Question

State how the following conversions can be carried out :

(i) Ethyl chloride to Ethyl alcohol.

(ii) Ethyl chloride to Ethene.

(iii) Ethene to Ethyl alcohol.

(iv) Ethyl alcohol to Ethene.

Answer 1

Chemical conversions are given below:

(i) Ethyl chloride to Ethyl alcohol.

By treating Ethyl chloride with aqueous Potassium hydroxide ($\mathrm{KOH}$) it will form Ethyl alcohol.

$\underset{\mathrm{Ethyl}\mathrm{chloride}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{l}\right)}+\underset{\mathrm{Potassium}\mathrm{hydroxide}}{\mathrm{KOH}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Ethyl}\mathrm{alcohol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{chloride}}{\mathrm{KCl}\left(\mathrm{s}\right)}$

(ii) Ethyl chloride to Ethene

By heating Ethyl chloride with alcoholic Potassium hydroxide ($\mathrm{KOH}$) it will form Ethene.

$\underset{\mathrm{Ethyl}\mathrm{chloride}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{l}\right)}+\underset{\mathrm{Potassium}\mathrm{hydroxide}}{\mathrm{KOH}\left(\mathrm{alc}\right)}\to \underset{\mathrm{Ethene}}{{\mathrm{CH}}_2={\mathrm{CH}}_2\left(\mathrm{g}\right)}$

(iii) Ethene to Ethyl alcohol

By passing Ethene into concentrated Sulphuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) at $80°\mathrm{C}$ and high pressure.

$\underset{\mathrm{Ethene}}{{\mathrm{CH}}_2={\mathrm{CH}}_2\left(\mathrm{g}\right)}\underset{80°\mathrm{C}}{\overset{\mathrm{conc}{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\underset{\mathrm{Ethanol}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}\left(\mathrm{l}\right)}$

(iv) Ethyl alcohol to Ethene

By heating Ethyl alcohol with conc. Sulphuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) at $170°\mathrm{C}$.

$\underset{\mathrm{Ethanol}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}\left(\mathrm{l}\right)}\underset{170°\mathrm{C}}{\overset{\mathrm{conc}{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\underset{\mathrm{Ethene}}{{\mathrm{CH}}_2={\mathrm{CH}}_2\left(\mathrm{g}\right)}$