Draw the addition products formed when one equivalent of HBr reacts with 2,4-hexadiene. (Draw a single product for each.)
The Correct Answer and Explanation is :
when one equivalent of HBr reacts with 2,4-hexadiene. I’ll generate the image of the correct products and provide an explanation.
Reaction of 2,4-Hexadiene with HBr
2,4-Hexadiene is a conjugated diene, meaning it has alternating double bonds (C=C) at positions 2 and 4 in a six-carbon chain. The reaction of one equivalent of HBr with a conjugated diene can proceed through two major pathways:
- 1,2-Addition (Direct Addition)
- HBr adds across the C2=C3 double bond, leading to a bromoalkene at position 3.
- 1,4-Addition (Conjugate Addition)
- HBr adds across the conjugated system, leading to a bromoalkene at position 4.
Markovnikov’s Rule
- The H+ (proton) prefers to add to the carbon already bonded to more hydrogens, while the Br− (bromide) attaches to the more substituted carbon in both cases.
Products
- 1,2-Addition Product: 3-Bromo-2-hexene
- 1,4-Addition Product: 4-Bromo-2-hexene
These two regioisomers form based on kinetic (1,2) vs. thermodynamic (1,4) control.
Now, I’ll generate the image of the correct products.
Here is the generated reaction diagram showing the addition of one equivalent of HBr to 2,4-hexadiene, forming the 1,2-addition product (3-Bromo-2-hexene) and the 1,4-addition product (4-Bromo-2-hexene).
Explanation of the Reaction
- 1,2-Addition (Kinetic Product)
- The proton (H⁺) adds to C2 (which has more hydrogens).
- The bromine (Br⁻) attaches to C3, forming 3-Bromo-2-hexene.
- This product forms faster under low temperature (kinetic control).
- 1,4-Addition (Thermodynamic Product)
- The proton adds to C2, leading to resonance stabilization.
- The Br⁻ ultimately attaches at C4, forming 4-Bromo-2-hexene.
- This product is more stable and dominates at higher temperatures (thermodynamic control).