Question

Order of the reactivity of halogens during halogenation of alkanes is:
A)\[\text{C}{{\text{l}}_{\text{2}}}\rangle {{\text{I}}_{\text{2}}}\rangle \text{B}{{\text{r}}_{\text{2}}}\]
B)\[\text{B}{{\text{r}}_{\text{2}}}\rangle {{\text{I}}_{\text{2}}}\rangle \text{C}{{\text{1}}_{\text{2}}}\]
C)\[{{\text{I}}_{\text{2}}}\rangle \text{C}{{\text{l}}_{\text{2}}}\rangle \text{B}{{\text{r}}_{\text{2}}}\]
 D)\[\text{C}{{\text{l}}_{\text{2}}}\rangle \text{B}{{\text{r}}_{\text{2}}}\rangle {{\text{I}}_{\text{2}}}\]

Answer 1

Hint: The halogenation of alkane is favored by the radical substitution reaction. The ability of radicals to undergo the reaction depends on the size of the radical. Larger the size of the radical weaker the bond form between the halogen and alkane. The trend for the halogen to undergo the halogenation reaction is to decrease the down halogen group.

Complete step by step answer:
Halogens ${{\text{X}}_{\text{2}}}$react with the alkanes in the presence of ultraviolet light ($\text{C}{{\text{l}}_{\text{2}}}\text{ }\!\!\And\!\!\text{ B}{{\text{r}}_{\text{2}}}$) an oxidizing agent $\text{(}{{\text{I}}_{\text{2}}}\text{)}$ to form haloalkanes. These reactions are free radical substitution reactions which give the mono, di, or polysubstituted haloalkanes.
For example,
\[\begin{align}
  & \begin{matrix}
   {} & {} & {} & {} & {} & \text{Cl} & {} \\
   {} & {} & {} & {} & {} & \text{ }\!\!|\!\!\text{ } & {} \\
   \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}} & \xrightarrow[\text{UV light}]{\text{C}{{\text{l}}_{\text{2}}}} & \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl} & \text{+} & \text{C}{{\text{H}}_{\text{3}}}- & \text{C} & -\text{C}{{\text{H}}_{\text{3}}} \\
   {} & {} & {} & {} & {} & {} & {} \\
\end{matrix} \\
 & \begin{matrix}
   \begin{matrix}
   \text{(Propane)} & {} & {} & {} & {} & \text{(1-Chloropropane)} & \text{(2-Chloropropane)} \\
\end{matrix} & {} & {} & {} \\
\end{matrix} \\
\end{align}\]

The carbon –halogen bond in alkyl halide results from the overlap of a carbon $\text{s}{{\text{p}}^{\text{3}}}$ hybrid orbital with an orbital of the halogen atom. Therefore, alkyl halides have a tetrahedral geometry with $\text{H-C-X}$ bond angles nearly ${{109}^{0}}$. As the size of the halogen atom increases on going down the group the bond length of haloalkanes increases and thus the bond dissociation energies
Halogens are more electronegative than carbon. Therefore,$\text{C-X}$ the bond is polar, with the carbon-bearing $\text{ }\!\!\delta\!\!\text{ +}$ charge and the halogen atom bears the slight $\text{ }\!\!\delta\!\!\text{ -}$charge. The $\text{C-X}$bond is polar
Halogenation is the replacement of one or more numbers of the hydrogen atom from the alkane by a halogen such as fluorine, chlorine, bromine, or Iodine .the halogenation of alkane is a simple substitution reaction. Here the $\text{C-H}$ bond is broken to form a new $\text{C-X}$ bond.
In halogen, the atomic radii increase along with the group and thus the distance between the last shell and the valence electrons goes on increasing. The shielding effect increases as thus the weaker the force of attraction between the nucleus and electrons.

 The halogenation of alkane takes place through a radical substitution reaction.
Fluorine radical formed is the most reactive radical. It reacts violently with alkanes. On the other hand the iodine due to its larger size is least reactive towards the halogenation of alkanes. It is so unreactive that it cannot abstract the proton. Thus iodination is carried out in presence of nitric acid.
The bromination or the formation of bromoalkane is a slow reaction compared to the chlorination of hydrocarbons. It is because the energy required for the abstraction of the proton is 4.5 times greater than required by chlorine. Bromine is selective. Bromine is found to be less reactive but more selective and chlorine radical is more reactive and less selective.
Thus the order of reactivities of halogen toward the halogenation is as:

${{\text{F}}_{\text{2}}}\rangle \text{C}{{\text{l}}_{\text{2}}}\rangle \text{B}{{\text{r}}_{\text{2}}}\rangle {{\text{I}}_{\text{2}}}$

So, the correct answer is “Option D”.

Note: The chlorination of alkane does not stop at the mono-product, it further forms the di and even tri substituted haloalkane. Chlorine is highly reactive but bromine forms a selective product.