Step 1: Understanding the Concept:
Alcohols react with Grignard reagents (\( \text{RMgX} \)) to liberate an alkane. This is the Zerewitinoff determination method. One mole of an alcohol with one active hydrogen atom (\( -\text{OH} \)) liberates one mole of methane from methyl magnesium bromide.
: Key Formula or Approach:
1. \( \text{Moles of methane} = \frac{\text{Volume at STP (mL)}}{22400 \text{ mL}} \)
2. \( \text{Molar Mass of Alcohol} = \frac{\text{Mass of sample (g)}}{\text{Moles}} \)
Step 2: Detailed Explanation:
Reaction: \( \text{R-OH} + \text{CH}_{3}\text{MgBr} \longrightarrow \text{CH}_{4}\text{(g)} + \text{Mg(OR)Br} \)
1. Calculate moles of Methane:
\[ \text{Moles of } \text{CH}_{4} = \frac{1.56 \text{ mL}}{22400 \text{ mL/mol}} \approx 6.964 \times 10^{-5} \text{ mol} \]
2. Determine moles of Alcohol:
Since the ratio is 1:1, moles of alcohol \( = 6.964 \times 10^{-5} \text{ mol} \).
3. Calculate Molar Mass of Alcohol:
Mass \( = 4.12 \text{ mg} = 0.00412 \text{ g} \).
\[ \text{Molar Mass} = \frac{0.00412 \text{ g}}{6.964 \times 10^{-5} \text{ mol}} \approx 59.16 \text{ g/mol} \]
4. Identify the Alcohol:
- \( \text{CH}_{3}\text{OH} \): 32 g/mol
- \( \text{C}_{2}\text{H}_{5}\text{OH} \): 46 g/mol
- \( \text{C}_{3}\text{H}_{7}\text{OH} \): \( (3 \times 12) + (7 \times 1) + 17 = 60 \text{ g/mol} \).
- \( \text{C}_{4}\text{H}_{9}\text{OH} \): 74 g/mol
The calculated molar mass (59.16) is closest to 60 g/mol, which corresponds to Propanol.
Step 3: Final Answer:
The unknown alcohol is \( \text{C}_{3}\text{H}_{7}\text{OH} \).