Fundamental Organic Chemistry

Prepared By

Dr. Essam M. Hussein

Assistant Professor of Organic Chemistry Chemistry Department

Faculty of Applied Science -Umm Al-Qura University

Carbon Compounds and Chemical bonds:

Organic chemistry is the chemistry of the compounds of

carbon.

Carbon compounds include DNA, RNA.

DNA deoxyribonucleic acid.

RNA Ribonucleic acid.

& Proteins & Hydrocarbons and carbohydrates.

Compounds are formed by covalent Bonds.

Covalent Bonds:

Covalent Bonds are formed by sharing electrons by atoms.

The first explanations of the nature of chemical bonds were

advanced by G. N. Lewis and W. Kossel

The atoms in covalent bond achieve noble gas configurations by

sharing electrons.

Covalent bonds form between the atoms form molecules.

Molecules may be represented by electron –dot formulas or by

dash formulas where each dash represents a pair of electron shared

by two atoms.

Examples: (1) H2 molecule

H 1s↑ + H 1s↑

2) Cl2 Cl Cl ClCl+ (dot formula)

Cl Cl (dash formula)or

H H

H H HH

H H

+or

(dot formula)

or

(dash formula)

3)CH4 C 4H H C

H

H

H

C

H

H HH

or

+ (dot formula)

(dash formula)

4) N2 N N+ N N

N N

(dot formula)

(dash formula)

6) CO2 C + O2 C OO

C OO

or

7) NH3 N + 3H N HHH

orN

HH

H

C +5) HN + C NH

C NH

(dot formula)

(dash formula)

HCN

8) NH4 N HHH

orN

HH

H

H

H

All the previous formulas are called "Lewis structures."

Note: in writing Lewis structures we show only the electrons of the valence shell.

NHH

H+ H

+ NHH

H

H

Lewis acidLewis base

Ex. 1

NHH

H+ N

H

H

H

Lewis acidLewis base

Ex. 2BF3

BF3

NHH

H+ N

H

H

H

Lewis acidLewis base

Ex. 3AlCl3

AlCl3

Polar Covalent bond and Electronegativity:

The electrons of the covalent bond between non identical

atoms are attracted to the atom having high electronegativity

(higher atomic number) in this case the covalent bond will be

polarized (called polar covalent bond).

H F or H F orH FEx: δ δ

Electronegativity:

The ability of atoms to attract covalence bond electrons to it’s side.

Writing Lewis Structures:

When we write Lewis structures (electron-dot formula).

We assemble the molecule or from the constituent atoms

showing only the valence electrons (the electrons of the outer

most shell).

By having the atoms share or transfer electrons we try to give

each atom the electronic configurations of the noble gas in the

horizontal row of the periodic table (octet rule).

e. g. give hydrogen atoms two electrons like He structure and

C, N, O and F eight electrons like Ne.

* The no. of valence electrons can be obtained from the

periodic table because it is equal to the group number of the

atom e. g. for carbon in IVA equal four and it has four

electrons.

* Fluorine (F) in group VIIA has seven.

* Hydrogen in group IA has one.

Example1:

Write the Lewis structure of CH3F

(1) We find the total no. of valence electrons of all the atoms.

C + 3H + F

4 + 3(1) + 7 = 14

(2) We use pairs of electrons to form bonds between all atoms that

are bonded to each other.

We represent these bonding pairs with lines, in our example four

pairs of electrons (8 of our 14 valance electrons).

(3) We then add the remaining electrons in pairs as to give each

hydrogen 2 electrons and every other atom 8 electrons in our

example the remaining 6 valence electrons were assigned to the

Fluorine atom in three non bonding pairs (unshared electrons).

So the Lewis structure of CH3F is

H C

H

H

C

H

H HF F

Example 2: Write the Lewis structure of ClO3-

(1) We find the total no. of valence electrons of all the atoms and the

extra electrons needed to give the ion a negative charge:

Cl + O3 + e-

7 + 3(6) + 1 = 26

(2) We use three pairs of electrons to form bonds between the

chlorine atom and three oxygen atoms.

Cl OO

O

(3) then add the remaining 20 electrons in pairs to give each atom octet.

So the Lewis structure of ClO3- is

Cl OO

O

Example 3: Write the Lewis structure of C2H6

(1) We find the total no. of valence electrons of all the atoms:2C + 6H2(4) + 6(1) = 14(2) We use three pairs of electrons to form bonds between the carbon and hydrogen atoms.

C C

H

H

H

H

H

H

7 X One pair (2 electron) = 14 electrons i. e. only 14 bonded electrons.

Formal charges:When we write Lewis structure it is often convenient to assign unit

positive or negative charge called "formal charge" to certain atoms

in the molecule or ion.

Calculations of formal charges is a bookkeeping method for

electrical charges, because the arithmetic sum of all of the formal

charges equals the total charge on the molecule or ion.

Simply to calculate the formal charge for atoms in molecules or

ions.

(1) Write the right Lewis structure.

(2) Use the equation

F = Z – S/2 – U for each atom in the molecule.

(3) To know the net hole charge of the molecule (should be zero)

or ions; calculate the arithmetic sum of the all atoms in

molecule or ion.

Ex. (1):

Ammonia NH3

* Write the Lewis structure of ammonia

N + 3H

5 + 3(1) = 8

NHH

H

unshared(nonbonding)

bonding(shared)

To calculate the formal charge use the equationF = Z – S/2 – UWhereF Formal chargeZ group no. of the atomS no. of shared (bonding electrons)U no. of unshared (nonbonding electrons)

For NH3 the Lewis structure is

NHH

H

The formal charge of H is

F = 1 – 2/2 – 0 = 0

For nitrogen

F = Z – S/2 – U

F = 5 – 6/2 – 2

F = 5 – 3 – 2 = 0

So the charge on NH3 molecule = sum. Of the formal charge of the

constituent atoms

= F(N) + F3(H)

= 0 + 3(0) = 0

Ex. (2): H2O(1) Write the Lewis structure of H2O 2H + O 2(1) + 6 = 8

OHH

F(H) = 1 – 2/2 – 0 = 0F(O) = 6 – 4/2 – 4 F(O) = 6 – 2 – 4 = 0So the charge on molecule = F(O) + F(H)=0 + 0 = zeroso H2O is a neutral molecule

Ex. (3): HNO3

Write the Lewis structure of HNO3

O NHO

Onot

NH

O

O O

1

2

3

because the hydrogen atom should attach to oxygen (mor electronegative atom).

F(H) = 1 – 2/2 – 0 = 0

F(O1) = 6 – 4/2 – 4 = 0

F(O2) = 6 – 2/2 – 6 = -1

F(O3) = 6 – 4/2 – 4 = 0

F(N) = 5 – 8/2 – 0 = +1

So the charge (net charge) on molecule =

F(O1) + F(O2) + F(O3) + F(N) +F(H)=

0 + 0 + (-1) + (+1)= zero

so nitric acid is a neutral molecule

Ex. (4): NO3-

(1) Write the Lewis structure of NO3

O NO

O1

2

3

F(O1) = 6 – 2/2 – 6 = -1

F(O2) = 6 – 2/2 – 6 = -1

F(O3) = 6 – 4/2 – 4 = 0

So the charge (net charge) on molecule =

F(O1) + F(O2) + F(O3) + F(N) =

0 + (-1) + (-1) + (+1)= -1

so NO3 has one negative charge.

O NO

O1

2

3

Ex. (5): NH4+

* Write the Lewis structure of ammonia

N + 4H – e =

5 + 4(1) – 1 = 8

NHH

H

H

The formal charges

F(H) = 1 – 2/2 – 0 = 0

F(N)= 5 – 8/2 =

= 5 – 4 = +1

So the charge on ion = 4(0) + (+1)

= 0+ 1 =+1

Orbital hybridization structure of alkanes

(sp3):sp3 hybridization: According to the quantum mechanism

the electronic configuration of a carbon in its lowest energy

state-called the ground state.

C6

1s22s22p2

1s22s12px12py

12pz1

Looking only on the valence electrons the carbon atom has

tow unpaired electrons i. e. it can only two bonds with

hydrogen in its ground state.

To account (explain) the four covalent bonds of carbon, we

should discuss "The hybridization of carbon orbitals".

C6 1s22s22p2

1S2

2S2

2P2

PyPx Pz

Ground state

excitation

1S2

2S1

2P3

Px Py Pzexcited state

Hybridization (sp3) i. e. ( one s + 3p)

i.e. Hybridization: mixing of orbitals of the same main shell.

(the same principle quantium no.). to give new hybridized

orbitals with the same energy.

Ex. (1): in methane CH4

C

H

H HHC

H

H

H

H

* In methane (CH4) each C-H σ bond formed by the overlap of

sp3 electron (from carbon) with 1s electron (from hydrogen)

* sp3 by hybridization and their compounds have tetrahedrol

structure and the bond angle of H-C-H (in methane) or H-C-C-

(in other alkanes) equals 109.28.

i.e. sp3 compounds are non planer.

Ex. (2): for sp3 hybridization

ROH, ROR, H2O, NH3

NH3 ammonia structure:

N5

1S21S22P3

1S2

2S2

2P2

PyPx Pz

Ground state

excitation and hybridization

1S2 SP3

excited state

Formation of N-H bonds

NHH

H

unshared(nonbonding)

σ- bonds

* The geometry of a molecule of ammonia is a triagonal

pyramid.

* The bond angles in a molecule of ammonia are 107º

* The hybridization of Nitrogen atom in ammonia explains

the presence of a lone pair of nonbonding (unshared)

electrons which explains the basicity of ammonia.

Sp2 hybridization and the structure of thene (ethylene):

* Before we going to sp2 hybridization, we should know that

the covalent bonds in carbon compounds are classified

into:

(1) Sigma bond (σ- bond):

Formed by head-head overlap of

1s with 1s electrons

1s with px electrons

1s with sp3 electrons

1s with sp2 electrons

1s with sp electrons

px with px electrons

px with sp3 electrons

px with sp2 electrons

px with sp electrons

(2) Pi bond (Л bond):

Formed only by sidewise overlap of py with py electrons and

pz with pz electrons.

sp2 hybridization of ethene:

HC C

H

H H

C1s2 2s2 2p2

1S2

2S2

2P2

PyPx Pz

Ground state

excitation

1S2

2S1

2P3

Px Py Pz

excitation (promotion of electron)

C1s2 2s2 2p2

1S2

2S2

2P2

PxPyPz

1S2

2S1

2P3

Pz Py Px

1S2

sp2

hybridization

Py

1S2

SP2

Py

C C

H

HH

H

sp2sp2sp2

sp2

sp2

sp2

1s1

1s11s1

1s1

PyPy

C C

H

H H

H

σ- bond σ- bond

σ- bondσ- bond

σ- bond

Л bond

* It is planer molecule

Sp hybridization:

This type of hybridization explained the band formation

of alkynes.

C1s2 2s2 2p2

1S2

2S2

2P2

PyPx Pz

Ground state

excitation

1S2

2S1

2P3

Px Py Pz

excitation (promotion of electron)

C1s2 2s2 2p2

1S2

2S2

2P2

PxPyPz

1S2

2S1

2P3

Pz Py Px

1S2

sp

hybridization

Py

1S2

SP

Py

sp sp spsp 1s11s1

Pz Pz

C C

Py

HHPz Pz

Py

Bond formation

CH

C

H

H

C CC

H

H

H

how many sp3-sp σ

sp2-sp3 σ

sp2-sp2 σ

sp-sp σ

C C

H

H

C CC H

H

H

C

H

H

H

SP

SP SP2

SP2

SP2

SP2

C CH Hσ- bondσ- bondσ- bond

Л bond

Л bond

Cleavage ( Breaking) of bonds:A covalent bond between two atoms can be broken by two

ways

(1) Homolytic bond fission:

A B hv orheat A B+

i.e. each atom takes one electron of the bond electrons

A + B A B

For C C Carbon-Carbon δ bond breaking

gave free radicals of SP2 hybridizationi.e

C C CHomalytic bond

fission

Free radical sp2

(planer)

(2) Heterolytic bond fission:

A + BA B

C C CHeterolytic bond

fission

carbocation

C+carboanion

sp3-sp3 σ

sp2-sp2 σ

sp-sp σ

sp3-sp2 σ

sp3-sp σ

sp2-sp σ

• In the carbocations the hybridization is sp2 ( the

carbocations are planer ).

• In the carbanions the hybridization is sp3.

C

Types of Reagents:

Reagents can be classified into:-

(1) Nucleophiles (Nucleophilic) Reagents:

They are nucleus seeking species, may be negatively

charged ions or groups or neutral atoms or groups and

can react with the donating of electrons.

Examples of negative charged nucleophiles:-

I , OH, RO, RS, CN R C C,

neutral nucleophiles are Lewis bases such as:

,, ROH NH3H2O

(2) Electrophilic ( electrophilic) Reagents:

They react with accepting a lone pair of electrons,

they are Lewis acids.

Examples of electrophiles:

, BF3, ZnCl2, AlCl3, FeCl3

CR

O

R

H CR

O

R

H

(3) Free Radicals:They are atoms or groups having one unpaired electrons, they are highly reactive with very short life time.e.g.

Cl, Br, C

Reaction Types:There are four general types of reaction which organic

compounds can under goes.

(a) Displacement (Substitution) Reactions.

(b) Addition Reactions.

(c) Elimination Reactions.(a)Rearrangement Reactions.

(a) Displacement (Substitution) Reactions.

It is displacement from carbon, the atom displaced be

either hydrogen or another atom or group and can be

classified into:

(1) Nucleophilic substitution reactions.

It is often an atom other than hydrogen that is displaced.

e.g

CN + BrR Br C R+N

(2) Electrophilic substitution reactions.It is often hydrogen that is displaced.

e.g. aromatic substitution.

+

H

NO2 +

NO2

H

(Electrophilic substitution)

It can be electrophilic, nucleophilic or radical in character depending on the type of species that initiates the process.e.g addition to simple carbon-carbon double bonds is either electrophilic or radical induced e.g. addition of HBr

(b) Addition Reactions:

HBrC C CH

C

Br

hydrohalogenation

Which can be initiated by the attach of either H or Br

Nucleophilic addition reaction:

e.g. the base catalyzed formation of cyanohydrins in liquid HCN

CNC O C

CN

O

slow

HCN

fastC

CN

OH

+ CN

( H +CN)C O C

CN

O

slow

HCN

fastC

CN

OH

+ CN

C O HCN

(C) Elimination Reactions:

The reversal of addition reactions.

HBrC CC

HC

Br

dehydrohalogenation

CH

C

OHdehydration H2O

(d) Rearrangement Reactions:

e.g. Pinacol- Pinacolone rearrangement

HC

HOC

H3C

OH

CH3H3C CH3 C

H3CC

H3C O

CH3H3C

Represntative carbon compounds:

(1) Aliphatic hydrocarbons:

(a) Saturated hydrocarbons(1) Open chain Saturated hydrocarbons:

Ex. CH4 methane, CH3-CH3 ethane and all alkanes with general formula CnH2n+2

(2) Cyclic Saturated hydrocarbons:

Ex

cyclo pentane cyclo propanecyclo hexane

cyclo butane

(b) Saturated hydrocarbons(1) Alkenes and cycloalkenes:

H2C CH2

ethylene

HC

CH2H3Cpropene

and all hydrocarbons with general formula CnH2n

Alkenes Ex.

Cycloalkenes :

i.e cyclic hydrocarbons contain double bond Ex.

cyclopropenecyclobutene cyclopentene cyclohexene

(2) Alkynes:

Unsaturated hydrocarbons contain triple bond CnH2n-2.

HC CHacetylene(ethyne)

C CHH3Cmethyl acetylene(propyne)

A Represntative Aromatic Hydrocarbons:

They are cyclic unsaturated hydrocarbons containing conjugated double bonds, example benzene is six membered ring with alternating single and double bonds called a Kekule structure.

C

CC

C

CC

H

H

H

H

H

H

Kekule structurefor benzene

or

bond linerepresentationfor Kekule structure

In fact, the carbon-carbon bonds of benzene are all the same length (1.39A), a value in between that of carbon-carbon single bond and carbon-carbon double bond.

There are two ways of dealing with this problem: with resonance theory or with molecular orbital theory.

Based on the principles of resonance theory we recognized that benzene can not be represented adequately by either structure but that, instead, it should be visualized as a hybrid of the two structures. We represent this hybrid by a hexagon with a circle in the middle.

Two contributing KekuleStructures for benzene

A representation of the resonancehybrid

In the molecular orbital explanation.p

Sp2 hybridized carbon

When the benzene ring is attached to some other group of atoms in a molecule, it is called a phenyl group and it is represented in several ways

or or C6H5- or ph or ф

The combination of a phenyl group and a –CH2- group is called a benzyl group

CH2 CH2

or

or C6H5-CH2

Functional groups:

A functional group is the part of a molecule where most of its reaction occur. It is the part that effectively determines the compound’s chemical properties and many of its physical properties as well.

• The functional group of alkanes is C-H and C-C

• alkenes has carbon-carbon double

C C as a functional group.

• The functional group of alkynes is carbon-carbon triple bond

C C

Alkyl groups and the symbol R:

They are groups that would be obtained by removed a hydrogen atom from an alkane

HHR Ralkane alkyl group

Examples:

AlkaneAlkyl groupabbreviation

CH4

(methane)

CH3-

(methyl group)

Me-

CH3CH3

(ethane)

CH3CH2- or

C2H5-

(ethyl group)

Et-

CH3CH2CH3

(propane)

CH3CH2CH2-

(propyl group)or

CH3CH1CH3

(isopropyl group)

Pr-

i-Pr-

Alkyl Halides or haloalkanes:

Alkyl halides are compounds in which a halogen atom ( fluorine, chlorine, bromine or iodine) replaces a hydrogen atom of an alkane.

Ex. CH3Cl methyl chloride C2H5Br ethyl bromide

• Alkyl halides are classified into Primary (1º), or secondary

(2º), or tertiary (3º) and this classification is based on the

type of the carbon to which the halogen is directly attached.

• Examples of primary, secondary and tertiary alkyl halides

are the following.

CH

CH

H

HH Cl

1º carbon

1º alkyl chloride

2º carbon

3º carbon

2º alkyl chloride

3º alkyl chloride

CH

CH

HCH

Cl

H

H

H

CH3C

ClH3C

H3C

Alcohols (R-OH):

The functional group of alcohols is the hydroxyl (OH) group attached to sp3- hybridized carbon atom.

C O H is the functional group of an alcohol

Examples:

• CH3OH methyl alcohol (methanol), C2H5OH ethyl alcohol (ethanol)

• Alcohols may be viewed in two ways

(1) as hydroxy derivatives of alkanes.

(2) as alkyl derivatives of water.

Example

ethane

H3C CH3

ethyl alcohol (ethanol)

109º 105º

water

CH2

O

H

H3C H

O

H

● As with alkyl halides, alcohols are classified into three groups Primary (1º), or secondary (2º), or tertiary (3º) alcohols

CH

CH

HOH

H

H

1º carbon

1º alcohol

Geraniol(1º alcohol) Benzyl alcohol (1º alcohol)

CH2OH

CH2OH

● Primary carbon is the carbon atom which is attached to

one other carbon.

● 1º carbon also has two other carbon atom attached to it.

● 3º carbon also has three other carbon atom attached to it.

CH

CH

HO

CH

H

H

H

H

2º alcohol

2º alcohol

3º alcohol

2º carbon

3º carbon

OH

CH

CH

HO

CH

H3C

H

H

H

Ethers:

Ethers have the general formula R-O-R or R-O-R’ where R, R’

are alkyl groups.

● ethers can be thought as derivatives of water which both

hydrogen atoms have been replaced by alkyl groups.

R

O

R

H3C

O

H3CGeneral formula for an ether

Dimethyl ether(a typical ether)

C O C

The functional group of an ether

CH2H2C

O Oethylene oxide Tetra hydro furan (THF)

Cyclic ether

Amines:● Amines may be considered as organic derivatives of

ammonia

N

H

H H N

H

R H

ammonia (an amine)amphetamine (1º amine)

NH2

CH3C6H5

• Amines are classified as Primary (1º), or secondary (2º), or tertiary (3º) amines, this classification is based on the number of the organic groups (alkyl groups) that attached to the nitrogen atom.

N

H

R H N

R1

R H N

R1

R R2

2º amine1º amine3º amine

Examples:

NH2

Isopropyl amine 1º amine

N

H

C2H5 C2H5

Diethyl amine2º amine

N

C2H5

C2H5 C2H5

Tri ethyl amine3º amine

NH

Cyclic 2º amine

● The bond angle C-N-C in tri ethyl amine is 108,7º which is

very close to H-C-H in methane, thus the nitrogen atom of an

amine can to be sp3 hybridized and the unshared electron pair

occupy an sp3 orbital.

Aldehydes and ketones:

Both contain the carbonyl group.C O

the carbonyl group.

• In aldehydes, the carbonyl group is bonded to least one hydrogen atom, i.e. aldehydes has the follwing general formula.

CH

OR

R may be hydrogen atom or alkyl group

• In ketones, the carbonyl group is bonded to two carbon atom, i.e. aldehydes has the following general formula.

CR1

OR

Where R and R1 may be the same or different

AldehydesKetones

Formaldehyeacetone

Acetaldehydeethyl methyl ketone

benzaldehydebenzophenoneC H

O

C6H5 C C6H5

O

C6H5

C H

O

H

C H

O

H3C

C CH3

O

H3C

C CH3

O

C2H5

• Aldehydes and Ketones have a trigonal planar arrangement

of groups around the carbonyl carbon. The carbon atom is

sp2 hybridized

e.g in formaldehyde OC

121º

121º

118º

Carboxylic acids:

It have the general formula C OH,

O

R the functional group

C OH,

O

Is called the carboxyl group (carbonyl + hydroxyl)

C OH

O

R or RCO2H or RCOOH Acarboxylic acid

C OH

O

or -CO2H or -COOHThe carboxyl group

Examples :

C OH

O

H or HCO2H or HCOOH formic acid(1)

(2) acetic acidor CH3CO2H or CH3COOHC OH

O

H3C

(3) or C6H5CO2H or C6H5COOH benzoic acid

C OHO

Amides:- They have the formula

RCONH2, RCONHR1 or RCONR1R2Examples:

OCH3C

H2NOC

H3C

H3CHNOC

H3C

(H3C)2Nacetamide N-Methylacetamide N,N-diMethylacetamide

Esters:- esters have the general formula

RCO2R1, RCOOR1 or OCR

R1O

Examples:

O

CH3C OC2H5, CH3CO2C2H5 or CH3COOC2H5

it is prepared by this reaction

Ethyl acetate

CH3CO2C2H5 CH3CO2H HOC2H5+- H2O

Physical properties and molecular structure:

(1) Electronegativity:

• The electron cloud that bonds two atoms is not symmetrical

except when the two atoms are the same and have the same

substituents.

• The cloud is necessarily distorted toward one side of the

bond depending on which atom maintains the greater

attraction of the cloud.

• Electronegativity is the ability of atoms to attract the

electron clouds to it’s side.

• Electronegativity can measured by Pauling method which is

based on bond energies of diatomic molecules.

•Fluorine is the most electronegative elements in the periodic

table.

• some examples of the electronegativities of some atoms.

F4.0I2.5Mg1.2

O3.5C2.5Na0.9

Cl3.0H2.1Cs0.7

N3.0P2.1

Br2.8B2.0

S2.6Si1.8

Bond distances:

It the distance between the nucleus which form the molecule

nucleus nucleus

Atomic Orbitals

Molecular Orbitals

Bond Distance

• Bond distance is determined by X- ray diffraction

( for solid), electron diffraction ( for gases).

• Bond distances in a molecule are characteristic

properties of the molecule which give some

information about the chemical properties of that

molecule.

•The bond distance of the vibration of the

molecule.

•Bond distance are measured in angstrom Aº.

The effect of the vibration for a bond between two

sp3.

C-C bond in

Bond length Aº

C-C bond inBond length Aº

Diamond 1.5441.540 ± 0.015

C2H61.5324 ± 0.0011

t-butyl chloride1.532

C2H5Cl1.5495 ± 0.0005

n- butane to n-hexane

1.531-1.534

C3H81.532 ± 0.003isobutene1.535 ± 0.001

Table of Bond distances.

Bond type

Length Aº

Typical compds.Bond type

Length Aº

Typical compds.

C-CC=C

sp3- sp31.54 CH3CH3sp2- sp21.34Ethylene

sp3- sp21.50CH3CHO, toluene

sp2- sp1.31Ketene

sp2- sp21.48ButadieneC≡C

sp2- sp1.43Acrylonitrilesp- sp1.20Acetylene

sp- sp1.38cyanoacetylene(C-H) sp3- H

1.11methane

Bond type

Length Aº

Typical compds.

Bond type

Length Aº

Typical compds.

C-HC-N

sp2- H1.10Benzenesp3- NAmines

sp- H1.08HCN, acetyleneC=N

C-Osp2- N1.28Oximes, imines

sp3- O1.41Dimethyl ethersp3- F1.38

C=Osp2- F1.35

sp2- O1.20Formaldehydesp- F1.27

sp- O1.16CO2

• As the percentage of s increas the bond distance is

shortened, C-D is shorter than C-H.

Bond angle:The angle between two molecular orbitals ( three atom, and

central atom).

C C

H

H120º

120º

SP2

S

S

CH

H

H

H109º28`

SP3

109º28`

Oxygen, Sulfur, Nitrogen bond angles:-

Angle Value Compd. Angle Value Compd.

H-O-H104º.27`waterC-S-C(CH3)2S

C-O-H107º-109º

methanolH-N-H106º.46`NH3

C-O-C111º.43`(CH3)2OH-N-H106ºCH3NH2

C-O-C124º.5`(ph)2OC-N-H112º(CH3)2N

H

H-S-H92º.1`H2SC-N-H108º.7`(CH3)3N

C-S-H99º.4`CH3SH

Bond Energies:

There are two kinds of bond energy.

(1) Dissociation energy D Which defined as the energy

necessary for cleavage a bond to give the constituent

radical.

e.g. D for water + H2O HO H +118 kcal/mol

and D for OH bond in water is

+ HO HO +100 kcal/mol

● The average is 109 kcal/mol is taken as bond energy.

● In diatomic molecules D=E.

Bond Mean valueE kcal/mol

Cal. from

Bond Mean valueE kcal/mol

Cal. from

O-H110-111H2OC-Cl79CCl4

C-H96-99CH4C-S66C2H5-SH

N-H93NH3C≡C199-200CH≡CH

S-H82H2SC=C145-151(CH2=CH2

C-O85-91CH3OHC-I52CH3I

C-C83-85C2H6

Table: Bond energy E values for some important bonds

From the previous Table

(1) There is a correlation of bond strengths with bond

distances in general, shorter bonds are stronger bonds.

Since we have seen that increasing S character shortens

bonds

bond strengths increase with increasing S character.

(2) Bonds become weaker as we move down the periodic

table.

Compare C-O and C-S and C-Cl and C-I.

(3) Double bonds are shorter and stronger than the

corresponding single bonds but not twice as strong because

Л overlap is less than σ overlap. This means that a σ bond is

stronger than a Л bond.

The difference in energy between a single bond, say C-C

and C=C is the amount of the energy necessary to cause

rotation a round the double bond.

Polarity of bonds: Two atoms joined by a covalent bond, share electrons and

their nuclei are held by the same electron cloud.

• In most cases the two nuclei do not share the electrons equally; the electron cloud is denser about one atom than the other i.e. one end is relatively negative and the other is relatively positive ( δ- & δ+).

Two Nuclie

Molecular Orbital Electron Cloud

i.e. there is a negative pole and a positive pole such bond is

called polar bond or to possess polarity.

• We can indicate polarity using δ- & δ+, δ+ is partial positive,

δ- is partial negative. Some time we say delta plus and delta

minus.

H Fδ+ δ- N

HH H

δ- δ-

δ+

δ+δ+

δ+δ+

OHH

polar bonds

• We can expect a covalent bond to be polar if it join atoms that differ in electronegativity.

• The most electronegative elements are those located in the upper right hand corner of the periodic table.

Electronegativity can arrange in the following order

F >O >Cl, N >Br >C,H

• Bond polarities are concerned with physical and chemical

properties.

e.g The polarity of bonds leads to polarity of molecules

which affect the melting point, boiling point and solubility.

The polarity of a bond determines the kind of reaction that

can take place at that bond and even affects reactivity of near

by bonds.

Polarity of molecules:-

• They are molecules which contain polar bonds, such a

molecule constitutes a dipole ( two equal and opposite

chages separate in space).

• A dipole is often symbolized by where the arraw points

from positive to negative.

Polar molecules possess a dipole moment μ which is equal

to the magnitude of the charge e, multiplied by distance d

between the centers of the charge.

∴ μ (Debye) = e (e.s.u) x d (Cm)

H20HF1.75CH3Cl1.86

O20H2O1.84CCl40

N20NH31.46CO20

Cl20CH40

Table of Dipole moments

• Molecules like H2, O2, N2, Cl2 and Br2 have zero dipole moments i.e. They are non- polar, since each molecule has two identical atoms e is zero hence μ = 0

• HF has large μ of 1.75 D, F is very high dectro- negativity, d is small and e is large μ is large too.

Melting point:-It is the temperature at which the thermal energy of the particles

is great enough to over come the intracrystalline forces that

hold them in position.

i.e. Solid ∆ liquid

Crystalline is arranged in very regular symmetrical way( highly ordered)

More random arrangement

Intermolecular forces:- The forces hold neutral molecules to each other seen to be

electrostatic in natural involving attraction of positive

charge for negative charge.

• There are two kinds of inter molecular forces.

(1) Dipole – Dipole interactions.

(2) Van der Waals forces.

(1) dipole – dipole interaction is the attraction of the positive

end of one polar molecule for the negative end of another

polar molecule.

+ -

- ++ - + -or

● As a result of dipole – dipole interaction, polar

molecules are generally held to each other more strongly

than non-polar molecules of comparable molecular weight.

Hydrogen bonding: it is strong kind of dipole-dipole attraction

in which a hydrogen atom serves as a bridge between two

electronegative atom holding one by a covalent bond and the

other by purely electrostatic forces.

● For hydrogen bonding both electronegative atoms must

come from the group F, O, N.

H F H F……5 kcal/ mol

50-100 kcal/ mol

O

H

H O

H

H……N

H

H

H

N

H

H

H

……&

N

H

H

H

O

H

H……

or

● These three elements owl their special effectiveness to the concentrated negative charge on their small atoms.

● Hydrogen bonding affects the b. p. and solubility

properties of compounds and plays a key role in determining the shapes of large molecules like proteins and nucleic acids.

Van der Waals forces:- The forces between the molecules of non- polar compound

which make them can solidify.

Van der Waals forces can explained by the following:

The average distribution of charge about say a methane

molecule is symmetrical so that there in no net dipole

moment. However the electrons move about so that at any

instant of time the distribution will probably be distorted and

a small dipole will exist. This momentary dipole will affect the

second methane molecule near by the negative end of the

dipole tends to repel electrons and the positive end tends to

attract electrons, the dipole thus induces an oriented dipole

in the neighboring molecule.- + - + - +

- + - + - +

• Although the momentary dipole and induced dipoles are

constantly changing the net result is attraction between the

two molecules.

• Van der Waals forces have a very short range, they act

only between the portion of different molecules that are in

close contact that is between the surfaces of molecules.

Boiling points:- The temperature at which the thermal energy of the

particles is great enough to overcome the cohesive force

that hold them in the liquid.

• In the liquid state the unit of a non- ionic compound is

the molecule. The weak intermolecular forces of non – ionic

liquid is dipole – dipole interactions and van der waals

forces are more readily overcome than the strong interionic

forces of ionic compounds and boils at very much lower

temperature.

e.g. the non – polar methane boils at -161.5° and even polar

HCl boils at - 85°.

• liquids whose molecules are held together by hydrogen

bonds are called associated liquids and boil at higher b. p.

than compounds of there molecular weight and dipole

moment.

e.g.

(1) HF boils at 100 degree higher than the non – associated

HCl.

(2) H2O boils at 160 degree higher H2S.

• Boiling points increase as molecular weights increase.

Dipole – Dipole forces:-Most organic molecules are not fully ionic but have instead

a permanent dipole moment resulting from a non uniform

distribution of the bonding electrons.

δ+ δ-

Solubility: non- ionic solute when a solid or liquid dissolves the

structural units ions or molecules become separated from each other and

the space in between become occupied by solvent molecules. The energy

required to break the bonds between solute particles is supplied by the

formation of bonds between the solute particles and the solvent

molecules i. e. the old attractive forces are replaced by new ones.

CH3C

OH3C

CH3C

OH3Cδ-δ+

The solubility characteristics of non – ionic compounds are

determined by their polarity. Non–polar or weakly polar

compounds dissolve in non-polar or weakly polar solvents,

highly polar compounds dissolve in highly polar solvents

“ like dissolves like” it is useful rule.

e.g. methane dissolve in carbon tetrachloride molecules to

each other is Van der Waals interaction are replaced by very

similar forces holding methane molecules to carbon

tetrachloride molecules.

• Neither methane nor carbon tetrachloride is readily soluble

in water. The highly polar water molecules are held to each

other by very strong dipole – dipole interactions hydrogen

bonds.

• The highly polar organic compound methanol CH3OH is

quite soluble in water, hydrogen bonds between water and

methanol molecule can replace the very similar hydrogen

bonds between different methanol molecules and different

water molecules.

Alcohols: R OH

Non polar part like methane hydrophobic (water hating) some times called lipophilic( solubility in non polar solvents).

polar part like water hydrophilic (water loving) ( solubility in polar solvents).

Tables of solubility of alcohols in water.

alcoholSolubility g/100g water

alcoholSolubility g/100g water

CH3OHXCH3(CH2)3OH7.9

CH3CH2OHXCH3(CH2)4OH2.3

CH3CH2CH2OHXCH3(CH2)7OH0.05

• From the above table it is clear that as the lipophilic

increase the solubility in water decrease.

Solubility: ionic solutes.

Protic and aprotic solvents and ion pairs:

In the dissolution of ionic compounds a great deal of energy

is necessary to overcome the powerful electrostatic forces

holding together in an ionic lattice. Only water or other highly

polar solvents that are able to dissolve ionic compounds.

Polar molecule has positive and negative ends,

consequently, there is electrostatic attraction between a

positive ion and the negative end of the solvents molecule

and between a negative ion and the positive end of the

solvent molecule, these attractions are called ion – dipole

bonds. Each ion – dipole bond is relatively weak but in the

aggregate they supply enough energy to overcome the

interionic forces in the crystal.

Ion – dipole interactions, solvated ( hydrated in case of H2O) cation & anion

To dissolve ionic compounds a solvent must also have

a high dielectric constant, that is have high insulating

properties to lower the attraction between oppositely charged

ions once they are solvated.

• Water owes its superiority as a solvent for ionic substances

in part to its high polarity and high dielectric constant.

• Cations are attracted to the negative pole of a polar solvent, in water the negative pole is oxygen.

OH

H δ+

δ-δ+

……

• Anions are attracted to the positive pole of a polar molecule, in water the positive poles are an hydrogen.

OH

H

δ+

δ+

δ-……

• Hydrogen bonding permits particularly strong solvation of

anions. Thus, water owes a large part of its special solvating

power to its OH group: it solvates cations strongly through

the unshared pairs on oxygen; it solvates anions strongly

through hydrogen bonding.

• CH3OH resembles H2O having an OH group dissolves ionic

compound. CH3OH is less polar than water, CH3- group is

bigger than the second –H of water.

Protic solvents: solvents like water and methanol are

called Protic solvents containing hydrogen that is attached to

oxygen or nitrogen and hence is appreciably acidic.

Aprotic solvents: polar solvents with moderately high dielectric constants which do not contain acidic hydrogen.

S

CH3

O

H3C

SN

O

HCH3

CH3

P N(CH3)2

O

(H3C)2N

N(CH3)2

DimethylsulfoxideDMSO

Dimethyl formamideDMF

Hexamethylphosphoro--triamide (HMOT)

They dissolve ionic compounds but in doing their action differ

from protic solvents, they can not form hydrogen bonds to anions.

They are highly polar with dipole moments several times as large

as that of water. The negative pole in aprotic solvents is an oxygen

atom that just out from the rest of the molecules. Through

unshared pairs of electrons on these negatively charged well

exposed atoms. Cations are solvated very

The positive pole is buried within the molecule and anions

are solvated very weakly.

water dissolves ionic compounds very well but poor

solvent for most organic compounds, this difficulty can be

overcome by addition of second solvent like methane,

methanol’s hydrophilic –OH makes it miscible with water,

through its lipophilic CH3 it brings about dissolution of

organic compounds.

CH3CH2OH does like CH3OH dissolve both ionic and non-

ionic compounds.

Polar covalent bonds: when two atoms of different electronegativities from a covalent

bond, the electrons are not shared equally between them.

The atom with greater electronegativity draws the electron

pair closer to it and a polar covalent bond result.

H Cl H Clδ-δ+

H ClH Cl

δ-δ+

∴ HCl has a partially negative end with charge δ- and a partially

positive end with δ+ charge and hence forms a dipole and has

+ -

a dipole moment μ = 1.08 D

∴ H Cl(positive end) (negative end)

Acids and bases:- The terms acid and base can be defined in a number of

ways, According to the Lowry- Bronsted definition acid is a

substance that gives up a proton and a base is substance

that accepts a proton.

+ H2SO4 H2O H3O HSO4+

Ex.

+

+

-

-c

+ HCl NH3 NH4 Cl+

Stronger acid

Stronger acid

Stronger base

Stronger base

weaker base

weaker base

weaker acid

weaker acid

According to the Lowry definition, the strength of an acid

depends upon its tendency to give up a proton and the

strength of a base depends upon its tendency to accept

a proton.

Ex. H2SO4 & HCl are strong acids because they tend to give

up a proton very readily. Conversely HSO4- and Cl- must be

weak bases.

If aqueous H2SO4 is mixed with aqueous NaOH, the acid H3O+

gives a proton to the base OH- to form the new acid H2O and

the new base H2O.

+ H3O OH H2O H2O+stronger acid

stronger base

weaker acid

weaker base

+ -

Also for NH4Cl and NaOH

+ NH4 OH H2O NH3++ -

stronger acid

stronger base

weaker acid

weaker base

Acids are arranged in the following order:

Acid strength:

H2SO4, H3O+ > NH4+ >H2O

for the corresponding conjugate bases have the opposite order

∴ Base strength

HSO4-, Cl- < H2O < NH3 <OH-

Like H2O many organic compounds that contain oxygen can

act as bases and accept protons.

+ C2H5OH H2SO4 C2H5 HSO4+OH

H-+

oxonium protonated ethyl alcohol

Ex. 1

Ex. 2

+(C2H5)O HCl (C2H5) Cl+O Hoxonium protonated diethyl ether

+ -

Lewis definition of bases and acids:-

Base is a substance that can furnish an electron pair to form

a coordination bond i.e. the base is electron pair donor e. g.

OH, C2H5OH NH3, (C2H5)2O,

Lewis acid is a substance that can take up an electron pair

to form a coordination bond i. e. is an electron pair acceptor

e. g. H, AlCl3, BF3, SnCl4equation of lewid acids bases reactions:

+ BF3NH3 B

F

F NH3

F

Acid base

(1)

(2)+ BF3

O(C2H5)2 B

F

F O(C2H5)2

F

Acid base

To be acidic in the Lowry- Bronsted sense, a

molecule must of course contain hydrogen and the degree

of acidic is determined largely by the kind of atoms that

holds the hydrogen and the ability to accommodate the

electron pair left behind by the departing hydrogenation.

The ability to accommodate the electron pair

depends upon.

(1) The atom’ s electronegativity.

(2) The atom’ s size.

Thus within a given row of the periodic table.

acidity increases as electronegativity increases

H-CH3 < H-OH2 < H-OH < H-F < H-SH < H-Cl

within a given family acidity increases as the atomic size

increase.

H-F < H-Cl< H-Br < H-I < H-OH < H-SH < H-SeH

Acids and Bases in Nan-aqueous solutions:-Many of the organic acid-base reactions occur in solutions.

These acids are much weaker acids than water and the

conjugate base of these very weak acids are powerful bases.

Reactions of organic compounds are carried out in liquid

ammonia. The most powerful base that can employed in

liquid ammonia is the amide ion –NH2.

NH2+stronger acid

ka= 10-25stronger base

C CH H NH3+C CHliquid

NH3 weaker acid

ka= 10-33weaker base

Ethane and all other alkanes are much weaker acids tha ammonia.

NH2+

stronger base stronger base

CH3CH2 NH3 +

weaker acid weaker acid

CH3CH3

The general order of the acidity of some weak acids.

RH < RCH=CH2 < H2 < NH3 < R-C≡CH < ROH < H2O < RCOOH

increasing acidity

The order of basicity of conjugate bases

R:- > RCH=CH:- > H:- > -:NH2 > RCH≡C:-

> RO:- > :OH- > RCO2

-

increasing basicity

Acid – base reactions and the preparation of

deuterium and tritium labeled compounds:Chemists often use compounds in which deuterium and

tritium atoms have replaced one or more hydrogen atoms of

the compounds as a method of “ Labeling ” or identifying

particular hydrogen atoms D is H & Tritium H.One way to introduce deuterium or tritium into a specific

location in a molecule is

1

32

1

Through acid – base reaction that takes place when a very

strong base is treated with D2O or T2O.

e.g.

OD+

isopropyllithiumstronger base

stronger acid

CH3CH Li D2O +

weaker base2-Deuteriumpropaneweaker acid

CH3

CH3CHD

CH3

Isomerism:

Isomerism: compounds have the same molecular formula

and different physical and chemical properties due to the

difference in structural formula.

e.g. C2H6O can be

EthanolH3CCH2OHb. p. 78°+ HI CH3CH2I+ Na C2H5ONa + H2

Dimethyl etherH3C-O-CH3

b. p. -24°+ HI 2CH3I+ Na No reaction

Toutomersion: it is rapid equilibrium between a mixture

Definite compound, it is a proton shifts from one atom of a molecule to another atom.

Keto–Enol toutomersion:-It takes place between a carbonyl compounds containing an α- hydrogen and its enol form.

C CR

R1

H O

R2 C CR

R1

OH

R2

keto form enol form

Ex.

H3C C

O

CH3 H2C C

OH

CH3

acetone enol form

The % of enol is affected by the structure, solvent, concentration and temperature.• Ethyl acetoacetate is in high % enol form because it is stabilized by internal hydrogen bonding which is unavailable to the keto form.

H3C C

O

CH2

CO2EtC

OH

O

C

HC OEtH3C

Constitutional Isomers:“isomer” is composed of two part “iso” in Greek words means the “same” and “meros” means part i.e. isomer means the same part.

Constitutional (structural) isomers:They are compounds having the same molecular formula but differ in the way of their atoms are bonded together (structural) formula.Ex. 1: urea and ammonium cyanate have the same molecular formula CH4N2O and different structures.

Ex.2: nitro methane and methyl nitrite CH3NO2

NH4NCO H2N C

O

NH2ammoniumcyanate

urea

H3C N

O

Onitromethane( liquid)

H3C O N Omethyl nitrite( gas)

HydrocarbonsHydrocarbons are compounds composed of carbon

and hydrogen only and they are classified as follows

Hydrocarbons

Aromatic hydrocarbons ( arenes)

Aliphatic hydrocarbons

AlkanesCH3-CH3

Ethane

AlkenesCH2=CH2

Ethylene

AlkynesCH≡CHacetylene

benzene

Alkanes Alkanes are saturated hydrocarbons and are classified

into.Alkanes

a cyclic alkanesGeneral formula (CnH2n+2)e.g.CH4 methaneC2H6 ethaneC3H8 propane

C4H10 butane

cyclic alkanesGeneral formula (CnH2n)e.g.

cyclo propane

cyclo pentanecyclo butane

A cyclic Alkanes

They are saturated hydrocarbons having the general

formula CnH2n+2and they are found as branched and

unbranched alkanes.

Nomenclature of unbranched Alkanes:

They are found as straight carbon chain ( called n-alkanes).

e.g. CH4 methane, CH3CH3 ethane, CH3CH2CH3 propane,

CH3CH2CH2CH3 n-butane, CH3CH2CH2CH2CH3 n-pentane,

CH3CH2CH2CH2CH2CH3 n-hexane, CH3(CH2)5CH3 n-heptane,

CH3(CH2)6CH3 n-octane, CH3(CH2)7CH3 n-nonane,

CH3(CH2)8CH3 n-decan.

For writing the accurate structure of sp3 hybridization e.g.

C

propane

C

H

CH

H

H

HH

H

H

CC

H

CH

C

H

HH

H

H H

H

H

n-butane

For simple representation we can use the Zigzag

structure (bond- line structure).

e.g. CH3(CH2)4CH3 n-hexane can be shown as.

n-hexane

n-butanen-heptane

propane

≡

Note: The alkanes take the suffix ane.

Alkyl groups:R H R

H

alkanesuffix-ane

alkylsuffix-yl

e.g. CH3- methyl, CH3CH2- ethyl, CH3CH2CH2- n-propyl,

CH3CH2CH2CH2- n-butyl, CH3CH2CH2CH2CH2- n-pentyl.

Classification of carbon atoms: Carbon atoms are classified according to their degree of

substitution by other carbons.

• A primary carbon is one that is directly attached to one

other carbon.C C

H

H

primary carbon

primary alkyl group

• Secondary carbon is directly attached to two other carbon atoms.

C C

H

C

secondary carbon

secondary alkyl group

• A tertiary carbon is directly attached to three carbon atoms.

C C

C

C

tertiary carbon

tertiary alkyl group

• A quaternary carbon is directly attached to four carbon atoms.

C C

C

C

C quaternary carbon

• There is no quaternary alkyl group.

Com. Names (trivial)

CH3- methyl

CH3CH2- ethyl

CH3CH2CH2- n-propyl

CH3CH2CH2CH2- n-butyl

CH4

ethane

methane

propane

n- butane

iso-butane

n- pentane

iso- pentane

neo- pentane

iso-butylCHCH2

H3C

H3C

Branched Alkyl groups: They are named by using the longest continuous chain that

begins at the point of attachment as the base name.

i.e. Numbering starts from the carbon where the hydrogen

atom was removed.

e.g CH3CH2CH2- n- propyl group,

CHH3C

CH3

IUPAC 1-methyl ethyl group(comm.) isopropyl

CHCH3H3C

CH3

CCH3H3C

CH3

isobutyl group

≡

tertiarybutyl group

CHCH2

H3C

CH3

H2C

CH2

H3C

H2C

CH2

n- pentane n- pentyl

CHCH2

H3C

H2C

CH3

iso- pentane iso- pentyl

CCH3

H3C

CH3

H3CC

CH2

H3C

CH3

H3Cneo- pentane neo- pentyl

≡

IUPAC (International Union of Pure and Applied

Chemistry) Nomenclature:-e.g. (1)

1- Methyl propyl group

21

3

H2C

CH

H3C

CH3

CHCH2

H3C

CH3

21

3

2- Methylpropyl group

e.g. 3

e.g. 2

CH3C

CH3

H3C

1,1- Dimethyl ethyl group(tert-butyl)

Systematic IUPAC Nomenclature of branched Alkanes:- Branched- chain alkanes are named according to the following rules:-(1) Locate the longest continues chain of carbon atoms; this chain determines the base name of the alkane.e.g. 1

1

1

2

2

3

3

4

4

5

5

6

6

7

hexane

e.g. 2

heptane

(2) Number the longest chain beginning with the end the chain nearest constituent. (give the position of the constituent the lowest).

2

1

1

3

4

5

6

7

2

3

4

5

6

substituent

ok

(3) Use the numbers obtained by application of rule 2 to

designate the location of the substituent group, the base

name is placed last, and the substituent group, preceded by

the number designating its location on the chain, is placed

first.

• Numbers are separated from words by a hyphen. e.g.

CC C CCC

C

≡112

2

334

4

5 566

2- Methylhexane

2- Methylheptane

CC C CCC

C

C ≡2 1 1

23 34

45 56

6

7 7

(4) When two or more substituents are present, give each substituents a number corresponding to its location on the longest chain.

CC C CCC

CC

C4- ethyl-2-methylhexane

123456

21 3 4 5 6

• The groups should be listed alphabetically (i.e. ethyl before

methyl).

• In deciding on alphabetical order disregard multiplying

prefixes such as “di” “tri” and disregard structure –defining

prefixes that are written in italics and are separated from the

name by a hyphen such as sec- and tert- and consider the

initial letter (s) of the substituent name.

Thus ethyl precedes dimethyl and tert- butyl precedes ethyl

but ethyl precedes isobutyl.

5- when two or more substituents are present on the same

carbon atom use that number twice e.g..

CC C CCC

C

C

C

3- ethyl-3-methyl hexane

6- when two or more substituents are identical, identical this

by the use of prefixes di, tri, tetra, …. and so on and then

make certain that each and every substituent has a number.

Commas are used separate numbers from the other.

CC CC

CC

≡ 2

2 113

3

4 4

2,3- dimethylbutane

2,3,4- trimethylpentane

CC CC

CC

C

C

CC CC

C

C

C

C C

2,2,4,4- tetramethylpentane

1

1

1

1

22

22

3

3

3

3

4

4

4

4

5

5

5

5

7- when two chains of equal length compete selection as base chain, choose the chain with the greater number of substituents.

1234567

2,3,5- tri methyl-4-propyl heptane

(four substituents)

(three substituents)

not 4- sec-butyl-2,3- dimethylheptane

CC CC

C

C

C C

C

CCC

C

8- when branching first occurs at an equal distance from

either end of the longest chain, choose the name that gives

the lower number at the first point of difference.

CC C C

C

C C

CC2,3,5- tri methyl hexane

(not 2,4,5- tri methyl hexane)

1

2

23456

1 3 4 5 6

Classification of hydrogen atoms:- The hydrogen atoms of an alkane are classified on the

basis of carbon atom to which they are attached. A hydrogen

atom attached to a primary carbon atom is a primary

hydrogen atom, attached to secondary carbon is a secondary

hydrogen, and attached to tertiary carbon is a tertiary

hydrogen.

Example: 2- methyl butane has primary (1°), secondary (2°), tertiary (3°) hydrogens.

CH2

CH CH3CH3

CH3

1° hydrogen atom

3° hydrogen atom2° hydrogen atom

≡

3° hydrogen atom 2° hydrogen atom

1° hydrogen atom

Cycloalkanes Nomenclature:• Cycloalkanes are also called alicyclic (aliphatic cyclic)

hydrocarbons.

Cycloalkanes are alkanes in which two carbons are bonded

together to form a ring, they are characterized by the

molecular formula CnH2n.

Examples:

H2C

H2C CH2

≡

cyclopropaneH2C

H2C CH2

CH2

cyclobutane

H2C

H2C CH2

CH2

H2C

≡

cyclopentane

H2C

H2C

CH2

CH2

CH2

H2C

cyclohexane

• Cycloalkanes are named under the IUPAC system, by

adding the prefix cyclo to the name of the unbranched alkane

with the same number of carbons as the ring.

Substituent groups are identified in the usual way.

• The positions of the substituents are specified by

numbering the carbon atoms of the ring in the direction that

gives the lowest number to the substituent groups at the first

point of difference.

Ethyl cyclopentane

3-Ethyl-1,1-di methylcyclopentane

2

1

3

4

5

6

2

1

3

4

5

6

21

3

4

5

6

1-ethyl-3-methylcyclohexane

(not 3-ethyl-1-methylcyclohexane)

(not 1-ethyl-5-methylcyclohexane)

• When the ring contains fewer carbon atoms than an alkyl

group attached to it, the compound is named as an alkane

and the ring is treated as cycloalkyl substituent:

2

1

3

4

5

3-cyclobutylpentane

1 32 4 5C C C C C C

6

3-cyclopropylhexane

5

5

6

4

4

3

3

2

2

1

1

1,1-dimethyl-2-propylcyclopentane

1-ethyl-2-methylcyclopentane

Functional groups in hydrocarbons and substituted

alkanes:-

• A functional group of a compound is the structural unit of

this molecule that is responsible for its chemical reactivity

under a particular set of conditions.

• The functional group may be single atom or a set of atom.

• The functional group of a hydrocarbon is any one of its

hydrogens, the reactivity of hydrocarbons may be illustrated

by replacement with chlorine.R H Cl2+alkane

hv R Cl HCl+alkyl chloride

CH

H

CH

H

HH

Cl2+

ethane

hvHCl+

chloroethane

CH

H

CH

H

HCl

Substituted Alkanes:-

• The hydrogen atoms of alkanes are not reactive.

• When a group replace a hydrogen in alkane, the group

becomes almost functional group.

Examples of functionally substituted alkanes:-

AlkaneSubst. AlkaneExample

R-HR-OHC2H5OH

Ethanol

R-HR-XC2H5Cl

Ethyl Chloride

R-HR-NH2C2H5NH2

Ethyl Amine

R-HR-O-RC2H5OC2H5

Diethyl ether

AlkaneSubst. AlkaneExample

R-HR-CNC2H5CN

Propane nitrile

R-HR-NO2C2H5NO2

Nitroethane

R-HR-SHC2H5SH

Ethanethiol

R-H CH2H2C

O

CH-R'R-HC

O

Epoxide Ethylene Oxide

Example 2 The C5H12 isomers

There are three isomeric alkanes having the molecular

formula. C5H12

CH3(CH2)2CH3 ≡

Comm. Name: n- pentaneIUPAC name: pentane

CH2

CH CH3CH3

CH3

≡

Comm. Name: isopentaneIUPAC name: 2-methyl butane

CCH3

H3C

CH3

H3C

≡

Comm. Name: neo- pentaneIUPAC name: 2,2- di methyl propane

Sources of hydrocarbons:-Hydrocarbons are the main constituents of petroleum.

• C1-C4 gas, Heating, cooking, petrochemical, raw material.

• C5-C12 liquids (Naphtha), Fuel, lighter fraction, petroleum

ether b. p. 30- 60°C, laboratory solvents.

• C12-C15 Kerosene, fuel.

• C15-C18 Fuel oil, diesel fuel.

• Over C40 lubricating oil, greases, paraffin waxes, asphalt.

The Biological effects of alkanes:-

• Gasoline, which is a mixture of alkanes should not be

swallows because it contains additives such as tetra ethyl

lead which is quite poisonous.

• Liquid alkanes can cause damage if they get into lungs.

• They dissolve the liquid molecules in the cell membranes.

• Liquid alkanes can also harm the skin by dissolving the

natural body oils and cause the skin to out.

• Mixtures of high molecular weight liquid alkanes are used

to soften and moisten skin. Example petroleum jel “vaseline”

which used to protect skin and it can protect babies from

diaper rash caused by the skin’ contact with urine.

Conformations of alkanes and cycloalkanes:- The different spatial arrangements that result from

rotation about single bond in a particular molecule are

called, Conformations, Conformers, rotational isomers or

rotamers.

(1)Conformational analysis of Ethane:

Ethane exists in two conformations known the staggered

conformation and the eclipsed conformation and can

represented as follow.

H

H

HH

H H

H

H HH

HHH

H H

H

HH

(a) Wedge- and dash (b) Newman projection

(c) Sawhorse

Staggered conformation of ehtane

Eclipsed conformation of ehtane

H

HH

H

HH

(a) Wedge- and dash (b) Newman projection

(c) Sawhorse

H

H

H

H

HHH

H H

H

H H

The H-C-C-H unit in ethane is characterized by a torsion angle ( dihedral angle).• Torsion angle is the angle between the H-C-C plane and the C-C-H plane and H-C-C-H unit.

HH180°

0°H

H 60°

H

H

Torsion angle = 0°Eclipsed

Torsion angle = 60°Gauche Torsion angle =180°

Anti

• The staggered and eclipsed conformations of ethane are interconvertable by rotation about its C-C single bond.• The angle between hydrogens and the two carbons of ethane is called the “ torsion angle” or “ dihedral angle”.• The torsion angle in staggered conformation is 60° while

eclipsed form it is 0 °.• The spatial relationship between substituents in the staggered conformation is called “ gauche”.• Between the staggered and eclipsed conformation there will be an infinite number of conformations that differ by tiny increments in their torsion angle and are known as “ skew” conformation.Conformational analysis of Butane:* Butane posses two different staggered conformations in the first the two methyl groups in gauche relation. in the second they are anti to each other

* In the gauche conformation of butane, a hydrogen form each methyl group will be very close to each other to the extent that they repel each other, the repulsive destabilization of a molecule resulting from crowding of atoms or groups is called “ Van der Waals strain”, “ Steric hindrance”

H

H CH3

CH3

HHCH3

H H

H

CH3H

Gauche conformation

CH3

H HCH3

HHCH3

H H

CH3

HH

Anti conformation

CH3

H HCH3

HHH

H3C

H

CH3

HH

H

H

CH3

CH3

HH

H3C

H

H

CH3

HH H

H CH3

CH3

HH

H

H3C HCH3

HH

Full eclipsed

eclipsedgauch

gauch

Anti

eclipsed

Conformation of cyclohexane:* Cyclohexane exists in a stable conformation known “chair” conformation, this conformation is free of angle strain becouse all of its bonds are staggered

H

H

H

H

H

H

H

H

H H

H

H

Chair conformation “more stable”

H

H

H

H

* Newman projection of the chair conformation of cyclohexane.

H

H

H

CH2

HH

H2C

H

H

H

H

H

H

H

H

H

H

H

Boat conformation

Less stable because a steric strain of two flagpole hydrogens and torsional strain due to eclipsing of the bond on four of its carbon.

H

H

H

HH

H

CH2

H

CH2

H

Torsional strain

Synthesis of Alkanes:(1)Hydrogenation of Alkenes

General Reaction

C C H HPt or Ni

SolventC C

HH

+

CH3CH=CH2 Ni

C2H2OHCH3CH2CH3H H+

C2H2OHH H+

Pt

propene propane

cyclohexenecyclohexane

Ex.(1)

(2)

(2) Reduction of Alkyl halides

General Reaction

Zn

CH3CH2CHCH3

+

Butane

alkyl halide

Ex.

R X 2H R H2 + ZnX2

alkane

BrH

CH3CH2CH2CH3

2-bromo butaneZn

(3) Lithium Dialkylcuprates: The Corey-Posner, Whitesides House synthesis:-

The over all synthesis:

+alkyl halide

R X several R

alkanealkyl halide

R' Xsteps (-2X)

R'

Reaction steps:

+

alkyl halide

(2)

R X diethyl etherR Li

alkane

alkyl lithiume

R' X R'

2Li + LiX(1)

R Li2 + CuI R2CuLi LiI+ lithium dialkylcuprate

(3) R2CuLi + R + +RCu LiX lithium dialkyl cuprate

Restirictions of the reaction:

R X

diethyl ether

alkane

R' X R'R LiCuI

R2CuLi R + +RCu LiX

R' X

Li

R-X is any alkyl halide, R’-X is a methyl, 1° alkyl or 2° cycloalkyl halide

Hexane

diethyl ether

Ex.1

CuI(CH3)2CuLi

LiCH3LiCH3I

CH3(CH2)3CH2ICH3(CH2)4CH3

+

methyl cyclohexane

diethyl ether

I

+ LiI

Ex.2

(CH3)2CuLi + CH3Cu

CH3

CH3 CH2

C CH3

CH3

CH3

Outline the synthesis of

Using Cory-House synthesis?

+ CuILiH3C C Cl

CH3

CH3

H3C CH2Cl

+ CuILiH3C C CH2Cl

CH3

CH3

CH3Br

+ CuILiC2H5 C Cl

CH3

CH3

CH3Br

or

Reactions of Alkanes:(1) Halogenation of Alkanes

+alkane alkyl halideR H X2 R X +HX

X=F fluorinationX=Cl chlorinationX=Br brominationX=I iodinationThe order of reactivity is F>Cl >Br >I* Fluorination is very vigorous exothermic reaction and is not used in the laboratory.* Iodination is an endothermic reaction and alkyl iodides are not prepared by direct iodination of alkanes.

+R H I2 R I +HI

Chlorination of methane:-

+ Cl2CH4Room temp.

No Reactionabsence of light

* In presence of heat or light

+

methylchloride

Cl2 +CH4or hv

440CH3Cl HCl

CH3Cl +

methylene chloride

Cl2 +or hv

440CH2Cl2 HCl

CH2Cl2 +

chloroform

Cl2 +or hv

440CHCl3 HCl

CHCl3 +

carbon tetrachloride

Cl2 +or hv

440CCl4 HCl

°

°

°

°

Mechanism of Chlorination of methane:-

Initiation step

Cl2or hv

heatCl2

chloride free radical

Propagation step

+ +CH4 abstraction

hydrogen radicalHCl

CH3Cl+ Cl2 +

Cl

methyl free radical

CH3

CH3 Cl

Termination step

CH3CH3

Cl2

CH3Cl+

Cl

CH3 Cl

2CH32

Structure and Stability of Free Radicals:-

R C

R

R

> > >R C

H

R

R C

H

H

H C

H

H

Tertiary

Free radical

Secondary

Free radical

Primary

Free radical

methyl

Free radical

Also carbocations

R C

R

R

> >>R C

H

R

R C

H

H

H C

H

H

Tertiary

carbocation

Secondary

carbocation

Primary

carbocation

methyl

carbocation

* Chlorination of methane, ethane, cyclobutane gives only mono halogen derivatives.

methyl chloride

CH3CH3

CH4 CH3Cl+ Cl2 +or hv

440HCl

+ Cl2 +or hv

440HClCH3CH2Cl

+ Cl2 +or hv

440HCl

Cl

methane

ethane

cyclobutane

ethyl chloride

cyclo butylchloride

* Chlorination is less selective process (i.e. Cl2 is highly reactive and less selective).* Chloriniation of alkanes containing nonequivalent hydrogen affords mixture of mono halogen derivatives (all possible derivatives)

Butane

++ Cl2 hv

1-chloro butaneCl

2-chloro butane

ClEx.1

2-methyl butane

++ Cl2 hv

1-chloro-2-methylpropane

Cl

ClEx.2

2-chloro-2-methylpropane

* Bromination process is milder than chlorination and more

selective.

e.g. bromination yields the tert-mono bromoalkane almost

exclusively

2-methyl pentane

+ Br2 hv

BrEx.1

2-bromo-2-methyl pentane

methyl cyclohexane

+ Br2 hv

Ex.2

1-bromo-1-methyl cyclohexane

Br

AlkenesAlkenes are hydrocarbons that contain a Carbon-Carbon

double. Their functional group is a Carbon-Carbon double

bond. They have the general formula when they have one

double bond and alicyclic is CnH2n.

Nomenclature of alkenes and cycloalkenes:

* There are old names for simple alkenes for example.

H2C CH2 HC CH2H3C C CH2H3C

CH3

IUPAC:

Ethylene

Ethene Propene 2-methylPropene

Propylene isobutyleneCom.:

The IUPAC Rules for naming Alkene:

(1) Determine the longer carbon chain that contains the

double bond, change the ending of the alkane to ene.

(2) Number the carbon chain that include the double bond

and begin numbering at the double bond;

Designate the location of the double bond by using the

number of the first atom of the double bond as a prefix.

1 32

2

413

4

1-Butene

(Not 3-Butene)

132

546

12

34

56

2-Hexene

(Not 4-Hexene)

(3)Indicate the location of the substituent groups by the

numbers of the carbon atoms to which they are attached.

Examples:

Cl1

2-methyl-2-butene

(Not 3-methyl-2-butene)

1

1

1

12

2

22

3

3

3

3 4

4

4

45

56

6

5,5-dimethyl-2-hexene

(Not 2,2-dimethyl-4-hexene)

1

1

12

2

2

243

33

34

44

5

56

6

2,5-dimethyl-2-hexene

(Not 2,5-dimethyl-4-hexene)

1-chloro-2-butene

(Not 4-chloro-2-butene)

(4) Number substituted cycloalkenes in the way that gives

the carbon atoms of double the 1- and2-positions that

gives the substituent groups the lower numbers at the first

point of difference. The position of the double is not

nescessary to specify (mentioned).

Examples:

1-methylcyclopentene

(Not 2-methylcyclopentene)

1

1

2

2

21 H2C

CH

CH2

CH

CH

HC

CH3H3C

1

3

4

5

63

45

6

3,5-dimethylcyclohexene

(Not 4,6-dimethylcyclohexene)

2

(5) Two frequently encountered groups are the vinyl group

and ally group.

C C

H

H

H

H2CHC

the vinyl group (ethenyl)

C C

CH2

H

H

H

H2CHC CH2

the allyl group

(2-propenyl)e.g

C C

Br

H

H

H

Bromo ethene

(vinyl bromide)

C C

CH2Cl

H

H

H

3-chloropropene

(allyl chloride)

(6)Designate the geometry of the double bond of

disubstituted alkene with perfixes Cis and trans (cis: two

identical groups on the same side, trans if the two identical

groups on opposite sides).

e.g.

C C

Cl

H

Cl

H

Cis 1,2-dichloroethene

C C

H

Cl

Cl

H

trans1,2-dichloroethene

Structure and Bonding in alkenes

C C

H

HH

H

sp2sp2sp2

sp2

sp2

sp2

1s1

1s11s1

1s1

PyPy

Ethene (ethylene)

Ex.

• Higher alkenes are related to ethylene by replacement of

hydrogen substituents by alkyl groups. There are two

different types of carbon-carbon bonds in propene,

CH3-CH=CH2, the double bond is of the δ+Π type, and the

bond to the methyl group is a δ bond formed by sp3- sp3

overlap.

C C

H

H

H

C

H

H

H

SP2 hybridized carbon

SP3 hybridized carbon

• Isomerisation in alkene (Cis & trans

isomerisation): The alkene hydrocarbon C4H8 can exist in four isomers

C C

H

C2H5

H

HC C

CH3

CH3

H

H

C C

H

CH3

H3C

H

C C

H

CH3

H

H3C

Cis 2-butenetrans 2-butene

1-butene 2-methylpropene

( No cis, trans isomerism here)

( two geometrical isomerism)

Ex.1

Ex.2C C

H

COOH

H

HOOC

C C

COOH

H

H

HOOC

Cis -isomermaleic acid

trans -isomerfumaric acid

Naming stereo isomeric alkenes by the E-Z system:

• When one of two different substituents on a carbon of the

C=C bond is identical with a substituent on the other

carbon, it will be easy to describ the isomers as cis-trans

Example.

C C

H

CHO

C6H5

H

Oleic acid (Cis)

C C

CH2(CH2)6COOH

H

H3C(H2C)6H2C

H

cinnamaldehyde (trans)

• When the four substituents on the two carbons of the C=C

bond are different, cis-trans description become ambiguity.

• In such cases we will use E/Z description ( E from German

means Entgegen = Opposite, Z means Zusammen =

together)

• For the assignment of E/Z we use the sequence rules, the

isomer having the two atoms or groups with high priority

on the same side of C=C bond will be described Z and

when they are on opposite sides of the molecule the

isomer will be E.

sequence rules:(1) Rule 1 The atom of higher atomic number getting high

priority, if two atoms are isotopes of the same element, the

atom of higher mass number has higher priority.

e.g the sequence is I > Cl > S > D > H

Example1.

1-bromo-1-chloro-2-iodo ethene

C C

I

H

Br

Cl

1

2

1

2

C C

I

H

Cl

Br

1

1

2

2

E

EZ

Z

Example 2: 1-bromo-2-chloro-1-flouro ethene.

C C

Br

F

H

Cl 1 11

1 2 22

2

C C

F

Br

H

Cl

Example 3: 1-bromo-1-chloro propene.

C C

H

CH3

Cl

Br

Z

(2) Rule 2: if the relative priority of two groups can not be

decided by rule 1 it shall be determined by a similar

comparison of the next atoms in the groups e.g. C2H5 group

has higher priority than CH3;

H3C CH

CH3

(isopropyl) has higher priority than C2H5 & -CH2Cl

Chloro methyl has higher priority than H3C CH

CH3

Precedence is determined at the first point of difference:

ethyl C2H5 [-C-( C, H, H)] outranks methyl CH3 [-C( H, H, H)]

& similarly tert-butyl outranks isopropyl and isopropyl

outranks ethyl.

-C(CH3)3 > -CH(CH3)2 > -CH2CH3

-C(C, C, C) > -C (C, C, H) > -C( C, H, H)

Ex.

C C

CH2CH3

CH3

Cl

Br

lower

higher

higher

lower

E

Also -CH(CH3)2 outranks -CH2CHOH

[-C-( C, C, H)] outranks [-C( C, H, H)]

Ex.

C C

CH(CH3)2

CH2CH2OH

Cl

Br lower

lower higher

higher

E

Also –CH2OH [-C-( O, H, H)] outranks –C(CH3)3 [-C( C, C, C)]

Ex.

C C

C(CH3)3

CH2OH

Cl

Br

lower lower

higherhigher

Z

(3) Rule 3: where there is a double or triple bond both atoms

are considered to be duplicated or triplicated thus C=A

equals C A

CA

C A C A

CA

A C

& equals

e.g. equals C O

H

C O

CO

H

e.g. for –OH, -CHO, -CH2OH groups the sequence is –OH,

-CHO, -CH2OH

HCC

CHC

≡ ≡

Phenyl C6H5-

e.g. has higher priority than H3CCH

CH3

(isopropyl)

& -CH=CH2 (vinyl group) equals C C

HC

H H

C

takes priority over C CH3

CH3

H

(isopropyl)

i.e. –CHO is treated as if it –C( O, O, H) so -CHO, [–C( O, O,

H)] outranks -CH2OH [–C( O, O, H)]

Ex. The compound

C C

CHO

CH2OH

Cl

Br lower

lower higher

higher

E

Also -CH=CH2 is treated as if it were

A vinyl group –CH=CH2

C C

H

C C

C C

H

C C

outranks an isopropyl group

Ex. The compoundC C

CH(CH3)2

CH=CH2

Cl

Br higherhigher

lower lowerZ

Relative stabilities of alkenes:

The heat of combustion of the four isomeric alkenes having

the same molecular C4H8 showed that

C C

H

CH3

H3C

H

C C

H

H

H3C

H3C

C C

CH3

H

H3C

H

C C

C2H5

H

H

H

1-butene cis-2-butene trans-2-butene 2-methylpropene

order of decreasing heat of combustion and increasing of stability

Stabilities and degree of substitution of the alkenes: Stability of alkenes increases with increasing of degree of

substitution on the C=C bond

i.e

C C

R

R

R

R

C C

R

R

H

R

C C

R

H

H

R

C C

H

R

H

R

C C

R

R

H

H

C C

H

R

H

H

C C

H

H

H

H

>>> >

> >

Stability

The degree of substitution in alkenes can extend in cyclic

alkenes

e.g

*

*, the ring carbon indicated by an asterisk counter

as a unique substituent on the double bond

> >

tetrasubstituted

alkene

trisubstituted

alkenedisubstituted

alkene

stability

Ex.

* This is because the C=C is electron attracting and will be

stabilized by the electron releasing effect of the alkyl

substituents.

C C

H

CH3

H3C

H

C C

H

H

H3C

H3C

C C

CH3

H

H3C

H

C C

C2H5

H

H

H

Steric factors (steric effect):

Bulky substituents attached in a cis configuration to C=C

will repel each other as a result of being too Close. In

contrast, to same bulky groups will be far a part in the

trans configuration. Accordingly cis form is less stable

than the trans form

> >>

Cis isomer

C C

H

C

H

C

CH3

H3C

H3C

H3C

CH3

CH3

C C

C

H

H

C

CH3

H3C

H3C

H3C

CH3

CH3

trans isomer

Synthesis of alkenes via Elimination Reactions:There are three main reactions (Elimination reactions) for the

synthesis of alkenes.(1) Dehydrohalogenation of alkyl halides:

C C C C

H

H

H

X

H

H base

-HX H

H

H

H

alkyl halidealkene

α

ß

(2) Dehydration of alcohols:

C C C C

H

H

H

H

OH

HH, heat

H

H

H

H

-H2O

(3) Debromination of vic-Dibromides:

C C C C

Br

H

H H

Br

H

H

H

H

H

-ZnBr2

Zn, CH3CO2H

Dehydrohalogenation of alkyl halides:E2 ( Bimolecular elimination Reaction) Beta elimination

Reaction.

C C C C

H

X

B

slow

E2 + BH X+

B, the used base (basic catalyst) and may be sodium ethoxide in ethanol (C2H5-ONa / C2H5OH ).

C C

H

X

E2

B

transition state concerted process

Or pot. tert.butyloxide in tert. Butyl alcohol ((CH3)3COK /

(CH3)3COH); Or pot. hydroxide in ethanol ( alcoholic KOH, KOH / C2H5OH) .

* RX should be secondary or tertiary alkyl halide.* Higher temperature favored E2 reactions.

E2 Reactions: the orientation of the double bond in

the product; Zaitsev’s Rule:E2 occurs to give the most stable, highly substituted alkene.

H3CCH

CH3

BrC2H5ONa

C2H5OHH2C

HC CH3, 55 °

EX. 1

EX. 2

H3CC

CH3

BrC2H5ONa

C2H5OHH2C C CH3, 55

CH3

CH3

EX. 3 CH3(CH2)15CH2CH2Br(CH)3CONa

15, 40(CH)3COH( )

H3C CH

C

H CH2

Br

CH3

H(b)

(a)

2-bromo-2-methylbutane

B

B =C2H5ONa

(a) (b)

C C

H

BrH

H3C

H3C

CH3

C2H5O

δ

δ

More stable transition stateresembles a tri substitutedalkene

less stable transition stateresembles a di substitutedalkene

(b)(a)

HC C

H3C CH3

CH3

C CH2

H3C

C2H5

C C

H3C

H

C2H5

H

H

OC2H5

Br

An Exception to Zaitsev’s Rule:The use of bulky base such as pot. tert-butoxide / tert-butyl

alcohol favors the formation

minor

+C2H5ONa

(C2H5OH

Br

major

δß

ß

ß

ß

ß

minor

+C2H5ONa

(C2H5OH

major

Br

δ

ßδ

minor

+C2H5ONa

(C2H5OH

major

Br

δ

ß

ß

+C2H5ONa

(C2H5OHBr

ß

Explain the formation of 4-methylcyclohexene in the

dehydrobromination 4-methyl bromocyclohexane of less

substituted alkene.

+ (CH3)3COHH3C C

CH3

O

CH3

H3C CH2

C

CH3

Br

CH3

+

2-methyl-1-butene2-methyl-2-butene

Synthesis of alkenes via the dehydration of alcohols (E1):

H2Oheat

H (strong acid)

alcohol

C C

OH

+

H alkene

* Dehydration of alcohols is an elimination and is favored at high temp..

* Acids used may be Bronsted acids such as sulfuric acid and phosphoric acid. Also Lewis acids such as Al2O3 is

often used

* The Dehydration process is closely related to the structure of alcohols and temp. and the conc. of the used acid.

H2OH2SO4conc.

H

H

H

ethanol

C C

H

H

OH

+H

H

H

H

ethylene

180 C

Ex.1

85%

80%

84%

Ex.2

cyclohexene

H3PO4

165-170 C

OH

cyclohexanol

Ex.3 H2SO4H3C C CH2H3C C

CH3

OH

CH3

CH3

2-methylpropene85 C

tert-butyl alcohol

20%

* In some case 1ry and secondary (2ry) alcohols undergo rearrangements of their carbon skeleton during dehydration.

Ex.H3PO4

H3C

CH3

CH3

3,3-Dimethyl-2-butanol

C C

H

CH3

OH

+

H3C

H3C

H3C

H3C

2,3-Dimethyl-2-butene

80 CH2C C

CH3

C2H5

2,3-Dimethyl-1-butene

85%

80%20%

* Thus the relative ease of the dehydration of alcohols can be arranged in the following order.

C OH

tertR

R

R > >C OH

sec.H

R

R C OH

1ryH

H

R

i.e

Is the starting carbon skeleton and

Is the producing carbon skeleton.

C C C

C

C

C

C C C

C

C

C

Mechanism of alcohol Dehydration E1 (unimolecular

elimination) Reaction:Example the dehydration of tert-butyl alcohol

H2SO4C C CC OH

C

80 CC

C

C

Step 1:Protonation of alcohol ( alkyloxonium ion formation)

+C OHfast

H3C

CH3

H3C O

H

H

H +C O

H3C

CH3

H3C O

H

H

H

H

Protonated alcohol ( alkyloxonium ion formation)

Step 2: slow+C

H3C

CH3

H3C O

H

HC O

H3C

CH3

H3C

H

H

carbocation

C CH2

H3C

H3C

+fast

O

H

H

H+C

H3C

C

H3C O

H

H

H HH

(b) Dehydration of primary Alcohols(E2):

* 1ry carbocation is not stable enough to be formed.

* The acid- catalyzed dehydration of primary alcohols takes

place by the following mechanism

+H2C OH

fastH2C O

H

H

HH2C OH2C

H

HH H

H2O + C CslowH2

C OH2C

H

HH

H2OH3O

H

H

H

H

+ +

Step 1:

+H2C OH

fastH2C O

H

H

HH2C OH2C

H

HH H

Step 2:

H2O + C CslowH2

C OH2C

H

HH

H2OH3O

H

H

H

H

+ +

Molecular Rearrangements in alcohols Dehydration

(carbocation stability):

It has been found that dehydration of some alcohols afforded

alkenes which were different from the expected ones.

A rearrangement process was proposed to account for the

formation of the unexpected alkenes (methide shift)

Ex.

C CHH3C

CH3

CH3OH

CH3

3,3-dimethyl-2-butanol

- H+

CHCH3C

CH3

CH3

CH2 C CHH3C

CH3

CH3

CH3

C C

CH3

CH3

H3C

H3C

+ H2C C CH

CH3CH3

CH3

-H2OC CHH3C

HCH3

CH3OH

CH3

CHCH3C

CH3

CH3

CH3

Methyl shift

2ry carbocation

- H+

3ry carbocation

(unexpected)

1,2-hydride shift:

In which a hydride ion H- migrates to an adjacent positively

charged carbon.

C CH

H

C C

H

~

* Hydride ion shift usually takes place during the dehydration

of primary alcohols

Ex.

CH3(CH2)4CH2CH2CH2OHH

CH3(CH2)4CH2CH2CH2

- H+ ~H

1ry carbocation

- H+

2ry carbocation

(unexpected)

CH3(CH2)4CH2CH2CH2OHH

CH3(CH2)4CH2CH2CH2

CH3(CH2)4CH2CH=CH2 CH3(CH2)4CH2CHCH3

CH3(CH2)4CH2CH=CH2 +CH3(CH2)4CH=CHCH3

~

~

~

C CH3CCH3

CH3

CH3 H

CH3C CH3C

CH3 H

CH3

CH3

methamidemigration

2ry carbocation

The following examples illustrated the carbocation

rearrangements.

Ex.1

3ry carbocation

Ex.2

C CH3CH

H

CH3 H

CH3C CH3C

CH3 H

CH3

H

hydridemigration

Ex.3

C CH2H3CCH3

CH3

CH3

C CH3C

CH3 H

CH3

H

methamidemigration

Ex.4: rearrangements of carbocations can also lead to a

change in ring size as the following.

-H2O

HCH

CH3

CH3

OH

,heat

HC

H3C

CH3

-H

CH3

CH3H

CH3

CH3

3 ry carbocation

Synthesis of Alkenes by the debromination of vicinal dibromides:

C C

X X

C C

X

X

Vicinal dihalide(avic-dihalide)

geminal dihalide(agem-dihalide)

* vic- dibromides undergo debromination with NaI/acetone or with a mixture of Zn dust/CH3CO2H

+ C CNaI +C C

Br Br

acetone+I2 NaBr2

Zn/CH3CO2H

or (ethanol)C C + ZnBr2

* The debromination by NaI takes place via E2 mechanism.

+ C C +C C

Br

Br

acetone+IBr BrI

+IBr I +I2 Br

Additions to Alkenes:The addition reactions are the characteristic to the C=C bond

with the general type shown below.

C C C CA Baddition+ A B

Ex. C C C CH X

H-X

C C C CH OSO3HHOSO3H

C C C CH OHH-OH

H

C C C CX XX-X

H

alkyl halide

alkyl hydrogen sulfate

dihalide

alcohol

Two characteristics of the double bond addition reactions:(1) The conversion of one Π bond and one δ into two

δ bonds

C C + C CX YX Y

δ bondΠ bond2 δ bond

Bonds formedBonds broken

(2) The Π electrons are particularly susceptible to the electrophiles

C CΠ bond

Ex. HX react with alkenes by donating a proton to the Π bond

C C +C CH

Br

H

X

Π bond

alkene

or

+ C CH C CH

H

Helectrophile

nucleophile

C C

BrH

+C C Br

H

Addition of hydrogen halides to alkene: Markavikov’s ruleHCl, HBr and HI add to the double bond of alkenes.

+ C CHX C C

H

X

* The addition is carried out by dissolving HX in acetic acid or CH2Cl2 or bubbling the HX gas in the alkene and the

alkene itself as the solvent.

* The order of the reactivity of HX is HI > HBr > HCl > HF

+H2C CH2HX H2C CH2

H

X

Markovnikov’s rule:

In the addition of HX to an alkene, the hydrogen atom

adds to the carbon atom of the double bond that has

(already has) the greater number of hydrogen atoms.

Ex.2

+H3CHC CH2HBr CH CH3H3C BrCH2CH2CH3

Br

1-bromopropane2-bromopropane(main product)

C CH2HBr C CH3H3C

Br

tert-butyl bromideH3C

H3C

CH3

isobutylene

Ex.1 HC CH2

Br

HC CH3H3CH3C

H

Br

The mechanism of Markovnikov’s addition:

Step1.

HC

C+ X

C

CH X

Π complex step2

+C C C Cfast

X

H H

X

δ complex

Theoretical Explanation of Markovnikov’s rule:

H3CHC CH2 +

Br

1 ry carbocation(less stable)

slowH

b

a

X(b)

(a)

H3CH2C CH2

H3CHC CH3

H3CHC CH3

Br

2 ry carbocation(mor stable)

fast

H3CHC CH2 +

Br

BrCH2CH2CH31 ry carbocation(less stable)

H

b

a

(b)

(a)

H3CH2C CH2

H3CHC CH3 H3CHC CH3

Br2 ry carbocation(mor stable)

Br

C CH2 +

Br

CH3CHCH2Br1 ry carbocation(less stable)

H

b

a

X(b)

(a)

H3CHC CH2

H3CC CH3 H3CC CH3

Br3 ry carbocation(mor stable)

Br

H3C

H3C CH3

CH3

CH3

CH3

isobutyl bromide( not formed)

tert-butyl bromide( actual product)

Ex.2

Modern statement of Markovnikov’s rule:

In the ionic addition of unsymmetrical reagent to a double

bond, the positive portion of the adding reagent attaches

itself to a carbon atom of the double bond so as to yield

the more stable carbocation as an intermediate.

Example:

C CH2 +I-Cl CH3CCH2-IH3CC CH2I

3 ry carbocation(mor stable)

Cl

H3C

H3C CH3CH3

Cl2-methylpropene 2-chloro-1-iodo-2-methyl propene

CH3

IH

CH3

H

+ ClI

Cl

Addition of sulfuric acid to alkenes:

+C C + C COH

O

O

HO S

H

OH

O

O

O S

C C

HO

O

O

HO S

Example:

C C

H

H

H

H3C

X

+ C CH2

isopropylhydrogen sulfate( product of the reaction)

OH

O

O

HO S H

HCH3

CH CH3O

O

O

HO S

CH3

CH CH2H3C

H

CH2CH2CH3O

O

O

HO S

OH

O

O

O S

OHO

O

O S

propylhydrogen sulfate( not formed)

Alcohols from alkyl hydrogen sulfates:

cold H2SO4CH2C

H

H3C CH3HC

HO3SO

H3C CH3HC

OH

H3CH2O

heat

Addition of water to alkenes: Acid catalyzed hydration.

General equation:

Ex.2H3O

CH2C

25 C+

H3C

H3C CH3C

OH

H3CH2O

CH3

Ex.1H3PO4CH2H2C300C

+ CH2OHH3CH2O

H3OCC + CC

OH

H2O

H

The mechanism of the hydration of 2-methylpropene.

Step 1.

Step 2.

+slow

H O

H

H

H2OCH3C

CH3

CH2

+CH3C

CH3

CH3

Step 3.

+fast

H

H

O

H

H

CH3C

CH3

CH3

CH3C

CH3

CH3

O

+fast

HO

H

H

+CH3C

CH3

CH3

OH

H

CH3C

CH3

CH3

O O

H

H

H

The reaction produces tert-butyl alcohol because step 1

leads to the formation of the more stable tert-butyl cation

rather than the much less stable isobutyl cation.

+slow

H O

H

H

H2OCH3C

CH3

CH2

+CH3C

CH3

CH3

for all practical purposes this reaction does not take

place because it 1ry carbocation.

The rearrangements associated with alkenes

hydrations.H2SO4 C CHCH2C

CH3

CH3

H3C

CH3

H2OH

C

CH3

H3C

CH3 HO3,3-dimethyl-1-butene 2,3-dimethyl-2-butanol

The mechanism for the above rearrangements:

H3O CHCCH2C CH3H3C

CH3

H

C

CH3

H3C

CH3 H3C

~CHC CH3H3C

CH3

H2OCH3

H3C

C CH CH3H3C

CH3 CH3

C CH CH3H3C

CH3 CH3

OH H

OC CH CH3H3C

CH3 CH3

OH H

H

H

+ OC CH CH3H3C

CH3 CH3

OH

H

H

+ H

Addition of Bromine and Chlorine to Alkenes: The reaction of

alkenes with Br2/CCl4 is a useful test for carbon-carbon

multiple bonds. Alkenes react rapidly with bromine at room

temp. and in the absence of light. If we add Br2 to C=C, the

red-bromine disappears.

Thus RT

C CCC +BrBr

Br2in the dark, CCl4

Alkene

(colorless) red-brown Vic-dibromide

(colorless)

Examples

-C C

HC

HC +

Cl

H

Cl

H

Cl2CH3H3C CH3H3C91)

2)

3)

C°

-9 C° C CCH2

HC +

Cl

H

Cl

H

Cl2C2H5 HC2H5

Trans-1,2-dibromo cyclohexane

CCl4/C2H5OH+ Br2

Br

H

Br

H

- 5C°

(as a racemic form)

+Br

C

C+ Br

C

CBr Br C

CBrBr

Π complex δ complex

SN2

Vic-di bromide

Step 2

Step 1

Mechanism of halogen Addition:

Π complex formation and δ complex.

+C

CBr Br

C

C

Br

Br

Stereo chemistry of halogen Addition to alkenes:

The addition of Br2 to alkenes is anti addition. The

product is trans in isomer in the cases of cyclic bromonium

cation and the anti addition of Br- to that intermediate via SN2

mechanism could be illustrated by the following reaction .

Ex.Br2/CCl4

H

Br

Br

H

cyclopentene

Trans1,2-dibromocyclopentane

Mechanism of the reaction:

Br+ Br Br Br

Π complex

+ BrBr

Br

H

H

Br

δ complex

(bromonium ion) Trans1,2-dibromocyclopentane

Halohydrin formation:

CC + X2+H2O C C

X HO

C C

X X

+

alkene halhydrin Vic-dihalide

The mechanism of Halohydrin formation:

XC

C+ X

C

CX XStep 1

Π complex

+

C

CX X C

CXX

Step 2

δ complex

(bromonium ion)

Step 3

H2O+C

C

HX C C

OH2

X

-C C

OH

X

Example

H2O

- H

CH2C

H3C

H3CBr2

CH2CH3C

CH3

Br

CH2CH3C

CH3

BrH2O

CH2CH3C

CH3

BrOH

Epoxidation of alkenes:Epoxides are cyclic ethers with three membered rings.

C C

OAn epoxide

Ex.

IUPAC name : Oxirane

common name : ethylene oxide

H2C CH2

O

Synthesis of (epoxidation):

CH

CH

R"COOOH ++CH

CH

R

R'

O

R'

R

R"COOH

* The used peracids are peracetic and perbenzoic acids, the mechanism of epoxidation is as shown

C

C

++C

CC

O R'

OO

O

H

CO R'

O

H

* The epoxidation reaction is a syn addition.Ex.1

+ RCOOOH

cis2-butene

CC

H

CH3H3C

H

CC

O

CH3H3C

HH

Ex.2

CHCl3

C6H5COOOH

cyclohexene

H

H

O

cyclohexene oxide

Acid-catalyzed hydrolysis of epoxide:Ex.

H2O+H3O CH2H2C

O

H

CH2H2C

O

- HCH2OHHOH2CCH2OHH2OH2C

1,2-ethanediol(ethyleneglycol)

Anti hydroxylation of alkenes:Ex.

RCOOOH

cyclopentene

H

H

O

cyclopentene oxide

+ RCOOH

Acid catalyzed hydrolysis of cyclopentene oxide yields a trans-diol, trans-1,2-cyclopentanediol.

H3OH

H

OH

H

H

O

H2OOH2

H

OH

H

OH2

H

OH

H - H

OH

H

OH

H

Alcohols from alkenes through Oxymercuration- Demercuration:

+

C C +HHO

CC Hg(OOCCH3)2+H2Ooxymercuration

THF

CC

HgOOCCH3OH

+ CH3COOH

OHNaBH4demercuration

Hg+ CH3COO

* Oxymercuration- Demercuration is highly regioselectiv.

The net orientation of water H- and –OH is in accordance

with Markovnikov’s rule.

General example:

CC

HHHg(OAc)2/THF

R H(1)

NaBH4/(2) OHCC

HH

R H

HOHSpecific example:

Ex.1

+

CH2H3C(H2C)2HCHg(OAc)2 CH2H3C(H2C)2HC

HgOOCCH3OHNaBH4/Hg

OH

THF-H2O

CH3H3C(H2C)2HC

HO

Ex.2

+Hg(OAc)2 NaBH4/ Hg

OHTHF-H2O

CH3 H3C OHHgOAc

H

H3C OHH

H

Alcohols from Alkenes Through Hydroboration-Oxidation:

The addition of the elements of water [ H,OH ] to C=C through

the use of diborane (B2H6) or THF:BH3.

The addition of diborane to C=C is hydroboration and the

second is oxidation and hydrolysis of organoborn

intermediate.

The method gave the alcohols ( Anti- Markovnikov addition of

water) which can not prepared through the acid-catalyzed

hydration of alkenes or oxymercuration-demercuration

Example: CH2HCH3C

propeneH2O / H3O+

3 mole

THF : BH3 hydroboration

2-propanol

CH3

HC

HO

H3C

H

CH2C

H

H3C

H

BH2C

CH2C

H

H3C

C

H

CH3

HHO

HO

HOHO HO HO

3 mole H2O2 / OH-3 CH3CH2CH2OH

1- propanol

1-hexanol

CH2H3CH2CH2CH2CHC(2) H2O2,OH -(1) THF:BH3 CH3(CH2)4CH2OHEx.2

Ex.33-methyl-2-butanol

HCC

(2) H2O2,OH -(1) THF:BH3CH3

CH3

H3C CHCH CH3

CH3

H3C

OH

Ex.4

trans-2-methylcyclopentanol

(2) H2O2,OH -

CH3 CH3

H

H

OH

(1) THF:BH3

The stereo chemistry of the oxidation of organoborans:

The oxidation step in the hydroboration-oxidantion synthesis

of alcohols takes place with retention of configuration, the

hydroxyl replaces the boron atom where it stands in the

organoboron compound the net result of the two steps is

the syn addition of –H and –OH.

Ex.

trans-2-methylcyclopentanol

syn addition+

H2O2,OH

CH3 CH3

H

H

OHBH3

CH3

H

H

B

Radical addition to alkenes:

The Anti Markovnikov addition of HBr:

When alkenes that contained peroxides or hydroperoxides

reacted with HBr, Anti-Markovnikov addition of HBr.

An organo peroxide

O OR R

An organic hydrogen peroxide

O OR H

CH2HC

propeneH3C

O OR R

HBrAbsence of

peroxide

CH3CHH3C

Br

2-bromopropane

CH2BrH2C

1-bromopropaneH3C

Ex.1 CH2BrH2C

1-bromopropane

O OR+ HBr H3C

RCH2

HC

propene

H3C

The mechanism of anti Markovnikov’s addition:

( HF, HCl, HI do not give anti Markovnikov’s addition) in the

presence of ROOR only HBr gives anti Markovinkov’s

addition, the mechanism is a readical chain reaction

initiated by peroxides: and involves the following steps.

Chain Initiation

Step 1

Step 2

O OR Rheat

2RO

+ HBrRO ROH +Br

Chain propagation

Step 3

Step 4

CH2BrHC+Br H3CCH2CHH3C

CH2BrH2CHBr H3C + BrCH2Br

HCH3C +

Reaction termination:

CH2BrHCH3C

CH2BrHCH3C2

CH

H3C CH2Br

Br2+ BrBr

Oxidations of alkenes: Syn Hydroxylation

Alkenes undergo the oxidation of C=C bond with KMnO4 or

OsO4 to give 1,2-diols (glycols)

Examples

(1)

(2)

1,2-ethylanediol

H2C CH2 H2C CH2KMnO4

OH -

OHOH

1,2-propanediol

HCH3C CH2

HC CH2

OsO4

Na2SO3

OHOH

H3C

Syn Hydroxylation of alkenes:

The mechanism for the formation of glycols by KMnO4 and

OsO4 involve the formation of cyclic intermediates followed

by the cleavage at the oxygen-metal bond to give the syn

hydroxylation product.

H2C CH2

OHHO

OH

H2O

Mn

+MnO2

H2C CH2 H2C CH2

OO

O

O

O

O

Mn

OO

H2C CH2

OHHO

H2O

Os

+OsO2

H2C CH2 H2C CH2

Na2SO3

OO

O

O

O

O

Os

OO

The Syn Hydroxylation can be seen by when cyclopentene

reacts with KMnO4 / OH- or OsO4 followed by treatment

with NaHSO3 or Na2SO3. The product in either case is cis-

1,2-cyclopentanediol.

OHH2O

Mn

+MnO2

OO

O

O

O

O

Mn

OO

OH OH

H2O

Os

+OsO2

Na2SO3

OO

O

O

O

O

Os

OO

OH OH

Oxidative cleavage of alkenes:

Hot KMnO4 oxidize alkene to pot. Salts of carboxylic acids.

Ex.H3O

OH2H3CHC CHCH3

MnO4H3C C O-K+

O

2H3C C OH

O

The terminal CH2 group of 1-alkene is oxidized to CO2 and

H2O by hot KMnO4. A disubstituted carbon atom of a

double bond becomes the C=O group of a ketone

Example:

H3O

OHH2O+ CO2C CH2

MnO4 /H3CH2C C CH3

O

H3CH2C

CH3

+

The oxidation cleavage of alkenes has been used to prove the

location of the double bond the alkenes chain or ring (via

retrosynthetic analysis)

Example:

H3O

OH

propanoic acid+

MnO4 /C8H16 H3CH2C C OH

O

HO C CH2CH2CH2CH3

O

pentanoic acid

+H3CH2C C OH

O

HO C CH2CH2CH2CH3

O

C C

H3CH2CHC CH(CH2)3CH3

Ozonolysis of alkenes:

A more widely used method for locoing of the double bond of

an alkene involves the use of O3 (ozone)

( either cis or trans 3-octene)

ozonide

CC

O

OO

C

O O

CO

O

CC

OO

initial ozonide

Ozone react with alkenes to form the unstable initial ozonide

which rearrange to ozonide. The mechanism of this

rearrangement can be depicted as follow

Ozones are very unstable and can not isolated but are

reduced with Zinc and water. The reduction products are

carbonyl compounds

aldehydes and / or ketones

C O + Zn(OH)2Zn / H2O

O C

ozonide

C

O O

CO

+

aldehydes and / or ketones

C O + Zn(OH)2Zn / H2O

O3O C +C C

ozonide

C

O O

CO

C

O

+

O

CC

OO

C

OO

The overall process of ozonolysis followed by reduction with

Zinc and water accounts to a disconnection of the carbon-

carbon double bond in the following fashion

C O + Zn(OH)2Zn / H2O

O3

R'

R

O C

H

R"

+C C

H

R"

R'

R

A –H attached to the double bond is not oxidized to –OH as it

is with KMnO4

Example:

C O + Zn(OH)2Zn / H2O

O3

H3C

H3C

O C

H

CH3

+C C

H

CH3

H3C

H3C

AlkynesThey are hydrocarbons containing -C≡C- (carbon-carbon

triple bonds).

* Noncyclic alkynes have the general formula CnH2n-2

* We called R-C≡CH monosubstituted or terminal alkynes,

R-C≡CR’ are called to have internal triple bonds

* R-C≡CH have acidic proton H-C≡C-H acetylene (ethylene) is

the simplest alkyne

Sources of Alkynes:

(1) From lime stone and Coke

+CaO 3C +CaC2 CO1800-2100

2-

+Ca2+ 2H2OC

C+Ca(OH)2

AcetyleneC CH H

(1) By thermal dehydrogenation of ethylene:

+ H2Acetylene

heatC CH HH2C CH2

Nomenclature of alkynes:

IUPAC rules for hydrocarbons were followed and alkynes

take the suffix yne

Ex. H-C≡C-H (ethyne), CH3C≡CH (propyne), CH3CH2C≡CH

(1-butyne), CH3C≡CCH3 (2-butyne), (CH3)3CC≡CCH3 (4,4-

dimethyl-2-pentyne).

* -C≡CH is named as ethynyl if it is a substituent

* We can use the older nomenclature as follow

HC≡CH (acetylene), CH3C≡CH ( Methyl acetylene),

Physical properties:

* They are resemble alkanes and alkenes in their physical

properties such as low density and low water solubility.

* They are non polar and dissolve in organic solvents such

as alkanes

Acidity of acetylene and terminal alkynes:Alkynes are more acidity than alkenes and alkanes

( Acetylene and terminal alkynes are stronger than other

hydrocarbons)

HC≡CH > H2C=CH2 > CH3CH3

pKa 26 45 62

Ka 10-26 10-45 10-62

* -OH ion is too weak base to convert acetylene to it’s anion,

The position of equilibrium in this equation lies to the left.

OH+ + H2O

weaker acid

HC CH HC C

weaker base stronger base stronger acid

Amide ion is a much stronger base than acetylide ion and

converts acetylene to it’s conjugate base quantitavely.

NH2+ + NH3

weaker acid

acetyleneHC CH HC C

weaker base stronger base stronger acidamide ion acetylide ion ammonia

Ka = 10-26

pKa = 26

Ka = 10-36

pKa = 36

Preparation of Alkynes by the alkylation of acetylene and

terminal alkynes:

Alkynes are prepared by combining smaller structural units

to build longer carbon chains.

e.g.

HC CH RC CH RC CR

NaNH2+ + NH3

acetyleneHC CH HC CNa

sodium amide sodium acetylide ammonia

+ NH3HC CNa1-bromobutane

1-hexyne

Br C CH

Dialkylation of acetylene can be achieved by carrying out the

sequence twice

NaNH2+ /NH3HC CH

CH3Br

Br C CHNaNH2/NH3

C C

Dehalogenation of dihalides:

Alkenes can be converted to alkyne e.g. general equation

X2+CH

HC

-2HX

NaNH2/NH3

R R'HC CHR R'

X XC CR R'

Br2+H2O

NaNH2 /NH3

Br

Br

The geminal dihalides and vicinal dihalides can be

dehyrohalogenation two reaction steps.

General equations:

orC R'RH2C Base

strong

C CR R'

X

X XX

CH

CR R'

X

C CR R'Base

The second dehydrohalogenation step is more difficult than

that the first so that rather strongly basic conditions or

high temperature are needed to convert dihalides to

alkynes

H2O

KOH1-propanol

NaNH2 /NH3

Cl

HCl

H

HCl

Chemical properties of alkynes:(1)Hydrogenation of alkynes:

(a) Catalytic hydrogenation

H2+Pt, Pd

R'C CR R'Ni or Rh

C C

H

R'

H

RH2/cat R

+ Ni2H2

(b) Metal-Ammonia Reduction of Alkynes:

It gives trans or E alkenes

Na / NH3

H

H

Addition of hydrogen halides to alkynes:

The addition of HX to alkynes is regioselective and follows

Markovinkov’s rule.

Ex.

+HBr

Br

The reaction mechanism is.

Step 1

+ +H Br Br

Step 2 is bromide ion captures the alkenyl cation

Br

+ Br

The carbocation formed by addition of a proton to an alkyne

is an alkenyl cation. The positively charged carbon of an

alkenyl cation is SP hybridized which is more

electronegative than its SP2 hybridization count part thus

alkenyl is less stable than alkyl carbocation

RCH2CH+R’ is more sable than RCH=C+R’

Alkyl cation alkenyl

Positive charge is positive charge is

On SP2hybridized carbon on SP hybridized carbon

* In the presence of excess HX, geminal dihalides are formed

Ex.

HXR'C CR R' C C

X

R'

H

RHX R

X

X

2HF

3-hexyne

F

F

* Free-radical addition of HBr to alkynes is a regioselectivity

oppositey to Markovinkov’s rule is observed

Ex.

+peroxideH Br

Br

Hydration of alkynes:

Addition of the elements of water to the triple bond ( HO- &H+)

to yield an enol which converted to the most stable form,

the keto form.

(not isolated)

H2OR'C CR R' C C

OH

R'

H

Rfast R

O

keto form

enol

H2O+H

H3C HC CH H C C

OH

H

H

Hfast

O

acetaldhydeHg+2

H2O+H2SO4

HgSO4

O

2-hexanone

Addition of halogens to alkynes:

+C CR R' C C

X

R'

X

R2X2

tetrahaloalkane

X X

+C CHH3C C CHCl2

Cl

H3C2Cl2

1,1,2,2-tetrachloro propane

Cl

A dihaloalkene is an intermediate and is the isolated product

when the alkyne and the halogen are present in equimolar

amounts. The stereochemistry of addition is anti.

Br2

Br

3-hexyne

Br(E) 3,4-dibromo-3-hexene

Ozonolysis of Alkynes:

Carboxylic acids are produced when alkynes are subjected to

ozonolysis.

Ozonolysis is some times used as a tool in structure

determination by edifying the carboxylic acid .

O3 HO C R'

O

+H2O

C CR C OHR

O

R'

H2O +O3

HO C H

O

formic acidCOOH

1-hexyne pentanoic acid