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Question: Which of the following has the highest bond dissociation energy? (A) \(C{{H}_{3}}-F\) (B) \(C{{...

Which of the following has the highest bond dissociation energy?
(A) CH3−FC{{H}_{3}}-F
(B) CH3−ClC{{H}_{3}}-Cl
(C) CH3−BrC{{H}_{3}}-Br
(D) CH3−IC{{H}_{3}}-I

Explanation

Solution

The bond dissociation energy depends on the strength of the bond. The energy required to cleave a C-X\text{C-X} depends on the polarity of the bond. The polarity of the bond depends on the electronegativity difference between the carbon and the halogen. The reactivity of the haloalkanes depends on the bond dissociation energy. More easily the bond is broken, more reactive is the haloalkanes.

Complete step by step answer:
The amount of energy necessary to break a chemical bond through homolytic fracture is known as bond dissociation energy.
The dissociation energy or enthalpy depends on the polarity of the bond. In haloalkanes, the carbon is bonded to the halogen atoms: F , Cl , Br , I\text{F , Cl , Br , I} .
We know that the halogens are the most electronegative elements in the periodic table.
Therefore, the polarity of the carbon-halogen bond decreases accordingly. Therefore, the CH3-F\text{C}{{\text{H}}_{\text{3}}}\text{-F} is more polar and CH3-I\text{C}{{\text{H}}_{\text{3}}}\text{-I} have a less polar bond with the dipole moment of 1.636D1.636\text{D}
The bond dissociation energies for the haloalkanes of the C-X\text{C-X} bond are as follows:

BondC−Cl\text{C}-\text{Cl}C−Br\text{C}-\text{Br}C−I\text{C}-\text{I}
Bond dissociation enthalpy (kJ mol−1\text{kJ mo}{{\text{l}}^{-1}} )326.4326.4284.5284.5213.4213.4

The electronegativity of the atoms goes on decreasing from F to the I. The electronegativity difference between the carbon and fluorine is higher compared to the electronegativity difference between the carbon and iodine.
Therefore, the C-I\text{C-I} bond is less polarized and it can be easily broken down.
Therefore, the C-I\text{C-I} bond can easily be cleaved while the C-Cl\text{C-Cl} bond is cleaved with difficulty. Fluoride forms the strongest bond with carbon among all halogens and therefore, is least reactive.
In fact, C-F\text{C-F} the bond is strong that the fluorides do not undergo the nucleophilic substitution reactions under ordinary conditions.
Therefore, the bond dissociation energy between the methyl halide is as shown below,
CH3-F ⟩ CH3-Cl ⟩ CH3-Br ⟩ CH3-I\text{C}{{\text{H}}_{\text{3}}}\text{-F }\rangle \text{ C}{{\text{H}}_{\text{3}}}\text{-Cl }\rangle \text{ C}{{\text{H}}_{\text{3}}}\text{-Br }\rangle \text{ C}{{\text{H}}_{\text{3}}}\text{-I}
So, the correct answer is “Option A”.

Note: The reactivity depends on the bond dissociation energy of the methyl halides. The trend of the reactivity of the haloalkanes is as follows:
 R−I ⟩ R−Br ⟩ R−Cl ⟩⟩ R−F\text{ R}-\text{I }\rangle \text{ R}-\text{Br }\rangle \text{ R}-\text{Cl }\rangle \rangle \text{ R}-\text{F}
The C-F\text{C-F} bond has high dissociation energy thus it does not undergo the reaction and therefore it is “less reactive.”