Homolytic bond-breaking

In a catalytic hydrogenation, the H—H bond breaks homolytically (Section 4.11). This means that the reduction reaction involves the addition of two hydrogen atoms to the organic molecule. 

In radical reactions, bonds break homolytically with one electron going one way and one the other. The radicals that are formed have an odd number of electrons, one of which must be unpaired. This makes them very reactive and they are not usually isolated. Even strong bonds can break into ions provided they are polarized, but to make radicals we need weak symmetrical bonds such as 0-0, Br-Br or I-I. Dibenzoyl peroxide, the Ph(C02)2 catalyst in this reaction, readily undergoes homolysis like this—the one-electron movements are represented by fish-hook arrows having one barb and odd electrons on atoms are represented by dots. 

The C—Cl (85 kcal/mol) and C — C (89 kcal/mol) bonds have similar strengths, so both bonds can break. The C—Cl bond breaks heterolytically, giving a base peak at m/z = 43. The C—C bond breaks homolytically the peaks at mIz = 63 and... 

The Co—C bond breaks homolytically, forming a 5 -deoxyadenosyl radical and reducing Co(III) to Co(II). 

Now let s consider a reaction in which bonds break homolytically to give radicals that combine to form a new molecule. When radicals combine, the process is called homogenic bond formation. Methane reacts with chlorine gas at elevated temperatures or in the presence of ultraviolet light as an energy source. In this reaction, a chlorine atom replaces a hydrogen atom. 

Step 1. Initiation. A chlorine molecule absorbs energy, from either ultraviolet hght or high temperatures, and the Cl—Cl bond breaks homolytically to give two chlorine atoms. They are electron-deficient, highly reactive radicals. This step statts the reaction and is called the initiation step. 

The species produced when a bond breaks homolytically are called free radicals. We can show the formation of free radicals by using an equation ... 

Solution Under depolymerization, a C—C bond breaks homolytically to give two radicals ... 

Consider now the behaviour of the HF wave function 0 (eq. (4.18)) as the distance between the two nuclei is increased toward infinity. Since the HF wave function is an equal mixture of ionic and covalent terms, the dissociation limit is 50% H+H " and 50% H H. In the gas phase all bonds dissociate homolytically, and the ionic contribution should be 0%. The HF dissociation energy is therefore much too high. This is a general problem of RHF type wave functions, the constraint of doubly occupied MOs is inconsistent with breaking bonds to produce radicals. In order for an RHF wave function to dissociate correctly, an even-electron molecule must break into two even-electron fragments, each being in the lowest electronic state. Furthermore, the orbital symmetries must match. There are only a few covalently bonded systems which obey these requirements (the simplest example is HHe+). The wrong dissociation limit for RHF wave functions has several consequences. 

If a bond breaks in such a way that each fragment gets one electron, free radicals are formed and such reactions are said to take place by homolytic or free-radical mechanisms. 

As might be expected, the results from both theory and experiment suggest that the solution is more than a simple spectator, and can participate in the surface physicochemical processes in a number of important ways [Cao et al., 2005]. It is well established from physical organic chemistry that the presence of a protic or polar solvent can act to stabilize charged intermediates and transition states. Most C—H, O—H, C—O, and C—C bond breaking processes that occur at the vapor/metal interface are carried out homolytically, whereas, in the presence of aqueous media, the hetero-lytic pathways tend to become more prevalent. Aqueous systems also present the opportunity for rapid proton transfer through the solution phase, which opens up other options in terms of reaction and diffusion. 

Homolytic bond dissociation (homolysis) electronically symmetrical bond breaking => produces radicals. 

When a covalent bond breaks to produce radicals, i.e. one electron of the bond pair goes to each atom, homolytic fission has occurred. These highly reactive chlorine radicals attack the methane molecules. 

The functionalization reaction as shown in Scheme 1(A) clearly requires the breaking of a C-H bond at some point in the reaction sequence. This step is most difficult to achieve for R = alkyl as both the heterolytic and homolytic C-H bond dissociation energies are high. For example, the pKa of methane is estimated to be ca. 48 (6,7). Bond heterolysis, thus, hardly appears feasible. C-H bond homolysis also appears difficult, since the C-H bonds of alkanes are among the strongest single bonds in nature. This is particularly true for primary carbons and for methane, where the radicals which would result from homolysis are not stabilized. The bond energy (homolytic dissociation enthalpy at 25 °C) of methane is 105 kcal/mol (8). 

In any reaction in chemistry, bonds in the reactants are broken and bonds in the products are made. The process of breaking bonds is known as bond fission and there are two types of bond fission homolytic fission and heterolytic fission. 

This bond-breaking process is known as homolytic fission because two species of the same charge (neutral) are formed. Such fission normally occurs when non polar covalent bonds are broken. You will notice that each fragment produced has an unpaired electron and can therefore be described as a free radical. 

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