Draw the addition products formed when one equivalent of HBr reacts with 2,4-hexadiene. Draw a single product for each gnore stereochemical or chiral isomers
The Correct Answer and Explanation is :
When one equivalent of HBr reacts with 2,4-hexadiene, the major products form via electrophilic addition, following Markovnikov’s Rule. Let’s analyze the reaction step by step.
Reaction Mechanism:
- Protonation of the Diene:
- HBr provides H⁺ (proton) and Br⁻ (bromide ion).
- The proton preferentially adds to the carbon of the double bond that already has more hydrogen atoms (Markovnikov’s Rule).
- This leads to the formation of a carbocation intermediate, with the positive charge on the more substituted carbon (more stable).
- Carbocation Rearrangement and Bromide Attack:
- The Br⁻ ion then attacks the carbocation, forming the final product.
- The reaction can occur at either C2=C3 or C4=C5 double bonds in 2,4-hexadiene, leading to two possible products.
Possible Products:
- Addition at the C2=C3 Double Bond:
- H⁺ adds to C2, forming a carbocation at C3.
- Br⁻ then attacks C3, leading to 3-Bromo-2-hexene.
- Addition at the C4=C5 Double Bond:
- H⁺ adds to C4, forming a carbocation at C5.
- Br⁻ then attacks C5, leading to 5-Bromo-4-hexene.
Since stereochemistry is ignored, we only consider regioisomers (positional isomers).
Now, I’ll generate an image of the reaction products.
Here is the reaction diagram illustrating the electrophilic addition of HBr to 2,4-hexadiene, showing the formation of the two major products: 3-Bromo-2-hexene and 5-Bromo-4-hexene. Let me know if you need further clarification!