Question

Ethylene from ethyl bromide is obtained by treating it with

• A. Hydrogen
• B. Alcoholic caustic potash
• C. Aqueous caustic potash
• D. Aqueous caustic soda

Answer 1

The correct option is B

Alcoholic caustic potash

Explanation for correct option:

Option (b) Alcoholic caustic potash

• When Alkyl Halide ($\mathrm{R}-\mathrm{X}$) is treated with Alcoholic Potash ($\mathrm{alc}.\mathrm{KOH}$), the molecule undergoes dehydrohalogenation to create alkene with the same number of carbon atoms as the starting Alkyl Halide.
• In this reaction, Bromine and Iodine halides are commonly employed to produce Alkene.
• To make the alcoholic Potash used in this procedure, dissolve Potassium Hydroxide ($\mathrm{KOH}$) in Ethanol (${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$).
• The alcoholic Potash utilised in the procedure has a proclivity to dissociate into Hydronium ion and Alkoxide ion with the chemical formula (${\mathrm{RO}}^{-}$).
• This Alkoxide eliminates the Hydrogen ($\mathrm{H}$) atom from Alkyl halide even more.
• This reaction is known as dehydrohalogenation because it involves the removal of a Hydrogen atom as well as a Halide atom from an Alkyl halide in order to form a double bond.
• The reaction is accompanied by the loss of Hydrogen atoms from the Carbon atom ($\mathrm{C}$) close to the Carbon with the Halide group, denoted as ($\mathrm{\beta }$) Carbon.
• As a result, this is sometimes referred to as an elimination reaction.
• The chemical reaction between Ethyl Bromide and alcoholic Potash produces-

$\underset{\mathrm{Ethyl}\mathrm{Bromide}}{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}}+\underset{\mathrm{Alcoholic}\mathrm{Potassium}\mathrm{Hydroxide}}{\mathrm{KOH}}\to \underset{\mathrm{Ethylene}}{{\mathrm{H}}_2\mathrm{C}={\mathrm{CH}}_2}+\underset{\mathrm{Potassium}\mathrm{Bromide}}{\mathrm{KBr}}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}$

• We may deduce from the above description that alcoholic Potash is utilised to produce Ethylene from Ethyl Bromide.
• So, option (b) is the right answer.

Explanation for incorrect options:

Option (a) Hydrogen

• This option is incorrect.
• Because use of Hydrogen might lead to an addition reaction but here elimination reaction is required.

Option (c) Aqueous caustic potash

• We generally use alcoholic $\mathrm{KOH}$ to form Alkene from Alkyl Halides, whereas aqueous $\mathrm{KOH}$ is used to form alcohols from Alkyl Halides.
• So, this option is also incorrect.

Option (d) Aqueous caustic soda

• Caustic soda is a frequent term for Sodium Hydroxide ($\mathrm{NaOH}$). Its popular name comes from its chemical identification as a sodium hydrate, as well as the fact that it is caustic or corrosive.
• Sodium Hydroxide is a caustic base and alkali that degrades proteins at room temperature and can inflict serious chemical burns.
• It is not employed while making Ethylene, so, this option is incorrect.

Hence, the correct option is (b).