Alkyl Halides - Elimination Reactions | Organic Chemistry 1

Further reactions of alkyl halides are studied in this chapter: elimination reactions (E reactions), competition between E1 and E2, competition between SN and E reactions

Alkyl Halide Reactions

Substitution reactions (SN reactions):
 

 

β Elimination reactions (E reactions):
 

 

In both reactions, the alkyl halide acts as an electrophile, reacting with an electron-rich reagent. In a substitution, the nucleophile attacks the carbon atom bearing the good leaving group, while in an elimination, the base removes a proton to form a π bond, and 2 carbons are involved in the reaction

Zaitsev Product

Substituted alkenes:

The degree of substitution of an alkene corresponds to the number of alkyl groups attached to the C=C double bond

  • A monosubstituted alkene has one carbon atom bonded to the carbons of the double bond
  • A disubstituted alkene has 2 carbon atoms bonded to the carbons of the double bond, and so on

 

Zaitsev rule:

Increasing the alkyl substitution stabilizes an alkene by an electron-donating inductive effect. Therefore, the major product in a β elimination reaction is the alkene with the most substituted double bond. This product is called the Zaitsev product
 

E2 Reactions

E2 reactions: 

Elimination reactions which proceed via a concerted mechanism ⇒ E2 reactions are bimolecular with simultaneous bond-making and bond-breaking steps: the carbon-hydrogen and carbon-halogen bonds break to form a new double bond. The kinetic rate involves 2 components: the base and the electrophile. Therefore the E2 reaction is favored by strong bases


Mechanism:

E2 reactions occur when the hydrogen and halogen atoms are oriented on 2 opposite sides of the molecule. This geometry is called anti periplanar and is preferred to syn periplanar geometry
 

E1 Reactions

E1 reactions: 

Elimination reactions which take place via an intermediate carbocation ⇒ E1 reactions are unimolecular with a bond-breaking step followed by a bond-making step. The kinetic rate only involves the starting material. Since the base does not appear in the rate equation, weak bases favor E1 reactions
 

Mechanism:

In the first step, the leaving group comes off to form a planar carbocation. In the second step, a β proton is removed by the base to give the alkene. Because of this two-step mechanism, E1 reactions do not require an anti periplanar geometry. The first step is slower and therefore determines the rate: it is the rate-determining step. The major product will be the Zaitsev product
 

Factors Favoring E1 or E2

Alkyl halide

Unlike the SN2 reactions which are inhibited by steric hindrance, the rate of the E1 and E2 reactions increases as the number of alkyl groups on the carbon bearing the leaving group increases. Indeed, the resulting alkene will be more substituted and therefore more stable. The nature of the alkyl halide does not allow us to determine which elimination mechanism occurs
 

Base

The strength of the base is the most important factor in determining the mechanism of an elimination reaction:

  • Strong bases (negatively charged bases such as HO- and RO-) favor E2 reactions
  • Weak bases (neutral bases such as H2O and ROH) favor E1 reactions

 

 

Solvent

  • Polar protic solvents (H2O, ROH, RCOOH) favor E1 reactions because the carbocation intermediates are stabilized by solvation
  • Polar aprotic solvents (CH3CN, ROR, RCOR, DMSO) favor E2 reactions because the negatively charged bases do not interact with the solvent and are therefore stronger

Summary E1 vs. E2 Reactions

E1 mechanism:

2 steps

planar intermediate carbocation

rate = k [RX]  ⇒  first-order kinetics

order of reactivity: R3CX > R2CHX

favored by weak bases

favored by polar protic solvents

E2 mechanism:

1 step

anti periplanar arrangement of H and X

rate = k [RX] [Base]  ⇒  second-order kinetics

order of reactivity: R3CX > R2CHX > RCH2X

favored by strong bases

favored by polar aprotic solvents

Competition Between SN1, SN2, E1, E2

SN vs. E reactions

Substitution reactions compete with β elimination reactions. The structure of alkyl halides and/or nucleophiles determines the type of reaction:

  • Strong nucleophiles that are weak bases favor substitution over elimination
  • Bulky, non-nucleophilic bases (tBuOK, LDA, DBU) favor elimination over substitution

     

 

How to determine the mechanism:

  1. Determine the nature of the alkyl halide (primary, secondary or tertiary)
  2. Determine the nature of the base or nucleophile (strong, weak, non-nucleophilic, bulky)
     

Primary alkyl halides

- strong nucleophile: SN2
- strong bulky base: E2

 

Secondary alkyl halides

- strong base/nucleophile: SN2 + E2
- strong bulky base: E2
- weak base/nucleophile: SN1 + E1

Tertiary alkyl halides:

- weak base/nucleophile: SN1 + E1
- strong base: E2

 

Check your knowledge about this Chapter

An alkyl halide typically undergoes a nucleophilic substitution reaction or an elimination reaction. In both cases, the alkyl halide acts as an electrophile, reacting with an electron-rich reagent. In a substitution, the nucleophile attacks the carbon atom bearing the good leaving group, while in an elimination, the base removes a proton to form a π bond, and 2 carbons are involved in the reaction

The degree of substitution of an alkene corresponds to the number of alkyl groups attached to the C=C double bond. Thus, a monosubstituted alkene has one carbon atom bonded to the carbons of the double bond. A disubstituted alkene has two carbon atoms bonded to the carbons of the double bond, and so on
 

Alkenes more highly alkylated are more stable, thus the order is tetra > tri > di > mono-substituted. Indeed, alkyl groups are electron-donating via hyperconjugation: they can donate electron density to the neighboring sp2-hybridized carbon atoms of a π bond. The resulting electron density delocalization has a stabilizing effect

The major product of a β elimination reaction is the alkene with the most substituted double bond

The Zaitsev product is the product of the β elimination reaction with the most substituted C=C double bond

An E2 reaction is an elimination reaction which proceeds via a concerted mechanism: E2 reactions are bimolecular with simultaneous bond-making and bond-breaking steps. The kinetic rate involves 2 components: the base and the electrophile. Therefore, the E2 reaction is favored by strong bases

E2 reactions are bimolecular with simultaneous bond-making and bond-breaking steps: the carbon-hydrogen and carbon-halogen bonds break to form a new double bond. E2 reactions occur when the geometry is anti-periplanar
 

E2 reactions occur when the hydrogen and halogen atoms are oriented on opposite sides of the molecule. This geometry is called anti-periplanar and is preferred to syn-periplanar geometry

An E1 reaction is an elimination reaction that takes place via an intermediate carbocation. E1 reactions are unimolecular with a bond-breaking step followed by a bond-making step. The kinetic rate only involves the starting material. Since the base does not appear in the rate equation, weak bases favor E1 reactions

The mechanism of an E1 reaction has two steps. In the first step, the leaving group comes off to form a planar carbocation. In the second step, a β proton is removed by the base to give the alkene. Because of this two-step mechanism, E1 reactions do not require an anti periplanar geometry
 

The rate of an E1 reaction is rate = k [R-X]. Indeed, the slowest step of an E1 reaction is the formation of the carbocation (first step). Therefore, it determines the rate and is called the rate-determining step

The main difference between the E1 and E2 reactions is their number of steps. The E1 reaction takes place in two steps and has a carbocation intermediate. In contrast, the E2 reaction takes place in one step and has no intermediate. Thus, E1 is a first-order reaction, while E2 is a second-order reaction requiring an anti periplanar geometry

The strength of the base is the most important factor in determining the mechanism of an elimination reaction:

  • Strong bases (negatively charged bases such as HO- and RO-) favor E2 reactions
  • Weak bases (neutral bases such as H2O and ROH) favor E1 reactions

Polar protic solvents (H2O, ROH, RCOOH) favor E1 reactions because the carbocation intermediates are stabilized by solvation while polar aprotic solvents (CH3CN, ROR, RCOR, DMSO) favor E2 reactions because the negatively charged bases do not interact with the solvent and are therefore stronger

Substitution reactions compete with β elimination reactions. The structure of the alkyl halides and/or nucleophiles determines the type of reaction:

  • Strong nucleophiles that are weak bases favor substitution over elimination
  • Bulky, non-nucleophilic bases (tBuOK, LDA, DBU) favor elimination over substitution
  1. Determine the nature of the alkyl halide (primary, secondary or tertiary)
  2. Determine the nature of the base or nucleophile (strong, weak, non-nucleophilic, bulky)

 

Primary alkyl halides

- strong nucleophile: SN2
- strong bulky base: E2

 

Secondary alkyl halides

- strong base/nucleophile: SN2 + E2
- strong bulky base: E2
- weak base/nucleophile: SN1 + E1

Tertiary alkyl halides:

- weak base/nucleophile: SN1 + E1
- strong base: E2