The decreasing order of reactivity of the following compounds towards $Na^{\oplus} O^{\ominus} Me$ is:

  • A
    $P > Q > R > S$
  • B
    $S > R > Q > P$
  • C
    $Q > P > R > S$
  • D
    $Q > R > S > P$

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The reaction of compound $P$ with $CH_3MgBr$ (excess) in $(C_2H_5)_2O$ followed by addition of $H_2O$ gives $Q$. The compound $Q$ on treatment with $H_2SO_4$ at $0^{\circ}C$ gives $R$. The reaction of $R$ with $CH_3COCl$ in the presence of anhydrous $AlCl_3$ in $CH_2Cl_2$ followed by treatment with $H_2O$ produces compound $S$ [$Et$ in compound $P$ is ethyl group].
$(1)$ The product $S$ is
$(2)$ The reactions,$Q$ to $R$ and $R$ to $S$,are
$A$. Dehydration and Friedel-Crafts acylation$B$. Aromatic sulfonation and Friedel-Crafts acylation
$C$. Friedel-Crafts alkylation,dehydration and Friedel-Crafts acylation$D$. Friedel-Crafts alkylation and Friedel-Crafts acylation

Given the answer question $(1)$ and $(2)$

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