Question

Predict the product(s) formed by addition of one mole of \(\mathrm{Br}_{2}\) to 2,4 -hexadiene.

Short answer

Answer: The products formed will be a mixture of 2-bromohexene and 4-bromohexene.

Step-by-step solution

Identify the reaction mechanism

The reaction taking place between 2,4-hexadiene and Br2 is an addition reaction, specifically a halogen addition reaction. The diene (2,4-hexadiene) contains double bonds, and in presence of a halogen like Br2, it will undergo an addition reaction to form the dibromide product.

Determine the position of Br2 addition on the double bonds

To predict the product(s) formed, let's first understand the structure of 2,4-hexadiene. It contains six carbon atoms with double bonds at C2 and C4, and single bonds between the other carbons. When Br2 is added to 2,4-hexadiene, the first step involves the attack of the halogen on one of the double bonds (either at C2=C3 or C4=C5). The double bond breaks, and the Br atom attaches to both carbons involved in the bond. This forms a bromonium ion intermediate. The second step involves the attack by the second Br atom on the bromonium ion, forming a dibromoalkane product. Two major products are possible: 1) The Br2 can add across the C2=C3 double bond, and the final product will be 2,4-dibromohexane. 2) The Br2 can add across the C4=C5 double bond, and the final product will be 2,3-dibromohexane.

Predict the final product

Given that only one mole of Br2 is added to the 2,4-hexadiene, the reaction will only have enough Br2 to react with one of the two double bonds. Therefore, the product formed will be a mixture of mono-substituted brominated products, specifically 2- and 4-bromohexene. The probability of forming a dibrominated product is very low in this case. So, the predicted product(s) formed when one mole of Br2 is added to 2,4-hexadiene will be a mixture of 2-bromohexene and 4-bromohexene.