312 112 Basic Organic Chemistry312 112 Basic Organic Chemistry

Atomic Structures Chemical Bondings

Hybridizations Acids-Bases

Hydrocarbons (Alkanes, Alkene, Alkynes)

Alkylhalides

Instructor: Chanokbhorn Phaosiri

e-mail: chapha@kku.ac.th

office: Sc4502-3

(Pharmacy English Program)(Pharmacy English Program)

Organic ChemistryOrganic ChemistryThe study of the compounds of carbonOver 10 million organic compounds have

been identifiedabout 1000 new ones are discovered or

synthesized and identified each day!C is a small atom it forms single, double, and triple bondsit is intermediate in electronegativity (2.5)it forms strong bonds with C, H, O, N, and

some metals

HOH H

OCH3

H

Estrone(Female Steroidal Hormone)

O

H3CO

H3CO

N CH3

Morphine(Analgesic)

OCH3

H3C

HOCH3

H3C

3

Vitamin E

OCH2 CHN

O

N

S

OCO2H

Penicillin V

COOH

O CH3

O

2-Acetyloxybenzoic acid(Aspirin, Bayer Aspirin)

OH

HNC

O

CH3

N-(4-Hydroxyphenyl)acetamide(Acetaminophen, Tylenol)

H3C

O

OH

2-[4-(2-Methylpropyl)-phenyl]propanoic acid

(Ibuprofen)

The Painkillers

1s22s22px12py

12pz1N(7)

1s22s22px12py

1C(6)

1s22s22px1B(5)

Electron Configurations

To write the electron configuration for Cl -

but Cl- has one more electron

Cl-

1s22s22p63s23px23py

23pz1

1s22s22p63s23px23py

23pz2

Cl (17)

To write the electronic structure for Na+

1s22s22p63s1Na (11)

Na+ 1s22s22p6

but Na+ has one less electron

Classification of Chemical Bonds

Difference in Electronegativity Between Bonded Atoms Type of Bond

less than 0.50.5 to 1.9

greater than 1.9

nonpolar covalentpolar covalent

ionic

Electronegativity (EN)

ElectronegativityElectronegativity: a measure of the force of an

atom’s attraction for the electrons it shares in a chemical bond with another atom

Pauling scale increases from left to right within a period

increases from bottom to top in a group

H Cl2.1 3.0

+ -A Polar Covalent Bond

Na F

N H

C Mg

C Cl

N O

C H

Lewis Structures

H ydrogen chloride

M ethane Ammonia

Water

H O

H

H

H NH C

H

H

H Cl

HH

H 2O (8)

N H 3 (8)CH 4 (8)

H Cl (8)

Lewis Structures

Carbonic acidMethanal

Acetylene

Ethylene

HC C

H

C C HH

HO

CC

H OH

OH

HH

O

C2H4 (12)

C2H2 (10)

CH2O (12) H2CO3 (24)

Draw Lewis Structures, showing all valence electrons, for these molecules

H2O2

CH3OH

CH3Cl

C2H2

(CH3CH2)2CO

CH3CN

CH3COOH

CH2NNH2

Draw Lewis Structures, showing by charges which bonds are ionic and by lines which bond are covalent.

NaOH

KHCO3

CH3ONa

CH3COONa

NH4Cl

Formal Charge# of valence electrons in

unbonded atom

all unsharedelectrons

one half of all shared

electrons+Formal

charge=

H

H N HNitrogen valences electrons = 5

Nitrogen nonbonding electrons = 2

Nitrogen bonding electrons = 6

Formal Charge = 5- 2-(6/2) = 0

Nitrogen valences electrons = 5

Nitrogen nonbonding electrons = 0

Nitrogen bonding electrons = 8

Formal Charge = 5- 0-(8/2) = +1

NH4 ?

if the number assigned to the bonded atom is less than that

assigned to the unbonded atom, the atom has a positive

formal charge.

if the number is greater, the atom has a negative formal

charge.

NH4+ CH3 NH3

+

HCO3- CO3

2-OH-

CH3 CO2-CH3

-

CH3 OH2+

BF4-

Assign formal charge in each structure as appropriate.

HOCO2

CH3CH2O

CH3CH2

NH3CH2CO2

Molecular Shapes

4 regions of e - density(tetrahedral, 109.5°)

••••

•• ••

••

••

••

••

HH

NH

C

H H H

H

HC C

H

O C

H

C

HH

OH

CH C HO

3 regions of e - density(trigonal planar, 120°)

2 regions of e - density(linear, 180°)

••

HH

O

Polar & Nonpolar Molecules

Bond Dipole Moments

H3C CH3 H3C NH2

H3C OH H3C Cl

C OO

OHH

Molecular Dipole Moments

Which of these molecules are polar? Indicate the direction of their polarity.

1. H2O

2. NH3

3. CH2Cl2

4. CH2=CH2

5. CH2=CHCl

6. CH3C N

Molecular Representations

(Line-Angle Formula)

Provide line-angle formula for these compounds.

2

2

Shapes of Atomic Orbitals All s orbitals have the

shape of a sphere, with its

center at the nucleus. Of

the s orbitals, a 1s is the

smallest, a 2s is larger,

and a 3s is larger still.

x

y

z

a pz orbital

xy

z

an s orbital

A p orbital consists of two

lobes arranged in a straight

line with the center at the

nucleus.

Sigma Bonding

Electron density lies between the nuclei.

A bond may be formed by s-s, p-p, s-p, or hybridized orbital overlaps.

The bonding Molecular Orbital (MO) is lower in energy than the original atomic orbitals.

The antibonding MO is higher in energy than the atomic orbitals.

H2: s-s overlap

Cl2: p-p overlap

Constructive overlap along the same axis forms a sigma bond.

s + p

s

px

node

Pi Bonding Pi bonds form after sigma bonds.

Sideways overlap of parallel p orbitals.

bonding MO

If all bonding occurred between s- and p-orbitals, then all bond angles would be 90o.

We know that isn’t true!

Hybridization

Linus Pauling

Atominc orbitals on the same atom combine to

form hybrid atomic orbitals. Why?

Hybrid orbitals are more directional, so they

have more effective bonding interctions.

2p

2s

Ground electron state of carbon

Higher-energy electronic state of carbon

sp3 hybrid state of carbon

2p

2s

2sp3

SS PP

+

4sp3

Conservation of orbitals

Tetrahedron

sp3 Hybridization

sp3 Hybrid Orbitals

each sp3 hybrid orbital has two lobes of

unequal size.

the four sp3 hybrid orbitals are directed

toward the corners of a regular

tetrahedron at angles of 109.5°.

sp3 CH4 (4 sigma bonds)Tetrahedron Bond angle 109.5o

bonds (H) - sp3 (C)

sp3 carbon sp3 carbon

+

sp3- sp3 sigma bond

6(s- sp3) sigma bonds

2p

2s

2p

2s2sp2

2p

sp2 Hybridization

3sp2

Trigonal Planar

each sp2 hybrid orbital has two lobes

of unequal size.

the three sp2 hybrid orbitals are

directed toward the corners of an

equilateral triangle at angles of 120°.

the unhybridized 2p orbital is

perpendicular to the plane of the sp2

hybrid orbitals .

sp2 hybrid orbitals

unhybridized2p orbital

sp2 Hybrid Orbitals

+

+ 4s of H

sp2- sp2 sigma bond

4(s- sp2) sigma bonds

sp2 carbons

1 + 1

p orbitals that remain on carbons overlap to form bond

C(2p)-C(2p) bond

H

H

H

HCC

H

H

H

HCC

2p

2s

2p

2s

2sp

2p

sp Hybridization

sp Hybrid Orbitals

– each sp hybrid orbital has two

lobes of unequal size.

– the two sp hybrid orbitals lie in a

line at an angle of 180°(linear).

– the two unhybridized 2p orbitals

are perpendicular to each other

and to the line through the two sp

hybrid orbitals.

x

yz

sp hybridorbitals

unhybridized 2p orbitals lie on the y and z axes

sp carbons

+

+ 2s of H

1 + 2

Multiple Bonds

• A double bond (2 pairs of shared electrons) consists of a sigma bond and a pi bond.

• A triple bond (3 pairs of shared electrons) consists of a sigma bond and two pi bonds.

• We study three types of hybrid atomic orbitals

spsp33 (1 s orbital + 3 p orbitals -> four sp3 orbitals)

spsp22 (1 s orbital + 2 p orbitals -> three sp2 orbitals)

spsp (1 s orbital + 1 p orbital -> two orbitals)

• Overlap of hybrid atomic orbitals can form two types of

bonds, depending on the geometry of the overlap

s bonds s bonds are formed by “direct” overlap

p bonds p bonds are formed by “parallel” overlap

Hybrid OrbitalsHybrid-ization

Types of Bonds to Carbon Example

sp3 four sigma bonds

sp2 three sigma bondsand one pi bond

sp two sigma bondsand two pi bonds

H H

H H

H- C C- H

H

C C

H

H H

H- C- C- H Ethane

Ethylene

Acetylene

Name

State the hybridization of each carbon atom.

3 Sigma bonds and one lone pair

:NH3

N

H HH

sp3 hybrid nitrogen

2p

2s

2p

2s

2sp3

2 sigma bonds and 2 lone pair

O

HH

sp3 hybrid oxygen H2O

2s 2s

2p 2p2sp3

sp2 hybrid oxygen

trigonal planar : 120o bond angles

2p

2s

2p

2s2sp2

2p

sp hybrid nitrogen

2p

2s

2p

2s

2sp

2p

Functional Groups

Functional groupFunctional group: an atom or group of atoms within a molecule that shows a characteristic set of physical and chemical properties.

Functional groups are important for three reason; they are

the units by which we divide organic compounds into classes.

the sites of characteristic chemical reactions.

the basis for naming organic compounds.

Hydrocarbons

Alkane/ Cycloalkane: single bonds.

Alkene/ Cycloalkene: double bonds.

Alkyne: triple bonds.

Aromatic: contains a benzene ring.

CH3CH2 C

O

H

CH3 C

O

CH3

H3C OH R OH Alcohols

RCHO Aldehydes

RCOR Ketones

Compounds Containing Oxygen

H3C O CH2CH3 R O R Ethers

Carboxylic Acid: RCOOH

Compounds Containing Oxygen

or or

O

CH3 - C-O- H CH3 COOH CH3 CO2 H

Acid Chloride: RCOCl Ester: RCOOR' Amide: RCONH2

C

O

OH

C

O

ClC

O

OCH3C

O

NH2

Compounds Containing Nitrogen

Amines: RNH2, RNHR', or R3N

Amides: RCONH2, RCONHR, RCONR2

Nitrile: RCN

N

O

CH3 CH3 C N

HOH H

OCH3

H

Estrone(Female Steroidal Hormone)

O

H3CO

H3CO

N CH3

Morphine(Analgesic)

OCH3

H3C

HOCH3

H3C

3

Vitamin E

OCH2 CHN

O

N

S

OCO2H

Penicillin V

Chemical Reactions

What? Change in electron configuration.Bonds broken/ Bonds formed.

Why? Attain a stable state.

How? Collisions between atoms, molecules, ions.

1) Sufficient kinetic energy.2) Proper orientation.

Energy

Reaction Time

A + B

(Reactants)

(Intermediate)

[A-B]# (Transition State)

C(Product)

Classifications of Reactions

addition

elimination

substitution

2

rearrangement

oxidation

reduction

Reaction Mechanism

Overall description of how a reaction occurs.

Which bonds are broken and formed.

Free Radical : Movement of one electron,

usually proceeds by chain reaction.

Polar : Movement of electron pairs from areas of high

electron density to areas of low electron density.

Ways to Form and Break Covalent Bonds.

1. Symmetrical (one electron goes to each atom in the bond).

Homolytic cleavage (Radical Mechanism)

2. Unsymmetrical (both electrons go to one atom in the bond).

Heterolytic cleavage (Polar Mechanism)

Acids and BasesAcids and Bases

Lewis Acids and Bases

Lewis base(donates an

electron pair)

Lewis acid(accepts an

electron pair)

-+ O-H H- OH

+ H- O-H

H

H O H

C H 3 - C +H

C H 3

+ C l - C H 3 - C - C lH

C H 3A carbocation(a Lewis acid)

Chloride ion(a Lewis base)

Lewis Acid(Accepts electron pair)

If H+ is doing the accepting

acid

If atom other than H+ is doing the accepting

electrophile

Lewis Base(Donates electron pair)

If donate electron pairto H+

base

If donates electron pair to atom other than H+

nucleophile

Complete the following Lewis acid-base reactions.

3 3

3 3

3

3

AlkanesAlkanesandand

CycloalkanesCycloalkanes

StructureHydrocarbonHydrocarbon: a compound composed only of carbon and

hydrogen.

Saturated hydrocarbonSaturated hydrocarbon: a hydrocarbon containing only

single bonds.

AlkaneAlkane: a saturated hydrocarbon whose carbons are

arranged in a chain.

Aliphatic hydrocarbonAliphatic hydrocarbon: another name for an alkane.

Nomenclature

• IUPAC system

International Union of Pure and Applied Chemistry

(IUPAC)

Prefix tells the number of carbon atoms

Suffix -aneane specifies an alkane

• Common names

undec-dodec-

tetradec-pentadec-hexadec-heptadec-

nonadec-eicos-

tridec-

11121314151617

octadec- 181920

Prefixmeth-eth-prop-but-pent-hex-

oct-non-dec-

1234567hept-89

10

Carbons

CarbonsPrefix

Alkanes have the general formula CnH2n+2

Condensed Structural Formula

Molecular FormulaName

heptane

hexane

pentanebutane

propaneethanemethane CH4 CH4

C2 H6 CH3 CH3C3 H8 CH3 CH2 CH3C4 H10 CH3 (CH 2 ) 2 CH3C5 H12 CH3 (CH 2 ) 3 CH3C6 H14 CH3 (CH 2 ) 4 CH3C7 H16 CH3 (CH 2 ) 5 CH3

Nomenclature1. The general name of an open-chain saturated

hydrocarbon is alkane.

2. For a branched-chain hydrocarbon, the alkane

corresponding to the longest chain is taken as the

parent chain and its name is the root name.

3. Groups attached to the parent chain are called

substituents.

Alkyl groups (R-)

isopropyl

propyl

ethyl

methyl

CondensedStructural FormulaName

CH3

- CH2 CH3

- CH3

- CH2 CH2 CH3

- CHCH3

4. If there is one substituent, number from the end of the

chain that gives it the lower number.

CH3CH2CH2CH2CHCH2CH2CH3

CH3

123456784-methyloctane

2-methylpentane

4-ethyloctane

4-propyloctane

n-octane

5-ethyl-3-methyloctane

2,2-dimethyl-4-methylpentane

6-ethyl-3,4-dimethyloctane

CH3CH2CHCH2CHCH3

CH3 CH3

2,4-dimethylhexane

CH3CH2CH2C

CH3

CH3

CCH2CH3

CH3

CH3

3,3,4,4-tetramethylheptane

CH3CH2CHCH2CH2CHCHCH2CH2CH3

CH2CH3

CH2CH3 CH2CH3

CH3

3,3,6-triethyl-7-methyldecane

5. Certain common nomenclatures are used in the IUPAC system.

CH3CH2CH2CHCH2CH2CH2CH3

CHCH3

CH3

4-isopropyloctane

Common Names Isobutane, “isomer of butane”

CH3CH2CH2CH2CHCH2CH2CH2CH2CH3

CH2CHCH3

CH3

5-isobutyldecaneor

5-(2-methylpropyl)decane

CycloalkanesGeneral formula CnH2n

Ring sizes from 3 to 30 and more are known

five- and six-membered rings are the most common

Line-angle drawingseach line represents a C-C bondeach angle represents a C

CC C

CC

CC

CH2 C

H2 C CCH

C

CH

H2

H2 CH 3

CH 3

Nomenclature of Cycloalkanes

1. No number is needed for a single substituent on a ring.

CH3

methylcyclopentane

CH2CH3

ethylcyclohexane

CH2CH2CH2CH2CH3

1-cyclobutylpentane

2. Name the two substituents in alphabetical order.

H3CCH2CH2CH3

1-methyl-2-propylcyclopentane

H3CH2C

CH3

1-ethyl-3-methylcyclopentane

CH3

CH3

1,3-dimethylcyclohexane

3. If there are more than two substituents,

CH3CH2CH2

H3C CH2CH3

4-ethyl-2-methyl-1-propylcyclohexanenot

1-ethyl-3-methyl-4-propylcyclohexanebecause2<3

not 5-ethyl-1-methyl-2-propylcyclohexane

because 4<5

CH3

CH3

CH3

1,1,2-trimethylcyclopentanenot

1,2,2-trimethylcyclopentanebecause1<2

not1,1,5-trimethylcyclopentane

because 2<5

Physical Properties

Low-molecular-weight alkanes (methane....butane)

are gases at room temperature

Higher-molecular weight alkanes (pentane, decane,

gasoline, kerosene) are liquids at room temperature

High-molecular weight alkanes (paraffin wax) are

semisolids or solids at room temperature

Physical Properties

Solubility: hydrophobic

Density: less than 1 g/mL

Boiling points increase with increasing

carbons (little less for branched chains).

Melting points increase with increasing carbons.

Sources of alkanesNatural gas 90-95% methane

Petroleum

C1-C2: gases (natural gas)

C3-C4: liquified petroleum (LPG)

C5-C8: gasoline

C9-C16: diesel, kerosene, jet fuel

C17-up: lubricating oils, heating oil

Origin: petroleum refining

Coal

Reactions of AlkanesCombustion/Oxidation

CH3CH2CH2CH3 + O2 CO2 + H2Oheat

8 10132

long-chain alkanes catalystshorter-chain alkanes

CH4 + Cl2 CH3Cl + CH2Cl2 CHCl3 CCl4+ +heat or light

Cracking and hydrocracking (industrial)

Halogenation

Alkanes are very unreactive compounds because they have

only strong s bonds and atoms with no partial charges.

However, alkanes do react with Cl2 and Br2

Chlorination of Methane

Requires heat or light for initiation.

The most effective wavelength is blue, which is absorbed

by chlorine gas.

Lots of product formed from absorption of only one

photon of light (chain reaction).

C

H

H

H

H + Cl2heat or light

C

H

H

H

Cl + HCl

Free-Radical Chain Reaction

Initiation generates a reactive intermediate.

Propagation: the intermediate reacts with a stable

molecule to produce another reactive intermediate (and

a product molecule).

Termination: side reactions that destroy the reactive

intermediate.

Initiation Step

A chlorine molecule splits homolytically into chlorine

atoms (free radicals).

Cl Cl + photon (h) Cl + Cl

Propagation Step (1)

The chlorine atom collides with a methane molecule and

abstracts (removes) a H, forming another free radical and

one of the products(HCl).

C

H

H

H

H Cl+ C

H

H

H

+ H Cl

Propagation Step (2)

The methyl free radical collides with another chlorine

molecule, producing the other product (methyl chloride)

and regenerating the chlorine radical.

C

H

H

H

+ Cl Cl C

H

H

H

Cl + Cl

Overall Reaction

C

H

H

H

H Cl+ C

H

H

H

+ H Cl

C

H

H

H

+ Cl Cl C

H

H

H

Cl + Cl

C

H

H

H

H + Cl Cl C

H

H

H

Cl + H Cl

Cl Cl + photon (h) Cl + Cl

Termination Steps

Collision of any two free radicals

Combination of free radical with contaminant or collision

with wall.

C

H

H

HCl+ C

H

H

H

Cl

Consider the relative stabilities of alkyl radicals,

The stable alkyl radical is formed faster, therefore 2-chlorobutane is formed faster.

3 2 2 3 2

3 2 2 2

3 2 3

Reactions of Cyclic Compounds

Chlorofluorocarbons remain very stable in the atmosphere

until they reach the stratosphere.

C ClFCl

F

hCFCl

F+ Cl

The chlorine radicals are ozone-removing agents.

Cl + O3 ClO + O2

ClO + O3 Cl + 2 O2

Hydrocarbons containing double bonds.

Alkenes

Noncyclic alkene: CnH2n

Cyclic alkene: CnH2n–2

CH3CH2=CH2

Physical PropertiesLow boiling points, increasing with mass.

Branched alkenes have lower boiling points.

Less dense than water.

Slightly polar

Pi bond is polarizable, so instantaneous dipole-dipole

interactions occur.

Alkyl groups are electron-donating toward the pi bond,

so may have a small dipole moment.

Polarity Examples

= 0.33 D = 0cis-2-butene, bp 4°C

C CH

H3C

H

CH3

trans-2-butene, bp 1°C

C CH

H

H3C

CH3

Common Names

Usually used for small molecules.

Examples:

CH2 CH2

ethylene

CH2 CH CH3

propylene

CH2 C CH3

CH3

isobutylene

IUPAC Nomenclature

Parent is longest chain containing the double bond.

-ane changes to -ene. (or -diene, -triene)

Number the chain so that the double bond has the

lowest possible number.

In a ring, the double bond is assumed to be between

carbon 1 and carbon 2.

Systematic Nomenclature of Alkenes

Longest continuous chain containing the functional group.

Cite the substituents in alphabetical order.

Name with the lowest functional group numberand then the lowest substituent numbers.

No numbering of the functional group is needed in a cyclic alkene.

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

1-butene

2-methyl-2-butene

3-methylcyclopentene

2-sec-butyl-1,3-cyclohexadiene

3-n-propyl-1-heptene

Isomers of Alkene

Cis-trans Isomerism

Similar groups on same side of double bond, alkene is cis.

Similar groups on opposite sides of double bond,

alkene is trans.

Cycloalkenes are assumed to be cis.

Trans cycloalkenes are not stable unless the ring has at

least 8 carbons.

Name these:

C CCH3

H

H

CH3CH2C C

Br

H

Br

H

trans-2-pentene cis-1,2-dibromoethene

An alkene is an electron-rich molecule

Nucleophile: an electron-rich atom or molecule that shares electrons with electrophiles

Examples of Nucleophiles

A nucleophile

Nucleophiles are attracted to electron-deficient atoms or molecules (electrophiles)

Examples of Electrophiles

Reactivity of C=C

Electrons in pi bond are loosely held.

Electrophiles are attracted to the pi electrons.

Carbocation intermediate forms.

Nucleophile adds to the carbocation.

Net result is addition to the double bond.

Curved Arrows in Reaction Mechanisms

Movement of a pair of electrons

Movement of one electron

Electrophilic Addition

• Step 1: Pi electrons attack the electrophile.

C C + E +C

E

C +

C

E

C + + Nuc:_

C

E

C

Nuc

• Step 2: Nucleophile attacks the carbocation.

Addition of Hydrogen Halides

What is the product?

Addition of HX (1)

Protonation of double bond yields the most

stable carbocation. Positive charge goes to

the carbon that was not protonated.

+ Br_

+

+CH3 C

CH3

CH CH3

H

CH3 C

CH3

CH CH3

H

H Br

CH3 C

CH3

CH CH3

Addition of HX (2)

CH3 C

CH3

CH CH3

H Br

CH3 C

CH3

CH CH3

H+

+ Br_

CH3 C

CH3

CH CH3

H+

Br_

CH3 C

CH3

CH CH3

HBr

Carbocation Stabilities

RegiospecificityMarkovnikov’s Rule: The proton of an acid adds to the

carbon in the double bond that already has the most H’s.

“Rich get richer.”

More general Markovnikov’s Rule:

In an electrophilic addition to an alkene, the electrophile

adds in such a way as to form the most stable intermediate.

HCl, HBr, and HI add to alkenes to form Markovnikov

products.

Addition of Halogens

Cl2, Br2, and sometimes I2 add to a double

bond to form a vicinal dibromide.

Anti addition, so reaction is stereospecific.

CC + Br2 C C

Br

Br

Mechanism for Halogenation (1)

• Pi electrons attack the bromine molecule.

• A bromide ion splits off.

• Intermediate is a cyclic bromonium ion.

CC + Br Br CCBr

+ Br

Halide ion approaches from side opposite the three membered ring.

CCBr

BrCC

Br

Br

Mechanism for Halogenation (2)

Test for Unsaturation

Add Br2 in CCl4 (dark, red-brown color) to an alkene

in the absence of light.

The color quickly disappears as the bromine adds to

the double bond.

“Decolorizing bromine” is the chemical test for the

presence of a double bond.

Formation of Halohydrin

If a halogen is added in the presence of water,

a halohydrin is formed.

Water is the nucleophile, instead of halide.

Product is Markovnikov and anti.

CCBr

H2O

CC

Br

OH H

H2O

CC

Br

OH

+ H3O+

Predict the Product

Predict the product when the given alkene reacts with chlorine in water.

CH3

D

Cl2, H2OOHCH3D

Cl

Addition of Hydrogen to Alkenes (Hydrogenation)

Alkene + H2 Alkane

Catalyst required, usually Pt, Pd, or Ni.

Finely divided metal, heterogeneous

Syn addition

Hydration of Alkenes

Reverse of dehydration of alcohol

Use very dilute solutions of H2SO4 or H3PO4 to drive

equilibrium toward hydration.

C C + H2OH+

C

H

C

OH

alkene alcohol

Mechanism for Hydration

+C

H

C+

H2O C

H

C

O H

H+

+ H2OC

H

C

O H

H+

C

H

C

OH

H3O++

C C OH H

H

++ + H2OC

H

C+

Orientation for Hydration

• Markovnikov product is formed.

+CH3 C

CH3

CH CH3 OH H

H

++ H2O+

H

CH3CH

CH3

CCH3

H2OCH3 C

CH3

CH CH3

HOH H

+

H2OCH3 C

CH3

CH CH3

HOH

Hydroboration

Borane, BH3, adds a hydrogen to the most

substituted carbon in the double bond.

The alkylborane is then oxidized to the alcohol which

is the anti-Mark product.

C C(1) BH3

C

H

C

BH2

(2) H2O2, OH-

C

H

C

OH

Addition of Borane (Hydroboration–Oxidation)

Anti-Markovnikov’s rule in product formation

Anti-Markovnikov Addition of an OH Group

Epoxidation

Alkene reacts with a peroxyacid to form an epoxide

(also called oxirane).

Usual reagent is peroxybenzoic acid.

CC + R C

O

O O H CCO

R C

O

O H+

Mechanism

One-step concerted reaction.

Several bonds break and form simultaneously.

OC

O

R

H

C

C

OOH

OC

O

RC

C

+

Opening the Epoxide Ring

Acid catalyzed.

Water attacks the protonated epoxide.

Trans diol is formed.

CCO

H3O+

CCO

H

H2O

CC

O

OH

H H H2O

CC

O

OH

H

Syn Hydroxylation of Alkenes

• Alkene is converted to a cis-1,2-diol,

• Two reagents:

– Osmium tetroxide (expensive!), followed by hydrogen

peroxide or

– Cold, dilute aqueous potassium permanganate, followed by

hydrolysis with base

Mechanism with OsO4

Concerted syn addition of two oxygens to form a cyclic ester.

C

COs

O O

OO

C

CO O

OO

Os

C

C

OH

OH+ OsO4

H2O2

Oxidative Cleavage

Both the pi and sigma bonds break.

C=C becomes C=O.

Two methods:

Warm or concentrated or acidic KMnO4.

Ozonolysis

Used to determine the position of a double bond in

an unknown.

Example

CCCH3 CH3

H CH3 KM nO4

(warm, conc.)C C

CH3

CH3

OHOH

H3C

H

C

O

H3C

H

C

CH3

CH3

O

C

O

H3COH

+

Cleavage with MnO4-

Permanganate is a strong oxidizing agent.

Glycol initially formed is further oxidized.

Disubstituted carbons become ketones.

Monosubstituted carbons become carboxylic acids.

Terminal =CH2 becomes CO2.

General formula: CnH2n–2 (acyclic); CnH2n–4 (cyclic)

Alkynes

Structure of Alkynes

1. The functional group is a triple bond.

a. The triple bond is composed of two π bonds anda σ bond.

i. The π bonds are oriented perpendicular to eachother.

ii. The two π bonds are electron-rich, theyundergo electrophilic addition like alkenes.

b. The general formula is: CnH2n-2

C C RR R = H, alkyl

The Structure of Alkynes

Acetylene

(Ethyne)H C C H

1.20 Å

* The strength of the carbon-carbon triple bond is about 837 kJ/mol.

* It is the strongest and the shortest known carbon-carbon bond.

Nomenclature: The alkynes are named according to 2 systems

1. They are considered to be derived from acetylene by replacement

of one or both hydrogen atoms by alkyl groups.

H C C C2H5

Ethylacetylene

H3C C C CH3

Dimethylacetylene

H3C C C CH(CH3)2Isopropylmethylacetylene

2. For more complicated alkynes, the IUPAC names are used.

IUPAC names for Alkynes

1. The rule are exactly the same as for the naming of alkenes, except

that the ending -yne replaces -ene.

2. The parent structure is the longest continuous chain that contains

the triple bond.

3. Numbering provides the lowest number for the triple bond.

H3C C C CH1 2 3 4 CH3

CH3

5

4-Methyl-2-pentyne

CH3CHCHC

Cl Br

CCH2CH2CH31 2 3 4 5 6 7 8

3-bromo-2-chloro-4-octyne

A substituent receives the lowest number if there is no functional group suffix,

or if the same number for the functional group suffix is obtained in both directions.

CH3CHC

CH3

CCH2CH2Br123456

1-bromo-5-methyl-3-hexyne

4-Bromo-2-hexyne

CH2 CHCH2CH2 CH

4-Penten-1-yne

CH2 C CHCCH3

H3CCH3

CH2 C CHCOH

H3CCH3

* Triple bonds have priority over double bonds.

* An -OH (hydroxyl group) has priority over the triple bond.

4,4-Dimethyl-1-pentyne 2-Methyl-4-pentyn-2-ol

Physical Properties of Alkynes1. The alkynes have physical properties that are essentially the same

as those of the alkanes and alkenes.

2. They are insoluble in water but quite soluble in the usual organic

solvents of low polarity.

3. They are less dense than water.

4. Their boiling points show the usual increase with the increasing

carbon number.

Preparation of Alkynes

Dehydrohalogenation of alkyl dihalides.

Treatment of a dihalide with strong base, leads to

elimination of HX (X = Cl, Br).

Just like alkene synthesis.

The reaction can take place twice with a dihalide to

form an alkyne.

C C

ClCl

HH KOH C C

H3CHC CH2Br2

H3CHC CH2

Br Br

H3CHC CHBr

KOH

+ KBrNaNH2

H3CC CH- KBr

+ H2O- H2O

Reaction of Alkynes

1. Addition Reactions

1.1 Addition of Hydrogen (Reduction to Alkenes)

RC CR'YZ RC CR'

Y Z

YZ RC CR'Y Z

Y Z

RC CR'

Na, NH3(liq)C C

H

R

R'

H

C CR

H

R'

H

H2

Lindlar catalyst

Anti

Syn

The trans alkene can be obtained by using Li or Na / liquid NH3 as the reducing agent.

1. When lithium or sodium are dissolved in liquid ammonia an intensely blue solution results.

2. These solutions are composed of the metal cations anddissolved electrons which produce the blue color.

3. When an alkyne is added to this solution, an electron adds to one of the π bonds to produce a radical anion.

C C RR C C RRe

(Li)+ Li+

4. The radical anion abstract a hydrogen from the solvent,ammonia leading to a radical.

C C RR NH2H C CR

HR

5. The radical then reacts with a second dissolved electron to form an anion which again abstracts a hydrogen to give thefinal product.

C CR

HR eC C

R

HR

C CR

HRC C

R

HR

HNH2H

+ NH2-

* The cis alkene can be obtained by using Lindlar’s catalyst.

* Lindlar’s catalyst is a form of Palladium (Pd) that has been

deactivated by treatment with leadacetate and quinoline.

Surface of metal catalyst

+ H2

H H

* When a Platinum (Pt) catalyst is used, the alkyne generally

react with two molar equivalents of hydrogen to give an alkane.

1.2 Halogenation

Mechanism

C CX2

C CX

X2C CX X

X X

(X = Cl, Br)

X

C CX2

(X = Cl, Br)C C

X

X

C CX

X

1.3 Hydrohalogenation

C CHX

C CX

HXC CH X

H X

(X = Cl, Br, I) H

Mechanism

C C H C CXH

X

C CH X

AlkenylAlkenyl CarbocationCarbocation

The second carbocation, with the halogen atom attached, is stabilized by the lone electrons on halogen atom.

The charge on the alkenyl carbocation is centered in an sp2 orbital, this is relatively high energy.

These two factors lead to the fact that the second hydrogen attack occurs faster than the initial attack.

H C C C CH X

H XX

H

X

C CH X X

Carbocation

1.4 Hydration

C C RHHgSO4

H2SO4H2O

C C RH

OH

H

* Terminal alkynes

* Other alkynes

C C RCH3HgSO4

H2SO4H2O

C C RCH3

OH

H

C C RR

OH

HH2OH2SO4

HgSO4C C RR

C C RCH3

HO

H

• Markovnikov product is found for the hydration with the OH group

adding to the more highly substituted carbon and the H

attaching to the less highly substituted carbon.

* The intermediate in the reaction, a vinyl alcohol, then rearranges to

a ketone in a process called tautomerism.

C CO

H

Enol tautomer(less favored)

C CO H

Keto tautomer(more favored)

Mechanism

C C

C CO H

HRH

R HHg+SO4

2-C CR

HOH2

Hg+SO42-

C CR

HOHH

-H+

Hg+SO42-

C CR

HOH

H3O+

HC C

R

HOH

1.5 Hydroboration/ Oxidation

2) H2O2, OHC C HCH3

OH

H

1) BH3, THFC C HCH3

* Hydroboration of symmetrically substituted internal alkynes gives

vinylic boranes.

* The vinylic boranes are oxidized by basic hydrogen peroxide to

yield ketones (via enols)

* The reaction occurs in anti-Markovnikov fashion.

2. Reactions as Acids

C CR H + Na C CR Na

Sodium acetylide

2 2 2 + H2

C CR H + LiNH2 C CR Li

Lithium acetylide

+ HNH2

C CR H + Ag+

Precipitate

C CR Ag

+ HNO3

C CR H + Ag+

Table 1. Acidity of Simple Hydrocarbons

Type Example Ka pKa

Alkyne 10-25 25

Alkene 10-44 44

Alkane 10-60 60

HC CH

H2C CH2

H3C CH3

Stronger acid

Weaker acid

Reactions of metal acetylides with primary alkyl halides

* Lithium or sodium acetylides can react with primary alkyl

halides

* The alkyl group becomes attached to the triply bonded carbon,

and a new, larger alkynes has been generated.

RC CR'X

Li RC C R' + LiBr

HC CCH3CH2Br

Li HC C CH2CH3 + LiBr

3. Oxidative cleavage

C C +

An internal alkyne

R'RKMnO4 or O3

CRO

OH CR'O

OH

A terminal alkyne

C C +HRKMnO4 or O3

CRO

OH CO O

Organic Synthesis* Prepare octane from 1 pentyne

1). NaNH2, NH3

2). BrCH2CH2CH3, THF

H2/Pt in Ethanol

* Prepare cis-2-hexene from 1 pentyne1). NaNH2, NH3

2). CH3Br, THF

H2/Pd in Ethanol

Designing a Synthesis

CH3CH2C CH CH3CH2CCH2CH2CH3

O?

Problem 1

Problem 2

Alkyl Halides

1. Aliphatic Alkyl Halides

2. Aromatic Alkyl Halides

Methyl halide

H C

H

X

H

1oalkyl halide

R C

H

X

H

R C

H

X

R

2oalkyl halide

R C

R

X

R

3oalkyl halide

X

Br CH3

CH3

CH3 CH3

Br

Cl

Cl

CH3

Br

Cl

CH3

CH3

Br

CH3ICl

Br X

C C

X

Br

Cl

C O C

halothane

F

C

H

F

H

Cl

F

H

H

C C F

Br

H

Cl

F

Fenflurane

Halomethane Dipole Moment, (D)

CH3F 1.82

CH3Cl 1.94

CH3Br 1.79

CH3I 1.64

+ +

ElectrophilicElectrophilic Substitution ReactionsSubstitution Reactions

+ NaCNI + NaICN

H Br

+ NaN3

N3 H

+ NaBr

C Br

C3H7

C2H5H3C

+ H2O C OH

C3H7

C2H5H3C

CHO

C3H7

C2H5CH3

+ HBr

C Cl

CH3

H3C

CH3

+ CH3OH C

+ HCl

CH3

H3C

CH3

OCH3

1. S1. SNN1 (1 (UnimolecularUnimolecular MechnismMechnism))

- Rate of the reaction depends on a concentration of a reagent.

C Br

C3H7

C2H5H3C

C

C3H7

C2H5 CH3

+ Br

ElectrophilicElectrophilic Substitution ReactionsSubstitution Reactions

- The first step is a cleavage of a leaving group.

CarbocationCarbocation

C

C3H7

C2H5 CH3+ H2O

C

C3H7

C2H5H3C

OH

H

C

C3H7

C2H5H3C

OH

+ Br

+ H2O

C

C3H7

C2H5CH3

OH

H

+ Br

CHO

C3H7

C2H5CH3

+ HBr

- The second step is a bond formation of carbocation and nucleophile.

2. S2. SNN2 (Bimolecular Mechanism)2 (Bimolecular Mechanism)

Transition StateTransition State

+ NaN3

BrH

Br

H

N3+ NaBr

HN3

ElectrophilicElectrophilic Substitution ReactionsSubstitution Reactions

Rate of the reaction depends on the conc. of a reagent and a substrate.

A cleavage of a leaving group and nucleophilic attack occurs in one step.

SSNN1 Mechanism 1 Mechanism

1. Rate of the reaction: tertiary > secondary > primary alky halides

2. Weak nucleophiles

3. Good leaving groups

HS- > CN- > I- > CH3O- > HO- > Cl- > NH3 > H2O

I- > Br- > Cl- > F-

SSNN2 Mechanism2 Mechanism

1. Rate of the reaction: primary/ secondary alky halides

2. Good nucleophiles.

3. Weak leaving groups.

+ NaCNCCH2Br

H CH3

+ NaBr

CCH2CN

H CH3

Elimination ReactionsElimination Reactions

Elimination ReactionsElimination Reactions

1) E11) E1

C Br

H3C

H3CH

C

CH3

H3C H

+ Br

- Rate of the reaction depends on a concentration of a reagent.

- The first step is a cleavage of a leaving group.

- Base will react with hydrogen.

C

H2C

H3C H

OHH

+ HOHC CH2

H3C

H

Br

CH3NaOH

CH2 CH3

+ HOH NaBr

2) E22) E2

A hydrogen atom and a halide will be removed.

OH

H

Br

HH

H

H

+ HOH

Elimination ReactionsElimination Reactions

Rate of the reaction depends on conc. of a reagent and a strong base.

Acid Formula pKaConjugate Base

ethanolwaterbicarbonate ionammonium ioncarbonic acidacetic acid

sulfuric acidhydrogen chloride

10.3315.715.9

4.766.369.24

-5.2-7

CH3 CH2 OH CH3 CH2 O-

H2 O HO-

HCO 3- CO3

2-

NH 4+ NH 3

H2 CO3 HCO 3-

CH3 CO2 H CH3 CO2-

H2 SO4 HSO4-

HCl Cl -

CH3CH2OHC Br

CH3

H3C

CH3

C OCH2CH3

CH3

H3C

CH3 (80%)

CH3CH2ONaC Br

CH3

H3C

CH3

C

CH2

H3C CH3 (97%)

BrNaCN

HBr

NaSCH3

Cl H

NaOH

CN SN2

SCH3

HSN2

E2