Chap 06 Alkene Reactions

7/30/2019 Chap 06 Alkene Reactions

1/79

Alkene Reactions


2/79

Pi bonds

Plane of molecule

Reactivity

above and

below the

molecular

plane!


3/79

Addition Reactions

A - B A

B

Important characteristics of addition reactions

Orientation (Regioselectivity)

If the doubly bonded carbons are not equivalent which one

get the A and which gets the B.

Stereochemistry: geometry of the addition.

Syn addition: Both A and B come in from the same side of

the alkene. Both from the top or both from the bottom.

Anti Addition: A and B come in from opposite sides (anti

addition).

No preference.


4/79

Reaction Mechanisms

Mechanism: a detailed, step-by-step description of how a reaction occurs.

A reaction may consist of many sequential steps. Each step involves a

transformation of the structure.

For the step C + A-B C-A + B

Reactants

Products

Transition State

Energy of

Activation.

Energy

barrier.

Three areas to be aware of.


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Energy Changes in a Reaction

Enthalpy changes, DH0, for a reaction

arises from changes in bonding in the

molecule.

If weaker bonds are broken and stronger ones

formed then DH0 is negative and exothermic.

If stronger bonds are broken and weaker ones

formed then DH0 is positive and endothermic.


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Gibbs Free Energy

Gibbs Free Energy controls the position ofequilibrium for a reaction. It takes into accountenthalpy, H, and entropy, S, changes.

An increase in H during a reaction favors

reactants. A decrease favors products.An increase in entropy (eg., more molecules being

formed) during a reaction favors products. Adecrease favors reactants.

DG0: if positive equilibrium favors reactants(endergonic), if negative favors products(exergonic).

DG0 = DH0 TDS0


7/79

Multi-Step ReactionsStep 1: A + B Intermediate

Step 2: Intermediate C + D

Step 1:

endergonic,

high energy of

activation.

Slow process

Step 2:exergonic, small

energy of

activation. Fast

Process.

Step 1 is the slow step, the

rate determining step.


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Characteristics of two step

Reaction 1. The Intermediate hassome stability. Itresides in a valley.

2. The concentration of

an intermediate is

usually quite low. The

Energies of Activationfor reaction of the

Intermediate are low.

3. There is a transition

state for each step. A

transition state is not a

stable structure.

4. The reaction

coordinate can be

traversed in either

direction: A+BC+D

or C+D A+B.


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Hammond Postulate

The transition state for a step isclose to the high energy end of the

curve.

For an endothermic step the

transition state resembles the

product of the step more than the

reactants.

For an exothermic step the

transition state resembles the

reactants more than the products.

Reaction coordinate.


10/79

Example

Endothermic

Transition state resembles the (higher

energy) products.

CH3 - H + Br CH3 + H - Br DH = 109 kJ

[H3C H Br]

Almostbroken.

Almost

formed.

Almost

formed

radical. Only a smallamount of radical

character remains.


11/79

Electrophilic Additions

Hydrohalogenation using HCl, HBr, HI

Hydration using H2O in the presence of H2SO4

Halogenation using Cl2, Br2

Halohydrination using HOCl, HOBr

Oxymercuration using Hg(OAc)2, H2O

followed by reduction


12/79

Electrophilic Addition

We now address regioselectivity.


13/79

Regioselectivity (Orientation)

The incoming hydrogen attaches to the carbon with the

greater number of hydrogens. This is regioselectivity.

It is called Markovnikov orientation.


14/79

Mechanism

Step 2

Step 1


15/79

Now examine Step 1 Closely

Rate Determining

Step. The rate at

which the

carbocation is

formed controlsthe rate of the

overall reaction.

The energy of

activation for

this process is

critical.

Electron rich, pi

system.

Showed thisreaction earlier as

an acid/base

reaction. Alkene

was the base.

New term: thealkene is a

nucleophile,

wanting to react

with a positive

species.

Acidic molecule, easily

ionized.

We had portrayed the HBrearlier as a Bronsted-

Lowry acid.

New term: the HBr is an

electrophile, wanting to

react with an electron richmolecule (nucleophile).

The carbocation intermediate is very

reactive. It does not obey the octet rule

(electron deficient) and is usually

present only in low concentration.


16/79

Carbocations

Electron deficient.

Does not obey octet rule.

Lewis acid, can receive

electrons.

Electrophile.

sp2 hybridized.

p orbital is empty and can receive

electrons.

Flat, planar. Can react on either

side of the plane.

Very reactive and present only in

very low concentration.


17/79

Step 2 of the Mechanism

Br

Br

Mirror objects

:Br-

:Br-


18/79

Regioselectivity (Orientation)

H - Br

H

+ Br

H

+ Br

HBr

2-Bromo-propane

HBr

1-Bromo-propane

Secondary

carbocation

Primary

carbocation

Secondary

carbocation more

more stable and

more easily formed.

Or


19/79

Carbocation Stabilities

Order of increasing stability:

Methyl < Primary < Secondary < Tertiary

Order of increasing ease of formation:

Methyl < Primary < Secondary < Tertiary

Increasing Ease of Formation


20/79

Factors Affecting Carbocation Stability -

Inductive

1. Inductive Effect. Electron redistribution due to

differences in electronegativities of substituents.

Electron releasing, alkyl groups, -CH3, stabilize the

carbocation making it easier to form.

Electron withdrawing groups, such as -CF3, destabilize

the carbocation making it harder to form.

HF

F

FH

d-

d+

d-

d-


21/79

Factors Affecting Carbocation

Stability - Hyperconjugation

2. Hyperconjugation. Unlike normal resonance or

conjugation hyperconjugation involves s bonds.

H

H

HH

HHH

H

HH

ethyl carbocation

Hyperconjugation spreads the positive charge onto the

adjacent alkyl group


22/79

Hyperconjugation Continued

Drifting of electrons from the filled C-H

bond into the empty p orbital of thecarbocation. Result resembles a pi bond.

Another description of the effect.


23/79

Factors Affecting Carbocation Stability -

Resonance

allylic carbocation

Utilizing an adjacent pi system.

H HH H

benzylic carbocation

H H H H

Positive charge delocalized through resonance.

Another

very

important

example.

Positive charge delocalized into the benzene ring. Increased stability of

carbocation.

Note: the allylic

carbocation can react at

either end!

The benzylic

carbocation will

react only at the

benzylic position

even thoughdelocalization

occurs!


24/79

Another Factor Affecting Carbocation

Stability Resonance

Utilizing an adjacent lone pair.

CH2

O

HCH2

O

H

Look carefully.

This is the

conjugate acid

of formaldehyde,

CH2=O.

P d ti f Chi l C t G l i t


25/79

Production of Chiral Centers. Goal is to see

all the possibilities.

The H willattach here.Regioselectivity Analysis:

the positive charge will go here

and be stabilized by resonance

with the phenyl group.

Ph Me

EtMe

Me is methyl group

Et is ethyl group

Ph is phenyl groupPh

Me

Et

Me

H

Ph Me

EtMe

H

mirror plane

Enantiomeric

carbocations.

Br-

PhMe

EtMe

HBr

Br-

Ph

Me

Et

MeH

Br

H+

H+

Br-

Ph MeEt

Me

HBr

Ph

Me

Et

MeH

Br

Br-

What has been made?

Two pairs of enantiomers.

React alkene with HBr.

Note that the ends of the double

bond are different.


26/79

Production of Chiral Centers - 2

PhMe

EtMe

HBr Ph

Me

Et

MeH

Br

Ph MeEt

Me

HBrPh

MeEt

MeH

Br

Racemic Mixture 1 Racemic Mixture 2

The product mixture consists of four stereoisomers, two pairs of

enantiomers

The product is optically inactive.

Distillation of the product mixture yields two fractions (different boiling

points). Each fraction is optically inactive.

Rule: optically inactive reactants yield opticallyinactive roducts either achiral or racemic .

diastereomers


27/79

Acid Catalyzed Hydration of

Alkenes

What is the orientation??? Markovnikov

M h i


28/79

Mechanism

Step 1

Step 2

Step 3

Note the electronicstructure of the

oxonium ion.


29/79

Carbocation Rearrangements

Expected product is not the major product; rearrangement of carbon

skeleton occurred.

The methyl group moved. Rearranged.


30/79

Also, in the hydration reaction.

The H moved.


31/79

Mechanism including the 1,2 shift

Step 1, formation of

carbocation

Step 2, the 1,2 shift of themethyl group with its pair of

electrons.

Step 3, the nucleophile

reacts with the

carbocation

Reason for Shift: Converting a

less stable carbocation (20) to a

more stable carbocation (30).


32/79

Addition of Br2 and Cl2

St h i t


33/79

Stereochemistry

Ant i Addi t ion (halogens enter on

oppos i te s ides); Stereoselective

Syn addition (on same side) does not

occur for this reaction.


34/79

Mechanism, Step 1

Step 1, formation of cyclic bromonium ion.

St 2


35/79

Step 2

Detailed Stereochemistry addition of Br2


36/79

Detailed Stereochemistry, addition of Br2

H3C CH3

C3H7 C2H5

Br

Br

(S) (R)

H3

C CH3

C3

H7

C2H5

Br

Br

Br

(S)

(S)

H3C

CH3C3H7

C2

H5

Br

Br

(R)

(R)

H3C

CH3

C3

H7

C2H5

Br

Br

(R) (S)

H3CCH3

C3H7 C2H5

Br

Br

Br

Br

(R)

(R)

H3C

CH3

C3H7

C2H5

Br

Br

(S)

(S)

H3C

CH3

C3H7

C2H5

Br

Br

enantiomers

Alternatively, the bromine

could have come in from the

bottom!

enantiomers

S,S

S,S

R,R

R,R

Only two compounds (R,Rand S,S) formed in equal

amounts. Racemic mixture.

Bromide ion

attacked the

carbon on the

right.

But can also

attack theleft-side

carbon.

f f


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Number of products formed.

(S)

(S)

H3C

CH3

C3H7

C2H5

Br

Br

enantiomers

enantiomers

S,S

S,S

R,R

R,R

We have formed only two products even though

there are two chiral carbons present. We know

that there is a total of four stereoisomers. Half

of them are eliminated because the addition is

anti. Syn (both on same side) addition does not

occur. (R)(R)

H3C

CH3

C3H7

C2H5

Br

Br

(R)

(R)

H3C

CH3

C3H7

C2

H5

Br

Br

(S)

(S)

H3C

CH3C3H7

C2H5

Br

Br


38/79

Attack of the Bromide Ion

(S) (R)

H3C CH3

C3H7 C2H5

Br

Br

(S)

(S)

H3C

CH3C3H7

C2H5

Br

Br

Starts as R Becomes S

The carbon was originally R with the Br on the top-side. It became S

when the Br was removed and a Br attached to the bottom.

In order to

preserve a

tetrahedralcarbon these

two substituents

must move

upwards.

Inversion.

P f A k


39/79

Progress of Attack

Things to watch for:

Approach of the red Br anion from the bottom.

Breaking of the C-Br bond.

Inversion of the C on the left; Retention of the C on the right.

U i Fi h P j ti


40/79

R1R2

R3R4

Br2

anti addition

R1R2

R3R4

Br

Br

+ enantiomer

Using Fischer Projections

Not a validFischer

projection since

top vertical

bond is coming

forward.

Convert to Fischer by

doing 180 deg rotationof top carbon.

+ enantiomer

Br

R1 R2

Br

R4 R3

=


41/79

There are many variations on the addition of X2 to

an alkene. Each one involves anti addition.

Br-

+ enantiomer

Br

R1 R2

Br

R4 R3

R2 R4

R1 R3

Br

I -

+ enantiomer

I

R1 R2

Br

R4 R3

+ enantiomer

Br

R1 R2

I

R4 R3

+

The iodide can attach to either

of the two carbons.

I -I -

Instead of iodide ion as nucleophile can

use alcohols to yield ethers, water to

yield alcohols, or amines.

R1

R2

R3R4

Br2


42/79

Regioselectivity

If Br2 is added to propene there is no regioselectivity issue.

Br2

Br

Br

If Br2 is added in the presence of excess alternative nucleophile, such as

CH3OH, regioselectivity may become important.

Br - Br

OCH3

Br

CH3O-H

Br

OCH3and/or

+ H+

+ Br-

+ H+

+ Br-


43/79

Regioselectivity - 2Consider, again, the cyclic bromonium ion and the resonance structures.

R

BrWeaker

bond

More positive charge

Stronger bond

Expect the nucleophile to attack here. Remember inversion occurs.

R i l ti it B i I


44/79

Regioselectivity, Bromonium Ion

Bridged bromonium ion from propene.

Example


45/79

Et

H Me

Me

Me

Cl2/H2O

H

p

Regioselectivity,

addition of Cl and OH

Cl, from the electrophile Cl2,

goes here

OH, the nucleophile, goes

here

Stereochemistry: anti addition

Note: non-reacting fragment unchangedEt

H Me

Me

Me

Et

H Me

Me

Me

+

Cl

H

H

OH

Cl

OH

Put in Fisher Projections. Be

sure you can do this!!

Et

H Me

Me OH

Me

H Cl

Et

H Me

HO Me

Me

Cl H

+

B i ti f b tit t d l h


46/79

Bromination of a substituted cyclohexene

Consider the following bromination.

C(CH3)3

Br2

Expect to form

two bromonium

ions, one on top

and the other on

bottom.

C(CH3)3

Br

+

C(CH3)3

Br+

+

Expect the rings

can be opened

by attack on

either carbon

atom as before.

But NO, only one

stereoisomer is

formed. WHY?

C(CH3)3

Br-Br

Br

C(CH3)3

Br

Br

+

Addition to substituted cyclohexene


47/79


HH

Br2

The tert butyl grouplocks the

conformation as

shown.

Br

Br

H

H

H H

+

The cyclic

bromonium ion can

form on either the top

or bottom of the ring.

How can the bromide ion come in?

Review earlier slide showing that the

bromide ion attacks directly on the side

opposite to the ring.


48/79

Progress of Attack

Things to watch for:

Approach of the red Br anion from the bottom.

Breaking of the C-Br bond.

Inversion of the C on the left; Retention of the C on the right.

Notice that the two bromines are

maintained anti to each other!!!



49/79


Br2

Br

Br

+

Observe

Ring is locked as shown. No ring flipping.

Attack as shown in red by incoming

Br ion will put both Br into equatorial

positions, not anti.

Br

Br

This stereoisomer is not

observed. The bromines

have not been kept anti to

each other but have become

gauche as displacement

proceeds.

Be sure to allow for the

inversion motion at the carbon

attacked by the bromide ion.



50/79


Br2

Br

Br

+

Attack as shown in green by the

incoming Br will result in both Br being

axial and anti to each other

Br

Br

This is the observed

diastereomer. We have kept

the bromines anti to eachother.

O ti R d ti


51/79

Oxymercuration-Reduction

Regioselective: Markovnikov Orientation

Occurs without 1,2 rearrangement, contrast the following

3,3-dimethylbut-1-ene

H2O

H2SO4

OH

formed via

rearrangement

1 Hg(OAc)2

2. NaBH4OH

No rearrangement

Alkene Alcohol


52/79

Mechanism

1

2

3

4


53/79

Hydroboration-Oxidation

Alkene Alcohol

Anti-Markovnikov orientation

Syn addition

1. BH3

2. H2O2

H

HO

H

HO


54/79

Borane, a digression

Isoelectronic with a carbocation

B B

H

HH

H

HH


55/79

MechanismSyn stereochemistry, anti-

Markovnikov orientationnow established.

Two reasons why anti-Markovnikov:

1. Less crowded transition state for B to approach the

terminal carbon.

2. A small positive charge is placed on the more highly

substituted carbon.

Just call the circled group R.

Eventually have BR3.

Next


56/79

Contd

Oxidation and Reduction Reactions


57/79

Oxidation and Reduction Reactions


58/79

We think in terms of Half Reactions

Write reactants and products of each halfreaction.

Cr2O7 2- + CH3CH2OH Cr3+ + CH3CO2H

Cr2O72- 2 Cr3+

Balance oxygen by adding water

+ 7 H2O

In acid balance H by adding H +

14 H+ +

Balance charge by adding electrons

6 e - +

Inorganic half reaction

If reaction is in base: first balance as above for acid and then add OH- to both sides to

neutralize H +. Cancel extra H2O.

Will be

oxidized.

Will be

reduced.


59/79

Contd

Now the organic half reaction

Balance oxygen by adding water

In acid balance H by adding H +

Balance charge by adding electrons

CH3CH2OH CH3CO2HH2O + + 4 H+ + 4 e-

Combine half reactions so as to cancel electrons

CH3CH2OH CH3CO2HH2O + + 4 H+ + 4 e-

Cr2O72- 2 Cr3+ + 7 H2O14 H

+ +6 e - +

3 x ( )

16 H+ +2 Cr2O72- + 3 CH3CH2OH 4 Cr

3+ + 3 CH3CO2H + 11 H2O

2 x ( )


60/79

Formation of glycols with SynAddition

Osmium tetroxide

Syn addition

KMnO4

cold, dilute, slightly alkaline

also KMnO4


61/79

Anti glycols

PhCO3H, a peracid

O

H+

O

H

H2

O

HO

OH

Using a peracid, RCO3H, to form an epoxide which is opened by aq. acid.

Peracid: for example, perbenzoic acid

O O

OH

The protonated epoxide is analagous

to the cyclic bromonium ion.

epoxide


62/79

An example

chiral, optically active

(S)-3-methylcyclohex-1-ene

PhCO3H

O + O

aq. acid

OH

OH

OH

OHOH

OH

OH

OH

Are these

unique?

Diastereomers, separable (in theory) by

distillation, each optically active


63/79

Ozonolysis

R3

R4

R1

R2

1. O3

2. (CH3)2SR4

R3

O

R1

R2

O+

Reaction can be used to break larger molecule down into smaller parts

for easy identification.


64/79

Ozonolysis Example

For example, suppose an unknown compound had the formula C8H12 and upon

ozonolysis yielded only 3-oxobutanal. What is the structure of the unknown?

The hydrogen deficiency is 18-12 = 6. 6/2 = 3 pi bonds or rings.

The original compound has 8 carbons and the ozonolysis product has only 4

Conclude: Unknown two 3-oxobutanal.

Unknown

C8H12

ozonolysysO

O

O

O

Simply remove the new oxygens and join to make double bonds.

But there is a second possibility.

O

O

Another Example


65/79

Another Example

2. An unknown compound (derived from the gall bladder of the gila monster) has the formulaC10H14 . When subjected to ozonolysis the following compound is isolated

O

O O

O

Suggest a reasonable structure for the unknown.

Hydrogen Deficiency = 8. Four pi

bonds/rings.

Unknown has no oxygens. Ozonolysis

product has four. Each double bondproduces two carbonyl groups. Expect

unknown to have 2 pi bonds and two rings.

To construct unknown cross out the oxygens and then connect. But there are many

ways the connections can be made.

a

b

c

d

a-b & c-d

a b

c

da-c & b-d

ac

d

b

a-d & b-c

ad

c

b

Look for a structure

that obeys the

isoprene rule.


66/79

Mechanism

OO

OO

O

OO

O

O O

O

O

Consider the resonance structures of ozone.

These two, charged at

each end, are the useful

ones to think about.

Electrophile

capability.Nucleophile

capability.


67/79

Mechanism - 2

O

O

OO

O

OO

O

O O

O

O


68/79

Mechanism - 3


69/79

Mechanism - 4


70/79

Hydrogenation

No regioselectivity

Syn addition

Heats of Hydrogenation


71/79


Consider the

cis vs trans

heats of

hydrogenationin more

detail


72/79

Heats of Hydrogenation - 2The trans alkene has a lower heat of hydrogenation.

Conclusion:

Trans alkenes with lower heats of hydrogenation are more stable than cis.

We saw same kind of reasoning when we talked about heats of combustion of

isomeric alkanes to give CO2 and H2O



73/79


Increasingsub

stitution

ReducedheatofHydrogenation

By same reasoning higher degree of substitution provide lower heat of

hydrogenation and are, therefore, more stable.


74/79

Acid Catalyzed Polymerization

Principle: Reactive pi electrons (Lewis base) can react with Lewis acid. Recall

Which now reacts with a Lewis base,

such as halide ion to complete additionof HX yielding 2-halopropane

Variation: there are other Lewis bases available. THE ALKENE.

+ HH

The new carbocation now reacts with a Lewis base such as

halide ion to yield halide ion to yield 2-halo-4-methyl

pentane (dimerization) but could react with another

propene to yield higher polymers.

the carbocation is an acid!

+


75/79

Examples ofSynthetic Planning

Give a synthesis of 2-hexanol from any alkene.

OH

Planning:

Alkene is a hydrocarbon, thus we have to introduce the OH group

How is OH group introduced (into an alkene): hydration

What are hydration reactions and what are their characteristics:

Mercuration/Reduction: Markovnikov

Hydroboration/Oxidation: Anti-Markovnikov and syn addition


76/79

What alkene to use? Must involve C2 in double bond.

Which reaction to use with which alkene?

Markovnikov rule can be

applied here. CH vs CH2.

Want Markovnikov!

Use

Mercuration/Reduction!!!

Markovnkov Rule cannot be

used here. Both are CH.

Do not have control over

regioselectivity.

Do not use this alkene.

For yourself : how would you make 1 hexanol, and 3-hexanol?

Another synthetic example


77/79

How would you prepare meso 2,3 dibromobutane from an alkene?

Analysis:

Alkene must be 2-butene. But wait that could be eithercis ortrans!

We want meso. Have to worry about stereochemistry

Know bromine addition to an alkene is anti addition (cyclic bromonium ion)


78/79

trans

Br2

Br

Br

H

Br

Br

rotate lower unit

Br H

Br H

meso

This worked! How about

starting with the cis?

cis

Br2

H Br

Br H

racemic mixture

+ enantiomer

This did not work, gave us

the wrong stereochemistry!

Addition Reaction General Rule


79/79

Characterize Reactant as cis ortrans, C or T

Characterize Reaction as syn oranti, S or A

Characterize Product as meso orracemicmixture, M or R

Relationship

C RA

cis

Br2H Br

Br H

racemic mixture

+ enantiomer

Characteristics can be

changed in pairs and C A R willremain true.

Want meso instead?? Have to

use trans. Two changed!!

A

T M

t

Br H

Br H

Br2

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Chap 06 Alkene Reactions

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