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In substitution the nucleophile takes exactly the leaving group's place — hydroxide gives an alcohol, an alkoxide an ether, cyanide a nitrile one carbon longer — and in elimination a C=C forms to the favoured beta carbon; the pathway choice says which product is major.
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 name substitution and elimination products, count how cyanide lengthens a chain, match reactions to product classes, say which bonds break and form, and choose the major product when pathways compete.
You can choose the pathway for a substrate and reagent. Writing the product is the last step: exactly which bonds the pathway breaks and forms.
The major product is the one formed in the greatest amount. A nitrile contains –C≡N; an ether R–O–R′. Chain extension makes a carbon chain longer.
Substitution. The nucleophile's atom bonds where the leaving group was, and nothing else changes:
Elimination. Remove the leaving group and a hydrogen from a neighbouring carbon, and put a C=C between them. Where there is a choice, a small base gives the more substituted alkene: 2-bromo-2-methylbutane → 2-methylbut-2-ene.
Which of the two happens is the previous lesson's table; this lesson writes the result.
Another way: steps
To predict a product:
Real reactions give mixtures. 2-bromobutane with ethoxide gives mostly but-2-ene (E2, the more substituted alkene), some but-1-ene, and a little 2-ethoxybutane by SN2. The prediction is the major product, and saying 'major' is honest about the rest. Knowing what the minor products would be is how chemists choose conditions that suppress them: a bulky base to reverse the alkene choice, a good nucleophile that is a weak base to avoid elimination altogether.
The product always retains every bond in the starting material. One bond breaks, one forms.
Cyanide substitution keeps the chain length. It adds a carbon.
A methyl halide can eliminate. It has no beta carbon.
A reaction gives one product. It gives a major one and minor ones.
Primary substrate, a good nucleophile: SN2.
Choose the pathway.
OH takes bromine's place on carbon 1.
Swap one group.
The product is propan-1-ol, CH3CH2CH2OH.
Name it.
Which pathway does this pair take?
E2: a tertiary carbon with a strong base.
What product forms by that pathway?
The alkene 2-methylpropene, (CH3)2C=CH2, forms.
1-chloropropane reacts by SN2 with ammonia, followed by loss of a proton. Name the organic product.
The product is p.
A 1-bromopropane, with $3$ carbons in an unbranched chain, reacts by SN2 with the cyanide ion. How many carbons does the product have?
Answer:
Match each reaction of 1-bromopropane to the class of its product.
| an alcohol | an ether | a nitrile | an alkene | |
|---|---|---|---|---|
| with sodium hydroxide, SN2 | ||||
| with sodium methoxide, SN2 | ||||
| with sodium cyanide, SN2 | ||||
| with potassium tert-butoxide, E2 |
In the SN2 reaction of iodomethane with methoxide to give methoxymethane, which bonds break and form?
2-bromobutane is heated with sodium ethoxide in ethanol. What is the major organic product?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
bromoethane is heated with sodium ethoxide, a strong, small base, and eliminates HBr by E2. Name the major alkene.
The major alkene is p.
You can write the product of a substitution or elimination. Tell someone why cyanide makes a chain longer. Next unit: additions to double and triple bonds.
8. Your turn: 2-bromo-2-methylpropane with sodium ethoxide, step 3