matriculation chemistry ( hydrocarbon ) alkene

ALKENESALKENES 12.12.22

General formula CnH2n , n 2.

Functional group double bond C=C

UnsaturatedUnsaturated hydrocarbon

Each carbon atom ( C=C ) is sp2 hybridizedhybridized.

Restricted rotation of carbon-carbon double bond causes cis-trans isomerism

IUPAC NomenclatureIUPAC Nomenclature

Determine the parent name by selecting the longest chain that contains the double bond and change the ending ‘-ane’ in alkane to ‘-ene’.

Step 1

When the chain contains more than three carbon atoms, a number is need to indicate the location of the double bond.

The chain is numbered starting from the end closest to the double bond..

Step 2

CH2

6

CH2

7

CH3

8

CH2

5

CH4

CH3

CH2

2

CH3

1

3-octene (not 5-octene)

Indicate the position of the substituent by the number of the carbon atoms to which they are attached.

C2

CH3

CH34

CH31

CH3

2-methyl-2-butene(not 3-methyl-2-butene)

Step 3

C2

CH3

CH2

4

CH3

1

CH3

CH5

CH3

6

CH3

2,5-dimethyl-2-hexene(not 2,5-dimethyl-4-

hexene)

The ending of the alkenes with more than one double bond should be change from - ene to

• diene – if there are two double bonds• triene – if there are three double bonds

Step 4

CH2

CH3

CH2

1

CH2

4

CH2

1

CH2

CH3

CH4

CH5

CH6

CH3

7

1,3-butadiene

1,3,5-heptatriene

MOREMORE

In cycloalkenes :

Number the carbon atoms with a double bond as 1 and 2, in the direction that gives the substituent encountered first with a small number.

Step 5

5

1

4

2

3

CH3

1-methylcyclopentene (not 2-methylcyclopentene)

3,5-dimethylcyclohexene(not 4,6-dimethylcyclohexene)

MORE

2

3

1

4

6

5CH3 CH3

Two frequently encountered alkenyl groups are vinyl group and allyl group.

RCH=CHR RCH=CH- alkene alkenyl

CH2=CH- CH2=CHCH2-

vinyl groupvinyl group allyl group allyl group

Step 6

-H

When two identical groups are attached:

a) on the same side of the double bond, the compound is cis

b) on the opposite sides of double bond, it is trans.

cis / transcis / trans

Step 7

C C

Cl

H H

Cl

C C

Cl

H Cl

H

cis-1,2-dichloroethene trans-1,2-dichloroethene

CH3

CH3

Give IUPAC names for the following alkenesGive IUPAC names for the following alkenes

2)

1)

CH3

CH2CH3

3)

4)

CH3CH=CHCH2C(CH3)2CH35)

Preparation of alkenes

a) Dehydration of alcohols

b) Dehydrohalogenation of alkyl halide

Dehydration of alcohols Alcohols react with strong acids in the

presence of heat to form alkenes and water.

C C

H OH

H2SO4 (conc.)C C + H2O

Concentrated sulphuric acid (H2SO4) or phosphoric acid (H3PO4) : as acidic catalysts and dehydrating agents.

• The major product is the most stable most stable alkene, which has greater number of alkene, which has greater number of alkyl group attach to C=C is follow alkyl group attach to C=C is follow Saytzeff’s rule.Saytzeff’s rule.

Saytzeff’s RuleSaytzeff’s Rule

An elimination usually gives the most stable alkene product, commonly the most highly substituted alkene.

ExamplesExamples

(1)

(2)

+CH3CH2OH

H2SO4 (conc.)

CH2 CH2

H2O

CH3 CH CH3

OH

H2SO4 (conc.)

CH2 CH CH3

+ H2O

CH3CH=CHCH

3(major product)CH CH

2

CH3

CH3

OH

H2

SO4

(conc.)

+ CH2=CHCH

2CH

3(minor product)

+ H2O

Mechanism for the Mechanism for the dehydration of alcoholdehydration of alcohol

Step 1:Protonation of alcoholProtonation of alcohol.

CH3

C CH2

CH3

OH

H

H

:++ O H

H

: ..

CH3

C CH2CH

3OH

H

: ..

+ +H

O H

H

:

Step 2:Formation of carbocationFormation of carbocation

CH3

C CH2CH

3OH

H

H:+

carbocation 3 2 3

H

CH C+ CH CH

+ O H

H

:..

Step 3:Formation of alkenesFormation of alkenes

C O H

H

+: ..C C

3H + CH

H

H

HH

H

a

b

Root a : major product

Root b : minor product

stable alkenes

C CCH

3

HH

CH3

+H

O H

H

:+

+ CH2=CHCH

2CH

3(minor product)

Rearrangement DuringRearrangement During Dehydration of AlcoholDehydration of Alcohol

Example 1:

C C CH3

CH3

CH3

CH3

OH

HH

2

SO4

(conc.)

(major product)

C C C H 3CH 3

C H 3

C H 3

+

(minor product)

C CH CH3

CH2

CH3

CH3

Step 1 and Step 2 are similar to the previous example.

Step 3:

• Now the rearrangement rearrangement occurs.

• The less stable, secondary carbocation rearrange to form more stable tertiary carbocation.

2o carbocation 3o carbocation (less stable) (more stable)

CH3

C C+

CH3

CH3

CH3

H rearrangementrearrangement

C+

C CH3

CH3

CH3H

CH3

The rearrangement rearrangement occurs through the migration of an alkyl group (methyl) from the carbon atom adjacent to the one with the positive charge.

Because a group migrates from the one migrates from the one carbon to the nextcarbon to the next, this kind of rearrangement is often called a 1,2 1,2 shift.shift.

(a)C CH CH

3CH

2

CH3

CH3

(minor product) less stable alkene

(b)

C C CH3CH3

CH3

CH3(major product) more stable alkene

(a)

O H

H

:..

C+

C CH3CH

CH3

HCH3

+

(b)

H

H

The final step can occur in two ways:

The final step can occur in two ways:

Path (a): leads to less stable, disubstituted alkene and produces the minor product of the reaction.

Path (b): leads to highly stable tetrasubstituted alkene and produces the major product according to the Saytzeff’s rule.

C C CH3CH3

CH3

H OH

HH2SO4 (conc)

C CH2CH2 CH3

CH3

C CH CH3CH3

CH3

+

(major product)

(minor product)

Try this!

(2) Dehydrohalogenation of Alkyl Halides(2) Dehydrohalogenation of Alkyl Halides• The eliminationelimination of a hydrogenhydrogen and a

halogenhalogen from an alkyl halide to form an alkene.

• Saytzeff’s ruleSaytzeff’s rule is used to determine the major product

C C

H X

+ KOHalcohol

refluxC C

+ HX

Examples:

(1)

CH CH2CH2CH3 CH3

Cl

+ KOHalcohol

reflux

CH CHCH2CH3 CH3

(major product)

+ CH CH2CH2 CH2CH3

(minor product)

(2)

KOHalcohol

refluxCH3 CH CH3

Br+ CH2 CH CH3

(3)

KOHalcohol

refluxCH

CH3

Br

+ CHCH

3

+ CH2CH3

(minor product)

(major product)

(4)

CH3 H

CH3 C C CH3 + KOH alcohol

CH3 Cl reflux

CH3 H CH3

CH2 C C CH3 CH3 C C CH3

CH3 CH3

2,3-dimethyl-1-butene 2,3-dimethyl-2-butene (minor product) ( major product) 2 alkyl groups 4 alkyl groups

Chemical Reaction of AlkenesChemical Reaction of Alkenes Comparison of The Reactivity Between Comparison of The Reactivity Between

Alkanes and AlkenesAlkanes and Alkenes Alkenes are more reactiveAlkenes are more reactive compared to alkanes. Alkanes have carbon-carbon single bonds (σ

bonds) while alkenes have carbon-carbon alkenes have carbon-carbon double bonds (double bonds (π π bondsbonds).).

The double bond is a site of high electron density (nucleophilicnucleophilic).

Therefore most alkenes reactions are electrophilic additionselectrophilic additions..

Electrophilic Additions Electrophilic Additions MechanismMechanism

C C

H

H H

H

+ X Y C C+

HH

H

X H

+ Y -

slow

C C+

HH

H

X H

+ Y- fast C C HH

H

X Y

H

Example:

C C

CH3

CH3 CH3

CH3

+ H Cl

C C+ CH3CH3

H

CH3CH3

+ Cl-

slow

i.

C C CH3CH3

CH3

H Cl

CH3

C C+

CH3CH3

H

CH3CH3

+ Cl-fast

Addition Reaction of AlkenesAddition Reaction of Alkenes

(1) Hydrogenation(1) Hydrogenation

• The reaction of an alkene with hydrogen in the presence of catalyst such as platinum, nickel and palladium to form alkane.

C C + H2Pt or Ni or Pd

C C

HH

Examples:(

1)

(2)

+ H2Pt /Ni / PdCH3 C CH2

CH3

CH3 CH CH3

CH3

+ H2

Pt /Ni / Pd

(2) (2) Halogenation of Alkenes

(i) In inert solvent (CH(i) In inert solvent (CH22ClCl22))

Alkenes react rapidly with chlorine or bromine in CH2Cl2 at room temperature to

form vicinal dihalides.vicinal dihalides.

C C + X2

CH2Cl2C C

X X

Example:

C C CH3CH3

Cl

CH3H

Cl

C C CH3CH3

CH3

H+ Cl2

CH2Cl2

• When bromine is used for this reaction, it can serve as a test for the presence of test for the presence of carbon-carbon double bondscarbon-carbon double bonds.

• If bromine is added to alkene, the reddish brown color of the bromine disappears almost instantly as long as the alkene is present in excess.

Unsaturation test

C C + Br2

CH2Cl2

room temperature

C C

Br BrObservation: The reddish brown bromine decolourised

(3) (3) Hydrohalogenation : Markovnikov’s : Markovnikov’s rulerule Hydrogen halides (HI, HBr, HCl and HF) add to the

double bond of alkenes to form haloalkanes.

The addition of HX to an unsymmetrical alkenesunsymmetrical alkenes, follows Markovnikov’s ruleMarkovnikov’s rule.

+ C C

H X

HXC C

• Markovnikov’s Rule:

– In the addition of HX to an alkenes, the hydrogen hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms.greater number of hydrogen atoms.

i. The additionaddition of HBr to propene, could conceivably lead to either 1-bromopropane or 2-bromopropane.

The main product, however, is 2-bromopropane

CH2 CH CH3 + HBr C C CH3H

H

H Br

H

ii. When 2-methylpropane reacts with HBr, the main product is tert-butyl bromide, not isobutyl bromide.

• The addition of HX to an unsymmetrical alkenes, yield the main product according to the Markovnikov’s rule.

CH3 C CH2

CH3

HBr+ CH3 C CH3

CH3

Br

CH 3 C

CH 3

CH CH 3+ H Br

CH3 C CH CH3

CH3

+

30 Carbocation

+ Br -

slow

MechanismMechanism

CH3 C CH CH3

CH3

++ Br -

fast

CH3 C CH CH3

CH3

Br H

(4) Addition of HBr to Alkenes in The(4) Addition of HBr to Alkenes in The

Presence of Peroxides Presence of Peroxides

(Anti Markovnikov’s rule)(Anti Markovnikov’s rule) When alkenes are treated with HBr in the presence

of peroxides, ROOR (eg: H2O2) the addition occurs

in an anti-Markovnikov manner

The hydrogen atom of HBr attached to the doubly bonded carbon with fewer hydrogen atoms.

Example:

CH2CH2CH3Br

CH2 CH CH3 HBr ROOR+

CH2C CH2CH3

CH3

+ HBrH2O2

Complete the following reactions:Complete the following reactions:

(1)

(2)

CH2HBr

H2O2+

(ii) (ii) In aqueous (Halohydrin Formation)

• If the halogenation of an alkene is carried out in aqueous solution, the major product of the overall reaction is a haloalcohol called a halohydrin.halohydrin.

C C + X2 + H2O C C

X OH

X2 = Cl or Br halohydrin

Example:

If the alkenes is unsymmetricalunsymmetrical, the halogenhalogen ends up on the carbon atom with the greater number of hydrogen atoms.greater number of hydrogen atoms.

C C

H

H H

H

+ Br2 + H2O C C HH

H

Br OH

H

2-bromobutanol Unsaturated test

Observation : the reddish brown bromine decolourised

Example:

CH3 C CH2

CH3

+ Br2 + H2O

1-bromo-2-methyl-2-propane

C C HCH3

CH3

OH Br

H

Complete the following reaction:

CH3

+ Br2 + H2O

(5) Hydration of Alkene(5) Hydration of Alkene The acid-catalyzed addition of water to the

double bond of an alkene (hydration of an alkene) is a method for the preparation of low molecular weight alcohols.

The acid most commonly used to catalyse the hydration of alkenes are dilute solution of sulphuric acid and phosphoric acid.

• The addition of waterThe addition of water to the double bond follows Markovnikov’s rule.

C C + H2OH3O

+

C C

H OH

Example:

CH3 C CH2

CH3

H2OH3O

+

+

C CH3CH3

OH

CH3

Complete the following reactions:Complete the following reactions:

(1)(1)

CH2H2O

H3O+

+

(2)

CH2 CH CH3 H2OH3O+

+

Mechanism

..

CH3 C CH CH3

CH3

+ H O

H

H+

CH3 C

CH3

CH CH3

H

+

STEP 1

..

..CH3 C

CH3

CH CH3

H

++ H2O

STEP 2

CH3 C

CH3

CH CH3

HO+H

H

..

STEP 3

CH3 C

CH3

CH CH3

HO+H

H

..

H2O.. ..

CH3 C CH CH3

CH3

OH H

+ H3O+..

(6) (6) Addition of Sulphuric Acid Addition of Sulphuric Acid to to AlkenesAlkenes

Alkenes dissolve in concentrated sulphuric acid to form alkyl hydrogen sulphatesalkyl hydrogen sulphates.

Alkyl hydrogen sulphates can be easily hydrolysed to alcohols by heating them with water.

The overall result of the addition of sulphuric acid to alkenes followed by hydrolysis is the Markovnikov addition of -H and -OH.

C C

CH3

CH3 CH3

CH3conc. H2SO4 C C CH3CH3

CH3

H OSO3H

CH3

alkyl hydrogen sulphate

H2O

heatC C CH3CH3

CH3

H OH

CH3

• Example:

CH2

CH CH3

conc.H2SO

4 CH3

CH CH2

OSO3H

H

H2O

heatCH

3CH CH

2

OH

H

Alkenes undergo a number of reactions in which the C=C is oxidized

7 ) Oxidation of Alkenes7 ) Oxidation of Alkenes

KMnO4 ozonolysis

basic,cold,dilute

acidic,hot,concentrated

i)i) With With coldcold and and dilutedilute potassium potassium permanganate, KMnOpermanganate, KMnO4 4

Potassium permanganate in alkaline solution can be used to oxidise alkenes to 1,2-diols (glycols).

C C + KMnO4(purple)

OH-,cold

C C

OH OH

+ MnO2

(brown precipitate)

Observation: Purple colour of KMnO4 decolourised and brown precipitate formed.

This reaction is called Baeyer’s test.

It is a test for the presence of C=Ctest for the presence of C=C where the purple colour of the KMnO4

decolourised, and brown precipitate of MnO2 is formed.

Example:

CH2 CH2 + KMnO4 cold

OH-,H2O

C C HH

OH

HH

OH

+ MnO2

ii) With hot hot potassium permanganate solutions to alkenes

When oxidation of the alkene is carried out in acidic solution of KMnOacidic solution of KMnO44, cleavage of the double bond occurs and carbonyl-containing products are obtained.

C

O

CH3CH3+ C

O

CH3CH3

C C CH3CH3

CH3

CH3

+ KMnO4 (ii) H3O+

(i) OH-,heat

If the double bond is tetrasubstitutedtetrasubstituted, the two carbonyl-containing products are ketonesketones

If a hydrogen is present at double a hydrogen is present at double bondbond, one of the carbonyl-containing products is a carboxylic acidcarboxylic acid;

If two hydrogens are present on one two hydrogens are present on one carboncarbon, COCO22 is formed.

2)

CH2 CH CH3 + KMnO4(i) OH

-,heat

(ii) H3O+

C

O

CH3HO+ CO2 + H2O

• The oxidative cleavage of alkenes can be used to establish the location of the establish the location of the double bonddouble bond in an unknown alkene.

Example:

• An unknown alkene with the formula CC77HH1414

undergoes oxidation with hot basic potasium permanganate solution to form propanoic acidpropanoic acid and butanoic acid.butanoic acid. What is the structure of this alkene?

C7H14 + KMnO4

(i) OH-,heat

(ii) H3O+

C

O

OHCH2

CH3 + C

O

OHCH2

CH2CH3

Answer:

C

O

OHOHCH2

CH3 C

O

OHOHCH2

CH2CH3

propanoic acid butanoic acid

CH2CH2CH3CCCH2CH3

H H

3-heptene

ExampleExample

• An unknown alkene undergoes oxidation in hot basic KMnO4 followed by acidific to give the following product:

CH3CCH2CH2CH2CH2C OH

Deduce the structural formula for the unknown alkene.

O O

iii) Ozonolysis of Alkenes

A more widely used method for locating

the double bond of an alkene is the use of ozone (O3).

Ozone reacts vigorously with alkenes to form unstable compounds called molozonidesmolozonides, which rearrange spontaneously to form compounds called ozonidesozonides..

C C + O3

C C

O

ozonide

O

O

Ozonides:Ozonides: very unstable compounds

can easily explode violently

they are not usually isolated but are reduced directly by treatment with water and in the presence of zinc and acid (normally acetic acid) to give carbonyl compounds (either aldehydes or ketones).

ozonide

C C

O O

O

+ ZnH2O,H

+

H

R

C = O

+ O = C

R

H

Example:

C CHCH3 CH3

CH3 (i) O3

(ii) Zn,H2O/H

+

CO

CH3

CH3

CO

CH3

H+

1. Write the structure of alkene that would produce the following products when treated with ozone followed by water, zinc and acid

CH3COCH3 and CH3CH(CH3)CHO

ExerciseExercise

Example Example • Deduce the structural formula of an

alkene that gives the following compound when it reacts with ozone in the presence of Zn / H+.

O=CH-CH2-CH2-CH(CH3)CH=O

2. Acid-catalyzed dehydration of neopentyl alcohol, (CH3)3CCH2OH,

yields 2-methyl-2-butene as the major product. Outline a mechanism showing all steps in its formation.

ExerciseExercise

• Compounds A, B, C and D are isomers with the molecular formula C4H8. AA and B B give a positive Baeyer testpositive Baeyer test, while C and D do not. AA exists as cis- and trans- isomerscis- and trans- isomers, while B does not have geometrical isomers. C has only secondary hydrogenC has only secondary hydrogen, while D has primary, secondary and D has primary, secondary and tertiary hydrogentertiary hydrogen. Give the IUPAC names for A, B, C, and D.