When bromoethane is treated with potassium sulphide, the main product formed is

Options

A. Ethanethiol
B. Ethanol
C. Mustrad gas
D. Thioethyl ethane

Explanation

To solve the question of what product is formed when bromoethane is treated with potassium sulfide, we can follow these steps:

### Step 1: Identify the Reactants
- **Bromoethane** (C2H5Br) is an alkyl halide where a bromine atom is attached to an ethyl group (C2H5).
- **Potassium sulfide** (K2S) is a salt that contains sulfide ions (S^2-), which act as nucleophiles.

### Step 2: Understand the Reaction Mechanism
- In this reaction, the nucleophile (sulfide ion, S^2-) from potassium sulfide will attack the carbon atom that is bonded to the bromine in bromoethane.
- This is a nucleophilic substitution reaction where the bromine atom is replaced by the sulfide ion.

### Step 3: Write the Reaction
- The reaction can be represented as follows:
\[ 2 \text{C}_2\text{H}_5\text{Br} + \text{K}_2\text{S} \rightarrow \text{C}_2\text{H}_5\text{S} + 2 \text{KBr} \]
- Here, two moles of bromoethane react with one mole of potassium sulfide to form **thioethyl ethane** (C2H5S) and two moles of potassium bromide (KBr) as a side product.

### Step 4: Identify the Main Product
- The main product formed from this reaction is **thioethyl ethane** (C2H5S), which is also known as ethyl sulfide.

### Conclusion
- The main product formed when bromoethane is treated with potassium sulfide is **thioethyl ethane**.

### Final Answer
- The answer is **thioethyl ethane** (option D).

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