In an E2 reaction, the reaction transforms two sp3sp3 size 12{ ital "sp" rSup { size 8{3} } } {} C atoms into two sp2sp2 size 12{ ital "sp" rSup { size 8{2} } } {} C atoms. This moves the substituents further apart, decreasing any steric interactions. So, more highly substituted systems undergo E2 eliminations more rapidly. This is the same reactivity trend as seen in E1 reactions.

-LG The C-LG bond is broken during the rate-determining step, so the rate does depend on the nature of the leaving group. However, if a leaving group is too good, then an E1 reaction may result.

B Since the base is involved in the rate-determining step, the nature of the base is very important in an E2 reaction. More reactive bases will favor an E2 reaction.

a) high concentration of a strong base b) poor leaving groups c) R-LG that do not lead to stable carbocations  (where the E1 mechanism will occur)

A high concentration of a strong base in a relatively nonpolar solvent is used to carry out the dehydrohalogenation reaction. Combinations include sodium methoxide in methanol, sodium ethoxide in ethanol, potassium isopropoxide in isopropanol, and potassium tert-butoxide in tert-butanol or dimethyl sulfoxide (DMSO).

Elimination reactions nearly always yield an isomeric mixture of alkenes. Under the reaction conditions, the elimination is regioselective and follows the Zaitsev rule when more than one route is available for the elimination of HX from an unsymmetrical alkyl halide. Consequently, the reaction proceeds in the direction that yields the most highly substituted alkene. In cases where cis or trans alkenes can be formed, the reaction exhibits stereoselectivity, where the more stable trans isomer is the major product. For example,

Experimental evidence indicates that the five atoms involved in the E2 elimination reaction must lie in the same plane. The anti-periplanar conformation is preferred. This conformation is necessary so that orbital overlap can occur in order for the ππ size 12{π} {} bond to be generated in the alkene. The sp3sp3 size 12{ ital "sp" rSup { size 8{3} } } {}-hybridized atomic orbitals on carbon that comprise the C-H and C-X σσ size 12{σ} {} bonds broken in the reaction develop into the ρρ size 12{ρ} {} orbitals comprising the ππ size 12{π} {} bond of the alkene formed:

There is a smooth transition between the reactants and products. Analogous to the SN2SN2 size 12{S rSub { size 8{N} } 2} {} reaction, no intermediate has been isolated or detected. Furthermore, no rearrangements occur under E2 conditions. This situation is in marked contrast to E1 elimination reactions, where carbocation intermediates are generated and rearrangements are frequently observed.

The alkyl halide adopts an anti-periplanar conformation in the transition state, and experimental evidence demonstrates that if the size of the base is increased, then it becomes more difficult for the large base to abstract an internal ββ size 12{β} {}-hydrogen atom. In such cases, the base removes a less hindered ββ size 12{β} {}-hydrogen, leading to the predominance of the thermodynamically less stable (terminal) alkene in the product mixture. This type of result is often referred to as anti-Zaitsev or Hofmann elimination. Thus, in the reaction of 2-bromo-2,3-dimethylbutane given above, 2,3-dimethyl-1-butene would be the major product (anti-Zaitsev) if the conditions involved use of a bulkier base. The anti-periplanar conformations are illustrated in the Newman projections below:

Dehydrohalogenation of alkyl halides in the presence of strong base (E2) is often accompanied by the formation of substitution ( SN2SN2 size 12{S rSub { size 8{N} } 2} {}) products. The extent of competitive substitution depends on the structure of the alkyl halide. Primary alkyl halides give predominantly substitution products (the corresponding ether), secondary alkyl halides give predominantly elimination products, and tertiary alkyl halides give exclusively elimination products. For example, the reaction of 2-bromopropane with sodium ethoxide proceeds as follows:

In general, for the reaction of alkyl halides with strong base:

Compound MW Amount mmol bp (°C) D np

2-Bromobutane 137.03 100 mL 0.92 91.2 1.26 1.4366

Methanol 32.04 3.5 mL 64.9 0.791 1.3288

2-Propanol 60.09 3.5 mL 82.4 0.785 1.3776

2-Methyl-2-propanol 74.12 3.5mL 82-83 0.786 1.3838

(tertbutanol)

3- Ethyl-3-pentanol 116.20 3.5 mL 140-142 0.839 1.4266

Sodium 22.98 60 mg 2.6 883 0.97

Potassium 39.10 60 mg 1.5 760 0.86

Alcohol Solvent Metal Alkoxide Base Produced

Methanol Sodium Sodium methoxide

2-Propanol Potassium Potassium 2-propoxide

2-Methyl-2-propanol Potassium Potassium 2-methyl-1-2-

(tertbutanol) propoxide (potassium

tert- butoxide)

3-Ethyl-3-pentanol Potassium Potassium 3-ethyl-3-

pentoxide

The combinations of reagents in Table 1.2 may be used to prepare the alkoxide base. Students should compare results to observe a total picture of the effect.

Temperature Conditions Base Temperature (°C)

NaOCH3NaOCH3 size 12{ ital "NaOCH" rSub { size 8{3} } } {} 100-110

KOCH(CH3)2KOCH(CH3)2 size 12{ ital "KOCH" \( ital "CH" rSub { size 8{3} } \) rSub { size 8{2} } } {} 130-140

KOC(CH3)3KOC(CH3)3 size 12{ ital "KOC" \( ital "CH" rSub { size 8{3} } \) rSub { size 8{3} } } {} 140-150

KOC(CH2CH3)3KOC(CH2CH3)3 size 12{ ital "KOC" \( ital "CH" rSub { size 8{2} } ital "CH" rSub { size 8{3} } \) rSub { size 8{3} } } {} 175-180

There will be a large beaker of isopropanol for you to dispense your unused metal. To destroy the metal, add small amounts to methanol.

Report: E2 Elimination (Total 30 Points)

On my honor, in preparing this report, I know that I am free to use references and consult with others. However, I cannot copy from other students’ work or misrepresent my own data.

…………………………………………………………….. (signature)

Print Name: _________________________________________

1. Predict the more stable alkene and supply with reasons for your choice: (6 points)

(a) 1-hexene or trans-3-hexene

(b) trans-3-hexene or cis-3-hexene

(c) 2-methyl-2-hexene or 2,3-dimethyl-2-pentene

2. Starting with the appropriate alkyl halide and base-solvent combination, outline a synthesis that would yield each of the following alkenes as the major or only product and include your reasoning: (8 points)

(a) 1-Butene

(b) 3-Methyl-1-butene

(c) 2,3-Dimethyl-1-butene

(d) 4-Methylcyclohexene

3. When cis-1-bromo-4-tert-butylcyclohexane reacts with sodium ethoxide in ethanol, it reacts rapidly to yield 4-tert-butylcyclohexene. Under similar conditions, trans-1-bromo-4-tert-butylcyclohexane reacts very slowly. Using conformational chair structures, explain the difference in reactivity between these stereo isomers. (4 points)

4. Write the major product of the following reactions with proper stereochemistry.

(4+ 2+ 2+ 4= 12 points)