Consider the following sequence of reactions: | vedclass

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(A) The reaction sequence is as follows:$1$. Nitrobenzene is reduced by $Sn + HCl$ to form aniline $(C_6H_5NH_2)$.$2$. Aniline reacts with $NaNO_2 + H

Vedclass · Accepted Answer

$154$

Vedclass · Answer

(A) The reaction sequence is as follows:$1$. Nitrobenzene is reduced by $Sn + HCl$ to form aniline $(C_6H_5NH_2)$.$2$. Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.$3$. Benzenediazonium chloride reacts with $Cu_2Cl_2$ (Sandmeyer reaction) to form chlorobenzene $(C_6H_5Cl)$.$4$. Chlorobenzene reacts with $Na$ in dry ether (Wurtz-Fittig reaction) to form biphenyl $(C_6H_5-C_6H_5)$.The final product $(A)$ is biphenyl $(C_{12}H_{10})$.Molar mass of biphenyl = $(12 	imes 12) + (10 	imes 1) = 144 + 10 = 154 \ g \ mol^{-1}$.

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