Product (B) of the reaction is :

(1) $\mathrm{Ph}-{\mathrm{CH}}_2-{\mathrm{NO}}_2$

(2) $\mathrm{Ph}-{\mathrm{CH}}_2-\mathrm{ONO}$

(3) $\mathrm{Ph}-\mathrm{CHO}$

(4) $\mathrm{Ph}-\mathrm{O}-\mathrm{N}=\mathrm{O}$

Reactant (A) in this reaction is : 

  (d) Both (b) and (c)

Correct order of reactivity of following compounds towards Grignard reagent ?

${\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\left(\mathrm{I}\right)$          $\mathrm{H}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}$
$<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\left(\mathrm{II}\right)$

(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) $1.<mspace\; width="1em"></mspace>{\mathrm{LiAlH}}_4,<mspace\; width="1em"></mspace>2.<mspace\; width="1em"></mspace>{\mathrm{H}}_2\mathrm{O}$

(b) ${\mathrm{H}}_2\mathrm{O},<mspace\; width="1em"></mspace>{\mathrm{H}}_2{\mathrm{SO}}_4,<mspace\; width="1em"></mspace>\mathrm{heat}$

(c) ${\mathrm{H}}_2\mathrm{O},<mspace\; width="1em"></mspace>\mathrm{NaOH},<mspace\; width="1em"></mspace>\mathrm{heat}$

(d) $\mathrm{PCC},<mspace\; width="1em"></mspace>{\mathrm{CH}}_2{\mathrm{Cl}}_2$

\(CH_3-C \equiv CH \xrightarrow [dil. H_2SO_4] {HgSO_4 } (A)\)
\(CH_3-C \equiv CH \xrightarrow [(2) H_2SO_4/HO^-] {(1) BH_3THF} (B)\)
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) ${\mathrm{KMnO}}_4,<mspace\; width="1em"></mspace>∆<mspace\; width="1em"></mspace>\mathrm{followed}<mspace\; width="1em"></mspace>\mathrm{by}<mspace\; width="1em"></mspace>{\mathrm{H}}^{+}$

(b) ${\mathrm{I}}_2/\mathrm{NaOH}<mspace\; width="1em"></mspace>\mathrm{followed}<mspace\; width="1em"></mspace>\mathrm{by}<mspace\; width="1em"></mspace>{\mathrm{H}}^{+}$

(c) H₂/Pt

(d) LiAlH₄

Product (A) of the reaction is :