Ethylene dibromide on heating with metallic sodium in ether yields.

Options

A. ethane
B. ethyene
C. 2-butene
D. 1 - butene

Answer

To solve the question, we need to analyze the reaction of ethylene dibromide with metallic sodium in ether. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the structure of ethylene dibromide
Ethylene dibromide has the molecular formula C2H4Br2. It can be represented as:
\[ \text{BrCH}_2\text{CH}_2\text{Br} \]

### Step 2: Understand the reaction conditions
The reaction involves heating ethylene dibromide with metallic sodium in the presence of ether. Sodium is a reducing agent and will facilitate the removal of bromine atoms.

### Step 3: Determine the reaction mechanism
When ethylene dibromide reacts with sodium, the sodium will replace the bromine atoms. Since there are two bromine atoms on the same carbon, the reaction will lead to the formation of a double bond between the two carbon atoms after the elimination of bromine.

### Step 4: Write the reaction
The reaction can be summarized as follows:
\[ \text{BrCH}_2\text{CH}_2\text{Br} + 2 \text{Na} \rightarrow \text{CH}_2=CH_2 + 2 \text{NaBr} \]
However, since we have two bromine atoms on the same carbon, the product will actually be:
\[ \text{BrCH}_2\text{CH}_2\text{Br} + 2 \text{Na} \rightarrow \text{CH}_3\text{CH} + 2 \text{NaBr} \]

### Step 5: Identify the product
The product formed is butene (specifically, 2-butene) as the double bond forms between the two carbon atoms after the elimination of the bromine atoms.

### Step 6: Name the product
The IUPAC name for the product, which is a four-carbon alkene with a double bond starting at the second carbon, is:
\[ \text{2-butene} \]

### Final Answer:
The product formed when ethylene dibromide is heated with metallic sodium in ether is **2-butene**.

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