Question

Which yields isopropyl methyl ether as the major product with little or no by products?

• A. ${\left(C{H}_3\right)}_{2}CH{O}^{-}N{a}^{+}+C{H}_{3}I⟶$
• B. $C{H}_{3}ONa+{\left(C{H}_3\right)}_{2}CHI$
• C. ${\left(C{H}_3\right)}_{2}CHOH+C{H}_{3}OH\stackrel{H_{2}S{O}_4}{\to }$
• D. All three gives isopropyl methyl ether as the major product

Answer 1

The correct option is A ${\left(C{H}_3\right)}_{2}CH{O}^{-}N{a}^{+}+C{H}_{3}I⟶$

Unsymmetrical ethers are prepared by reaction of an alkyl halide with alcoholic sodium or potassium alkoxide.
Primary alkyl halides are preferred as they are more susceptible to $S_{N}2$ reaction.
Thus the reaction of Sodium isopropoxide with methyl iodide yields isopropyl methyl ether as the major product with little or no by products.
During the reaction of sodium methoxide with isopropyl iodide, hydroiodic acid is eliminated from isopropyl iodide which forms propene as a by product.
During the reaction of isopropyl alcohol with methyl alcohol in presence of sulfuric acid, dehydration of isopropyl alcohol gives propene.