Orbital Overlap In Ethene Alkenes

Alkenes

1435-1436 2014-2015

Learning Objectives

Chapter two discusses the following topics and the student by the end of this chapter will:

 Know the structure, hybridization and bonding of alkenes

 Know the common and IUPAC naming of alkenes

 Know the geometry of the double bond i.e. cis/trans isomerization

 Know the physical properties of alkenes

 Know the different methods used for preparation of alkenes (elimination reactions ; dehydrogenation, dehydration and alkenes stability (Zaitsev’s rule) play an important role in understanding these reactions

 Know the addition reactions of alkenes and the effect of Markovnikov’s rule in determining the regioselectivity of this reaction.

 They are unsaturated hydrocarbons – made up of C and H atoms and contain one or more C=C double bond somewhere in their structures.

Their general formula is C

_n

H

_2n

- for non-cyclic alkenes

Their general formula is C

_n

H

_2n-2

- for cyclic alkenes Structure Of Alkenes

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Trigonal planar

2s² 2p³ x sp² 3 2p

sp

²

Hybridization Of Orbitals In Alkenes

The electronic configuration of a carbon atom is 1s

²

2s

²

2p

²

Thus

promotion hybridization

2p² 2s¹

 In ethylene (ethene), each carbon atom use an sp² orbital to form a single

C-C bond. Because of the two sp² orbitals overlap by end- to- end the resulting bond is called σ bond. The pi (π) bond between the two carbon atoms is formed by side- by-side overlap of the two unhybridized p- orbitals (2p–2p ) for maximum overlap and hence the strongest bond, the 2p orbitals are in line and perpendicular to the molecular plane.

 This gives rise to the planar arrangement around C=C bonds. Also s orbitals of hydrogen atoms overlap with the sp² orbitals in carbon atoms to form two

C-H bonds with each carbon atom.

 The resulting shape of ethene molecule is planar with bond angles of 120º and C=C bond length is 1.34 Å

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Orbital Overlap In Ethene

two sp² orbitals overlap to form a sigma bond between the two carbon atoms

Orbital Overlap In Ethene

two 2p orbitals overlap to form a pi bond between the two carbon atoms

s orbitals in hydrogen atoms overlap with the sp² orbitals in carbon atoms to form C-H bonds

the resulting shape is planar with bond angles of 120º and C=C (1.34 Å)

sp² hybridized carbon atoms

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Nomencalture Of Alkenes And Cycloalkenes 1. Alkene common names:

Substituent groups containing double bonds are:

 H

₂

C=CH–   Vinyl group

 H

₂

C=CH–CH

₂

–   Allyl group

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Br

Cl

Common: Allyl bromide Vinyl chlorride

H₂C CH₂ CH₃-CH CH₂ C CH₂ CH₃

H₃C

Common: Ethylene Propylene Isobutene

2. IUPAC Nomenclature Of Alkenes

 Find the longest continuous Carbon chain containing the double bond this determines the root name then add the suffix -ene.

 Number the C- chain from the end that is nearer to the double bond. Indicate the location of the double bond by using the number of the first atom of the double bond just before the suffix ene or as a prefix.

 Indicate the positions of the substituents using numbers of carbon atoms to which they are bonded and write their names in alphabetical order (N.B.

discard the suffixes tert-, di, tri,---when alphabetize the substituents) and if more than one substituent of the same type are present use the prefixes di- or tri or tetra or penta,--- to indicate their numbers.

CH₃CH CHCH₂CH₂CH₃

Hex-2-ene or 2-Hexene (not 4-Hexene)

1 2 3 4 5 6

8

H₂C CH CH₂CH₃

But-1-ene or 1-Butene (not 3-Butene)

1 2 3 4

CH₃ C

H₃C CH

2-Methyl-but-2-ene or 2-Methyl-2-butene (not 3-Methyl-2-butene)

3

1 CH₃

2 4

CH₃ Cl

3-Chloro-2-hexene

(not 2-Chloro-1-methyl-1-pentene) 2

1

3 4

5

6 2

3 4

5 6 7

8

CH₃-CH₂-CH₂-CH=CH-C-CH₃

2,2-Dibromo-3-heptene (not 6,6-Dibromo-4-heptene)

5 1

7 6 4 3 2

(CH₃CHCHCH₂OCH₃)

1-Methoxy-but-2-ene (not 4-Methoxy-but-2-ene)

OCH₃ 2

3 1

8 1

2

5 6 7

3 4

2,3,7-Trimethyl-non-3-ene (not 2-Isopropyl-6-methyl--2-octene)

9 CN

4-Cyano-2-ethyl-1-pentene (not 2-Ethyl-4-cyano-1-pentene)

1

2 3

4 5

5-Methylcyclopenta-1,3-diene

An ''a'' is added due to inclusion of di put two consonants consecutive

1 3 2

CH₃

=

Br Br 6-Methyl-2-octene

1

4

 In cycloalkenes the double bond carbons are assigned ring locations #1 and #2.

Which of the two is #1 may be determined by the nearest substituent rule.

 If the substituents on both sides of the = bond are at the same distance, the

numbering should start from the side that gives the substituents with lower alphabet the lower number.

10 3-tert-Butyl-7-isopropyl-cycloheptene

(not 3-Isopropyl-7-tert-butylcycloheptene) CH₃

6 2

3 4

5

1-Methyl cyclopentene (not 2-Methylcyclopeneten)

CH₃ H₃C

3,5-Dimethyl-cyclohexene (not 4,6-Dimethylcyclohexen) (not 1,5-Dimethyl-2-cyclohexen)

1

1

 When the longer chain cannot include the C=C, a substituent name is used.

6

2 3

5 1

1

CH CH₂

Vinyl-cyclohexane

CH CH₂

3-Vinyl-cyclohexene

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Geometrical Isomerism In Alkenes

 G. I. found in some, but not all, alkenes

 It occurs in alkenes having two different groups / atoms attached to each carbon atom of the = bond

G. I. x G. I. X G. I.  G. I. 

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A B

C D

A=C or B=D No Cis or transe

(G. I. X)

A≠C B≠D,

A =B or C=D Cis / A =D or C= B transe G. I. 

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 It occurs due to the Restricted Rotation of C=C bonds so the groups on either end of the bond are ‘fixed’ in one position in space; to flip between the two groups a bond must be broken.

X

Geometrical isomers can not convert to each at room temperature.

Geometrical Isomerism In Alkenes

A) Cis / trans isomerism in alkenes

 Exhibited by alkenes having two H’s and two other similar groups or atoms attached to each carbon atom of the = bond (or generally the alkene have only two types of atom or groups i.e. ABC=CAB)

 Cis prefix used when hydrogen atoms on both carbon atoms are on the SAME side of C=C bond

 Trans prefix used when non-hydrogen groups / atoms are on the opposite sides of C=C bond

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Types Of Geometric Isomerism

Cis-But-2-ene Trans-But-2-ene

Cis

Groups / atoms are on the Same Side of the double

bond

Trans

Groups / atoms are on Opposite Sides across the

double bond

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H

Cl Cl

H H

Cl H

Cl

cis-1,2-Dichloro-ethene trans-1,2-Dichloro-ethene

Trans-Oct-4-ene

H H

Cis-Oct-4-ene

=

H₃C

H H

H

CH₃ H

H

CH₃ H H

H CH₃

trans-trans-2,4-hexadiene cis-cis-2,4-hexadiene trans-4-ethyl-3-heptene

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If the groups attached to the C=C are different, we distinguish the two isomers by adding the prefix Z (from German word Zusammen) if the higher-priority groups are together in the same side or E (from German word Entgegen) if the higher- priority groups are opposite sides depending on the atomic number of the atoms attached to each end of the C=C.

Atoms with higher atomic numbers receive higher priority I> Br > Cl > F > O > N > C > H

B) Z/ E isomerism in alkenes

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18

CH₃

Cl Br

CH₃ I

Cl

C H

CH₃ O

O H

CH₃

I

Cl

H

CH₂

Z-2-bromo-1-chloro-1-iodopropene E-2-bromo-1-chloro-1-iodopropene

Z Z

Exercise

Q1-Which of the following compounds can exhibit cis / trans isomerism a) 2-Methylpropene

b) 1-Butene

c) 2-Methyl-2-pentene d) 2-Butene

e) 3-Methyl-2-hexene

Q2- Name the following compounds according to IUPAC system

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b) c) a)

20

1 2 Br 3

4

3-Bromo propene

1

2 4

2-Ethyl-4-methyl pentene CH₃

Cl

CH₃ 1 3 2

4

5 6

4-Chloro-3,6-dimethylcyclohexene 3-Chloro-2,5-dimethylcyclohexene CH₃

Cl

CH₃ 6 2 1

3

4 5

C C

H

Br Br

H

C C

H

H Cl

Cl Cis-1,2-Dibromoethene

Geometrical isomerism 1,1-Dichloroethene

Trans-trans-2,4-heptadiene Trans-1,3,5-heptatriene

Cis-cis-2,5-heptadiene Home work

not geometrical isomerism 1

2 3

4 5

6 7

C C

H

H₃C CH₃

CH₂CH₃ 1

2 3

4 5 E-3-Methyl-2-pentene

Physical Properties of Alkenes

 Alkenes are nonpolar compounds thus:

 Insoluble in water

 Soluble in nonpolar solvents ( hexane, benzene,…)

 The boiling point of alkenes increase as the number of carbons increase.

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Preparation Of Aalkenes

1- Dehydration of alcohols ( removal of OH group and a proton from two adjacent carbon atoms ) using mineral acids such as H

₂

SO

₄

or H

₃

PO

₄

CH₃CH₂OH

C

H₂ CH₂

+

^H2O

OH H

+

^H2O

H⁺/ heat

cyclohexanol cyclohexene

Ethanol Ethene

C o nc.H₂S O₄ 1 8 0 ^oC

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Zaitsev’sRule

 If there are different protons can be eliminated with the hydroxyl group or with halogen atom, in this case more than one alkene can be formed, the major product will be the alkene with the most alkyl substituents attached to the double bonded carbon.

H₃C

CH₃ OH

H / Heat

H₂C

CH₃

H₃C

CH₃ ^{+ H2O}

+ H₂O 1- Butene Minor

2- Butene Major

Zaitsev rule: an elimination occurs to give the most stable, more highly

substituted alkene

2- Dehydrohalogenation of alkyl halides using a base

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or NaOH

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3. Dehalogenation of vicinal dihalides

For example: Dehalogenation of 1,2-Dibromobutane leads to the formation of 1- Butene. In the presence of catalyst ( AcOH acetic acid).

Br

Zn/AcOH

Br

Reactions Of Alkenes

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Reactions of Alkenes

Oxidation Reactions

Addition(Electrophilic) reaction:

- Hydrogenation - Halogenation

- Hydrohalogenation - Halohydrin formation -Hydration

KMnO₄

Ozonolysis

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An electrophile, an electron-poor species,(from the Greek words meaning electron loving). It is a species (any molecule, ion or atom) that accept a pair of electrons to form a new covalent bond.

A nucleophile, an electron-rich species, ,(from the Greek words meaning nucleus loving). It is a species (any molecule, ion or atom) that donate an electron pair to form a new covalent bond.

C , H , Br , Cl , I , etc., AlCl₃ , BF₃ , FeCl₃ , FeBr₃ , etc.

C , OH , Br , Cl , I , etc., H₂N, HS, etc., H₂O ,

CH₃OH , RNH₂ , R₂NH , R₃N , , etc.

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Electrophilic Addition Reaction

1- Additions To The Carbon-Carbon Double Bond

1.1 Addition Of Hydrogen: Hydrogenation (Reduction)

A A

A A

+

^H2

H H

A A A

A

An alkene An alkane

Pt or Ni or Pd

C

H₂ CH₂

+

^{H2 Pt H3C CH3}

C H₃

CH₂

+

^{H2 Pt H3C CH3}

CH₃

CH₃

+

^{H2 Pt}

CH₃

CH₃

Cis-1,2-Dimethyl cyclohexane

1.2.Addition of Halogens( Halogenation)

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A A

A A

+

^X2

X X

A A A

A (X= Cl or Br)

C H₃

CH₃

+

^{Cl2 CCl4 H3C} _CH

3

Cl Cl

+

^{Br2 CCl4}

Br

Br CH₃

CH₃

+

^{Br2 CCl4}

CH₃

Br Br

CH₃

Trans-1,2- Dibromo-1,2-Dimethyl cyclohexane

1.3. Addition of Hydrogen Halides

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 However, if the double bond carbon atoms are not structurally equivalent, i.e.

unsymmetrical alkenes as in molecules of 1- propene, 1-butene, 2-methyl-2-butene and 1- methylcyclohexene, the reagent may add in two different ways to give two isomeric products. This is shown for 1-propene in the following equation.

 Only one product is possible from the addition of these strong acids to symmetrical alkenes such as ethene, 2-butene and cyclohexene.

A A A

A

+ HX

A A

H X A A

(x= Cl or Br or I)

+ HCl

H₃C CH₃ Cl

H + HI

H I

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Stability of carbocation

C

CH₃ H₃C

CH₃

CH H₃C

CH₃

CH₂CH₂ CH₃

3^o 2^o 1^o

HBr

CH₃CHCH₃

CH₃CH₂CH₂ CH₃CH=CH₂

Br

Br

CH₃CHCH₃

CH₃CH₂CH₂Br 2^o Carbocation Br

1^o Carbocation

maijor

minor

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Markovnikov’s rule stats that : In addition of unsymmetrical reagent to unsymmetrical alkenes the positive ion adds to the carbon of the alkene that bears the greater number of hydrogen atoms and the negative ion adds to the other carbon of the alkene.

 However when the addition reactions to such unsymmetrical alkenes are carried out, it was found that 2-bromopropane is nearly the exclusive product. Thus it said the reaction proceeded according to Markovnikov’s rule

+ HCl

H₃C CH₃ CH₃

H₃C CH₃ CH₃

Cl

1.4. Addition of HOX halogen in aqueous solution (

^-

OH, X

⁺

):

Halohydrin formation

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 Only one product is possible from the addition of HOX acids (formed from mixture of H2O and X2) to symmetrical alkenes such as ethene and cyclohexene.

A A A

A

+ H₂O / X₂

A A

OH X A

A (x= Cl or Br )

H₃C CH₃ Cl

OH + H₂O / Cl₂

Symmetrical akenes

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 However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially.

Cl

+ H₂O / Cl₂ OH

Unsymmetrical akenes

CH₂Br OH

+ H₂O / Br₂

1.5. Addition of H

₂

O: Hydration

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 Only one product is possible from the addition of H₂O in presence of acids as catalysts to symmetrical alkenes such as ethene and cyclohexene.

 However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially.

CH₃ CH₃

OH H

+ H₂O

H Unsymmetrical akenes

Symmetrical akenes

A A A

A

+ H₂O

A A

H OH A A

H₃C CH₃ OH

H

+ H₂O

H

H

2.1- Ozonolysis: Oxidation with ozone (Oxidative cleavage):

This reaction involves rupture of the C=C to give aldehydes or ketones according to the structure of the original alkene.

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A A A

A

+ O₃

A A

O O A A

O

Zn /H₂O

- H₂O₂ O

A A

+ O

A

A ( A= H or R)

i) O₃

ii) Zn /H₂O O ⁺ O

i) O₃

ii) Zn /H₂O O ⁺ O H

i) O₃ ii) Zn /H₂O

O

O

2-Oxidation Reaction:

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2- Oxidation with KMnO

₄

(Oxidative addition):

Cis- diol

OH

OH

KMnO 4 / OH/ H 2 O

Thank You for your kind attention !

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