Anti-Markovnikov rule describes the regiochemistry where the substituent is bonded to a less substituted carbon, rather than the more substitued carbon. This process is quite unusual, as carboncations which are commonly formed during alkene, or alkyne reactions tend to favor the more substitued carbon. This is because substituted carbocation allow more hyperconjugation and indution to happen, making the carbocation more stable.

## Introduction

This process was first explained by Morris Selig Karasch in his paper: 'The Addition of Hydrogen Bromide to Allyl Bromide' in 1933.1 Examples of Anti-Markovnikov includes Hydroboration-Oxidation and Radical Addition of HBr. A free radical is any chemical substance with unpaired electron. The more substituents the carbon is connected to, the more substituted is that carbon. For example: Tertiary carbon (most substituted), Secondary carbon (medium substituted), primary carbon (least substituted)

Anti-Markovnikov Radical Addition of Haloalkane can ONLY happen to HBr and there MUST be presence of Hydrogen Peroxide (H2O2). Hydrogen Peroxide is essential for this process, as it is the chemical which starts off the chain reaction in the initiation step. HI and HCl cannot be used in radical reactions, because in their radical reaction one of the radical reaction steps: Initiation is Endothermic, as recalled from Chem 118A, this means the reaction is unfavorable. To demonstrate the anti-Markovnikov regiochemistry, I will use 2-Methylprop-1-ene as an example below:

### Initiation Steps

Hydrogen Peroxide is an unstable molecule, if we heat it, or shine it with sunlight, two free radicals of OH will be formed. These OH radicals will go on and attack HBr, which will take the Hydrogen and create a Bromine radical. Hydrogen radical do not form as they tend to be extremely unstable with only one electron, thus bromine radical which is more stable will be readily formed.

### Propagation Steps

The Bromine Radical will go on and attack the LESS SUBSTITUTED carbon of the alkene. This is because after the bromine radical attacked the alkene a carbon radical will be formed. A carbon radical is more stable when it is at a more substituted carbon due to induction and hyperconjugation. Thus, the radical will be formed at the more substituted carbon, while the bromine is bonded to the less substituted carbon. After a carbon radical is formed, it will go on and attack the hydrogen of a HBr, which a bromine radical will be formed again.

### Termination Steps

There are also Termination Steps, but we do not concern about the termination steps as they are just the radicals combining to create waste products. For example two bromine radical combined to give bromine. This radical addition of bromine to alkene by radical addition reaction will go on until all the alkene turns into bromoalkane, and this process will take some time to finish.

## References

1. K. Peter C. Vollhardt, Neil E. Schore; Organic Chemistry: Structure and Function Fifth Edition; W. H. Freeman and Campany, 2007
2. Micheal Vokin; Nuffield Advance Chemistry Student's Book Forth Edition; Person Education Limited, 2004

## Problems

Please give the product(s) of the reactions below:

1. $$\ce{CH_3-C(CH3)=CH-CH_3 + HBr + H_2O_2 \rightarrow}$$
2. $$\ce{CH_3C(CH_3)=CH-CH_3 + HI + H_2O_2 \rightarrow }$$
3. $$\ce{CH_3C(CH_3)=CH-CH_3 + HCl + H_2O_2 \rightarrow }$$
4. $$\ce{CH_3CH=CH-CH3 + HBr + H_2O_2 \rightarrow }$$
5. $$\ce{CH_3C(CH_3)=CH-CH_3 + HBr \rightarrow }$$