\[C{{H}_{4}}+C{{l}_{2}}\xrightarrow{h\nu }C{{H}_{3}}Cl\]
Chain initiation step involves the fission of $$C{l_2}$$  molecule into chlorine free radical.
\[C{{l}_{2}}\xrightarrow{h\nu }C{{l}^{\bullet }}+C{{l}^{\bullet }}\]

In general, electron releasing groups activate and electron withdrawing groups deactivate the benzene ring towards electrophilic substitution. Hence, the correct order is

  ( $$+ M$$  group ) ( More activating group )

Unsaturated hydrocarbons decolourise alk. $$KMn{O_4}$$  solution; $${C_2}{H_4}\left( {{H_2}C = C{H_2}} \right)$$     is an alkene.

Products of ozonolysis :
\[\underset{\text{2-Butene}}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,\xrightarrow[\frac{Zn}{{{H}_{2}}O}]{{{O}_{3}}}\]      \[C{{H}_{3}}-\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,=O+O=\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}\]
2-Butene is obtained by dehydration of 2-butanol.
\[C{{H}_{3}}-C{{H}_{2}}\underset{\begin{smallmatrix} |\,\,\,\,\,\, \\ OH\,\, \end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{3}}\xrightarrow{{{H}^{+}}}\]      $$C{H_3} - CH = CH - C{H_3} + {H_2}O$$

Bromine is more selective
∴ abstract that hydrogen which forms stable free- radical
∴   ( \[{{3}^{\circ }}\]  free-radical ) is most stable.

Key Idea It is an example of Friedel-Craft reaction. First $$\pi $$ - electrons of cyclohexene attack at $${H^ + }ion$$   of $$HF$$  and form carbocation. This carbocation further reacts with benzene and forms addition product.

Thus, the correct option is (C).

\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CH=C{{H}_{2}}\xrightarrow{HCl}\]       \[C{{H}_{3}}-\underset{\begin{smallmatrix} \,\,\,\,\,\,| \\ \,\,\,\,\,\,Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}C{{H}_{2}}CHC{{H}_{3}}}}\,\]

Alkylation and acylation of benzene in the presence of anhydrous aluminium chloride is known as Friedel-Crafts reaction.

With Lithium in liquid ammonia, trans-alkene is almost an exclusive product.