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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

  • Hence Option C is incorrect

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