General Organic Chemistry
is the study of the structure, properties, composition, reactions, and synthesis of carbon-containing compounds. It encompasses a vast array of substances, including hydrocarbons and their derivatives, and forms the backbone of many industries from pharmaceuticals to petrochemicals.
Hybridisation
"Hybridization in carbon compounds is the blending of atomic orbitals to form new, hybrid orbitals apt for bonding.
Common types include
sp- linear (50% s character),
sp
2
- trigonal planar (33% s character), and
sp
3
- tetrahedral (25% s character)
The type of hybridization directly impacts bond lengths and bond enthalpies:
sp-
hybridised carbons form shorter and stronger bonds compared to sp
3
-hybridized carbons due to the greater s character, leading to variations in bond lengths and energies."
Classification of Organic Compounds
Alkanes (Saturated Hydrocarbons): Single bonds,
C
n
H
2n+2
Example:
Methane (CH
₄
).
Alkenes (Unsaturated Hydrocarbons): Double bonds,
C
n
H
2n
.
Example:
Ethene (C
2
H
4
).
Alkynes (Unsaturated Hydrocarbons): Triple bonds,
CnH
2n-2
.
Example:
Ethyne (C
2
H
2
).
Aromatics: Based on the benzene ring. Example:
Benzene (C
6
H
6
).
Functional Groups
Alcohol:
-OH.
Example:
Methanol (CH
₃
OH).
Aldehyde:
-CHO.
Example:
Formaldehyde (CH
₂
O).
Ketone:
>C=O
(carbonyl group in middle of chain).
Example:
Propanone (CH
₃
COCH
₃
).
Carboxylic Acid:
-COOH.
Example:
Ethanoic Acid (CH
₃
COOH).
Amine:
-NH
₂
.
Example:
Methylamine (CH
₃
NH
₂
).
Ester:
-COOR
.
Example:
Methyl ethanoate (CH
₃
COOCH
₃
).
Haloalkane:
-X (X=F, Cl, Br, I).
Example:
Chloromethane (CH
₃
Cl).
Also Check –
Tungstic Acid formula
IUPAC Nomenclature
It refers to the systematic naming of organic compounds as recommended by the International Union of Pure and Applied Chemistry. The system ensures that every unique chemical structure has its distinct name, eliminating confusion. To break down an IUPAC name, let's use the compound:
4-ethyl-3-methylheptane
as an example.
Root Name (Parent Chain):
This is based on the number of carbons in the longest continuous chain.
In our example:
"heptane"
indicates a seven-carbon alkane chain.
Substituents (Branches):
These are groups that replace hydrogen atoms in the main chain.
In our example:
"ethyl" and "methyl"
are substituents.
Locants (Numbers):
These indicate the carbon positions of the substituents on the main chain. The chain is numbered such that the substituents get the lowest possible numbers.
In our example:
"4-ethyl"
means the ethyl group is on the fourth carbon, and "3-methyl" indicates the methyl group is on the third carbon.
Functional Group:
This is a group of atoms responsible for the characteristic reactions of a particular compound. In the given example, there isn't a functional group, but if there was, it would affect the suffix and possibly the numbering of the chain.
Examples include
"-ol"
for alcohols and
"-al"
for aldehydes.
Multiple Bonds (if present): Indicated by prefixes like
"ene"
for double bonds and
"yne"
for triple bonds. Not present in our example.
Also Check –
Charles Law Formula
Reaction Mechanism in Organic Reactions
Bond Fission:
It's the process of breaking a bond.
There are two types:
Homolytic Fission:
Each atom gets one electron from the covalent bond, leading to the formation of two free radicals.
Example: The breaking of
Cl-Cl
bond under UV light to produce
two Cl
·
radicals.
Cl−Cl → 2Cl
⋅
Heterolytic Fission:
One atom takes both electrons from the bond, producing a cation and an anion.
Example: In the reaction of HCl with water, the H-Cl bond breaks with chlorine retaining both electrons, forming
H
⁺
and Cl
⁻
.
H−Cl →H
+
+ Cl
–
Electrophiles (E
+
):
These are electron-poor species that seek electrons. They can be positively charged or neutral species with a positive center.
Example: In the reaction of bromine with ethene,
Br
₂
acts as an electrophile, attacking the double bond.
Br 2 +CH 2 =CH 2 →CH 2 Br−CH 2 Br
Nucleophiles (Nu
⁻
):
These are electron-rich species that donate electron pairs. They can be negatively charged or neutral species with a lone pair of electrons.
Example: In SN
2
reactions, a nucleophile like
OH
⁻
can attack an electrophilic
CH
3
CH
2
Br + OH
−
→ CH
3
CH
2
OH + Br
–
Electron Displacement:
Movement of electrons within or between molecules. There are main types:
Inductive Effect (I-effect):
Polarization of the sigma bond because of the electronegativity difference.
Example: In CH₃Cl, the C-Cl bond is polar due to chlorine being more electronegative. Polarization in CH₃Cl. Due to electronegativity, chlorine pulls electrons, making the C-Cl bond polar.
Resonance Effect (R-effect):
Delocalization of π electrons over the entire molecule. Example: In benzene, the π electrons are delocalized over all six carbon atoms, creating a resonance structure.
C
6
H
6
(showing alternating double bonds.)
Hyperconjugation:
Delocalization of sigma electrons with adjacent empty or partially filled p-orbitals or π-orbitals.
Example: In
propene (CH
₃
-CH=CH
₂
),
the sigma bond between the terminal carbon and its hydrogens can delocalize with the π-bond.
Electromeric Effect (E-effect):
A temporary effect where the pi electrons are shifted towards one atom in the presence of an attacking reagent.
Example: When HBr approaches propene, the pi electrons can shift towards a carbon, creating a temporary positive charge on the other.
CH
2
=CH−CH
3
+ HBr → CH
3
−CH(Br)−CH
3
Also Check –
Amino Acid Formula
Qualitative Analysis
Carbon and Hydrogen:
When an organic compound is heated in the presence of dry CuO, carbon is oxidized to CO₂ and hydrogen is oxidized to H₂O.
C + O
2
→ CO
2
2H+O
2
→H
2
O
Nitrogen:
When an organic compound is heated with concentrated sulfuric acid, the nitrogen present is converted to ammonium sulfate. By treating it with sodium hydroxide, ammonia gas is evolved.
(NH
4
)
2
SO
4
+ 2NaOH → Na
2
SO
4
+ 2H
2
O + 2NH
3
↑
Sulfur:
An organic compound containing sulfur, when heated with sodium, forms sodium sulfide. Adding acid will release H₂S gas.
2Na + S → Na
2
S
Na
2
S + 2HCl → 2NaCl + H
2
S ↑
Halogens:
Organic compounds with halogens, upon fusion with sodium, produce sodium halides which can be tested with silver nitrate for precipitate.
RX + 2AgNO
3
→ 2AgX↓ + RNO
3
(X = Cl, Br, I)
Phosphorus:
Organic compounds containing phosphorus, when heated with an oxidizing agent, produce phosphoric acid which can be detected by its reaction with ammonium molybdate.
H
3
PO
4
+12(NH
4
)
2
.MoO
4
+21HNO
3
→(NH
4
)
3
PO
4
.12MoO
3
+21NH
4
NO
3
+12H
2
O
Quantitative Analysis
Carbon and Hydrogen:
Combustion of an organic compound:
CxHy + (x+y/4)O
2
→ xCO
2
+ y/2H
2
O
Where the weight of CO
2
and H
2
O evolved helps determine the percentages of carbon and hydrogen.
Weight of CO
2
evolved x 12/44 = Weight of Carbon in the sample.
% Carbon = (Weight of Carbon/Weight of sample) x 100
For Hydrogen:
Weight of H
2
O evolved x 2/18 = Weight of Hydrogen in the sample.
% Hydrogen = (Weight of Hydrogen/Weight of sample) x 100
Nitrogen
Dumas Method:
2R − NH
2
+3O
2
→ 2N
2
+ 2R−H
2
O
The volume or weight of N
2
helps to calculate nitrogen's percentage.
% Nitrogen = (Volume of N2 evolved/22400) x (1.4 x 100/Weight of sample)
Kjeldahl's Method:
R−NH
2
+H
2
SO
4
→ (NH
4
)
2
SO
4
(NH
4
)
2
SO
4
+2NaOH → 2NH
3
+ Na
2
SO
4
+ 2H
2
O
The evolved NH
3
is absorbed in an acid and its amount is determined by titration.
Weight of NH
3
= Volume of acid x Normality of acid x 0.017
% Nitrogen = (Weight of NH
3
/Weight of sample) x 14 x 100
Halogens:
After digesting with nitric acid:
RX + AgNO
3
→ AgX↓+ R−NO
3
The precipitate of silver halide AgX (where X is Cl, Br, or I) is weighed to determine the percentage of the halogen.
% Halogen = ( At.wt of halogen/molecular wt of AgX )×(weight of AgX/wt of organic sample) ×100
Phosphorous:
Organic phosphorus is oxidized to H
3
PO
4
. This is then precipitated as ammonium phosphor molybdate.
H
3
PO
4
+12(NH
4
)
2
. MoO
4
+21HNO
3
→(NH
4
)
3
PO
4
.12MoO
3
+21NH
4
NO
3
+12H
2
O
The ammonium phosphomolybdate weight gives the percentage of phosphorus.
Weight of phosphorus = (Weight of ammonium phosphomolybdate x 3.1)/Weight of sample
% Phosphorus = Weight of phosphorus/Weight of sample x 100
Oxygen
Oxygen's percentage is typically determined by difference.
% Oxygen = 100 - % Carbon - % Hydrogen - % Nitrogen - % Halogens (and any other elements present).
