A trinitro compound, 1,3,5-tris-(4-nitrophenyl)benzene, on complete reaction with an excess of Sn / HCl gives a major product, which on treatment with an excess of NaNO_2 / HCl at 0^\circ C provides P as the product. P , upon treatment with excess of H_2 O at room temperature, gives the product Q . Bromination of Q in aqueous medium furnishes the product R . The compound P upon treatment with an excess of phenol under basic conditions gives the product S . The molar mass difference between compounds Q and R is 474 ~g ~mol^-1 and between compounds P and S is 172.5 ~g ~mol^-1 . The number of heteroatoms present in one molecule of R is \underline\hspace2cm . [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]

Compounds Containing Nitrogen PYQ — JEE Advanced 2023 · Shift 2 · Q16 with Solution

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Correct answer: 9

The user wants me to solve a multi-step organic chemistry synthesis problem and determine the number of heteroatoms in the final product R.

Step-by-step Solution:

Identify the Starting Material and its Structure The starting compound is 1,3,5-tris-(4-nitrophenyl)benzene. This consists of a central benzene ring with three (4-nitrophenyl) groups attached at positions 1, 3, and 5.
- Structure of Starting Material (A): A central C6H3 ring is connected to three C6H4NO2 groups.
- Molecular Formula of A: C6H3(C6H4NO2)3 = C24H15N3O6.
Reaction 1: Reduction with Sn/HCl The starting compound A reacts completely with excess Sn/HCl. This is a standard reagent for the reduction of a nitro group (–NO2) to a primary amine group (–NH2).

$\text{Ar-NO}_2 \xrightarrow{\text{Sn/HCl}} \text{Ar-NH}_2$

Since there are three nitro groups, all of them are reduced.
- Product (B): 1,3,5-tris-(4-aminophenyl)benzene.
- Molecular Formula of B: C6H3(C6H4NH2)3 = C24H21N3.
Reaction 2: Diazotization The major product B is treated with an excess of NaNO2/HCl at 0°C. This is the diazotization reaction, which converts primary aromatic amines to diazonium salts (–N2+Cl–).

$\text{Ar-NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0^\circ\text{C}} \text{Ar-N}_2^+\text{Cl}^-$

All three amino groups are converted.
- Product (P): 1,3,5-tris-(4-diazoniumchloridophenyl)benzene.
- Molecular Formula of P: [C6H3(C6H4N2)3]3+(Cl–)3 or C24H15N6Cl3.
Reaction 3: Hydrolysis of Diazonium Salt Product P is treated with excess H2O at room temperature. Aromatic diazonium salts hydrolyze to form phenols, releasing nitrogen gas.

$\text{Ar-N}_2^+\text{Cl}^- + \text{H}_2\text{O} \rightarrow \text{Ar-OH} + \text{N}_2 + \text{HCl}$

All three diazonium groups are converted to hydroxyl groups.
- Product (Q): 1,3,5-tris-(4-hydroxyphenyl)benzene.
- Molecular Formula of Q: C6H3(C6H4OH)3 = C24H18O3.
Reaction 4: Bromination of Phenol Derivative Product Q is brominated in an aqueous medium. The hydroxyl (–OH) group is a strongly activating, ortho-para directing group. In an aqueous medium (like bromine water), bromination is exhaustive, substituting all available ortho and para positions.

In each of the three (4-hydroxyphenyl) rings, the para position (relative to –OH) is occupied by the bond to the central benzene ring. Therefore, substitution occurs at the two available ortho positions (positions 3' and 5' of the outer rings).
- Substitution: 2 Br atoms are added to each of the 3 outer rings, replacing 2 H atoms each.
- Total substitution: 6 H atoms are replaced by 6 Br atoms.
- Product (R): 1,3,5-tris-(3,5-dibromo-4-hydroxyphenyl)benzene.
- Molecular Formula of R: Let's derive it from Q (C24H18O3). We replace 6 H with 6 Br. The formula becomes C24H12Br6O3.
Verification of Molar Mass Difference (Q and R) The problem states the molar mass difference between Q and R is 474 g mol-1. Let's verify this. The change from Q to R is the replacement of 6 H atoms by 6 Br atoms.
- Molar mass of H = 1 g/mol.
- Molar mass of Br = 80 g/mol.
- Mass difference = M(R) - M(Q) = 6 × M(Br) - 6 × M(H) = 6 × 80 - 6 × 1 = 480 - 6 = 474 g mol-1. This matches the given data, confirming the structure of R.
Reaction 5 (for verification): Azo Coupling Product P reacts with excess phenol under basic conditions to give S. This is an azo coupling reaction, where the diazonium salt acts as an electrophile and attacks the electron-rich phenol ring, typically at the para position.

$\text{Ar-N}_2^+ + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{basic}} \text{Ar-N=N-C}_6\text{H}_4\text{OH} + \text{H}^+$

For each diazonium group, the reaction is [Ar-N2]+Cl– + C6H5OH → Ar-N=N-C6H4-OH + HCl. The mass change is M(C6H5OH) - M(HCl) = (126+61+16) - (1+35.5) = 94 - 36.5 = 57.5 g/mol. Since P has three diazonium groups, the total mass difference is 3 × 57.5 = 172.5 g mol-1. This matches the provided data, confirming our structures.
Determine the Number of Heteroatoms in R The question asks for the number of heteroatoms in one molecule of R. Heteroatoms are defined as atoms other than carbon (C) and hydrogen (H).
- Molecular Formula of R: C24H12Br6O3.
- The heteroatoms present are Bromine (Br) and Oxygen (O).
- Number of Bromine atoms = 6.
- Number of Oxygen atoms = 3.
- Total number of heteroatoms = 6 + 3 = 9.

Final Answer

The number of heteroatoms in one molecule of R is 9.

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View written solutionFree

Correct answer: 9

The user wants me to solve a multi-step organic chemistry synthesis problem and determine the number of heteroatoms in the final product R.

Step-by-step Solution:

Identify the Starting Material and its Structure The starting compound is 1,3,5-tris-(4-nitrophenyl)benzene. This consists of a central benzene ring with three (4-nitrophenyl) groups attached at positions 1, 3, and 5.
- Structure of Starting Material (A): A central C6H3 ring is connected to three C6H4NO2 groups.
- Molecular Formula of A: C6H3(C6H4NO2)3 = C24H15N3O6.
Reaction 1: Reduction with Sn/HCl The starting compound A reacts completely with excess Sn/HCl. This is a standard reagent for the reduction of a nitro group (–NO2) to a primary amine group (–NH2).

$\text{Ar-NO}_2 \xrightarrow{\text{Sn/HCl}} \text{Ar-NH}_2$

Since there are three nitro groups, all of them are reduced.
- Product (B): 1,3,5-tris-(4-aminophenyl)benzene.
- Molecular Formula of B: C6H3(C6H4NH2)3 = C24H21N3.
Reaction 2: Diazotization The major product B is treated with an excess of NaNO2/HCl at 0°C. This is the diazotization reaction, which converts primary aromatic amines to diazonium salts (–N2+Cl–).

$\text{Ar-NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0^\circ\text{C}} \text{Ar-N}_2^+\text{Cl}^-$

All three amino groups are converted.
- Product (P): 1,3,5-tris-(4-diazoniumchloridophenyl)benzene.
- Molecular Formula of P: [C6H3(C6H4N2)3]3+(Cl–)3 or C24H15N6Cl3.
Reaction 3: Hydrolysis of Diazonium Salt Product P is treated with excess H2O at room temperature. Aromatic diazonium salts hydrolyze to form phenols, releasing nitrogen gas.

$\text{Ar-N}_2^+\text{Cl}^- + \text{H}_2\text{O} \rightarrow \text{Ar-OH} + \text{N}_2 + \text{HCl}$

All three diazonium groups are converted to hydroxyl groups.
- Product (Q): 1,3,5-tris-(4-hydroxyphenyl)benzene.
- Molecular Formula of Q: C6H3(C6H4OH)3 = C24H18O3.
Reaction 4: Bromination of Phenol Derivative Product Q is brominated in an aqueous medium. The hydroxyl (–OH) group is a strongly activating, ortho-para directing group. In an aqueous medium (like bromine water), bromination is exhaustive, substituting all available ortho and para positions.

In each of the three (4-hydroxyphenyl) rings, the para position (relative to –OH) is occupied by the bond to the central benzene ring. Therefore, substitution occurs at the two available ortho positions (positions 3' and 5' of the outer rings).
- Substitution: 2 Br atoms are added to each of the 3 outer rings, replacing 2 H atoms each.
- Total substitution: 6 H atoms are replaced by 6 Br atoms.
- Product (R): 1,3,5-tris-(3,5-dibromo-4-hydroxyphenyl)benzene.
- Molecular Formula of R: Let's derive it from Q (C24H18O3). We replace 6 H with 6 Br. The formula becomes C24H12Br6O3.
Verification of Molar Mass Difference (Q and R) The problem states the molar mass difference between Q and R is 474 g mol-1. Let's verify this. The change from Q to R is the replacement of 6 H atoms by 6 Br atoms.
- Molar mass of H = 1 g/mol.
- Molar mass of Br = 80 g/mol.
- Mass difference = M(R) - M(Q) = 6 × M(Br) - 6 × M(H) = 6 × 80 - 6 × 1 = 480 - 6 = 474 g mol-1. This matches the given data, confirming the structure of R.
Reaction 5 (for verification): Azo Coupling Product P reacts with excess phenol under basic conditions to give S. This is an azo coupling reaction, where the diazonium salt acts as an electrophile and attacks the electron-rich phenol ring, typically at the para position.

$\text{Ar-N}_2^+ + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{basic}} \text{Ar-N=N-C}_6\text{H}_4\text{OH} + \text{H}^+$

For each diazonium group, the reaction is [Ar-N2]+Cl– + C6H5OH → Ar-N=N-C6H4-OH + HCl. The mass change is M(C6H5OH) - M(HCl) = (126+61+16) - (1+35.5) = 94 - 36.5 = 57.5 g/mol. Since P has three diazonium groups, the total mass difference is 3 × 57.5 = 172.5 g mol-1. This matches the provided data, confirming our structures.
Determine the Number of Heteroatoms in R The question asks for the number of heteroatoms in one molecule of R. Heteroatoms are defined as atoms other than carbon (C) and hydrogen (H).
- Molecular Formula of R: C24H12Br6O3.
- The heteroatoms present are Bromine (Br) and Oxygen (O).
- Number of Bromine atoms = 6.
- Number of Oxygen atoms = 3.
- Total number of heteroatoms = 6 + 3 = 9.

Final Answer

The number of heteroatoms in one molecule of R is 9.

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Compounds Containing Nitrogen question

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