-ketal monooxoanion is one of the first steps in the mechanism (hemi-ketal formation is depicted in the insert box of above image). The dioxoanion formation is depend on the concentration of $\ce{NaOH}$ solution (Ref.3). The hydride transfer from mono- or di-oxoanion follows, which is the rate determining step (determined by the isotope effect). The observed rate is based on following equation:

$$k_\mathrm{obs} = \frac{a[\ce{OH-}]}{1 + b[\ce{OH-}]}$$

where $a= \pu{(1.38 \pm 0.02)\times 10^{-3} M-1s-1}$ and $b= \pu{0.99 \pm 0.02 M-1}$ (the reaction follows this equation for $\pu{0.05-2.0 M}$ $\ce{NaOH}$). There are no other product has been observed.

I understand your concern of intra- versus inter-molecular hydride transfer. If intermolecular transfer occurs, at least 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid should form as minor components. Since hydride transfer is the rate determining, intramolecular hydride transfer is much faster than that of intermolecular version. Therefore no chance of minor product formation is possible since all other steps are in equilibrium.

Note: I assume your suggestion of forming phthalic acid and catechol as byproduct is mistaken for 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid (phthalic acid). Keep in mind that there is noway of breaking two aromatic-CHO bonds and making two aromatic-O bonds to form catechol molecule.

References:

1. C. Gardner Swain, Arnet L. Powell, William A. Sheppard, Charles R. Morgan, "Mechanism of the Cannizzaro reaction," J. Am. Chem. Soc. 1979, 101(13), 3576–3583 (https://doi.org/10.1021/ja00507a023).
2. Arthur Lachman, "The Benzyl Rearrangement v. Cannizzaro's reaction," J. Am. Chem. Soc. 1923, 45(10), 2356–2363 (https://doi.org/10.1021/ja01663a017).
3. Robert S. McDonald, Christine E. Sibley, "The intramolecular Cannizzaro reaction of phthalaldehyde," Canadian Journal of Chemistry 1981, 59(7), 1061-1067 (https://doi.org/10.1139/v81-157).