Question

complete the curved arrow mechanism of the following double elimination reaction when 2,3‑dibromopentane is treated with two equivalents of sodium amide and heated in mineral oil.

Answer 1

The curved arrow mechanism for the double elimination reaction of 2,3-dibromopentane with sodium amide is as follows:

The reaction begins with 2,3-dibromopentane and one equivalent of sodium amide. The sodium amide abstracts a proton from the α-carbon, as shown by the curved arrow. This creates a carbanion which resonates between the two equivalent α-carbon positions.

NaNH2 + CH2CHBrCH2CH2Br → CH2=CHBrCH2CH2Br- + Na+

A second equivalent of sodium amide then abstracts a proton from the second α-carbon, creating another carbanion.

NaNH2 + CH2=CHBrCH2CH2Br- → CH2=CHCH2CH2Br- + Na+

The two carbanions then eliminate two bromide ions in a concerted manner to form the pentene product.

CH2=CHCH2CH2Br- + CH2=CHBrCH2CH2Br- → CH2=CHCH=CHCH2CH2 + 2Br-

1. Overall, two equivalents of sodium amide abstract two α-protons and the resulting carbanions eliminate two bromide ions to form the pentene product via an E2 mechanism.