Chapter 8: Alkene Structure and Preparation via Elimination Reactions

2. Nomenclature of Alkenes• the longest continuous carbon chain containing both carbons of the C=C bond is the parent chain• assign the C=C bond the lowest locant value• in cyclic molecules the C=C bond is always given the 1,2-designation

3. Stereochemistry of Alkenes

• since the C=C bond requires overlap of the p-orbitals, bond rotation is prohibited• this gives rise to two different stereoisomers. Relationship?

1. Nature of the pi bond

HH

HCC

H

• a C=C double bond is stronger than a C–C single bond• the pi bond component, however, is generally weaker than the sigma bond component• unlike sigma bonds where rotation takes place readily, rotation about a pi bond wouldrequire breaking the pi bond (loss of overlap of the P-orbitals) and therefore does nottake place under normal conditions.

Cl

draw the otherdraw the other

P: 8.1, 8.2, 8.50(a,b)

[Sections: 8.1-8.13]

H3C

HCH3

HH3C

H3CH

HH3C

H

CH3H

H3C CH3

H2C CH2

bondlength

bondstrength

4. Alkene Stereochemistry and Eyesight

5. Alkene Nomenclature With More Than Two Substituents on the Double Bond

Cl CH3

Cl CH3

H3C

• look at one "end" of the double bond• assign priority of one group or atom over the otherbased on atomic number• repeat the process for the other end of the double bond• if the two high-priority groups are on: the same side of the double bond = Z isomer the opposite sides of the double bond = E isomer• cis isomers can also be called Z isomers, trans = E

HO H

ClH2N OH

Cl

BrCl

Z or E? Draw the other isomer

P: 8.5, 8.6, 8.51

Plan of Attack for naming alkenes

Cl CH3

H3C H

same endsame end

same side

same side

7. Relative Stability of Alkene Isomers

cis-2-buteneO2

CO2 + H2O + kJ/mol

trans-2-buteneO2

CO2 + H2O + kJ/mol

E

2,712 kJ/mol 2,710 kJ/mol 2,707 kJ/mol 2,700 kJ/mol

4 CO2 + 4 H2O (heats of combustion)

6. Alkenes and Pheromones

OH

1-octene-3-ol

mosquitosand fly

9-methyl-α-himalachenemale sex pheromone

oriental fruit moth

O

O

(E)-8-dodecen-1-yl acetate 7%

O

O

(Z)-8-dodecen-1-yl acetate 93%

sex phermonone mixture

H

H

H

H

H

H

R

H

R

H

R

H

H

R

R

H

R

R

H

H

R

R

R

H

R

R

R

R< < < < < <

8. Alkene Substitution Patterns

unsub monodisubstituted

tri tetra

stability

A B C D

Predict the relative stabilities of the following isomeric alkenes

9. Bimolecular Elimination Mechanism

H

HH Br

HH

HO

E

• identify starting materials and products• exothermic or endothermic?• multistep or concerted?• RDS = unimolecular or bimolecular?

rate law:• dependent upon concentrations of compounds during (and prior to) the RDS

rate = k

reaction name:

P: 8.4, 8.7, 8.8, 8.53

P: 8.13, 8.57

Predict the products of the following E2 reactions:

KOHEtOH, heat

Br

• E2 elimination results in the formation of all possible alkene products, including stereoisomers• E2 elimination using strong, small bases (HO–, MeO–, EtO–) results in formation of the most stable alkene product (Zaitsev's rule)• the most stable alkene is the most substituted

Predict all of the products of the following E2 reactions. Circle the major product:

OTs

Br

OTs

NaOtBu

tBuOH

NaOCH2CH3

ethanol, heat

NaNH2NH3, heat

LDA

THF

• identify the base• identify the leaving group• locate all β-hydrogens• draw products from removal of a β-hydrogen and the leaving group• draw stereoisomers where relevant• check to make sure no products are duplicated!

Plan of Attack for E2 reactions

Br

• 1° and 2° substrates could go via either pathway, and often the two reaction mechanisms compete with one another such that a little substitution occurs even though elimination is the preferred pathway, and a little elimination occurs even though substitution is the preferred pathway• for 3° substrates, SN2 is impossible, so E2 is the only possible reaction route• there are ways to influence reactivity towards one direction or another:

The SN2 reaction prefers strong nucleophiles that are weak bases, polar aprotic solvents and minimal, or no, heatThe E2 pathway prefers strong bases, polar protic solvents, and heat

Always Nucleophiles

X–, HS–, RS–, H2SRSH, CN–

Always BasestBuO–, LDANaH, NaNH2

Either/Or

HO–, RO–

Br

OCH3

NaOH

DMSO

NaOH

CH3OH, heat

H

HH Br

HH

HO

10. Substitution (SN2) or Elimination (E2)?

P: 8.64, 8.78, 8.79, 8.81

solvent heat

NH3

O

N Li

11. Unimolecular Elimination Mechanism

Most commonly used E1 type reaction is the dehydration of alcohols

OH

HCl

H3PO4or

H2SO4heat

• alcohols may react via either the SN1 or the E1 reaction• in both cases an acid is required to protonate the OH group to convert it to a good leaving group• the intermediate carbocation is then either trapped by a nucleophile (SN1) or loses a β-hydrogen (E1)• elimination is guaranteed to occur if the acid is H3PO4 or H2SO4 since the counterions from these acids are not nucleophilic• heat also favors the elimination process

OH

H3PO4heat

Draw all the products expected. Circle the major product.

P: 8.10-8.12, 8.27-8.34, 8.54, 8.58, 8.62, 8.67,8.68, 8.69, 8.73, 8.76

Br

EtOH

Brtert-butanol

heat

13. The Four Basic Reaction Mechanisms

SN2

Nu R LG Nu R

I- > Br- > Cl- or OTs- or H2O [from protonation of OH]

Leaving groups

NucleophilesX-, HS-, H2S, RSH, CN-, N3

- always act as nucleophilesHO-, RO-, NH3 can act as nucleophiles or bases

BasesHO-, RO-, NH3 can act as bases or nucleophiles

tBuO-, LDA, NaH, NaNH2, DBU, DBN

always act as bases

• methyl > 1° > 2° (NO 3°)• no intermediates, one step reaction (concerted)• rate = k [Nu][R-X]• inversion of configuration• favored by strong nucleophiles• favored by polar aprotic solvents

SN1

R LG Nu R

• 3° > 2° (NO 1° or methyl)• carbocation intermediate (multistep reaction)• carbocation subject to rearrangement• rate = k [R-X]• usually weak nucleophiles (solvent is often the nucleophile as well = solvolysis reaction)• favored by polar protic solvents

R Nu

E2 • can occur with all types of substrates• no intermediates, one step reaction (concerted)• rate = k [Base][R-X]• favored by strong bases (small bases favor Zaitsev elimination; large bulky bases favor Hofmann elimination)• favored by polar protic solvents• heating favors elimination over substitution

E1 • 3° > 2° (NO 1° or methyl)• carbocation intermediate (multistep reaction)• carbocation subject to rearrangement• rate = k [R-X]• usually very weak base present (solvent molecule or, for dehydration reactions, H2PO4

- or HSO4-)

• favored by polar protic solvents• heating favors elimination over substitution

BLG

Hβ

RR

RR R

R

R

R

LG

Hβ

RR

RR

R

Hβ

RR

R R

R

R

R

-H+