Question

Explain markownikoves and anti markownikoves addition with an example

Answer 1

$\huge \sf {\orange {\underline {\pink{\underline {A᭄ɴsᴡᴇʀ࿐ \ :- }}}}}$

A᭄ɴsᴡᴇʀ࿐ :−

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.

hope it's helpful to you ☺️♥️☺️

Answer 2

$\huge \red {\colorbox{gold}{♡Required \: Answer♡}}$

When a protic acid (HX) is added to an asymmetric alkene, the acidic hydrogen attaches itself to the carbon having a greater number of hydrogen substituents whereas the halide group attaches itself to the carbon atom which has a greater number of alkyl substituents.

To simplify the rule, it can also be stated as – “Hydrogen is added to the carbon with the most hydrogens and the halide is added to the carbon with least hydrogens”.

An example of a reaction that observes Markovnikov’s rule is the addition of hydrobromic acid (HBr) to propene, which is shown below.

Examples of Markovnikov and Anti-Marknovnikov Addition Reactions

The Hydration of Alkenes

When alkenes are treated with certain aqueous acids (usually sulfuric acid), the resulting electrophilic addition reaction yields an alcohol as the product. The regioselectivity of such reactions can be predicted by Markownikoff’s rule. Therefore, these reactions can be classified as Markovnikov reactions. In the hydration of alkenes, the H+ ion acts as an electrophile and attacks the alkene to generate a carbocation intermediate (the intermediate with greater stability is protonated). The subsequent nucleophilic attack on the carbocation by water molecules forms an oxonium ion, which is deprotonated to afford the required alcohol product.

The Hydroboration/Oxidation of Alkenes

When alkenes are treated with borane (BH3) in the presence of hydrogen peroxide or sodium hydroxide, an alcohol is obtained as the final product. In this electrophilic addition reaction, the boron atom acts as an electrophile. This reaction does not obey Markovnikov’s rule and can, therefore, be classified as an anti-Markovnikov reaction.