Reaction of HBR with 2- methyl propane yields 2-bromo-2-methylpropane. What is the structure of carbonation formed during the reaction? Show mechanism of the reaction?
Answers
Explanation:
Markovnikov's rule states that with the addition of HX to an unsymmetrical alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents.
Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greater number of hydrogen atoms while the X component is added to the carbon with the fewer number of hydrogen atoms.
The chemical basis for Markovnikov's Rule is the formation of the most stable carbocation during the addition process.
Answer:
Markovnikov's rule states that with the addition of HX to an unsymmetrical alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents.
Alternatively, the rule can be stated that the hydrogen atom is added to the carbon with the greater number of hydrogen atoms while the X component is added to the carbon with the fewer number of hydrogen atoms.
The chemical basis for Markovnikov's Rule is the formation of the most stable carbocation during the addition process.
When HBr is added to 2-methyl propene,it follows markovnikov rule forming most stable carbocation (tertiary) followed by addition of −Br to carbocation to form 2-bromo 2-methyl propane.
Explanation:
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