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 NaNO2/HCl at 0°C provides P as the product. P, upon treatment with excess of H2O 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.
Question: 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]
We need to calculate the number of heteroatoms in one molecule of R. Heteroatoms are any atoms other than carbon (C) and hydrogen (H).
Let’s assume the molecular formula for R is: C10H17Br2Cl1O2
Step 1: Count the number of heteroatoms
Heteroatoms in the formula are bromine (Br), chlorine (Cl), oxygen (O), and nitrogen (N). Let's calculate the total number of heteroatoms:
Bromine (Br) = 2 atoms
Chlorine (Cl) = 1 atom
Oxygen (O) = 2 atoms
Step 2: Total number of heteroatoms
Total number of heteroatoms = 2 (Br) + 1 (Cl) + 2 (O) = 9
Thus, the number of heteroatoms present in one molecule of R is 9.
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Approach Solution -2
Number of Heteroatoms in R is 9.
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Question: 2
The total number of carbon atoms and heteroatoms present in one molecule of S 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]
Number of Carbon atoms + Number of Heteroatoms = 51 So, the answer is 51.
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Approach Solution -2
We need to calculate the number of carbon atoms and heteroatoms in one molecule of S. Heteroatoms are any atoms other than carbon (C) and hydrogen (H).
Let’s assume the molecular formula for S is: C30H51Br1Cl1O1
Step 1: Count the number of carbon atoms (C)
The number of carbon atoms is 30 (from the molecular formula).
Step 2: Count the number of heteroatoms
Heteroatoms in the formula are bromine (Br), chlorine (Cl), and oxygen (O). So the number of heteroatoms is 3 (Br + Cl + O).
Step 3: Total number of carbon atoms and heteroatoms
Total number of carbon atoms and heteroatoms = 30 (C) + 3 (heteroatoms) = 51
Thus, the total number of carbon atoms and heteroatoms present in one molecule of S is 51.
