Organic Chemistry · Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations
The SN1 Reaction
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In 30 seconds
The SN2 reaction swaps a leaving group for a nucleophile in one concerted shove. The SN1 reaction takes the opposite strategy: the leaving group departs first, forming a Carbocation A carbon with a positive charge and only six valence electrons (empty p orbital) Full entry → intermediate, and only then does the nucleophile attack. The name decodes as S = substitution, N = nucleophilic, 1 = Unimolecular Only one molecule appears in the rate-determining step Full entry → — only the alkyl halide appears in the slow, rate-determining step. Because the first step is an Ionization Heterolytic breaking of C–X, both electrons go to X Full entry →, SN1 works only when the carbocation formed is reasonably stable: tertiary and resonance-stabilized (benzylic, allylic) substrates are ideal, and polar protic solvents help by solvating the ions. Two consequences follow: the rate law contains no nucleophile term, and the planar carbocation can be attacked from either face, giving mostly racemized products from chiral starting materials. Carbocations can also rearrange to more stable isomers. The next topic examines stereochemistry, solvent effects, and reactivity in detail.
Why this matters
- Solvolysis The solvent (water, alcohol) acts as the nucleophile Full entry → chemistry: Dissolving an alkyl halide in water or an alcohol converts it to an alcohol or ether — relevant to drug stability.
- Carbocations everywhere: Terpene and steroid biosynthesis runs on carbocation intermediates and rearrangements; SN1's cation chemistry explains "unexpected" products.
- Rearrangement prediction: Exams routinely ask for the rearranged product; knowing when a hydride or methyl shift occurs is high-yield.
- Acid-catalyzed reactions: Alcohols under acidic conditions react through protonated forms whose chemistry mirrors SN1.
- Mechanistic reasoning: SN1 vs SN2 decisions test weighing substrate, nucleophile, solvent, and leaving group — the core analytical habit of the course.
The college version
Core Concepts
The two-step mechanism
SN1 proceeds through a carbocation intermediate in two steps:
- Ionization (slow, rate-determining): the C–X bond breaks heterolytically; the leaving group departs with both electrons: R–X → R⁺ + X⁻. The carbocation is planar (sp²) at the charged carbon, with an empty p orbital.
- Nucleophilic capture (fast): any nucleophile present — often the solvent itself — attacks the cation: R⁺ + Nu⁻ → R–Nu. If the nucleophile is neutral, a final deprotonation gives the neutral product.
Because step 1 is slow and step 2 is fast, everything after ionization is invisible to the rate.
The rate law
Only the substrate appears in the rate law:
rate = k[R–X]
The reaction is first order overall. Doubling [R–X] doubles the rate; changing the nucleophile's concentration or identity has no effect on the rate, only on the product. This separates SN1 from SN2.
Carbocation stability decides feasibility
The ionization step is energetically expensive, so SN1 proceeds only when the cation is reasonably stable:
3°> 2°> 1°> CH3+
Alkyl groups stabilize the positive charge by Hyperconjugation σ-bond electrons delocalize into the empty p orbital Full entry → (σ-bond electrons delocalize into the empty p orbital) and inductive donation; resonance adds extra stability for benzylic and allylic cations. As an approximate guide, tertiary halides solvolyze roughly 10⁵–10⁶ times faster than primary halides. Methyl and primary halides almost never react by SN1; secondary halides do so only under strongly ionizing conditions.
The role of leaving group and solvent
A good leaving group (I⁻ > Br⁻ > Cl⁻; also H₂O from a protonated alcohol, and tosylate) lowers the barrier to ionization. Polar protic solvents are essential partners: water and alcohols solvate both the cation and the departing anion, stabilizing the ions formed in the slow step — the opposite of SN2, which is slowed by protic solvents.
Carbocation rearrangements
If a 1,2 shift (hydride or alkyl) produces a more stable cation, it happens before the nucleophile attacks. A 2° cation adjacent to a 3° carbon can rearrange via a 1,2-Hydride shift H with its electron pair migrates to the adjacent cationic carbon Full entry →; a 1,2-Methyl shift A methyl group migrates when no hydride is available Full entry → is slower but occurs when no hydride is available. The result is a product with a different carbon skeleton — a classic exam trap that becomes predictable once you check cation stability.
Solvolysis: the solvent as nucleophile
In solvolysis, the solvent acts as the nucleophile: R–X in water gives R–OH, in methanol R–OCH₃. The solvent is in huge excess, so the reaction is cleanly first order.
How It Works / Step-by-Step Process
Mechanism of (CH₃)₃CBr + H₂O → (CH₃)₃COH (tert-butyl alcohol):
- The C–Br bond breaks heterolytically; Br⁻ departs with the electron pair, leaving a planar tert-butyl cation, (CH₃)₃C⁺.
- A water molecule's lone pair attacks the empty p orbital from either face, forming (CH₃)₃C–OH₂⁺.
- A second water removes a proton, giving tert-butyl alcohol and H₃O⁺.
Curved arrows in words: C–Br bond pair → bromine; water lone pair → empty p orbital; O–H bond pair → second water.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| SN1 (2 steps, carbocation) | SN2 (1 step, no intermediate) | SN1 has an intermediate and first-order kinetics; SN2 is concerted and second order |
| Rate depends on substrate only | Rate depends on both (SN2) | The nucleophile is absent from the SN1 rate law |
| 3° fast in SN1 | 3° fast in SN2 | Tertiary halides are inert to SN2 (steric) but ideal for SN1 (cation stability) |
| A 1,2-shift always happens | Shifts only to a more stable cation | If the cation is already 3° (most stable), no rearrangement occurs |
| Carbocation (intermediate) | Transition state | An intermediate is an energy minimum; a transition state is the energy maximum |
| SN1 needs a strong nucleophile | SN1 needs any nucleophile | The nucleophile doesn't affect the rate; even weak ones like water work (solvolysis) |
Eli explains
The same idea, in plain words
Explain it like I’m 10
SN1 is like a relay race where the first runner (the leaving group) runs away before the second (the nucleophile) even starts. The first runner leaves slowly — that's the hard part — and the second grabs the open spot quickly. Because the spot is flat and open on both sides, the new runner can come from either direction. Sometimes the team rearranges into a stronger lineup before the second runner arrives — that's the carbocation rearrangement.
Worked example
Example 1: Rate-law arithmetic for a solvolysis
tert-Butyl bromide solvolyzes in water: (CH₃)₃CBr + H₂O → (CH₃)₃COH + HBr, with rate = k[(CH₃)₃CBr]. Using an illustrative k = 4.5 × 10⁻³ s⁻¹ and [(CH₃)₃CBr] = 0.050 M, what is the rate? What happens if the water concentration is doubled?
Formula first:
rate = k[(CH3)3CBr]
Substitute with units (dimensional analysis):
rate = (4.5 × 10-3 s-1)(0.050 M) = 2.25 × 10-4 M · s-1
The units work out because s⁻¹ × M = M·s⁻¹. Doubling [H₂O] changes nothing — water is the solvent, absent from the rate law. Only changes in [(CH₃)₃CBr] affect the rate.
Example 2: Predicting the SN1 product
Predict the major product when 2-bromo-2-methylpropane reacts with water.
Reasoning: Tertiary substrate, good leaving group, polar protic solvent — textbook SN1.
Steps: ionization → (CH₃)₃C⁺; water attacks; deprotonation gives the neutral product:
(CH3)3CBr H2O⟶ (CH3)3COH
Answer: tert-butyl alcohol (2-methyl-2-propanol). No rearrangement is possible because the cation is already 3°.
Example 3: Spotting a rearrangement
2-Bromo-3-methylbutane (CH₃CH(Br)CH(CH₃)₂) is treated with ethanol. Predict the major substitution product.
Reasoning: Ionization gives the 2° cation CH₃CH⁺CH(CH₃)₂; a 1,2-hydride shift gives the 3° cation CH₃CH₂–C⁺(CH₃)₂.
Answer: ethanol attacks the rearranged cation, giving 2-methyl-2-butanol, CH₃CH₂C(OH)(CH₃)₂ — the rearranged product; the unrearranged 3-methyl-2-butanol is minor. Always check for a more stable cation first.
Key takeaways
- SN1 = substitution, nucleophilic, unimolecular: two steps, one intermediate (carbocation), ionization rate-determining.
- Rate law: rate = k[R–X] — first order; the nucleophile does not appear.
- Carbocation stability: 3° > 2° > 1° > methyl; benzylic and allylic cations are resonance-stabilized.
- Requires a good leaving group and a polar protic solvent (water, alcohols) that stabilizes ions.
- Planar cation → attack from either face → racemization (details in next topic).
- Watch for 1,2-hydride and 1,2-methyl shifts that form more stable cations; rearranged products are common.
- Solvolysis (solvent = nucleophile) is the classic SN1 experiment: R–X + H₂O → R–OH.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Write the rate law for the solvolysis of (CH₃)₃CBr in water. What happens if [(CH₃)₃CBr] is tripled? If [H₂O] is doubled?
Show answer
rate = k[(CH₃)₃CBr]; tripling [(CH₃)₃CBr] triples the rate; doubling [H₂O] changes nothing (water is the solvent and absent from the rate law).
Why is 2-bromo-2-methylpropane an ideal SN1 substrate but 1-bromopropane never reacts by SN1?
Show answer
tert-Butyl bromide gives a stable 3° carbocation; 1-bromopropane would give a very unstable primary cation, so SN1 never happens for primary halides.
What intermediate forms in the first step of SN1, and what is its geometry?
Show answer
A carbocation — planar (sp²), with an empty p orbital at the charged carbon.
When does a carbocation rearrange, and what two kinds of 1,2-shifts exist?
Show answer
A cation rearranges when a 1,2-hydride or 1,2-methyl shift forms a more stable cation (e.g., 2° → 3°); if already most stable, no shift occurs.
Predict the product of (CH₃)₃CBr + CH₃OH.
Show answer
tert-Butyl methyl ether, (CH₃)₃COCH₃ — SN1 solvolysis with methanol as the nucleophile.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Carbocation
- A carbon with a positive charge and only six valence electrons (empty p orbital)
- Unimolecular
- Only one molecule appears in the rate-determining step
- Ionization
- Heterolytic breaking of C–X, both electrons go to X
- Hyperconjugation
- σ-bond electrons delocalize into the empty p orbital
- 1,2-Hydride shift
- H with its electron pair migrates to the adjacent cationic carbon
- 1,2-Methyl shift
- A methyl group migrates when no hydride is available
- Solvolysis
- The solvent (water, alcohol) acts as the nucleophile
- First-order kinetics
- Rate ∝ [R–X] only
Sources & references
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