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\( + \)

SN1 Reactions
Carbocation Stability & Rearrangements

Master the two-step mechanism of unimolecular nucleophilic substitution, understand why carbocation stability drives the reaction, and predict rearrangements that shift atoms to lower-energy intermediates.

Organic Chemistry 1

What is an SN1 Reaction?

SN1 stands for Substitution, Nucleophilic, Unimolecular. The "1" means the rate-determining step involves only one molecule: the substrate. Unlike SN2 reactions, the nucleophile does not participate in the slow step.

The Two-Step Mechanism

\( \text{R–LG} \)
Substrate
Slow
\( \text{R}^+ \)
Carbocation
\( \text{LG}^- \) leaves
Fast
\( \text{R–Nu} \)
Product
\( \text{Nu:}^- \) attacks
Step 1: Ionization (RDS)
Step 2: Nucleophilic Attack

Why Does Carbocation Stability Matter?

Since the rate-determining step is the formation of the carbocation, the entire reaction speed depends on how stable that carbocation is. A more stable carbocation means a lower activation energy and a faster SN1 reaction.

Stable Carbocation
Lower energy intermediate: reaction proceeds readily via SN1.
Unstable Carbocation
Higher energy barrier: SN1 is disfavored. SN2 or elimination may dominate instead.

The Carbocation Stability Hierarchy

Least Stable
Most Stable
Methyl
\( \text{CH}_3^+ \)
0 alkyl groups
Primary (\(1^\circ\))
\( \text{RCH}_2^+ \)
1 alkyl group
Secondary (\(2^\circ\))
\( \text{R}_2\text{CH}^+ \)
2 alkyl groups
Tertiary (\(3^\circ\))
\( \text{R}_3\text{C}^+ \)
3 alkyl groups
\(<\)
\(<\)
\(<\)
Increasing Stability & SN1 Reactivity

Stabilization by Hyperconjugation

\( \text{C}^+ \)
empty p orbital
C
H
\( e^- \) donation
R
R
Adjacent \( \text{C–H} \) \( \sigma \) bond overlaps with empty p orbital

Resonance & Special Stabilization

Beyond alkyl substitution, certain carbocations gain extra stability through resonance. These special cases can make even primary or secondary carbocations surprisingly stable.

Allylic
The \(+\) charge is delocalized across a \(\pi\) system: \( \text{C}{=}\text{C}{\text{-}}\text{C}^+ \) has two resonance structures.
Benzylic
Adjacent to an aromatic ring: the charge delocalizes into the ring, giving multiple resonance forms.
Heteroatom
Lone pairs from O, N, or S can donate into the empty p orbital, providing powerful stabilization.

SN1 Energy Diagram

Reaction Coordinate
Energy
\( \text{R–LG} \)
TS1
\( \text{R}^+ \)
Intermediate
TS2
\( \text{R–Nu} \)
\( E_{a1} \)
(large)

SN1 vs SN2: Key Differences

Feature SN1 SN2
Steps 2 steps (via carbocation) 1 step (concerted)
Rate Law \(k[\text{sub}]\) \(k[\text{sub}][\text{Nu}]\)
Best Substrate \(3^\circ > 2^\circ\) \(\text{methyl} > 1^\circ\)
Stereochemistry Racemization Inversion
Nucleophile Weak is fine Strong required
Solvent Polar protic Polar aprotic
Rearrangements? Yes No

Carbocation Rearrangements

Carbocations are electron-deficient and will rearrange to become more stable whenever possible. A neighboring hydrogen or alkyl group can shift to the carbocation center, creating a more substituted (and thus more stable) carbocation.

1,2-Hydride Shift
A hydrogen atom (with its bonding electrons) migrates from an adjacent carbon to the carbocation center.
1,2-Methyl Shift
A methyl group (with its bonding electrons) migrates from an adjacent carbon to the carbocation center.

1,2-Hydride Shift in Detail

\(2^\circ\) Carbocation
\( \text{C}^+ \)
C1
C
C2
H
R
R
R
H shift
\(3^\circ\) Carbocation
C
C1
H
\( \text{C}^+ \)
C2
R
R
R
Less stable (\(2^\circ\))
More stable (\(3^\circ\))
The H migrates with its bonding electrons to the adjacent \( \text{C}^+ \)

1,2-Methyl Shift

When no hydrogen is available on the adjacent carbon (or when shifting a methyl group creates a more stable carbocation), an entire methyl group can migrate instead. The principle is the same: move to create the most stable intermediate.

\(2^\circ\) Carbocation
\( (\text{CH}_3)_3\text{C–C}^+\text{HCH}_3 \)
\( \text{CH}_3 \) shift
\(3^\circ\) Carbocation
\( (\text{CH}_3)_2\text{C}^+\text{–CH}(\text{CH}_3)_2 \)
A \( \text{CH}_3 \) group migrates with its bonding electrons

When Do Rearrangements Occur?

1
Carbocation forms after the leaving group departs in Step 1 of SN1.
2
Check adjacent carbons: Is a more stable carbocation possible via a 1,2-shift?
3
If yes: a hydride or methyl shift occurs before the nucleophile attacks.
4
Nucleophile attacks the rearranged (more stable) carbocation.
5
Product forms at the new position: the nucleophile ends up on a different carbon than the original leaving group!

Step-by-Step: Predicting SN1 Products

1
Identify the leaving group and classify the substrate (\(1^\circ, 2^\circ, \text{or } 3^\circ\)).
2
Draw the carbocation that forms after the LG departs.
3
Check for rearrangement: Can a 1,2-hydride or methyl shift create a more stable carbocation?
4
Add the nucleophile to the (possibly rearranged) carbocation center.
5
Predict stereochemistry: Expect a racemic mixture (both R and S) since the nucleophile attacks a planar carbocation from either face.

Example 1: Tertiary Substrate

\( (\text{CH}_3)_3\text{C–Br} \)
\(3^\circ\) substrate
\( \text{-Br}^- \)
\( (\text{CH}_3)_3\text{C}^+ \)
\(3^\circ\) (stable!)
\( +\text{H}_2\text{O} \)
\( (\text{CH}_3)_3\text{COH} \)
Product
No rearrangement needed: already at maximum stability (\(3^\circ\))

Example 2: Rearrangement Changes the Product

\( \text{CH}_3\text{CH(Br)}\text{CH(CH}_3)_2 \)
\(2^\circ\) substrate
\( \text{-Br}^- \)
\( \text{CH}_3\text{C}^+\text{H}\text{CH(CH}_3)_2 \)
\(2^\circ\) carbocation
H shift
\( \text{CH}_3\text{CH}_2\text{C}^+(\text{CH}_3)_2 \)
\(3^\circ\) carbocation
\( \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)_2 \)
Final Product
The H on C3 shifts to C2, moving the + charge to the more substituted carbon
The nucleophile attacks at C3 (not C2!), giving a different product than expected

Common Mistakes to Avoid

✕ Forgetting Rearrangements
Always check if a 1,2-shift can produce a more stable carbocation before drawing the product.
✕ Applying SN1 to \(1^\circ\) Substrates
Primary carbocations are too unstable for SN1. These substrates react via SN2 instead.
✕ Predicting Inversion
SN1 gives racemization (mixture of R and S), not inversion. Inversion is an SN2 feature.

SN1 Summary at a Glance

Property SN1 Detail
Mechanism Two steps: ionization then nucleophilic attack
Rate Law \( \text{Rate} = k[\text{substrate}] \) (first-order)
Substrate Preference \(3^\circ > 2^\circ \gg 1^\circ \text{ (rare)}, \text{ methyl (never)}\)
Nucleophile Weak nucleophiles suffice (\(\text{H}_2\text{O}\), ROH)
Solvent Polar protic (stabilizes carbocation + \(\text{LG}^-\))
Stereochemistry Racemization (planar carbocation)
Rearrangements Yes: 1,2-hydride and methyl shifts

Factors That Favor SN1

Substrate
\(3^\circ > 2^\circ\) (benzylic, allylic also favorable). Must form a stable carbocation.
Leaving Group
Good LG required: \(\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-\). Weak bases are better LGs.
Solvent
Polar protic solvents (water, alcohols) stabilize both the carbocation and the departing anion.
Nucleophile
Strength does not matter: weak or strong nucleophiles both work since the RDS has no nucleophile.

Stereochemistry: Racemization

Planar \( \text{C}^+ \)
\( sp^2 \) hybridized, trigonal planar
\( \text{Nu:}^- \)
(top face)
\( \text{Nu:}^- \)
(bottom face)
(R)-Product
~50%
(S)-Product
~50%
Racemic Mixture (50:50 R and S)

Question 1 of 5

Which of the following carbocations is most stable?

Question 2 of 5

The rate of an SN1 reaction depends on the concentration of:

Question 3 of 5

A secondary carbocation rearranges to a tertiary carbocation via a 1,2-hydride shift. What is the driving force for this rearrangement?

Question 4 of 5

What stereochemical outcome is expected when an SN1 reaction occurs at a chiral center?

Question 5 of 5

Which solvent type best promotes SN1 reactions?