Consider the following reactions:C_2 H_2 xrig | vedclass

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(D) $1$. The reaction of $C_2 H_2$ (acetylene) with a red hot iron tube at $500^{\circ}C$ is a cyclic polymerization reaction that produces $C_6 H_6$

Vedclass · Accepted Answer

$A=C_6 H_6, B=C_6 H_5 NO_2$

Vedclass · Answer

(D) $1$. The reaction of $C_2 H_2$ (acetylene) with a red hot iron tube at $500^{\circ}C$ is a cyclic polymerization reaction that produces $C_6 H_6$ (benzene). Therefore,$A = C_6 H_6$.$2$. The reaction of benzene $(A)$ with a mixture of concentrated $HNO_3$ and concentrated $H_2 SO_4$ at $70^{\circ}C$ is a nitration reaction,which produces $C_6 H_5 NO_2$ (nitrobenzene). Therefore,$B = C_6 H_5 NO_2$.$3$. The reduction of nitrobenzene $(B)$ with $LiAlH_4$ can lead to the formation of azobenzene $(C_6 H_5-N=N-C_6 H_5)$ under specific conditions.$4$. Thus,$A = C_6 H_6$ and $B = C_6 H_5 NO_2$.

Consider the following reactions:
$C_2 H_2 \xrightarrow[500^{\circ}C]{\text{red hot iron tube}} A$
$A \xrightarrow[70^{\circ}C]{\text{conc. } HNO_3, \text{conc. } H_2 SO_4} B$
$B \xrightarrow{LiAlH_4} C_6 H_5-N=N-C_6 H_5$
$A$ and $B$ are:

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Consider the following reactions:$C_2 H_2 \xrightarrow[500^{\circ}C]{\text{red hot iron tube}} A$$A \xrightarrow[70^{\circ}C]{\text{conc. } HNO_3, \text{conc. } H_2 SO_4} B$$B \xrightarrow{LiAlH_4} C_6 H_5-N=N-C_6 H_5$$A$ and $B$ are: