Question

In the following reactions, the major products (A) and (C) are respectively

Answer 1

Hint: The stability of the carbocation is important in predicting the stability of the molecule. The stability order of the carbocation is tertiary carbocation ˃ secondary carbocation ˃ primary carbocation.

Complete Step-by-Step Answer:
 - In the given question, two reactions are given in which we have to find the major product of the reaction.
- Firstly, we will write the first reaction in which firstly the hydroxyl ion will be released due to bond breakage due to heat and a secondary carbocation will be formed.

- Now, as we know, secondary carbocation is less stable than the tertiary carbocation because more alkyl groups are attached with the tertiary carbocation which tends to increase the electron density at the carbocation.
- Now, we will rearrange the position of carbocation by the method of 1,2 hydride shift in which the position of hydrogen atom changes as shown below:

- Now, after the formation of the tertiary carbocation, the hydrogen atom will be released from the adjacent carbon due to which the pi- the bond will be formed.
-And the major product formed is 2-methyl but -2-ene.

-Now, in the second reaction, the compound A is the 2-methyl but -2-ene which undergoes anti- Markovnikov reaction.
-In Anti - Markovnikov reaction, the negatively charged element goes to that carbon atom which has the maximum number of the alkyl group.
-So, here firstly the pi - the bond will break and the addition of the bromine will take place through Anti - Markovnikov rule.

- Here, the bromine attached to the tertiary carbocation is more stable than secondary.

Therefore, the major product A and C are 2-methyl but -2-ene and 2-methyl 2-Bromobutane.

Note: The Anti - markovnikov rule is applicable only in the presence of hydrogen bromide and peroxide. If peroxide is absent in the reaction the negatively charged group attaches through electrophilic addition mechanism.