Organic Compounds Containing Halogens

When a nucleophile/base meets an alkyl halide, four mechanisms are possible. Use the decision tree below.

SN2 — Bimolecular substitution.

• One-step mechanism, concerted.
• Nucleophile attacks from back side of C-X → inversion at the carbon (Walden inversion).
• Rate = k[substrate][Nu] — bimolecular.
• Favours: primary (1°) > methyl > secondary (2°). Tertiary (3°) too crowded — won't go SN2.
• Strong nucleophile required (OH⁻, CN⁻, RO⁻).
• Polar aprotic solvent (DMSO, DMF, acetone).

SN1 — Unimolecular substitution.

• Two-step: carbocation forms first (rate-limiting), then nucleophile attacks.
• Racemization at the carbon (carbocation is planar → 50/50 mix of products).
• Rate = k[substrate] — only depends on substrate.
• Favours: tertiary (3°) > 2° > 1° (carbocation stability).
• Weak nucleophile (water, alcohols).
• Polar protic solvent (water, methanol).

E2 — Bimolecular elimination.

• One-step, concerted.
• Removes H from β-carbon AND leaving group from α-carbon simultaneously, with anti-periplanar geometry.
• Forms alkene.
• Favours: strong, bulky base (t-BuOK).
• Saytzeff product (more substituted alkene) usually, unless bulky base → Hofmann (less substituted).

E1 — Unimolecular elimination.

• Two-step: carbocation, then base removes β-H.
• Rate = k[substrate].
• Same conditions as SN1 (carbocation-forming) → often compete.

Decision tree:

Is the substrate methyl/primary?
  YES → SN2 dominates with strong nucleophile.
        E2 dominates with strong bulky base.

Is the substrate tertiary?
  YES → SN1/E1 (compete) with weak nucleophile in polar protic solvent.
        E2 with strong base.

Is the substrate secondary?
  STRONG nucleophile, weak base → SN2.
  STRONG, bulky base → E2.
  WEAK nucleophile in polar protic solvent → SN1/E1 mix.

Worked example. (CH₃)₃C-Br + H₂O → ?

Tertiary substrate + weak nucleophile (water) + polar protic solvent → SN1 / E1 compete.

• SN1 product: (CH₃)₃C-OH (t-butanol).
• E1 product: (CH₃)₂C=CH₂ (isobutylene).

At room temperature, SN1 dominates. Higher temperature favors E1.

Worked example. CH₃CH₂Br + NaOH (in ethanol, heat) → ?

Primary substrate + strong base + heat → E2 favored (or SN2). With alcoholic KOH/NaOH heat: elimination dominates → CH₂=CH₂. With aqueous NaOH: substitution dominates → CH₃CH₂OH.