Which of the following carboxylic acids could be resolved by reaction with an enantiomerically pure chiral amine?

Match List-$I$ with List-$II$ :

List-$I$ (Chemical Reaction)	List-$II$ (Reagent used)
$(a)$ $CH_{3}COOCH_{2}CH_{3} \rightarrow CH_{3}CH_{2}OH$	$(i)$ $CH_{3}MgBr / H_{3}O^{+}$ ($1$ equivalent)
$(b)$ $CH_{3}COOCH_{3} \rightarrow CH_{3}CHO$	$(ii)$ $H_{2}SO_{4} / H_{2}O$
$(c)$ $CH_{3}C \equiv N \rightarrow CH_{3}CHO$	$(iii)$ $DIBAL-H / H_{2}O$
$(d)$ $CH_{3}C \equiv N \rightarrow CH_{3}COCH_{3}$	$(iv)$ $SnCl_{2}, HCl / H_{2}O$

Choose the most appropriate match :

Which of the following reactions produce carboxylic acids?
$(a)$ $MnO_2$ with $CH_3COCH_3$
$(b)$ $Ph-CCl_3 \xrightarrow[(ii) H_2O/H^+]{(i) aq. NaOH} PhCOOH$
$(c)$ $C_2H_5Br \xrightarrow[(iii) H_2O/H^+]{(i) Mg, (ii) CO_2} C_2H_5COOH$
$(d)$ $CH_3CH=CHCH_3 \xrightarrow[\Delta]{K_2Cr_2O_7/H_2SO_4} 2CH_3COOH$

How will you bring about the following conversions in not more than two steps?
$(i)$ Propanone to Propene
$(ii)$ Benzoic acid to Benzaldehyde
$(iii)$ Ethanol to $3-$Hydroxybutanal
$(iv)$ Benzene to $m-$Nitroacetophenone
$(v)$ Benzaldehyde to Benzophenone
$(vi)$ Bromobenzene to $1-$Phenylethanol
$(vii)$ Benzaldehyde to $3-$Phenylpropan$-1-$ol
$(viii)$ Benzaldehyde to $\alpha-$Hydroxyphenylacetic acid
$(ix)$ Benzoic acid to $m-$Nitrobenzyl alcohol

What is the product $C$ in the following reaction?
$CH_3Br$ $\xrightarrow{KCN} A$ $\xrightarrow{H_3O^+} B$ $\xrightarrow{LiAlH_4} C$

Match Column-$I$ with Column-$II$ :
| Column-$I$ | Column-$II$ |
| :--- | :--- |
| $(A)$ Cyclopentanone + $2,4-$$DNP$ $\xrightarrow{\text{mild } H^+}$ Cyclopentanone$-2,4-$$DNP$ derivative | $(P)$ Electrophilic substitution |
| $(B)$ $4-$phenyl$-2-$methylbutan$-2-$ol $\xrightarrow{\text{Conc. } H_2SO_4}$ $1,1-$dimethyl$-1,2,3,4-$tetrahydronaphthalene | $(Q)$ Nucleophilic substitution |
| $(C)$ $4-$chlorocyclohexanethiol $\xrightarrow{\text{Base}}$ $7-$thiabicyclo[$2.2$.$1$]heptane | $(R)$ Nucleophilic addition |