Ethyl bromide reacts with lead sodium alloy to form:

Options

A. Tetraethyl lead
B. Ethyl sodium
C. Ethane
D. Ethene

Answer

To solve the question of what ethyl bromide reacts with lead-sodium alloy to form, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are ethyl bromide (C2H5Br) and lead-sodium alloy (Pb-Na).

**Hint:** Look for the chemical formula of ethyl bromide and understand what lead-sodium alloy consists of.

### Step 2: Understand the Reaction Mechanism
When ethyl bromide reacts with lead-sodium alloy, the sodium in the alloy acts as a reducing agent. It helps in the removal of the bromine atom from ethyl bromide, leading to the formation of an ethyl radical.

**Hint:** Recall that sodium can replace halogens in haloalkanes through a reduction reaction.

### Step 3: Formation of Tetraethyl Lead
The ethyl radical generated can then react with lead (Pb) from the alloy. The reaction leads to the formation of tetraethyl lead (Pb(C2H5)4). This compound is formed when four ethyl groups bond to a single lead atom.

**Hint:** Consider how many ethyl groups can bond to one lead atom and what the resulting compound would be.

### Step 4: By-products of the Reaction
As a by-product of this reaction, sodium bromide (NaBr) is formed. This occurs as the bromine atom is replaced by the ethyl group.

**Hint:** Think about the halogen that is being replaced and what salt is formed from this substitution.

### Step 5: Write the Balanced Reaction
The overall balanced reaction can be summarized as:
\[ 4 \text{C}_2\text{H}_5\text{Br} + \text{Pb-Na} \rightarrow \text{Pb(C}_2\text{H}_5\text{)}_4 + 4 \text{NaBr} \]

**Hint:** Ensure that the number of atoms of each element is the same on both sides of the equation.

### Conclusion
The major product formed from the reaction of ethyl bromide with lead-sodium alloy is tetraethyl lead (option A).

**Final Answer:** Tetraethyl lead (A).