Question

What is the number of moles of water molecules required to prepare n moles of methane from n moles of methyl magnesium iodide?

• A. n
• B. 2n
• C. 0.5n
• D. 0.1n

Hint:

Logic Tip: The highly polar $C-Mg$ bond makes the alkyl group a strong nucleophile and a very strong base. It will abstract a single proton ($H^+$) from exactly one molecule of water to form the alkane. The $1:1$ ratio applies universally to simple monohalide Grignard reagents and water.

Answer 1

The Correct Option is A

Concept:
Grignard reagents ($R-MgX$) are highly reactive towards compounds containing active/acidic hydrogen atoms, such as water, alcohols, and amines. When a Grignard reagent reacts with water, it undergoes hydrolysis to form an alkane corresponding to the alkyl group of the Grignard reagent.
Step 1: Write the balanced chemical equation for the reaction.
Methyl magnesium iodide ($CH_3MgI$) reacts with water ($H_2O$) to produce methane ($CH_4$) and basic magnesium iodide ($Mg(OH)I$). $$CH_3MgI + H_2O \xrightarrow{\text{dry ether CH_4 + Mg(OH)I$$
Step 2: Determine the stoichiometric ratio.
From the balanced equation, we can see the molar ratio of the reactants and products: $$1 \text{ mole of } CH_3MgI + 1 \text{ mole of } H_2O \longrightarrow 1 \text{ mole of } CH_4$$ This means that exactly 1 mole of water is required to convert 1 mole of methyl magnesium iodide into 1 mole of methane.
Step 3: Calculate the requirement for n moles.
Since the stoichiometric ratio is $1:1:1$, scaling the reaction by a factor of $n$ yields: $$n \text{ moles of } CH_3MgI + n \text{ moles of } H_2O \longrightarrow n \text{ moles of } CH_4$$ Therefore, $n$ moles of water molecules are required.