In a set of reactions,p-nitrotoluene yielded | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. The starting material is $p$-nitrotoluene. $2$. Electrophilic aromatic substitution with $Br_2/FeBr_3$ introduces a bromine atom at the ortho

Vedclass · Accepted Answer

$2,4-$dibromotoluene

Vedclass · Answer

(C) $1$. The starting material is $p$-nitrotoluene. $2$. Electrophilic aromatic substitution with $Br_2/FeBr_3$ introduces a bromine atom at the ortho position relative to the methyl group (since the $-CH_3$ group is ortho/para directing and the $-NO_2$ group is meta directing). This gives $B$ as $2$-bromo-$4$-nitrotoluene. $3$. Reduction of the nitro group with $Sn/HCl$ converts the $-NO_2$ group to an $-NH_2$ group,yielding $C$ ($2$-bromo-$4$-aminotoluene). $4$. Diazotization with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ converts the $-NH_2$ group into a diazonium salt,$D$ ($2$-bromo-$4$-methylbenzenediazonium chloride). $5$. The Sandmeyer reaction with $CuBr/HBr$ replaces the diazonium group with a bromine atom,resulting in $E$ ($2,4$-dibromotoluene).

In a set of reactions,$p$-nitrotoluene yielded a product $E$. The product $E$ would be:
$p$-nitrotoluene $\xrightarrow[FeBr_3]{Br_2} B$ $\xrightarrow{Sn/HCl} C$ $\xrightarrow[HCl]{NaNO_2} D$ $\xrightarrow[HBr]{CuBr} E$

Similar Questions

The decreasing order of reactivity of the following compounds towards $Na^{\oplus} O^{\ominus} Me$ is:

Which one of the following reactions is most suitable for the preparation of $n$-propyl benzene?

The strongest ortho-para and strongest meta-directing groups respectively are

Identify the product of the following Wurtz-Fittig reaction.

Which of the following reactions cannot be used to prepare an aromatic nitrile $(ArCN)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In a set of reactions,$p$-nitrotoluene yielded a product $E$. The product $E$ would be:$p$-nitrotoluene $\xrightarrow[FeBr_3]{Br_2} B$ $\xrightarrow{Sn/HCl} C$ $\xrightarrow[HCl]{NaNO_2} D$ $\xrightarrow[HBr]{CuBr} E$