Hydrocarbons

The reaction follows Markownikoff rule which states that when unsymmetrical reagent adds across unsymmetrical double or triple bond the negative part adds to carbon atom having lesser number of hydrogen atoms.

$FeC{l_3}\,\,$ is Lewis acid. In presence of $FeC{l_3}\,\,$ side chain hydrogen atoms of toluene are substituted.

Hofmann's rule : When theoretically more than one type of alkenes are possible in eliminations reaction, the alkene containing least alkylated double bond is formed as major product. Hence



NOTE : It is less stearically $\beta $-hydrogen is removed

$2,3$-dichloro butane will exhibit optical isomerism due to the presence of two asymmetric carbon atom.

Ease of replacement of $H$-atom ${3^ \circ } > {2^ \circ } > {1^ \circ }.$

Since it contains only two types of $H$-atoms hence it will give only two mono chlorinated compounds viz.



and

NOTE : Among isomeric alkanes, the straight chain isomer has higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower is the boiling point. Further due to the presence of $\pi $ electrons, these molecules are slightly polar and hence have higher boiling points than the corresponding alkanes.

Thus $B.pt.$ follows the order

alkynes $>$ alkene $>$ alkanes (straight chain) > branched chain alkanes.

In neopentane all the $H$ atoms are same $\left( {{1^ \circ }} \right).$

Alkenes combine with hydrogen under pressure and in presence of a catalyst $(Ni, Pt$ or $Pd)$ and form alkanes.

Acetylene reacts with the other three as :

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.