Consider the reaction shown below. Compound B | vedclass

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

Question

(A) Step $1$: The reaction of $p$-aminobenzenesulfonic acid (sulfanilic acid) with $NaNO_2$ and $HCl$ at $273-278 \ K$ results in the formation of $p$

Vedclass · Accepted Answer

$HO_3S-C_6H_4-N=N-C_6H_4-N(CH_3)_2$

Vedclass · Answer

(A) Step $1$: The reaction of $p$-aminobenzenesulfonic acid (sulfanilic acid) with $NaNO_2$ and $HCl$ at $273-278 \ K$ results in the formation of $p$-benzenediazonium sulfonate (Compound $A$).
Step $2$: Compound $A$ undergoes an electrophilic aromatic substitution (coupling reaction) with $N,N$-dimethylaniline at $273 \ K$.
Step $3$: The diazonium group $-N_2^+$ acts as an electrophile and attacks the para-position of $N,N$-dimethylaniline, which is activated by the electron-donating $-N(CH_3)_2$ group.
Step $4$: The final product $B$ is $4-(dimethylamino)azobenzene-4'-sulfonic acid$, represented as $HO_3S-C_6H_4-N=N-C_6H_4-N(CH_3)_2$.

Consider the reaction shown below. Compound $B$ is:

Similar Questions

$CH_3CH_2I$ $\xrightarrow{NaCN} A$ $\xrightarrow{OH^- \text{ (partial hydrolysis)}} B$ $\xrightarrow{Br_2/NaOH} C$. Identify the final product $C$.

Consider the following sequence of reactions: $C_2H_5-Br$ $\xrightarrow{AgCN} A$ $\xrightarrow{H_2/Ni} B$. Product $B$ is:

Nitrobenzene gives $N$-phenylhydroxylamine by

Given below are two statements:
Statement $I$: Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement $II$: Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements,choose the correct answer from the options given below:

Which of the following amines reacts most rapidly with $CH_3I$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module