The product C of the following reaction seque | vedclass

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

Question

(D) $1$. Aniline reacts with $Br_2/H_2O$ to form $2,4,6$-tribromoaniline as the major product $(A)$.$2$. $2,4,6$-tribromoaniline reacts with $NaNO_2/H

Vedclass · Accepted Answer

$1,3,5-$Tribromo$-2-$nitrobenzene

Vedclass · Answer

(D) $1$. Aniline reacts with $Br_2/H_2O$ to form $2,4,6$-tribromoaniline as the major product $(A)$.$2$. $2,4,6$-tribromoaniline reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt, $2,4,6$-tribromobenzenediazonium chloride $(B)$.$3$. The reaction of the diazonium salt with $HBF_4$ followed by $NaNO_2/Cu, \Delta$ is a variation of the Balz-Schiemann or Sandmeyer-type reaction where the diazonium group $(-N_2^+Cl^-)$ is replaced by a nitro group $(-NO_2)$.$4$. Thus, the final product $C$ is $1,3,5$-tribromo$-2-$nitrobenzene.

The product $C$ of the following reaction sequence is:

Similar Questions

Which of the following statements is true?
$1)$ Tertiary amines react with acid chlorides.
$2)$ $N$-ethylbenzene sulphonamide is soluble in alkali.
$3)$ $N,N$-diethylbenzene sulphonamide is soluble in alkali.
$4)$ Tertiary amines react with Hinsberg's reagent.

The product of the following reaction is:

The following is a confirmatory test for aromatic primary amines. Identify reagent $(A)$ and $(B)$.

Which of the following is the strongest base?

$A$ solution contains $1 \ g \ mol$ each of $p-$toluene diazonium chloride and $p-$nitrophenyl diazonium chloride. To this,$1 \ g \ mol$ of alkaline solution of phenol is added. Predict the major product and explain your answer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module