Characteristics of E2 Mechanism

1. single step

2. bimolecular

rate = k [CH₃CH₂O^-] [t-BuBr]

base R-X

3. reaction rate is dependent on L group ability

Cl^- < Br^- < I^-

leaving group ability increasing ®

elimination reaction rate increasing ®

4. E2 elimination requires that H and L be anti to occur.

In order for a p bond to form, how must the p orbitals be arranged?

The 2 transition states shown are in specific conformations. What are they called?

anti elimination syn elimination

“backside displacement” “frontside displacement”

Which transition state would you expect to be more stable and why?

only alkene that can form

only alkene that can form

Can anti elimination occur in both conformations?

cis fast

trans very slow

In the following reactions, it is possible for an H to be anti to give each of the products without difficulty. Why are these product ratios observed?

83% (trans:cis = 3.1:1) 17%

Formation of the more substituted (______________________) alkene in an elimination reaction: Saytzeff elimination.

If a bulky base is used, it becomes easier for the molecule to form the less substituted alkene (Hofmann elimination).

79.3 % trans:cis=3:1 20.7%

If the substrate and base are bulky, then the Hofmann elimination is preferred.

28% 72%

There is also a strong preference for the Hofmann elimination for certain leaving groups. (related to steric hindrance or different mechanism E1cB)

L = -N⁺(CH₃)₃ , -S⁺(CH₃)₂

2% 98%

E1 Mechanism

dehydration: loss of water

Mechanism

Characteristics of E1 (similar to S_N1)

1. “2” step or multistep mechanism

2. unimolecular

rate = k[substrate] (substrate = alcohol)

Orientation of E1

If an elimination occurs by the E1 mechanism, the most stable alkene(s) tend to form (Saytzeff elimination).

minor, also forms, due to

carbocation rearrangement

(Ch 8)

Guidelines for Determining Mechanism and Products

S_N1 vs. S_N2 vs. E1 vs. E2

1. higher temperatures favor elimination (will indicate by including D)

2. nonpolar solvents favor elimination (alkanes, carbon tetrachloride)

3. polar solvents

hydroxylic (protic): S_N1

polar aprotic: S_N2

4. Solvolysis reaction: solvent is also a reagent (nucleophile or base), favors nucleophilic substitution

5. In most cases, nucleophilicity and basicity parallel each other.

Exceptions

Increasing nucleophilicity ®

Decreasing basicity ®

F^- < Cl^- < Br^- < I^-

HO^- < HS^- < HSe^- < HTe^-

H₂N^- < H₂P^- < H₂As^- < H₂Sb^-

The following often act as bases: HO^-, RO^-, H₂N^-, R₂N^-, RHN^-

The following often act as nucleophiles: I^-, sulfur compounds, CºN^-

The larger the nucleophile/base, the more often it acts as a base.

more often Nu: more often base

88% 12%

n- = straight chain alkyl group

CH3-X Methyl	RCH2-X 1°	R2CH-X 2°	R3C-X 3°
gives SN2 reactions	Mostly SN2 if large strong base and heat, then E2	mostly SN2 with weak base, (ROH solvent, SN1, polar aprotic SN2) if strong base and heat, then E2	no SN2, SN1 and E1 in solvolysis, heat if strong base, then E2, heat

Chapter 6 Miscellaneous

Nucleophilic substitution reactions are often not strictly S_N1 or S_N2.

Elimination reactions are often not strictly E1 or E2.

E1cB Mechanism

elimination, unimolecular, conjugate base

The E1cB mechanism occurs when L is a poor leaving group and the structure of the substrate has an acidic H. It will be acidic because certain structural features are present that stabilize the conjugate base anion.

Chapter 7 Equilibria

as defined earlier in course

Reaction Mechanism : step by step description of bond breaking and bond making processes that occur when reagents react to form products.

Mechanisms cannot be proven, only disproved.

A proposed mechanism may be considered “well accepted”.

Reaction 1

Reaction 2

K = [products] / [reactants]

K > 1 products favored K < reactants favored

For Reaction 1 K₁ = ([CH₃OH][Cl^-]) / ([CH₃Cl][HO^-])

For Reaction 2 K₂ = [B]/[A]

K₂ = 0.25, at 275° C A : B = 4 :1

For Reaction 1 K₁ = 1 X 10¹⁶

< 0.1% of CH₃Cl and HO^- are left in reaction mixture

The reaction is considered to go to completion.

DG° = -RTlnK

DG° 1 = - 22 kcal/mole heat liberated

(-) value

exergonic

favorable reaction

DG° 2 = +1.51 kcal/mole heat absorbed

(+) value

endergonic

unfavorable reaction

If DG° = -4.1 kcal/mole, then at 25° C, the reaction goes 99.9% to completion

DG° = DH° - TDS°

DH° = standard enthalpy of reaction DS° = standard entropy of reaction

kcal/mole cal/Kmole = eu

T = temperature (K)

In solution

DH° = DH°_{f reactants} - DH° _{f products}

DH° » S BDE bonds broken - S BDE bonds formed

If DH° = (-) heat liberated

exothermic

favorable

stronger bonds formed than broken

product has stronger bonds overall than reactant (starting material)

If DH° = (+) heat absorbed

endothermic

unfavorable

no bonds or weaker bonds formed than broken

product has weaker bonds overall than reactant (starting material)

Reaction 1 DH° = -18 kcal/mole

partially due to BDE H₃C-OH > BDE CH₃C-Cl

92 kcal/mole 85 kcal/mole (-7 kcal/mole)

remainder is due to how well solvated reactants are relative to products,

DH° solvation

DG° = DH° - TDS°

In gas phase

DH° = S BDE bonds broken - S BDE bonds formed

catalyst : often a transition metal such as Pd, Ni, Pt

BDE BDE

bonds broken bonds formed

kcal/mole kcal/mole

Table 7.2, p. 279

CH2=CH2 p	H-H	CH3-CH3
66 kcal/mole	104 kcal/mole	100.3 kcal/mole

DH° = [66 + 104] - [2 x 100.3] = -30.6 kcal/mole

entropy (S) : measure of randomness of particles (amount of disorder)

DS° : change in entropy in a process

more random, more favorable DS° = (+) value

more structured (restricted in motion), less favorable DS° = (-) value

H₂CO₃ ® H₂O + CO₂ DS° = very (+)

1 molecule ® 2 molecules } freedom of motion increases

HCl (aq) + NaHCO₃ (aq) ® H₂CO₃ (aq) + NaCl

¯

H₂O (l) + CO₂ (g)

H^. + H^. ® H₂ DS° = very (-)

2 molecules ® 1 molecule } motion restricted

less favorable S less favorable S

motion of atoms motion of atoms

restricted restricted