
The halogen which is most reactive in the halogenation of alkanes under sunlight is:
(A) fluorine
(B) chlorine
(C) bromine
(D) iodine
Answer
448.5k+ views
Hint: An attempt to this question can be made by writing down the mechanism for halogenation of alkanes under sunlight. After that, compare the energy needed for homolytic fission of halogen molecules to arrive at the answer.
Complete step by step solution:
We will now study the reaction mechanism for the halogenation of alkanes under sunlight.
\[{{X}_{2}}{{\xrightarrow{uv}}_{{}}}\overset{\bullet }{\mathop{X}}\,+\overset{\bullet }{\mathop{X}}\,\]
Here X= F, Cl, Br, I.
-This initial reaction is called the initiation step of the reaction mechanism. The free radicals that are formed are in a high energy state and react quickly to gain stability.
-The halogen-free radical polarizes the C-H bond of alkane leading to homolytic fission.
$C{{H}_{3}}-H+\overset{\bullet }{\mathop{X}}\,{{\xrightarrow{{}}}_{{}}}\overset{\bullet }{\mathop{C{{H}_{3}}}}\,+HX$
$\overset{\bullet }{\mathop{C{{H}_{3}}}}\,+{{X}_{2}}{{\xrightarrow{{}}}_{{}}}C{{H}_{3}}X+\overset{\bullet }{\mathop{X}}\,$
-The above reaction is called the propagation step as it keeps the reaction going forward to form products as well free radicals.
-When the methyl free radical reacts either with the halogen free radical or second molecule of methyl free radical, it leads to termination of the reaction. Hence the reaction is called the chain termination step.
$\begin{align}
& \overset{\bullet }{\mathop{C}}\,{{H}_{3}}+\overset{\bullet }{\mathop{X}}\,{{\to }_{{}}}C{{H}_{3}}X \\
& \overset{\bullet }{\mathop{C}}\,{{H}_{3}}+\overset{\bullet }{\mathop{C}}\,{{H}_{3}}{{\to }_{{}}}{{C}_{2}}{{H}_{6}} \\
\end{align}$
We will now try to understand the initiation step i.e. the homolytic fission of halogen molecules. Out of the four halogen elements, Fluorine requires the least amount of energy for fission as F2 is extremely unstable due to high repulsion of electrons and small size.
Hence fluorine is the most reactive halogen atom in the halogenation of alkane as it requires the least energy for fission. The reaction enthalpy for fluorination of alkane is -431 kJ/mol.
Therefore, the correct answer is option (A).
Note: It is important to know that although fluorination of alkane is a feasible reaction theoretically, in practice we do not carry out this reaction as it is highly exothermic and catches fire easily hence destroying the products formed. In addition to that, the C-F bond is very strong and we cannot carry out further reactions with this haloalkane. This is the reason why the chlorination of alkane is preferred over fluorination.
Complete step by step solution:
We will now study the reaction mechanism for the halogenation of alkanes under sunlight.
\[{{X}_{2}}{{\xrightarrow{uv}}_{{}}}\overset{\bullet }{\mathop{X}}\,+\overset{\bullet }{\mathop{X}}\,\]
Here X= F, Cl, Br, I.
-This initial reaction is called the initiation step of the reaction mechanism. The free radicals that are formed are in a high energy state and react quickly to gain stability.
-The halogen-free radical polarizes the C-H bond of alkane leading to homolytic fission.
$C{{H}_{3}}-H+\overset{\bullet }{\mathop{X}}\,{{\xrightarrow{{}}}_{{}}}\overset{\bullet }{\mathop{C{{H}_{3}}}}\,+HX$
$\overset{\bullet }{\mathop{C{{H}_{3}}}}\,+{{X}_{2}}{{\xrightarrow{{}}}_{{}}}C{{H}_{3}}X+\overset{\bullet }{\mathop{X}}\,$
-The above reaction is called the propagation step as it keeps the reaction going forward to form products as well free radicals.
-When the methyl free radical reacts either with the halogen free radical or second molecule of methyl free radical, it leads to termination of the reaction. Hence the reaction is called the chain termination step.
$\begin{align}
& \overset{\bullet }{\mathop{C}}\,{{H}_{3}}+\overset{\bullet }{\mathop{X}}\,{{\to }_{{}}}C{{H}_{3}}X \\
& \overset{\bullet }{\mathop{C}}\,{{H}_{3}}+\overset{\bullet }{\mathop{C}}\,{{H}_{3}}{{\to }_{{}}}{{C}_{2}}{{H}_{6}} \\
\end{align}$
We will now try to understand the initiation step i.e. the homolytic fission of halogen molecules. Out of the four halogen elements, Fluorine requires the least amount of energy for fission as F2 is extremely unstable due to high repulsion of electrons and small size.
Hence fluorine is the most reactive halogen atom in the halogenation of alkane as it requires the least energy for fission. The reaction enthalpy for fluorination of alkane is -431 kJ/mol.
Therefore, the correct answer is option (A).
Note: It is important to know that although fluorination of alkane is a feasible reaction theoretically, in practice we do not carry out this reaction as it is highly exothermic and catches fire easily hence destroying the products formed. In addition to that, the C-F bond is very strong and we cannot carry out further reactions with this haloalkane. This is the reason why the chlorination of alkane is preferred over fluorination.
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