Solveeit Logo
Login

Question

Question: Identify “A” and “B” in the following reaction: \(C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrighta...

Identify “A” and “B” in the following reaction:
CH3−CH=CH2→HBrA→alcKOHBC{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrightarrow{{alcKOH}}B

Explanation

Solution

In order to solve the given problem first we will understand the hydrohalogenation reaction of the given compound which is a propene. After seeing the hydrohalogenation of propene we can get compound “A” of the equation further after recognizing compound “A” we will see its reaction with alcoholic potassium hydroxide to find the other product “B” in the equation.

Complete step by step answer:
Given that the starting substance of the equation is propene whose chemical formula is CH3−CH=CH2C{H_3} - CH = C{H_2} and it is an Alkene.
In the first part of the reaction it undergoes hydrohalogenation with hydrogen bromide whose chemical formula is HBrHBr.
In order to solve the first part of the question let us see the hydrohalogenation of propene.
As we know that in the process of hydrohalogenation the double bond of alkene breaks down to form alkane. So let us see the chemical reaction.
CH3−CH=CH2→HBrCH3−CH2−CH2−Br(A)C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}C{H_3} - C{H_2} - C{H_2} - Br\left( A \right)

In the above reaction we can see that when propene (an alkene ) reacts with Hydrogen bromide, it converts into bromo propane as a product.
So the compound “A” is CH3−CH2−CH2−Br(A)C{H_3} - C{H_2} - C{H_2} - Br\left( A \right) and its chemical name is bromo propane.
Now as we know the compound “A” is bromopropane so let us see the reaction of bromopropane with alcoholic potassium hydroxide.
As we know that when an halo propane undergoes reaction with alcoholic potassium hydroxide, the reaction is β\beta (beta) elimination reaction in which we get an alkene as product along with some by product.

So let us see the β\beta - elimination reaction of bromopropane in the presence of alcoholic potassium hydroxide.
CH3−CH2−CH2−Br(A)→alcKOHCH3−CH=CH2(B)+KBr+H2OC{H_3} - C{H_2} - C{H_2} - Br\left( A \right)\xrightarrow{{alcKOH}}C{H_3} - CH = C{H_2}\left( B \right) + KBr + {H_2}O

In the above reaction which is part 2 of the main reaction we can see that when compound “A” or bromopropane reacts with alcoholic potassium hydroxide after the beta elimination reaction propene(B) is formed as the main product along with some by product. So, compound “B” is CH3−CH=CH2(B)C{H_3} - CH = C{H_2}\left( B \right) and its name is propene.
Hence, the compounds “A” and “B” are bromopropane and propene respectively.

Note: In order to solve such types of problems, students must break the overall larger reaction into smaller sub parts and single reactions. Students must remember some common name reactions, some of which are mentioned above. Also students must avoid the by products in such reactions as the problem involves only the main product.