I came across a question:

Predict the major product when conc. $\ce{NaOH}$ is added to a mixture of $\ce{PhCHO}$ and $\ce{HCHO}$.

According to me, as $\ce{HCHO}$ is less hindered and $\ce{>C=O}$ of $\ce{HCHO}$ is more reactive, so the major products must be $\ce{CH3ONa}$ and $\ce{HCOONa}$. The hydride transfer should be from one Formaldehyde molecule to the other.

However, the answer was given as $\ce{HCOONa}$ and $\ce{PhCH2ONa}$. Is there a mistake on my part?

  • 4
    $\begingroup$ Rate determining step of Cannizaro is the hydride transfer and not nucleophilic attack $\endgroup$ – Safdar Sep 24 '20 at 12:59
  • $\begingroup$ Why would the hydride prefer transferring to the PHCHO? $\endgroup$ – Soumyadwip Chanda Sep 24 '20 at 13:06
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    $\begingroup$ So what happens to the benzaldehyde> $\endgroup$ – Safdar Sep 24 '20 at 13:18
  • 2
    $\begingroup$ See this earlier Q&A. $\endgroup$ – ron Sep 24 '20 at 13:39
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    $\begingroup$ Effectiively there is no free formaldehyde to accept hydride. $\endgroup$ – user55119 Sep 24 '20 at 14:45