a. Trans-2-butene + HBr + peroxide
b. (Z)-3-methyl-2-pentene + HBr
c. (Z)-3-methyl-2-pentene + HBr + peroxide
d. Cis-3-hexene + HBr
e. Cis-2-pentene + Br2
f. 1-hexene + Br2
The correct answer and explanation is :
Let’s break down each reaction and explain what happens:
a. Trans-2-butene + HBr + peroxide
This reaction undergoes anti-Markovnikov addition due to the presence of peroxides (a radical initiator). In the presence of peroxides, the bromine radical (Br•) adds to the less substituted carbon atom of the double bond, resulting in the formation of 1-bromo-2-butene. This follows the mechanism of a radical chain reaction, which leads to the anti-Markovnikov product rather than the typical Markovnikov product.
- Product: 1-Bromo-2-butene
b. (Z)-3-methyl-2-pentene + HBr
In this reaction, the addition of HBr occurs via Markovnikov’s rule because no peroxides are present. The hydrogen adds to the carbon atom of the double bond that has the greater number of hydrogen atoms (the more substituted carbon receives the bromine atom). The result is a mixture of products, but the major product is the one where the Br adds to the more substituted carbon (i.e., at the 2 position of the pentene).
- Product: 3-Bromo-2-methylpentane
c. (Z)-3-methyl-2-pentene + HBr + peroxide
In this case, the reaction proceeds via anti-Markovnikov addition due to the peroxide’s presence. The bromine radical (Br•) will add to the less substituted carbon of the double bond, leading to the formation of 1-bromo-3-methylpentene. This is the typical product for reactions with HBr and peroxides.
- Product: 1-Bromo-3-methylpentene
d. Cis-3-hexene + HBr
This reaction proceeds according to Markovnikov’s rule, where the hydrogen atom adds to the carbon atom with the greater number of hydrogen atoms, while the bromine attaches to the more substituted carbon. For cis-3-hexene, the bromine will add to the carbon atom at the 2 position of the hexene, leading to the formation of 3-bromo-hexane.
- Product: 3-Bromo-hexane
e. Cis-2-pentene + Br2
The addition of Br2 to cis-2-pentene involves an anti-addition mechanism. The Br2 molecule adds across the double bond in a way that places the two bromine atoms on opposite sides of the original plane of the double bond. This gives a trans-dibromide as the major product.
- Product: 2,3-Dibromo-pentane (trans-configuration)
f. 1-hexene + Br2
Similar to the previous reaction, Br2 adds across the double bond in anti-addition fashion, resulting in a trans-dibromo product. Since 1-hexene is a terminal alkene, the bromine atoms will add to carbons 1 and 2 of the molecule, creating a trans-2,3-dibromohexane.
- Product: trans-2,3-Dibromo-hexane
Summary:
- a. 1-Bromo-2-butene (anti-Markovnikov, with peroxide)
- b. 3-Bromo-2-methylpentane (Markovnikov)
- c. 1-Bromo-3-methylpentene (anti-Markovnikov, with peroxide)
- d. 3-Bromo-hexane (Markovnikov)
- e. 2,3-Dibromo-pentane (trans, anti-addition)
- f. trans-2,3-Dibromo-hexane (trans, anti-addition)
This detailed breakdown should give a clear view of how the reagents and conditions influence the product distribution in each reaction.