Question

‘X’ can be

(A) ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{/OEt, }\Delta $
(B) $\text{Red}\,\text{P/}\,\text{HI}$
(C) $\text{HS-C}{{\text{H}}_{2}}\text{-C}{{\text{H}}_{2}}\text{-SH(dithiol), Dry}\,\text{HCl/Raney}\,\text{Ni/}{{\text{H}}_{2}}$
(D) All of these

Answer 1

Hint: Carbonyl compounds can be converted into respective hydrocarbons by treating with á$\text{Zn/HCl and }{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{(hydrazine)/}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ONa}$, these reagents known as Clemmensen and Wolff-Kishner reagent respectively.
- Raney nickel $\text{(Ni)}$ is like palladium on carbon $\text{(Pt/C)}$ reducing agent, which is mainly used for the hydrogenation of alkene and alkynes. Raney nickel is mainly used for the reduction of $\text{C-S}$ bond to $\text{C-H}$ bond. This reagent can be used as an alternative means for the conversion ketone to alkane.
- Red phosphorus with hydrogen iodide acts as a strong reducing agent by producing nascent hydrogen atoms.

Complete step by step answer:
(A) Aldehyde and ketone are reduced by hydrazine $\text{(N}{{\text{H}}_{\text{2}}}\text{-N}{{\text{H}}_{\text{2}}}\text{)}$ in the presence $\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{ONa}$ and form hydrocarbon, this reaction is known as Wolff-Kishner reduction. So $\text{buten-2-one}$ reacts with hydrazine in the basic medium gives $\text{n-butane}$. This reaction is represented by following reaction-

(B) $\text{Red}\,\text{P/}\,\text{HI}$, acts as a powerful reducing agent, it will give alkane ($\text{n-butane}$) after the complete reduction carbonyl group of $\text{buten-2-one}$ . This reaction is represented by following reaction –
          

(C) Thiols will react in the acidic medium with the carbonyl group of $\text{buten-2-one}$ (ketone) to form thioketal. After reacting with Raney nickel removal of the sulphur group takes place and finally forms $\text{n-butane}$. This reaction is represented in the following reaction.

The correct answer is option “D” .

Note: In $\text{Red}\,\text{P/}\,\text{HI}$ , Red phosphorus acts as a catalyst and this reducing agent reduces carbonyl compounds to respective alkane. In this reduction process red phosphorus reacts with $\text{HI}$ and forms $\text{P}{{\text{I}}_{\text{3}}}$ . It is the reaction controlling agent. It removes iodine which is necessary because iodine is highly reactive. It will react with hydrocarbon and form haloalkane.