HCDO xrightarrow[{(50%)}]{{OH^{-}}} Product o | vedclass

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Question

(A) The Cannizzaro reaction involves the disproportionation of an aldehyde lacking $\alpha$-hydrogen atoms in the presence of a concentrated base.In t

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$DCOO^- + CH_2DOH$

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(A) The Cannizzaro reaction involves the disproportionation of an aldehyde lacking $\alpha$-hydrogen atoms in the presence of a concentrated base.In the given reaction,$HCDO$ (deuterated formaldehyde) undergoes disproportionation.The hydride/deuteride ion transfer occurs from the carbonyl carbon to another molecule of the aldehyde.Specifically,the $H^-$ or $D^-$ is transferred. In $HCDO$,the transfer of $H^-$ leads to the formation of $DCOO^-$ and $CH_2DOH$.The reaction is: $2HCDO + OH^- \rightarrow DCOO^- + CH_2DOH$.

$HCDO \xrightarrow[{(50\%)}]{{OH^{-}}}$ Product of this Cannizzaro reaction is:

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