A sample of 450 mg of unknown alcohol is added to `CH_(3)MgBr` when 168 of `CH_(4)` at STP is obtained the unknown alcohol is

Options

A. Methanol
B. Ethanol
C. Propan -1- ol
D. Butan -1- ol

Explanation

To solve the problem, we need to determine the molecular weight of the unknown alcohol based on the amount of methane (CH₄) produced when it reacts with methylmagnesium bromide (CH₃MgBr).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between the unknown alcohol (R-OH) and CH₃MgBr produces methane (CH₄) and an alkoxide (R-O⁻MgBr). The reaction can be summarized as:
\[ R-OH + CH₃MgBr \rightarrow R-O⁻MgBr + CH₄ \]
This indicates that one mole of alcohol produces one mole of methane.

2. **Calculate Moles of Methane Produced**:
Given that 168 mL of CH₄ is produced at STP (Standard Temperature and Pressure), we can convert this volume to moles. At STP, 1 mole of gas occupies 22.4 L (or 22400 mL).
\[ \text{Moles of CH₄} = \frac{\text{Volume of CH₄}}{\text{Volume of 1 mole at STP}} = \frac{168 \, \text{mL}}{22400 \, \text{mL}} = 0.0075 \, \text{moles} \]

3. **Relate Moles of Alcohol to Moles of Methane**:
Since 1 mole of alcohol produces 1 mole of methane, the moles of alcohol will also be 0.0075 moles.

4. **Convert Mass of Alcohol to Grams**:
The mass of the unknown alcohol is given as 450 mg. We need to convert this to grams:
\[ \text{Mass of alcohol} = \frac{450 \, \text{mg}}{1000} = 0.450 \, \text{g} \]

5. **Calculate Molecular Weight of Alcohol**:
The molecular weight (M) of the alcohol can be calculated using the formula:
\[ M = \frac{\text{Mass of alcohol}}{\text{Moles of alcohol}} = \frac{0.450 \, \text{g}}{0.0075 \, \text{moles}} = 60 \, \text{g/mol} \]

6. **Identify the Unknown Alcohol**:
The molecular weight of the unknown alcohol is 60 g/mol. We can now check the possible alcohols:
- Methanol (CH₃OH): Molar mass = 32 g/mol
- Ethanol (C₂H₅OH): Molar mass = 46 g/mol
- Propanol (C₃H₇OH): Molar mass = 60 g/mol
- Butanol (C₄H₉OH): Molar mass = 74 g/mol

The only alcohol with a molecular weight of 60 g/mol is Propanol (C₃H₇OH).

### Final Answer:
The unknown alcohol is **Propan-1-ol (C₃H₇OH)**.