Epoxide Synthesis and Reactions

Question 1

Epoxide Formation via peroxyacid: Reactants

Answer 1

Alkene + mCPBA
—————————>

(mCPBA can be substituted for any RCO₃H)

Question 2

Halohydrin Epoxide Formation: Reactants

Answer 2

1. Alkene + Br₂, H₂O
—————————————->
2. OH^-

Question 3

Halohydrin Epoxide Formation: Process

Answer 3

1) Alkene attacks Br₂ to form bromonium ion

2) H₂O does backside attack on most sub carbon
==> Halohydrin with inverted stereo at OH carbon

3) Br^- deprotonates the added H₂O to form the finalized halohydrin

4) OH^- deprotonates the OH grp of the halohydrin to form an alkoxide

5) Alkoxide does intramolecular backside attack on the carbon with the halogen
==> Kicks halogen off and forms the epoxide

Question 4

What is the one reaction epoxides undergo? Why?

Answer 4

Epoxides undergo ring openings! Although they seem like they’d be inert like ethers, the 3 member ring creates a lot of ring strain and so reactions breaking the ring open are energetically favorable

Question 5

What are the 2 ways an epoxide ring opening can occur?

Answer 5

1) Acid Catalyzed ring opening

2) Base catalyzed ring opening

Question 6

Epoxide Ring Opening (acid = nuc.): Reactants

Answer 6

Epoxide + HX
——————–>

Question 7

Epoxide Ring Opening (acid is not nuc.): Reactants

Answer 7

Epoxide + H₃O⁺
——————————>
ROH

Question 8

Epoxide Ring Opening (acid = nuc.): Process

Answer 8

1) Epoxide oxygen deprotonates acid (HX)

2) X^- does backside attack on the TERTIARY carbon of the epoxide (different from a regular SN2!!!!!)

==> ring is broken open putting an OH on the non 3° side and an X on the 3° side with INVERTED STEREO (due to backside attack!)

Question 9

Epoxide Ring Opening (acid is not nuc.): Process

Answer 9

1) Epoxide oxygen deprotonates the H₃O⁺

2) ROH does backside attack at the TERTIARY carbon of the epoxide (different from a regular SN2!!!!!)
==> Backside attack = stereo inversion!

3) H₂O deprotonates the added ROH to from an ROR

==> Ring is broken open putting an OH on the non 3° side and an -OR on the 3° side with INVERTED STEREO (due to backside attack)

Question 10

Base Catalysed Epoxide Ring Opening: Reactants

Answer 10

MeO^- (or any alkoxide for that matter)
————>
MeOH

Question 11

Base Catalysed Epoxide Ring Opening: Process

Answer 11

1) MeO^- alkoxide will do backside attack on the LEAST SUB carbon of the epoxide

==>Creates MeOR on the least sub side (with INVERTED STEREO) and an alkoxide (O^-) on the most sub side

2) Alkoxide reacts with the SOLVENT, deprotonating it
==> Turns alkoxide into OH group and turns the solvent into the original base catalyst (alkoxide)

–> Ring is broken open with inverted -OR group on LEAST sub side and an OH group on the most sub side