Ethene is a symmetrical alkene which contains two carbon atoms. In the presence of HBr, ethene will give bromoethane as the product. This reaction is an addition reaction and mechanism is explained in this tutorial.
In this tutorial, we will learn followings of HBr and propene reactions.
Ethene is an symmetrical alkene with two carbon atoms. With HBr, ethene readily reacts and give ethyl bromide (bromoethane) as the product.
HBr molecule is added across the double bond of ethene. In this case, Markovnikov rule is not required because ethene molecule is symmetrical around the double bond. Otherwise, we can say number of hydrogen atoms belong to double bond carbon atoms are equal.
HBr molecule is polarized because there is a electronegativity difference between hydrogen and bromine atoms. Also, electrons density around the double bond is higher. (due to higher electrons density, those places can be attracted by positive charges.
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-) exist with bromide ions in the reactants medium. So there is a chance to get join hydroxyl ions to the carbocation 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.
Because ethene is a symmetrical alkene, there is no effect of adding an organic peroxide to change the location of HBr molecule is added. Ethene gives same product with HBr alone and HBr with organic peroxide.
Questions asked by students. Ask your question free and find the answer.
Ethene and ethane are separate type of compounds. Ethhene contains a double bond which has a higher electrons density.
Due to higher electron density around the double bond, it can attack positively charged parts like hydrogen atom in the HBr molecule. But, in ethane there is not such double bond like ethene. Therefore, ethane cannot react with HBr.
In the double bond, there is a pi bond. Usually pi bonds are weaker than sigma bonds. Therefore, electrons of pi bonds are free to move here and there than sigma electrons. Due to those pi bond electrons motion, they create a higher electron density range around the double bond. But, in ethane there are no such pi electrons and electrons density.
If you have dilute sulfuric acid, you can directly prepare ethanol from ethene. But, when dilute sulfuric acid is not available, you can follow this method to prepare ethanol. As the first step, prepare bromomethane from ethene and HBr reaction. Then, bromoethane is reacted with aqueous KOH or NaOH to get ethanol.
With HBr, ethene gives 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, ethene gives 1,2-dibromoethane 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. Therefore, Markovnikov's rule is applied in propene and HBr reaction.
Yes. This reaction happen and give the product as CH3CH2Br.