Problem 6 - Basic Principles of Organic Chemistry

Question

a. Write the initiation and propagation steps involved in the radical bromination of methylbenzene (toluene) with bromine. Write the low-energy valence-bond structures of the intermediate phenylmethyl radical. b. Calculate \(\Delta H^{0}\) for the following reactions of the radical, using the \(\mathrm{C}-\mathrm{Br}\) bond strength of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}(55 \mathrm{kcal})\), and any other necessary bond energies. Assume that stabilization arising from electron delocalization is 38 kcal for a phenyl group (Section \(6-5 \mathrm{~A}\) ) and \(5 \mathrm{kcal}\) for the triene structure 3 . What can you conclude from these calculations about the stability of 3 and the likelihood of its formation in this kind of bromination?

Step-by-Step Solution

Answer

The phenylmethyl radical is moderately stable due to delocalization. Bromination to form C₆H₅CH₂Br is slightly endothermic, indicating moderate likelihood of occurrence.

Step 1: Initiation Step

In the radical bromination of methylbenzene (toluene), the initiation step involves homolytic cleavage of the bromine molecule (\( \text{Br}_2 \)) under the influence of heat or light to form two bromine radicals. This can be represented as: \[ \text{Br}_2 \rightarrow 2\cdot\text{Br} \]

Step 2: Propagation Step 1

The bromine radical (\( \cdot\text{Br} \)) abstracts a hydrogen atom from methylbenzene, forming hydrogen bromide and a phenylmethyl radical (C₆H₅CH₂\( \cdot \)). This is represented as: \[ \text{C}_6\text{H}_5\text{CH}_3 + \cdot\text{Br} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\cdot + \text{HBr} \]

Step 3: Propagation Step 2

The phenylmethyl radical (C₆H₅CH₂\( \cdot \)) then reacts with another bromine molecule to form bromomethane (C₆H₅CH₂Br) and regenerate the bromine radical. This is represented as: \[ \text{C}_6\text{H}_5\text{CH}_2\cdot + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{Br} + \cdot\text{Br} \]

Step 4: Low-energy Valence-bond Structures

The intermediate phenylmethyl radical (C₆H₅CH₂\( \cdot \)) can be represented by resonance structures where the radical electron is delocalized over the aromatic ring. The primary structure shows the radical electron localized on the benzyl carbon (CH₂\( \cdot \)). To show this, draw resonance structures which include movement of the radical electron into the benzyl ring, showing delocalization of the unpaired electron.

Step 5: Calculate Enthalpy Change (\(\Delta H^{0}\))

Use bond dissociation energies (BDE) to find \( \Delta H^0 \) for the formation of C₆H₅CH₂Br from C₆H₅CH₂\( \cdot \) and \( \cdot\text{Br} \). Given BDE for C-Br is 55 kcal/mol and delocalization stabilization is summed up to 43 kcal/mol (38 from phenyl group and 5 from triene). The energy balance of breaking a radical C-H bond (~105 kcal/mol) and forming C-Br bond (55 kcal/mol) minus stabilization gives \( \Delta H^0 = 105 - (55 + 43) \). \[ \Delta H^0 = 7 \text{ kcal/mol} \].

Step 6: Conclusion on Stability and Formation

The positive but small \(\Delta H^0\) (7 kcal/mol) indicates the reaction is endothermic but only slightly. The structure 3 is stabilized by sizeable delocalization energy, suggesting moderate stability. The formation of 3 is feasible, supported by mild endothermicity and significant electron delocalization stabilization.

Key Concepts

Initiation Step

In radical bromination, the initiation step is crucial. This process begins with the homolytic cleavage of a bromine molecule, where energy from heat or light breaks the bond to create two bromine radicals: \[ \text{Br}_2 \rightarrow 2\cdot\text{Br} \] Homolytic cleavage means that the bond between the bromine atoms splits evenly, with each atom taking one electron. These bromine radicals are highly reactive, setting the stage for the propagation steps that follow.

Propagation Step

Propagation steps involve the chain reaction where radicals react with stable molecules to form new radicals and products. In the case of methylbenzene bromination, it consists of two main stages:

• The first propagation step: A bromine radical abstracts a hydrogen atom from methylbenzene: \[ \text{C}_6\text{H}_5\text{CH}_3 + \cdot\text{Br} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\cdot + \text{HBr} \] This forms a phenylmethyl radical and hydrogen bromide.
• The second step: The phenylmethyl radical reacts with another bromine molecule to regenerate a bromine radical and form bromomethane: \[ \text{C}_6\text{H}_5\text{CH}_2\cdot + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{Br} + \cdot\text{Br} \]

These steps sustain the reaction until the reactants are consumed.

Phenylmethyl Radical Stability

Phenylmethyl radical stability is a key factor in radical chemistry. This radical, formed during propagation, can be represented with resonant structures. Radical stability is increased when the unpaired electron is delocalized across the aromatic ring, meaning that the electron isn't confined to just one atom. This movement, shown in resonance structures, spreads the electron density and enhances stability by reducing energy levels. Understanding resonance is essential to account for why radicals like phenylmethyl can exist long enough to participate in further reactions.

Bond Dissociation Energy

Bond Dissociation Energy (BDE) is the energy required to break a bond homolytically. For radical bromination, the BDE of the C-Br bond is 55 kcal/mol. This value helps to calculate the enthalpy change \( \Delta H^{0} \) for the reactions involved. Consider the energy needed to break the C-H bond (approximately 105 kcal/mol) and the formation of a C-Br bond, subtracting the stabilization from delocalization (43 kcal/mol in this case). The net \( \Delta H^{0} \) results in a slightly endothermic process, indicating the reaction can proceed with moderate ease due to the energy balance.

Delocalization Stabilization

Delocalization stabilization plays a pivotal role in the stability of radical intermediates. When electrons are delocalized over several atoms, energy is minimized. For the phenylmethyl radical, the combined stabilization from the phenyl group (38 kcal/mol) and additional structures (5 kcal/mol) sums up to 43 kcal/mol. This significant stabilization implies that the phenylmethyl radical is more stable than it would be without delocalization. It supports the reaction by allowing intermediates to form more readily and persist long enough to contribute to product formation.