Addition of $Br_{2}$ to which of the following gives meso-dibromide product?

Q: Addition of $Br_{2}$ to which of the following gives meso-dibromide product?

To determine why the addition of $ Br_2 $ to trans-2-butene gives a meso-dibromide product, we need to understand the stereochemistry involved in the bromination of alkenes. 1. **Explanation for Option 2 (Trans-2-butene):** - Trans-2-butene is an alkene with the two methyl groups on opposite sides of the double bond. When $ Br_2 $ is added to trans-2-butene, the reaction proceeds via an anti-addition mechanism. This means that the two bromine atoms add to opposite sides of the double bond. - The anti-addition of bromine to trans-2-butene results in a meso compound because the product is symmetrical and has an internal plane of symmetry. The two chiral centers created in the reaction are mirror images of each other, leading to a meso compound, which is optically inactive. 2. **Explanation for why other options are incorrect:** - **Option 1 (Cis-2-butene):** - Cis-2-butene has both methyl groups on the same side of the double bond. When $ Br_2 $ is added, the anti-addition still occurs, but the resulting dibromide is not meso. Instead, it forms a pair of enantiomers because the product lacks an internal plane of symmetry. - **Option 3 (1-butene):** - 1-butene is a terminal alkene, and the addition of $ Br_2 $ results in a dibromide with no chiral centers. The product is simply 1,2-dibromo-butane, which is not a meso compound because it does not have chiral centers or an internal plane of symmetry. - **Option 4 (Propene):** - Propene, like 1-butene, is a terminal alkene. The addition of $ Br_2 $ to propene results in 1,2-dibromo-propane, which also lacks chiral centers and does not form a meso compound. In conclusion, the addition of $ Br_2 $ to trans-2-butene uniquely results in a meso-dibromide product due to the anti-addition mechanism and the symmetrical nature of the resulting compound. This is why Option 2 is the correct answer.

Question

Accepted Answer

To determine why the addition of $ Br_2 $ to trans-2-butene gives a meso-dibromide product, we need to understand the stereochemistry involved in the bromination of alkenes.

1. **Explanation for Option 2 (Trans-2-butene):**
   - Trans-2-butene is an alkene with the two methyl groups on opposite sides of the double bond. When $ Br_2 $ is added to trans-2-butene, the reaction proceeds via an anti-addition mechanism. This means that the two bromine atoms add to opposite sides of the double bond.
   - The anti-addition of bromine to trans-2-butene results in a meso compound because the product is symmetrical and has an internal plane of symmetry. The two chiral centers created in the reaction are mirror images of each other, leading to a meso compound, which is optically inactive.

2. **Explanation for why other options are incorrect:**

- **Option 1 (Cis-2-butene):**
     - Cis-2-butene has both methyl groups on the same side of the double bond. When $ Br_2 $ is added, the anti-addition still occurs, but the resulting dibromide is not meso. Instead, it forms a pair of enantiomers because the product lacks an internal plane of symmetry.

- **Option 3 (1-butene):**
     - 1-butene is a terminal alkene, and the addition of $ Br_2 $ results in a dibromide with no chiral centers. The product is simply 1,2-dibromo-butane, which is not a meso compound because it does not have chiral centers or an internal plane of symmetry.

- **Option 4 (Propene):**
     - Propene, like 1-butene, is a terminal alkene. The addition of $ Br_2 $ to propene results in 1,2-dibromo-propane, which also lacks chiral centers and does not form a meso compound.

In conclusion, the addition of $ Br_2 $ to trans-2-butene uniquely results in a meso-dibromide product due to the anti-addition mechanism and the symmetrical nature of the resulting compound. This is why Option 2 is the correct answer.