Looking for classes? Ksquare Career Institute, Bengaluru →
Aniline in a set of the following reactions yielded a coloured product Y.
$C_6H_5NH_2 \xrightarrow[(273-278\ K)]{NaNO_2/HCl}$ X $\xrightarrow{N,N\text{-dimethylaniline}}$ Y
The structure of Y would be
$C_6H_5NH_2 \xrightarrow[(273-278\ K)]{NaNO_2/HCl}$ X $\xrightarrow{N,N\text{-dimethylaniline}}$ Y
The structure of Y would be
A
$CH_3NH-C_6H_4-N=N-C_6H_4-NHCH_3$
B
$C_6H_5-N=N-C_6H_4-N(CH_3)_2$ (para)
C
$CH_3NH-C_6H_4-NH-C_6H_4-NHCH_3$
D
$H_3C-C_6H_4-N=N-C_6H_4-NH_2$
Detailed Solution
Step 1 (diazotisation): aniline reacts with $NaNO_2$/HCl at 273–278 K to give benzenediazonium chloride: $C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$. So X = $C_6H_5N_2^+Cl^-$.
Step 2 (coupling): the diazonium ion is a weak electrophile; it attacks the electron-rich ring of N,N-dimethylaniline at the para position to the $-N(CH_3)_2$ group (electrophilic substitution).
$C_6H_5N_2^+Cl^- + C_6H_5N(CH_3)_2 \rightarrow C_6H_5-N=N-C_6H_4-N(CH_3)_2 + HCl$
The product, p-(N,N-dimethylamino)azobenzene, is a yellow azo dye; the extended conjugation through the –N=N– group makes it coloured.
Hence Y is $C_6H_5-N=N-C_6H_4-N(CH_3)_2$.
Step 2 (coupling): the diazonium ion is a weak electrophile; it attacks the electron-rich ring of N,N-dimethylaniline at the para position to the $-N(CH_3)_2$ group (electrophilic substitution).
$C_6H_5N_2^+Cl^- + C_6H_5N(CH_3)_2 \rightarrow C_6H_5-N=N-C_6H_4-N(CH_3)_2 + HCl$
The product, p-(N,N-dimethylamino)azobenzene, is a yellow azo dye; the extended conjugation through the –N=N– group makes it coloured.
Hence Y is $C_6H_5-N=N-C_6H_4-N(CH_3)_2$.