Question

Ethyl bromide gives ethylene when reacted with
(a) Ethyl alcohol
(b) Dilute \[{H_2}S{O_4}\]
(c) Aqueous \[KOH\]
(d) Alcoholic \[KOH\]

Answer 1

Hint: The synthesis of an alkene can be done with the help of dehalogenation reaction of alkyl halide and from vicinal dihaloalkane with the formation of with \[HX\]and \[M{X_2}(M = Zn)\]as a side product.

Complete Step by Step Answer:
The haloalkane and vicinal dihaloalkane can be as \[RX\]and \[XR - RX\]respectively. Where \[X = Br,I,Cl\].
The haloalkanes and vicinal dihaloalkanes belong to the class of hydrocarbons which contain bromo (\[ - Br\]), iodo(\[ - I\]), and sometime chloro (\[ - Cl\]) functional group in their structure.
In the structure of haloalkane, a single halogen atom is connected to the alkyl group. For example, Iodopropane (\[C{H_3} - C{H_2} - C{H_2} - I\]), 1-bromopropane (\[C{H_3} - C{H_2} - C{H_2} - Br\]), etc.
Whereas, in the structure of vicinal dihaloalkane is composed of two halogen atoms which are connected with two adjacent carbon atoms. For example, 1,2-dibromoethane ethene (\[Br - C{H_2} - C{H_2} - Br\]), and 1,2-dichloromoethane ethene (\[Cl - C{H_2} - C{H_2} - Cl\]), etc.
Both alkyl and vicinal dihaloalkane can form alkene upon removal of halogen i.e., dehalogenation reaction.
When ethyl bromide (\[C{H_3} - C{H_2} - Br\]) reacts with alcoholic \[KOH\]formation of ethylene (\[C{H_2} = C{H_2}\]) is occurs.
\[C{H_3} - C{H_2} - I + KOH(Alcoholic) \to C{H_2} = C{H_2} + KI + {H_2}O\]
Hence, the above explanation indicates that Ethyl alcohol, dilute \[{H_2}S{O_4}\], and aqueous \[KOH\]are not used for the synthesis of an alkene.
Therefore, from the above discussion, it is quite clear that option (d) i.e., alcoholic \[KOH\], will be the correct answer.

Note: Ethyl bromide is abbreviated as \[EtBr\]. It is a colourless liquid. It is a light and air sensitive compound. Upon exposure of light and air, it changes its colour to yellowish. It can be used in refrigerant, fumigant and as a solvent in various syntheses.