Question

The major product ‘Y’ in the following reaction is:

(A)

(B)

(C)

(D)

Answer 1

Hint:. In the first step of the reaction, in presence of sodium ethoxide HCl will be removed. So, dehydrohalogenation reactions will take place. In the second place, there will be an addition reaction.

Complete step by step answer:
Let’s see the complete reaction in order to find the answer.
- We have an alkyl halide as a starting material. It is a secondary chloride. It reacts with Sodium ethoxide (NaOEt) which is a strong base. So, a strong base will remove the halogen atom and a hydrogen atom from the $\alpha$-carbon. Thus, we can say that dehydrohalogenation reactions will occur here. Here, the hydrogen atom of tertiary carbon will get removed. The mechanism of the reaction can be given by

Then, the resultant alkene is allowed to react with HBr. We know that hydrogen atom is positively charged and bromine atom is negatively charged in HBr.

- So, hydrogen atoms will be attacked in a way that more stable carbocation is formed. We can see that there is a possibility of formation of a secondary or tertiary carbocation. So, as tertiary carbocation is more stable, it will be formed. This will react with negatively charged bromine ion to give 2-bromo-2-methyl butane as a product. The mechanism can be given as

Thus, we can say that Y in the given reaction is 2-bromo-2-methyl butane.
So, the correct answer is “Option C”.

Note: Remember that in dehydrohalogenation reaction, the elimination of hydrogen takes place in a way that more substituted alkene gets formed. In addition reaction, the more electronegative atom gets attached to the carbon having the least number of hydrogen atoms which is also called Markovnikov’s rule.