Number of Heteroatoms in Substituted Benzene Derivative R

To find the number of heteroatoms present in one molecule of R, we follow the reaction steps described:

1. Formation of P: Starting compound: 1,3,5-tris-(4-nitrophenyl)benzene ($\text{C}_{24}\text{H}_{15}\text{N}_3\text{O}_6$).

   • Reaction with excess $\text{Sn/HCl}$ reduces all three nitro groups ($-\text{NO}_2$) to amino groups ($-\text{NH}_2$), giving 1,3,5-tris-(4-aminophenyl)benzene.
   • Diazotization with excess $\text{NaNO}_2/\text{HCl}$ at $0\text{ }^\circ\text{C}$ converts the three amino groups into diazonium chloride groups ($-\text{N}_2^+\text{Cl}^-$), forming compound P: 1,3,5-tris(4-diazoniophenyl)benzene trichloride ($\text{C}_{24}\text{H}_{15}\text{N}_6\text{Cl}_3$).

2. Formation of Q:

   • Hydrolysis of P with excess $\text{H}_2\text{O}$ replaces each $-\text{N}_2^+\text{Cl}^-$ group with a phenolic hydroxyl group ($-\text{OH}$).
   • Therefore, compound Q is 1,3,5-tris(4-hydroxyphenyl)benzene with molecular formula $\text{C}_{24}\text{H}_{18}\text{O}_3$.

3. Formation of R:

   • Bromination of Q in an aqueous medium results in electrophilic aromatic substitution on the highly activated phenolic rings.
   • Each 4-hydroxyphenyl group has two vacant ortho positions relative to the $-\text{OH}$ group available for bromination.
   • The mass increase during substitution of $x$ hydrogen atoms by $x$ bromine atoms is given by: $\Delta M = x \times (M_{\text{Br}} - M_{\text{H}}) = x \times (80 - 1) = 79x \text{ g mol}^{-1}$
   • Given the molar mass difference between Q and R is $474\text{ g mol}^{-1}$: $79x = 474 \implies x = \frac{474}{79} = 6$
   • Thus, 6 bromine atoms are introduced (2 on each of the three phenolic rings), giving compound R with molecular formula $\text{C}_{24}\text{H}_{12}\text{O}_3\text{Br}_6$.

4. Counting Heteroatoms in R:

   • Heteroatoms are defined as atoms other than carbon ($\text{C}$) and hydrogen ($\text{H}$).
   • In one molecule of R ($\text{C}_{24}\text{H}_{12}\text{O}_3\text{Br}_6$), the heteroatoms are:
     • Oxygen ($\text{O}$) atoms = $3$
     • Bromine ($\text{Br}$) atoms = $6$

$\text{Total number of heteroatoms} = 3 + 6 = 9$