Which one of the following is the intermediate in the preparation of a ketone by hydration of an alkyne in the presence of sulfuric acid and mercury (II) sulphate?

1. 

2. 

3. 

4. 

 $\stackrel{{\mathrm{RCO}}_2\mathrm{H}}{\to }$ product; Product of this reaction is:

(a) 

(b) 

(c) both (a) and (b)

(d) none of these

Above conversion can be achieved by:

(a) Wolf-kishner reduction

(b) Clemmensen reduction

(c) LiAlH₄

(d) NaBH₄

Product (A) is:

1. 

2. 

3.  

4.   

Reagent to carry out above conversion,P,Q,R respectively are:

(1) H₂C=CH-CH₂-Br,(HO^{$\mathrm{\theta }$}),[HO^{$\mathrm{\theta }$},$∆$], Wacker-process

(2) H₂C=CH-CH₂-Br(HO^{$\mathrm{\theta }$}), Wacker-process, HO^{$\mathrm{\theta }$},$∆$

(3) Wacker process, H₂C=CH-CH₂-Br(HO^{$\mathrm{\theta }$}),HO^{$\mathrm{\theta }$}($∆$)

(4) Wacker process,HO^{$\mathrm{\theta }$}($∆$), H₂C=CH-CH₂-Br(HO^{$\mathrm{\theta }$})

Above compounds can be differentiated by following reagent:

1. 2-4 DNP (Brady reagent)

2. Tollen's reagent

3. Bromine water reagent

4. NaHSO₃

Consider the following reaction, 

Product (D) will be :

1.		2.
3.		4.

Product (B) in this reaction is :

1. 

2.     

3.    

4.    

$\mathrm{Ph}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{CH}}_3\underset{\mathrm{HCl}}{\overset{{\mathrm{NaNO}}_2}{\to }}\left(\mathrm{A}\right)\underset{\mathrm{heat}}{\overset{{\mathrm{AC}}_2\mathrm{O}}{\to }}\left(\mathrm{B}\right)\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\left(\mathrm{c}\right)$

Product (C) of the above reaction is :

(1) $\mathrm{Ph}-{\mathrm{CO}}_2\mathrm{H}$

(2) $\mathrm{Ph}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{CO}}_2\mathrm{H}$

(3) $\mathrm{Ph}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}$

(4) $\mathrm{Ph}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{CO}}_2\mathrm{H}$

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