A hydrocarbon X has molecular formula C8H18. Substitution of a hydrogen of X by bromine atom gives one single bromoalkane. Hence, compound X is a)2,2,3,3-Tetramethylbutane. b)2,3,4-Trimethylpentane. c)2,5-dimethylhexane
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Answer:
option A.
Explanation:
it is substitution rxn.
ease of substitution of halogen group is,
F>CL>BR>I.
order of substitution of hydrogen in alkane is,
tertiary hydrogen>sec hydrogen>primary hydrogen.
option A has primary and quaternary carbon atom.but quaternary carbon atom has no hydrogen to remove.so we need to consider as,this molecule contains primary carbon atom only.so it is remove and result in single monobromo alkane.
option B has primary and tertiary carbon atom.so both undergo rxn and results two derivatives.
option C has primary,secondary and tertiary carbon atom.so all three undergo rxn and results three derivatives.
hope you understand.
if you have any doubt ,ask me i send pic of the reaction.
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