Chemical Reactions of Haloalkanes

Nucleophilic substitution replaces the halogen (leaving group) with an incoming nucleophile, $R\text{-}X + Nu^- \rightarrow R\text{-}Nu + X^-$, through one of two limiting mechanisms.

$S_N2$ (bimolecular, concerted). The nucleophile attacks the carbon from the face directly opposite the leaving group (backside attack) in a single step, passing through a trigonal-bipyramidal transition state where the incoming and outgoing groups are momentarily $180°$ apart. Because both reactants are involved in the one rate-determining step, the rate depends on both concentrations. Backside attack forces inversion of configuration (Walden inversion) at the reacting carbon, and steric crowding around that carbon matters enormously — primary substrates react fastest, tertiary substrates are essentially blocked.

$S_N1$ (unimolecular, stepwise). The C–X bond ionises first, heterolytically, forming a flat, $sp^2$ carbocation in the slow, rate-determining step; the nucleophile then attacks this cation in a fast second step. Only the substrate concentration appears in the rate law, since the nucleophile plays no role until after the slow step. Because the carbocation is planar, attack can occur from either face with roughly equal probability, so a chiral starting carbon gives a racemic mixture, not a single inverted product. $S_N1$ is favoured by tertiary substrates (their carbocation is stabilised by hyperconjugation and the $+I$ effect of three alkyl groups), weak nucleophiles, and polar protic solvents that stabilise the developing ions.

where: $k$ is the rate constant for that pathway, $[R\text{-}X]$ is the haloalkane concentration, and $[Nu^-]$ is the nucleophile concentration; note $S_N1$'s rate law is independent of $[Nu^-]$ because the nucleophile enters only after the rate-determining ionisation.

Four factors, one decision. Substrate class (1° pushes toward $S_N2$, 3° pushes toward $S_N1$; 2° is genuinely borderline), nucleophile strength (strong, small nucleophiles favour $S_N2$), solvent polarity (polar protic solvents stabilise ions and favour $S_N1$), and leaving-group ability (a good leaving group speeds up both, but especially $S_N1$'s ionisation step) all point together toward one mechanism or the other; no single factor decides alone.

Elimination (dehydrohalogenation). A strong base — classically alcoholic KOH — can instead pull off a $\beta$-hydrogen while the halogen leaves, forming an alkene: $R\text{-}CH_2\text{-}CH_2\text{-}X + KOH(\text{alc.}) \rightarrow R\text{-}CH=CH_2 + KX + H_2O$. When more than one $\beta$-hydrogen is available, the more substituted (more stable) alkene generally dominates — Saytzeff's rule. Substitution and elimination are always in competition; a strong, bulky base favours elimination, while a small, strong nucleophile in a less basic, more polar medium favours substitution.

Stereochemistry as a diagnostic. A carbon bearing four different groups is chiral. Since $S_N2$ inverts and $S_N1$ racemises, tracking the optical rotation of a chiral substrate through a reaction is direct experimental proof of which mechanism actually operated — this is the classroom link between reaction kinetics and stereochemistry.

Problem. Optically active (R)-2-bromobutane, a secondary halide, is treated with $NaOH$, a strong, small nucleophile, in conditions where the reaction is found experimentally to be second order overall (first order in each reactant). Predict the mechanism, the stereochemical outcome, and what would happen to the rate if $[NaOH]$ were doubled.

Step 1 — Classify the substrate and nucleophile. 2-Bromobutane is a secondary alkyl halide, and $OH^-$ from $NaOH$ is a strong, small, effective nucleophile — both facts favour $S_N2$ over $S_N1$ for a substrate that is not so hindered as to block backside attack.

Step 2 — Confirm with the kinetics. The problem states the reaction is second order overall, matching $\text{rate} = k[R\text{-}X][Nu^-]$ from Equation 1: this is the $S_N2$ rate law, so the mechanism is concerted, not stepwise.

Step 3 — Apply the mechanism to predict stereochemistry. $OH^-$ attacks directly opposite the C–Br bond; as $Br^-$ departs from the front, the three other groups flatten and flip through the transition state. Since attack is exclusively backside, the configuration inverts: (R)-2-bromobutane gives predominantly (S)-2-butanol.

Step 4 — Answer the rate question using the diagnostic from the derivation. Because this is $S_N2$, doubling $[NaOH]$ enters the rate law linearly, so the rate doubles. Answer: the mechanism is $S_N2$, the product is predominantly (S)-2-butanol by Walden inversion, and doubling $[NaOH]$ doubles the reaction rate.

Competitive stretch

tert-Butyl bromide, $(CH_3)_3C\text{-}Br$, is treated with water (weak nucleophile, polar protic solvent); separately, the same substrate is treated with alcoholic KOH. With water, the tertiary substrate and weak nucleophile favour $S_N1$: the C–Br bond ionises to the stable tertiary carbocation $(CH_3)_3C^+$, and water attacks from either face to give racemic tert-butanol after loss of a proton. With alcoholic KOH, $OH^-$ acts predominantly as a strong base rather than a nucleophile on this hindered tertiary substrate, abstracting a $\beta$-hydrogen to eliminate HBr and give isobutylene, $(CH_3)_2C=CH_2$, by Saytzeff's rule where a choice of alkene exists.

• Treating every secondary halide as automatically one mechanism. Secondary substrates are genuinely borderline; state $S_N1$ or $S_N2$ only after weighing the nucleophile strength and solvent given, not from the substrate class alone.
• Saying $S_N1$ gives full retention of configuration. It does not — the planar carbocation is attacked from both faces, giving predominantly a racemic mixture, not the original stereochemistry.
• Mixing up which rate law depends on the nucleophile. $S_N2$'s rate depends on both $[R\text{-}X]$ and $[Nu^-]$; $S_N1$'s depends only on $[R\text{-}X]$ — doubling nucleophile concentration doubles $S_N2$ rate but leaves $S_N1$ rate unchanged.
• Reversing "alcoholic KOH gives elimination, aqueous KOH gives substitution." This exam shortcut is frequently stated backwards; alcoholic (less nucleophilic, more basic) conditions favour elimination.

Ask the tutor: want to work through a batch of substrate/nucleophile/solvent combinations and predict the mechanism, rate law and stereochemical outcome for each?