STANDARDChemistry-Reactions of Alkenes
STANDARD Chemistry Anti-Markovnikov Addition MCQ Question
Type: MCQ-conceptual-Medium
The product (A) and (B) are respectively are (i) CH₃—CH—CH—CH₃, C₂H₅ONa → (A) | Br CH₃ (ii) CH₃ | CH₃—C=CH—CH₃, HBr Peroxide → (B) (Product A & B are separated by comma (,))
A
CH₃ | CH₃—CH—CH—CH₃(CH₂)₂CH₃
B
CH₃—CH₂—(CH₃)₂—CH—CH₂—CH—CH₃ | CH₃ CH₂—Br CH₃
C
CH₃CH₂—CH—CH₂CH₃ | ONa CH₃ | BrCH₂—C—CH=CH₂ | Br
D
CH₃CH₂CH₂CH₂CH₂CH₃ | CH₃—C—CH₂CH₂CH₃ | Br
Correct Answer
Option B
Detailed Explanation
In reaction (i), the elimination of HBr follows Saytzeff's rule, leading to the formation of the more substituted alkene. In reaction (ii), the presence of peroxide leads to anti-Markovnikov addition of HBr.
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