The sequence from the following that would re | vedclass

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

Question

(C) To obtain $3,4,5-$Tribromoaniline,we start with $p-$nitroaniline. $1.$ The amino group $(-NH_2)$ is a strong activating group,so treatment with ex

Vedclass · Accepted Answer

$p-$Nitroaniline $\xrightarrow[(ii) NaNO_2, HCl, CuBr]{(i) Br_2(excess), acetic acid}$ $\xrightarrow[(iii) Sn, HCl]{}$

Vedclass · Answer

(C) To obtain $3,4,5-$Tribromoaniline,we start with $p-$nitroaniline. $1.$ The amino group $(-NH_2)$ is a strong activating group,so treatment with excess $Br_2$ in acetic acid leads to bromination at the $2$ and $6$ positions relative to the $-NH_2$ group (which are the $3$ and $5$ positions relative to the $-NO_2$ group),yielding $2,6-$dibromo$-4-$nitroaniline. $2.$ The amino group is then converted to a diazonium salt using $NaNO_2/HCl$. $3.$ The diazonium group is replaced by a bromine atom using $CuBr$ (Sandmeyer reaction),resulting in $3,4,5-$tribromonitrobenzene. $4.$ Finally,the nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Sn/HCl$ to yield $3,4,5-$tribromoaniline.

The sequence from the following that would result in giving predominantly $3,4,5-$Tribromoaniline is $:$

Similar Questions

What is obtained by the reduction of acetaldoxime?

Which reagent reacts with acetamide to produce methylamine?

Which of the following do not give aliphatic primary amine as a product?

When propanamide reacts with $Br_2$ and $NaOH$,which of the following compounds is formed?

What are $X$ and $Y$ in the following reactions?
$X \xrightarrow[(ii) H_2O]{(i) LiAlH_4} R CH_2 NH_2 \xleftarrow{H_2 / Ni} Y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module