Which of the following represents the decreasing order of reactivity towards reaction with ammonia?

Give the conversion of the following:
$(i)$ Phthalimide from $o-xylene$
$(ii)$ Ethanol from Ethanoic anhydride

What is the major product of the above Friedel-Crafts reaction?

$(i) \ F_3CCOOH$
$(ii) \ CH_3COOH$
$(iii) \ C_6H_5COOH$
$(iv) \ CH_3CH_2COOH$
Correct order of $pK_a$ value is:

Match List-$I$ with List-$II$.
List-$I$:
$(a)$ $\text{Benzene} \xrightarrow[\text{CuCl}]{\text{CO, HCl, Anhyd. AlCl}_3} \text{Benzaldehyde}$
$(b)$ $R-CO-CH_3 + NaOX \rightarrow$
$(c)$ $R-CH_2-OH + R'COOH \xrightarrow{\text{Conc. H}_2\text{SO}_4}$
$(d)$ $R-CH_2COOH \xrightarrow[(ii) H_2O]{(i) X_2/\text{Red P}} R-CH(X)COOH$
List-$II$:
$(i)$ $\text{Hell-Volhard-Zelinsky reaction}$
(ii) $\text{Gattermann-Koch reaction}$
(iii) $\text{Haloform reaction}$
(iv) $\text{Esterification}$
Choose the correct answer from the options given below.

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-$II$ |
| :--- | :--- |
| $P$. $\text{Ph}_2\text{C(OH)-C(OH)Me}_2 + \text{H}_2\text{SO}_4$ | $1$. $\text{I}_2, \text{NaOH}$ |
| $Q$. $\text{Ph}_2\text{C(NH}_2\text{)-CH(OH)Me} + \text{HNO}_2$ | $2$. $[\text{Ag(NH}_3)_2]\text{OH}$ |
| $R$. $\text{PhC(OH)(Me)-C(OH)(Ph)Me} + \text{H}_2\text{SO}_4$ | $3$. $\text{Fehling solution}$ |
| $S$. $\text{Ph}_2\text{C(Br)-CH(OH)Me} + \text{AgNO}_3$ | $4$. $\text{HCHO, NaOH}$ |
| | $5$. $\text{NaOBr}$ |