Chapter 6: Q62P (page 283)
When 3-methyl-1-butene reacts with HBr, two alkyl halides are formed: 2-bromo-3-methylbutane and 2-bromo-2-methylbutane. Propose a mechanism that explains the formation of these two products.
Short Answer

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Chapter 6: Q62P (page 283)
When 3-methyl-1-butene reacts with HBr, two alkyl halides are formed: 2-bromo-3-methylbutane and 2-bromo-2-methylbutane. Propose a mechanism that explains the formation of these two products.

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a. Identify two alkenes that react with HBr to form 1-bromo-1-methylcyclohexane without undergoing a carbocation rearrangement.
b. Would both alkenes form the same alkyl halide if DBr were used instead of HBr? (D is an isotope of H, so D+ reacts like H+.)
What will be the major product obtained from the reaction of Br2 with 1-butene if the reaction is carried out in
a. dichloromethane?
b. water?
c. ethyl alcohol?
d. methyl alcohol?
Why are Na+ and K+ unable to form covalent bonds?
What is the product of the addition of I-Cl to 1-butene? (Hint: Chlorine is more electronegative than iodine [Table 1.3].)

:a. How does the first step in the reaction of propene with Br2 differ from the first step in the reaction of propene with HBr?
b. To understand why Br- adds to a carbon of the bromonium ion rather than to the positively charged bromine, draw the product that would be obtained if Br- did add to bromine.
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