Answer to Question #105432 in Organic Chemistry for Rahul pratp

Question

What is difference between Markownikoff's and Markownikoff's rule.explain with suitable examples

Answer 1

The rule states that with the addition of a protic acid HX to an asymmetric alkene, the acid hydrogen (H) gets attached to the carbon with more hydrogen substituents, and the halide (X) group gets attached to the carbon with more alkyl substituents. Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greatest number of hydrogen atoms while the X component is added to the carbon with the least number of hydrogen atoms.

The same is true when an alkene reacts with water in an addition reaction to form an alcohol which involve formation of carbocations. The hydroxyl group (OH) bonds to the carbon that has the greater number of carbon–carbon bonds, while the hydrogen bonds to the carbon on the other end of the double bond, that has more carbon–hydrogen bonds.

The chemical basis for Markovnikov's Rule is the formation of the most stable carbocation during the addition process. The addition of the hydrogen ion to one carbon atom in the alkene creates a positive charge on the other carbon, forming a carbocation intermediate. The more substituted the carbocation, the more stable it is, due to induction and hyperconjugation. The major product of the addition reaction will be the one formed from the more stable intermediate. Therefore, the major product of the addition of HX (where X is some atom more electronegative than H) to an alkene has the hydrogen atom in the less substituted position and X in the more substituted position

Mechanisms that do not involve a carbocation intermediate may react through other mechanisms that have other regioselectivities not dictated by Markovnikov's rule, such as free radical addition. Such reactions are said to be anti-Markovnikov, since the halogen adds to the less substituted carbon, the opposite of a Markovnikov reaction. Similar to a positive charged species, the radical species is most stable when the unpaired electron is in the more substituted position. The anti-Markovnikov rule can be illustrated using the addition of hydrogen bromide to propene in the presence of benzoyl peroxide. The reaction of HBr with substituted alkenes was prototypical in the study of free-radical additions. Early chemists discovered that the reason for the variability in the ratio of Markovnikov to anti-Markovnikov reaction products was due to the unexpected presence of free radical ionizing substances such as peroxides. The explanation is that HBr produces a Br radical, which then reacts with the double bond. Since the bromine atom is relatively large, it is more likely to encounter and react with the least substituted carbon since this interaction procedure less static interactions between the carbon and the bromine radical. In this case the terminal carbon is a reactant which produces a primary addition product instead of a secondary addition product, in the case of propene.