Question

Mono addition of HBr to unsymmetrical dienes generally produces four products - two 1,2-addition products and two 1,4-addition products. Draw the reactant diene that forms the following four products, and identify the addition route that forms each product.

Answer 1

Concepts and reason

The given reaction involves the addition of hydrogen bromide to unsymmetrical dienes. The hydrogen halides undergo electrophilic addition reaction with hydrogen halides via 1,2 and 1,4 mechanisms.

Fundamentals

In dienes, the double bonds are present alternatively that is a double bond followed by a single bond and then next double bond. The high reactivity of dienes is due to the presence of the $\pi$ bonds present.

The following steps are involved in the addition reaction:

• Firstly, the formation of carbocation takes place via the addition of electrophile to the double bond. The carbocation formed is resonance stabilized, being an allylic carbocation.

• The next step involves the attack of the nucleophile on carbocation to give two different addition products via 1,2 and 1,4 addition respectively.

The allyl halide products are formed via following addition as shown below:

The removal of HBr as shown below gives the structure of required diene:

Ans:

Answer 2

Answer 3

СНЗ

Answer 4

Reactant diene

1) 1,4 addition

2) 1,2 addition

3) 1,2 addition

4)1,4 addition