2011 Ch 7 & 8

8/3/2019 2011 Ch 7 & 8

1/67

Chs. 7 & 8: Alkenes

8/3/2019 2011 Ch 7 & 8

2/67

Alkenes (olefins): hydrocarbons containing C=C. Unsaturated less than 4 single (sigma) bonds to C.

Simplest alkene = ethene

C-C bond length = 1.34 average bond angle = ~120o

.

H

H

H

HC=C

ethene (ethylene)

8/3/2019 2011 Ch 7 & 8

3/67

H

H

H

H

pi bond - sideways overlap of p orbitals

ethene (ethylene)

sigma bond - electron density along

bond axis

Orbital picture shows & bonding

electron density is along axis of bond and above and below planeof molecule in bond.

8/3/2019 2011 Ch 7 & 8

4/67

I. Nomenclature : IUPAC rules

Alkenes

1. Select the longest carbon chain containing the double bond asthe parent. Name as ethene, propene, butene, pentene, etc.

(Alkane name minus ane plus ene)

2. # the C chain of parent so that C=C has the lowest possiblenumber (locant).

5 4 3 2 1Ex: CH

3CH

2CH=CHCH

3

If more than one way of numbering gives lowest # to C=C, choosethe way that gives the lowest #s to branch points.

8/3/2019 2011 Ch 7 & 8

5/67

CH3CH3

CH3

CH3 CH3

CH3

123

4

56

NOT

65

4

3

21

8/3/2019 2011 Ch 7 & 8

6/67

3. Write full name #ing substitutents & listing them alphabetically.

CH3CH3

CH3

CH3 CH3

2-pentene 2-methyl-3-hexene

8/3/2019 2011 Ch 7 & 8

7/67

4. If two or more double bonds are present, choose as the parent thelongest C chain containing both and use the prefixes di, tri, tetra, etc.

before ene.

CH2CH3

H

H

H

H

H

CH3H

H

1,3-pentadiene 2-ethyl-1,3-butadiene

8/3/2019 2011 Ch 7 & 8

8/67

Cycloalkenes

1. Give double bond lowest locant 1.

2. Number substituents thru the double bond so that they have thelowest possible locants.

3. List the substituents alphabetically.

4-methylcyclohexne

1-ethyl-4-methyl-1,3-cyclohexadiene

5,5-dimethyl-1,3-cyclohexadiene

8/3/2019 2011 Ch 7 & 8

9/67

Common names

CH2=CH2 ethylene

CH3-CH=CH2 propylene

H2C=CH- vinyl (IUPAC: ethenyl)

H2C=CH-CH2- allyl (IUPAC: 2-propenyl)

8/3/2019 2011 Ch 7 & 8

10/67

II. Cis/trans stereoisomerism

- Occurs in alkenes because there is no rotation about C=C

CH3H

H CH3

HCH3

H CH3

is not the same as

8/3/2019 2011 Ch 7 & 8

11/67

Geometric isomers (cis/trans) exist for every alkene which hasnonidentical groups attached to both C atoms of double bond

CH3 CH3

CH3CH2 H

CH3 H

CH3CH2 H

H CH3

CH3CH2 H

-no cis/trans isomerism

cis-2 pentene trans 2-pentene

cis: substituents on same side of double bond

trans: substituents on opposite sides of double bond

8/3/2019 2011 Ch 7 & 8

12/67

As molecules get larger and more complex, cis/trans nomenclaturegets complicated.

H CH3

CH2CH3CH3

CH3

For cpds like:

use Z,E nomenclature1. Assign priority to groups on each C of double bond

2. If high priority groups on both C's are on same side of the double bond,use Z (Zusammen - together).

3. If high priority groups on both C's are on opposite side of the double bond,use E (Entgenen -apart).

CC

high high

lowlow Z

CC

high low

highlow E

8/3/2019 2011 Ch 7 & 8

13/67

Assignment of priority: Cahn-Ingold-Prelog Sequence Rules

1. Assign priorities of the atoms directly attached to C atoms of thedouble bond according to atomic #.higher atomic # = higher priority

C(6) > H(1)

N(7) > C(6)

O(8) > N(7)

D(2) > H(1) Isotopes = special case since they have the same atomic #Use atomic wt. # instead.

2. If priority cannot be established by rule #1, make a similar

comparison of the next atoms in each group.

8/3/2019 2011 Ch 7 & 8

14/67

3. Multiply-bonded atoms are equivalent to the same # of singlybonded atoms

CH CH3

H

C

H

H

Ex:

= -C

H

C

C

C H

H

-C C - H = - C C H

C C

C C

=- C N

N

N

C

C

-C N:

8/3/2019 2011 Ch 7 & 8

15/67

CH3

O

H

O

= -C

O

O

C

H

H

H

C

= -C

O

O

H

C

8/3/2019 2011 Ch 7 & 8

16/67

Designate the following molecules as E or Z

H

Cl

Br

CH3

H

CH2OH

OHO

H

Br

H

CH3

CH3

Z Z E

8/3/2019 2011 Ch 7 & 8

17/67

III. Stability of alkenes

a) cis vs trans

Generally, cis isomers are less stable than trans isomers

E = 2-3 kJ/mol

CH3

H

H

CH3

CH3

H

CH3

H

trans 2-butene cis 2-butene

-less stable: higher E due to stericstrain. The larger the groups on thedouble bond, the greater the Edifference between cis and trans.

8/3/2019 2011 Ch 7 & 8

18/67

Energy difference is reflected in

1) interconversion equilibria and 2) heats ofhydrogenation

Interconversion equilibria:

G = -RTlnK = -(8.314J/K)(298K)ln(.24/.76) 2.8kJ

CH3

H

H

CH3CH3

H

CH3

H acidcat.

at 298K: 76% 24%

8/3/2019 2011 Ch 7 & 8

19/67

heats of hydrogenation:

H2, cat H2, cat

trans 2-butane butane cis 2-butene

8/3/2019 2011 Ch 7 & 8

20/67

Hhyd (cis)= -120 kJ/mol

Hhyd (trans) = -116 kJ/mol

Energy difference = 4 kJ/mol (close to 2.8 kJ, 4kJ is from H datawhile 2.8 kJ = G data)

b) stability of mono - vs polysubstituted alkenes

General trend:

R2C=CR2 > R2C=CHR > RCH=RCH R2C=CH2 > RCH=CH2

more stable less stable

lower Hhyd higher Hhyd

8/3/2019 2011 Ch 7 & 8

21/67

Explanation of observed stability trend:

-hyperconjugation occurs when a pi orbital can overlap (partially)with a neighboring sp3 orbital. only happens when C=C issubstituted. In the figure, the pi orbital can interact with 3 sp3

orbitals.

-greater bond strength for sp2-sp3 bonds than for sp3-sp3 bonds.

Compare CH3-CH=CH-CH3 (disub.) and CH3-CH2-CH=CH2 (monosub.)

2 sp3-sp2 bonds vs 1 sp3-sp3 and 1 sp3-sp2

8/3/2019 2011 Ch 7 & 8

22/67

Reactions of Alkenes 4 types well examine

1. electrophilic addition

2. hydrogenation (reduction)

3. oxidation4. radical addition

8/3/2019 2011 Ch 7 & 8

23/67

Electrophilic Addition

I. Addition of acid: H-B -acid catalyzed

Y z

+ Y-Z

nucleophile electrophile

R2C=CHR RR

B HR

H+ H-B

H-B = HX (X=Cl, Br, I)H-HSO4

H-OH

8/3/2019 2011 Ch 7 & 8

24/67

A) Mechanism

R2C=CHR

RR

B HR

H

BH R

R

H

R

H

B

R

R

H

R

HB

++

+

++

Rate determining step

1.

2.

8/3/2019 2011 Ch 7 & 8

25/67

For H-Br + (CH3)2C=CH2

8/3/2019 2011 Ch 7 & 8

26/67

Outline the mechanism:

8/3/2019 2011 Ch 7 & 8

27/67

Rate law

from RDS: rate = k[alkene][H-B]

first order in alkene

first order in H-B

second order overall

8/3/2019 2011 Ch 7 & 8

28/67

B) intermediate of rxn = carbocation

structure of R3C+:

Relative stabilities of carbocations effect of alkylgroups

R3C+ > R2CH+ > RCH2+ > CH3+most stable least stable

3o > 2o > 1o > CH3+

RR

R

120o

-sp2 hybridized-empty p orbital-6 valence electrons

+

8/3/2019 2011 Ch 7 & 8

29/67

How do alkyl groups stabilize the positive charge?

1. Hyperconjugation

- Overlap between vacant p orbital and an adjacent C-H bond (sp3 orbital)

- Not possible in methyl cation

- The more R groups, the more hyperconjugative interactions are possible.

H

H

H

H

H

H+

+

=

8/3/2019 2011 Ch 7 & 8

30/67

2. Inductive effects

- alkyl groups donate e- density by induction toward (+) charge.

- the more alkyl groups, the greater the stabilizing effect.

C) Orientation of addition

There are two possible products from the addition of H-B

Rxn is regioselective: only one of two possibledirections of addition is observed.

R2C=CHR R

R

B H

R

H

R2C=CHR R

R

H B

R

H

+ H-B

OR

+ H-B

observed product

8/3/2019 2011 Ch 7 & 8

31/67

Markovnikovs Rule: For electrophilic addition of HB toan alkene, the acid H will become attached to the doublebond C already bonded to the greater # of Hs.

Predict the products: H-Br

ether

H-Cl

ether

8/3/2019 2011 Ch 7 & 8

32/67

To understand:

Q: What does it mean to obtain only one of two possibleproducts?

A: Favored product is formed faster.RDS for two possibilities:

CH3

CH3

CH3

H

CH3

CH3

Br

CH3

H

H

CH3

CH3

H

CH3

H

BrH-Br

or

product

CH3

CH3

CH3

H

CH3

CH3CH

3

H

H

CH3

CH3

CH3

H CH3

CH3

H

CH3

H

H-Br+

+ Br-

H-Br

++ Br-

8/3/2019 2011 Ch 7 & 8

33/67

Shown is the energy profile for the RDS

The rate is determined by the activation energy energydifference between reactant and transition state of RDS.

The rate is proportional to e (-Ea/RT). The greater the Ea,the slower the reaction.

8/3/2019 2011 Ch 7 & 8

34/67

Since the Markovnikov product is the favored product, itmust be formed faster, that is, with a lower Ea. Thereforeits TS must be lower in energy than the TS of the non-

Markovnikov product.

Could this have been predicted?

Hammond Postulate:

For an endothermic reaction step, the TS resembles theproduct.

For an exothermic reaction step, the TS resembles thereactant.

8/3/2019 2011 Ch 7 & 8

35/67

For the RDS in the addn of H-B, the TS resembles thecarbocation product of that step, so the more stable thecarbocation, the more stable the TS, the lower the E

a

,the faster the reaction.

So, compare the carbocations arising from theMarkovnikov and non-Markovnikov pathways.

CH3

CH3CH3

H

H

CH3

CH3

H

CH3

H

+

+

Markovnikovpathway

3o carbocation

2o carbocationNon-Markovnikovpathway

8/3/2019 2011 Ch 7 & 8

36/67

stability of carbocations: 3o

> 2o

> 1o

E of the Mark. TS < E of non-Mark. TS

Ea (Mark. pathway) < Ea (non-Mark. pathway)

So Mark. pathway is faster than non-Mark pathway

8/3/2019 2011 Ch 7 & 8

37/67

Predict the products:

Write the mechanism for the acid catalyzed reactionbetween 1-methylcyclohexene and water (the 3rd

reaction)

H-Cl

ether

H2SO4H2O

H2O, H+

All three reactions pass through the same intermediate carbocation.What is its structure?

8/3/2019 2011 Ch 7 & 8

38/67

D. Rearrangement of carbocations

In some cases, more than one product may be obtainedas a result of rearrangement of the carbocationintermediate.

CH3

CH3

HH

H

H(CH3)2CH CH3

Cl

H

CH3

CH3

Cl

CH2CH3

H-Cl

ether

+

predicted Mark. product

rearranged product -major product

Example:

8/3/2019 2011 Ch 7 & 8

39/67

Provide a mechanism for the formation of the second product.

CH3

CH3

H

H

H

H

CH3

CH3CH

2CH

3

CH

CH3

CH3CHCH3

CH3

CH3

H

CHCH3CH3

CH3CH2CH3

CH3

CH3

H

CHCH3CH3

CH3CH2CH3

Second Q: What carbocation is originally produced (product of Mark. addition of H+)?

H-Cl

First ask: What intermediate carbocation produced the rearranged product?

A.

A.

Third Q: How do you get from initial R+ to the rearranged R+?

A.

Justification: conversion of a 2o carbocation to a 3o carbocation with increase in stability

2o 3o

1,2 hydride shift

+

+

+

+

+

+

8/3/2019 2011 Ch 7 & 8

40/67

Other possible rearrangement: 1,2-alkyl shift

H-Cl

H-Cl

Cl-

Example:

C

8/3/2019 2011 Ch 7 & 8

41/67

II. Addition of X2 (X2 = Cl2, Br2; F2 is too reactive;

I2 is slow & not as useful)

A. Mechanism

R2C=CR2X2 R

X

R

X

R

R (a vicinal dihalide)

R

R

R

R

Br Br

Br

R RR R

Br

R R

R R

Br

Br

R

RBr

Br

R

R

++

cyclic bromonium ion

+

+

+

8/3/2019 2011 Ch 7 & 8

42/67

B. Stereochemistry of addition = anti

due to cyclic intermediate: Br-

must add to side oppositeto first Br.

Br

R R

R RBr

R

RBr

Br

R

R

Br

R R

R RBr

RR

Br

Br

R

R

+ +

OR

+ +

8/3/2019 2011 Ch 7 & 8

43/67

Predict the product and give a mechanism:

Br2

CCl4

8/3/2019 2011 Ch 7 & 8

44/67

C. The Nu: in the 2nd step of the rxn can be any nucleophile.

R2

C=CR2 R2C CR2

BrBr

Cl

NO3-

Br

BrBr

ClBr

I

Br

NO3Br

OHhalohydrin

+

(NaCl)

I-

(NaI)

(NaNO3

)

H2O

formation

8/3/2019 2011 Ch 7 & 8

45/67

Halohydrin

H2C=CH2 HO-CH2CH2-Br

Cl2, H2O

HO-CH2CH2-Cl

C C

X OH

2-bromoethanol- a bromohydrin

Br2, H2O

2-chloroethanol- a chlorohydrin

8/3/2019 2011 Ch 7 & 8

46/67

For asymmetrical alkenes, 2 products are possible.

R2C=CHR

Br2, H2O

R

Br

R

OH

R

H R

OH

R

Br

R

Hor

observed product

-halogen becomes attached to the double bond C already bonded to the greater # of H atoms

8/3/2019 2011 Ch 7 & 8

47/67

To understand:

Br

R R

R H

Br

R R

R H+

resembles 3o

carbocationhas more (+) character = point ofattack by nucleophile

resembles 2o carbocation

+

+ +

8/3/2019 2011 Ch 7 & 8

48/67

When nucleoophile (H2O) attacks in 2nd step, it attacks the morehighly substituted C atom.

Br

H R

R ROH2

H

R

Br

O

R

R

H H

H

R

Br

O

R

R

H

+ +

+

-H+

..

:

:

: :

8/3/2019 2011 Ch 7 & 8

49/67

Mechanism for halohydrin formation: noteregiochemistry and stereochemistry

Br2

OH2

OH

CH3Br

H

Br Br

H CH3Br

Br

H CH3Br

OH2

HCH3Br

O H

H

H CH3Br

O H

H BrOH

CH3

Br

H

1.

+

+

2.

+

+

3.

+

+ HBr

8/3/2019 2011 Ch 7 & 8

50/67

III. Addition of H2O hydration 2 methods

A. oxymercuration - demercuration

-regiochemistry = Markovnikov

R2C=CHR

Hg(OAc)2, H2O1.

2.

- non-stereospecific

8/3/2019 2011 Ch 7 & 8

51/67

Example: OH

CH3CH3

CH3CH2

H

H

HgOAc

HH

HgOAc OH2

OH

HH

HgOAc

H

OH

HH

HgOAc

H

OH

HH

NaBH4

Hg(OAc)2, H2O

AcO-Hg-OAc -OAc

-OAc

1.

2.

NaBH4

(exact mechanismnot known)

not stereospecific - syn & anti addition is observed

+

+

+

+

+

8/3/2019 2011 Ch 7 & 8

52/67

B. Hydroboration oxidation non-Mark. Addition

- syn stereochemistry

BH3, THF H2O2, OH-

HBH2

OH

D

H

H H

H2O2, OH-

2.

8/3/2019 2011 Ch 7 & 8

53/67

D CH2CH3

OH

D

H

CH2CH3

D CH2CH3

H

D

BH2

CH2CH3

H

D CH2CH3

H2B

Mechanism:BH3, THF H2O2, OH-

H-BH2

(- H and BH2 are added to the same side of the dbl bond - synstereochemistry- BH2 is added to the less substituted C atom - results in less steric

crowding in transition state - non-Mark. regiochemistry)

1.

-----

CH3

HCH

3

D

OH

CH2CH

3

CH3

D

CH3

CH2CH

3

CH3

D

CH3

CH2CH

3

BH3 CH

3

D

CH3

CH2CH3

H2B H

CH3

H

CH3

D

BH2

CH2CH

3

CH3

H

CH3

D

BH2

CH2CH

3

CH3

H

CH3

D

OH

CH2CH

3

BH3, THF H2O2, OH-

1. + -----

2.

-BH2 and -H add to the same side of the dbl bond - syn stereochemistry

-BH2 is added to the less substituted C atom and -H is added to the more highly substituted C atom

because that reduces crowding in the transition state. Since -BH2 is later replaced by -OH,

non-Markonikov regiochemistry results.

H2O2, OH-

8/3/2019 2011 Ch 7 & 8

54/67

IV. Addition of carbene - R2C:

R2C:

CR2

:CCl2

I-CH2-Zn-I

+

cyclopropyl ringcarbene is generated"in situ"

Example:

KOH

or KOt-bu

(dichlorocarbene)

CH2I2Zn(Cu) (carbenoid - adds :CH2)

- Simmons-Smith reaction

CHCl3

8/3/2019 2011 Ch 7 & 8

55/67

Reaction of carbenes with alkenes is stereospecific:

CHCl3

KOH

CH2I2, Zn(Cu)

CH3

H CH3

H CHCl3

KOH

CH2I2, Zn(Cu)

CH3

H H

CH3

H

CH3

H

CH3

CH3

H

H

CH3

Cl

Cl

H

CH3

CH3

H

H

CH3

H

CH3

Cl

Cl

cis

trans

cis

cis

trans

trans

8/3/2019 2011 Ch 7 & 8

56/67

Hydrogenation (reduction)

H

R

R

H

R

R

R2C=CR2

catalyst =

H2

Addition is syn

Pt, PtO2, Pd/C, Ni

+

H2, cat.

8/3/2019 2011 Ch 7 & 8

57/67

Under the relatively mild rxn conditions (Rm temp.)which reduce C=C, other multiple bonds are not

reduced.

Pd/C

phenyl groups:

carbonyl groups:

cyano groups:

H2

Pd/C

H2

PtO2

H2

PtO2

H2

8/3/2019 2011 Ch 7 & 8

58/67

Oxidation

A. Epoxidation1. via halohydrin

CH3 Br2

CH3Br

OHH

CH3

O

CH3Br

OH

H

OH-

CH3Br

:O:-H

CH3

O

H2O

mechanism of epoxide formation:

..

via halohydrin

epoxide(oxirane)

::

::

epoxide(oxirane)

8/3/2019 2011 Ch 7 & 8

59/67

2. via peroxyacid

CH3

CH3

O

R O

O

OH

epoxide

(oxirane)

8/3/2019 2011 Ch 7 & 8

60/67

B. Hydroxylation glycol formation (glycol = cpd with vicinal hydroxygroups)

H

OHOH

H

O

HH

H

OHH

OH

OsO4, pyridine

NaHSO3 (aq)

RCOOOH

1.

2.

H3O+

cis

trans

To get cis glycol:

To get trans glycol:

8/3/2019 2011 Ch 7 & 8

61/67

C. Oxidative Cleavage - 3 types

1. ozonolysis

2. permanganate oxidation

3. periodate oxidation of glycols

8/3/2019 2011 Ch 7 & 8

62/67

R4

R3R1

R2 O

R1

R2 O

R3

R4Zn,H+

O3

+

1.

2.

1. ozonolysis of alkenes

R = alkyl, H

8/3/2019 2011 Ch 7 & 8

63/67

H

HR

H

KMnO4, H+

KMnO4, neutral R OH

O

H

R3R1

R2

KMnO4, H+

KMnO4, neutral

O

R2

R1

R3OH

O

R = alkyl

or + CO2

or+

=CH2

=CHR

=CR2

CO2, carbon dioxideRCOOH, carboxylic acid

R2C=O, ketone

2. permanganate oxidation of alkenes

8/3/2019 2011 Ch 7 & 8

64/67

R4

R3R1

R2

OsO4, pyr

NaHSO3 (aq) R4

OHOH

R2

R1 R3

HIO4

O

R1

R2

OR4

R3

1.

2.

3. periodate oxidation of glycols (result like ozonolysis)

alkene glycol

8/3/2019 2011 Ch 7 & 8

65/67

Radical Polymerization use fish hook arrows in mechanism

nH2C=CH2

RO-OR

H

H

H

H

RO *n

RO OR

H2C=CH2

H2C=CH2

R R

3 steps in any radical reaction:

1) initiation: generation of a free radical 2RO.

2) propagation:

RO-CH2-CH2-CH2-CH2.

RO-CH2-CH2. +

Ex: Combination of 2 radicals

R-R

attack of alkene by a free radical generates another free radical

- generates another free radical

RO. + RO-CH2-CH2.

3) termination: ends the polymer and doesn't produce a radical

+

8/3/2019 2011 Ch 7 & 8

66/67

Polymerization of unsymmetrical alkenes to predict product, needto take into account the stability of radicals.

observed product not observed

H2C=CHR

H2C=CHR CH2-CH-CH2-CHRRO

R

H2C=CHR

H2C=CHR CH-CH2-CH-CH2RO

R R

Mechanism for formation of first product:

1. RO - OR 2RO.

2. RO. + RO-CH2-CHR

.

RO-CH2-CHR

.

3. +

.

Mechanism for formation of second product:

1. RO - OR 2RO.

2. RO. + RO-CHR-CH2

.

RO-CHR-CH2

.

3. +.

H2C=CHRRO CH

2-CHR *n

RO CHR-CH

2 *n

RO-ORn or ??

Q

8/3/2019 2011 Ch 7 & 8

67/67

H2C=CHR

RH

H HRO

HR

H H

RO

Why do radicals in step 2 & step 3 always add to the less hindered C atom?

A:

RO.

+or

. 2o radical

. 1o radical

Order of radical stability:

3o > 2o > 1o >.CH3

so the more stable radical is formed.

Q:

So the more highly substituted radical is formed

Documents

2011 Ch 7 & 8