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

  • A
    $C_2H_5HSO_4$
  • B
    $CH_2=CH_2$
  • C
    $CH_3CHO$
  • D
    $HCHO$

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$HC \equiv CH$ $\xrightarrow[{excess}]{{CH_3MgBr}} \mathop {(A)}\limits_{\text{salt}} + \mathop {(B)}\limits_{\text{hydrocarbon}}$
$(A)$ $\xrightarrow[{excess}]{{CH_3I}} (C)$ $\xrightarrow[{(ii) H^{+} }]{{(i) NaNH_2}} (D)$

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