Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Bond Dissociation Energy (BDE)

Welcome to a fascinating exploration of chemical bonds and the energy required to break them! The passage introduces us to a fundamental concept in organic chemistry: Bond Dissociation Energy (BDE).

BDE is the amount of energy you need to supply to break a bond homolytically in the gaseous state. Homolytic cleavage means the bond breaks evenly, and each atom takes away one electron, forming highly reactive species called free radicals.

Here is the golden rule you must always remember: The stability of the resulting radical is inversely proportional to the BDE.

If breaking a bond leads to a highly stable radical, the molecule is "happy" to break that bond, meaning it requires less energy (lower BDE). Conversely, if breaking the bond creates a highly unstable radical, the molecule will resist, requiring a massive amount of energy (higher BDE).

Let's apply our golden rule to the four molecules given in the first question. We need to break the indicated $\text{C-H}$ bonds and analyze the stability of the resulting radicals.

1. Molecule P (${\text{H–CH(CH}}_3{\text{)}}_2$): Breaking this bond gives a $2^{\circ }$ (secondary) radical. This radical is moderately stable because it is supported by hyperconjugation from the adjacent methyl groups.

2. Molecule Q (${\text{H–CH}}_2\text{Ph}$): Breaking this bond yields a benzyl radical. This is the jackpot of stability! The unpaired electron can delocalize around the entire benzene ring through resonance. Because resonance provides immense stabilization, this radical is the most stable of the bunch.

3. Molecule R (${\text{H–CH=CH}}_2$): This gives a vinyl radical. The unpaired electron sits on an $s{p}^2$ hybridized carbon. Because $s{p}^2$ carbons are more electronegative than $s{p}^3$ carbons, they do not like holding an electron deficiency. This makes the radical quite unstable.

4. Molecule S ($\text{H–C}\equiv \text{CH}$): This yields an ethynyl radical. The unpaired electron is on an $sp$ hybridized carbon. An $sp$ carbon has 50% $s$-character, making it highly electronegative. It holds its electrons very tightly, making this radical extremely unstable.

The Stability Order: $\text{Q}>\text{P}>\text{R}>\text{S}$

Since BDE is inversely proportional to stability, the BDE order is the exact opposite:

The BDE Order: $\text{Q}<\text{P}<\text{R}<\text{S}$

Matching this with the given numerical values ($88,95,110,132{\text{ kcal mol}}^{-1}$): - $\text{Q}$ (lowest BDE) $\to 88$ (iv) - $\text{P}$ $\to 95$ (iii) - $\text{R}$ $\to 110$ (ii) - $\text{S}$ (highest BDE) $\to 132$ (i)

This perfectly matches Option (A).

Now, let's dive into the mechanism of the free radical halogenation of methane:

This reaction proceeds via a chain mechanism consisting of initiation, propagation, and termination steps. We need to calculate the enthalpy change ($\Delta H^{\circ }$) for these steps using the provided BDE values. Remember, breaking bonds requires energy (positive $\Delta H$), and forming bonds releases energy (negative $\Delta H$).

The Initiation Step Light energy is used to break the $\text{Cl-Cl}$ bond homolytically:

Since we are only breaking a bond, energy is absorbed. The BDE of $\text{Cl-Cl}$ is $58{\text{ kcal mol}}^{-1}$, so $\Delta H^{\circ }=+58{\text{ kcal mol}}^{-1}$. This step is strictly endothermic. Option (A) claims it is exothermic, so it is incorrect.

Propagation Step 1 The highly reactive chlorine radical attacks methane:

- Energy required to break the $\text{C-H}$ bond: $+105{\text{ kcal mol}}^{-1}$ - Energy released forming the $\text{H-Cl}$ bond: $-103{\text{ kcal mol}}^{-1}$

This step is slightly endothermic. Option (B) claims it is exothermic, so it is incorrect.

Propagation Step 2 The newly formed methyl radical attacks a chlorine molecule:

- Energy required to break the $\text{Cl-Cl}$ bond: $+58{\text{ kcal mol}}^{-1}$ - Energy released forming the $\text{C-Cl}$ bond: $-85{\text{ kcal mol}}^{-1}$

This step is highly exothermic. Option (C) claims it is endothermic, so it is incorrect.

The Overall Reaction The overall reaction enthalpy is simply the sum of the enthalpies of the two propagation steps (the initiation step is not part of the net stoichiometry because the radicals act as catalysts in the chain cycle).

The overall reaction is exothermic by $25{\text{ kcal mol}}^{-1}$. This perfectly matches Option (D)!