Check out the pattern of bonds broken/ bonds formed in this reaction:
- We’re always forming a C-C Pi bond . Always!!!
- We’re breaking a C-Br bond in this case.
- We’re breaking a C-H bond on C3. NOTE. MAKE SURE YOU CAN SEE THIS. So many students I know have a hard time “seeing” the hidden hydrogens. This is extremely important!
- Also note how the double bond forms between C2 and C3, NOT between C2 and C1 (Zaitsev’s rule!)
- Finally, note that we’re forming a bond between the H and the oxygen. Oxygen is a strong base here.
Here’s a few more important things to note:
- rate is dependent on concentration of base *and* the alkyl halide. So it’s “bimolecular”. That’s why we call it “elimination, bimolecular” or “E2”.
- we’re always using a strong base, and it always involves a good leaving group
- where both trans and cis products are possible, the trans product will be favored.
Unlike the SN2 (steric hindrance) and the SN1/E1 (carbocation stability) there’s no “big barrier” to this reaction. However, that doesn’t just mean “anything goes”. There’s actually one key aspect of the E2 which makes it unique. In order for elimination to occur, the leaving group and the hydrogen that is removed are always anti.
This is the most testable aspect of the E2: its stereochemistry. More tomorrow.
Thanks for reading! James