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…

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 &#176; 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 &#160; g &#160; mol - 1 and between compounds P and S is 172 . 5 &#160; g &#160; mol - 1 . The number of heteroatoms present in one molecule of R is [Use: Molar mass (in g mol-1): H = 1 , C = 12 , &#160; N = 14 , O = 16 , Br = 80 , Cl = 35 . 5 Atoms other than C and H are considered as heteroatoms]

Aicharya · Accepted Answer

Nitrobenzene part converts to aniline part using $Sn / HCl$ . Aniline forms diazonium salt with nitrous acid and hydrogen chloride. Diazonium salt forms phenol with warm water.Phenol forms poly substituted bromo compound with bromine water.
The hetero atoms present the product $R$ are 6 bromine atoms and 3 oxygen atoms. Number of Hetero atoms in $R$ is $9$ .

A trinitro compound, 1 , 3 , 5 -tris-( 4 -nitrophenyl) benzene, on complete reaction with an excess of Sn /…

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