JEE Advanced 2023 Paper 2 · Q17 · Amines & Diazonium

Question

The total number of carbon atoms and heteroatoms present in one molecule of $S$ is ___. [Use: molar mass (in g mol$^{-1}$): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5. Atoms other than C and H are considered as heteroatoms.]

SolveKar AI · Accepted Answer

Step 1: 1,3,5-tris-(4-nitrophenyl)benzene with Sn/HCl reduces all three $-$NO$_2$ groups to $-$NH$_2$, giving 1,3,5-tris-(4-aminophenyl)benzene. Step 2: NaNO$_2$/HCl at $0\,^\circ$C converts each $-$NH$_2$ to a diazonium $-$N$_2^+$Cl$^-$, so $P = $ 1,3,5-tris-(4-diazoniumphenyl)benzene tris-chloride. Step 3: For $S$, $P$ reacts with excess phenol under basic conditions. Each $-$N$_2^+$ undergoes azo coupling with phenol at its para position, attaching $-$N=N$-$C$_6$H$_4$$-$OH to each of the three aryl rings of $P$. So $S$ is 1,3,5-tris-\{4-[($4'$-hydroxyphenyl)diazenyl]phenyl\}benzene. Carbon count: central benzene ring contributes $6$ C; three inner $C_6H_4$ rings contribute $3 	imes 6 = 18$ C; three terminal $C_6H_4OH$ (phenol) rings contribute $3 	imes 6 = 18$ C. Total C $= 6 + 18 + 18 = 42$. Heteroatom count: each azo linkage $-$N=N$-$ has $2$ N atoms, giving $3 	imes 2 = 6$ N; each terminal phenol $-$OH contributes $1$ O, giving $3$ O. Total heteroatoms $= 6 + 3 = 9$. Verification via mass difference: $S - P$ replaces three Cl$^-$ counter-ions with three $-$C$_6$H$_4$OH groups via loss of HCl per coupling: per ring, $-$N$_2^+$Cl$^-$ + HOC$_6$H$_5$ $	o$ $-$N=N$-…[truncated]

JEE Advanced 2023 Paper 2 · Q17 · Amines & Diazonium

Solution