A. Nucleophilic substitution
where R is alkyl and X is usually a halogen or sulfonate; may sometimes be –OH2+ with alcohols.
SN2: With primary and secondary RX + strong nucleophile
Usually a negatively charged nucleophile such as I¯, ¯OH, ¯SH, ¯OR, but uncharged nitrogen nucleophiles NH3/RNH2/R2NH also work.
One step only:
- Polar aprotic solvents (e.g., DMSO, DMF, CH3CN, acetone)
- Cold temperatures reduce chances of E2 elimination
- Primary and secondary R only; simple tertiary R–X never does SN2
- Resonance stabilization of the transition state can also speed up SN2.
SN1: With tertiary and secondary RX + weak nucleophile
Usually an uncharged nucleophile such as H2O, ROH, H2S, RSH, but nonbasic negative nucleophiles such as Cl¯, Br¯, I¯ also work. With uncharged nucleophiles, there will be an acid–base step at the end to lose H+ and give an uncharged final product.
Note: If ¯OH or ¯OR are used instead, E2 elimination will dominate over SN1.
Two steps for substitution (heterolysis followed immediately by coordination), then an acid-base step.
- Polar protic solvents – often use the nucleophile as the solvent.
- Cold temperatures reduces chances of E1 elimination
- Tertiary & secondary R. Never primary unless resonance stabilizing R+
Examples of nucleophilic substitution
In the SN2 example, note the primary alkyl group, the strong nucleophile (¯SCH3) and the polar aprotic solvent (DMF) – all point to SN2 as the mechanism.
In the SN1 example, note the resonance-stabilized secondary carbocation, and the weak nucleophile (water), which also serves as the polar protic solvent. R, Nuc and solvent all point to SN1 as the mechanism.
B. Elimination reactions
General reaction type:
- R can be H, alkyl or aryl – easiest with aryl, hardest with H, but all will do the reaction.
- For E2, X is a halogen or sulfonate
- For E1, X is a halogen, sulfonate or –OH2+ (protonated alcohol).
- Zaitsev’s Rule applies in most cases, i.e., the most substituted/conjugated alkene is formed the most.
- If R3 and/or R4 contain hydrogens, then elimination may lead to other isomers with the double bond going into R3 and/or R4.
E2 elimination: RX + strong base
- Shown above with ¯OH as base, but ¯OR works similarly.
- Works with most alkyl halides, tertiary is easiest but others work fine.
- However, there must be an H anti to the X on the neighboring carbon (Anti Rule).
- This anti constraint means that only one stereoisomer of the product is formed, alkene 1.
- Favored by heat.
- Zaitsev’s Rule applies with most bases, unless they are sterically hindered (e.g., KOtBu). Common bases include KOH (for ¯OH), NaOCH3, NaOEt, KOtBu and some amine bases.
E1 elimination: Tertiary RX + weak base, or ROH + acid
E1 elimination of tertiary alkyl halides (X = Cl, Br, I) or sulfonates (e.g., X = OTs) involves just two steps:
- Only works well with tertiary or resonance-stabilized carbocation intermediates.
- Other products are possible, due to competing carbocation rearrangements.
- This reaction is less used in synthesis than E2, because E2 gives good yields, is more selective, and has fewer side reactions (such as rearrangements).
- Favored by heat.
- Zaitsev’s Rule applies.
However, with E1 elimination of alcohols, an initial acid-base step is needed to turn the –OH into –OH2+ so it can be a leaving group, as shown.
All the above bullet points (from the alkyl halide E1) apply, but also:
- Secondary alcohols also work well. Primary alcohols may do the reaction moderately well, but they usually do this reaction via the E2 mechanism.
- Usual reagents include H2SO4 and H3PO4. More often used in synthesis then the alkyl halide reaction, since strong bases cannot be used in the presence of strong acids. However, the reaction still suffers from rearrangements and lack of stereoselectivity.
Note how in the E2 case, three alkene products are possible, but the major one is the most stable one, as predicted via Zaitsev’s Rule.
C. Addition of HX or H2O to alkenes
Shown with propene as an example. Follows Markovnikov’s Rule: The H goes onto the less substituted carbon, while the X or OH goes onto the more substituted carbon. Mechanisms involving alkynes are quite similar, though in that case H2O addition does involve a subsequent tautomerization step. Note that the pi bond disappears, and two new sigma bonds are formed (C-H and C-X). The added H has not been shown explicitly in this first scheme.
Addition of HCl or HBr
Simply electrophilic addition followed by coordination. Halogen goes onto the more substituted carbon.
Addition of H2O (hydration)
Done using H2SO4 then H2O. The mechanism now includes an additional acid-base step at the end, in order to produce an uncharged form of the alcohol.
D. Electrophilic addition of Cl2 or Br2
Direct addition of the elemental halogen in the absence of light, often in an inert solvent such as CCl4. Although X–X has no permanent dipole, the electron-rich alkene induces a δ+ in the nearer halogen that then attaches first.
Unlike the mechanisms in C above, the first halogen that adds uses its lone pair to form a ring cation called a chloronium or bromonium ion. This is equivalent to a combined electrophilic addition and coordination step in one. The halonium ion blocks syn attack, and forces the other halogen (which is halide, X¯) to attack from the other side and give an anti product.
- Only the anti product is formed.
- Reaction fails for I2 or F2.
- Other mechanisms such as epoxidation, halohydrin formation and oxymercuration are somewhat similar.