Key ideas
A haloalkane can react with a nucleophile or a base in four ways: SN1, SN2, E1 and E2. The structure of the substrate, the nucleophile, the leaving group and the solvent decide which one wins.
- SN2 is one step with backside attack, so the configuration is inverted. It is fastest for methyl and primary substrates, slow for secondary and does not happen at tertiary ones.
- SN1 goes through a carbocation, so it is fastest for tertiary substrates, gives a racemic product and its rate depends only on the substrate.
- Carbocation stability is 3° > 2° > 1° > methyl. Better leaving groups are weaker bases: I⁻ > Br⁻ > Cl⁻ > F⁻.
- Polar protic solvents favor SN1, polar aprotic solvents favor SN2. A strong, bulky base favors E2, and heat favors elimination over substitution.
Key equations
SN1: rate = k[RX]SN2: rate = k[RX][Nu]carbocation: 3° > 2° > 1° > CH₃⁺leaving group: I⁻ > Br⁻ > Cl⁻ > F⁻
Worked example
Compare CH₃Br, CH₃CH₂Br, (CH₃)₂CHBr and (CH₃)₃CBr with iodide in acetone (SN2): the rates fall in the order CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr. In ethanol, 2-bromo-2-methylpropane reacts by SN1 and E1, and doubling the nucleophile concentration changes nothing because the rate is k[RX].
Common mistakes
- Expecting SN2 at a tertiary carbon, where the backside is too crowded.
- Using the SN1 rate law, which has no nucleophile term, for an SN2 reaction.
- Choosing fluoride as a good leaving group. It is a poor one because fluoride is the strongest base of the halides.
Practice
Test yourself with practice questions on this topic: Substitution, elimination and carbocations