The correct answer is Pd-BaSO4/H2 then Br2/CCl4.

Here's the step-by-step explanation:

1. 2-butyne is a terminal alkyne, which means it has a triple bond at the end of the carbon chain.

2. The first step is to add Pd-BaSO4/H2. This is a Lindlar catalyst and hydrogen (H2), which will selectively reduce the alkyne (triple bond) to a cis-alkene (double bond). This means that both of the hydrogens will add to the same side of the molecule.

3. The second step is to add Br2/CCl4. This will perform a halogenation reaction, where the bromine atoms will add across the double bond. Because the alkene is cis, the two bromine atoms will add to the same side of the molecule.

4. The product of these two steps is a meso compound, which is a type of stereoisomer that has an internal plane of symmetry. This means that although it has chiral centers, the molecule is overall achiral (not chiral). This is because the two halves of the molecule are mirror images of each other.

The other options will not produce a meso product. Na/NH3(l) will reduce the alkyne to a trans-alkene, and Cl2/CCl4 and OsO4/NaHSO3 will not react with an alkyne.

Question

The correct answer is Pd-BaSO4/H2 then Br2/CCl4.

Here's the step-by-step explanation:

1. 2-butyne is a terminal alkyne, which means it has a triple bond at the end of the carbon chain.

2. The first step is to add Pd-BaSO4/H2. This is a Lindlar catalyst and hydrogen (H2), which will selectively reduce the alkyne (triple bond) to a cis-alkene (double bond). This means that both of the hydrogens will add to the same side of the molecule.

3. The second step is to add Br2/CCl4. This will perform a halogenation reaction, where the bromine atoms will add across the double bond. Because the alkene is cis, the two bromine atoms will add to the same side of the molecule.

4. The product of these two steps is a meso compound, which is a type of stereoisomer that has an internal plane of symmetry. This means that although it has chiral centers, the molecule is overall achiral (not chiral). This is because the two halves of the molecule are mirror images of each other.

The other options will not produce a meso product. Na/NH3(l) will reduce the alkyne to a trans-alkene, and Cl2/CCl4 and OsO4/NaHSO3 will not react with an alkyne.

Knowee AI · Accepted Answer

The correct answer is Pd-BaSO4/H2 then Br2/CCl4.

Here's the step-by-step explanation:

1. 2-butyne is a terminal alkyne, which means it has a triple bond at the end of the carbon chain.

2. The first step is to add Pd-BaSO4/H2. This is a Lindlar catalyst and hydrogen (H2), which will selectively reduce the alkyne (triple bond) to a cis-alkene (double bond). This means that both of the hydrogens will add to the same side of the molecule.

3. The second step is to add Br2/CCl4. This will perform a halogenation reaction, where the bromine atoms will add across the double bond. Because the alkene is cis, the two bromine atoms will add to the same side of the molecule.

4. The product of these two steps is a meso compound, which is a type of stereoisomer that has an internal plane of symmetry. This means that although it has chiral centers, the molecule is overall achiral (not chiral). This is because the two halves of the molecule are mirror images of each other.

The other options will not produce a meso product. Na/NH3(l) will reduce the alkyne to a trans-alkene, and Cl2/CCl4 and OsO4/NaHSO3 will not react with an alkyne.

Which of the following sets of reagent when applied sequentially on 2-butyne will produce a meso product?Cl2/CCl4 then Br2/CCl4Na/NH3(l) then Br2/CCl4Pd-BaSO4/H2 then Br2/CCl4pd:BaSO4/H2 then OsO4/NaHSO3

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