CHE-240 Unit 3

Physical and Chemical Properties and Reactions of Alkenes and Alkynes

CHAPTER EIGHT TERRENCE P. SHERLOCK BURLINGTON COUNTY COLLEGE 2004 Chapter 8 1

Reactivity of C=C

• Electrons in pi bond are loosely held.

• Electrophiles are attracted to the pi electrons.

• Carbocation intermediate forms.

• Nucleophile adds to the carbocation.

• Net result is addition to the double bond. => Chapter 8 2

Electrophilic Addition

• Step 1: Pi electrons attack the electrophile.

C C + E + E C C +

• Step 2: Nucleophile attacks the carbocation.

E C C + + Nuc: _ E C Nuc C

Chapter 8 3 =>

Types of Additions

Chapter 8 => 4

Addition of HX (1)

Protonation of double bond yields the most stable carbocation. Positive charge goes to the carbon that was not protonated.

CH 3 CH 3 C CH CH 3 H Br CH 3 CH 3 C + CH CH 3 H

X

CH 3 CH 3 C H CH CH 3 +

Chapter 8

+ Br _

=> 5

Addition of HX (2)

CH 3 CH 3 C CH CH 3 H Br CH 3 CH 3 C + CH CH 3 H + Br _ CH 3 CH 3 C + Br _ CH CH 3 H

Chapter 8

CH 3 CH 3 C CH CH 3 Br H

=> 6

Regiospecificity

• Markovnikov’s Rule: The proton of an acid adds to the carbon in the double bond that already has the most H’s. “Rich get richer.” • More general Markovnikov’s Rule: In an electrophilic addition to an alkene, the electrophile adds in such a way as to form the most stable intermediate.

• HCl, HBr, and HI add to alkenes to form Markovnikov products. => Chapter 8 7

Free-Radical Addition of HBr

• In the presence of peroxides, HBr adds to an alkene to form the “anti Markovnikov” product.

• Only HBr has the right bond energy.

• HCl bond is too strong.

• HI bond tends to break heterolytically to form ions. => Chapter 8 8

Free Radical Initiation

• Peroxide O-O bond breaks easily to form free radicals.

heat

R O O R R O + O R

• Hydrogen is abstracted from HBr.

R O

+

H Br

Chapter 8

R O H

+

Br Electrophile

9 =>

Propagation Steps

• Bromine adds to the double bond.

Br

+

C C C Br C

• Hydrogen is abstracted from HBr.

C Br C

+

H Br

Chapter 8

C C Br H

+

Br Electrophile =>

10

Anti-Markovnikov ??

CH 3 CH 3 C CH CH 3

+

Br

X

CH 3 CH 3 C CH CH 3 Br CH 3 CH 3 C Br CH CH 3

• Tertiary radical is more stable, so that intermediate forms faster. => Chapter 8 11

Hydration of Alkenes

C C alkene + H 2 O H + H C OH C alcohol

• Reverse of dehydration of alcohol • Use very dilute solutions of H 2 SO 4 H 3 PO 4 to drive equilibrium toward hydration. or Chapter 8 => 12

Indirect Hydration

• Oxymercuration-Demercuration  Markovnikov product formed  Anti addition of H-OH  No rearrangements • Hydroboration  Anti-Markovnikov product formed  Syn addition of H-OH => 13 Chapter 8

Hydroboration

• Borane, BH 3 , adds a hydrogen to the most substituted carbon in the double bond.

• The alkylborane is then oxidized to the alcohol which is the anti-Mark product.

C C (1) BH 3 C H C BH 2 (2) H 2 O 2 , OH -

Chapter 8

C C H OH

=> 14

2

Borane Reagent

• Borane exists as a dimer, B 2 H 6 , in equilibrium with its monomer.

• Borane is a toxic, flammable, explosive gas.

• Safe when complexed with tetrahydrofuran.

THF O + B 2 H 6

Chapter 8

2 O + H B H H THF .

BH 3

=> 15

Predict the Product

Predict the product when the given alkene reacts with borane in THF, followed by oxidation with basic hydrogen peroxide.

CH 3 D (1) BH 3 , THF (2) H 2 O 2 , OH -

syn addition Chapter 8

H D CH 3 OH

=> 16

Hydrogenation

• Alkene + H 2  Alkane • Catalyst required, usually Pt, Pd, or Ni.

• Finely divided metal, heterogeneous • Syn addition 17 => Chapter 8

Addition of Halogens

• Cl 2 , Br 2 , and sometimes I 2 add to a double bond to form a vicinal dibromide.

• Anti addition, so reaction is stereospecific.

Br C C + Br 2 C C

=>

Br

Chapter 8 18

Mechanism for Halogenation

• Pi electrons attack the bromine molecule.

• A bromide ion splits off.

• Intermediate is a cyclic bromonium ion.

C C + Br Br C Br C

+

Br

=> Chapter 8 19

Mechanism (2)

Halide ion approaches from side opposite the three-membered ring.

C Br C Br Br C C Br

=> 20 Chapter 8

Examples of Stereospecificity

Chapter 8 => 21

Test for Unsaturation

• Add Br 2 in CCl 4 (dark, red-brown color) to an alkene in the presence of light.

• The color quickly disappears as the bromine adds to the double bond.

• “Decolorizing bromine” is the chemical test for the presence of a double bond. => 22 Chapter 8

Formation of Halohydrin

• If a halogen is added in the presence of water, a halohydrin is formed.

• Water is the nucleophile, instead of halide.

• Product is Markovnikov and anti.

C Br C Br C C Br C C

+

H 3 O + O H 2 O H O H

Chapter 8

H 2 O H

=> 23

Predict the Product

Predict the product when the given alkene reacts with chlorine in water.

CH 3 D Cl 2 , H 2 O

Chapter 8

OH D CH 3 Cl

=> 24

Epoxidation

• Alkene reacts with a peroxyacid to form an epoxide (also called oxirane).

• Usual reagent is peroxybenzoic acid.

C C + R O C O O H C O C + R O C O H

=> Chapter 8 25

One-Step Reaction

• To synthesize the glycol without isolating the epoxide, use aqueous peroxyacetic acid or peroxyformic acid.

• The reaction is stereospecific.

O CH 3 COOH OH H H OH

=> Chapter 8 26

Syn Hydroxylation of Alkenes

• Alkene is converted to a

cis

-1,2-diol, • Two reagents:  Osmium tetroxide (expensive!), followed by hydrogen peroxide

or

 Cold, dilute aqueous potassium permanganate, followed by hydrolysis with base => 27 Chapter 8

POWER POINT IMAGES FROM “ORGANIC CHEMISTRY, 5 TH EDITION” L.G. WADE ALL MATERIALS USED WITH PERMISSION OF AUTHOR PRESENTATION ADAPTED FOR BURLINGTON COUNTY COLLEGE ORGANIC CHEMISTRY COURSE BY: ANNALICIA POEHLER STEFANIE LAYMAN CALY MARTIN Chapter 8 28