CH_3-CO-CH_3 xrightarrow{I_2/NaOH} underset{t | vedclass

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(A) Step $1$: $CH_3-CO-CH_3$ (Acetone) reacts with $I_2/NaOH$ (Iodoform reaction) to give $CHI_3$ (Iodoform,yellow compound $A$) and $CH_3COONa$ (Sodi

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Iodoform,Acetylene & Acetaldehyde

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(A) Step $1$: $CH_3-CO-CH_3$ (Acetone) reacts with $I_2/NaOH$ (Iodoform reaction) to give $CHI_3$ (Iodoform,yellow compound $A$) and $CH_3COONa$ (Sodium acetate $B$).Step $2$: $2CHI_3 + 6Ag \xrightarrow{\Delta} CH \equiv CH$ (Acetylene or Ethyne $C$) $+ 6AgI$.Step $3$: $CH \equiv CH$ undergoes hydration in the presence of $HgSO_4/H_2SO_4$ to give $CH_3CHO$ (Acetaldehyde $D$).Therefore,$A = 	ext{Iodoform}$,$C = 	ext{Acetylene}$,$D = 	ext{Acetaldehyde}$.

$CH_3-CO-CH_3$ $\xrightarrow{I_2/NaOH} \underset{\text{yellow compound}}{(A)} + (B)$ $\xrightarrow{Ag - \text{powder}, \Delta} (C)$ $\xrightarrow{Dil. H_2SO_4/HgSO_4} (D)$
$A, C$ and $D$ are :-

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$CH_3-CO-CH_3$ $\xrightarrow{I_2/NaOH} \underset{\text{yellow compound}}{(A)} + (B)$ $\xrightarrow{Ag - \text{powder}, \Delta} (C)$ $\xrightarrow{Dil. H_2SO_4/HgSO_4} (D)$ $A, C$ and $D$ are :-