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 = $(12 	imes 12) + (10 	imes 1) = 144 + 10 = 154 \ g \ mol^{-1}$.

List-$I$ (Reaction)	List-$II$ (Major product)
$A. \ CH_3CH=CH_2 \xrightarrow{HI}$	$I. \ 1\text{-bromo-}2\text{-phenylethane}$
$B. \ C_6H_5CH_2CH_3 \xrightarrow{Br_2/\Delta}$	$II. \ 1\text{-iodopropane}$
$C. \ C_6H_5CH=CH_2 \xrightarrow{HBr}$	$III. \ 2\text{-iodopropane}$
	$IV. \ 1\text{-bromo-}1\text{-phenylethane}$

Similar Questions

When $32.25 \ g$ of ethyl chloride undergoes dehydrohalogenation,it yields $50\%$ of the theoretical alkene product. What is the mass of the product obtained in $g$? (Atomic mass: $Cl = 35.5, C = 12, H = 1$)

Assertion : Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene.
Reason : Alkyl halides are less reactive than acyl halides.

Identify the major product formed from the following reaction sequence:

Which of the following products can be obtained from the above reaction?

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