`CH_(3)-C-=C-CH_(3)underset((2)Br_(2))overset((1)H_(2)//Pt)rarrX`

Options

A. (d)-2, 3-Dibromobutane
B. (l)-2, 3-Dibromobutane
C. (dl)-2, 3-Dibromobutane
D. meso-2, 3-Dibromobutane

Answer

To solve the given reaction step by step, let's break it down into two main parts: hydrogenation and bromination.

### Step 1: Identify the Starting Compound
The starting compound is CH₃-C≡C-CH₃, which is 2-butyne. It has a triple bond between the second and third carbon atoms.

### Step 2: Hydrogenation
The first step is hydrogenation, which is the addition of hydrogen (H₂) to the triple bond in the presence of a catalyst (Pt). This process converts the triple bond into a double bond, resulting in the formation of an alkene.

- The reaction can be represented as:
\[ \text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 + \text{H}_2 \xrightarrow{\text{Pt}} \text{CH}_3-\text{C}=\text{C}-\text{CH}_3 \]

- The product formed is 2-butene (specifically, it can be either cis or trans depending on the addition of hydrogen).

### Step 3: Bromination
The next step is bromination, which involves the addition of bromine (Br₂) to the double bond formed in the previous step. This is an electrophilic addition reaction.

- The reaction can be represented as:
\[ \text{CH}_3-\text{C}=\text{C}-\text{CH}_3 + \text{Br}_2 \rightarrow \text{Br}-\text{C}-\text{C}-\text{Br} \]

- Since bromine adds across the double bond, we can expect that the bromine atoms will add to opposite sides of the double bond, leading to the formation of a racemic mixture of 2,3-dibromobutane.

### Step 4: Final Product
The final product of the reaction is a racemic mixture of 2,3-dibromobutane, which means that both enantiomers (D and L forms) are produced.

### Summary of the Reaction
The overall reaction can be summarized as:
\[ \text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 \xrightarrow{\text{H}_2/\text{Pt}} \text{CH}_3-\text{C}=\text{C}-\text{CH}_3 \xrightarrow{\text{Br}_2} \text{2,3-Dibromobutane (racemic mixture)} \]