Classify each reaction as homolysis or heterolysis. Heterolysis Homolysis Prev A. H3C-CH3 B. C. H3C- 3 of 17 CI -CH3 Br: Next CH3 CH3 H3C ❓ :Br: CH3 1849 +
The Correct Answer and Explanation is:
To classify each reaction as homolysis or heterolysis, we need to understand how the bond is breaking:
- Homolysis: The bond breaks evenly, and each atom takes one electron from the bond, forming two radicals.
- Heterolysis: The bond breaks unevenly, with both electrons going to one atom, forming a cation and an anion.
Let’s examine each case:
A. H₃C–CH₃ → H₃C• + •CH₃
This reaction breaks the bond between two carbon atoms in such a way that each carbon gets one electron, forming two radicals.
✅ This is HOMOLYSIS.
B. H₃C–Br → H₃C⁺ + :Br⁻
In this case, the bond between carbon and bromine breaks and both electrons go to Br, making it a bromide ion (:Br⁻), while carbon becomes a carbocation (H₃C⁺).
✅ This is HETEROLYSIS.
C. H₃C–CH₃ → H₃C⁺ + :CH₃⁻
Here, the C–C bond breaks and both electrons go to one carbon, making it a carbanion (:CH₃⁻) and a carbocation (H₃C⁺).
✅ This is HETEROLYSIS.
Explanation
Bond dissociation reactions are crucial in understanding organic chemistry mechanisms. When a covalent bond breaks, the manner in which the electrons are distributed determines whether the process is homolytic (homolysis) or heterolytic (heterolysis).
In homolysis, the shared pair of electrons in a bond is divided equally between the two atoms. This results in the formation of free radicals, which are species with an unpaired electron. Homolytic cleavage is common in non-polar bonds and under conditions like heat or UV light, which provide energy to break bonds symmetrically. For example, in reaction A (H₃C–CH₃ → H₃C• + •CH₃), the carbon-carbon bond breaks evenly, and each carbon retains one electron, forming two methyl radicals. This is typical of radical reactions such as halogenations.
In heterolysis, both bonding electrons go to one of the atoms, creating a cation and an anion. This usually occurs in polar bonds where one atom is more electronegative. Reaction B (H₃C–Br → H₃C⁺ + :Br⁻) and reaction C (H₃C–CH₃ → H₃C⁺ + :CH₃⁻) are heterolytic cleavages. In B, the more electronegative bromine atom takes both electrons, resulting in a bromide ion and a methyl carbocation. In C, one carbon becomes positively charged while the other takes the electrons, forming a carbanion.
Understanding these types of bond dissociation is essential for predicting reaction mechanisms, intermediates, and reactivity in organic chemistry.
