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Question

(C) Formaldehyde $(HCHO)$ undergoes the Cannizzaro reaction when treated with concentrated $NaOD$: $2HCHO + NaOD \rightarrow HCOONa + CH_3OD$ Subseque

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$H-C(=O)-OCH_3$

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(C) Formaldehyde $(HCHO)$ undergoes the Cannizzaro reaction when treated with concentrated $NaOD$: $2HCHO + NaOD \rightarrow HCOONa + CH_3OD$ Subsequent treatment with concentrated $H_2SO_4$ causes acidification of the salt to form formic acid $(HCOOH)$ and then promotes esterification between $HCOOH$ and $CH_3OD$: $HCOONa + H_2SO_4 \rightarrow HCOOH + NaHSO_4$ $HCOOH + CH_3OD \xrightarrow{H_2SO_4} HCOOCH_3 + HDO$ The final product is methyl formate $(H-C(=O)-OCH_3)$.

What will be the product obtained when formaldehyde is treated with conc. $NaOD$ followed by treatment with conc. $H_2SO_4$?

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