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Elimination (E2) of alkyl halides to give alkenes

Description: Addition of strong base to an alkyl halide results in elimination to form a new C–C π bond.

Notes: X here can be Cl, Br, I or also a sulfonate (such as OTs, OMs, or OTf)

The more substituted alkene is formed except when an extremely bulky base is used.

Furthermore in order to occur the hydrogen that is removed must be “anti” to the leaving group.



Notes: Lots of little things to note here. First, it’s common to use alkoxide bases such as NaOEt and KOCH3 for these reactions (for our purposes, whether it’s Na or K doesn’t matter). It’s common to show the conjugate acid as the solvent (e.g. EtOH for NaOEt) since this will tend to favor elimination over competing substitution reactions where appropriate (as will heat, “Δ “) .

Note the stereochemistry in example 2. It is imperative that the leaving group and the hydrogen be trans to each other in order for elimination to occur, which can’t occur at C–1 of example 2 since a methyl group is there. Finally note that the more substituted alkene (Zaitsev product) is formed (where possible) except in example 4, where the bulky base potassium t-butoxide (KOtBu) favors formation of the less substituted alkene (“Hoffmann product”)

Mechanism: These reactions proceed through a one-step E2 mechanism.

Deprotonation of the carbon by strong base (arrow A) leads to formation of a carbon-carbon PI bond (arrow B) with expulsion of a leaving group (arrow C). Note that the hydrogen and the leaving group must be 180 degrees from each other (in the “anti” conformation).

Notes: Note that when cis- and trans- products are possible (as in this example) the trans will be the favored product due to less steric hindrance.