draw the two major products obtained in the reaction shown. [Solved]

Question 1

Question 2

Question 3

Question 4

Answer to question 1

Solution:

1^st question.

The addition of bromine to reactant 1,3-diene (R₁) results roughly a 50:50 mixture of two major products, 3,4-dibromo-1-ene (P₁) and 1,4-dibromo-2-ene (P₂). Br₂ addition is anti addition.

2^nd question.

The addition of HBr to reactant 1,3-diene (R₂) results 1,2- addition product (P₃) and 1,4-addition product (P₄). 

Refer the below image for the products.

Answer to question 2

Answer to question 3

Step 1

This is an addition reaction wherein, Br2 adds to the alpha,beta-unsaturated system to give the final product.

Step 2

First step is the reaction of Br2 with the diene as shown in the picture. The tertiary carbocation intermediate generated can then directly react with the Br- species to form a 1,2-addition product to generate kinetic product having less substituted double bond in it. On the other hand, the carbocation may undergo a rearrangement to give another secondary carbocation intermediate which then reacts with Br- to form 1,4-addition product having more substituted double bond. This is thermodynamic product. Both the intermediates and the products are shown in the picture.

Thus, the addition reaction of Br2 with the given diene results in the formation of two products, the 1,2-addition product (kinetic product) and the 1,4-addition product (thermodynamic product).

Answer to question 4

Bromination of 2,3-dimethyl-buta-1,3-diene: The addition of B r 2 in the presence of C H 2 C l 2 will yield a dibromo product as shown. The reaction is;