In the reaction `CH_3-CH_2-OHunderset((ii)LiAiH_4)overset((i)TsCl)rarr(X)` , (X) will be

Options

A. `CH_2=CH_2`
B. `CH_3-CH_2-CH_2-CH_3`
C. `CH_3-CH_3`
D. `CH_3-CH=CH-CH_3`

Answer

To solve the problem step by step, we will analyze the given reaction involving ethanol, tosyl chloride, and lithium aluminum hydride.

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Ethanol (CH₃-CH₂-OH)
- Tosyl chloride (TsCl)

### Step 2: Reaction with Tosyl Chloride
When ethanol reacts with tosyl chloride (TsCl), it undergoes a substitution reaction where the hydroxyl group (-OH) of ethanol is converted into a tosylate group (-OTs). This is a good leaving group.

The reaction can be written as:
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{TsCl} \rightarrow \text{CH}_3\text{CH}_2\text{OTs} + \text{HCl} \]

### Step 3: Formation of Tosylate
The product of the reaction with tosyl chloride is:
- Ethyl tosylate (CH₃-CH₂-OTs)

### Step 4: Reaction with Lithium Aluminum Hydride
Next, we treat the ethyl tosylate with lithium aluminum hydride (LiAlH₄), which is a strong reducing agent. In this step, the tosylate group (-OTs) acts as a good leaving group, and a hydride ion (H⁻) from LiAlH₄ will attack the carbon atom that was bonded to the tosylate group.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{OTs} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_3 + \text{OTs}^- \]

### Step 5: Identify the Final Product
The final product after the reaction with LiAlH₄ is:
- Ethane (CH₃-CH₃)

### Conclusion
Thus, the product (X) formed from the reaction is ethane (CH₃-CH₃).

### Final Answer
(X) = Ethane (CH₃-CH₃)

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