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$\text{C}_6\text{H}_5\text{CHO} + \text{HCHO} \xrightarrow{\text{50% NaOH, then H}_3\text{O}^+}$ ?
(1). (PhCH$_2$OH + PhCOOH)
(2). (CH$_3$OH + benzoate)
(3). (HCOOH + PhCH$_2$OH)
(4). (CH$_3$OH + HCO$_2$H)
Crossed Cannizzaro: HCHO is oxidized to HCOO$^-$ (formate) and PhCHO is reduced to PhCH$_2$OH (benzyl alcohol).
BETA
