Multistep Reactions: 1-Pentene Reacts with Br2/Water Followed by NaH

Question

Predict the product and propose a mechanism when 1-pentene reacts with bromine and water followed by the addition of sodium hydride.

Answer 1

1-pentene will initially add Br, by breaking the Br-Br bond and forming a cyclic bromonium cation, so that there is a 3-membered ring between C1, C2 and Br. The other Br becomes Br^-. The O atom of a water molecule then attacks at the more substituted carbon (C-2, not C-1, because there's more distribution of + charge on C-2), anti (opposite sides of the C1-C2 bond) to the Br, giving an intermediate in which the + charge now resides on O. Then another water molecule pulls off an H attached to the +O atom, giving a racemic mixture of 1-bromo-2-pentanol.

 

With NaH, a very strong base, the H^- ion removes the H from the OH group of the alcohol, to give the -O alkoxide ion of the alcohol (H₂ is the other product). Then, by an internal S_N2 mechanism, the O^-atom attacks C-1, simultaneously ejecting the Br as Br^- ion. This gives an epoxide as the final product (3-membered ring with C1-C2 and O). There is no specific stereochemistry in this particular case, but that is by no means a general statement.

 

The cyclic ether (epoxide) would be named either 1,2-epoxypentane, or propyloxirane