Elimination Reactions

Question 1

E1

Answer 1

• two separate steps: bond breaking to LG, then deprotonation by BASE

Question 2

E2

Answer 2

• one step: deprotonation and bond breaking to LG simultaneous

Question 3

Key Features of E2

Answer 3

1. “Anti-coplanar” - H must have 180deg dihedral angle LG
2. pi bond (alkene) forms from between the two new p-orbitals
3. E/Z orientation about C=C bond determined by anti-coplanar TS
4. Zaitsev or Hofmann alkene, depends on unhindered/hindered base

Question 4

Key Features of E1

Answer 4

1. “Coplanar C-H” - C-H must be parallel to adjavent p-orbital
2. pi bond (alkene) forms from between the two new p-orbitals
3. Always mixture of E/z alkenes due to planar intermediate (C+)
4. Always competition with SN1 - proceed through Same C+
5. ALWAYS makes Zaitsev alkene

Question 5

E1 Substrate Selectivity

Answer 5

• primary effect is the stability of the carbocation (resonance > tertiary carbon > secondary&raquo_space;» primary > methyl)
• primary and methyl are way too slow and are rarely considered
• vinyl, ethynyl, aryl halides are not reactive (both E1 and E2)

Question 6

E2 Substrate Selectivity

Answer 6

• primary effect is need for anti-coplanar C-H and C-X
• tertiary halide > secondary halide > primary halide
• more anti-coplanar C-H bonds are better

Question 7

E1 Base Selectivity

Answer 7

• ONLY FOR E2, NO EFFECT ON RATE
• Weak bases are ok
• competition w/ solvolysis (addition of neutral Nu) - substitution by solvent molecule always joint with E1 elimination (share same intermediate carbocation)

Question 8

E2 Base Selectivity

Answer 8

• Need a strong base, or conjugate acid with a HIGH pKa or a small Ka (need to abstract very weakly acidic C-H bond
• Typical Strong Bases: NaOCH2CH3, KO(CH3)3, NaHLiN(iPr)2, LiNH2
• Anionic Strong bases!!!
• HOMe (WEAK!)

Question 9

E1 Solvent Selectivity

Answer 9

polar protic > polar aprotic&raquo_space; non-polar

Question 10

E2 Solvent Selectivity

Answer 10

polar aprotic&raquo_space;> polar protic > non-polar

- Aprotic enhances, protic hinders reactivity of strong, anionic bases

Question 11

Leaving Group Selectivity (E1 and E2)

Answer 11

I > Br > Cl&raquo_space;»F

- best LG form stable anions (weak bases) or neutral species

Question 12

What makes a good leaving group?

Answer 12

• biggest/least basic substituents make best leaving group
• Good leaving groups: Cl, Br, I, OH2, SO4
• Poor leaving groups: F, H, C, CR3 NR3 OH