Ethylene reacts with `S_(2)Cl_(2)` to give

Options

A. Mustard gas
B. Lewisite
C. Thiophene
D. Ethanethiol

Answer

To solve the question regarding the reaction of ethylene with \( S_2Cl_2 \) (di-sulfur di-chloride), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Ethylene, which has the molecular formula \( C_2H_4 \) or \( CH_2=CH_2 \).
- Di-sulfur di-chloride, which has the formula \( S_2Cl_2 \).

### Step 2: Understand the Reaction Type
This reaction is known as the Levenstein method, which involves the reaction of alkenes (like ethylene) with di-sulfur di-chloride to produce sulfur-containing compounds.

### Step 3: Write the Balanced Reaction
The balanced reaction can be represented as:
\[ 16 \, C_2H_4 + 8 \, S_2Cl_2 \rightarrow 8 \, ClCH_2CH_2S + 8 \, S \]
This indicates that 16 molecules of ethylene react with 8 molecules of \( S_2Cl_2 \) to produce 8 molecules of a chlorinated compound and sulfur.

### Step 4: Identify the Product
The primary product formed in this reaction is:
\[ ClCH_2CH_2S \]
This compound is a precursor to mustard gas, which is a well-known chemical weapon.

### Step 5: Conclusion
The reaction of ethylene with \( S_2Cl_2 \) yields a chlorinated compound that can lead to the formation of mustard gas, which is a yellow oily liquid known for its harmful effects.

### Final Answer
The reaction of ethylene with \( S_2Cl_2 \) produces mustard gas.

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