Identify the product obtained in following reaction. $\mathrm{n} \mathrm{CH}_3 \mathrm{MgI}+\mathrm{H}_2…
Identify the product obtained in following reaction.
$\mathrm{n} \mathrm{CH}_3 \mathrm{MgI}+\mathrm{H}_2 \mathrm{O} \stackrel{\text { dry ether }}{\longrightarrow} \text { product }$
$\mathrm{n} \mathrm{MgI}$ and $\mathrm{n} \mathrm{CH}_4$
$\frac{\mathrm{n}}{2} \mathrm{C}_2 \mathrm{H}_6$
$\mathrm{n} \mathrm{CH}_3 \mathrm{OH}$ and $\mathrm{n} \mathrm{MgI}$
$\mathrm{n} \mathrm{CH}_4$ and $\mathrm{n} \mathrm{MgI}(\mathrm{OH})$
Solution
In this reaction, the Grignard reagent $\mathrm{nCH}_3 \mathrm{MgI}$ reacts with water ($\mathrm{H}_2 \mathrm{O}$) in dry ether. The Grignard reagent acts as a strong nucleophile and reacts with water to produce methane ($\mathrm{CH}_4$) and magnesium hydroxide iodide ($\mathrm{MgI(OH)}$). Therefore, the products of the reaction are $\mathrm{nCH}_4$ and $\mathrm{nMgI(OH)}$.