How can I illustrate Markovnikov's rule by the reaction of propene with hydrobromic acid?

Question

Delilah Anderson · Accepted Answer

Treat the acid as #D-Br#, i.e. #D# = deuterium. Consider the intemediate after the olefin has reacted with #D^+#.

Markownikow's inclusion: #H_2C=CHCH_3 + D-Br rarr H_2DC-C^+HCH_3 + Br^-# Addition against Markownikow: #H_2C=CHCH_3 + D-Br rarr H_2C^+(-CHDCH_3) + Br^-# As you know, the secondary carbocation is more stable than the primary carbocation. Because the carbocation intermediate is then going to react with the electrophile, #Br^-#, we should see more #H_2(D)C-CHBrCH_3# than the anti-Markownikow product, #H_2(Br)C-CHDCH_3#. J. Chem. Ed. articles are generally excellent. For additional background, see Hughes, P. J. Chem. Ed. 2006, 83, 1152. Was Markovnikov's Rule an Inspired Guess? I am unable to access this article, but your librarian should be able to provide it.

Isaac Aronson · Answer

undefined In the reaction of propene with hydrobromic acid, Markovnikov's rule is illustrated as the hydrogen atom of HBr adds to the carbon atom with more hydrogen substituents, while the bromine atom adds to the carbon atom with fewer hydrogen substituents, resulting in the formation of 2-bromopropane.