1. Starting compound and reduction

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

It has three para-nitrophenyl groups attached to a central benzene ring.

On complete reduction with excess $\mathrm{Sn/HCl}$, each $-\mathrm{NO_2}$ group is converted into $-\mathrm{NH_2}$.

So the major product is:

2. Formation of $\mathbf{P}$

Treatment with excess $\mathrm{NaNO_2/HCl}$ at $0^\circ \mathrm{C}$ diazotizes each amino group:

Hence $\mathbf{P}$ is the tris-diazonium chloride derivative.

So $\mathbf{P}$ contains 3 diazonium groups.

3. Formation of $\mathbf{Q}$ from $\mathbf{P}$

Diazonium salt on treatment with excess water at room temperature gives phenol:

Thus $\mathbf{Q}$ is:

So $\mathbf{Q}$ has 3 phenolic rings, each para-substituted.

4. Formation of $\mathbf{R}$ by bromination of $\mathbf{Q}$

A phenolic ring strongly activates bromination at the ortho positions to $-\mathrm{OH}$.

In each $p$-substituted phenol ring, the para position is already occupied by linkage to the central benzene ring, so bromination occurs at the two ortho positions.

Thus each outer ring gets 2 bromine atoms.

There are 3 such rings, so total bromines introduced:

Mass increase from $\mathbf{Q}$ to $\mathbf{R}$ per bromination is replacement of H by Br:

For 6 brominations:

This matches the given data, confirming the structure.

5. Formation of $\mathbf{S}$ from $\mathbf{P}$ with phenol/basic medium

Diazonium salts couple with phenol in basic medium to give azo compounds.

Each diazonium group couples with one phenol molecule.

Since $\mathbf{P}$ has 3 diazonium groups, $\mathbf{S}$ is formed by coupling with 3 phenol molecules.

The mass difference between $\mathbf{P}$ and $\mathbf{S}$ is given as $172.5\,\mathrm{g\,mol^{-1}}$.

Let us verify per coupling.

For one coupling:

• one phenol unit added: $\mathrm{C_6H_6O}$, mass $= 6(12)+6(1)+16=94$
• one H is lost from phenol ring during coupling: $-1$
• diazonium group loses $\mathrm{N_2}$? No, in azo coupling the diazonium group becomes $-\mathrm{N=N}-$, so the two nitrogens remain in product.
• chloride ion is not part of covalent structure of product.

So net increase per diazonium group from diazonium salt to azo-coupled product is:

For 3 groups:

So indeed all 3 diazonium groups couple with phenol.

Hence $\mathbf{S}$ is the tris-azo coupled product with 3 phenol units.

6. Count carbon atoms in $\mathbf{S}$

Initial skeleton in starting trisubstituted benzene after diazotization still has:

• central benzene ring: $6$ carbons
• three outer phenyl rings: $3 \times 6 = 18$ carbons

So in $\mathbf{P}$ (and retained in $\mathbf{S}$):

Now 3 phenol molecules couple, each contributes one phenyl ring:

Total carbons in $\mathbf{S}$:

7. Count heteroatoms in $\mathbf{S}$

Heteroatoms are atoms other than C and H.

In $\mathbf{S}$:

• each azo linkage contributes $2$ nitrogen atoms
• there are $3$ azo linkages

So nitrogen atoms:

Each coupled phenol unit retains one oxygen atom:

Total heteroatoms:

8. Total number of carbon atoms and heteroatoms in $\mathbf{S}$

Therefore, the required integer is:

9. Comparison with stored correct answer

Stored correct answer = $51$.

Our derived answer also is $51$, so they agree.