The major product formed in the following rea | vedclass

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

Question

(C) The reaction proceeds as follows:$(i)$ $Sn/HCl$ reduces the $-NO_2$ group to an $-NH_2$ group,forming $p$-toluidine.(ii) $Br_2$ $(1 \ eq.)$ in the

Vedclass · Accepted Answer

$1-$methyl$-3,4-$dibromobenzene

Vedclass · Answer

(C) The reaction proceeds as follows:$(i)$ $Sn/HCl$ reduces the $-NO_2$ group to an $-NH_2$ group,forming $p$-toluidine.(ii) $Br_2$ $(1 \ eq.)$ in the presence of the strongly activating $-NH_2$ group leads to ortho-bromination,yielding $2$-bromo-$4$-methylaniline.(iii) $NaNO_2/HCl$ at $273-278 \ K$ converts the $-NH_2$ group into a diazonium salt,$-N_2^+Cl^-$.(iv) $Cu_2Br_2/HBr$ (Sandmeyer reaction) replaces the diazonium group with a $-Br$ atom,resulting in $1,2$-dibromo-$4$-methylbenzene (also known as $3,4$-dibromotoluene).

The major product formed in the following reaction is:

Similar Questions

Identify $X$ in the following reaction sequence:

Starting from propanoic acid,the following reactions were carried out: $CH_3-CH_2-COOH$ $\xrightarrow{SOCl_2} X$ $\xrightarrow{NH_3} Y$ $\xrightarrow{Br_2 + KOH} Z$. What is the compound $Z$?

Which one of the following on reduction with lithium aluminium hydride yields a secondary amine?

Sulphonation is most favourable at the carbon number $...$

$p-$Methylbenzonitrile can be prepared from which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module