This reaction is Friedel-Craft’s acylation.
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${P} \xrightarrow[\substack{\text { 2. } \mathrm{H}^{+}, \mathrm{H}_2 \mathrm{O} \\ \text { 3. } \mathrm{H}_2 \mathrm{SO}_4, \Delta}]{\text { 1. MeMgBr }}Q \xrightarrow[\text { 2. } \mathrm{Zn}, \mathrm{H}_2 \mathrm{O}]{1 . \mathrm{O}_3} {R} \xrightarrow[\text { 2. } \Delta]{1 . \mathrm{OH}^{-}} {S}$
$1.$ The structure of the carbonyl compound ${P}$ is
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$2.$ The structures of the products ${Q}$ and ${R}$, respectively, are
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$3.$ The structure of the product ${S}$ is
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Give hte answer question $1,2$ and $3.$
$(a)$ Octahedral $Co(III)$ complexes with strong field ligands have very high magnetic moments
$(b)$ When $\Delta_{0}< P$, the $d-$electron configuration of $Co(III)$ in an octahedral complex is $t_{\text {eg }}^{4} e_{g}^{2}$
$(c)$ Wavelength of light absorbed by $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$ is lower than that of $\left[\mathrm{CoF}_{6}\right]^{3-}$
$(d)$ If the $\Delta_{0}$ for an octahedral complex of $\mathrm{Co}(\mathrm{III})$ is $18,000 \;\mathrm{cm}^{-1},$ the $\Delta_{\mathrm{t}}$ for its tetrahedral complex with the same ligand will be $16,000\;\mathrm{cm}^{-1}$
