Elimination reactions are a class of organic reactions where atoms or groups of atoms are removed from a molecule, resulting in the formation of a double or triple bond. There are three main types of elimination reactions:
1. E1 Elimination (Unimolecular Elimination)
- Mechanism:
- The E1 reaction occurs in two steps:
- Formation of a Carbocation: The leaving group departs, forming a carbocation intermediate.
- Deprotonation: A base abstracts a proton (β-hydrogen) from the carbocation, leading to the formation of a double bond.
- Rate-Determining Step: The first step (formation of the carbocation) is the slowest and determines the rate of the reaction.
- Kinetics: First-order (depends only on the concentration of the substrate).
- Example:
[
(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-
]
[
(CH_3)_3C^+ \rightarrow (CH_3)_2C=CH_2 + H^+
] - Tert-butyl bromide undergoes E1 elimination to form isobutene.
- Characteristics:
- Favored by stable carbocations (tertiary > secondary > primary).
- Occurs in polar protic solvents (e.g., water, alcohols).
- Often competes with (S_N1) reactions.
2. E2 Elimination (Bimolecular Elimination)
- Mechanism:
- The E2 reaction occurs in a single step:
- The base abstracts a β-hydrogen while the leaving group departs simultaneously, forming a double bond.
- Rate-Determining Step: The single step involving both the substrate and the base.
- Kinetics: Second-order (depends on the concentration of both the substrate and the base).
- Example:
[
CH_3-CH_2-Br + NaOH \rightarrow CH_2=CH_2 + NaBr + H_2O
] - Ethyl bromide undergoes E2 elimination to form ethylene.
- Characteristics:
- Favored by strong bases (e.g., (NaOH), (KOH)).
- Occurs in polar aprotic solvents (e.g., DMSO, acetone).
- Stereospecific: Requires anti-periplanar geometry (H and leaving group must be 180° apart).
3. E1cB Elimination (Unimolecular Conjugate Base Elimination)
- Mechanism:
- The E1cB reaction occurs in two steps:
- Deprotonation: A base abstracts a β-hydrogen, forming a carbanion intermediate.
- Loss of Leaving Group: The carbanion expels the leaving group, forming a double bond.
- Rate-Determining Step: The second step (loss of the leaving group) is the slowest and determines the rate of the reaction.
- Kinetics: First-order (depends only on the concentration of the substrate).
- Example:
[
CH_3-CH_2-CH_2-Br + NaOH \rightarrow CH_3-CH=CH_2 + NaBr + H_2O
] - Propyl bromide undergoes E1cB elimination to form propene.
- Characteristics:
- Favored when the carbanion intermediate is stabilized (e.g., by electron-withdrawing groups).
- Common in substrates with poor leaving groups or strong electron-withdrawing groups.
Comparison of E1, E2, and E1cB Mechanisms:
| Feature | E1 | E2 | E1cB |
|---|---|---|---|
| Mechanism | Two-step (carbocation intermediate). | One-step (concerted). | Two-step (carbanion intermediate). |
| Rate-Determining Step | Formation of carbocation. | Concerted elimination. | Loss of leaving group. |
| Kinetics | First-order. | Second-order. | First-order. |
| Base Strength | Weak base. | Strong base. | Strong base. |
| Solvent | Polar protic. | Polar aprotic. | Polar protic or aprotic. |
| Stereochemistry | Not stereospecific. | Anti-periplanar geometry. | Not stereospecific. |
| Substrate Preference | Tertiary > secondary > primary. | Secondary or tertiary. | Substrates with stabilized carbanions. |
Key Points:
- E1:
- Favored by stable carbocations.
- Common in tertiary alkyl halides.
- E2:
- Favored by strong bases.
- Common in secondary and tertiary alkyl halides.
- E1cB:
- Favored by substrates with electron-withdrawing groups.
- Common in compounds with poor leaving groups.
Example Problems:
- E1 Example:
- Predict the product of the E1 elimination of 2-bromo-2-methylpropane. [ (CH_3)_3C-Br \rightarrow (CH_3)_2C=CH_2 + HBr ]
- Product: Isobutene.
- E2 Example:
- Predict the product of the E2 elimination of 1-bromopropane with (NaOH). [ CH_3-CH_2-CH_2-Br + NaOH \rightarrow CH_3-CH=CH_2 + NaBr + H_2O ]
- Product: Propene.
- E1cB Example:
- Predict the product of the E1cB elimination of 2-chloro-2-nitropropane. [ (CH_3)_2C(Cl)-NO_2 + NaOH \rightarrow (CH_3)_2C=CH_2 + NaCl + H_2O ]
- Product: 2-Nitropropene.
These elimination reactions are essential tools in organic synthesis for forming alkenes and alkynes, and understanding their mechanisms helps predict reaction outcomes and design synthetic pathways.
Elimination reactions are a fundamental class of organic reactions in which atoms or groups of atoms are removed from a molecule, resulting in the formation of a double bond (alkene) or triple bond (alkyne). These reactions typically involve the loss of two substituents from adjacent carbons, leading to the creation of π-bonds. There are three main types of elimination reactions: E1, E2, and E1cB. Let’s explore each type in detail.
1. E1 Reaction (Unimolecular Elimination)
Mechanism:
- The E1 reaction proceeds via a two-step mechanism:
- Formation of a carbocation intermediate: The leaving group departs first, forming a carbocation.
- Deprotonation: A base abstracts a proton (H⁺) from a β-carbon (adjacent to the carbocation), forming a double bond.
Characteristics:
- Rate-determining step: Formation of the carbocation (first step).
- Rate law: Rate = $k[\text{substrate}]$ (unimolecular).
- Substrate preference: Tertiary > Secondary > Primary (due to carbocation stability).
- Base strength: Weak bases can be used because the carbocation is already formed.
- Solvent: Polar protic solvents (e.g., ethanol, water) stabilize the carbocation.
Example:
- Dehydration of an alcohol to form an alkene:
$$
\text{(CH₃)₃C-OH} \xrightarrow{\text{H₂SO₄, heat}} \text{(CH₃)₂C=CH₂ + H₂O}
$$
- Protonation of the hydroxyl group forms a good leaving group ($H_2O$).
- Loss of water generates a tertiary carbocation.
- Deprotonation by a weak base (e.g., $H_2O$) forms the alkene.
Stereochemistry:
- No specific stereochemical requirement; the product depends on the stability of the resulting alkene (Zaitsev’s rule).
2. E2 Reaction (Bimolecular Elimination)
Mechanism:
- The E2 reaction proceeds via a concerted one-step mechanism:
- The base abstracts a proton (H⁺) from a β-carbon simultaneously with the departure of the leaving group, forming a double bond.
Characteristics:
- Rate-determining step: Simultaneous abstraction of a proton and departure of the leaving group.
- Rate law: Rate = $k[\text{substrate}][\text{base}]$ (bimolecular).
- Substrate preference: Tertiary > Secondary > Primary (steric hindrance favors E2 over SN2).
- Base strength: Strong bases (e.g., $OH^-$, $RO^-$) are required.
- Solvent: Polar aprotic solvents (e.g., DMSO, acetone) favor E2.
Example:
- Dehydrohalogenation of an alkyl halide:
$$
\text{CH₃CH₂CHBrCH₃} + \text{NaOH} \xrightarrow{\text{Heat}} \text{CH₃CH=CHCH₃ + NaBr + H₂O}
$$
- The strong base ($OH^-$) abstracts a proton from a β-carbon.
- Simultaneously, the bromine atom leaves as $Br^-$, forming the double bond.
Stereochemistry:
- Antiperiplanar geometry: The proton and leaving group must be in an antiperiplanar arrangement (opposite sides of the bond) for optimal overlap of orbitals during the transition state.
3. E1cB Reaction (Unimolecular Conjugate Base Elimination)
Mechanism:
- The E1cB reaction proceeds via a two-step mechanism involving the formation of a carbanion intermediate:
- Deprotonation: A strong base abstracts a proton from a β-carbon, forming a carbanion.
- Loss of the leaving group: The carbanion expels the leaving group, forming a double bond.
Characteristics:
- Rate-determining step: Deprotonation (first step).
- Rate law: Rate = $k[\text{substrate}][\text{base}]$ (depends on substrate and base).
- Substrate preference: Substrates with poor leaving groups (e.g., $OH^-$) or stabilized carbanions (e.g., conjugated systems, enols).
- Base strength: Strong bases are required.
- Solvent: Polar aprotic solvents favor E1cB.
Example:
- Elimination in a β-halo ketone:
$$
\text{CH₃COCH₂Br} + \text{OH}^- \xrightarrow{\text{Heat}} \text{CH₃COCH=CH₂ + Br}^- + \text{H₂O}
$$
- The strong base ($OH^-$) abstracts a proton from the β-carbon adjacent to the carbonyl group, forming a resonance-stabilized carbanion.
- The carbanion expels the bromine atom, forming the double bond.
Stereochemistry:
- No specific stereochemical requirement; the product depends on the stability of the carbanion intermediate.
Comparison of E1, E2, and E1cB:
| Reaction Type | Mechanism | Rate Law | Substrate Preference | Base Strength | Leaving Group |
|---|---|---|---|---|---|
| E1 | Two-step (carbocation intermediate) | Rate = $k[\text{substrate}]$ | Tertiary > Secondary > Primary | Weak base | Good leaving group |
| E2 | Concerted one-step | Rate = $k[\text{substrate}][\text{base}]$ | Tertiary > Secondary > Primary | Strong base | Good leaving group |
| E1cB | Two-step (carbanion intermediate) | Rate = $k[\text{substrate}][\text{base}]$ | Poor leaving group, stabilized carbanion | Strong base | Poor leaving group |
Summary of Elimination Reactions:
- E1: Carbocation intermediate, weak base, polar protic solvent.
- E2: Concerted mechanism, strong base, polar aprotic solvent.
- E1cB: Carbanion intermediate, strong base, poor leaving group.
Final Answer:
$$
\boxed{\text{The three types of elimination reactions are E1 (carbocation intermediate), E2 (concerted mechanism), and E1cB (carbanion intermediate).}}
$$