Question

69. By treating methyl alcohol with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ at $140°\mathrm{C}$, which of the following products is formed?

• A. ethene
• B. dimethyl ether
• C. methyl ethyl ether
• D. ethyne

Answer 1

The correct option is B

dimethyl ether

Explanation:

• When Methyl alcohol $\left({\mathrm{CH}}_3\mathrm{OH}\right)$ is treated with concentrated Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ at $140°\mathrm{C}$, Dimethyl ether $\left({\mathrm{CH}}_3{\mathrm{OCH}}_3\right)$ and a water $\left({\mathrm{H}}_2\mathrm{O}\right)$molecule are formed.
• This reaction is an example of a dehydration reaction in which a water molecule is eliminated from methyl alcohol.
• The reaction involved is: $2{\mathrm{CH}}_3\mathrm{OH}\underset{140°\mathrm{C}}{\overset{\mathrm{conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}{\mathrm{CH}}_3{\mathrm{OCH}}_3+{\mathrm{H}}_2\mathrm{O}$

Hence, option (B) is the correct answer. By treating Methyl alcohol with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ at $140°\mathrm{C}$, Dimethyl ether is formed.