CaC_2 + H_2O to A xrightarrow{H_2SO_4/HgSO_4} | vedclass

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(A) The reaction of calcium carbide with water produces acetylene $(C_2H_2)$:$CaC_2 + 2H_2O 	o C_2H_2 + Ca(OH)_2$Acetylene undergoes hydration in the

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$C_2H_2$ and $CH_3CHO$

Vedclass · Answer

(A) The reaction of calcium carbide with water produces acetylene $(C_2H_2)$:$CaC_2 + 2H_2O 	o C_2H_2 + Ca(OH)_2$Acetylene undergoes hydration in the presence of $dil. H_2SO_4$ and $HgSO_4$ (Kucherov reaction) to form an unstable enol intermediate,which tautomerizes to acetaldehyde $(CH_3CHO)$:$C_2H_2 + H_2O \xrightarrow{H_2SO_4/HgSO_4} [CH_2=CHOH] 	o CH_3CHO$Therefore,$A = C_2H_2$ and $B = CH_3CHO$.

$CaC_2 + H_2O \to A \xrightarrow{H_2SO_4/HgSO_4} B$. Identify $A$ and $B$ in the given reaction.

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Product $(C)$ is: $Ph-CH=CH_2$ $\xrightarrow[CCl_4]{Br_2} (A)$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} (B)$ $\xrightarrow[(ii) CH_3-Cl]{(i) NaNH_2} (C)$

$CH_3-C \equiv C-O-CH_3 \xrightarrow{dil. \ H_2SO_4, HgSO_4} \text{major product}$

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Acetylene reacts with a $42\% H_2SO_4$ solution containing $1\% HgSO_4$ to give

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