Draw the 2 products of this halogenation reaction. Use a wedge or dash bond to indicate stereochemistry of substituents on asymmetric centers. ignore inorganic byproducts.

Question

Answer 1

When this alkene reacts with bromine (Br2), a halogenation reaction takes place. The bromine molecule adds across the double bond of the alkene. During the process, two bromine atoms (Br) are added to the carbon atoms that were part of the double bond.

The stereochemistry of the products is determined by the way the bromine atoms add to the carbon atoms. In the case of halogenation reactions, the addition typically follows the anti-Markovnikov rule, which means that one bromine atom adds to the carbon atom with the fewer hydrogen substituents (the less substituted carbon), and the other bromine atoms adds to the more substituted carbon.

In this specific reaction, one bromine will add to the carbon with the CH3 (methyl) group (less substituted carbon), and the other bromine atom will add to the carbon with the CH2CH3 propyl group (more substituted carbon).

The products formed after the halogenation reaction are 4-bromo-4-propylhexane and 4-bromo-4-hexylbromide. The structural formulas of the products, considering the stereochemistry, are as follows:

1. 4-bromo-4-propylhexane (CH3CH2CH(Br)CH2CH2CH3):
2. the bromine atom (Br) is attached to the carbon with the CH3 (methyl) group.
3. the remaining groups (CH2CH2CH3) are on the opposite side of the molecule.
4. 4-bromo-4-hexylbromide (CH3CH2CH2CH(Br)CH2CH3):
5. the bromine atom (Br) is attached to the carbon with the CH2CH3 (propyl) group.
6. the remaining groups (CH3CH2CH2CH3) are on the opposite side of the molecule.

In both products, the bromine atoms are on opposite sides of the molecule, resulting. in an anti-addition stereochemistry. This means that the two bromine atoms are positioned on opposite faces of the molecule with respect to the plane of the double bond.