The following question was on one of the tests we had recentlyenter image description here

Now i worked out the intermediates and the product as follows:

The first step is an intramolecular Cannizzaro reaction in which the aldehyde is oxidised to acid and the ketone reduced to base

In the next step decarboxylation takes place and we are left with benzyl alcohol (B)

Oxidation of (B) using Pot. Permanganate gives benzoic acid and decarboxylation of benzoic acid using soda lime gives us benzene as the final product which is the correct answer to the question.

Now i am quite satisfied with the validity of all the other steps excluding the acidic decarboxylation of the alpha-hydroxy acid

Can someone explain to me the probable mechanism for that reaction?

Or correct any other mistake i may have made in any of the other steps/conversions

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    $\begingroup$ You can turn PhCH(OH)CO2H to benzaldehyde (PhCHO) by direct heating (dx.doi.org/10.1039/CT9048500827), but I'm not sure if you can do an acid-catalysed decarboxylation... $\endgroup$ – orthocresol Dec 26 '18 at 16:01
  • $\begingroup$ @orthocresol Then by what way would the following set of steps give benzene as the final product? Or is that not the final product at all? (The answer key for our tests has been wrong before!) $\endgroup$ – Rutwik Dec 26 '18 at 17:06
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    $\begingroup$ I don't know, to be honest. I am always rather skeptical of some of these questions, because they are often not grounded in fact, but rather in some made-up synthetic sequence which nobody has actually carried out before. Not all of them are fake, of course. But I cannot tell which are and which aren't without actually looking up the reaction in a database. Suffice it to say that I couldn't find any transformations of PhCH(OH)CO2H to PhCH2OH (or PhCHO, which you can oxidise to PhCO2H as well) with just acid as the sole reagent. $\endgroup$ – orthocresol Dec 26 '18 at 17:08
  • $\begingroup$ @orthocresol ok fair enough! However if any alternative probable pathway comes to mind please do let me know $\endgroup$ – Rutwik Dec 26 '18 at 17:12
  • $\begingroup$ I prefer to call the Cannizzaro rearrangement as a benzylic acid rearrangement to give mandelic acid (A). Attack of hydroxide at the aldehyde with subsequent hydride transfer. I agree with @orthocresol about made up conditions. I doubt that H3O+ will work to give benzaldehyde (B). But strong H+ sources (H2SO4, polyphosphoric acid, etc.) can initiate loss of water and CO (decarbonylation) to get B. Oxidation of B to C gives benzoic acid, which, when heated with CaO as Eilhard Mitscherlich did by dry distillation (1834). No NaOH needed. $\endgroup$ – user55119 Dec 26 '18 at 18:56

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