The reaction sequence is as follows: C_6H_5SO | vedclass

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(A) $1$. The reaction of benzenesulfonic acid with $NaOH$ followed by acidification gives phenol $(P = C_6H_5OH)$.$2$. The oxidation of phenol with $N

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(A) $1$. The reaction of benzenesulfonic acid with $NaOH$ followed by acidification gives phenol $(P = C_6H_5OH)$.$2$. The oxidation of phenol with $Na_2Cr_2O_7$ and $H_2SO_4$ (Jones reagent or similar strong oxidizing conditions) yields $p$-benzoquinone as the major product $(Q = C_6H_4O_2)$.$3$. In $p$-benzoquinone $(C_6H_4O_2)$,the structure consists of a six-membered ring with two carbonyl groups at para positions and two double bonds in the ring.$4$. Counting the bonds:- $\sigma$ bonds: $6$ ($C$-$C$ in ring) + $4$ ($C$-$H$) + $2$ ($C$=$O$) + $2$ ($C$-$O$) = $14$ $\sigma$ bonds.- $\pi$ bonds: $2$ ($C$=$C$) + $2$ ($C$=$O$) = $4$ $\pi$ bonds.- Wait,let's re-count carefully: The structure of $p$-benzoquinone has $6$ $C$-$C$ $\sigma$ bonds,$4$ $C$-$H$ $\sigma$ bonds,$2$ $C$=$O$ $\sigma$ bonds,and $2$ $C$-$O$ $\sigma$ bonds (total $14$). The $\pi$ bonds are $2$ from $C$=$C$ and $2$ from $C$=$O$ (total $4$).- The ratio is $14:4 = 7:2$. Let's re-examine the structure. $p$-Benzoquinone has $12$ $\sigma$ bonds and $4$ $\pi$ bonds. Ratio $12:4 = 3:1$.

The reaction sequence is as follows: $C_6H_5SO_3H$ $\xrightarrow[(ii) H^+]{(i) NaOH} P$ $\xrightarrow{Na_2Cr_2O_7, H_2SO_4} Q$. The ratio of $\sigma$ bonds to $\pi$ bonds in $Q$ is:

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