Question

Discuss the mechanism of the following reactions: (a) Dehydrohalogenation of alkyl halides.

Answer 1

Hint: On heating alkyl halide with hot conc potassium hydroxide solution, hydrogen halide molecule gets eliminated and there is formation of an alkene.

Complete answer:
In order to answer our question, we need to know about dehydrohalogenation reactions in case of alkyl halides. Dehydrohalogenation is a reaction where the alkyl halide gets treated with potassium hydroxide solution, which is prepared in a very concentrated solution, and also, the temperature of the solution is kept high. It is done so in order to reach the optimum conditions for the reaction to occur and so that the reaction occurs fast.
On adding this potassium hydroxide solution, elimination of a molecule of hydrogen halide occurs and in order to fill the octet, there is a formation of a double bond, which gives the product as an alkene. Let us see the reaction:

Here, we can see that as potassium hydroxide is a strong base, so the hydrogen gets replaced by $O{{H}^{-}}$. But, on a single carbon atom, two molecules of hydrogen and one molecule of oxygen will get unstable. So, when it is heated, these molecules come out as ${{H}_{2}}O$, which is water. Since one valency is missing it is accompanied by a double bond. So, we get the final compound as ethene.

Summing it up, we can write the whole reaction as:

Note:
It is to be noted that apart from water, some amount of $KBr$ is also formed, as the potassium ion reacts with the outgoing bromide ion.