Organic reaction mechanisms

$S_N2$ — concerted, one transition state. The nucleophile attacks directly opposite the leaving group; as the new bond forms the old one breaks, passing through a pentacoordinate transition state that forces inversion of configuration (Walden inversion) at the reacting carbon. Because both substrate and nucleophile appear in the rate-determining (only) step, the kinetics are second order:

where: $k_2$ is the second-order rate constant (units $\text{M}^{-1}\text{s}^{-1}$), $[\text{RX}]$ is substrate concentration, and $[\text{Nu}^-]$ is nucleophile concentration. Unhindered carbons (methyl greater than primary greater than secondary; tertiary is essentially blocked), strong/small nucleophiles, and polar aprotic solvents (no hydrogen-bond cage around the nucleophile) all push toward $S_N2$.

$S_N1$ — stepwise, through a planar cation. The C–LG bond breaks first, in the slow, rate-determining step, to give a planar $sp^2$ carbocation; the nucleophile then captures it fast, from either face. Only the substrate appears in the rate law:

where: $k_1$ is the first-order rate constant ($\text{s}^{-1}$). Attack on the flat cation from either lobe gives racemisation, with a small net-inversion bias because the departing leaving group briefly shields one face. Tertiary, benzylic, or allylic substrates (stable cations), weak nucleophiles, and polar protic solvents (which solvate and stabilise the ions) favour $S_N1$.

$E2$ and $E1$ — the elimination partners. $E2$ is concerted: a base removes a $\beta$-hydrogen while the leaving group departs the adjacent carbon in the same step, so the C–H and C–LG bonds must be anti-periplanar (180° dihedral) for the emerging $p$-orbitals to line up into the new $\pi$ bond; $\text{Rate}=k[\text{RX}][\text{base}]$. $E1$ shares $S_N1$'s carbocation: after ionisation, a base removes any accessible $\beta$-H, so free rotation destroys the anti-periplanar requirement and stereospecificity is lost. $E1cb$ is a third route for a poor leaving group with an acidic $\beta$-H: deprotonation first gives a carbanion, which then ejects the leaving group.

Selectivity and competition. Small bases give the more substituted, more stable Zaitsev alkene; bulky bases (e.g. potassium tert-butoxide) cannot reach a crowded $\beta$-H and are forced to the less hindered Hofmann alkene. Substitution and elimination compete on the same substrate: primary substrates favour $S_N2$ unless the base is bulky or hot; secondary substrates are genuinely mixed; tertiary substrates cannot do $S_N2$/$E2$ at all, only $S_N1$/$E1$ (or $E2$ with a strong small base). Elimination is entropically favoured (two product molecules from one) and grows with temperature.

Predict the mechanism and stereochemical outcome for the reaction of (R)-2-bromobutane with sodium ethoxide ($\text{NaOEt}$, 0.50 M) in ethanol, and estimate how the initial rate would respond if the ethoxide concentration were doubled.

Step 1. Classify the substrate: 2-bromobutane is a secondary alkyl bromide, so both $S_N2$/$E2$ and $S_N1$/$E1$ pathways are geometrically possible — the decision comes down to the nucleophile/base and solvent.

Step 2. Classify the reagent: ethoxide is a strong, only moderately bulky base and a good nucleophile. With a secondary substrate and a strong small base in a protic solvent, elimination is favoured as the major pathway, but $S_N2$ remains a real competing channel.

Step 3. Elimination channel ($E2$): ethoxide removes a $\beta$-hydrogen anti-periplanar to the C–Br bond while the C–Br bond breaks in the same step, forming the new $\pi$ bond directly. Elimination toward C-1 gives 1-butene (Hofmann); elimination toward C-3 gives 2-butene. Because ethoxide is only moderately bulky, the more substituted alkene — (E)-2-butene, the Zaitsev product — dominates, with the E isomer favoured over Z since its anti-periplanar transition state is lower in energy.

Step 4. Substitution channel ($S_N2$): backside attack by ethoxide at C-2 inverts the stereocentre, turning the (R) substrate into (S)-2-ethoxybutane.

Step 5. Kinetics: since ethoxide participates in the rate-determining step of both the $E2$ and $S_N2$ channels, $\text{Rate}=k[\text{RX}][\text{EtO}^-]$ for each. Using the flooding-derived form from the derivation above, doubling $[\text{EtO}^-]$ from $0.50\ \text{M}$ to $1.00\ \text{M}$ doubles $k_{\text{obs}}=k_2[\text{EtO}^-]$, and therefore doubles the initial rate of both product-forming channels — the product ratio (Zaitsev-2-butene versus (S)-2-ethoxybutane) is unchanged, because both rates scale by the same factor. The major product is (E)-2-butene from $E2$, with (S)-2-ethoxybutane from competing $S_N2$, and the total rate doubles on doubling ethoxide concentration while the product ratio stays fixed.

• Assuming $S_N1$ gives exactly 50:50 racemisation — real systems often show slight net inversion because the departing leaving group briefly shields one face of the carbocation.
• Drawing an eclipsed or syn transition state for $E2$ — only the anti-periplanar (180°) arrangement gives correct $p$-orbital overlap for the forming $\pi$ bond.
• Confusing nucleophilicity with basicity — the same species can be a strong base but a weak nucleophile (or vice versa), and this, not "strength" alone, decides substitution versus elimination.
• Forgetting that doubling a reagent's concentration only doubles the rate, not necessarily the product ratio, when that reagent enters the rate-determining step of two competing pathways equally.

Ask the tutor: if you want to work through the full free-energy diagram for a specific substrate/nucleophile pair, or see why a bulky base flips Zaitsev to Hofmann selectivity, ask the AI tutor to build it with you step by step.