The compound formed by the reaction of ethyl chloride with a sodium-lead alloy at 400°C is:

Options

A. Tetraethyl lead
B. Sodium ethoxide
C. Ethyl sodium
D. No reaction

Answer

To determine the compound formed by the reaction of ethyl chloride (C2H5Cl) with a sodium-lead alloy at 400°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are ethyl chloride (C2H5Cl) and a sodium-lead alloy (Na + Pb).

2. **Understand the Reaction Conditions**: The reaction occurs at a high temperature of 400°C, which is significant for the reactivity of the components involved.

3. **Reaction Mechanism**: In this reaction, sodium (Na) will react with the chlorine (Cl) from ethyl chloride, leading to the formation of sodium chloride (NaCl). The ethyl group (C2H5) will then combine with lead (Pb) from the alloy.

4. **Write the Reaction**: The overall reaction can be represented as:
\[ 4 \text{C}_2\text{H}_5\text{Cl} + 4 \text{Na} + \text{Pb} \rightarrow \text{Pb(C}_2\text{H}_5\text{)}_4 + 4 \text{NaCl} \]
Here, tetraethyl lead (Pb(C2H5)4) is formed along with sodium chloride (NaCl) as a byproduct.

5. **Identify the Product**: The compound formed is tetraethyl lead, which is represented as Pb(C2H5)4.

6. **Conclusion**: Therefore, the compound formed by the reaction is tetraethyl lead.

### Final Answer:
The compound formed by the reaction of ethyl chloride with a sodium-lead alloy at 400°C is **tetraethyl lead**.

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