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°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 ______ . [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]

Question

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°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 ______ .  [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]

cdquestions Admin · Accepted Answer

Chemical Reaction Sequence and Heteroatom Calculation  Chemical Reaction Sequence  This image shows a sequence of chemical reactions starting with a compound containing nitro groups. The process involves various reagents and results in several intermediate products: Step 1: Reduction The compound with two nitro groups ($ NO_2 $) undergoes a reduction using tin (Sn) and hydrochloric acid (HCl), converting the nitro groups into amino groups ($ NH_2 $). The resulting intermediate is shown as the product $ P $. Step 2: Diazotization The amino group is diazotized with sodium nitrite ($ NaNO_2 $) and hydrochloric acid at 0-5°C, forming a diazonium chloride ($ N_2Cl $) group. The resulting compound is labeled as $ P $. Step 3: Hydrolysis In the third step, hydrolysis of the diazonium group leads to the hydroxyl group ($ OH $) in the intermediate $ Q $. Step 4: Chlorination Next, chlorination occurs, resulting in the substitution of the hydroxyl group ($ OH $) with a chloro group ($ Cl $) to form $ P $. Step 5: Bromination In the final step, the compound undergoes bromination using $ Br_2 / H_2O $ to introduce three bromine atoms ($ Br $) to the aromatic ring, resulting in the product labeled as $ R $. Calculation of Heteroatoms in R The final compound $ R $ has three hydroxyl groups ($ OH $), and three bromine atoms ($ Br $), giving us: 3 hydroxyl groups contribute 3 oxygen atoms 3 bromine atoms contribute 3 bromine atoms The total number of heteroatoms (atoms that are not carbon) in the structure of $ R $ is 9, as given in the problem statement.

Correct Answer: 9

Solution and Explanation

Chemical Reaction Sequence

Step 1: Reduction

Step 2: Diazotization

Step 3: Hydrolysis

Step 4: Chlorination

Step 5: Bromination

Calculation of Heteroatoms in R

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