How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom:
$(i)$ Methyl benzoate
$(ii)$ $m$-Nitrobenzoic acid
$(iii)$ $p$-Nitrobenzoic acid
$(iv)$ Phenylacetic acid
$(v)$ $p$-Nitrobenzaldehyde.

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(N/A) $(i)$ Methyl benzoate: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{KMnO_4, OH^-}$ Benzoic acid $\xrightarrow{CH_3OH, H^+}$ Methyl benzoate.
$(ii)$ $m$-Nitrobenzoic acid: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{conc. HNO_3, conc. H_2SO_4}$ $p/o$-Nitrotoluene (separate $m$-isomer) $\xrightarrow{KMnO_4, OH^-}$ $m$-Nitrobenzoic acid.
$(iii)$ $p$-Nitrobenzoic acid: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{conc. HNO_3, conc. H_2SO_4}$ $p$-Nitrotoluene $\xrightarrow{KMnO_4, OH^-}$ $p$-Nitrobenzoic acid.
$(iv)$ Phenylacetic acid: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{Cl_2, h\nu}$ Benzyl chloride $\xrightarrow{KCN}$ Benzyl cyanide $\xrightarrow{H_3O^+}$ Phenylacetic acid.
$(v)$ $p$-Nitrobenzaldehyde: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{conc. HNO_3, conc. H_2SO_4}$ $p$-Nitrotoluene $\xrightarrow{CrO_2Cl_2, CS_2, H_3O^+}$ $p$-Nitrobenzaldehyde.

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Assertion : The $pK_a$ of acetic acid is lower than that of phenol.
Reason : Phenoxide ion is more resonance stabilised.

The products obtained in the given reactions are shown below. The number of possible products for $x$ and $y$ are:

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The end product of the above reaction is:

The desired product $X$ $(\text{Ph-C(Me)(Ph)-COOH})$ can be prepared by reacting the major product of the reactions in List-$I$ with one or more appropriate reagents in List-$II$. (Given,order of migratory aptitude: $\text{aryl} > \text{alkyl} > \text{hydrogen}$)
| List-$I$ | List-$II$ |
| :--- | :--- |
| $P$. $\text{Ph}_2\text{C(OH)-C(OH)Me}_2 + \text{H}_2\text{SO}_4$ | $1$. $\text{I}_2, \text{NaOH}$ |
| $Q$. $\text{Ph}_2\text{C(NH}_2\text{)-CH(OH)Me} + \text{HNO}_2$ | $2$. $[\text{Ag(NH}_3)_2]\text{OH}$ |
| $R$. $\text{PhC(OH)(Me)-C(OH)(Ph)Me} + \text{H}_2\text{SO}_4$ | $3$. $\text{Fehling solution}$ |
| $S$. $\text{Ph}_2\text{C(Br)-CH(OH)Me} + \text{AgNO}_3$ | $4$. $\text{HCHO, NaOH}$ |
| | $5$. $\text{NaOBr}$ |

The product of the reaction is:

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