The computer-generated representation shown does not distinguish between single and multiple bonds. Explain why these are different and predict the carbon-oxygen bond length in carbon dioxide. Explain this reaction in terms of a free-radical mechanism.
Does not show termination steps. Requires the Flash plugin, standard with most new browsers. Free radical halogenation is a reaction that substitutes a chlorine or a bromine for a hydrogen on an alkane. This reaction is a photochemical one. That is, it occurs only when performed in the presence of uv light abbreviated hv.
Typically, free radical reactions are described in three steps: Note the use of a single headed arrow when describing the movement of a single electron. The reaction begins with an initiation step, which is the separation of the halogen X2 into two radicals atoms with a single unpaired electron by the addition of uv light.
This is called the initiation step because it initiates the reaction. The initiation step, or the formation of the chlorine radicals, is immediately followed by the propogation steps--steps directly involved in the formation of the product. As an example, isobutane C4H10 will be used in the chlorination reaction.
This first propogation step forms the tertiary radical. In the last step, the tertiary radical then reacts with another one of the chlorine molecules to form the product. Notice that another chlorine radical is regenerated, so this reaction can, in theory, go on forever as long as there are reagents.
This is called a chain reaction. A sidenote on free radical stabilities: Hydrogens attached to more highly substituted carbons ie. These neighboring alkyl groups have the ability to donate some of their electron density to the electron-deficient radical carbon a radical is short one electron of filling the atom's valence octet.
Thus the hydrogen on the tertiary carbon here is abstracted in preference to the 9 other hydrogen atoms attached to a primary carbon a carbon that is attached to only one other carbon atom because it forms a more stable radical.
Here, the tertiary radical is stabilized by electron donation from neighboring alkyl groups. Selectivity of free-radical halogenation A point of note about free radical processes is that the intermediates are so highly reactive and short lived that usually you obtain a mixture of products, even though there is preference for forming more highly substituted free radical intermediates.
In this example with isobutane, for instance, there would certainly be some abstraction of hydrogens attached to the primary carbons, leading to a different product than the above product can you draw it out?
Bromine reacts exactly the same way as chlorine; however, it is far more selective. If propane CH3CH2CH3for example, was the substrate, 2-bromopropane would be the dominant product, and there would be only a small amount of 1-bromopropane.
Free radical chlorination, though, would not be quite as selective, and there would be a greater amount of the chlorination of the primary carbon than in the bromination reaction. Side reactions that can stop the chain reaction are called termination steps.
These termination steps involve the destruction of the free-radical intermediates, typically by two of them coming together.
So why can't the other halogens such as fluorine or iodine be used? Iodine reacts endothermically energetically uphill and too slowly to be of much good in these free radical processes, while fluorine is at the other pole--it reacts too violently and too quickly to be selective, and can, if uncontrolled, even break carbon-carbon bonds.
To understand why this is so, derive the DH's for the 4 reactions and compare them you will find that flourination is highly exothermic, while iodonation is endothermic; chlorination and bromination, however, are right in the middle.The chloromethyl radical formed can then interact with a chlorine molecule in a new propagation step CH 2 Cl + Cl 2 CH 2 Cl 2 + Cl and so dichloromethane is formed and a chlorine radical regenerated.
benjaminpohle.com is a platform for academics to share research papers. 11 Worksheet 1 Substitution of alkanes by radicals 1 Methane and chlorine react together to give chloromethane and hydrogen chloride.
a Write a balanced symbol equation for this reaction.  b Explain the mechanism for this reaction using at least four steps. You should include: • an initiation step  • two propagation steps [2. May 04, · STEP TWO benjaminpohle.com 3 + Cl2 CH3Cl + Cl. The chlorine free radicals are very reactive and remove an H from the methane leaving a methyl free radical The methyl free radical reacts with a Cl2 molecule to produce the main product and another Cl free radical All propagation steps have a free radical in the reactants and in the products.
(iii) Write an equation for a propagation step in which chloromethane reacts.
(iv) Write an equation for a propagation step in which trichloromethane is formed. (4). A useful reaction mechanism. consists of a series of elementary steps (defined below) that can be written as chemical equations, and whose sum gives the net balanced reaction equation; must agree with the experimental rate law; can rarely if ever be proved absolutely.