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Identify the product A and B formed in the following reaction: CH_(3)-CH_(2)-CH=CH-CH_(3)+CHltoA+B |
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Answer» Solution :Since both the carbon atoms of the DOUBLE bond has the same NUMBER of hydrogen atoms (i.e., 1), therfore, Markovnikov's rule cannot be applied. In other words, both the products (A and B) will be formed. `UNDERSET("Pent-2-ene")(CH_(3)CH_(2)-CH=CH-CH_(3)+HCL) to underset("3-Chloropentane (A)")(CH_(3)CH_(2)-underset(Cl)underset(|)(C)H-CH_(2)CH_(3))+underset("2-Chloropentane (B)")(CH_(3)CH_(2)CH_(2)-underset(Cl)underset(|)(C)H-CH_(3))` `underset('I' " Stabilized by four hyperconjugationstructures (slightly LESS stable)")(CH_(3)-CH_(2)-overset(+)(C)H-CH_(2)+CH_(3))""underset("II Stabilized by five hyperconjugation structures (slightly more stable)")(CH_(3)CH_(2)-CH_(2)-overset(+)(C)H-CH_(3))` the amount of 2-chloropentane (B) will be slightly more than that of 3-chloropentane (A), approx. 55:45. |
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