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Elimination removes a beta hydrogen and a leaving group to form a C=C: E2 in one step with a strong base, E1 through a carbocation with a weak one; both usually give the more substituted alkene, and a bulky base gives the less substituted one.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will count beta hydrogens, tell E1 from E2 by the conditions, predict the major alkene for small and bulky bases, rank alkene stability, and count the alkenes a substrate can give.
You know how SN1 and SN2 replace a leaving group. A base can do something else: remove a neighbouring hydrogen, so that a double bond forms where the leaving group was.
Elimination removes H and a leaving group from neighbouring carbons, forming a C=C. The beta carbon is next to the carbon carrying the leaving group, and its hydrogens are beta hydrogens. Zaitsev's rule: the more substituted alkene usually forms. The Hofmann product is the less substituted one.
E2 — a strong base removes a beta hydrogen as the leaving group departs, in one step. The H and the leaving group must be anti-periplanar, on opposite sides of the C–C bond. Rate = k[substrate][base].
E1 — the leaving group departs first, giving a carbocation (as in SN1); a weak base then removes a beta hydrogen. Rate = k[substrate]. It needs a substrate that makes a stable cation, and usually heat.
Which alkene? Alkenes with more carbon groups on the C=C are more stable, and both mechanisms usually give the most substituted alkene the substrate allows. From 2-bromobutane: mostly but-2-ene, less but-1-ene.
Another way: table
E1 and E2 compared.
| E2 | E1 | |
|---|---|---|
| base | strong | weak |
| steps | one | two, via a carbocation |
| rate | k[substrate][base] | k[substrate] |
| alkene | usually more substituted | usually more substituted |
A small base such as ethoxide reaches even a crowded beta hydrogen and gives the Zaitsev alkene. A bulky base such as potassium tert-butoxide cannot; it takes the most exposed hydrogen, usually on a CH3, and gives the less substituted, Hofmann alkene. So from 2-bromo-2-methylbutane, ethoxide gives mainly 2-methylbut-2-ene and tert-butoxide mainly 2-methylbut-1-ene. Chemists choose the base to choose the product.
Elimination never depends on the base. A strong base means E2; a bulky one changes the alkene.
The hydrogen comes off the carbon carrying the leaving group. It comes off the beta carbon.
E1 needs a strong base. It needs a stable carbocation.
The least substituted alkene is always favoured. Usually the most substituted is.
Sodium ethoxide is a strong base: E2.
Strong base, one step.
The beta carbons are the two CH3 groups, equivalent: 6 beta hydrogens.
Count them.
Only one alkene can form, propene, CH2=CHCH3.
Equivalent beta carbons, one product.
Strong base present?
No: E1, via the tert-butyl cation.
What alkene forms?
2-methylpropene, (CH3)2C=CH2, the only one possible.
How many beta hydrogens — hydrogens on carbons next to the carbon carrying bromine — does 2-bromo-2-methylpropane, (CH3)3CBr have?
Answer:
2-bromo-2-methylpropane is treated with hot dilute sulfuric acid in water, a weak base, and an alkene forms. Which elimination mechanism is it?
2-bromo-2-methylbutane undergoes E2 with different bases. Match each base to the major alkene.
| 2-methylbut-2-ene, the more substituted alkene | 2-methylbut-1-ene, the less substituted alkene | |
|---|---|---|
| sodium ethoxide | ||
| potassium tert-butoxide | ||
| sodium hydroxide |
Rank these alkenes from most stable to least: but-1-ene CH2=CHCH2CH3, 2-methylbut-2-ene (CH3)2C=CHCH3, 2,3-dimethylbut-2-ene (CH3)2C=C(CH3)2, (E)-but-2-ene CH3CH=CHCH3.
Number the steps in order (write the number in the box):
A chemist wants but-1-ene, not but-2-ene, from 2-bromobutane by E2. Which base should they choose?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Ignoring E and Z isomers, how many different alkenes can elimination of HBr from bromoethane, CH3CH2Br give?
Elimination can give a different alkenes.
You can predict an elimination's mechanism and product. Tell someone how to make the less substituted alkene. Next: what makes a leaving group good.
8. Your turn: 2-bromo-2-methylpropane warmed in ethanol, no base, step 3