How To Make APAAR ID For NEET 2025 Registration?

Chemical Properties of Alkenes MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:24 PM | #NEET

Quick Facts

  • Markonikov and Anti-markonikov Reaction is considered one the most difficult concept.

  • 27 Questions around this concept.

Solve by difficulty

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 \rightarrow CH3CH = CH2 + KBr + H2O

(b) 

(c) 

Which of the following statements is correct?

Identity Z in the sequence of reactions :

\text{CH}_{3}\text{CH}_{2}\text{CH}=\text{CH}_{2} \xrightarrow{{\text{HBr}/\text{H}_2 \text{O}_2 }}\text{Y}\xrightarrow{{\text{C}_{2}\text{H}_5 \text{ONa}}}\text{Z}

NEET 2025: SyllabusMost Scoring concepts NEET PYQ's (2015-24)

NEET PYQ's & Solutions: Physics | ChemistryBiology

In the reaction with HCl, an alkene reacts in accordance with the Markovnikov's rule, to give a product 1-chloro-1-methylcyclohexane. The possible alkene is:

Concepts Covered - 3

Hydrohalogenation of Alkenes

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 the other.

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. This is basically the reverse of the last step in the  E1 reaction. The resulting molecule will have a single carbon-carbon bond with a positive charge on one of them (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.

Halogenation of Alkenes

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:


Markonikov and Anti-markonikov Reaction

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 \pi-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:

  1. Chain initiation: 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.

     
  2. Chain propagation: The Bromine Radical will go on and attack theless 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 HBr, and thus a bromine radical will be formed again.
  3. Chain termination: In termination step, two bromine radicals 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.

Study it with Videos

Hydrohalogenation of Alkenes
Halogenation of Alkenes
Markonikov and Anti-markonikov Reaction

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Halogenation of Alkenes

Chemistry Part II Textbook for Class XI

Page No. : 388

Line : 43

Markonikov and Anti-markonikov Reaction

Chemistry Part II Textbook for Class XI

Page No. : 390

Line : 1

E-books & Sample Papers

Get Answer to all your questions

Back to top