Consider the following sequence of reactions: | vedclass

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(B) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ in the presence of $HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl

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$N,N$-Dimethyl$-4-$(phenylazo)aniline

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(B) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2$ in the presence of $HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$,which is product '$A$'.Step $2$: Benzenediazonium chloride undergoes a diazo coupling reaction with $N,N$-dimethylaniline in a weakly acidic medium. The diazonium group $( -N=N- )$ attacks the para-position of the $N,N$-dimethylaniline ring to form $N,N$-dimethyl$-4-$(phenylazo)aniline,which is product '$B$'.The reaction is: $C_6H_5N_2^+Cl^- + C_6H_5N(CH_3)_2 \rightarrow C_6H_5-N=N-C_6H_4-N(CH_3)_2 + HCl$.

Consider the following sequence of reactions:
$C_6H_5NH_2$ $\xrightarrow{NaNO_2, 0-5^{\circ}C} 'A'$ $\xrightarrow{N,N-\text{Dimethylaniline}} 'B'$
The product $B$ is

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Consider the following sequence of reactions:$C_6H_5NH_2$ $\xrightarrow{NaNO_2, 0-5^{\circ}C} 'A'$ $\xrightarrow{N,N-\text{Dimethylaniline}} 'B'$The product $B$ is