The major product formed in the reaction is
- A
- B
- C
- D
Medium
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- 1A trinitro compound, $1, 3,5$ tris-($4$-nitrophenyl) benzene, on complete reaction with an excess of $\mathrm{Sn} / \mathrm{HCl}$ gives 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 $S$.View Solution
The molar mass difference between compounds $Q$ and $R$ is $474 \mathrm{~mol}^{-1}$ and between compounds $P$ and $S$ is $172.5 \mathrm{~g} \mathrm{~mol}^{-1}$.
($1$)The number of heteroatoms present in one molecule of $R$ is. . . . . .
[Use: Molar mass (in g mol ${ }^{-1}$ ): $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Br}=80, \mathrm{Cl}=35.5$
Atoms other than $\mathrm{C}$ and $\mathrm{H}$ are considered as heteroatoms]
($2$)The total number of carbon atoms and heteroatoms present in one molecule of $S$. . . . . .
[Use: Molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ ]: $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Br}=80, \mathrm{Cl}=35.5$
Atoms other than $\mathrm{C}$ and $\mathrm{H}$ are considered as heteroatoms
Give the answer quetion ($1$) and ($2$)
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