Correct order of reactivity of following compounds towards Grignard reagent ?

${\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}$
$\left(\mathrm{I}\right)$          $\mathrm{H}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}$
$\left(\mathrm{II}\right)$

(1) I > II >III

(2) II > I > III

(3) II > III > I

(4) I > III > I

What reagent and/or reaction conditions would you choose to bring about the following conversion?

(a) $1. {\mathrm{LiAlH}}_4, 2. {\mathrm{H}}_2\mathrm{O}$

(b) ${\mathrm{H}}_2\mathrm{O}, {\mathrm{H}}_2{\mathrm{SO}}_4, \mathrm{heat}$

(c) ${\mathrm{H}}_2\mathrm{O}, \mathrm{NaOH}, \mathrm{heat}$

(d) $\mathrm{PCC}, {\mathrm{CH}}_2{\mathrm{Cl}}_2$

\(CH_3-C \equiv CH \xrightarrow [dil. H_2SO_4] {HgSO_4 } (A)\)
\(CH_3-C \equiv CH \xrightarrow [(2) H_2SO_4/HO^-] {(1) BH_3THF} (B)\)
Products (A) and (B) are differentiated by:

1. 2-4 DNP

2. NaOI

3. Na-metal

4. NaHSO₃

End Product (C) in above reaction is :

(b)

(c)

Compound (X) C₄H₈O, which reacts with 2, 4-DNP derivative and gives negative haloform test is.
1. ${\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

2. ${\mathrm{CH}}_3-\underset{\underset{{\mathrm{CH}}_3}{\mathrm{l}}}{\mathrm{CH}}-\mathrm{CHO}$

3. 

4. ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\stackrel{\stackrel{\mathrm{OH}}{\mathrm{l}}}{\mathrm{CH}}-{\mathrm{CH}}_3$

Product (B) of the reaction is :

1. 

2. 

3. 

4. 

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3\to \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-{\mathrm{CO}}_2\mathrm{H}$
Above conversion can be achieved by :

(a) ${\mathrm{KMnO}}_4, ∆ \mathrm{followed} \mathrm{by} {\mathrm{H}}^{+}$

(b) ${\mathrm{I}}_2/\mathrm{NaOH} \mathrm{followed} \mathrm{by} {\mathrm{H}}^{+}$

(c) H₂/Pt

(d) LiAlH₄

Product (A) of the reaction is :

$R-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}\stackrel{\mathrm{R}-{\mathrm{NH}}_2}{\to }R-\mathrm{CH}=\mathrm{N}-R$. This reaction gives best yield at:

(a) pH 1 - 2

(b) pH 4 - 5

(c) pH 10 - 11

(d) pH 13 - 14

An aromatic compound A of the molecular formula C₈H₁₀O on reaction with iodine and dilute NaOH gives a yellow precipitate. The structure of the compound is expected to be-

1. 

2. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CHOHCH}}_3$

3. 

4.