The number of heteroatoms present in one molecule of R 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]

The starting compound is 1,3,5-tris(4-nitrophenyl)benzene.

Step-1 Sn/HCl reduces each $$\!NO_2$$ group to $$\!NH_2$$, giving 1,3,5-tris(4-aminophenyl)benzene.

Step-2 NaNO$$\_2$$/HCl at 0 °C converts each $$\!NH_2$$ into a diazonium group (-$$N_2^+Cl^-$$).  The product obtained is denoted as $$P$$.

Step-3 Pure water at room temperature hydrolyses every diazonium group to an $$OH$$ group (Sandmeyer hydrolysis), giving $$Q =$$ 1,3,5-tris(4-hydroxyphenyl)benzene.

Step-4 Bromination of $$Q$$ in aqueous medium produces $$R$$.
Phenol rings undergo electrophilic substitution chiefly at the two ortho positions (positions 2 and 6 with respect to the $$OH$$) because the para position (4) is already occupied by the C-C bond that connects the side ring to the central benzene.

Mass-difference data supplied in the problem: $$\text{molar-mass}(R)\;-\;\text{molar-mass}(Q)=474\;\text{g mol}^{-1}$$

Each bromination step replaces an H atom (1 g mol$$^{-1}$$) by a Br atom (80 g mol$$^{-1}$$).
Mass increase per Br introduced = $$80-1=79\;\text{g mol}^{-1}$$.

Number of Br atoms introduced = $$\frac{474}{79}=6$$.

There are three phenolic rings in $$Q$$, so every ring receives $$\frac{6}{3}=2$$ bromine atoms, exactly the two ortho positions expected. Hence $$R$$ is 1,3,5-tris(2,6-dibromo-4-hydroxyphenyl)benzene.

Counting hetero-atoms in one molecule of $$R$$ (all atoms other than C and H):
• Oxygen: 3 (one $$O$$ in each $$OH$$ group)
• Bromine: 6 (two per phenolic ring × three rings)
Total hetero-atoms = $$3+6=9$$.

Therefore, the number of hetero-atoms present in one molecule of $$R$$ is 9.