Notes and Study Materials -Organic Chemistry – Some Basic Principles and Techniques

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Classification and Nomenclature Reference Book
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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 Compounds

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

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

Estimation of Carbon and Hydrogen

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.

Carius method

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 = \frac {(atomic~mass~of~X~×~m_1~g)}{(molecular~mass~of~AgX)}

Percentage of halogen = \frac {(atomic~mass~of~X~×~m_1~×~100)}{(molecular~mass~of~AgX~×~m)}

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 = mg

We know that 32 g sulphur is present in 1 mol of BaSO4

Therefore, 233 g BaSO4 contains 32 g sulphur:

⇒   M1 g of BaSOcontains \frac {32~×~m_1}{233} g  of sulphur

Percentage of sulphur = \frac {32~×~m_1~×~100}{233~×~m}

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.

Dumas Method

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.

Kjeldahl Method

Estimation of Nitrogen by Kjeldahl Method

The reactions are given below:

  1. C + H + S → CO2 + H2O + SO2
  2. N → (NH4)2SO4
  3. (NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3 + 2H2O
  4. 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

The ability of carbon to combine with large number of elements especially O, N, S, X etc, to undergo catenation to form chains of varying lengths and shapes and existence of isomers has led to the formation of more than five million organic compounds. These have been classified into the following main groups

 

ACYCLIC OR OPEN CHAIN COMPOUNDS

CYCLIC OR CLOSED CHAIN COMPOUNDS

                     

HOMOCYCLIC COMPOUNDS

The ring system is made up of one type of atoms generally carbon
  • Alicyclic : The cyclic compounds resembling open chain aliphatic compounds. For example: Cycloalkanes
  • Aromatic : The benzene, naphthalene and their derivatives etc are homocyclic aromatic compounds           
              

HETEROCYCLIC COMPOUNDS

The ring system is made up of two or more than two types of atoms. They may be
  • Alicyclic
               
  • Aromatic
                        

CLASSIFICATION BASED ON FUNCTIONAL GROUPS

On the basis of functional groups which confer characteristic properties on them, the organic compounds have been classified as follows
Class
Functional group
Halides
X (Cl, Br, I) Halo
Esters
Olefins Alkenes
>C = C<
Acid halides
Acetylenes/ Alkynes
Anhydrides
Alcohols
—OH (Hydroxy)
Amines
—NH2
Aldehydes
Ketones                                   
Sulphonic acid
—SO3H
Acids
Amides

HOMOLOGOUS SERIES

A group of a particular class of compounds where a preceding or succeeding member differ by one–CH2. The members of the series are known as homologues. The homologues
  • have the same general formula CnH2n+2 or  CnH2n+1 X
  • molecular weight differing by 14 of two successive members
  • can be prepared by general methods of preparation
  • have almost similar chemical properties
  • show regular gradation in physical properties such as mpt, bpt, density etc

NOMENCLATURE

The most widely accepted and the latest system of naming organic compounds is IUPAC (International Union of Pure and Applied Chemists) system, according to which the name essentially consists of three parts.

WORD ROOT

It indicates the nature of the basic carbon skeleton. From C1 to C4 common names have been retained and from C5 upwards Greek number roots have been used
Chain length
Word root
Chain length
Word root
C1
Meth-
C7
Hept-
C2
Eth-
C8
Oct-
C3
Prop-
C9
Non-
C4
But-
C10
Dec-
C5
Pent-
C11
Undec-
C6
Hex-
C12
Dodec-
The generic word root for any carbon chain is “alk”.

SUFFIX

These are of two types

 

PRIMARY SUFFIX
It is added to the word root to designate saturation or unsaturation in a carbon chain
Type of Carbon chain
Primary Suffix
Generic name
Saturated
– ane
Alkane
Unsaturated with one C=C
– ene
Alkene
Unsaturated with one C≡C
– yne
Alkyne

 

SECONDARY SUFFIX
It is added to indicate the functional group present in the compound. The terminal ‘e’ is dropped, if secondary suffix begins with a vowel (a, e, i, o, u, y) but it is retained if the secondary suffix begins with a consonant.
Functional Group
Secondary Suffix
Generic name
– OH
– ol
Alkanol
– CHO
– al
Alkanal
>C = O
– one
Alkanone
– COOH
– oic acid
Alkanoic acid
– COX
– oyl halide
Alkanoyl halide
– CONH2
– amide
Alkanamide
– COOR
– alkyl — – oate
Alkyl alkanoate
– (CO)2O
– oic anhydride
Alkanoic anhydride
– CN
– nitrile
Alkane nitrile
– SH
– thiol
Alkanethiol
– NH2
– amine
Alkanamine

PREFIX

They are of two types

 

PRIMARY PREFIX
It  is for cyclic nature of the compound and primary prefix cyclo is used immediately before the word root. eg.:
Primary prefix
Word root
Prim. suffix
Sec. suffix
IUPAC name
Cyclo
hex
ane
–        
Cyclohexane
SECONDARY PREFIX
The certain atoms and groups which are not considered as functional groups but are treated as substituents are called secondary prefixes. They are added before the word root in case of acyclic compounds and before the primary prefix in case of cyclic compounds in alphabetical order.
The important secondary prefixes are
Substituent
Sec. prefix
– X (F, Cl, Br, I)
   Halo
– NO2
    Nitro
– NO Nitroso
 
                                 
Diazo
  • OR
Alkoxy
–R (CH3, C2H5, C3H7, etc)
Alkyl

 

Thus the complete IUPAC name of an organic compound consists of the following parts
  • Secondary prefix
  • Primary prefix
  • Word root
  • Primary suffix
  • Secondary suffix
For example:
IUPAC name is 4-Bromocyclohex-2-ene-1-ol   or 4-Bromo-2-Cyclohexenol
Primary prefix = Cyclo
Secondary prefix = 4-bromo
Word root = hex
Primary suffix = ene
Secondary suffix = ol

ALKYL GROUPS

Univalent groups formed by the removal of one hydrogen atom from an alkane are known as alkyl groups or alphyl groups. Their names are obtained by changing the suffix –ane of parent hydrocarbon by –yl.

 

Alkane
Group
Shorthand notation
IUPAC Name
Methane
Methyl CH3
Me
Methyl
Ethane
ethyl C2H5
Et
Ethyl
Propane
n-propyl  CH3CH2CH2
Isopropyl
n-Pr,   or Pr
Iso-
1-propyl
2-propyl
Butane
n-butyl  CH3CH2–CH2–CH2
s-butyl
Iso-butyl (CH3)2CH–CH2–  
t-butyl (CH3)3C–
–n-Bu, ,Bu
s-Bu,  or ,BuS    
Iso-Bu, or Bui
t-Bu, But
1-butyl
1-methyl propyl
2-methyl propyl
1,1-dimethyl ethyl

LINE ANGLE FORMULA

Bonds are represented by lines, carbon atoms are assumed to be present at the start and finish of a line. Nitrogen, oxygen and halogens are labelled, but hydrogens are only shown when bonded to a drawn atom. Each atom is assumed to have sufficient hydrogen atoms around it to make it neutral. For example:
n-Hexane CH3(CH2)4CH3
2-Cyclohexenone   
But-1, 3-diene                                  
3-methyl but-2-ol       

NOMENCLATURE OF COMPLEX HYDROCARBONS

The following rules are followed
  • Longest chain rule : The longest continuous chain of carbon atoms is picked up which forms the base name of the compound.
  • Numbering : The longest chain is numbered by arabic numerals beginning with the end nearest a substituent.
  • If two or more side chains are in equivalent positions, then the one cited first in the name is assigned the lower number.
  • If two or more of the same alkyl groups are present, use the prefixes di, tri etc to avoid repetition
  • Alphabetical order : The side chains are cited in alphabetical order
  • 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.
  • 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.
  • 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.
  • When the side chains have the identical name the priority is given to side chain having lowest locant
  • The numerical prefixes bis, tris tetrakis are used to indicate the multiplicity of substituted substituent

NOMENCLATURE OF COMPLEX ALKENES AND ALKYNES

  • Selection of longest chain containing maximum number of double or triple bonds (sometimes longest chain rule is violated)
It contains longest chain of 7C atoms, but both the double bonds are not included. Hence longest chain of 6C catoms is picked up)
  • 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.
  • 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
 
(Note that di, tri, tetra.. are numerical affix)
  • Side chains containing multiple bonds are named as follows
Allyl CH2= CH2=CH–CH2
Ethylidene CH3–CH= 
Vinyl CH2=CH–

NOMENCLATURE OF CYCLOALKANES (ALICYCLIC COMPOUNDS)

  • 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
  • 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
  • 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
  • When the acylic ring is directly attached to benzene ring, it is named as derivative of benzene
  • 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

There are three ways that rings can be joined.
  • Fused rings         
  • Bridged rings 
  • Spirocyclic compounds   

 

The carbon atoms common to both the rings are called bridge head atoms. The chain of carbon atoms connecting the bridge head atoms is called a bridge.

 

Numbering of C-atoms in fused rings and bridge rings : The numbering starts from bridge head carbon, proceeds along the longest bridge passing through the second bridge head atom, proceeds to the next longest bridge and completed along the shortest path.

 

Numbering of C-atoms in spiro compounds : The numbering starts from the carbon atom, next to spiro atom, present in the smaller ring giving minimum number to atoms containing functional groups.

FUSED RINGS

Fused rings share two adjacent carbon atoms and the bond between them eg. :

BRIDGED RINGS

These share two non adjacent carbon atoms (the bridge head carbons) and one or more carbon atoms between them
More examples of bridge Compounds

SPIROCYCLIC COMPOUNDS

The two rings share one carbon atom

 

More examples of spiro 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

 

Numbering of C-atoms : The numbering starts from the C-atom joining the rings.
     
   

NOMENCLATURE OF COMPOUNDS CONTAINING TERMINATING FUNCTIONAL GROUPS

If only one group such as –COOH, –CHO, –COOR, –CONH2, –COCl or – is present in the molecule it is always given number 1 and 1 is never written when there is no ambiguity.
2-Ethylhexanoic acid
  
2-Methylpropanamide
Methylpentanoate

NOMENCLATURE OF COMPOUNDS CONTAINING TWO OR MORE THAN TWO SIMILAR TERMINAL GROUPS

PRESENCE OF ONLY TWO SIMILAR TERMINAL GROUPS

The carbon atoms of such groups are included in the principal chain. For example

PRESENCE OF MORE THAN TWO SIMILAR TERMINAL GROUPS ATTACHED TO THE MAIN PRINCIPAL CHAIN

In this case special suffixes are used and carbons of terminal groups are not counted in the 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

In such case the longest chain with two similar terminal groups is selected and carbons of groups are counted in the principal chain.

NOMENCLATURE OF COMPOUNDS CONTAINING SUBSTITUENTS (NOT REGARDED PRINCIPAL FUNCTIONAL GROUPS)

                      
2-Bromo-3-Chlorobutane                     3- Nitromethylhexane
(follow alphabetical order)

NOMENCLATURE OF COMPOUNDS CONTAINING MORE THAN ONE TYPE OF FUNCTIONAL GROUPS

In such case the compound is regarded as derivative of senior functional group and the other functional groups are regarded as substituents. The numbering of the parent chain is done in such a way so that the functional group of highest priority gets the lower number and the chain contains the maximum number of functional groups.

 

The seniority of functional groups (highest priority) follow the following order:-
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)
 

 

The terminal e of the primary suffix is replaced by the suffix name of functional group.

 

Alphabetical order for substituents : These should be placed in alphabetical order.
Naming of substituted substituents : In this case the subsidiary substituents are named as prefixes. For example
 

NOMENCLATURE OF AROMATIC COMPOUNDS

Generally Benzene and its derivatives are known as aromatic compounds. They are of two types
  • 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.
In the first case the compounds are named as derivatives of benzene and in the second case as derivatives of aliphatic compounds (except arenes).
The IUPAC name of benzene is cyclohex-1,3,5-triene, but now aromatic compounds have their popular common name adopted by IUPAC. In IUPAC system the position of functional groups are indicated by arabic numerals i.e. 1, 2, 3 instead of o, m and p.

AROMATIC HYDROCARBONS (ARENES)

CONTAINING ONE RING
CONTAINING MORE THAN ONE RING

ARYL GROUPS

HALOGEN DERIVATIVES

PHENOLS

Nuclear substituted hydroxy derivatives are known as phenols

 

Side chain substituted hydroxy derivatives are known as alcohols

AROMATIC ETHERS

AMINES

    

NITRO COMPOUNDS

ALDEHYDES

KETONES

ACIDS

ACID DERIVATIVES

SULPHONIC ACIDS

 

HYBRIDISATION AND SHAPES OF ORGANIC MOLECULES

HYBRIDISATION

Sigma bonds are the most common bonds in organic chemistry. All single bonds are sigma (σ) bonds and formed by the overlapping between s-s, s-p and p-p (head on) atomic orbitals present on different atoms. A pi () bond results from the overlap of two p-orbitals that are oriented perpendicular to the axis of the nuclei. A p bond is not cylindrically symmetrical. A σ bond is stronger than p bond  due to better overlap. All multiple bonds contain one σ bond and others bond(s).

 

To have more efficient overlapping and to provide more symmetrical structure to the molecule the atomic orbitals on the same atom interact to provide hybrid atomic orbitals and the interaction is known as hybridisation. The hybrid atomic orbitals have enhanced electron density.

HYBRIDISATION OF CARBON

The ground state electronic configuration of carbon is . The electronic configuration of carbon in excited state is .

sp3 HYBRIDISATION

If we superimpose one s and three p atomic orbitals we get 4sp3 hybrid orbitals.

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).
These hybrid orbitals can overlap with four s atomic orbitals provided by four hydrogen atoms to form methane molecule.

sp2 HYBRIDISATION

If we superimpose one s and two p atomic orbitals we get 3sp2 hybrid orbitals

 

 
Each sp2 hybrid orbital has 33% s character and 67% p character. They lie in the same plane with their axis directed towards the corner of an equilateral triangle and are 120º apart from each other. The unhybridized pz atomic orbital is perpendicular to the plane of sp2 hybrid orbitals.

BONDING IN ETHYLENE

Consider two sp2 hybridised carbon atoms approaching to each other and four hydrogen atoms which provide four s atomic orbitals

sp HYBRIDISATION

If we superimpose one s and one p atomic orbitals we get 2sp hybrid orbitals.
 
Each sp hybrid orbital has 50% s character and 50% p character. They are diagonally present with their axis forming an angle of 180º. The unhybridized 2py and 2pz atomic orbitals are perpendicular to each other and perpendicular to hybrid orbitals also.

BONDING IN ACETYLENE

HYBRIDISATION OF NITROGEN

The ground state electronic configuration of nitrogen is
7N = 1s2, 2s2 2px1 2py1 pz1

 

One s and three p atomic orbitals superimpose and give 4sp3 hybrid orbitals. These are tetrahedrally present.

sp2 HYBRIDISATION

When nitrogen attaches itself to two other atoms it is present in the sp2 hybridised form. Consider the formation of methylimine CH2 = NH in which carbon and nitrogen both are in sp2 hybrid state

sp HYBRIDISATION

When nitrogen is attached to only one atom its hybridisation is sp. In  both carbon and nitrogen are in sp hybridised form

HYBRIDISATION OF OXYGEN

The electronic configuration of oxygen is .

sp3 HYBRIDISATION

When oxygen is attached to two atoms the hybridisation is sp3.
  

sp2 HYBRIDISATION

When oxygen is attached to one atom as in case of aldehydes and ketones e.g. in Formaldehyde carbon and oxygen, both are in sp2 hybrid form.

 

BOND LENGTHS

Some importants bond lengths are as follows
C–C sp3 – sp3 1.54 Å C–O sp3 – O 1.41 Å
sp3 – sp2 1.50 Å sp2 – O 1.34 Å
sp3 – sp 1.46 Å C=O sp2 – O 1.20 Å
sp2 – sp2 1.48 Å sp – O 1.16 Å
sp2 – sp 1.43 Å C–N sp3 – N 1.47 Å
sp – sp 1.38 Å sp2 – N 1.36 Å
C=C sp2 – sp2 1.34 Å C=N sp2 – N 1.28 Å
sp2 – sp    1.31 Å CºN sp – N  1.16 Å
sp – sp     1.28 Å
CºC sp – sp 1.21 Å
C–H sp3– H 1.11 Å
sp2 – H 1.10 Å
sp – H 1.08 Å

BOND ANGLES IN SELECTED MOLECULES

AROMATICITY AND AROMATIC COMPOUNDS

Aromatic indicates a stable system which undergoes substitution rather than addition, retaining the closed p-electron system. Many such systems contain only six p electrons, but generally they contain (4n+2) p electrons, where n is an integer.
Non-benzenoid heterocyclic compounds with 6electrons are aromatics
The hetero atom contributes to non bonded electrons, to complete the sextet.
In general, higher polycyclic aromatic compounds are somewhat less stable than benzene.
n = 0 cyclopropenyl cation contains 2– electrons and is aromatic

ANTIAROMATICITY

The less stability of monocyclic compounds containing (4n) electrons than their acyclic analogues is called anti aromaticity. For example
Cyclobutadiene    is less stable than 1,3-Butadiene
Here Resonance is the cause of destabilisation (hence the concept of antiaromaticity)
         
More examples of antiaromatic compounds
       
The electronic configuration of carbon in excited state is .

 

 

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