As suggested by Markovnikov, the Anti Markovnikov mechanism is one of the few reactions following free radical mechanisms in organic chemistry in place of electrophilic addition. This reaction is seen only with HBr (Hydrogen Bromide), but not with HCl (Hydrochloric Acid) or HI (Hydrogen Iodide).
Explanation
Alkanes belong to an unsaturated hydrocarbon group, which means one molecule of an alkane contains at least one double bond. Due to the existence of 'pi' electrons, alkenes show the anti-markovnikov's addition reactions in which the electrophile attacks the carbon-carbon double bond to create additional products. When Hydrogen Bromide (HBr) is added to unsymmetrical alkenes in the peroxide presence, 1-bromopropane is formed, contrary to 2-bromopropane.
Better, this reaction is called either the Anti-Markovnikov addition or the Kharash effect after the name of M. S. Kharash, who discovered it first. Also, this reaction is known as either Peroxide or Kharash effect.
When any polar molecule is added to an unsymmetrical alkene in the presence of any organic peroxide, the negative part of the molecule is added to that carbon atom connected to more Hydrogen atoms than the other unsaturated carbon atom. This is known as the peroxide effect.
CH3-CH=CH2+HBr \[ \xrightarrow[]{(ORGANIC \, PEROXIDE)} \] CH3-CH(H)-CH2(Br)
The Mechanism of Anti-Markovnikov's Addition Rule With an Example
The most common type of Anti-Markovnikov addition mechanism is free radical addition. This type of mechanism is applicable only for HBr - not HCl or HI - with either hydrogen peroxide (H2O2) or benzoyl peroxide (C14H10O4). Peroxide is an essential part. It acts as a catalyst that breaks HBr into Br and H radicals (any chemical species with one unpaired electron is referred to as a radical).
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Here The Phases (a) and (b) are explained as,
Br added to 1° carbanion for stability.
2° carbanion reacts with another molecule of HBr to give the major product
The Br (Bromine) radical attacks the alkene first. It attacks the less substituted carbon because the carbon radical so formed will be at more substituted carbon for greater stability. Then, the radical carbon attacks the hydrogen of another HBr (Hydrogen Bromide) molecule, liberating another Br (Bromine) radical, and hence the reaction is carried forward.
Anti Markovnikov Halogenation
Halogenation of alkanes refers to the addition of a halogen to the C = C double bond of an alkane. An Anti-Markovnikov halogenation is a free radical reaction of the hydrogen bromide to an alkene.
In a Markovnikov addition of HBr (Hydrogen Bromide) to propene, the H (Hydrogen) adds to the C atom with more H atoms. The resultant product is 2-bromopropane pictured below.
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In the presence of peroxides, H adds to the C atom with a lower count of H atoms. This refers to anti-Markovnikov addition. The resultant product is 1-bromopropane.
The reason for the Anti-Markovnikov addition is, it is the Br (Bromine) atom that attacks the alkene. It attacks the C (Carbon) atom with the most H (Hydrogen) atoms. So the H adds to the C atom with the less number of H atoms.
Anti-Markovnikov Rule
Anti-Markovnikov rule narrates the regiochemistry where the substitute is bonded to fewer carbon substitutes instead of more carbon substitutes. One such process is quite unusual, as carbocations commonly formed during alkene, or alkyne reactions, tend to favour more substituted carbon. This happens because the substitution of carbocation allows for more hyperconjugation and induction to make the carbohydrate more stable.
In his paper, Morris Selig Karasch explained this process for the first time, 'Addition of Hydrogen Bromide to Allyl Bromide' in 1933.1 Examples of Anti-Markovnikov rule include the Radical Addition of HBr and Hydroboration-Oxidation. A free radical is any chemical substance with an unpaired electron. Here, the resultant carbon forms based on more carbon substituents. Some of the examples of the Anti-Markovnikov rule are Primary carbon (least substituted), Secondary carbon (medium substituted), and Tertiary carbon (most substituted).
Anti-Markovnikov Radical addition of Haloalkane will only happen to HBr, and Hydrogen Peroxide ( H2O2) MUST be there. Hydrogen Peroxide is necessary for this process because it is the chemical that starts off the chain reaction at the initiation step itself. HI (Hydrogen Iodide) and HCl (Hydrochloric Acid) can't be used in radical reactions. In their radical reaction, one of the radical reaction steps is initiation Endothermic, as recalled from Chem 118A, which means the reaction is unfavourable.
To demonstrate the example of the anti-Markovnikov rule of regiochemistry, let us use 2-Methylpropene as an example below.
Initiation Steps
Hydrogen Peroxide is an unstable molecule. If we shine it or heat it with sunlight, two free OH radicals will be formed. These OH radicals will move on and attack HBr, which will take the Hydrogen and produce a Bromine radical. Hydrogen radicals do not form because they appear to be highly unstable with only one electron. Thus bromine radical will be readily formed, which is more stable.
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Propagation Steps
The Bromine Radical will move on and attack the less substituted carbon of alkene. This happens because, after the bromine radicals attack the alkene, a carbon radical will be formed. A carbon radical is highly stable when it relies on a more substituted carbon due to hyperconjugation and induction. Thereby, the radical will be established at the more substituted carbon, whereas the bromine is bonded to the less substituted carbon. Once a carbon radical is formed, it will move on and attack the Hydrogen of an HBr, which a bromine radical will again be formed.
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Termination Steps
There are the Termination Steps also. But we are not concerned about the termination steps as they are just the radicals combining to create the waste products. For example, two bromine radicals combine to produce bromine. This radical addition of bromine to an alkene by a radical addition reaction will move on until all the alkene turns into bromoalkane. And this process will take a bit more time to finish.
FAQs on Anti Markovnikov Addition Reaction
1. Does Anti-Markovnikov addition behaviour extend only to alkenes?
Anti-Markovnikov behaviour can extend to other chemical reactions than additions to alkenes, but relatively they are rare.
For example, the hydration of phenylacetylene results in acetophenone in a Markovnikov addition. But hydration that happens in the presence of a complex ruthenium catalyst gives phenylacetaldehyde in an Anti-Markovnikov addition.
Also, a few other anti-Markovnikov reactions involving transition metal catalysts are known.
2. What is more reactive: Alkynes or Alkenes?
To make it clear, Alkynes are more reactive than alkenes.
The reactivity is represented by,
Alkenes > Alkynes > Alkanes.
Alkenes have only one pi-bond between two (or more) carbon atoms and an sp2-sp2 hybridised orbital bonding (a sigma bond). The pi-bond easily can be broken to free the valence shell electrons to combine with other atoms, but as the number of pi bonds is lesser than Alkynes, they result in less reactivity.
Alkynes have two pi-bonds between two or more carbon atoms, along with an sp-sp hybridised orbital bonding (a sigma bond). The pi-bonds easily can be broken to free the valence shell electrons to combine with other atoms. Thus, its reactivity is more than alkanes but lesser than alkenes. The reason suggested for it is the presence of a triple bond between two carbon atoms. Also, the bond length decreases when compared to alkenes. So it becomes difficult for alkynes to stay highly reactive.
Alkanes have no pi-bonds between the carbon atoms. They only have an sp3-sp3 hybridised orbital bonding (a sigma bond). If Alkynes are to react, first, the sigma bonds with Hydrogen (s-sp3 sigma bonds) have to be broken, and then only the other atoms can combine. But, this entails more energy. Alkanes, thereby, are the least reactive among the three.
3. How does the Markovnikov rule work?
When an asymmetric alkene is treated with a protic acid HX or another polar reagent, the acid hydrogen (H) group or electropositive component attaches to the carbon with more hydrogen substituents, whereas the halide (X) group or electronegative component attaches to the carbon with more alkyl substituents.
