E2: push the electrons yourself
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Drag from the electrons that move (a lone pair or a bond) to where they go, or tap one and then the other. Open the demo full screen
You just pushed every arrow of the E2.
That's what every Arrow Lab exercise feels like. Keep going:
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Organic Chemistry 1 · Free interactive demo
Ethoxide pulls off a hydrogen on the carbon next to the C–Br carbon, and in the same concerted step that C–H bond pair swings in to make the C=C double bond while bromide leaves. Taking the hydrogen from the internal carbon gives but-2-ene, the more substituted alkene (Zaitsev).
Draw all three arrows above: base to hydrogen, C–H bond to the new π bond, and C–Br bond to bromine. Each arrow is checked as you draw it, and Hint reveals the next move a little at a time.
Turn your own notes into exercises like this one
With the Arrow Lab student plan you can snap a photo of a reaction from your lecture notes or textbook. Arrow Lab reads the starting materials and reagents, you check that reading, and it builds the mechanism as an exercise like the one above: you draw each arrow, get hints when you are stuck, and see right away whether each move is correct.
Scan Standard, Scan Lite and Scan each include monthly scan credits, up to 20 a day. A credit is a unit of work, not a problem: a typical scan uses about 1 credit, working backwards from a product about 2, and a long or tricky mechanism 2–4.
The E2 mechanism, step by step
E2 is bimolecular: the base and the substrate meet in one transition state, so the rate depends on both. The three arrows are drawn one at a time here, but they all happen together. With a small base such as ethoxide, the hydrogen usually comes off the carbon that leaves the more substituted, more stable alkene; a bulky base can favour the less substituted one instead.
The 2 steps of the demo, in the exercise's own words. Try drawing each one above before you read it.
Remove H and Br together
Remove the drawn hydrogen on the inner β carbon, not a hydrogen on the terminal methyl group.
But-2-ene and ethanol
Ethoxide gains H to become ethanol. The C–H pair forms the π bond, and bromide balances sodium. The drawing shows (E)-but-2-ene, the major isomer; (Z)-but-2-ene also forms as a minor product.
How the mechanism is drawn
2-Bromobutane reacts with sodium ethoxide in ethanol under conditions that favor E2. Draw but-2-ene using the drawn β-H, which is anti to Br in this conformation. The product is drawn as (E)-but-2-ene, the major stereoisomer; (Z)-but-2-ene forms from the other anti-periplanar conformer.