Electrophilic Substitution Reactions of Haloarenes
On mixing a c...
Question
On mixing a certain alkane with chlorine and irradiating it with ultraviolet light it forms only one monochloroalkane. The alkane is:
A
Propane
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B
Pentane
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C
Isopentane
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D
Neopentane
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Solution
The correct option is D
Neopentane
Explanation for correct Answer:
Option D Neopentane
The haloalkanes also known as alkyl halides are formed when an alkane is reacted with a halogen in the presence of ultraviolet (UV) light or heat and this reaction is called halogenation of alkanes.
In this process, the H-atom present in the hydrocarbon molecule is replaced by one or more atoms of halogens and it is a three-step process.
The first process is the initiation step in which the UV light falls and the halogen radical is formed.
In the second step, the radical forms attack the hydrogen present on the alkane to form an alkyl radical and a halogen atom.
In the last step, the halogen radical binds with another radical to form the halogen molecule again.
On mixing a Neopentane with chlorine and irradiating it with ultraviolet light it forms 1-Chloro-2-2-dimethylpropane.
Hence Option D is correct
Explanation for incorrect Answer:
Option A Propane
On mixing a Propane with chlorine and irradiating it with ultraviolet light it forms 1-Chloropropane, 2-Chloropropane
Hence Option A is incorrect
Option B Pentane
On mixing a Pentane with chlorine and irradiating it with ultraviolet light it forms 1-Chloropentane, 2-Chloropentane, and 3-Chloropentane.
Hence Option B is incorrect
Option C Isopentane
On mixing Isopentane with chlorine and irradiating it with ultraviolet light it forms 1-Chloro-3-methyl butane, 2-Chloro-3-methyl butane, 2-Chloro-2-methyl butane, and 1-Chloro-2-methyl butane
