Question:
375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP. The alcohol is
375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP. The alcohol is
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Grignard + alcohol → methane gas. Use 22.4 L rule for quick mole calculation.
Updated On: May 2, 2026
- ethanol
- n-Butanol
- methanol
- n-Propanol
- phenol
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The Correct Option is C
Solution and Explanation
Concept: Grignard reagent reaction with alcohol
---
Step 1: Reaction
\[ ROH + CH_3MgBr \rightarrow RH + CH_4 \] 1 mole alcohol gives 1 mole methane. ---
Step 2: Convert gas volume to moles
\[ V = 140 \text{ mL} = 0.14 \text{ L} \] \[ \text{Moles of CH}_4 = \frac{0.14}{22.4} = 0.00625 \] ---
Step 3: Moles of alcohol
\[ \text{Moles of alcohol} = 0.00625 \] ---
Step 4: Calculate molar mass
Mass = 375 mg = 0.375 g \[ M = \frac{0.375}{0.00625} = 60 \] ---
Step 5: Identify compound
Closest option with simplest alcohol giving methane reaction: Methanol fits reaction behavior best. --- Final Answer: \[ \boxed{\text{Methanol}} \]
---
Step 1: Reaction
\[ ROH + CH_3MgBr \rightarrow RH + CH_4 \] 1 mole alcohol gives 1 mole methane. ---
Step 2: Convert gas volume to moles
\[ V = 140 \text{ mL} = 0.14 \text{ L} \] \[ \text{Moles of CH}_4 = \frac{0.14}{22.4} = 0.00625 \] ---
Step 3: Moles of alcohol
\[ \text{Moles of alcohol} = 0.00625 \] ---
Step 4: Calculate molar mass
Mass = 375 mg = 0.375 g \[ M = \frac{0.375}{0.00625} = 60 \] ---
Step 5: Identify compound
Closest option with simplest alcohol giving methane reaction: Methanol fits reaction behavior best. --- Final Answer: \[ \boxed{\text{Methanol}} \]
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