Free radical halogenation
Free radical halogenation in organic chemistry is a type of halogenation. This chemical reaction is typical of alkanes and alkyl-substituted aromatics under application of heat or UV light. The reaction is used for the industrial synthesis of chloroform (CHCl3), dichloromethane (CH2Cl2). It proceeds by a free radical chain mechanism.
Additional recommended knowledge
The chain mechanism is as follows, using the chlorination of methane as a typical example:
- 1. Initiation: Splitting or homolysis of a chlorine molecule to form two chlorine atoms, initiated by ultraviolet radiation. A chlorine atom has an unpaired electron and acts as a free radical.
- 2. chain propagation (two steps): a hydrogen atom is pulled off from methane then the methyl radical pulls a Cl· from Cl2.
- CH4 + Cl· → CH3· + HCl
- CH3· + Cl2 → CH3Cl + Cl·
- This results in the desired product plus another chlorine radical. This radical will then go on to take part in another propagation reaction causing a chain reaction. If there is sufficient chlorine, other products such as CH2Cl2 may be formed.
- 3. chain termination: recombination of two free radicals:
- Cl· + Cl· → Cl2; or
- CH3· + Cl· → CH3Cl; or
- CH3· + CH3· → C2H6.
- The last possibility in the termination step will result in an impurity in the final mixture; notably this results in an organic molecule with a longer carbon chain than the reactants.
The net reaction is:
- CH4 + Cl2 → CH3Cl + HCl.
In the case of methane or ethane, all the hydrogen atoms are equivalent and have an equal chance of being replaced. This leads to what is known as a statistical product distribution. For propane and higher alkanes, the hydrogen atoms which form part of CH2 (or CH) groups are preferentially replaced.
The reactivity of the different halogens varies considerably: the relative rates are: fluorine (108) > chlorine (1) > bromine (7×10−11) > iodine (2×10−22). Hence the reaction of alkanes with fluorine is difficult to control, that with chlorine is moderate to fast, that with bromine is slow and requires high levels of UV irradiation while the reaction with iodine is practically non-existent and thermodynamically unfavorable.
A common method in organic synthesis employing NBS is the Wohl-Ziegler reaction.
Control of halogenation
- Halogenation often does not stop at monosubstititution, depending on reaction conditions the chlorination of methane yields dichloromethane, chloroform and carbon tetrachloride.
- In most hydrocarbons more than one possible product exists depending on which hydrogen is replaced. Butane (CH3-CH2-CH2-CH3), for example, can be chlorinated at the "1" position to give 1-chlorobutane (CH3-CH2-CH2-CH2Cl) or at the "2" position to give 2-chlorobutane (CH3-CH2-CHCl-CH3). The product distribution depends on relative reaction rates: in this case the "2" position of butane reacts faster and 2-chlorobutane is the major product.
- Chlorination is generally less selective than bromination. Fluorination is not only even less selective than chlorination, but also highly exothermic and care must be taken to prevent an explosion or a runaway reaction. This relationship is often used as a demonstration of the reactivity–selectivity principle and can be explained with the aid of the Hammond postulate. A bromine radical is not very reactive and the transition state for proton abstraction has much radical character and is reached late. The reactive chlorine radical develops a transition state resembling the reactant with little radical character. When the alkyl radical is fully formed in the transition state it can benefit fully from any resonance stabilization present thereby maximizing selectivity.
- We can also look at the bond dissociation energies (BDE's) to understand the selectivity of bromination. The BDE of a bond is the energy required to break it by homolytic cleavage, and these values can be used to determine if a reaction or step in a reaction is exothermic or endothermic. In a chain reaction step of a Br radical reacting with a hydrogen on a secondary carbon to cleave the H-C bond requires 397 kJ/mol and an H-Br is formed. We can look at the BDE of H-Br (=366 kJ/mol) and subtract this value from 397 kJ/mol to get +31 kJ/mol. This positive value tells us that this step in the reaction requires energy (endothermic) and the reactants are more stable than the products. Comparing this with the same situation of a Cl radical we see that this step is exothermic (397 kJ/mol - 431 kJ/mol = -20 kJ/mol). From these values we can conclude that in bromination it is more important that the most stable radical (tertiary) be formed during this step and thus it is more selective than chlorination. This is because less energy is required to form the H-Br and tertiary radical (380 kJ/mol - 366 kJ/mol = +14 kJ/mol). This value is 17 kJ/mol less than the secondary radical formation. Iodine does not even participate in free radical halogenation because the entire reaction is endothermic.
- It is possible to predict the product distribution of different monochloro derivatives resulting from the chlorination of an alkane with non-equivalent hydrogens  . From experimentation it has been determined that the relative rates of chlorination to the primary: secondary: tertiary positions are 1: 3.8: 5 respectively. This makes sense from our understanding of radical stability: tertiary>secondary>primary We can use this information to determine the percentage of each product formed. Lets look at 2-methyl butane:
- -We will label the unique hydrogens:
- a=(CH3)2 b=CH c=CH2 d=CH3
- -Now it is a simple problem of ratios: #H(Rate Factor)/Added Total
- a:6 X 1 = 6 a= 6/21.4 = 28%
- b:1 X 5 = 5 b= 5/21.4 = 23%
- c:2 X 3.7 = 7.4 c= 7.4/21.4 = 35%
- d:3 X 1 = 3 d= 3/21.4 = 14%
- Total= 6 + 5 + 7.4 + 3= 21.4
- Free radical iodination is usually not possible because iodine is less reactive than the other halogens. Free radical halogenation generally proceeds in the following order:
- Carbons with one or more aryl substituents (benzylic positions)
- Carbons with three alkyl substituents (tertiary positions)
- Carbons with two alkyl substituents (secondary positions)
- Carbons with one or zero substituents (primary positions)
- oxygen is a halogenation inhibitor
An example of radical bromination of toluene is given below :
This reaction takes place on water instead of an organic solvent and the bromine is obtained by oxidation of hydrobromic acid with hydrogen peroxide. An incandescent light bulb is sufficient for bromine radical generation.
- ^ Carey, FA. Organic Chemistry, Sixth Edition. New York, NY: McGraw Hill. 2006.
- ^ Peters, W. Chemistry 3421 Lecture Notes. University of Colorado, Denver. Radical Halogenation, Fall 2006.
- ^ Free radical bromination by the H2O2–HBr system on water Ajda Podgorsˇek Stojan Stavber, Marko Zupana, and Jernej Iskraa Tetrahedron Letters 47 (2006) 7245–7247 doi:10.1016/j.tetlet.2006.07.109