Consider the reaction shown below. Compound B | vedclass

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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 $p$

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

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:

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