Alkyl Halides
SN1 vs SN2 (mechanisms, kinetics, stereochemistry, influencing factors), E1 vs E2, substitution/elimination competition, reactions of carbocations.
Complete Theory
4Bimolecular nucleophilic substitution (SN2) is a reaction in which a nucleophile attacks an alkyl substrate, displacing a leaving group. It is a concerted single-step process: nucleophile attack and leaving-group expulsion occur simultaneously.
Kinetic characteristics: the rate depends on the concentration of both reactants: (second order, bimolecular).
Stereochemistry: Walden inversion. The nucleophile attacks from the side opposite the leaving group (back-side attack), causing inversion of configuration at the stereogenic carbon. It is like an umbrella turning inside out.
Factors affecting SN2 rate:
Kinetic characteristics: the rate depends on the concentration of both reactants: (second order, bimolecular).
Stereochemistry: Walden inversion. The nucleophile attacks from the side opposite the leaving group (back-side attack), causing inversion of configuration at the stereogenic carbon. It is like an umbrella turning inside out.
Factors affecting SN2 rate:
- Substrate (steric effect): CH₃X (methyl, fastest) > 1° > 2° >> 3° (tertiary, no SN2). Bulky groups prevent back-side attack. Order: .
- Nucleophile: stronger and negatively charged gives faster rate. Nucleophilicity order (in polar aprotic solvents): .
- Leaving group: more stable as an anion is better. Order: .
- Solvent: polar aprotic solvents (acetone, DMSO, DMF, CH₃CN) accelerate SN2 because they do not solvate the nucleophile. Polar protic solvents (H₂O, ROH) slow SN2 by solvating the nucleophile.
Unimolecular nucleophilic substitution (SN1) occurs in two steps, with formation of a carbocation intermediate:
Real-world applications: SN1 is involved in solvolysis reactions (hydrolysis of tertiary halides with H₂O to form alcohols). Carbocation rearrangements are important in synthesis (pinacol rearrangement, Wagner-Meerwein) and in the biosynthesis of terpenes and steroids.
- Dissociation (slow, rate-determining step): the C—X bond breaks heterolytically, forming a carbocation R⁺ and anion X⁻. Rate depends only on substrate concentration: (first order).
- Nucleophilic attack (fast step): the nucleophile (often the solvent) quickly attacks the carbocation from either side, yielding a racemic mixture (racemization) if the carbon is chiral.
- Kinetics: first order, (independent of [Nu⁻])
- Substrate: 3° > 2° >> 1° ≈ CH₃X (no SN1). Carbocation stability is crucial: tertiary > secondary >> primary ≈ methyl
- Stereochemistry: racemization (nucleophile attacks from both sides of the planar carbocation). If the starting substrate is enantiomerically pure R, an R/S mixture is obtained
- Rearrangements: the carbocation intermediate may undergo rearrangements (1,2-hydride or alkyl migration) to form a more stable carbocation, leading to unexpected products
- Leaving group: same order as SN2: better anion = easier dissociation
- Solvent: polar protic solvents (H₂O, alcohols, acids) favor SN1 because they stabilize the carbocation
- Nucleophile: nucleophile strength is not critical (fast step). Even weak nucleophiles (H₂O, ROH) react if the carbocation is stable enough
Real-world applications: SN1 is involved in solvolysis reactions (hydrolysis of tertiary halides with H₂O to form alcohols). Carbocation rearrangements are important in synthesis (pinacol rearrangement, Wagner-Meerwein) and in the biosynthesis of terpenes and steroids.
Elimination reactions compete with substitution and lead to alkene formation by HX loss (dehydrohalogenation). Two main mechanisms:
Bimolecular elimination (E2):
Real-world applications: dehydrohalogenation of alkyl halides is a common method for preparing alkenes in the lab and industry. The E2 reaction is used in monomer synthesis (e.g., vinyl chloride from dichloroethane).
Bimolecular elimination (E2):
- Concerted single-step process: a strong base (e.g., HO⁻, RO⁻) abstracts a eta proton while the leaving group X⁻ departs. The C=C bond forms simultaneously.
- Kinetics: (second order)
- Antiperiplanar stereochemistry: the H and the leaving group must be anti (180° apart) to allow orbital overlap leading to bond formation
- Substrate: 1° < 2° < 3° (E2 elimination is favored by more substituted halides because the alkene formed is more stable — Saytsev rule)
- Requires a strong base and steric hindrance blocking SN2
- Two steps: (1) substrate dissociates to carbocation (same as SN1), (2) deprotonation of the carbocation by a weak base (often the solvent).
- Kinetics: (first order, independent of [B⁻])
- Direct competition with SN1: every carbocation can be attacked by a nucleophile (SN1) or lose a proton (E1). At higher temperatures, E1 is favored.
- Follows the Saytsev rule: the more substituted (more stable) alkene is the major product
- May give carbocation rearrangements
Real-world applications: dehydrohalogenation of alkyl halides is a common method for preparing alkenes in the lab and industry. The E2 reaction is used in monomer synthesis (e.g., vinyl chloride from dichloroethane).
The competition between substitution (SN1/SN2) and elimination (E1/E2) is a central aspect of alkyl halide reactivity. Factors determining the reaction course:
1. Substrate structure:
4. Solvent: polar protic solvents favor SN1/E1 (stabilize the carbocation). Polar aprotic solvents favor SN2/E2.
Empirical rules: 1° + strong base/nucleophile → SN2; 3° + strong base → E2; 3° + H₂O/ROH → SN1/E1 (solvolysis, mixture). 2° is the intermediate case: with strong base (HO⁻) in ethanol → E2 ≥ SN2.
Real-world applications: understanding SN/E competition is crucial for designing efficient syntheses. For example, to obtain an ether from an alcohol (Williamson synthesis) a 1° halide with a strong alkoxide is used: SN2 is favored and elimination minimal.
1. Substrate structure:
- 1° substrates: SN2 (if nucleophile is good) or E2 (if base is strong and bulky). SN1/E1 do not occur for 1° (carbocations too unstable).
- 2° substrates: can undergo SN2, SN1, E2 or E1 depending on base, nucleophile, and solvent. Competition is greatest (most studied case).
- 3° substrates: SN2 impossible (steric hindrance). E2 (strong base) or SN1/E1 (neutral/acidic conditions) prevail.
- Strong, small base (HO⁻, CH₃O⁻) → E2 on 3°, SN2 on 1°
- Strong, bulky base (t-BuOK) → E2 (Hofmann elimination, less substituted product)
- Weak nucleophile, weak base (H₂O, ROH) → SN1/E1 (solvolysis conditions)
4. Solvent: polar protic solvents favor SN1/E1 (stabilize the carbocation). Polar aprotic solvents favor SN2/E2.
Empirical rules: 1° + strong base/nucleophile → SN2; 3° + strong base → E2; 3° + H₂O/ROH → SN1/E1 (solvolysis, mixture). 2° is the intermediate case: with strong base (HO⁻) in ethanol → E2 ≥ SN2.
Real-world applications: understanding SN/E competition is crucial for designing efficient syntheses. For example, to obtain an ether from an alcohol (Williamson synthesis) a 1° halide with a strong alkoxide is used: SN2 is favored and elimination minimal.
Worked Examples
2Example 1SN2: Walden inversion in 2-bromooctane
Given
(R)-2-bromooctane + OH⁻ (in DMSO)
SN2 reaction
Find
Configuration of the product
Stereochemical explanation
Step-by-step solution
1OH⁻ (strong nucleophile) attacks the stereogenic carbon (C₂) from the side opposite to bromine (back-side attack). The bromine (leaving group) leaves from the other side.
2During attack, the carbon configuration inverts: Walden inversion. If the starting substrate is (R)-2-bromooctane, the product (2-octanol) has S configuration.
3If the reaction were SN1, racemization would occur (R/S 50:50). Instead, SN2 gives complete stereochemical inversion (100% S). This stereospecificity is experimental proof of the SN2 mechanism.
✓ Final result: (R)-2-bromooctane + OH⁻ → (S)-2-octanol (complete inversion)
Example 2Carbocation rearrangement in SN1
Given
3-methyl-1-butanol + conc. HCl (SN1 via halide)
Find
Major product of the reaction
Step-by-step solution
13-methyl-1-butanol (CH₃)₂CHCH₂CH₂OH is protonated by HCl, water leaves, forming a primary carbocation (CH₃)₂CHCH₂CH₂⁺, very unstable.
2The primary carbocation undergoes a rearrangement: 1,2-hydride shift from C₃ (CH) to C₂ (carbocation), forming a more stable secondary carbocation: (CH₃)₂C⁺HCH₂CH₃ (actually (CH₃)₂C⁺CH₂CH₃ after further rearrangement).
3Subsequent rearrangement (1,2-methyl shift): the secondary carbocation becomes tertiary: (CH₃)₃C⁺CH₂CH₃.
4Cl⁻ attacks the tertiary carbocation: product = 2-chloro-2-methylbutane (not the expected 1-chloro-3-methylbutane!).
✓ Final result: 2-chloro-2-methylbutane (rearrangement product via tertiary carbocation)
Exercises with Solutions
3Exercise 1SN2 vs SN1Medium
Problem to solve
Predict whether the following substrates react preferentially via SN1 or SN2 with OH⁻ in water: (a) CH₃Br, (b) (CH₃)₃CBr, (c) CH₃CH₂Br.
Given data
CH₃Br (bromomethane)(CH₃)₃CBr (bromo-tert-butane)CH₃CH₂Br (bromoethane)
Step-by-step solution
1CH₃Br is a methyl substrate (1° without hindrance). OH⁻ is a strong nucleophile. No steric hindrance to back-side attack → SN2.
2(CH₃)₃CBr is tertiary. SN2 is impossible due to steric hindrance (three methyl groups block attack). OH⁻ in water: the tertiary carbocation is stable, but OH⁻ is a strong base → in water gives mainly E2 (elimination) or SN1 (solvolysis). With OH⁻ in water at room temperature: SN1 and E1 compete.
3CH₃CH₂Br is primary. OH⁻ is a strong nucleophile. SN2 is favored: back-side attack not hindered. SN1 does not occur (primary carbocation too unstable).
✓ Final answer: (a) SN2 (methyl), (b) SN1/E2 (tertiary, OH⁻ strong base), (c) SN2 (primary)
Exercise 2E2 and Saytsev ruleHard
Problem to solve
What elimination products form from the reaction of 2-bromobutane with t-BuOK in t-BuOH?
Given data
2-bromobutane + t-BuOK (potassium tert-butoxide) in t-BuOH
Step-by-step solution
12-bromobutane (CH₃CHBrCH₂CH₃) has bromine on C₂. t-BuOK is a strong and very bulky base.
2Due to the bulky base, H abstraction preferentially occurs from the least hindered position: primary H from the terminal CH₃ (C₁), rather than secondary H from CH₂ (C₃).
3Abstraction from C₁: forms 1-butene (terminal alkene, less substituted) — Hofmann elimination. Abstraction from C₃: forms 2-butene (internal alkene, more substituted) — Saytsev elimination. With t-BuOK, 1-butene (Hofmann) prevails.
✓ Final answer: Mixture: 1-butene (Hofmann, major with t-BuOK) and 2-butene (Saytsev, minor)
Exercise 3SN/E competitionVery Hard
Problem to solve
Predict the products (and relative proportions) of the reaction of 2-bromopropane with CH₃CH₂ONa in ethanol at 50°C.
Given data
2-bromopropane (CH₃CHBrCH₃) + CH₃CH₂ONa (sodium ethoxide) in ethanol
Step-by-step solution
1CH₃CH₂ONa is a strong base and a good nucleophile. The substrate is secondary (2°). Possible pathways: SN2, E2. SN1/E1 are not favored (strong base).
2SN2: ethoxide attack on C₂ (bonded to Br). Product: CH₃CH(OCH₂CH₃)CH₃ (2-ethoxypropane, an ether). SN2 on a 2° substrate is possible but slower than on a 1°.
3E2: ethoxide abstracts a eta H. 2-bromopropane has two equivalent eta H types (both CH₃ groups) → single product: propene (CH₃CH=CH₂).
4At 50°C with ethoxide in ethanol, for a 2° substrate: E2 slightly prevails over SN2. Approximate ratio: 60% propene (E2), 40% 2-ethoxypropane (SN2). At higher T, E2 increases.
✓ Final answer: Mixture: ~60% propene (E2), ~40% 2-ethoxypropane (SN2)
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