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Some Basic Concepts of Chemistry
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Atomic Structure
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Classification of Elements and Periodicity in Properties
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Chemical Bonding and Molecular Structure
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States of Matter
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Thermodynamics
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Chemical and Ionic Equilibrium
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Redox Reactions
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Hydrogen
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S – Block Elements (Alkali & Alkaline Earth Metals)
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p-Block Elements – Boron Family
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p-Block Elements – Carbon Family
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Organic Chemistry - Some Basic Principles and Techniques
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Hydrocarbons
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Environmental Chemistry
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Solid State
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Solutions
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Electrochemistry
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Chemical Kinetics
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Surface Chemistry
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General Principles & Processes of Isolation of Elements
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p-Block Elements (Group 15, 16, 17 and 18)
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d & f-Block Elements
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Coordination Compounds
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Haloalkanes and Haloarenes
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Alcohols, Phenols and Ethers
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Aldehydes , Ketones and Carboxylic Acids
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Organic Compounds Containing Nitrogen
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Biomolecules
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Polymers
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Chemistry in Everyday Life
Notes and Study Materials -Organic Chemistry – Some Basic Principles and Techniques
About this unit
Tetravalency of carbon; Shapes of simple molecules – hybridization (s and p). Classification of organic compounds based on functional groups: – C = C – , – C ? C – and those containing halogens, oxygen, nitrogen and sulphur; Homologous series. Isomerism: structural and stereoisomerism. Nomenclature (Trivial and IUPAC): Covalent bond fission – Homolytic and heterolytic: free radicals, carbocations and carbanions; stability of carbocations and free radicals,electrophiles and nucleophiles. Electronic displacement in a covalent bond: Inductive effect, electromeric effect, resonance and hyperconjugation Common types of organic reactions: Substitution, addition, elimination and rearrangement.
Table of Content
- Qualitative analysis of Organic Compound
- Classification and Nomenclature Of Organic Compound
- Hybridisation and Shapes of Organic Molecules
- Basic Concepts of Organic Chemistry
What is Qualitative Analysis?
Quantitative analysis is an analysis method used to determine the number of elements or molecules produced during a chemical reaction. Organic compounds are comprised of carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur and halogens. The various methods used for the measurement of percentage composition of elements in an organic compound are explained here.
Table of Content
- Detection of C and H
- Test for Phosphorous
- Liebig’s Combustion Method
- Carius Method
- Estimation of Sulphur
- Estimation of Phosphorus
- Estimation of Nitrogen by Dumas Method
- Estimation of Nitrogen by Kjeldahl Method
- Estimation of Oxygen by Aluise’s method
Qualitative analysis is the analysis of the species present in a given compound. For example, if a compound is taken, the qualitative analysis would be more focused on finding the elements and the ions present in the compound rather than study as to how much they are present.
Detection of C and H
C and H are detected by heating the compound with CuO in a dry test tube. They are oxidised to CO2 and H2O respectively. If the CO2 turns lime water milky, and H2O turns anhydrous CuSO4 blue, then the presence of C and H is confirmed.
Detection of Carbon and Hydrogen
Test for Phosphorous
The organic compound is heated with an oxidising agent to oxidise phosphorous to phosphate. The solution is then boiled with concentrated HNO3 and treated with ammonium molybdate. Yellow precipitate confirms the presence of phosphorous.
The reaction is given below,
Na3PO4 + 3HNO3 → H3PO4 + 3NaNO3
H3PO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O
Quantitative analysis is more towards finding out how much of the elements are present, that is, their amounts.
Estimation of C and H
Liebig’s Combustion Method
A known mass of the compound is heated with CuO. The carbon present is oxidised to CO2 and hydrogen to H2O. The CO2 is absorbed in KOH solution, while H2O is absorbed by anhydrous CaCl2 and they are weighed.
Percentage of C = 12/44 x ((Mass of CO2)/(Mass of compound)) x 100
Percentage of H = 2/18 x ((Mass of H2O)/(Mass of compound)) x 100
Liebig’s combustion method
Estimation of Halogens
Carius Method
A known mass of the compound is heated with Conc. HNO3 in the presence of AgNO3 in a hard glass tube called Carius tube. C and H are oxidised to CO2 and H2O. The halogen forms the corresponding AgX. It is filtered, dried and weighed.
Estimation of Halogens by Carius method
Percentage of X =
((Atomic mass of X)/(Molecular mass of AgX))x((Mass of AgX)/(Mass of the compound))x100
Calculations:
Let the mass of the given organic compound be m g.
Suppose the mass of AgX formed = m1 g.
We know that 1 mol of AgX consists of 1 mol of X.
So, in m1 g of AgX , mass of halogen =
Percentage of halogen =
Estimation of Sulphur
A known mass of the compound is heated with conc. HNO3 in the presence of BaCl2 solution in Carius tube. Sulphur is oxidised to H2SO4 and precipitated as BaSO4. It is then dried and weighed.
Percentage of S= ((Atomic mass of S)/(Molecular mass of BaSO4)) x (( Mass of BaSO4)/(Mass of the compound)) x 100
Calculations:
Suppose the mass of organic compound = mg
Let the mass of barium sulphate formed = m1 g
We know that 32 g sulphur is present in 1 mol of BaSO4
Therefore, 233 g BaSO4 contains 32 g sulphur:
⇒ M1 g of BaSO4 contains of sulphur
Estimation of Phosphorus
A known mass of the compound is heated with HNO3 in a Carius tube, which oxidises phosphorous to phosphoric acid. It is then precipitated as ammonium phosphomolybdate ((NH4)3PO4.12MoO3) by adding NH3 and ammonium molybdate ((NH4)2MoO4). It is filtered, dried and weighed.
Percentage of P=
((Atomic mass of P)/(Molecular mass of (NH4)3PO4.12MoO3)) x ((Mass of (NH4)3PO4.12MoO3)/(Mass of compound)) x 100
Estimation of Nitrogen
Estimation of Nitrogen by Dumas Method
A known mass of the compound is heated with CuO in an atmosphere of CO2, which yields free nitrogen along with CO2 and H2O.
CxHyNz + (2x+ 0.5y) CuO → xCO2 + 0.5y H2O + 0.5z (N2) + (2x+ 0.5y)Cu
The gases are passed over a hot copper gauze to convert trace amounts of nitrogen oxides to N2. The gaseous mixture is collected over a solution of KOH which absorbs CO2, and nitrogen is collected in the upper part of the graduated tube.
Estimation of Nitrogen by Dumas Method
Let the volume of N2 collected be V1 mL.
Then, volume of N2 at STP = (P1V1 x 273)/ (760 x T1) = V mL
Where P1 and V1 are the pressure and volume of N2.
P1= Atmospheric pressure – aqueous tension
22.4 L of N2 weighs 28 g,
Therefore, V ml of N2 weighs
=(28 x V)/22400 grams
Percentage of N would be,
= (28/22400) x (V/ Mass of compound) x 100
Estimation of Nitrogen by Kjeldahl Method
A known mass of an organic compound is (0.5 g) is mixed with K2SO4 (10 g) and CuSO4 (1.0 g) and conc.H2SO4 (25 mL), and heated in a Kjeldahl’s flask.
CuSO4 acts as a catalyst, while K2SO4 raises the boiling point of sulphuric acid. The nitrogen in the compound is quantitatively converted to (NH4)2SO4. The resulting mixture is reacted with an excess of NaOH solution and the NH3 so evolved, is passed into a known but the excess volume of standard acid.
The acid left unreacted is estimated by titration with some standard alkali. Thus the percentage of nitrogen can be calculated.
Estimation of Nitrogen by Kjeldahl Method
The reactions are given below:
- C + H + S → CO2 + H2O + SO2
- N → (NH4)2SO4
- (NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3 + 2H2O
- 2NH3 + H2SO4 → (NH4)2SO4
Calculation of the percentage of N
Let the mass of the organic compound be mg.
Volume of H2SO4 (Molarity M) = V mL
The volume of NaOH of molarity M used for titration excess of H2SO4= V1 mL
mEq of excess H2SO4 = mEq of NaOH
= MV1 mEq
Total mEq of H2SO4 taken = 2MV
mEq of H2SO4 used for neutralisation of NH3 = (2MV – MV1)
Therefore,
mEq of NH3 = (2MV-MV1)
1000 mEq or 1000 mL of NH3 solution contains = 17 g of NH3 (or) 14 g of N
Therefore,
(2MV-MV1) mEq of NH3 contains
=(14 x (2MV-MV1)) / 1000 g of N
Percentage of N= ((14x(2MV-MV1)) x (100/(1000xm)))
Estimation of Oxygen – Aluise’s method
A known mass of the compound is decomposed by heating it in the presence of N2 gas. The mixture of gases so produced is passed over red hot coke. This is done so that all the O2 is converted to CO. This mixture is heated with I2O5 in which CO is oxidised to CO2 liberating I2.
The reactions are given below,
Organic compound → Other gaseous products + O2
2C + O2 → 2CO
I2O5 + 5CO → 5CO2 + I2
Percentage of O = ((Molecular mass of O2/Molecular mass of CO2) x (Mass of CO2/Mass of the compound) x 100
CLASSIFICATION AND NOMENCLATURE OF ORGANIC COMPOUNDS
CLASSIFICATION OF ORGANIC COMPOUNDS
ACYCLIC OR OPEN CHAIN COMPOUNDS
CYCLIC OR CLOSED CHAIN COMPOUNDS
HOMOCYCLIC COMPOUNDS
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Alicyclic : The cyclic compounds resembling open chain aliphatic compounds. For example: Cycloalkanes
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Aromatic : The benzene, naphthalene and their derivatives etc are homocyclic aromatic compounds
HETEROCYCLIC COMPOUNDS
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Alicyclic
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Aromatic
CLASSIFICATION BASED ON FUNCTIONAL GROUPS
Class
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Functional group
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Halides
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X (Cl, Br, I) Halo
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Esters
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Olefins Alkenes
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>C = C<
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Acid halides
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Acetylenes/ Alkynes
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Anhydrides
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Alcohols
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—OH (Hydroxy)
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Amines
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—NH2
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Aldehydes
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|
Ketones
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Sulphonic acid
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—SO3H
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Acids
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Amides
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|
HOMOLOGOUS SERIES
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have the same general formula CnH2n+2 or CnH2n+1 X
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molecular weight differing by 14 of two successive members
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can be prepared by general methods of preparation
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have almost similar chemical properties
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show regular gradation in physical properties such as mpt, bpt, density etc
NOMENCLATURE
WORD ROOT
Chain length
|
Word root
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Chain length
|
Word root
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C1
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Meth-
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C7
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Hept-
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C2
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Eth-
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C8
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Oct-
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C3
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Prop-
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C9
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Non-
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C4
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But-
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C10
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Dec-
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C5
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Pent-
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C11
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Undec-
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C6
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Hex-
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C12
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Dodec-
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SUFFIX
Type of Carbon chain
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Primary Suffix
|
Generic name
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Saturated
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– ane
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Alkane
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Unsaturated with one C=C
|
– ene
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Alkene
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Unsaturated with one C≡C
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– yne
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Alkyne
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Functional Group
|
Secondary Suffix
|
Generic name
|
– OH
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– ol
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Alkanol
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– CHO
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– al
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Alkanal
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>C = O
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– one
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Alkanone
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– COOH
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– oic acid
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Alkanoic acid
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– COX
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– oyl halide
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Alkanoyl halide
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– CONH2
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– amide
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Alkanamide
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– COOR
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– alkyl — – oate
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Alkyl alkanoate
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– (CO)2O
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– oic anhydride
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Alkanoic anhydride
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– CN
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– nitrile
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Alkane nitrile
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– SH
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– thiol
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Alkanethiol
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– NH2
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– amine
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Alkanamine
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PREFIX
Primary prefix
|
Word root
|
Prim. suffix
|
Sec. suffix
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IUPAC name
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Cyclo
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hex
|
ane
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–
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Cyclohexane
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Substituent
|
Sec. prefix
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– X (F, Cl, Br, I)
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Halo
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– NO2
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Nitro
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– NO Nitroso
|
|
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Diazo
|
|
Alkoxy
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–R (CH3, C2H5, C3H7, etc)
|
Alkyl
|
-
Secondary prefix
-
Primary prefix
-
Word root
-
Primary suffix
-
Secondary suffix
ALKYL GROUPS
Alkane
|
Group
|
Shorthand notation
|
IUPAC Name
|
Methane
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Methyl CH3–
|
Me
|
Methyl
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Ethane
|
ethyl C2H5–
|
Et
|
Ethyl
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Propane
|
n-propyl CH3CH2CH2–
Isopropyl
|
n-Pr,
Iso-
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1-propyl
2-propyl
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Butane
|
n-butyl CH3CH2–CH2–CH2–
s-butyl
Iso-butyl (CH3)2CH–CH2–
t-butyl (CH3)3C–
|
–n-Bu,
s-Bu, or
Iso-Bu, or Bui
t-Bu, But
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1-butyl
1-methyl propyl
2-methyl propyl
1,1-dimethyl ethyl
|
LINE ANGLE FORMULA
NOMENCLATURE OF COMPLEX HYDROCARBONS
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Longest chain rule : The longest continuous chain of carbon atoms is picked up which forms the base name of the compound.
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Numbering : The longest chain is numbered by arabic numerals beginning with the end nearest a substituent.
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If two or more side chains are in equivalent positions, then the one cited first in the name is assigned the lower number.
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If two or more of the same alkyl groups are present, use the prefixes di, tri etc to avoid repetition
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Alphabetical order : The side chains are cited in alphabetical order
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Longest chain with maximum number of side chains : If two or more chains of the same length are possible, choose the one with maximum number of side chains.
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Locant sum : Sum of the locants must be minimum. But out of two sets of the sum of the locants, the set having the lowest number when compared by term is preferred. For example out of (2+6+7=15) and (3+4+7=14), the first set is correct.
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The name of a complex radical is considered to begin with the first letter of its complete name i.e. including the numerical affix (di, tri, tetra etc are numerical affix) for alphabetical order.
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When the side chains have the identical name the priority is given to side chain having lowest locant
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The numerical prefixes bis, tris tetrakis are used to indicate the multiplicity of substituted substituent
NOMENCLATURE OF COMPLEX ALKENES AND ALKYNES
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Selection of longest chain containing maximum number of double or triple bonds (sometimes longest chain rule is violated)
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If both, the double and triple bonds, are present the compound is regarded as derivative of alkyne. In such cases the terminal ‘e’ of -ene is dropped if it is followed by suffix starting with a,i,o,u,y. For example :
-
If double and triple bonds are at equidistant from either side, the preference is given to double bond.
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If the compound contains two or more double or triple bonds a terminal “a” is added to the word root.
-
The terminal ‘a’ is not added to the word root when the complete primary suffix do not start with a numerical affix
-
Side chains containing multiple bonds are named as follows
NOMENCLATURE OF CYCLOALKANES (ALICYCLIC COMPOUNDS)
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The base name is decided by the number of carbon atoms which the cyclic or acyclic portion contains. If the ring contains more or equal number of carbon atoms as alkyl then it is regarded as derivative of cycloalkane
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Carbons are numbered to give lowest numbers to substituted carbons. For example
-
When there are more acyclic than cyclic carbons the cyclic part becomes cycloalkyl substituent
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When acyclic portion contains a multiple bond or a functional group, the cyclic portion is treated as substituent.
-
In case when both contain the same functional group, the base name is decided by the number of c-atoms.
-
When both contain the different functional groups, the base name is decided by principal characteristic group
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When the acylic ring is directly attached to benzene ring, it is named as derivative of benzene
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Presence of certain groups
Functional gp.
|
Suffix
|
–COOH
|
Carboxylic acid
|
–COOR
|
Alkyl carboxylate
|
–COX
|
Carbonyl halide
|
–CONH2
|
Carboxamide
|
|
Carbonitrile
|
–CHO
|
Carbaldehyde
|
NOMENCLATURE OF POLYCYCLIC ALKANES
-
Fused rings
-
Bridged rings
-
Spirocyclic compounds
FUSED RINGS
BRIDGED RINGS
SPIROCYCLIC COMPOUNDS
NOMENCLATURE OF COMPOUNDS CONTAINING IDENTICAL CYCLIC UNITS JOINED BY A SINGLE BOND
No. of cyclic hydrocarbon units
|
Two
|
Three
|
Four
|
Prefix
|
bu
|
ter
|
quarter
|
NOMENCLATURE OF COMPOUNDS CONTAINING TERMINATING FUNCTIONAL GROUPS
NOMENCLATURE OF COMPOUNDS CONTAINING TWO OR MORE THAN TWO SIMILAR TERMINAL GROUPS
PRESENCE OF ONLY TWO SIMILAR TERMINAL GROUPS
PRESENCE OF MORE THAN TWO SIMILAR TERMINAL GROUPS ATTACHED TO THE MAIN PRINCIPAL CHAIN
Functional groups
|
Suffix
|
–COOH
|
– Carboxylic acid
|
–CHO
|
– Carbaldehyde
|
–COX
|
– Carbonylhalide
|
–CONH2
|
– Carboxamide
|
–COOR
|
– Alkyl carboxylate
|
–CN
|
– Carbonitrile
|
PRESENCE OF MORE THAN TWO SIMILAR TERMINAL GROUPS NOT DIRECTLY ATTACHED TO THE PRINCIPAL CHAIN
NOMENCLATURE OF COMPOUNDS CONTAINING SUBSTITUENTS (NOT REGARDED PRINCIPAL FUNCTIONAL GROUPS)
NOMENCLATURE OF COMPOUNDS CONTAINING MORE THAN ONE TYPE OF FUNCTIONAL GROUPS
Group
|
Prefix name
|
2º Suffix name
|
– SO3H
|
Sulpho
|
Sulphonic acid
|
– COOH
|
Carboxy
|
Oic acid
|
– COOR
|
alkoxy carbonyl
|
Oate
|
– COX
|
Halo carbonyl/Halo formyl
|
–oyl halide
|
– CONH2
|
Carbamoyl
|
amide
|
– CHO
|
Aldo or formyl
|
al
|
– CN
|
Cyano
|
nitrile
|
– NC
|
Isocyano
|
Isonitrile
|
>C = O
|
Keto or oxo
|
one
|
– OH
|
Hydroxy
|
ol
|
– SH
|
Mercapto
|
thiol
|
– NH2
|
Amino
|
Amine
|
– OR
|
Alkoxy
|
–
|
|
Epoxy
|
–
|
> C = C <
|
–
|
ene
|
|
–
|
yne
|
– N = N
|
Azo
|
–
|
– NO2
|
Nitro
|
–
|
– NO
|
Nitroso
|
–
|
– X (Cl, Br, I)
|
Halo (Cl, Br, I)
|
NOMENCLATURE OF AROMATIC COMPOUNDS
-
Nuclear substituted : The functional group is directly attached to the benzene nucleus e.g. phenol, toluene, chlorobenzene etc.
-
Side chain substituted : The functional group is present in the side chain e.g. Benzyl alcohol, Benzylamine etc.
AROMATIC HYDROCARBONS (ARENES)
ARYL GROUPS
HALOGEN DERIVATIVES
PHENOLS
AROMATIC ETHERS
AMINES
NITRO COMPOUNDS
ALDEHYDES
KETONES
ACIDS
ACID DERIVATIVES
SULPHONIC ACIDS
HYBRIDISATION AND SHAPES OF ORGANIC MOLECULES
HYBRIDISATION
HYBRIDISATION OF CARBON
sp3 HYBRIDISATION
Each hybrid orbital contains single electron, has 25% s character and 75% p character. They are directed towards the four corners of a regular tetrahedron with the carbon located in the centre. The angle between any two sp3 hybrid orbitals is 109º 28′ (109.5º).
sp2 HYBRIDISATION
BONDING IN ETHYLENE
sp HYBRIDISATION
BONDING IN ACETYLENE
HYBRIDISATION OF NITROGEN
sp2 HYBRIDISATION
sp HYBRIDISATION
HYBRIDISATION OF OXYGEN
sp3 HYBRIDISATION
sp2 HYBRIDISATION
BOND LENGTHS
BOND ANGLES IN SELECTED MOLECULES
AROMATICITY AND AROMATIC COMPOUNDS
ANTIAROMATICITY