QJEEM02Cha968: JEE Main 2002 Chemistry - Class 11 JEEM02Cha968 Solution: Acetylene does not react with :

1. Acetylene reacts with sodium :

$\mathrm{2CH} \equiv \mathrm{CH} + 2\mathrm{Na} \longrightarrow 2\mathrm{CH} \equiv \mathrm{C}^{-}\mathrm{Na}^{+} + \mathrm{H}_2$

Here, acetylene reacts with sodium to produce sodium acetylide and hydrogen gas.

2. Acetylene reacts with ammoniacal silver nitrate :

$\mathrm{CH} \equiv \mathrm{CH} + 2\mathrm{AgNO}_3 \longrightarrow \mathrm{Ag}_2\mathrm{C}_2 + 2\mathrm{HNO}_3$

In this reaction, acetylene reacts with ammoniacal silver nitrate to form a white precipitate of silver acetylide and nitric acid.

3. Acetylene reaction with hydrochloric acid in the presence of a $\mathrm{Hg}^{2+}$ catalyst (oxymercuration-demercuration) :

$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{Hg(OAc)_2} + \mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{OH} + \mathrm{HgOAc}$

$\mathrm{CH}_2=\mathrm{CH}-\mathrm{OH} + \mathrm{NaBH}_4 \longrightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{H} + \mathrm{B(OH)_3} + \mathrm{NaOAc}$

Acetylene does not readily react with $\mathrm{HCl}$ . However, in the presence of a catalyst like $\mathrm{Hg}^{2+}$ , a specific type of electrophilic addition known as oxymercuration can occur. The result is vinyl chloride. Note that $\mathrm{NaBH}_4$ is used to reduce the mercurinium ion intermediate and to replace the $\mathrm{Hg}$ group with a hydrogen atom.

4. Acetylene does not react with sodium hydroxide :

$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{NaOH} \longrightarrow \text { no reaction }$

According to the principles of the Hard and Soft Acids and Bases (HSAB) theory, soft acids prefer to bind with soft bases and hard acids prefer to bind with hard bases. In this case, $\mathrm{OH}^-$ is a hard base, while the $\mathrm{C}$ in acetylene is a soft acid, so they are not particularly reactive with each other.

5. Acetylene reacts with sodium amide :

$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{NaNH}_2 \longrightarrow \mathrm{CH} \equiv \mathrm{C}^{-}\mathrm{Na}^{+} + \mathrm{NH}_3$

Here, sodium amide acts as a strong base and nucleophile, deprotonating the acetylene to form sodium acetylide and ammonia.