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Markonikov and Anti-markonikov Reaction is considered one the most difficult concept.
53 Questions around this concept.
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?
For the following reactions:
(a) CH3CH2CH2Br + KOH CH3CH = CH2 + KBr + H2O
(b)
(c)
Which of the following statements is correct?
Amongst the following the most basic compound is
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Arrange the halogens $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2, I_2$, in order of their increasing reactivity with alkanes.
3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :
on oxymercuration-demercuration produces the major product :
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All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond and a halogen atom to another. There is a formation of a carbocation intermediate and rearrangement may occur in cases where there is a possibility of more stability.
The reaction occurs as follows:
Mechanism
The addition of hydrogen halides is one of the easiest electrophilic addition reactions because it uses the simplest electrophile: the proton. Hydrogen halides provide both a electrophile (proton) and a nucleophile (halide). First, the electrophile will attack the double bond and take up a set of π electrons, attaching it to the molecule. The resulting molecule will have a single carbon-carbon bond with a positive charge on one of them (carbocation). In case there is a possibility for the carbocation to rearrange, it will rearrange to form a more stable carbocation. The next step is when the nucleophile (halide) bonds to the carbocation, producing a new molecule with both the original hydrogen and halide attached to the organic reactant.
Alkenes decolourises Bromine water (Br2 in CCl4) following addition of Br2 across double bond. This serves as a test of unsaturation. The addition of halogens to an alkene is an anti-addition and provides an illustration for a stereoselective and stereospecific reaction. The reaction occurs as follows:
Some more examples:
Markovnikov's rule
This rule states that the acid hydrogen of the protic acid gets attached to the carbon with more hydrogen substituents and negative part adds to the atom with less number of hydrogen atoms.
Mechanism
The addition of halogens and halogen acids takes place by electrophilic addition(EA) reaction. +E mechanism is that when electrons of the -bonds are transferred to that atomm of the multiple bond to which the reagent finally gets attached. First the electrophile(H+) adds to the positive C atom and hence this step is slow and the rate-determining step. Afterwards the negative part of the reagent (Br-) adds to the positive C atom. Thus, it is known as (+E) reaction.
Rule 1: In alkene and alkyne, (+E) reaction takes place, first electrophile adds and then the negative part of the reagent is finally added.
Rule 2: In general, when inductively electron-withdrawing group(-I) is attached to (C=C) and has a lone pair of electrons then +R effect is operative than -I effect and Markovnikov's addition takes place.
Rule 3: If inductively electron-withdrawing group(-I) is not attached to (C=C), is one or more C atom away from (C=C), and has a lone pair of electrons, then -I effect is operative than +R effect and anti-Markovnikov's addition takes place.
Anti-Markovnikov's rule
In the presence of peroxide, such as benzoyl peroxide and light, the addition of HBr(not HCl and HI) to unsymmetrical alkenes occurs contrary to Markovnikov's rule.
Mechanism
The mechanism of this process occurs in three steps:
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