Show how each of the following compounds can be converted to benzoic acid.
$(i)$ Ethylbenzene
$(ii)$ Acetophenone
$(iii)$ Bromobenzene
$(iv)$ Phenylethene (Styrene)

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(N/A) The conversion of the given compounds to benzoic acid is as follows:
$(i)$ Ethylbenzene: Oxidation with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
$C_6H_5CH_2CH_3$ $\xrightarrow{KMnO_4/KOH, \Delta} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$
$(ii)$ Acetophenone: Oxidation with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
$C_6H_5COCH_3$ $\xrightarrow{KMnO_4/KOH, \Delta} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$
$(iii)$ Bromobenzene: Reacts with $Mg$ in dry ether to form phenylmagnesium bromide,which reacts with $CO_2$ (dry ice) followed by acidification to give benzoic acid.
$C_6H_5Br$ $\xrightarrow{Mg, \text{ether}} C_6H_5MgBr$ $\xrightarrow{CO_2} C_6H_5COOMgBr$ $\xrightarrow{H_3O^+} C_6H_5COOH$
$(iv)$ Phenylethene (Styrene): Oxidation with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
$C_6H_5CH=CH_2$ $\xrightarrow{KMnO_4/KOH, \Delta} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$

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Match List-$I$ with List-$II$.
List-$I$ (Name of reaction) List-$II$ (Reagent used)
$A$. Hell-Volhard-Zelinsky reaction $I$. $NaOH + I_2$
$B$. Iodoform reaction $II$. $(i) CrO_2Cl_2, CS_2; (ii) H_2O$
$C$. Etard reaction $III$. $(i) Br_2 / \text{red phosphorus}; (ii) H_2O$
$D$. Gatterman-Koch reaction $IV$. $CO, HCl, \text{anhyd. } AlCl_3$

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Choose the best sequence of reactions for the transformation given. Semicolons indicate separate reaction steps to be used in the order shown.

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Match Column-$I$ with Column-$II$ based on the $pK_a$ values of the given carboxylic acids:
Column-$I$Column-$II$
$(A)$ $CF_3COOH$$(i)$ $3.41$
$(B)$ Benzoic acid$(ii)$ $4.76$
$(C)$ $CH_3COOH$$(iii)$ $0.23$
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