Elimination Reactions Notes

Elimination Reactions

Elimination reactions involve the loss of fragments or groups from a molecule, leading to the formation of multiple bonds.

Types of Elimination Reactions

• α-elimination (1,1-elimination):
  • Two atoms or groups are removed from the same atom.
• β-elimination (1,2-elimination):
  • Loss of atoms or groups on adjacent atoms.
• γ-elimination:
  • Loss of atoms or groups from the 1st and 3rd positions.
  • Results in cyclic compounds.

Dehydrohalogenation of Alkyl Halides

Example reactions:

• CH3CH2CH2Cl + KOH ewline CH3CH=CH_2 (in C2H5OH)
• CH3CH2CH2CH2Cl + KOH
  ewline
  CH3CH2CH=CH_2 (in C2H5OH, no rearrangement)
• CH3CH2CHCH3 + KOH ewline CH3CH=CHCH3 + CH3CH2CH=CH2 (Cl at position 2 leading to 80% of but-2-ene and 20% of but-1-ene)
  • Cl
  • C2H5OH

Mechanisms: E1 and E2

• With weak bases at low concentrations, and moving from primary to secondary to tertiary halides, the reaction becomes first order.
• With a strong base, the reaction follows second-order kinetics.

E2 Mechanism

$-\underset{|}{C} - \underset{|}{C} - + :B \longrightarrow -C=C- + HB + X$
H X

• Rate order: $RI > RBr > RCl > RF$
• Second-order (bimolecular).
• Rate = $K[CH<em>3CHBrCH</em>3][EtO^-]$
• E2 reaction has one transition state.
• Free Energy Diagram of E2 Reaction

E1 Mechanism

$\underset{|}{C} - \underset{|}{C} - \longrightarrow [\underset{|}{C} - \underset{|}{C}^+ ] + X$
H X

• Step 1 (slow):

$[CH<em>3]</em>3CCl + H<em>2O \longrightarrow [CH</em>3]_3C^+ + Cl^-$

• Aided by the polar solvent, a chlorine departs with the electron pair that bonded it to the carbon.
• This slow step produces the relatively stable 3° carbocation and a chloride ion.
• The ions are solvated (and stabilized) by surrounding water molecules.

Step 2 (fast):

$[CH<em>3]</em>3C^+ + H<em>2O \longrightarrow [CH</em>3]<em>2C=CH</em>2 + H_3O^+$

• A molecule of water removes one of the hydrogens from the carbon of the carbocation.
• These hydrogens are acidic due to the adjacent positive charge.
• At the same time, an electron pair moves in to form a double bond between the carbon atoms.
• This step produces the alkene and a hydronium ion.

Orientation and Reactivity

• Example:

CH3CH2CHCH3 + KOH ewline CH3CH=CHCH3 + CH3CH2CH=CH2
Cl

• (80% major) but-2-ene + (20% minor) but-1-ene

• The ease of alkene formation follows the sequence:

$R<em>2C=CR</em>2 > R<em>2C=CHR > R</em>2C=CH<em>2, RHC=CHR > RHC=CH</em>2$

• This is also the order of alkene stability.

• Therefore, the more stable the alkene formed, the faster it is formed.

• $-\underset{|}{C} - \underset{|}{C} - \longrightarrow -\underset{|}{C} = \underset{|}{C} -$

• B

• The double bond is partially formed in the transition state, and therefore the transition state resembles an alkene (Hammond-Leffler postulate).

• Factors that stabilize alkenes will stabilize this nascent alkene - Zaitsev elimination.

The Hammond–Leffler Postulate

• Formation of the carbocation, (+ve ΔG° and ΔH°); therefore, this step is endothermic.
• According to the Hammond–Leffler postulate, the transition-state structure for a step that is uphill in energy should show a strong resemblance to the structure of the product of that step.

Temperature

• Increasing the reaction temperature favors elimination (E1 and E2) over substitution.
• Elimination reactions are entropically favored over substitution.
  • Reason 1: As $\Delta G° = \Delta H° - T \Delta S°$, an increase in temperature further enhances the entropy effect.
  • Reason 2: The products of an elimination reaction are greater in number than the reactants:
    • $R-CH<em>2CH</em>2Br + NaOCH<em>2CH</em>3 \longrightarrow R-CH=CH<em>2 + NaBr + CH</em>3CH_2OH$

Substitution vs. Elimination

• All nucleophiles are potential bases, and all bases are potential nucleophiles.

• Substitution reactions are always in competition with elimination reactions.

• Different factors can affect which type of reaction is favored.

• S_N1 vs E1 example:

$CH<em>3C(CH</em>3)<em>2Cl + H</em>2O \longrightarrow CH<em>3C(CH</em>3)<em>2OH + CH</em>2=C(CH<em>3)</em>2$

• (major (S_N1)) + (minor (E1))

Primary Substrate

• With a strong base, e.g., EtO⁻:
  • Favor S_N2

$R-CH<em>2-Br + NaOEt \longrightarrow R-CH</em>2-OEt + R-CH=CH_2$

• S_N2: 90%
• E2: 10%

Secondary Substrate

• An S_N2 reaction occurs if a good nucleophile that is a weak base is used in a polar aprotic solvent (I⁻, Br⁻, SCN⁻, N3⁻, CN⁻, Amines, etc.).
• An S_N1 reaction along with an E1 reaction occurs if a poor nucleophile that is a weak base is used in a protic solvent.

Tertiary Substrate

• With a strong base, e.g., EtO⁻:
  • E2 is highly favored

$R<em>3C-Br + NaOEt \longrightarrow R</em>2C=CH_2$

• E2: 91%
• with weak base, S_N1 is preferred over E1 mechanism.

Base: Small vs. Bulky

• Unhindered "small" base/Nu:
  • Example:

$R-CH<em>2-Br + NaOMe \longrightarrow R-CH</em>2-OMe + R-CH=CH_2$

S_N2: 99%

E2: 1%

• Hindered "bulky" base/Nu:
  • Example:

$R-CH<em>2-Br + KOtBu \longrightarrow R-CH</em>2-OtBu + R-CH=CH_2$

• S_N2: 15%

• E2: 85%

• When the base used for E2 elimination is a bulky base, exceptions to the Zaitsev rule occur.

• A high proportion of the less substituted alkene (Hofmann Elimination).

• Example:

Basicity vs. Polarizability

$CH<em>3CH</em>2Br + CH<em>3CO</em>2Na \longrightarrow CH<em>3CH</em>2CO<em>2CH</em>3$

• (weak base, in acetic acid) S_N1: 100% E1: 0%

$CH<em>3CH</em>2Br + NaOEt \longrightarrow CH<em>3CH</em>2OEt + CH<em>2=CH</em>2$

• (strong base) S_N2: 20% E2: 80%

Strong Bases/Strong Nucleophiles

• A good base is usually a good nucleophile.
• Strong bases—substances with negatively charged O, N, and C atoms—are strong nucleophiles.
• Participate in S_N2-type substitutions.
• Examples are: RO⁻, OH⁻, RLi, RC≡C:⁻, and NH₂⁻.

Weak Nucleophiles & Bases

• Typically neutral molecules.
• Participate in S_N1-type concurrently with E1.
• Examples: H2O, ROH, H2S, RSH.

Strong Bases / Poor Nucleophiles

• Some strong bases are poor nucleophiles because of steric hindrance.
• Participate in E2 ONLY.
• Examples are t-BuO⁻, t-BuLi, and LiN[CH(CH₃)₂].

Weak Bases / Good Nucleophiles

• Anions with a full negative charge; it is a good nucleophile.
• Weak Base, as the large electron cloud is highly polarizable.
• Participate in S_N2-type substitutions.
• Examples: any NaOR, any RLi, CN⁻, NaCCR (acetylide anion), NaNH2, NaNHR, NaNR2, NaI, LiBr, KI, NaN3.

Solvent Effects

• Solvated species are more stable and less reactive than the unsolvated "naked" anions.

• Polar, protic solvents such as water and alcohols solvate anions by hydrogen bonding.

• Polar, aprotic solvents such as DMSO, DMF, acetonitrile, etc., do not solvate anions but provide good solvation of the accompanying cations.

• In polar protic solvents, the order of nucleophilicity is I⁻ > Br⁻ > Cl⁻ > F⁻

  • F⁻ is a small ion with a high charge density and is tightly solvated.
  • I⁻ is a large ion with a low charge density and is loosely solvated.

• In polar aprotic solvents, the order of nucleophilicity is reversed: F⁻ > Cl⁻ > Br⁻ > I⁻.

Nucleophilicity Trends

• For a given element, negatively charged species are more nucleophilic (and basic) than are equivalent neutral species.
• For a given period of the periodic table, nucleophilicity (and basicity) decreases on moving from left to right.
• For a given group of the periodic table, nucleophilicity increases from top to bottom (i.e., with increasing size), although there is a solvent dependence due to hydrogen bonding. Basicity varies in the opposite manner.

Summary

Notes:

• Note 1: Bulky bases (e.g., NaOt-Bu) will lead to increased E2 relative to S_N2.
• Note 2: Need a relatively strong base to do E2 (e.g., NaOH, NaOR, and stronger).
• Note 3: Higher temperatures favor elimination. Look out for carbocation rearrangements.
• Note 4: Need a relatively strong base for E2 (NaOH, NaOR, or stronger). If conditions favor carbocation formation, weakly basic nucleophiles like NaCN and NaN3 may do S_N1.
• Note 5: Higher temperatures favor elimination.