Propene belongs to the alkene compound group. Propene readily reacts with HBr and produce an alkyl bromide compound. Reaction of propene and HBr belongs to the nucleophilic addition reaction type. This is explained in detail in the mechanism section of the reaction.
In this tutorial, we will learn followings of HBr and propene reactions.
Propene is an unsymmetrical alkene with three carbon atoms. With HBr, propene readily reacts and give 2-bromopropane as the major product and 1-bromopropane as the minor product.
CH3CH=CH2 + HBr → CH3CHBrCH3
CH3CHBrCH3 (2-bromopropane) is given as the major product.
HBr molecule is added across the double bond of propene. Marconikov rule is used to find the locations (to which carbon atom in the double bond) of hydrogen and bromine atoms are added.
Markovnikov's rule explains us how hydrogen and bromine atoms are added to carbon atoms of unsymmetrical alkenes.
Hydrogen atom of HBr prefer to bond to the carbon which contains most number of hydrogen atoms while bromine prefer to bond to other carbon atom which have less number of hydrogen atoms.
This rule will be explained in detail in the example.
As propene and HBr reaction, propene reacts with in same fashion with HI and give products.
HBr molecule is polarized because there is a electronegativity difference between hydrogen and bromine atoms. Also, electrons density of double bond is higher in alkene.
The electrophilic addition of HBr to an unsymmetrical alkene always occurs through the most stable carbocation formation and it is indirectly expressed in Markovnikov's rule.
In the mechanism, you may see, bromide ion is added to a carbon atom as the final step. In the aqueous state, hydroxyl ions (OH-) are also with bromide ions in the reactants medium. So there is a chance to get join hydroxyl ions instead of bromide ions. Also, if aqueous solution contains anions such as chloride, iodide with bromide, one of these anions will be added to the carbon atom in the second step of the mechanism.
When organic peroxides are with reactant, reaction takes place opposite to the Markovnikov rule. This is known as, anti-Markovnikov rule or peroxide effect or Kharasch effect.
Due to this effect, Br atom joints to the carbon which carries the more hydrogen atoms while H joins to the carbon which has less hydrogen atoms in the double bond.
So 1-bromopropane is given as the major product when propene reacts with HBr in the presence of organic peroxides.
Questions asked by students. Ask your question free and find the answer.
Addition of hydrogen bromide to unsymmetrical alkenes in the presence of organic peroxide happens according to the anti-Markovniv rule.
There is a research done to study Oxidation of alkenes with hydrogen peroxide, catalyzed by boron trifluoride. Synthesis of vicinal methoxyalkanols
With HBr, propene gives 2-bromopropane as the product. This product only contain one bromine atom. But with Br2 / CCl4, two bromine atoms are added across the double bond to give dialkyl halide. With Br2 / CCl4, propene gives 1,2-dibromopropane as the dialkyl halide compound.
Propene is an unsymmetrical molecule. But, ethene is a symmetrical molecule. In HBr and unsymmetrical alkene reactions, Markovnikov's rule is applied to find the loctions of hydrogen and bromine atoms which are joined across the double bond.
Propene reacts with HBr and produce CH3CHBrCH3 (2-bromopropane). It is an alkyl haalide compound and this reation is an addition reation. Also, you can treat HBr in the presence of organic peroxides to change the location where bromine atom is joint.
Yes. This reaction happen and give the product as CH3CHBrCH3.