1) Occurrence of carbon :- i) Carbon is found in the atmosphere, inside the earth’s crust and in

all living organisms.

ii) Carbon is present in fuels like wood, coal, charcoal, coke, petroleum,

natural gas, biogas, marsh gas etc.

iii) Carbon is present in compounds like carbonates,

hydrogen carbonates etc.

iv) Carbon is found in the free state as diamond, graphite, fullerenes etc.

2) Bonding in carbon – Covalent bond :- The atomic number of carbon is 6, its electronic arrangement is 2,4, it

has 4 valence electrons. It can attain stability by gaining 4 electrons, losing 4 electrons or sharing 4 electrons with other atoms.

It does not gain 4 electrons because it is difficult for the 6 protons to hold 10 electrons.

It does not lose 4 electrons because it needs a large amount of energy to lose 4 electrons.

So it shares 4 electrons with other atoms to attain stability resulting in the formation of covalent bonds.

Since carbon atom needs 4 electrons to attain stability, its valency is 4 and it is tetravalent.

C CX

X

X X__

I

I

3) Formation of covalent bonds :- Covalent bond is chemical bond formed by the sharing of electrons between atoms. The sharing of one pair of electrons results in the formation of single covalent bond, sharing of two pairs of electrons results in the formation of double covalent bond and sharing of three pairs of electrons results in the formation of triple covalent bond.

Eg :- Formation of single covalent bond in Hydrogen

molecule - H2

The atomic number of hydrogen is 1, its electronic arrangement is 1, it

has 1 valence electron. It needs 1 electron more to attain stability. So two hydrogen atoms share 1 pair of electrons resulting in the formation

of a single covalent bond in hydrogen molecule H2.

H x + x H H X X H H – H H2

Formation of double covalent bond in oxygen molecule - O2

The atomic number of oxygen is 8, its EC is 2,6, it has 6 VE, it needs 2 electrons more to attain stability. So two oxygen atoms share two pairs of electrons resulting in the formation of a double covalent bond in oxygen molecule O2

O + O O O O = O O2

Formation of triple covalent bond in Nitrogen molecule - N2

The atomic number of nitrogen is 7, its EC is 2,5, it has 5 VE, it needs 3 electrons more to attain stability. So two nitrogen atoms share three pairs of electrons resulting in the formation of a triple covalent bond in nitrogen molecule N2

N + N N N N Ξ N N2

XX

XX

X

XX

X

X

XX X

X

X

X

X

XX

X X

XX

X

X

XX

XX

XX XX

XX

X

XX

XX

X

XX

XX

4) Electron dot structures :-

Methane molecule – CH4 Ethane molecule – C2H6

H H H

H C H H C C H

H H H

H H H

I I I

H – C – H H – C – C – H

I I I

H H H

X

X

X X X

X

X

X

X

X X

X

5) Formation of a very large number of carbon compounds :-

Carbon forms a very large number of compounds. The number of carbon compounds is more than three million. It is more than the number of compounds formed by all other elements. This is because :-

i) Carbon atom can form bonds with other carbon atoms to form long

chains, branched chains and closed rings. This property is called

catenation.

ii) Since the valency of carbon is 4, it can form bonds with other

carbon atoms or with atoms of other elements like hydrogen,

oxygen, nitrogen, halogens etc.

I

C

I I I I I I I I I I C

C – C – C – C – C – C C – C – C – C C C

I I I I I I I I I I C C

C C

I

Long chain Branched chain Closed ring

__

_ _

_ __ _

6) Hydrocarbons, Saturated and Unsaturated hydrocarbons :-

i) Hydrocarbons :- are compounds containing carbon and hydrogen atoms. ii) Saturated hydrocarbons :- are hydrocarbons having all single covalent bonds between the carbon atoms. Eg : Alkanes :- have all single covalent bonds between the carbon atoms and their names end with – ane.

H I

Methane – CH4 H – C – H I H H H I I

Ethane – C2H6 H – C – C – H I I H H

iii) Unsaturated hydrocarbons :- are hydrocarbons having a double or triple covalent bond between two carbon atoms. Eg : Alkenes and Alkynes. Alkenes :- have a double covalent bond between two carbon atoms. and their names end with – ene. H H H H I I I I

Ethene - C2H4 C = C Propene – C3H6 H – C = C – C – H

I I I I H H H H Alkynes :- have a triple covalent bond between two carbon atoms and their names end with – yne.

Ethyne – Ethyne – C2H2 H – C Ξ C – H

H I

Propyne - C3H4 H – C Ξ C – C – H

I H

7) Isomerism :- Carbon compounds having the same molecular formula but different structural formulae are called isomers. This property is called isomerism.

Eg:- Butane – C4H10 has 2 isomers. They are Normal butane and Iso butane. H H H H H H H I I I I I I I H – C – C – C – C – H H – C – C – C – H Iso butane I I I I I I H H H H H H H – C – H Normal butane I H

Pentane – C5H12 has 3 isomers. They are Normal pentane, Iso pentane and Neo pentane. Neo pentane

Iso pentane H H I I H – C – H Normal pentane H – C – H H H I I H H H H H H H H H – C – C – C – H I I I I I I I I I I H – C – C – C – C – C – H H – C – C – C – C – H H H I I I I I I I I I H – C – H H H H H H H H H H I H

8) Functional groups :- An atom or a group of atoms which decides the properties of a carbon compound is called a functional group. i) Halide ( Halo group) :- - Cl, - Br, etc. ( Names end with – ane ) Eg :- CH3Cl – Chloro methane, C2H5Br – Bromo ethane

ii) Alcohol :- - OH ( Names end with – ol ) Eg :- CH3OH – Methanol, C2H5OH – Ethanol H iii) Aldehyde :- - CHO - C ( Names end with – al ) O

Eg :- HCHO – Methanal, CH3CHO – Ethanal O II iv) Carboxylic acid :- - COOH - C - OH (Names end with – oic acid ) Eg :- HCOOH – Methanoic acid, CH3COOH – Ethanoic acid v) Ketone :- - CO - C - (Names end with – one ) II O Eg :- CH3COCH3 – Propanone , CH3COC2H5 - Butanone

9) Homologus series :- Homologus series is a group of carbon compounds having similar structures, similar chemical properties and whose successive members differ by a – CH2 group. Eg :- Alkanes, Alkenes, Alkynes etc. Alkanes :- have general molecular formula CnH2n+2 . Their names end with – ane and the members are as follows :- Methane - CH4

Ethane - C2H6 Propane - C3H8 Butane - C4H10

Pentane - C5H12 H IMethane :– CH4 H – C – H I H

H H H H H I I I I I

Ethane :– C2H6 H – C – C – H Propane – C3H8 H – C – C – C – H I I I I I H H H H H

Alkenes :-

Alkenes have general molecular formula CnH2n . Their names end with – ene and the members are as follows :-

Ethene - C2H4

Propene - C3H6

Butene - C4H8

Pentene - C5H10

H H I I

Ethene :- C2H4 C = C I I H H

H H H H H H H I I I I I I I

Propene :- C3H6 H – C = C – C – H Butene :- C4H8 H – C = C – C – C – H

I I I H H H

Alkynes :- Alkynes have general molecular formula CnH 2n – 2 .Their names end

with – yne and the members are as follows :-

Ethyne - C2H2

Propyne - C3H4

Butyne - C4H6

Ethyne :- C2H2 H – C C – H

H

I

Propyne :- C3H4 H – C C – C – H

I

H

H H

I I

Butyne :- C4H6 H – C C – C – C – H

I I

H H

Saturated and Unsaturated Compounds

Saturated compounds (alkanes) have the maximum number of hydrogen atoms attached to each carbon atom

Unsaturated compounds have fewer hydrogen atoms attached to the carbon chain than alkanes

Unsaturated compounds contain double or triple bonds

1st CLASSAlkenes – contains one or more C-C double bonds

H2C=CH2

ethene (ethylene)

2nd CLASSAlkynes – contains one or more C-C triple bonds

HCCH ethyne (acetylene)

3 Classes of Unsaturated Hydrocarbons

Unsaturated Hydrocarbons

HH

CC

C - H C - H

C - H C - H

H - C H - C H - C H - C CC

HH

3rd CLASS Arenes- aromatic hydrocarbons

Benzene

(not chemically reactive under any of the conditions described) Arenes are found in proteins, nucleic acids, and pharmaceuticals like aspirin

Alkenes

Carbon-carbon double bonds Names end in –en-e

H2C=CH2 ethene (ethylene)

H2C=CH-CH3 propene

(propylene)

cyclohexene

VSEPR predicts 120o for bond angles in ethene and propene

H2C=CH2 H2C=CH-CH3

121.7° 124.7°The actual for these molecules are close to the predicted; however, in other alkenes the predicted angles will have a larger deviation from that predicted in the VESPR model b/c there is limited rotation around a double bond

Shapes of Alkenes

Carbon-carbon triple bonds Names end in -yne

HCCH ethyne(acetylene)

HCC-CH3 propyne

Alkynes

VSEPR predicts 180o for bond angles in ethyne

H CC H

ethyne

180°

Shapes of Alkynes

The double bond takes precedence over substituents in numbering the parent chain.

1. Use the infix “en” for all alkenes and cycloalkenes2. Use the suffix “e” for all alkenes and cycloalkenesChange the infix “an” to the corresponding alkane to “en”

• butane butene

• propane propene

• octane octene

Naming Alkenes

Naming Alkenes

3. For open chain alkenes, identify the parent chain as the longest sequence ofcarbons that includes the double bond. CH2

CH3CH2CCH2CH2CH2CH3

There is a longer chain of 7C but it does not include the double bond.

4. For an open chain alkenes, number the parent chain for whichever end gives the lower number to the first carbon of double bonds.

These rules give precedence to the location of the double bond over the location of the first substituent on the parent chain.

CH3

CH3CHCH2CH=CH2

double bond is at position 14-methyl-1-pentene

Naming Alkenes

5. For cycloalkenes always give position 1 to one of the two carbons at the double bond

CH3

3-methylcyclohexene

Naming Alkenes

6. Place the # that locates the 1st carbon of the double bond as a prefix, and separate this number from the name by a hyphen

1 2 3 4

CH2=CHCH2CH3 1-butene

CH3CH=CHCH3 2-butene

*Remember to separate # from numbers by commas, but use hyphens to connect a number to a word

Naming Alkenes

7. When a compound has two double bonds, it is named as a diene with 2 numbers in the name to specify the locations of the double bonds. 6

CH3 5 1

CH2=CCH=CH2 4 2

2-methyl-1,3-butadiene 3

1,4 - cyclohexadieneThis pattern can be easily extended to trienes, tetraenes, etc.

Naming Alkenes

Learning Check HA2

Write the IUPAC name for each of the following unsaturated compounds:

CH3

A. CH3C=CHCH3 B.

CH3

Solutions HA2

Write the IUPAC name for each of the following unsaturated compounds:

CH3

A. CH3C=CHCH3 B.

2-methyl-2-butene 3-methylcyclopentene

CH3

3rd Geometric isomerism

•No free rotation at the double bond in a ring

•Have identical constitution including the location of the double bond but differ in geometry

•Differ only in the direction taken by their end of chain methyl group

•Common at the molecular level , particularly in edible fats, oils and in related compounds that make up most of a cell membrane

Isomers of Alkenes

Alkenes

• Alkene Nomenclature– Cis isomer:

• two groups (on adjacent carbons) on the same side of the C = C double bond

– Trans isomer:• two groups (on adjacent carbons)

on opposite sides of the C = C double bond

Alkenes can exist as isomers in 3 waysConstitutional isomers:1st different carbon skeletons CH3

CH2=CHCH2CH3 CH2=CCH3

1-butene 2-methylpropene2nd H atoms attached differently to the skeleton CH3

CH3CH=CHCH3 CH2=CCH3

2-butene 2-methylpropene

Isomers of Alkenes

Alkenes

• Alkene Nomenclature• Different geometric isomers are

possible for many alkenes.– Compounds that have the same

molecular formula and the same groups bonded to each other, but different spatial arrangements of the groups

• cis isomer• trans isomer

•When there are two identical groups at one end of a double bond, geometric isomers are not possible

•Cyclic compounds can also have geometric isomers

•This cis-trans isomerism is found in the many cyclic structures of carbohydrates.

•Geometric differences alone make most carbohydrates unusuable in human nutrition.

Geometric Isomers

Alkene

cis-2-butene

CHCH33 CHCH33

C = CC = C

HH HH

CHCH33 HH

C = CC = C

HH CHCH33

trans-2-butene

Geometric Isomers

Double bond is fixed Cis/trans Isomers are possible

CH3 CH3 CH3 H

C = C C = C

H H H CH3

(bp 3.7°C) (bp 0.9°C)

cis trans

Naming Alkenes

8. Name all other substituents in a manner similar to the alkanes.

9. Use a prefix to indicate the geometric isomer present, if necessary.

Learning Check

Draw the structures for the following compounds:

cis-6-methyl-3-heptene

Alkynes

• Alkynes:– unsaturated hydrocarbons that contain a

C C triple bond

• Alkyne Nomenclature:– Identify the longest continuous chain

containing the triple bond

– To find the base name, change the infix of the corresponding alkane from “an” to “yn”

Alkynes

• Alkyne Nomenclature:– Use a number to designate the

position of the triple bond• number from the end of the chain

closest to the triple bond– just like with alkenes

– Name substituents like you do with alkanes and alkenes

Learning Check HA3

Write the IUPAC name for each of the following unsaturated compounds:

CH3CH2CCCH3

Solutions HA3

Write the IUPAC name for each of the following unsaturated compounds:

A. CH3CH2CCCH3

2-pentyne

Alkynes

Name the following compounds:

CH3CH2C CCHCH3

CH2CH3

CH3CH2C C

Cl

Alkynes

Draw the following alkynes.

4-chloro-2-pentyne

3-propyl-1-hexyne

Physical Properties

• alkenes and alkynes are nonpolar compounds• the only attractive forces between their

molecules are London dispersion forces• their physical properties are similar to those of

alkanes with the same carbon skeletons• alkenes and alkynes are insoluble in water but

soluble in one another and in nonpolar organic liquids

• alkenes and alkynes that are liquid or solid at room temperature have densities less than 1 g/mL; they float on water

Reactions of Alkenes & Alkynes• More reactive than alkanes or

aromatics, why?

• Generally undergo addition reactions

• Presence of easily accessible electrons

• Unsaturated: can fit more atoms around the carbons

Reaction of Alkenes

Additions reactions of the double bondThe new double bond is broken and inits place single bonds are formed to the new atoms or groups of atoms

The double bond becomes a single bond

H H H H

H– C= C–H + X-Y H–C–C–H

X Y

Hydrogenation

Adds a hydrogen atom to each carbon atom of a double bond

H H H H

Ni

H–C=C–H + H2 H–C–C–H

H H ethene ethane

Reacts with H2 in the presence of transition metal catalyst (Pd, Pt, Ru, Ni)

Products of Hydrogenation

Adding H2 to vegetable oils produces

compounds with higher melting points

Margarines

Soft margarines

Shortenings (solid)

Trans FatsIn vegetable oils, the unsaturated fats

usually contain cis double bonds.

During hydrogenation, some cis double bonds are converted to trans double bonds (more stable) causing a change in the fatty acid structure

If a label states “partially” or “fully hydrogenated”, the fats contain trans fatty acids.

Learning Check HA4

What is the product of adding H2 (Ni catalyst) to 1-butene?

Solution HA4

What is the product of adding H2 (Ni catalyst) to 1-butene?

Ni

CH2=CHCH2CH3 + H2

CH3CH2CH2CH3

Learning Check HA5

Write the product of the following addition reactions:

CH3CH=CHCH3 + H2

+ Br2

Solution HA5

Write the product of the following addition reactions:

CH3CH=CHCH3 + H2 CH3CH2CH2CH3

+ Br2Br

Br

Addition of Bromine

a) Br2 (in CCl4) is added to an unknown liquidb) The unknown is saturated b/c Br2 does not lose its red color.c) The unknown was unsaturated. The deep red color of Br2 is decolorized as it reacts with the double bond.

Orientation of Addition

• Both alkene & reagent are symmetric: one possible product

• One is symmetric and the other is asymmetric: one possible product

• Both alkene & reagent are asymmetric:

two possible products

Markovinkov’s Rule

• When an unsymmetrical reactant of the type X-Y adds to an unsymmetrical alkene, the carbon with the greater number of hydrogens gets more H

• Used to predict the product of many alkene addition reactions however it does not explain Why???

Markovinkov’s Rule

Try This !!!

Addition of Hydrogen Halides

Adds a H atom and Cl to each carbon atom of a double bond

H H H H

H–C=C–H + HCl H–C–C– H

H Cl

ethene chloroethane

Addition of Hydrogen HalidesMarkovnikov’s rule – when

unsymmetrical reagent adds to an unsymmetrical carbon, the carbon with the greater # of hydrogens gets more H

Cl

CH3C=CH2 + HCl CH3CCH3

CH3 CH3

Learning Check

CH3CH=CH2 +HI

+ HBr

=CH2 + HBr

Question 3

• What is the major product of the following reaction?

+ HBr

Addition of Water (Hydration)

• Water does not react with an alkene in the absence of an acid catalyst

• Water is a weak donor of H+ b/c it holds it protons too strongly

Addition of H2O

• Addition of water is called hydrationhydration– hydration is acid catalyzed, most

commonly by H2SO4

– hydration follows Markovnikov’s rule; H adds to the less substituted carbon and OH adds to the more substituted carbon

CH3CH=CH2 H2OH2SO4 CH3CH-CH2

HOH

Propene 2-Propanol+

CH3

CH3C=CH2

CH3

H2OH2SO4 CH3C-CH2

HOH2-Methyl-2-propanol2-Methylpropene

+

Alkene Addition

10) Chemical properties of Carbon compounds :-a) Combustion :- Carbon compounds burn in oxygen to form water, carbon dioxide, heat and light.

Eg :- C + O2 CO2 + heat + light

CH4 + 2O2 2H2O + CO2 + heat + light

C2H5OH + 3O2 3H2O + 2CO2 heat + light

b) Oxidation :- Carbon compounds like alcohols are oxidised to carboxylic acids on heating with oxidising agents like alkaline Potassium permanganate

– KMnO4 or acidic potassium dichromate - K2Cr2O7 . Eg:- Alcohols are oxidised to Carboxylic acids

alkaline KMnO4 + heat

C2H5OH CH3COOH

Ethanol acidic K2Cr2O7 + heat Ethanoic acid

c) Addition reaction :-

Unsaturated hydrocarbons undergo addition reaction with hydrogen in the presence of nickel or palladium as catalyst to form saturated hydrocarbons. Eg:- Ethene undergoes addition reaction with hydrogen to form ethane in the presence of nickel or palladium as catalyst. Ni or Pd catalyst

C2H4 + H2 C2H6

H H H H I I Ni or Pd catalyst I I

C = C + H2 H – C – C – H I I I I H H H H The addition of hydrogen to unsaturated hydrocarbons to form saturated hydrocarbons is called hydrogenation. Hydrogenation is used to convert unsaturated oils and fats to saturated oils and fats.

d) Substitution reaction :- Saturated hydrocarbons undergo substitution reaction with halogens to form substitution products. Eg :- Methane undergoes substitution reaction with chlorine in the presence of sunlight to form substitution products.

CH4 + Cl2 CH3Cl + HCl CH3Cl + Cl2 CH2Cl2 + HCl

CH2Cl2 + Cl2 CHCI3 + HCl CHCI3 + Cl2 CCl4 + HCl

11) Some important carbon compounds :-

a) ETHANOL :- C2H5OH - Ethyl alcohol Properties :- i) Ethanol is a colourless liquid with a pleasant smell and burning taste. ii) It is soluble in water.iii) Ethanol reacts with sodium to form sodium ethoxide and hydrogen.

2C2H5OH + 2Na 2C2H5ONa + H2

iv) Ethanol reacts with hot conc. H2SO4 to form ethene and water. Conc.

H2SO4 is a dehydrating agent and removes water from ethanol.

conc. H2SO4

C2H5OH C2H4 + H2O Uses :- i) Ethanol is used for making alcoholic drinks. ii) It is used as a solvent.iii) It is used for making medicines like tincture iodine, cough syrups, tonics etc.

b) ETHANOIC ACID :- CH3COOH – Acetic acid

Properties :- i) Ethanoic acid is a colourless liquid with a pungent smell and sour taste. ii) It is soluble in water.iii) A solution of 5% to 8% ethanoic acid in water is called Vinegar.iv) Esterification :- Ethanoic acid reacts with ethanol to form the ester ethyl ethanoate in the presence of conc. H2SO4. conc.H2SO4

CH3COOH + C2H5OH CH3COOC2H5 + H2O The reaction between carboxylic acid and alcohol to form an ester is called esterification. v) Saponification :- When an ester reacts with sodium hydroxide solution, the sodium salt of the carboxylic acid and the parent alcohol are formed. This reaction is called saponification. Eg :-Ethyl ethanoate reacts with sodium hydroxide to form sodium acetate and ethanol. CH3COOC2H5 + NaOH CH3COONa + C2H5OH vi) Ethanoic acid reacts with bases to form salt and water. CH3COOH + NaOH CH3COONa + H2Ovii) Ethanoic acid reacts with carbonates and hydrogen carbonates to form salt, water and carbon dioxide. 2CH3COOH + Na2CO3 2CH3COONa + H2O + CO2

CH3COOH + NaHCO3 CH3COONa + H2O + CO2

12) Soaps and detergents :-a) Soaps :- Soaps are long chain sodium or potassium salts of carboxylic acids. Eg:- Sodium stearate – C17H35COONa

Structure of soap molecule :- A soap molecule has two parts. A long hydrocarbon part which is hydrophobic (water repelling) and soluble in oil and grease and a short ionic part which is hydrophyllic (water attracting) and insoluble in oil and grease.

COO Na +

Hydrocarbon part Ionic part

(Water repelling) (Water attracting)

Cleansing action of soap :- When soap is dissolved in water it forms spherical structures called micelles. In each micelle the soap molecules are arranged radially such that the HC part is towards the centre and the ionic part is towards the outside. The HC part dissolves the dirt, oil and grease and forms an emulsion at the centre of the micelles which can be washed away by water.

b) Detergents :- Detergents are long chain sodium salts of sulphonic acids. Soaps do not wash well with hard water because it forms insoluble precipitates of calcium and magnesium salts in hard water. Detergents wash well with hard water because it does not form insoluble precipitates of calcium and magnesium salts in hard water.

c) Differences between soaps and detergents :- Soaps Detergents

i) Soaps are sodium salts of Detergents are sodium salts of fatty acids. sulphonic acids. ii) Soaps clean well in soft water but Detergents clean well with both do not clean well in hard water. hard and soft water. iii) Soaps do not clean as well as Detergents clean better than soaps.

detergents.

iv) Soaps are biodegradable and Some detergents are non biodegradable do not cause pollution. and cause pollution.