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The major pathway of an alkyl halide comes from crossing the carbon's class with the reagent's character: good nucleophiles substitute at uncrowded carbons, strong or bulky bases eliminate at crowded ones, weak reagents wait for tertiary carbons to ionize, and heat favours elimination.
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 predict the major pathway for a substrate and reagent, classify reagents, order the steps of a prediction, pick out conditions that favour elimination, and explain a failed ether synthesis.
You know the four mechanisms and what favours each. Real reactions put them in competition, and the substrate and reagent decide which wins.
Nucleophilicity is how readily a species attacks carbon; basicity, how readily it takes a proton. A bulky base such as potassium tert-butoxide is a strong base too crowded to attack carbon. A polar aprotic solvent such as acetone or DMSO dissolves ions without O–H bonds.
| good nucleophile, weak base (I−, CN−) | strong base (HO−, EtO−) | bulky base (tBuO−) | weak both (H2O, EtOH) | |
|---|---|---|---|---|
| methyl | SN2 | SN2 | SN2 | no reaction |
| primary | SN2 | SN2 (some E2) | E2 | no reaction |
| secondary | SN2 | E2 | E2 | slow SN1/E1 |
| tertiary | SN1 or E1 | E2 | E2 | SN1 (E1 with heat) |
Read across: a methyl carbon, with no beta hydrogens and an open back, always substitutes. A tertiary carbon, blocked at the back, never undergoes SN2: a strong base eliminates, a weak reagent waits for it to ionize. The middle rows are where the reagent decides.
Another way: steps
To predict:
Chemists run the table backwards to plan a synthesis. To make an ether (CH3)3C–O–CH2CH3, reacting 2-bromo-2-methylpropane with ethoxide fails — a tertiary carbon with a strong base eliminates — but reacting bromoethane with tert-butoxide works, because a primary carbon with a good nucleophile substitutes. The same two pieces joined the other way round give a different outcome, and the table says which way works.
One mechanism applies to every alkyl halide. The substrate and reagent decide.
A strong base always substitutes. With crowded carbons it eliminates.
Tertiary halides undergo SN2 with strong nucleophiles. Never.
Temperature does not matter. Heat favours elimination.
Secondary carbon; ethoxide is a strong base.
Classify both.
The table gives E2.
Secondary with a strong base.
The major product is propene, with some ether from SN2.
Major, not only.
Classify the substrate and the reagent.
Primary; azide is a good nucleophile and weak base.
What pathway does the table give?
SN2, giving 1-azidobutane.
What is the major pathway for 1-bromobutane with potassium tert-butoxide?
Match each reagent to its character.
| good nucleophile, weak base | strong base | strong, bulky base | weak nucleophile and weak base | |
|---|---|---|---|---|
| sodium iodide | ||||
| sodium ethoxide | ||||
| potassium tert-butoxide | ||||
| water | ||||
| sodium cyanide |
Put the steps for predicting the major pathway in order.
Number the steps in order (write the number in the box):
Mark every condition that shifts an alkyl halide's reaction towards elimination rather than substitution.
This task has no paper form; do it on a device.
A student tries to make the ether (CH3)3C–OCH2CH3 by treating 2-bromo-2-methylpropane with sodium ethoxide, and gets a gas instead. What happened?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Give the major pathway for each case.
| major pathway | |
|---|---|
| 1-bromopropane with sodium cyanide | |
| 2-bromobutane with sodium ethoxide | |
| 2-bromo-2-methylpropane in water | |
| 1-bromopropane with potassium tert-butoxide |
You can choose among SN1, SN2, E1 and E2 for a real case. Tell someone why tert-butyl bromide and ethoxide give an alkene. Next: predicting the product itself.
8. Your turn: 1-iodobutane with sodium azide in DMSO, step 3