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(D) $1$. The first reaction is the cyclic polymerization of acetylene $(C_2H_2)$ in a red hot iron tube at $500^{\circ}C$,which yields benzene $(C_6H_

Vedclass · Accepted Answer

$A=C_6H_6, B=C_6H_5NO_2$

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(D) $1$. The first reaction is the cyclic polymerization of acetylene $(C_2H_2)$ in a red hot iron tube at $500^{\circ}C$,which yields benzene $(C_6H_6)$. Thus,$A = C_6H_6$.$2$. The second reaction is the nitration of benzene using a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ at $70^{\circ}C$,which yields nitrobenzene $(C_6H_5NO_2)$. Thus,$B = C_6H_5NO_2$.$3$. The third reaction involves the reduction of nitrobenzene $(B)$ with $LiAlH_4$ to form azobenzene $(C_6H_5-N=N-C_6H_5)$,which confirms the identity of $B$.$4$. Therefore,$A = C_6H_6$ and $B = C_6H_5NO_2$.

Consider the following reactions:
$C_2H_2 \xrightarrow[500^{\circ}C]{\text{red hot iron tube}} A$
$A \xrightarrow[70^{\circ}C]{\text{conc. } HNO_3, \text{conc. } H_2SO_4} B$
$B \xrightarrow{LiAlH_4} C_6H_5-N=N-C_6H_5$
$A$ and $B$ are:

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Consider the following reactions:$C_2H_2 \xrightarrow[500^{\circ}C]{\text{red hot iron tube}} A$$A \xrightarrow[70^{\circ}C]{\text{conc. } HNO_3, \text{conc. } H_2SO_4} B$$B \xrightarrow{LiAlH_4} C_6H_5-N=N-C_6H_5$$A$ and $B$ are: