The product formed in the following reaction | vedclass

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(B) The reaction involves the oxidative cleavage of a vicinal diol (or $\alpha$-hydroxy carbonyl compound) by periodic acid $(HIO_4)$.$1$. The startin

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$HCHO, 2HCO_2H$

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(B) The reaction involves the oxidative cleavage of a vicinal diol (or $\alpha$-hydroxy carbonyl compound) by periodic acid $(HIO_4)$.$1$. The starting material is glyceraldehyde $(CHO-CH(OH)-CH_2OH)$.$2$. The first $HIO_4$ molecule cleaves the bond between the $C_1$ (aldehyde) and $C_2$ (hydroxyl) carbons. This produces $HCOOH$ (formic acid) and $HOCH_2CHO$ (glycolaldehyde).$3$. The second $HIO_4$ molecule cleaves the glycolaldehyde $(HOCH_2CHO)$ into $HCHO$ (formaldehyde) and $HCOOH$ (formic acid).$4$. Summing the products: $HCOOH + HCOOH + HCHO = 2HCOOH + HCHO$.Therefore,the final products are $HCHO$ and $2HCO_2H$.

Reactant	$K_{eq.}$
$PhCHO$	$a$
$CH_3CH_2COCH_3$	$b$
$PhCOCH_3$	$c$
$CH_3CHO$	$d$

The product formed in the following reaction is:
$CHO-CH(OH)-CH_2OH \xrightarrow{2HIO_4} ?$

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The correct order of decreasing value of $K_{eq.}$ for the addition of $HCN$ to the following carbonyl compounds is:
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The product formed in the following reaction is: $CHO-CH(OH)-CH_2OH \xrightarrow{2HIO_4} ?$