`CH_3-CH_2SHunderset((ii)"ethylene oxide")overset((i)CH_3O^(ᶱ))to`Product, Product is :

Options

A. `CH_3-CH_2-S-CH_2-CH_2-OH`
B. `CH_3-CH_2O^(Theta)`
C. `CH_3-CH_2-O-CH_2-CH_2-OH`
D. None of these

Explanation

To solve the given reaction sequence, we will follow these steps:

### Step 1: Identify the Reactants
The reactants given are:
- **CH₃-CH₂-SH** (Ethanethiol)
- **CH₃O⁻** (Methoxide ion)
- **Ethylene oxide** (C₂H₄O)

### Step 2: Deprotonation of Ethanethiol
The first step involves the reaction of ethanethiol (CH₃-CH₂-SH) with the methoxide ion (CH₃O⁻). The methoxide ion acts as a base and abstracts a proton (H⁺) from the thiol group (-SH) of ethanethiol.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{SH} + \text{CH}_3\text{O}^- \rightarrow \text{CH}_3\text{CH}_2\text{S}^- + \text{CH}_3\text{OH} \]

After this step, we have the thiolate ion:
- **CH₃-CH₂-S⁻** (Ethanethiolate ion)

### Step 3: Nucleophilic Attack on Ethylene Oxide
Next, the thiolate ion (CH₃-CH₂-S⁻) acts as a nucleophile and attacks the ethylene oxide (C₂H₄O). Ethylene oxide is a three-membered cyclic ether that is strained and reactive.

**Mechanism:**
1. The nucleophile (thiolate ion) attacks one of the carbon atoms in the epoxide ring.
2. This leads to the opening of the ring due to the strain.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{S}^- + \text{C}_2\text{H}_4\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{S-CH}_2\text{CH}_2\text{OH} \]

### Step 4: Final Product Formation
After the nucleophilic attack and ring opening, we end up with the final product:
- **CH₃-CH₂-S-CH₂-CH₂-OH** (2-(Ethyldisulfanyl)ethanol)

### Final Answer
The final product of the reaction sequence is:
**CH₃-CH₂-S-CH₂-CH₂-OH**

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