Given below are two statements:
Statement $I:$ Ethyl pent$-4-$yn$-$oate on reaction with $CH_{3}MgBr$ gives a $3^{\circ}$ alcohol.
Statement $II:$ In this reaction,one mole of ethyl pent$-4-$yn$-$oate utilizes two moles of $CH_{3}MgBr$.
In the light of the above statements,choose the most appropriate answer from the options given below:

What will be the product obtained from the following reaction?
$CH_3CH_2C \equiv N + C_2H_5OH + H_2O \xrightarrow{\text{conc. } H_2SO_4}$

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

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

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

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