SN2: 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 SN2.
That's what every Arrow Lab exercise feels like. Keep going:
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Organic Chemistry 1 · Free interactive demo
Hydroxide attacks the shown enantiomer of 2-bromobutane from the side opposite the bromine, and in one concerted step the C–Br bond breaks as the C–O bond forms. The three groups on that carbon flip over like an umbrella in the wind, so the product has the opposite configuration.
Draw both arrows above: the lone pair that forms the new bond and the bond that leaves as bromide. 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 SN2 mechanism, step by step
SN2 is bimolecular: nucleophile and substrate meet in a single transition state, so the rate depends on both and the geometry is forced. Backside attack is why a chiral substrate inverts, why hindered carbons react slowly, and why a strong nucleophile and a good leaving group both matter.
The 2 steps of the demo, in the exercise's own words. Try drawing each one above before you read it.
Backside attack
Hydroxide attacks carbon from the side opposite Br while the C–Br bond breaks.
Alcohol after inversion
Oxygen becomes neutral and bromide takes the broken bond pair. Backside attack reverses the tetrahedral arrangement at carbon like an umbrella turning inside out. Compare methyl, ethyl and H relative to the incoming C–O bond and the original C–Br bond, not their positions on the page. Together, these changes show inversion at carbon.
How the mechanism is drawn
The shown enantiomer of 2-bromobutane reacts with NaOH; draw SN2, not competing E2. The solid wedge points toward you and the hashed bond points away; Br and ethyl lie in the page. Compare the whole starting structure and product to see the inversion.