The final product (IV) obtained in the reacti | vedclass

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

Question

(D) The reaction sequence is as follows:$1$. Toluene is oxidized by $KMnO_4/OH^-$ followed by $H^+$ to form benzoic acid $(I = C_6H_5COOH)$.$2$. Benzo

Vedclass · Accepted Answer

$C_6H_5NH_2$

Vedclass · Answer

(D) The reaction sequence is as follows:$1$. Toluene is oxidized by $KMnO_4/OH^-$ followed by $H^+$ to form benzoic acid $(I = C_6H_5COOH)$.$2$. Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride $(II = C_6H_5COCl)$.$3$. Benzoyl chloride reacts with $NH_3$ to form benzamide $(III = C_6H_5CONH_2)$.$4$. Benzamide undergoes the Hofmann bromamide degradation reaction with $OBr^-$ (formed from $Br_2/NaOH$) to yield aniline $(IV = C_6H_5NH_2)$.Thus,the final product $(IV)$ is aniline $(C_6H_5NH_2)$.

The final product $(IV)$ obtained in the reaction sequence
Toluene $\xrightarrow[{2. \ H^{+}}]{{1. \ KMnO_4/OH^{-}}} I$ $\xrightarrow{{SOCl_2}} II$ $\xrightarrow{{NH_3}} III$ $\xrightarrow{{OBr^{-}}} IV$ is

Similar Questions

Assertion : Aniline does not undergo Friedel-Crafts reaction.
Reason : $-NH_2$ group of aniline reacts with $AlCl_3$ (Lewis acid) to give acid-base reaction.

When $(P)$ undergoes Hoffmann exhaustive methylation (twice) then the product obtained will be

Give the structure of $A$ in the following reaction.
$2\text{-Nitro-4-methylaniline} \xrightarrow[(ii) \ H_3PO_2, H_2O]{(i) NaNO_2 + HCl, 273 - 278 \ K} A$.

Reduction of nitrobenzene with $Zn$ and alcoholic $KOH$ solution results in the formation of the following compound:

Identify the reagent $R$ used in the reaction stated below.
$C_6H_5N_2Cl + R \rightarrow C_6H_6$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The final product $(IV)$ obtained in the reaction sequenceToluene $\xrightarrow[{2. \ H^{+}}]{{1. \ KMnO_4/OH^{-}}} I$ $\xrightarrow{{SOCl_2}} II$ $\xrightarrow{{NH_3}} III$ $\xrightarrow{{OBr^{-}}} IV$ is