The document discusses carbon and its compounds. It begins by describing the occurrence of carbon in nature and its bonding properties. Carbon has four valence electrons and forms covalent bonds by sharing electrons in order to attain stability. It then discusses the formation of single, double, and triple covalent bonds between carbon atoms and other elements like hydrogen. This allows carbon to form a huge variety of compounds through catenation and bonding with different functional groups. As a result, over three million carbon compounds are known. Isomerism is also discussed, where compounds with the same molecular formula but different structural formulas are called isomers.

1. 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. 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.
X I
_ _
X C X
C
X
I

3. 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.
Hx + x H H XX H H–H H2

4. 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
X X XX XX XX
X X XX
X X XX
XX XX XX XX
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
X X X X
X X XX
X X XX
X X X XX
X X X
N + N N N NΞN N2

6. 4) Electron dot structures :-
Methane molecule – CH4 Ethane molecule – C2H6
H
X H
X H
X
X
X X X X
X
X X X
H C H H C C H
H H H
H H H
I I I

7. 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

8. 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

9. 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–CΞC–H
H
I
Propyne - C3H4 H–C ΞC–C–H
I
H

10. 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

11. 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

12. 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
I
Methane :– 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

13. 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

14. 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 - C4H 6
Ethyne :- C2H2 H–C C–H
H
I
Propyne :- C3H4 H–C C–C–H
I
H
H H
I I
Butyne :- C H H–C C–C–C–H

15. 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

16. 3 Classes of Unsaturated
Hydrocarbons
1st CLASS
Alkenes – contains one or more C-C double
bonds
H2C=CH2
ethene (ethylene)
2nd CLASS
Alkynes – contains one or more C-C triple
bonds
HC≡CH
ethyne (acetylene)

17. Unsaturated Hydrocarbons
3rd CLASS Arenes- aromatic hydrocarbons
H
Benzene
C
H-C C-H
H-C C-H
C
H
(not chemically reactive under any of the
conditions described)
Arenes are found in proteins, nucleic acids, and
pharmaceuticals like aspirin

18. Alkenes
Carbon-carbon double bonds
Names end in –en-e
H2C=CH2 ethene (ethylene)
H2C=CH-CH3 propene
(propylene)
cyclohexene

19. Shapes of Alkenes
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

20. Alkynes
Carbon-carbon triple bonds
Names end in -yne
HC≡CH
ethyne(acetylene)
HC≡C-CH3 propyne

21. Shapes of Alkynes
VSEPR predicts 180o for bond angles in
ethyne
H C≡C H
ethyne
∠ 180°

22. Naming Alkenes
The double bond takes precedence over substituents in
numbering the parent chain.
2. Use the infix “en” for all alkenes and cycloalkenes
3. Use the suffix “e” for all alkenes and cycloalkenes
Change the infix “an” to the corresponding alkane to “en”
• butane butene
• propane propene
• octane octene

23. Naming Alkenes
1. For open chain alkenes, identify the
parent chain as the longest sequence of
carbons that includes the double bond.
CH2
CH3CH2CCH2CH2CH2CH3
There is a longer chain of 7C but it does
not include the double bond.

24. Naming Alkenes
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 1
4-methyl-1-pentene

25. Naming Alkenes
5. For cycloalkenes always give position 1 to one
of the two carbons at the double bond
CH3
3-methylcyclohexene

26. 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

27. 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 - cyclohexadiene
This pattern can be easily extended to trienes,
tetraenes, etc.

28. Learning Check HA2
Write the IUPAC name for each of the
following unsaturated compounds:
CH3
CH3
A. CH3C=CHCH3 B.

29. Solutions HA2
Write the IUPAC name for each of the
following unsaturated compounds:
CH3
CH3
A. CH3C=CHCH3 B.

30. Isomers of Alkenes
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

31. 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

32. Isomers of Alkenes
Alkenes can exist as isomers in 3 ways
Constitutional isomers:
1st different carbon skeletons CH3
CH2=CHCH2CH3 CH2=CCH3
1-butene 2-methylpropene
2nd H atoms attached differently to the skeleton
CH3
CH3CH=CHCH3 CH2=CCH3
2-butene 2-methylpropene

33. 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

34. Geometric Isomers
•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.

35. Alkene
CH3 CH3 CH3 H
C=C C=C
H H H CH3
trans-2-butene
cis-2-butene

36. 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)

37. Naming Alkenes
– Name all other substituents in
a manner similar to the
alkanes.
– Use a prefix to indicate the
geometric isomer present, if
necessary.

38. Learning Check
Draw the structures for the
following compounds:
cis-6-methyl-3-heptene

39. 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”

40. 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

41. Learning Check HA3
Write the IUPAC name for each of the
following unsaturated compounds:
CH3CH2C≡CCH3

42. Solutions HA3
Write the IUPAC name for each
of the following unsaturated
compounds:
• CH3CH2C≡CCH3
2-pentyne

43. Alkynes
Name the following compounds:
CH3CH2C CCHCH3
CH2CH3
CH3CH2C C
Cl

44. Alkynes
Draw the following alkynes.
4-chloro-2-pentyne
3-propyl-1-hexyne

45. 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

46. Reactions of Alkenes &
Alkynes
• More reactive than alkanes or
aromatics, why?
• Generally undergo addition
reactions
• Presence of easily accessible π
electrons
• Unsaturated: can fit more atoms
around the carbons

47. Reaction of Alkenes
Additions reactions of the double bond
The new double bond is broken and in
its 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
XY

48. 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)

49. Products of Hydrogenation
Adding H2 to vegetable oils produces
compounds with higher melting
points
 Margarines
 Soft margarines
 Shortenings (solid)

50. Trans Fats
In 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.

51. Learning Check HA4
What is the product of adding H2
(Ni catalyst) to 1-butene?

52. Solution HA4
What is the product of adding H2
(Ni catalyst) to 1-butene?
Ni
CH2=CHCH2CH3 + H2
CH3CH2CH2CH3

53. Learning Check HA5
Write the product of the following
addition reactions:
CH3CH=CHCH3 + H2
+ Br2

54. Solution HA5
Write the product of the following addition
reactions:
CH3CH=CHCH3 + H2 CH3CH2CH2CH3
+ Br2 Br
Br

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

56. 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

57. 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???

58. Markovinkov’s Rule
Try This !!!

59. 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

60. Addition of Hydrogen
Halides
Markovnikov’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

61. Learning Check
CH3CH=CH2 +HI
+ HBr
=CH2 + HBr

62. Question 3
• What is the major product of the
following reaction?
+ HBr

63. 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

64. Addition of H2O
• Addition of water is called hydration
– 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
OH H
H 2 SO4
CH3 CH=CH 2 + H2 O CH3 CH-CH2
Propene 2-Propanol

65. CH3 CH3
H2 SO4
CH3 C=CH2 + H2 O CH3 C-CH2
HO H
2-M ethylpropene 2-Methyl-2-propanol

66. Alkene Addition

67. 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

68. 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

69. 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.

70. 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 + H2O
vii) 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

71. 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.

73. 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.