Question

Ethene can be prepared by the reaction of ethanol with

• A. Hot conc. $H_2{SO}_4$
• B. Alkaline ${KMnO}_4$
• C. Sodium metal
• D. ${NaHCO}_3$

Answer 1

The correct option is A

Hot conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$

Explanation for correct option:

(A) Hot conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$

• A dehydration reaction is a reaction in which water molecule is eliminated from reactant molecule.
• Ethene can be prepared by the reaction of ethanol $\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$ with hot conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$.
• The chemical reaction is as follows:

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\stackrel{\mathrm{Hot}\mathrm{conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }{\mathrm{CH}}_2={\mathrm{CH}}_2+{\mathrm{H}}_2\mathrm{O}$

• The above reaction is an example of a dehydration reaction as water molecule is eliminated here.
• Here, concentrated sulfuric acid acts as a dehydrating agent and eliminates water from ethanol to form ethene.

Explanation for incorrect options:

(B) When ethanol reacts with alkaline ${\mathrm{KMnO}}_4$, the formation of ethanoic acid takes place in place of ethene. The chemical reaction is as follows:-

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\stackrel{\mathrm{Alk}.{\mathrm{KMnO}}_4}{\to }{\mathrm{CH}}_3\mathrm{COOH}$

(C) Ethene can not be prepared by the reaction of ethanol with sodium metal. The chemical equation for the reaction of ethanol with sodium is as follows:-

$2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+2\mathrm{Na}\to 2{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{ONa}+{\mathrm{H}}_2$

(D) Ethene can not be prepared by the reaction of ethanol with ${\mathrm{NaHCO}}_3$ because ethanol is quite insoluble in ${\mathrm{NaHCO}}_3$. Ethanol is a weak acid and sodium bicarbonate $\left({\mathrm{NaHCO}}_3\right)$ is a weak base. Thus, Ethanol and ${\mathrm{NaHCO}}_3$ do not react with each other.

Hence, the correct option is (A). Ethene can be prepared by the reaction of ethanol $\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$ with hot conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$.