UNIT 12 ORGANIC CHEMISTRY SOME BASIC PRINCIPLES
UNIT 12
ORGANIC CHEMISTRY
SOME BASIC PRINCIPLES
SYLLABUS
• Tetravalency of carbon
• Hybridisation
• (s and p ) bonds
• Shapes of simple molecules
• Functional groups : C=C , CC and functional groups containing halogen, oxygen, nitrogen and sulphur.
• Homologous series
• Isomerism(structural)
• General introduction to naming of organic compounds
• Trivial names and IUPAC nomenclature. Illustration with examples.
• Electronic displacement in a covalent bond
• Inductive effect
• Electromeric effect
• Resonance
• Hyperconjugation
• Fission of covalent bond
• Free radicals
• Electrophiles
• Nucleophiles
• Carbocation
• Carbanions
• Common types of organic reactions.
• Substitution reactions
• Addition reactions
• Elimination reactions
• Rearrangement reactions
• Illustrations with examples
ORIGIN OF ORGANIC CHEMISTRY
In early days of chemistry, compounds were classified into two types. Compounds derived from non-living matter like minerals and rocks were ‘inorganic compounds’ and those obtained from plant and animal sources were ‘organic compounds’. It was believed that a ‘vital force’ present in living system was essential for the synthesis of organic compounds. However, in 1828 Wholer synthesised the organic compound urea (present in urine) from inorganic compound ammonium cyanate by simple heating.
Lavoisier by qualitative and quantitative analysis of numerous organic compounds showed that organic compounds were made up of relatively few elements and all of them contained carbon.
Now thousands of organic compounds of plant and animal origin have been synthesised in the laboratory without the
help of ‘vital force’. Therefore as such there is no difference in inorganic and organic compounds. However, compounds such as carbonates , bicarbonates, carbon monoxide, carbon dioxide etc. though containing carbon are not included as organic compounds.
Organic chemistry is the chemistry of carbon compounds.
UNIQUE CHARACTER OF CARBON
The number of organic compounds is more than five million. This number is far more than the number of compounds of all other elements put together. The existence of such a large number of organic compounds is due to the following properties of carbon.
(i) Tetracovalency of carbon : The atomic number of carbon is 6. Its ground state electronic configuration is 1s22s22Px12Py12Pz0. Since , it is tetravalent in its compounds, there is promotion of one electron from filled 2s orbital to 2Pz empty orbital. The configuration of the atom in the excited state is 1s22s12Px12Py12Pz1 i.e., it has four valence electrons. The carbon atom is not in a position to either lose or gain four electrons to have a noble gas configuration. Thus, it has very little tendency to form ionic compounds. Rather it can share each of the valence electrons with the electrons of other atoms to complete its octet. For example, a carbon atom joins with four atoms to form four covalent bonds and a molecule of methane is formed.
In a similar manner, carbon can also complete its octet by sharing its valence electrons with electrons of other atoms. The characteristic of carbon atom by virtue of which it forms four covalent bonds is generally referred to as tetracovalency of carbon.
(ii) Catenation : One of the remarkable property of carbon atom is its unique capacity to link with other carbon atoms. This property of linking of atoms of one element with one another forming chains of identical atoms is called catenation. Carbon exhibits catenation to maximum extent because of strong carbon-carbon bond and tetracovalency. Due to the property of catenation carbon atom can form various types of straight chain, branched chain and ring structures, thus giving rise to a large variety of compounds.
3. Isomerism : The organic compounds exhibit isomerism, i.e., for a particular molecular formula two or more different organic compounds may be possible. The phenomenon is called isomerism. For example, corresponding to the molecular formula C4H10 , the compounds as shown below are possible. Compound I is called butane and compound II is called isobutane.
4. Formation of multiple bonds : Due to small size, carbon atom can easily form multiple bonds (double or triple bonds) with carbon, oxygen, nitrogen etc. For example,
5. Strong carbon-carbon bonds : Because of the small atomic radius of carbon atom (77 pm), the mutual overlap of its half-filled atomic orbitals is very effective and therefore, the bond formed are quite strong. Thus, the chains of carbon atoms normally do not break but atoms or groups such as H, Cl, OH etc attached to the carbon atom get replaced under suitable conditions. For example,
HYBRIDISATION AND SHAPES OF MOLECULES
An important aspect of organic chemistry is the understanding of fundamental concepts of molecular structure. This helps in understanding and prediction of the properties of the organic compounds. The tetravalency of carbon and formation of covalent bonds by it are explained in terms of electronic configuration and the hybridisation of s and p orbitals. The formation and shapes of molecules like methane (CH4), ethene (C2H4) , ethyne (C2H2) are explained in terms of the use of sp3 , sp2 and sp hybrid orbitals by carbon atoms in the respective molecules. The formation of sigma, pi and multiple bonds can be explained in terms of these hybridisation schemes. The formation of ethane having two carbon atoms can be easily explained in terms of sp3 hybrid orbital overlaps as shown in Fig.
C2H6 formation
(b) orbital overlaps between C(sp3) and C(Sp3) and H(1s) orbitals
(c) Line bond diagram
Hybridisation influences the bond length and bond enthalpy (strength) in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 orbital. The sp2 hybrid orbital is intermediate between sp and sp3 and hence , the length and enthlpy of bonds it forms , is also intermediate between them. The change in hybridisation is also associated with a change in electronegativity . The greater the s character of the hybrid orbitals, the greater electronegativity. Thus an sp hybridised carbon atom having hybrid orbitals with 50% s character is more electronegative than sp2 or sp3 hybridised carbon atoms. This relative electronegativity is reflected in several physical and chemical properties of the molecules concerned.
SOME CHARACTERISTIC FEATURES OF PI BONDS
In a bond formation, parallel orientation of two p orbitals on adjacent atoms is necessary for a proper sideways overlap. Thus in H2C=CH2 molecule all the atoms must be in the same plane. The p-orbitals should be parallel mutually and both perpendicular to the plane of the molecule. Rotation of one CH2 fragment with respect to the other interferes with maximum overlap of the p-orbitals and therefore , such rotation of carbon-carbon double bond ( C=C ) is restricted. Consequently, a disubstituted ethene like C2H2Cl2 can exist in two forms (I and II) shown below because of the restriction of rotation around the C=C bond. In I , called the cis-form , the Cl atoms are on the same side, while in II - the trans form, the Cl atoms are on opposite sides.
The electron charge cloud of the -bond is placed above and below the plane of bonding atoms. This results in the electron being easily available to attacking (oxidising) reagents. It is for this reason, that C2H4 readily reacts with oxidising agents like potassium permanganate or potassium dichromate at ordinary temperature, whereas C2H6 (which has only - electrons) is totally unaffected by these reagents at normal temperature. In general -bonds provide the most reactive centres in unsaturated molecules.
The rotation of one CH3 fragment against another in a C2H6 molecule does not affect the overlap in -bond. Consequently, free rotation around a -bond is permissible. The carbon-carbon -bond rotations also result in different arrangement of atoms in space giving different inter-converting forms of the molecule known as conformations.
Problems
1. How many and bonds are present in each of the following molecules ?
(a) C6H6 (b) C6H12 (c) CH2Cl2
(d) CH2=C=CH2 (e) HCONHCH3
2. What is the type of hybridisation of each carbon in the following compounds ?
(a) CH2=C=O (b) CH3CH=CH2
(c) (CH3)2CO (d) CH2=CHCN
(e) C6H6
3. What is the shape of the following compounds ?
(a) H2C=O (b) CH3F (c) HCN
STRUCTURAL REPRESENTATIONS OF ORGANIC MOLECULES
COMPLETE, CONDENSED AND BOND-LINE STRUCTURAL FORMULAE
The structure of organic compounds are represented in several ways. The Lewis structure is one of the ways. The Lewis structures can simplified by representing the two–electron covalent bond by a ‘stick’ (dash , ) . Such structural formula focuses on the electrons involved in bond formation , using a dash to represent a single bond, double dash for double bond and triple dash for triple bond. The lone pairs of electrons on heteroatoms ( e.g., oxygen, nitrogen, sulphur, halogens etc.) may or may not be shown. Thus ethane (C2H6) , ethene (C2H4) , ethyne (C2H2) and methanol (CH3OH) can be represented by the following structural formulae. Such structural representations are called complete structural formulae.
These structural formulae can be further abbreviated by omitting some or all of the covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula. Thus , ethane, ethene, ethyne and methanol can be written as :
CH3CH3 or C2H6 H2C=CH2 or C2H4
Ethane Ethene
HCCH or C2H2 CH3OH
Ethyne methanol
Similarly,
CH3 CH2 CH2 CH2 CH2 CH2 CH2 COOH can be further condensed to CH3(CH2)6COOH. For further simplification , bond-line structural representation can be used. In this representation, the carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are drawn in zig-zag fashion. The only atoms specifically written are those that are neither carbon nor hydrogen bound to carbon. The termini denote methyl (-CH3) groups (unless indicated otherwise by functional group), while the line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy the valency of the carbon atoms. Thus, CH3 CH2 CH2 CH2 CH2 CH2 CH2 COOH can be represented in bond-line formula as :
CH3CH2CH(CH3)CH2CH2CH2CH2CH3 is represented as :
There are many organic compounds in which carbon atoms are not joined in chain but are joined in rings. A compound containing one or more rings is called a cyclic compound , which is represented by drawing the appropriate ring (polygon) without showing carbon and hydrogen atoms. The corner of polygon represents a carbon atom and the sides of the polygon denote a carbon-carbon bond. If an atom or group of atoms other than hydrogen is attached to carbon, then the atom or group of atoms is shown in the structure. Examples of cyclic compounds are given below.
Problems
4. Expand the following condensed formulae into complete structural formulae.
(a) CH3(CH2)3OH (b) CH3CH2CO CH2CH3
(c) CH3CH=CH(CH2)3 CH3 (d) HOCH2CH2NH2
5. For each of the following compounds , write a more condensed formula and also their bond-line formula.
6. Expand each of the following bondline formulae to show all the atoms including carbon and hydrogen.
7. Draw the polygon formulae for molecular formula C5H10.
THREE-DIMENSIONAL REPRESENTATION OF ORGANIC MOLECULES
The three-dimensional (3-D) structure of organic molecules can be represented on paper by using certain conventions. For example , by using solid and dashed wedge formula , the 3-D image of a molecule from a two dimensional picture can be perceived. In these formulae, solid–wedge is used to indicate a bond projecting from the plane of paper towards the observer and the dashed-wedge is used to depict the bond going away from the plane of the paper. The bonds lying in the plane of paper are depicted using a normal line as shown in the Fig for CH4
Wedge-and-dash representation of CH4
The two carbon- hydrogen bonds shown by normal lines are in plane of the paper, while the carbon-hydrogen bond shown with a solid wedge is intended to be in front of this plane. The hydrogen connected to carbon by dashed line is behind this plane.
PROJECTION FORMULAE
In organic molecules, the 3-D arrangement of atoms in space can be represented in plane by projection formulae. These formulae are obtained by actually projecting the 3-D model of the molecule using a ray of light. The two dimensional image of the molecule is drawn on a paper according to certain conventions. While projecting the molecule it may be oriented in different ways with respect to plane of paper. There are three different types of projection formulae : Fischer projection, Newmann projection and Sawhorse projection formulae.
MOLECULAR MODELS
Molecular models are physical devices that are used for a better visualization and perception of three dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available. Commonly three types of molecular models are used :
• Frame work model
• Ball and stick model
• Space filling model
In the frame work model only the bonds connecting the atoms of a molecule and not the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of atoms. In ball and stick model , both the atoms and bonds are shown. Balls represent atoms and stick denotes a bond. Compounds containing C=C ( e.g., ethane) can best be represented by using springs in place of sticks. These models are referred to as ball and spring model. The space filling model emphasizes the relative size of each atom based on its van der Waal’s radius. Bonds are not shown in this model. It conveys the volume occupied by each atom in the molecule. In addition to these models , computer grahics can also be used for molecular modelling
CLASSIFICATION OF ORGANIC COMPOUNDS
The organic compounds are classified on the basis of structures as follows :
I. Acyclic or open chain compounds
These are also called as aliphatic compounds and consist of straight or branched chain compounds, for example ;
II Alicyclic or closed chain or ring compounds
Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring (homocyclic). Sometimes atoms other than carbon are also present in the ring (heterocyclic). Some examples of this type of compounds are :
These exhibit some of the properties similar to those of aliphatic compounds.
Aromatic compounds
Aromatic compounds are special types of compounds. These include benzene and other related ring compounds (benzanoid) . Like alicyclic compounds , aromatic compounds may also have hetero atom in the ring . Such compounds are called heterocyclic aromatic compounds. Some of the examples of various types of aromatic compounds are :
Benzanoid aromatic compounds
Non-benzanoid compound
Heterocyclic aromatic compounds
Organic compounds can also be classified on the basis of functional groups , into families or homologous series.
FUNCTIONAL GROUPS
A functional group is an atom or group of atoms in a molecule that gives a molecule its characteristic chemical properties
For example, the family of alcohols have its characteristic properties due to the presence of -OH (hydroxyl group), called functional group. Similarly, aldehydes owe their characteristic properties to the functional group -CHO (aldehyde) and so on.
We often use the symbol R to represent the hydrocarbon portion to which the functional group is attached. Thus R can be CH3 , CH3CH2 , (CH3)2CH or any other group of C and H atoms with one free valence by which the functional group is attached. The following table shows some functional groups and the corresponding classes of compounds.
Each functional group undergoes characteristic reactions. By recognising the functional group in a molecule, it is possible to predict the reactions which that molecule will undergo. The concept of functional group is important to organic chemistry for three reasons.
(i) Functional group serve as basis for nomenclature of organic compounds.
(ii) Functional groups serve to classify organic compounds into classes/ families. All compounds with the same functional group belong to the same class.
(iii) A functional group is a site of chemical reactivity in a molecule. Compounds in the same class have similar chemical properties.
A molecule can contain more than one functional group. It is then said to be polyfunctional, and the properties of each functional group may be modified by the presence of others.
HOMOLOGOUS SERIES
The organic compounds have been classified into various families or classes known as homologous series. A homologous series may be defined as : ‘ a series of similarly constituted compounds in which the members possesses the same functional group and have similar characteristics’ The two consecutive members of a series differ in their molecular formula by -CH2 group. The different members of a series are known as a homologue.
Characteristics of Homologous series
1. All members of a series can be represented by the same general formula. For example, general formula of alkanes is CnH2n +2 where n is the number of carbon atoms,
2. Any two consecutive members differ in their formula by a common difference of -CH2.
3. Different members in a series have a common functional group. For example, the members of the alcohol family have -OH group as their functional group.
4. The members of a particular family have almost identical chemical properties. Their physical properties such as melting point, boiling point, density, solubility etc. show regular gradation with increase in molecular mass.
5. The members of a particular series can be prepared almost by the identical methods known as general methods of preparation.
HYDROCARBONS
Organic compounds containing only carbon and hydrogen are called hydrocarbons. They occur in petroleum, coal and natural gas. Depending upon the structure of carbon skeleton present in the hydrocarbons, they can be classified as follows.
Hydrocarbons are broadly classified into two main classes.
(a) Open chain hydrocarbons
(b) Closed chain hydrocarbons.
Open chain hydrocarbons
These are called acyclic hydrocarbons. They contain open chain carbon atoms in their molecules. The chain may be straight or branched as shown below:
These compounds are also called aliphatic hydrocarbons. Aliphatic hydrocarbons are further divided into three classes ; alkanes, alkenes and alkynes. Alkanes are saturated hydrocarbons. They contain only carbon-carbon and carbon-hydrogen single bonds. Alkenes and alkynes are unsaturated hydrocarbons. They contain at least one carbon-carbon double bond or triple bond respectively.
Closed chain or Cyclic hydrocarbons
These hydrocarbons having closed chains or carbocyclic rings in their molecules. They are further divided into two categories.
(i) Alicyclic hydrocarbons : These compounds contain ring or closed chain of carbon atoms but they resemble with open chain hydrocarbons in many respects. Like open chain hydrocarbons, they may be further classified into cycloalkanes, cycloalkenes and cycloalkynes. Some examples are given below.
(ii) Aromatic hydrocarbons : These hydrocarbons are collectively called arenes. They contain one or more hexagonal carbocyclic rings. The name aromatic is derived from Greek word aroma meaning sweet smell, because most of the compounds belong to this class has sweet fragrance. One of the earliest aromatic hydrocarbon known is benzene. It is represented as follows:
Some other examples are :
NOMENCLATURE OF ORGANIC COMPOUNDS
In the case of aliphatic compounds, two systems are generally used for nomenclature. They are :
(a) Trivial system (b) IUPAC system
Trivial system
In early days , the organic compounds were named after the source from which they were obtained. For example, urea got its name because the compound was obtained from the urine of mammals. Similarly, methyl alcohol was called wood spirit, since it could be obtained as one of the products during destructive distillation of wood ; formic acid derived its name from Greek word formicus (red ants). These names were without any systematic basis and are known as common names or trivial names. Such system of nomenclature is known as trivial system.
IUPAC system
The system of naming has been improved from time to time by International Union of Pure and Applied Chemistry and the system is called IUPAC system of naming. The system of nomenclature is very useful in the study of organic compounds.
IUPAC SYSTEM OF NAMING ORGANIC COMPOUNDS
In IUPAC system, the name of the organic compound consists of three parts
(i) Word root (ii) Suffix (iii) Prefix
(i) Word root : The word root denotes the number of carbon atoms present in the chain. For example,
Word root for carbon chain lengths
Chain length Word root Chain length Word root
C1 meth- C6 hex
C2 eth- C7 hept
C3 prop C8 oct-
C4 but- C9 non-
C5 pent C10 dec-
ii) Suffix : The word root is linked to the suffix which may
be primary, secondary or both.
(a) Primary suffix : It indicates the nature of linkages in the carbon atoms. For example, if the carbon atoms are linked by single covalent bonds (C C), the primary suffix -ane is used. Similarly, for a double bond (C=C) between carbon atoms, the suffix -ene, while for tripple bonded carbon atoms (C C) , the suffix -yne is used.
(b) Secondary suffix : It indicates the presence of functional group present in the organic compound.
For example :
Functional group Suffix Functional group Suffix
Alcohol (OH) ol carboxylic acid (COOH) oic acid
Aldehyde
(CHO) al acid amide (CONH2) amide
Ketones (>C=O) one acid chloride (COCl) oylchloride
Nitriles (CN) nitrile esters
(COOR) oate
(iii) Prefix : There are many groups which are not functional groups in IUPAC name of a compound. These are regarded as substituents or side chains.and are represented as prefixes and are put before the word root, while naming a particular compound. For example,
substituent prefix substituent prefix
F fluoro NO nitroso
Cl chloro N=N diazo
Br bromo OCH3 methoxy
I iodo CH3 methyl
NO2 nitro CH2CH3 ethyl
NH2 amino CH2CH2CH3 npropyl
Thus the IUPAC name of an organic compound may be represented as :
Prefix + word root + primary suffix + secondary suffix
For example :
In this case,
Word root : But
Prefix : Chloro (at 3-position)
Primary suffix : ene ( chain contains C=C at position 1)
Secondary suffix : -ol ( for -OH group at position 1)
Therefore the name of the compound is :
3-Chlorobut-1-en-1-ol
Different classes of Organic compounds
1.HYDROCARBONS
These are organic compounds containing carbon and hydrogen atoms only. They may be further classified into two classes : saturated and unsaturated hydrocarbons.
Saturated Hydrocarbons - Alkanes
The general formula : CnH2n + 2 Suffix : -ane
These are organic compounds which contain only carbon-carbon single bonds. These were named as paraffins (Latin meaning little affinity) due to their least chemical reactivity. According to IUPAC system, these are named as alkanes (-ane is the suffix with word root).
Formula IUPAC Common name
CH4 methane methane
CH3CH3 ethane ethane
CH3CH2CH3 propane propane
CH3 CH2 CH2CH3 butane n-butane
CH3 CH2 CH2 CH2CH3 pentane n-pentane
CH3CH2 CH2 CH2 CH2CH3 hexane n-hexane
CH3CH2 CH2 CH2 CH2 CH2CH3 heptane n-heptane
CH3 CH2 CH2 CH2 CH2 CH2 CH2CH3 octane n-octane
CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2CH3 nonane n-nonane
CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2CH3 decane n-decane
The simple branched chain compounds are named according to common names. In the common system, all the isomeric alkanes (having same molecular formula) have the same parent name. The names of various isomers are distinguished by prefixes.
(i) Prefix n-(normal) is used for those alkanes in which all the carbon atoms form a continuous chain with no branching.
CH3CH2CH2CH3 CH3CH2CH2CH2CH3
n-Butane n-Pentane
(ii) Prefix iso- is used for those alkanes in which one methyl group is attached to the next to end carbon atom (second last) of the continuous chain.
(iii) Prefix neo- is used for those alkanes which have two methyl groups attached to the second carbon atom of the continuous chain.
Classification of Carbon atoms in alkanes
The carbon atoms in an alkane molecule may be classified into four types as primary(10), secondary(20), tertiary(30) and quaternary(40).
(i) A carbon atom attached to one (or no other) carbon atom is called primary carbon atom (written as 10 carbon ).
(ii) A carbon attached to two carbon atoms is called secondary carbon(written 20 carbon).
(iii) A carbon attached to three carbon atoms is called tertiary carbon(written as 30 carbon)
(iv) A carbon attached to four carbon atoms is called quarternary carbon(written as 40 carbon).
Hydrogen atoms attached to primary, secondary and tertiary carbon atoms are referred to as primary, secondary and tertiary hydrogen atoms respectively. The following example differentiates the various carbon atoms.
Alkyl Radicals
The alkanes are also represented by the general formula RH, where R is called alkyl group. The alkyl group therefore contains only one H-atom less than the alkane. The alkyl group is named by substituting the suffix ‘ane’ of the name of corresponding alkane by ‘yl’ (alkane ane + yl = alkyl ). For example,
CH3 CH3CH2 C3H7
methyl ethyl propyl
It may be noted that there is one methyl radical and one ethyl radical. However, there are two propyl radicals depending upon the nature of H atom removed. There are two types of carbon atoms in propane, the two end carbon atoms are primary(attached to only one carbon atom), while the middle carbon atom is secondary(attached to two other carbon atoms). The removal of hydrogen may occur as
(i) The six primary H atoms shown by dotted circle in the formula are equivalent (10). Removal of any one of these six hydrogen atoms gives an alkyl radical called normal group or simply as n-propyl.
(ii) Removal of one of the two secondary H-atoms shown in solid circles in the formula for propane gives different propyl group called iso-propyl group.
Similarly removal of one H from butane gives four butyl radicals (two from n-butane and two from isobutane as :
The prefixes sec- or tert- before the name of the group indicate that the H-atom was removed from a secondary or tertiary carbon atom respectively.
UNSATURATED HYDROCARBONS
These are hydrocarbons which contain carbon to carbon double bonds (>C=C<) or carbon to carbon triple bonds (C C ) in their molecules. These are called alkene and alkynes respectively. ALKENES General formula : CnH2n Suffix : ene In IUPAC system, the name of the alkene is derived by replacing suffix -ane of the corresponding alkane by -ene. For example, CH3CH3 Ethane (alkane) H2C=CH2 Ethene (alkene) In higher alkene, the position of the double bond is indicated by assigning numbers 1,2,3,4…. to the carbon atoms present in the molecule. The position is mentioned by the first carbon of the double bond. CH2=CHCH2CH3 CH3CH=CHCH2CH3 1-Butene 2-Pentene (not 3-Butene) (not 3-Pentene) Alkene IUPAC name Common name CH2=CH2 ethene ethylene CH3CH=CH2 propene propylene CH3CH2CH=CH2 1-Butene Butylene CH3CH=CHCH3 2-Butene Butylene CH3CH2CH2CH=CH2 1-Pentene Pentylene CH3CH2CH=CHCH3 2-Pentene Pentylene ALKYNES General formula : CnH2n 2 Suffix : yne Alkynes are named by replacing suffix -ane of alkane by -yne. In higher members the position of the triple bond is indicated by giving numbers 1,2,3,….. to the carbon atom in the molecule. The numbering of the chain is always done from end in such a manner that the triple bonded carbon atoms gets the least number as indicated in alkenes. For example CH3CCH CH3CH2CCH CH3CCCH3 Propyne 1-Butyne 2-Butyne (not 2-Propyne) (not 3-Butyne) Alkyne IUPAC name Common name CHCH ethyne acetylene CH3CCH propyne methylacetylene CH3CH2CCH 1-Butyne ethyl acetylene CH3CH2 CH2 C CH 1-Pentyne n-propyl acetylene CH3CH2 C C CH3 2-Pentyne ethylmethylacetylene ALKYL HALIDES (Halogen derivatives of alkanes) General formula : CnH2n +1 X Prefix : halo In IUPAC system, alkyl halides are named as halogen substituted alkane, i.e., haloalkanes. Thus the name of the halogen compound is given by prefixing, fluoro, chloro or iodo to the name of the alkane. The position of the halogen atom is also indicated giving the number of C atom to which the halogen is attached. Formula IUPAC name Common name CH3Cl Chloromethane methyl chloride CH3 CH2Cl Chloroethane ethyl chloride CH3CH(I)CH3 2-Iodopropane isopropyl iodide CH3CH(Br)CH2CH3 2-Bromobutane Sec-butylbromide CH3CH2CH2CH2 I 1- Iodobutane n-Butyl iodide MONOHYDRIC ALCOHOLS - ALKANOLS General formula : CnH2n +1 OH or ROH Functional group : -OH Suffix : -ol In IUPAC system, these are called alkanols and their names have been derived by changing -e of the corresponding alkane by -ol (alkane e + ol = alkanol ) Formula IUPAC name Common name CH3OH methanol methyl alcohol CH3 CH2OH ethanol ethyl alcohol CH3CH2CH2OH 1-propanol n-Propyl alcohol CH3CH2CH2CH2OH 1-Butanol n-Butyl alcohol CH3CH(OH)CH2CH3 2-Butanol Sec-Butyl alcohol ETHERS - ALKOXYALKANES General formula : CnH2n +1 O CnH2n +1 or R O R’ Functional group : O Prefix : alkoxy In an ether molecule, two alkyl radicals are linked on either side to a divalent oxygen( O). They are called simple ethers if R and R’ are the same (eg. CH3 O CH3) and mixed ethers when R and R’ are different (eg. CH3 O C2H5). In IUPAC system, ethers are called alkoxyalkanes. The alkoxy group is taken with smaller of the alkyl group. Formula IUPAC name Common name CH3 O CH3 methoxymethane Dimethyl ether CH3 O C2H5 methoxyethane ethylmethyl ether C2H5 O C2H5 ethoxyethane diethyl ether ALDEHYDES - ALKANALS General formula : CnH2n +1 CHO where n can be zero also eg. HCHO These are called alkanals in the IUPAC system of nomenclatrure and the names of individual members are derived by changing -e of the corresponding alkane by -al (alkane e + al = alkanal). In aldehydes the -CHO group is present at the end of the chain and there is no need to designate its position as 1. While counting the carbon atoms in the parent chain in CH3CH2CH2CHO corresponds to 4 C-atom and not 3. Formula IUPAC name Common name HCHO methanal formaldehyde CH3CHO ethanal acetaldehyde CH3CH2 CHO propanal propionaldehyde CH3CH2 CH2CHO butanal n-butyraldehyde KETONES - ALKANONE General formula : CnH2n +1 CO CnH2n +1 or RCOR’ The ketones are called simple ketones when R and R’ are the same. (e.g. CH3CO CH3 ) and mixed ketones when R and R’ are different ( e.g. CH3 CO C2H5 ) . In IUPAC system, ketones are called alkanones. The names of individual members are derived by replacing -e of the corresponding alkane by -one (alkane e + one = alkanone ) Formula IUPAC name Common name CH3 CO CH3 propanone dimethyl ketone CH3CO CH2CH3 2-Butanone ethylmethyl ketone CH3CH2 CO CH2 CH3 3-Pentanone diethyl ketone MONOCARBOXYLIC ACID - ALKANOIC ACID General formula : CnH2n +1 COOH (where n can be zero) In IUPAC system, they are called alkanoic acids and are named by replacing the terminal -e of the corresponding alkane by -oic acid (alkane e + oic acid = alkanoic acid). Formula IUPAC name Common name HCOOH methanoic acid formic acid CH3 COOH ethanoic acid acetic acid CH3CH2COOH propanoic acid propionic acid CH3CH2CH2COOH butanoic acid n-Butyric acid AMINES General formula : CnH2n +1 NH2 Functional group : NH2 : amino These are the functional derivatives of ammonia. In IUPAC system, these are named as alkanamines. The name is derived by replacing -e of alkane by amine. For example, CH3CH2 NH2 CH3CH2 CH2 NH2 ethanamine propanamine RULES FOR IUPAC NOMENCLATURE OF BRANCHED CHAIN ALKANES The following rules are used for naming the branched chain alkanes : (i) Select the longest continuous chain of carbon atoms in the molecule. The selected chain containing the maximum number of carbon atoms is regarded as the parent chain and it gives the name of the parent hydrocarbon. Carbon atoms which are not included in the parent chain are identified as substituents or branched chains. (ii) Number the carbon atoms in the parent chain as 1,2,3,…. etc. starting from the end which gives lower number to the carbon atoms carrying the substituents. In the above example, if X represents a substituent, then the correct number of the chain is given in structure A. The numbering of carbon chain as given in structure B is wrong as it gives higher number to carbon atom carrying the substituents. The number that indicates the position of the substituent or side chain is called locant. The position of the locant in the above structure is 2. In case , there are two or more substituents attached to the parent chain, then the end of parent chain which gives lowest sum of the locants is preferred for numbering. This is also called lowest sum rule. For example, the compounds given below should be numbered as indicated in structure A (substituted carbon atom is 2). For naming the compound, the position of the substituent or side chain is indicated by the number of carbon atom to which it is attached. The position number is written before the name of the alkyl group which is separated by using hyphens. For example: If two different substituents are in equivalent positions from the two ends of the chain, the lower number is given to one coming first in the alphabetical listing. Thus the following compound is 3-ethyl-6-methyloctane and not 6-ethyl-3-methyloctane. (iii) If the same substituent or side chain occurs more than once, the prefixes di-, tri-, tetra etc are attached to the names of the substituent. It may be noted that the positions of the substituents are indicated separately and the numericals representing their positions are indicated separately and the numericals representing their positions are separated by commas. For example, (iv) If two or more different substituents or side chains are present in the molecule, they are named in the alphabetical order along with their appropriate positions. It must be remembered that the prefixes di, tri- etc are ignored while comparing the substituents. (v) The branched alkyl groups can be named by following the above mentioned procedures. However, the carbon atom of the branch attaches to the root alkane is numbered 1 as exemplified below. The name of such branched chain alkyl group is placed in parenthesis while naming the compound. While writing the trivial names of substituents in alphetical order , the prefixes iso- and neo- are considered to be the part of the fundamental name of alkyl group. The prefixes sec – and tert- are not considered to be the part of the fundamental name. The use of iso and related common prefixes for naming alkyl groups is also allowed by the IUPAC nomenclature as long as these are not further substituted. In multisubstituted compounds, the following rules may also be remembered : • If there happens to be two chains of equal size, then that chain is to be selected which contains more number of side chains. • After selection of the chain, numbering is to be done from the end closer to the substituent. CYCLIC HYDROCARBONS The saturated hydrocarbons with ring of carbon atoms in the molecule are called cycloalkanes. These have the general formula CnH2n. The cyclic compound is named by prefixing cyclo to the name of the corresponding straight chain alkanes. For example, For simplicity, the cycloalkanes are also represented by simplified structural formulae without showing carbon and hydrogen atoms as shown above. Nomenclature of Alicyclic Compounds Following rules are observed while naming alicyclic compounds. 1. A saturated alicyclic compound is named by adding the prefix cyclo to the name of the corresponding straight chain alkane. For example, 2. If side chains are present, they are named and positioned exactly in the same way as open chain compounds. For example, 3. When an alicyclic compound contains a multiple bond or a functional group , the corresponding primary and secondary suffixes are attached to the word root and their positions are indicated by appropriate numbers. The numbering of carbon atoms present in the ring is done in such a way so as to assign the minimum possible number to the functional group or multiple bond . For example, 4. When a cyclic ring is attached to a chain containing greater number of carbon atoms or more than one cyclic rings are attached to a single chain, the compound is named as cycloalkylalkane. For example, However, if the chain attached to a cyclic ring contains lesser carbon atoms , the compound is named as a derivative of cycloalkane. For example, 5. When aromatic ring is attached to a cycloalkane ring , the compound is named as derivative of benzene. For example, RULES FOR NAMING COMPOUNDS CONTAINING MULTIPLE BONDS In naming the compounds containing double and triple bonds, the following rules are followed : (i) Select the longest continuous chain containing the carbon atoms involved in the multiple bonds (double and triple). This gives the parent name of alkene or alkyne. For example, in the structures given below, the longest chain has five carbon atoms. While writing the name of the alkene or alkyne, the suffix ‘ane’ of the corresponding alkane is replaced by ‘ene’ or ‘yne’ respectively. The selected chain may or may not be the longest chain in the structure. But it must contain double or triple bonded carbon atoms. For example, in the above structure the longest chain containing double bonded carbon atoms is of five atoms and not six. ii) The numbering of carbon atoms in the parent chain is done in such a way that the carbon atom carrying the multiple bond gets the lowest number. The position of the multiple bond is then indicated by using the number of the first C-atom of the multiple bond. Here we give lowest number to the carbon atom having double or triple bond and not to any side chain (as in alkanes) iii) All rules for naming side chains or substituents are then followed (as in alkanes). iv) If the multiple bond occurs two or more times in the chain, then it is named as diene or diyne or triene or triyne etc. The position of the double bonds are also prefixed with numbers. Examples : CH3CH2CH=CH2 1-Butene CH3CH2CCH 1-Butyne CH2=CHCH=CH2 1,3-Butadiene CHCCH2CCH 1,4-Pentadiyne If the parent chain in the hydrocarbon includes both ‘ene’ and ‘yne’, the former gets preference in numbering. However, the lowest sum rule must also be kept in mind. But the family will be of -yne. For example : CH C CH2CH=CH2 CHCCH=CHCH3 Pent1 en 4 yne Pent3 en 1 yne The name of the second compound cannot be Pent-2-en-4 -yne as in that case the sum of the locants would be 6 (=4 + 2 ). This is against lowest sum rule. It may be noted that if the word ‘ene’ or ‘yne’ comes in between , then ‘e’ of ‘ene’ or ‘yne’ should be dropped. RULES FOR NAMING ORGANIC COMPOUNDS CONTAINING A FUNCTIONAL GROUP (i) Select the largest continuous chain containing the carbon atom having the functional group. This is illustrated below: (ii) The numbering of atoms in the parent chain is done in such a way that carbon atom bearing the functional group gets the lowest number. (iii) All the rules for naming the side chains or substituents are then followed as in the case of alkanes If the functional groups such as -CHO, -COOH, -CONH2, -COOR, -CN etc are present in the molecule, numbering of the parent chain in such a case must start from carbon atom having the functional group. However, this number is normally not indicated in the IUPAC name of the compound. If a halogen atom is present in addition to functional group, the halogen is treated as substituent and is written as halo. For example, CH3COCH2CH2CH2Cl 5-Chloro-2-pentanone If two or more same functional groups are present, these are indicated by using di ( 2) , tri (3), tetra (4) as prefixes to the name of the functional group. Rules for Naming Compounds having functional groups, multiple bonds and side chain (or substituents) If the organic compound contains a functional group, multiple bond, side chain or the substituent, the following order of preference must be followed : Functional group > Double bond > Triple bond > Side chain.
For example,
CH3CH=CHCH2OH : But-2-en-1-ol
CH3CH=CHCH2CHO : Pent-3-enal
CH2=CHCH2COCH3 : Pent-4-en-2-one
CH3CH2CH=CHCH(CH3)CHO : 2-Methyl hex-3-enal
OHCCH2CH=CHCHO Pent-2-en-1,5-dial
HOOCCH=CHCOOH
But-2-en-1,4-dioic acid or But-2-enedioic acid
5. Rules for naming organic compounds containing two or more functional groups
When an organic compound contains two or more functional groups, one group is regarded as the principal functional group . The other group is treated as the secondary functional group and may be treated as substituent. The following order of preference is used for selecting the principal functional group :
Carboxylic acids > acid anhydrides > esters > amides > nitriles > aldehydes > ketones > alcohols > amines > ethers.
For example, if an organic group contains -COOH and -OH groups, then -COOH group is regarded as the principal group and -OH group is the substituent and is called hydroxy.
The names of secondary groups which are used as (substituents) prefixes are given below :
Functional group Prefix Functional group Prefix
-COOH Carboxy -CHO Formyl
-COOR Alkoxy carbonyl >CO oxo or keto
-COCl Chlorocarbonyl -OH hydroxy
-CONH2 Carbamoyl -SH Mercapto
--CN Cyano -NH2 Amino
-OR Alkoxy =NH imino
Some examples are :
CH3COCH2CH2COOH : 4-Oxopentanoic acid
(-COOH is the principal group)
CH3CH(NH2)CH2CH(OH)CH3 : 4-Aminopentan-2-ol
(-OH is the principal group)
CH3COCH2CHO : 3-Oxobutanal
(-CHO is the principal group)
Writing Structural formula From The Name of the Compound
For writing the structure of an organic compound from its name, the following procedure is adopted :
i) Write a straight chain of carbon atoms according to the number of carbon atoms in the parent compound. At this stage, do not attach any hydrogen atoms to complete the tetravalence of carbon atom as
ii) Number the chain of carbon atoms from any end
iii) In case , the name of the compound contains ene or -yne, then fill the double or triple bond at indicated positions.
iv) Attach substituents and functional groups at the desired positions.
v) Make each carbon atom tetravalent by attaching a suitable number of hydrogen atoms.
Problem
8. Write the structural formulae of the following compounds:
i) But-2-en-1-oic acid
ii) 2,4,4-Trimethyl-3-isopropylpent-1-ene.
9. Write the structural formulae of :
(a) 2-Chlohexane (b) Pent-4-en-2-ol
(c) 3-nitrocyclohexene (d) Cyclohex-2-en-1-ol
(e) 6-hydroxyheptanal
NOMENCLATURE OF SOME AROMATIC COMPOUNDS
For IUPAC nomenclature of substituted benzene compounds , the substituent is placed as prefix to the word benzene as shown in the following examples. However, common names (written in bracket below) of many substituted benzene compounds are also universally used.
If benzene ring is disubstituted , the position of the substituents is defined by numbering around the ring such that the substituents are located at the lowest numbers possible. For example, the following compound is named as 1,3-dibromobenzene and is not 1,5-dibromobenzene.
In the trivial system of nomenclature the terms ortho (-o) , meta (-m) and para(-p) are used as prefixes to indicate the relative positions 1,2- ; 1,3- and 1,4- respectively. Thus, 1,3-dibromobenzene is named as m-dibromobenzene (meta is abbreviated as m-) and other isomers of dibromobenzene 1,2- and 1,4- are named as ortho (or just o-) and para (or just p-)-dibromobenzene respectively.
For tri- or higher substituted derivatives, these prefixes cannot be used and the compounds are named by identifying substituent positions on the ring following the lowest locant rule. In some cases, common names of benzene derivatives is taken as the base compound.
The base compound substitent is numbered 1 and then the direction of numbering is chosen such that the next substituent gets lowest number. The substituents appear in the name in alphabetical order. Some examples are given below.
When a benzene ring is attached to an alkane with a functional group, it is considered as substituent, instead of a parent. The names for benzene as substituent is phenyl (C6H5- also abbreviated as Ph)
When functional group is present in the side-chain attached to the benzene ring, the compound is referred to as side-chain substituted compound. Many of these compounds are better known by their common names. In IUPAC system , they are named as phenyl derivatives of the corresponding aliphatic compounds. The positions of the substituents on the side chain as well as that on benzene ring are indicated by Arabic numerals. Some examples are as follows.
Aryl Groups
Problems
10. Write the structural formula of :
(a) o-Ethylanisole
(b) p-Nitroaniline
(c) 2,3-dibromo-1-phenylpentane
(d) 4-Ethyl-1-fluoro-2-nitrobenzene
11. Draw the structures of the functional groups corresponding to :
(i) aldehyde (ii) nitro (iii) carboxylic acid (iv) ether
12.
13.
14. Give the IUPAC names of :
(i) CH3CH(CH3)CH2CH(CH3)CH2CH3
(ii) CH3CH2C(Cl)(Br)CH2CH2CH3
(iii) CH3CH(CH3)CH2CHO
(iv) ICH2CH2COOH
(v) CH3CH(CH3)COCH2CH2Cl
15. Draw the structures of the following compounds.
(i) 3-Hexenoic acid
(ii) 3-Chloro-2-methylbutanol
(iii) 5,5-Diethyl-3-nonanol
(iv) 1-Bromo-3-chlorocyclohexene
(v) 4-Nitropentyne
(vi) 1,3-Diaminopropane
16. Give the IUPAC names of the following compounds
i) CH3CH(CH3)CH2C(CH3)3
ii) HOCH2CH2OH
iii) CH3CH2CH(C2H5)C(CH3)=CH2
iv) CH3CH(CH3)CH2OH
v) CH3C(OH)(CH3)C(OH)(CH3)2
vi) (CH3)3C Cl
vii) CH3CH2CH2NHCH3
viii) CH2OHCHOHCH2OH
ix) CH3CN
x) CH3CH(CH3)CH3
xi) (CH3)4C
xii) CH3CH2CH(CH3)CH2CH3
xiii) CH3CH2CH(CH3)CH2CH2CH2CH3
xiv) (CH3)3CCH2CH3
xv) CH3CH(CH3)CH2CH3
xvi) CH3CH(CH3)CH2CH2CH2CH3
xvii) CH3CH(CH3)CH(CH3)CH3
xviii) CH3CH(CH3)CH2CH(CH2CH3)CH2CH3
xix) CH3CH(CH3)CH(CH3)CH3
xx) CH3C(CH3)2CH(CH3)CH2CH3
17. Give the IUPAC name of :
i) CH2=CH2 vii) CH CH
ii) CH3CH=CH2 viii) CH3C CH
iii) CH3CH=CHCH3 ix) CH3CH2C CH
iv) (CH3) 2C=CH2 x) CH3C CCH3
v) CH3CH2CH=CHCH3 xi) CH3CH2CH2CH2C CH
vi) CH3CH2CH2CH=CH2 xii) CH3CH(CH3)CH=CHCH3
18. Give the IUPAC name of the following compounds
19. Write the structural formulae of the following
compounds.
a) But-2-en-1-oic acid
b) 5-Chloro-2-pentanone
20.
21. Give the structural formulae and IUPAC names of the following compounds.
a) Isobutyl bromide e) Ethyl bromide
b) Tertiary butyl chloride f) Benzyl chloride
c) o-Bromotoluene g) 4-Nitro-2-pentene
d) Isopropyl bromide
22. Write all possible structures for a compound having molecular formula C3H7Cl.
23. Write the IUPAC names of isomers of butyl chloride.
24. Write the structures of all possible isomers of CH3CH(Cl)(Br)
25. Write all isomers of C4H9Br
26. Give the structural formulae and IUPAC names of the following :
(i) aniline (iv) ethyl amine
(ii) n-propyl amine (v) n-Butyl amine
(iii) ethyl methyl amine (vi) diethyl amine
27. Give the IUPAC name of the following compound
28. Give the IUPAC name of the following :
29. Give the IUPAC names of following compounds :
30. Write bond line formula for :
i) isopropyl alcohol ii) 2,3-dimethylbutanal
iii) heptan-4-one
31. Give the IUPAC name of the following :
32. Draw formula for five members of each homologous series beginning with the following compounds :
(a) HCOOH (b) CH3COCH3 (c) CH2=CH2.
33. Give the condensed and bond line structural formula and identify the functional group(s) , if any for :
(a) 2,2,4-trimethylpentane
(b) 2-hydroxy-1,2,3-propanetricarboxylic acid
(c) hexane dial
34. Identify the functional groups in the following compounds.
ISOMERISM
The phenomenon of existence of two or more compounds possessing the same molecular formula but different properties is known as isomerism. Such compounds are individually referred to as isomers. The different types of isomerism is shown below.
Structural Isomerism
Compounds having the same molecular formula but different structures (manners of linking the atoms) are classified as structural isomers. Some typical examples of different types of structural isomerism are given below.
(i) Chain isomerism : When two or more compounds have similar molecular formula but different carbon skeletons, these are referred to as chain isomerism. For example, C5H12 represents three compounds :
(ii) Position isomerism : When two or more compounds differ in their substituent atom or group on the carbon skeleton, they are called position isomers and the phenomenon is termed as positioin isomerism. For example, the molecular formula C3H8O represents two alcohols.
(iii) Functional Isomerism : Two or more compounds having the same molecular formula but different functional groups are called functional group isomerism.
For example, the molecular formula C3H6O represents an adehyde and a ketone ;
and C3H8O represents an ether and alcohol.
(iv) Metamerism : It arises due to unequal distribution of alkyl groups on either side of the functional group in the molecule. For example, C4H10O represents methoxypropane (CH3OC3H7 ) and ethoxyethane (C2H5OC2H5)
(v) Tautomerism : It is a special type of functional isomerism in which the isomers exist simultaneously in dynamic equilibrium with each other. It may be defined as the phenomenon in which a single compound exists in two readily interconvertible forms which differ markedly in the relative position of at least one atomic nucleus , generally hydrogen.
Tautomerism is of various types. Among these , the keto-enol tautomerism is the most important. In keto-enol tautomerism, a compound exists in two interconvertible forms, one containing a keto group (known as keto form) and other containing an alcoholic group and an alkenyl group (known as enolic form). The two forms continuously change into each other through the oscillation of a proton and -electrons. For example, acetyl acetone shows keto-enol tautomerism as shown below.
Some more examples of keto-enol tautomerism are as follows.
STEREOISOMERS
The compounds that have the same constitution and sequence of covalent bonds but differ in the relative positions of their atoms or groups in space are called stereoisomers. Stereoisomerism is of two types :
• Geometrical isomerism
• Optical isomerism
Poblem
34. Write the structural isomers of molecular formula C3H6O.
SOME BASIC CONCEPTS IN ORGANIC CHEMISTRY
A organic reaction is supposed to take place when an attacking reagent attacks on a compound containing carbon (referred to as substrate) and leads to the formation of products. Thus, in general an organic reaction may be represented as given below.
Substrate + Attacking reagent Products
During the course of reaction, some of the covalent bonds present in the substrate molecule get broken and some new bonds are formed.
The organic reactions are of varied nature and may occur in several steps. When an attacking reagent attacks on a substrate molecule, some highly reactive species called reaction intermediates are formed on account of the change (breaking) of some of the covalent bonds present in the substrate. These reactive species immediately combine with similar species or molecules present around them and establish new bonds with them leading to the formation of products. The sequence of steps depicting the cleavage of old bonds in substrate molecules and formation of new bonds leading to the formation of products through reaction intermediates is called reaction mechanism.
The mechanism of an organic reaction involves several basic concepts. We shall study some important concepts which help a lot in understanding the mechanism of an organic reaction.
ELECTRON DISPLACEMENTS IN COVALENT BONDS
The behaviour of an organic compound is influenced to a large extent by electron displacements taking place in its covalent bonds. The electron availability in a particular bond or at a particular atom in a molecule is a dominant factor which determines the reactivity of the compound. These displacements may be of permanent nature in a molecule or may be temporary taking place only in the presence of another molecule (reactant). The acidity and basicity of organic compounds, their stability , their reactivity towards other substances etc. can be easily predicted by considering such electronic displacements. The electron availability depends upon five factors, These are:
i) Inductive effect
ii) Electromeric effect
iii) Resonance, mesomeric or conjugative effect
iv) Hyper conjugation effect
v) Steric effect
One or more these factors may operate in a molecule, in opposition or in support to each other. To understand the reactivity and physical properties of a molecule, it is, therefore, essential to determine the direction and magnitude of these effects qualitatively.
1. INDUCTIVE EFFECT
The permanent polarity produced in a molecule due to the displacement of the bonding electron pair towards the more electronegative atom of the pair C X , is called inductive effect or simply I-effect. When the substituent (X) bonded to the carbon is electron attracting (more electronegative) , it is said to exert a negative inductive effect, or I-effect. On the other hand, if the substituent bonded to carbon is electron releasing (less electronegative) then, it is said to exert a positive inductive effect , or + I-effect.
The inductive effect is transmitted from one carbon to another along the chain of carbon atoms and falls rapidly along the chain as one goes away from the substituent atom.
Let us consider chloroethane (CH3CH2Cl) in which C Cl bond is polar covalent bond. It is polarised in such a way that that the carbon - 1 gains some positive charge (+ ) and chlorine atom some negative charge ( ) (the fractional electronic charges on the two atoms in polar covalent bond are denoted by the symbol and different electron density within a covalent bond by arrow that points from + to end of the polar bond).
In turn carbon-1 , which has developed partial positive charge (+ ) draws some electron density towards it from the adjacent C C bond. Consequently, some positive charge (+) develops on carbon-2 also, where + symbolises relatively smaller positive charge as compared to that on carbon-1. In other words, the polar C Cl bond induces polarity in the adjacent bonds. Such polarisation of one - bond caused by the polarisation of adjacent - bond is referred to as inductive effect. This effect is passed on to the subsequent bonds also but the effect decreases rapidly as the number of intervening bonds increases and becomes vanishingly small after four bonds. The inductive effect is related to the ability of substituent(s) to either withdraw or donate electron density to the attached carbon atom. Based on this ability , the substituents can be classified as electron with-drawing or electron donating groups relative to hydrogen.
Some common atoms or groups which cause inductive effect (I effect) are :
(a) Electron donating groups (+I effect) groups
Methyl (-CH3), Ethyl (-C2H5) , Isopropyl [-CH(CH3)2 ], tert-butyl [ (CH3)3C-] . Tertiary alkyl groups exert greater +I effect than secondary, which in turn exerts a greater +I effect than primary alkyl group.
(b) Electron withdrawing groups ( or I effect)
Halogens (X= Cl, Br, I), Nitro (-NO2), Cyano (-CN), carboxy (-COOH), Ester (-COOR), alkoxy(-OR), Aryloxy (-OAr or –OC6H5)
2. ELECTROMERIC EFFECT
The temporary or time-variable effect involving the complete transfer of a shared pair of -electrons to one of the atoms joined by multiple bond, double or triple, at the requirement of an attacking reagent is known as electromeric effect. As soon as the attacking reagent is removed , the transferred -electron pair again forms the bond and the molecule reverts to its ground state electronic condition. For this reason, the electronic effect may be called polarisability of multiple bond. There is of course , permanent polarization of the sigma bond in the multiple bond, but this inductive effect is negligible with respect of the electromeric effect.
The complete transfer of the shared pair of the -electrons of a multiple bond to one of the shared atoms in the presence of an attacking reagent is called electromeric effect or E-ffect.
Due to the transfer of -electrons, the atoms involved in multiple bond formation acquire positive and negative charges. The atom which loses electrons acquires a +1 charge while the one which gains electrons acquires a 1 charge as shown below.
The electromeric effect is shown by curved arrow ( ) . It is of the following two types.
i) + E - effect : When the electrons of the -bond are transferred to that atom of the double bond to the reagent which the attacking reagent gets attached , the effect is called +E-effect . For example,
ii) E - effect
When the electrons of double bond are transferred to that atom of the double bond to which the attacking reagent is not attached, the effect is called E - effect. For example,
The electromeric effect facilitates the occurrence of a reaction because it comes into existence at the requirement of the attacking reagent. It is to be noted that if both inductive and electromeric effects are operative in the same molecule but in opposite direction, the electromeric effect usually dominates the inductive effect.
Application
The electromeric effect , i.e., the close approach of a reagent to a multiple bond enhances the reactivity of the reactant molecule and explains its addition product. Thus addition of ammonia to acetaldehyde may be cited. The greater electronegativity of O atom and + I effect of the methyl group in acetaldehyde help in transfer the -electrons on O atom at the requirement of ammonia molecule which co-ordinates with the double-bonded C atom.
Distinction between Inductive effect and Electromeric Effect
Inductive effect Electromeric effect
i) Permanent effect i) Temporary effect
ii) It is the permanent polarization of a single bond. ii) It is the polarizability of a multiple bond.
iii) -electrons are involved in this case. iii) -electrons are involved in this case.
iv) Electron displacement takes
place in inductive effect. iv) Complete transfer of -electrons takes place in electromeric effect.
v) It determines both the physical properties and chemical reactivity of the molecules concerned. v) It has nothing to do with the physical property , but it enhances the chemical reactivity of molecules concerned.
vi) Electronegativity of the bonded atoms controls the direction of the inductive effect. vi) The Effect has no specific direction ; electronegativity of the bonded atoms and I-effects of the attached groups may determine the direction of the electromeric effect. In fact, its direction is that which favours the reaction.
Problems
35. Which bond is more polar in the following pairs of molecules :
(a) H3CH , H3CBr (b) H3CNH2 , H3C OH
(c) H3C OH , H3CSH
36. In which CC bond of CH3CH2CH2Br , the inductive effect is
expected to be the least ?
3. RESONANCE STRUCTURES
There are many organic molecules whose behaviour cannot be explained by a single Lewis structure. An example is that of benzene. Its cyclic structure containing alternating CC single and C=C double bonds shown below is inadequate for explaining its characteristic properties.
As per the above representation benzene should exhibit two different bond lengths , due to C C single and C=C double bonds. However, as determined experimentally benzene has a uniform CC bond ldistances of 139 pm, a value intermediate between the CC single and C=C double bonds. Thus , the structure of benzene cannot be represented adequately by the above structure. Further, benzene can be represented equally well by energetically identical structures I and II.
Therefore , according to the resonance theory, the actual structure of benzene cannot be adequately represented by any of the structures, rather it is a resonance hybrid of the two structures (I and II) called resonance structures. The resonance structures (cannonical structures or contributing structures) are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability.
Another example of resonance is provided by nitromethane (CH3NO2) which can be represented by two Lewis structures, (I and II ). There are two types of N O bonds in these structures.
However, it is known that the two N O bonds of nitromethane are of the the same length (intermediate between a N O single bond and a N=O double bond). The actual structure of nitromethane is therefore a resonance hybrid of two cannonical forms I and II.
The energy of actual structure of the molecule (the resonance hybrid) is lower than that of any of the cannonical structures. The difference in energy between the actual structure and lowest energy resonance structure is called the resonance stabilisation energy or simply the resonance energy. The more the number of important contributing structures , the more is the resonance energy. Resonance is particularly important when contributing structures are equivalent in energy.
The following rules are applied while writing resonance structures.
The resonance structures have :
• The same positions of nuclei.
• The same number of unpaired electrons
• Among the resonance structures , the one which has more number of covalent bonds, less speration of opposite charges, a negative charge if any on more electronegative atom, a positive charge if any on more electropositive atom and more dispersal of charge is more stable than others.
Problems
37. Write the resonance structures of (a) CH3COO and
(b) C6H5NH2. Show the movement of electrons by curved arrows.
38. Write the resonance structures of CH2=CHCHO. Indicate
relative stability of the contributing structures.
39. Give reasons why the following two structures I and II cannot be the major contributors to the real structure of CH3COOCH3.
40. Which of the two ; O2NCH2CH2O or CH3CH2O is expected to be more stable and why ?
41. Draw the resonance structures of the following compounds. Show the electron shift using curve arrow notation.
(a) C6H5OH (b) C6H5NO2 (c) CH3CH=CHCHO
(d) C6H5CHO (e) C6H5+CH2 ( f) CH3CH=CH+CH2
iv) HYPERCONJUGATION
Hperconjugation describes the orbital interactions between the -systems and the adjacent -bonds of the substituent group(s) in organic compounds.
The hyperconjugation effect is due to the delocalisation of sigma C H (sp3-s) bonding electrons into the empty p-orbital of an adjacent positively charged carbon atom.
Hyperconjugation can be illustrated by taking the example of ethyl carbonium ion (CH3C+H2) . The positively charged C-atom of ethyl carbonium ion has an empty p-orbital.
Illustrating hyperconjugation in ethyl carbonium ion
When one of C H bonds of the methyl group aligns in the plane of this empty p-orbital, the C H bond electrons get delocalised into the empty p-orbital giving rise to hyperconjugation.
Hyperconjugation stabilises the cation (here ethyl carbonium ion) by dispersing the positive charge. In general greater the number of alkyl groups attached to a positively charged carbon atom, greater is the hyperconjugation interaction and stabilisation of the cation. Thus the relative stability of some carbocations follows the order,
(CH3)3C+ > (CH3)2+CH > +CH3
Hyper conjugation is also called no-bond resonance. Larger the number of hyperconjugation structures (no-bond resonance structures) that can be written for a species, greater is the stability of the species.
Hyperconjugation structures or no-bond resonance structures of ethyl carbonium ion
Hyperconjugation structures of ethyl carbonium ion are shown below.
Isopropyl carbonium ion gives six hyperconjgation structures and tert-butyl carbonium ion gives nine hyperconjugation structures.
Hyperconjugation structures of propene
Hyperconjugation is possible in alkenes and alkylarenes. The hyperconjugation structures of propene are shown below.
v) STERIC HINDRANCE
Two atoms or groups in an organic molecule at a distance less than or equal to the sum of the the van der Waal’s radii repel each other due to spatial crowding. This repulsion is referred to as steric hindrance ( steric strain or van der Waal’s strain). Molecules with steric strain are relatively less stable as compared to those having no steric strain. For example, steric strain is present in the cis-isomer of but-2-ene but not in the trans isomer as shown below.
Steric effect greatly influence the structure and reactivity of organic compounds.
Problem
42.
43. Explain why (CH3)3C+ is more stable than CH3C+H2 and
+CH3 is the least stable cation.
FISSION OF COVALENT BOND
Organic reactions are molecular in nature, and involve breaking and formation of one or more covalent bonds. The breaking process , also called bond cleavage or bond fission, takes place in two different ways.
(i) Homolytic bond fission (or bond cleavage)
(ii) Heterolytic bond fission (or bond cleavage)
HOMOLYTIC BOND FISSION
The bond-breaking process in which each atom or a group separates out with one electron of each of the bond–electron pair, is called homolytic fission (homo – same , lysis – to split). For example, in the reaction,
the shared pair of electrons is equally divided between two fragments formed due to homolytic fission.
The two fragments formed by the homolytic fission are electrically neutral, but each has one unpaired electron. The fragments obtained by homolytic cleavage are called free-radicals. Free radicals are very reactive.
Homolytic cleavage in benzoyl peroxide
Alkyl radicals (R•) are also classified as primary, secondary and tertiary and their stability parallels to that of alkyl cabocations. Thus, alkyl radical stability increases as we proceed from methyl to tertiary.
which can be explained on the basis of hyperconjugation. Organic reactions , which proceed by homolytic fission are called free radical or homopolar or nonpolar reactions.
HETEROLYTIC BOND FISSION
When a molecule splits up in a manner such that the bonded electron pair (the shared pair of electrons) goes with one of the two bonded atoms, then the fission is called heterolytic bond fission (hetero – different) or heterolytic bond cleavage. Heterolytic fission results in the formation of a cationic and anionic species . For example,
The charged species having positive charge on the carbon , viz., RC+H2 is called a carbocation (earlier called carbonium ion) and that having a negative charge , viz., RCH2 on carbon is known as carbanion. The positive atom has six electrons around it, while the negative atom will have eight electrons in its shell (with at least one lone pair of electrons).
Some carbocations are :
Characteristics of Carbocations and carbanions
• Carbocations are trigonal planar and positively charged carbon is sp2 hybridised. Carbocations are highly unstable and reactive species.
• Alkyl groups directly attached to the positively charged carbon stabilize the carbocation due to inductive and hyperconjugative effects.
• The relative stabilities of carbocation are :
Tertiary > secondary > primary > methyl
• The relative stabilities of carbanions are :
Methyl > primary > secondary > tertiary
NUCLEOPHILES AND ELECTROPHILES
Nucleophiles or Nucleophilic reagents
These are nucleus-loving or electron-donating species. These may be neutral molecules (:Nu) with a lone pair of electrons or negatively charged species (:Nu ).
Nucleophiles attack the substrate at a location where the electron density is low. For example,
Some typical nucleophiles are :
Electrophile or Electrophilic reagent
Electrophilic reagents or electrophiles (E or +E) are electron-loving or electron-seeking species. These always attack an electron-rich location in the molecule of a substrate, viz.,
Some typical electrophiles are :
Problem
44. Using curved arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
(a) CH3 SCH3 (b) CH3 CN (c) CH3 Cu
45. Giving justification , categorize the following molecules/ ions as
nucleophiles or electrophiles.
46. Identify the electrophilic centre in the following :
(a) CH3CH=O (b) CH3CN (c) CH3I
47. Identify the reagents shown in bold in the following equations
as nucleophiles or electrophiles.
(a) CH3COOH + HO CH3COO + H2O
(b) CH3COCH3 + CN (CH3)2C(CN)(OH)
(c) C6H6 + CH3C+O C6H5COCH3
CLASSIFICATION OF ORGANIC REACTIONS
Organic reactions can be classified into the following common reaction types.
(i) Substitution reactions
(ii) Addition reactions
(iii) Elimination reactions
(iv) Condensation reactions
(v) Rearrangement reactions
(vi) Isomerisation reactions
A. SUBSTITUTION REACTIONS
Reactions in which an atom or group of atoms is replaced by some other atom or another group of atoms without causing any change in the structure of the remaining part of the molecule are called substitution reactions. The entering species is termed as substituent. A substitution reaction may be represented as :
RX + Y R Y + X
Substituent Substituted
Product
Depending upon the nature of the substituent , the substitution reactions may be classified as follows:
(i) Nucleophilic substitution reaction (SN)
In these reactions, an atom or group of atoms in a molecule is replaced by a nucleophile. In nucleophilic substitution reaction, a weaker nucleophile is replaced by a stronger nucleophile. For example, hydrolysis of bromomethane by using aqueous KOH is a nucleophilic substitution reaction.
(ii) Electrophilic substitution reaction
Aromatic substitution reactions are examples of electrophilic substitution. For example, nitration of benzene is an electrophilic substitution reaction involving nitronium ion (+NO2) as an electrophile.
(iii) Free radical substitution reaction
In these reactions, an atom or group of atoms is replaced by a free radical. For example, reaction of chlorine with methane in the presence of light is a typical free radical substitution reaction.
and so on until all the hydrogens of methane molecule are replaced by chlorine. Such reactions are considered to proceed through chain mechanism.
B. ADDITION REACTIONS
Those organic reactions in which the reactants combine to form a single product having all the atoms of combining units are termed as addition reactions. For example, the following reaction of but-1-ene with hydrogen bromide giving 2-bromobutane is an example of an addition reaction.
The addition reaction may be electrophilic, nucleophilic or free radical types.
C. ELIMINATION REACTIONS
In an elimination reaction, two atoms or groups of a molecule are eliminated to give the product. Depending upon the relative positions of the atoms or groups eliminated , the reactions are classified as (alpha) , (beta) or (gamma) elimination reactions.
- Elimination
In - elimination , two atoms or groups are removed from the -position giving electron deficient reactive intermediates, which further react to give stable product(s).
-Elimination
In this two groups are removed from adjacent positions ( , ) . Elimination of HBr from 1-Bromobutane in the presence of alcoholic potassium hydroxide giving but-1-ene is an example of such elimination.
-Elimination
In such reactions the atoms or groups being removed are three bonds away, i.e., at and positions. It involves formation of three membered rings as shown below.
D. CONDENSATION REACTION
In these reactions , two different or same organic reactants unite to give a product with or without elimination of another simple molecule. For example, two molecules of acetaldehyde condense in presence of alkali yielding 3-hydroxybutanal (aldol condensation) , which loses a water molecule to give but-2-enal.
E. REARRANGEMENT REACTIONS
These reactions involve migration of a group from one atom to another in the same molecule. For instance, under acidic condition, 1,2-migration of a –CH3 group in 2,2-dimethylpropan-1-ol gives rearranged product 2-methylbut-2-ene.
E. ISOMERISATION REACTIONS
These are reactions effecting interconversion of isomers wherein the molecular formulae and carbon skeletons of reactant and product always remain the same. The interconversion of trans-but –2-ene into cis - but¬-2 -–ne is referred to as geometrical isomerisation reaction.
There are some other organic reactions , which do not involve ionic or free radical intermediates. These reactions occur in single step via a cyclic transition state. The bond breaking and making occurs simultaneously in these reactions. Such reactions are collectively called pericyclic reactions.
Problem
48. Classify the following transformations according to the reaction
type :
(a) CH3CH=CHCH3 + Br2 CH3CHBrCHBrCH3
(b) (CH3)2C=C(CH3)2+Br2 (CH3)2C=C(CH3)CH2Br +HBr
(c) CH2=CHCH2CH3 CH3CH=CHCH3
(d) C6H5CHO + CH3COCH3 C6H5CHOHCH2COCH3
(e) (CH3)3CCl + OH (CH3)2C=CH2 + H2O + Cl
49. Classify the following reactions in one of the reaction type :
(a) CH3CH2Br + HS CH3CH2SH
(b) (CH3) 2C=CH2 + HCl (CH3) 2CClCH3
(c) CH3CH2Br + HO CH2=CH2 + H2O
(d) (CH3) 3CCH2OH + HBr (CH3)2CBrCH2CH3
METHODS OF PURIFICATION OF ORGANIC COMPOUNDS
Organic compounds when isolated from natural sources or prepared by organic reactions are seldom pure; they are usually contaminated with small amounts of other compounds which are produced along with the desired product. Before carrying out the qualitative and quantitative analysis of the organic compounds, in order to characterise them it is very important to purify them. The various steps involved in the characterisation of organic compounds are :
• Purification of the compound
• Qualitative analysis for determining the elements present.
• Quantitative analysis of elements.
• Determination of molecular mass.
• Calculation of empirical formula.
• Elucidation of the structure by various methods including chemical methods.
50. What is the relationship between the members of the following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ?
51. For the following bond cleavages , use curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical , carbocation and carbanion.
PURIFICATION
Different methods are employed for the purification of a particular compound depending upon its nature and also on the nature of the impurities present. The various methods used for this purpose are :
i) Crystallisation
ii) Sublimation
iii) Differential extraction.
iv) Distillation
v) Fractional distillation.
vi) Distillation under reduced pressure.
vii) Steam distillation.
viii) Chromatography.
1. Crystallisation
This technique is used for separating or purifying a solid substance , which is soluble in some solvent (like water, alcohol, ether, benzene, chloroform, carbon tetrachloride etc). The efficiency of the process depends mainly upon the choice of the suitable solvent. A good solvent is one which dissolves a large amount of the substance at elevated temperature and deposits the same on cooling.
Fractional Crystallisation
In the case of the compounds the contaminated impurities which differ much in their solubilities, then their separation can be effected by fractional crystallisation. On cooling the hot saturated solution, the least soluble component separates out first ; leaving behind a liquor which will be rich in more soluble component. Thus, the first crop of the crystal will be richer in least soluble component ; whereas the last one will be richer in most soluble component present. A number of such repetitions will afford a pure sample of the compound.
2. SUBLIMATION
Certain solids undergo sublimation i.e., on heating they change directly from the solid to vapour state and the vapours on cooling change back into solid form.
This process is used for the separation of volatile solids, which sublime on heating from non-volatile solids. The impure substance is heated in a china dish covered with a perforated filter paper over which an inverted funnel is placed. The stem of the funnel is plugged with a little cotton. Vapours of the solid, which sublime, pass through the holes in the filter paper and condense on the cooler walls of the funnel. The non-volatile impurities are left behind in the dish.
Iodine, camphor naphthalene, benzoic acid etc. are purified by this method.
3. DIFFERENTIAL EXTRACTION
Organic compounds, whether solids or liquids can be removed from the aqueous solutions by shaking the solution in a separatory funnel with a suitable organic solvent which is immiscible with water, but in which the organic compound is highly soluble. This process is known as extraction or solvent extraction. Some commonly employed solvents are ether , benzene and chloroform. The process is carried as follows :
The funnel is stoppered and its contents shaken for some time when the organic solvent dissolves out the solute. The mixture is allowed to settle and in this way solvent and water form two separate layers. The lower aqueous layer is run out by opening the tap and the solvent layer is collected separately. The whole process may be repeated to remove the solute completely from the aqueous solution. The solute is finally recovered from the organic solvent by distilling off the latter.
4. DISTILLATION
This method is widely used for the purification of liquids from non-volatile impurities. It can also be used to separate liquids, if their boiling points are quite different (about 40 K) from each other .In distillation , impure liquid is heated in a flask (Fig 2). The more volatile component having a lower boiling point begins boiling sooner than the compound having a higher boiling point.
Simple distillation
The vapours of a substance formed are condensed and liquid is collected in conical flask
5. FRACTIONAL DISTILLATION
Simple distillation does not give a pure liquid, if the boiling points of the liquids in a mixture are very close to each other (difference less than 40 K). In such a case simple distillation gives a distillate which is not pure but contains both the liquids. Such mixtures can be purified by using fractional distillation.
Fractional distillation makes use of a fractionating column of suitable length and shape. A fractionating column provide a large surface area where vapours can condense and come into contact with incoming vapours.
For fractional distillation, a suitable fractionating column is placed between the flask and the condenser (Fig a). Fractionating columns are available in various sizes and designs as shown in Fig b.The mixture on heating forms vapours of both the higher and lower boiling liquids. As these vapours go up in the fractionating column, the vapours of the higher boiling liquid condenses as the column provides a surface that is cooler than its boiling point. The vapours of the lower boiling liquid remain in the vapour form and rise up in the column. Thus, the lower boiling liquid is collected as the distillate and higher boiling liquid is left in the flask.
Fig a Fractional Distillation
The vapours of lower boiling fraction reach the top of the column first followed by vapours of higher boiling fractions
Fig b Different types of fractionating columns.
6. DISTILLATION UNDER REDUCED PRESSURE
Many organic compounds decompose on heating even before they begin to boil. Such liquids cannot be purified by simple distillation. The pressure inside the flask is reduced with the help of a vaccum pump (Fig 4). Thus the liquid boils at much below its normal boiling point and distils over without undergoing decomposition.
Distillation under reduced pressure
A liquid boils at a temperature below its vapour pressure by reducing the pressure
7. STEAM DISTILLATION
The process of steam distillation is employed in the purification of a substance from non-volatile impurities provided the substance itself is volatile in steam and insoluble in water. This method is based on the fact that :
(i) A liquid boils at a temperature when its vapour pressure becomes
equal to the atmospheric pressure.
(ii) The vapour pressure of the mixture of two immicible liquids is equal to the sum of the vapour pressures of individual liquids.
In actual practice, the steam is continuously passed through the impure organic liquid. Steam heats the liquid but itself gets practically condensed to water. After some time, the mixture of the liquid and water begins to boil because the vapour pressure of the liquid mixture becomes equal to the atmospheric pressure. Obviously , this happens at a temperature which is lower than the boiling point of the substance or that of water. For instance, a mixture of aniline (b.p = 453 K with decomposition) and water (b.p = 373 K), under normal atmospheric pressure, boils at 371 K. At this temperature, the vapour pressure of water is equal to 760 mm. Thus in steam distillation, the liquid gets distilled at a temperature lower than its boiling point, any chances of decomposition are avoided. The apparatus set up for steam distillation is shown in Fig 5.
Steam distillation
Steam volatile component volatilizes, the vapours condense in the condener and liquid collects in conical flask
The impure liquid ( or solid along with some water ) is taken in a round bottomed flask, which is kept in a slightly slanting position. It is attached to a steam generator. Steam is passed into the flask which may be gently heated on a sand bath to prevent undue condensation of steam in it. The distillate consisting of water and the organic compound is collected in the receiver. The pure substance is separated from the water with the help of a separating funnel or by extraction with a suitable solvent.
CHROMATOGRAPHY
Chromatography is an important technique extensively used to separate mixtures into components , purify compounds and also to test the purity of compounds. The name chromatography is based on the Greek word chroma, for colour, since the method was first used for separation of coloured substances found in plants. In this technique the mixture of substances is applied on to a stationary phase , which may be a solid or a liquid. A pure solvent , a mixture of solvents or a gas is allowed move slowly over the stationary phase. The components of the mixture gets gradually separated from one another. Based on the principle involved , chromatography is classified into different categories. Two of these are :
i) Adsorption chromatography.
ii) Partition chromatography.
COLUMN CHROMATOGRAPHY
Column chromatography involves separation of a mixture into a long glass tube , known as the column, packed with an adsorbent (stationary phase). The column is fitted with a stopcock at its lower end (Fig).
The mixture dissolved in a minimum amount of solvent is dipped on the top of the adsorbent in the column. An appropriate solvent or a mixture of solvents is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed , partial or complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column (Fig)
THIN LAYER CHROMATOGRAPHY
Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent. A thin layer (about 0.2 mm thickness) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate. The solution of the mixture to be separated is applied as a small spot about 2 cm above one end of the TLC plate. The glass plate is then placed in a closed jar containing the solvent ( Fig a)
Fig a Thin layer chromatography
Chromatogram being developed
Fig (b) Developed chromatogram
As the solvent in the jar moves up the plate , the components of the mixture moves up along the plate to different distances depending on their degree of adsorption and separation takes place. The relative adsorption of each component of the mixture is expressed in terms of its retention factor i.e., Rf value.
The spots of coloured compounds are visible on TLC plate due to their original colour. The spots of the colourless compounds which are invisible to the eye but which fluoresce can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine. Spots of compounds , which absorb iodine , will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate. For example, amino acids may be detected by spraying the plate with ninhydrin solution.
PARTITION CHROMATOGRAPHY
Partition chromatography is based on continuous partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography , a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.
A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or mixture of solvents (fig).
Paper chromatography
Chromatography paper in two different shapes
This solvent acts as mobile phase. The solvent rises the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.
QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS
The qualitative analysis of an organic compound involves the detection of all elements present in it. Carbon and hydrogen are generally present in all the organic compounds. The other elements which may be present in the organic compounds are oxygen, nitrogen , sulphur , halogens , phosphorus etc.
1. DETECTION OF CARBON AND
HYDROGEN
A small amount of the dry and powdered substance is mixed with about double the amount of pure and dry copper oxide. The mixture is heated in a well dried glass tube provided with a delivery tube having a bulb in the centre, the other end of the delivery tube is dipped in a tube containing lime water . The bulb in the delivery tube is packed with glass wool containing anhydrous copper sulphate (white).
When the mixture is heated , the carbon present in the compound is oxidised to carbon dioxide which turns lime water milky. The hydrogen present in the organic compound is oxidised to water which turns anhydrous copper sulphate in the bulb blue.
C + 2 CuO CO2 + 2 Cu
CO2 + Ca(OH) 2 CaCO3 + H2O
lime water white ppt (milky)
2 H + CuO H2O + Cu
CuSO4 + 5 H2O CuSO4. 5 H2O
Anhydrous Hydrated copper sulphate
(white) (blue)
2. DETECTION OF NITROGEN - LASSAIGNE’S TEST
A small piece of a dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. Then a small amount of organic substance is heated strongly till it becomes red hot. The red hot tube is plunged into distilled water contained in a china dish and boiled for some time. It is then cooled and filtered. The filtered liquid is known as sodium fusion extract or Lassaigne’s extract.
The Lassaigne’s extract is usually alkaline. If not, it may be made alkaline by adding few drops of dilute sodium hydroxide. To a part of sodium fusion extract, a small amount of a freshly prepared ferrous sulphate solution is added and the contents are warmed. A few drops of ferric chloride solution are then added to the contents and the resulting solution is acidified with dilute hydrochloric acid. The appearance of a bluish green or a blue colouration confirms the presence of nitrogen in the organic compound. The following chemical reactions occur during the test.
Na + C + N NaCN
(from the organic compound)
FeSO4 + 2 NaCN Fe(CN)2 + Na2SO4
Fe(CN) 2 + 4 NaCN Na4[Fe(CN) 6]
sodium ferrocyanide
3 Na4[Fe(CN) 6] +4 FeCl3 Fe4[Fe(CN) 6]3 + 12 NaCl
Ferric ferrocyanide (blue)
Note
If the organic compound contains nitrogen and sulphur together, sodium thiocyanate may be formed in the sodium fusion extract which gives blood red colouration with ferric chloride due to the formation of ferric thiocyanate.
Na + C + N + S NaCNS
Sodium thiocyanate
3 NaCNS + FeCl3 Fe(CNS) 3 + 3 NaCl
Ferric thiocyanate (blood red )
3.DETECTION OF SULPHUR
Sulphur in the organic compound can be detected by Lassaigne’s test. Sodium fusion extract is prepared. The extract contains sodium sulphide formed by reaction between sulphur (present in the compound) and sodium.
2 Na + S Na2S
The lassaigne’s extract is divided into two parts and the following tests are performed.
i) Lead acetate test : One part of the extract is acidified with acetic acid and then lead acetate solution is added. Formation of a black precipitate confirm the presence of sulphur in the organic compound.
Na2S +Pb(CH3COO)2 2 CH3COONa + PbS (black)
ii) Sodium nitropruside test : A few drops of sodium nitropruside solution are added to another part of the lassaigne’s extract. The appearance of purple colour confirms the presence of sulphur.
Na2S + Na2[Fe(CN)5 NO] Na4[Fe(CN)5 NOS]
sodium fusin ext sodium nitropruside Purple colour
4. DETECTION OF HALOGENS
Halogens in an organic compound can be detected by the following tests :
i) Beilstein’s test
A clean copper wire is heated in the non-luminous flame till it does not impart blue colour to the flame. The hot end of the copper wire is touched with the organic substance and is again introduced into the flame. The reappearance of blue or green colour indicates the presence of halogens in the organic compound. This test is given by substances such as urea, thiourea etc. which do not contain halogens.
It does not tell us to which halogen is present.
ii) Lassaigne’s test
Sodium fusion extract is prepared. During fusion, sodium will combine with the halogens (from the organic compound) to form sodium halide.
The extract is acidified with nitric acid and boiled to expel the gases. It is then cooled and a small amount of silver nitrate is added. The formation of white or yellow precipitate confirms halogens.
(a) A white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound.
NaCl + AgNO3 AgCl + NaNO3
white ppt
(b) A dull yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound.
NaBr + AgNO3 AgBr + NaNO3
dull yellow ppt
(c) A bright yellow precipitate completely insoluble in ammonium hydroxide solution indicates the presence of iodine in the organic compound.
NaI+ AgNO3 AgI + NaNO3
yellow ppt
Function of Nitric acid
In case , nitrogen and sulphur are present with the halogens in the organic compound, the Lassaigne’s extract contains sodium sulphide(Na2S) and sodium cyanide (NaCN) along with sodium halide. These will form precipitates with silver nitrate solution. Therefore, nitric acid is added to decompose sodium cyanide and sodium sulphide.
NaCN + HNO3 HCN + NaNO3
Na2S + 2 HNO3 H2S + 2 NaNO3
Special Test For Bromine and Iodine
Acidify the Lassaigne’s extract with nitric acid and add few drops of chlorine water. Shake this solution with carbon disulphide or carbon tetra chloride. The liberated halogens ( bromine or iodine) will dissolve in it giving an orange or brown colour for bromine and violet colour for iodine.
2 NaBr + Cl2 2 NaCl + Br2
(Brown colour in CCl4)
2 NaI + Cl2 2 NaCl + I2
(violet colour in CCl4)
5. DETECTION OF PHOSPHORUS
A very few organic compounds contain phosphorus. Its presence is detected by fusing the compound with an oxidising mixture of sodium carbonate and potassium nitrate or with sodium peroxide alone, when alkali phosphate is formed. The fused mass is extracted with water and filtered. The filtrate containing sodium phosphate is heated with Con. HNO3 and an excess of ammonium molybdate solution is added. A yellow precipitate or colouration is formed if phosphorus is present . The yellow precipitate is ammonium phosphomolybdate (NH4)3[P Mo12O40] or (NH4)3PO4. 12 MoO3
QUANTITATIVE ANALYSIS
1. ESTIMATION OF CARBON AND HYDROGEN
Carbon and hydrogen in the organic compound are estimated together .
Principle
A known mass of the organic compound is heated with dry copper oxide in an atmosphere of air or oxygen free from moisture and carbon dioxide. The carbon and hydrogen of the organic compound are oxidised to carbon dioxide and water respectively.
CxHy + ( x + y/4 ) O2 x CO2 + ( y/2) H2O
Carbon dioxide produced is collected in potash bulb (containing KOH) whereas water is absorbed in calcium chloride tube (containing calcium chloride). The respective masses of carbon dioxide and water are determined by difference.
Knowing the mass of carbon dioxide and water vapours formed and mass of the compound taken, the percentage of carbon and hydrogen can be calculated. The apparatus used for the estimation of C and H is shown in Fig 8. It consists of :
(a) Arrangement for the supply of air or oxygen.
(b) Combustion tube.
(c) Absorption unit comprising CaCl2 tube for absorption of water and potash bulbs for absorption of carbon dioxide.
Calculations
Problems
52. 0.378 g of an organic compound gave on combustion 0.264 g of carbon dioxide and 0.162 g of water. Calculate the percentage of carbon and hydrogen.
53. On complete combustion , 0.246 g of an organic compound gave 0.198 g of carbon dioxide and 0.1014 g water. Determine the percentage composition of carbon and hydrogen in the compound.
2. ESTIMATION OF NITROGEN
There are two methods for the estimation of nitrogen. They are :
(i) Duma’s Method
Principle
A known mass of the organic compound is heated strongly with excess of copper oxide in an atmosphere of carbon dioxide. The carbon and hydrogen are oxidised to carbon dioxide and water respectively and the nitrogen is set free as dinitrogen. If any oxide of nitrogen is produced, it is reduced to dinitrogen by passing over hot reduced copper spiral. The dinitrogen is collected over concentrated solution of potassium hydroxide and its volume is measured at room temperature and atmospheric pressure.
The chemical reactions can be represented as :
CxHyNz + ( 2 x + y/2 ) CuO x CO2 + y/2 H2O + z / 2 N2
+ ( 2 x + y/2 ) Cu
The apparatus consists of :
a) Arrangement for the supply of carbon dioxide.
b) Combustion tube.
c) Schiff’s nitrometer - An arrangement for the collection of dinitrogen liberated from the organic compound. The KOH solution present in the nitrometer absorbs carbon dioxide and water produced during the combustion.
The apparatus used is shown in Fig 9.
Apparatus for estimation of nitrogen by Duma’s method
Calculations
Mass of organic compound = w g
Volume of moist nitrogen collected = V cm3
Barometric pressure = P mm
Room temperature = T K
Pressure of dry nitrogen = ( P p ) mm
p = aqueous tension
Step I :
To reduce the volume of nitrogen to STP
V1 = V cm3 V2 = ?
P1 = ( P p ) mm P2 = 760 mm
T1 = T T2 = 273 K
Step II : Calculation of percentage of nitrogen
Problems
54. 0.25 g of an organic compound gave 30 cm3 of moist nitrogen at 288 K and 745 mm pressure. Calculate the percentage of nitrogen (aqueous tension at 288 K = 12.7 mm ).
55. In Duma’s method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K and 715 mm pressure.
(ii) KJELDAHL’S METHOD
A known mass of the organic compound is heated with con. Sulphuric acid. The nitrogen in the organic compound is quantitatively converted into ammonium sulphate (Fig 9 )
Digestion (heating in Kjeldahl’s flask)
Kjeldahl’s distillation
The resulting liquid is distilled with excess of sodium hydroxide solution and the ammonia evolved is passed into a known , but excess volume of the standard acid (HCl or H2SO4 ). The acid left unused is estimated by titration with some standard alkali. The amount of acid used against ammonia can thus , be known and from this , the percentage of nitrogen in the compound can be calculated.
Reactions
Calculations
Kjeldahl’s method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.
Problems
56. 0.5 g of an organic compound was kjeldahlised and the ammonia obtained was passed through 100 cm3 of N/5 H2SO4. The excess of acid required 154 cm3 of N/10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.
ESTIMATION OF HALOGENS
Principle
A known mass of the organic substance containing halogen is heated with fuming nitric acid and few crystals of silver nitrate in a Carius tube. (Fig 10).
Heating Carius tube in a furnace for estimation of halogens
The silver halide so formed is separated , washed , dried and weighed. From the weight of silver halide obtained , the percentage of halogen can be calculated. Carbon and hydrogen are oxidised to carbon dioxide and water respectively.
Calculation
Problem
57. 0.40 g of an organic compound gave 0.30 g of silver bromide by Carius method. Find the percentage of bromine in the compound .
ESTIMATION OF SULPHUR
Principle
Estimation of sulphur in the organic compound is carried out by Carius method.
A known mass of the organic compound is heated with fuming nitric acid in a sealed tube. The sulphur present is quantitatively converted into sulphuric acid. Sulphuric acid thus formed is precipitated as barium sulphate by adding barium chloride solution. The precipitated barium sulphate is filtered, washed , dried and weighed. From the weight of barium sulphate formed, the percentage of sulphur can be calculated. The reactions involved can be represented as :
Calculations
Mass of organic substance taken = w g
Mass of barium sulphate formed = a g
ESTIMATION OF PHOSPHORUS
A known mass of the organic compound is heated with fuming nitric acid. The phosphorus present in the organic compound is oxidised to phosphoric acid (H3PO4) . The phosphoric acid thus formed is treated with magnesia mixture to get the precipitate of magnesium ammonium phosphate (MgNH4PO4). The precipitate is separated, dried and ignited to magnesium pyrophosphate (Mg2P2O7)
The chemical reactions involved are :
P + HNO3 H3PO4
(organic compound)
H3PO4 + Magnesia mixture MgNH4PO4
MgNH4PO4 Mg2P2O7 + 2 NH3 + H2O
Calculation
Mass of organic compound = w g
Mass of Mg2P2O7 obtained = a g
1 mole of Mg2P2O7( 222 g) contains 2 moles of phosphorus atoms (62 g).
ESTIMATION OF OXYGEN
The percentage of oxygen in an organic compound is usually found by difference between total percentage composition (100) and the sum of the percentage of all other elements. However, oxygen can also be estimated directly as follows:
A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine.
The percentage of oxygen can be derived from the amount of carbon dioxide or iodine produced.
Presently the estimation of elements in an organic compound is carried out by using micro-quantities of substances and automatic experimental techniques. The elements carbon , hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance ( 1 – 3 mg) and displays the values on a screen within a short time
Problems
58. 0.530 g of an organic compound gave 0.900 g of barium sulphate in Carius method. Calculate the percentage of sulphur.
59. 0.25 g of an organic compound containing phosphorus gave 0.111 g of Mg2P2O7 . Calculate the percentage of phosphorus in the compound.
60. 0.244 g of an organic compound containing carbon, hydrogen and oxygen was analysed by combustion method. The increase in the mass of calcium chloride tube and potash bulbs at the end of the operation was found to be 0.15 g and 0.1837 g respectively. Calculate the percentage composition of the compound.
QUESTIONS
1. Explain with an example, crystallisation.
2. What is fractional distillation.
3. Explain with an example, the use of ‘sublimation’ in the purification of organic compound.
4. Explain with an example ’simple distillation’.
5. Write a note on separation of a mixture by distillation.
6. Describe briefly the process that you will use to separate a mixture of two liquids having different boiling points.
7. Explain with an example, the use ‘fractional distillation’ in the purification of organic compounds.
8. How will you separate a mixture of two miscible liquids ?
9. Explain with an example ‘fractional distillation’.
10. Explain with an example, the use of distillation under reduced pressure in the purification of organic compound.
11. Explain the use of ‘steam distillation’ in the purification of organic compounds.
12. How would you separate a mixture of two organic compounds having different solubilities ?
13. Explain with an example, the use of differential extraction in the purification of organic compounds.
14. Explain with an example the ‘extraction with a solvent’.
15. Suggest a scheme for the separation of a mixture containing water and benzene.
16. Explain the use of column chromatography in the purification of organic compounds.
17. Explain with an example ‘chromatography’.
18. Describe an experiment which is used to separate substances using the chromatographic technique. Draw a diagram of apparatus.
19. When is the process of fractional distillation employed ?
20. How does fractional distillation differ from ordinary distillation ?
22. How can you detect the presence of carbon and hydrogen in an organic compound.
23. Describe the chemistry of Lassaigne’s test for detection of following in an organic compound (i) Nitrogen (ii) sulphur (iii) Chlorine
24. Will you get a precipitate , if you add silver nitrate solution to chloromethane ? If not , why ?
25. How can you estimate carbon and hydrogen in an organic compound ?
26. Describe Kjeldahl’s method for quantitative estimation of nitrogen in an organic compound.
27. Discuss Duma’s method for the estimation of nitrogen in an organic compound.
28. Describe Carius method for estimation of (i) Halogens (ii) Phosphorus
29. Describe Victor Meyer’s method for determining molecular mass of a volatile compound.
30. Describe volumetric method for determining the molecular mass of an acid (or a base)
31. 0.2613 g of an organic compound on combustion in oxygen gave 0.8844 g of carbon dioxide and 0.1809 g of water. Find the percentage of carbon and hydrogen in the substance.
33. Name stationary and moving phases in thin layer chromatography.
34. Write an essay on paper chromatography.
21. Explain with an example, crystallisation.
22. What is fractional distillation.
23. Explain with an example, the use of ‘sublimation’ in the purification of organic compound.
24. Explain with an example ’simple distillation’.
25. Write a note on separation of a mixture by distillation.
26. Describe briefly the process that you will use to separate a mixture of two liquids having different boiling points.
27. Explain with an example, the use ‘fractional distillation’ in the purification of organic compounds.
28. How will you separate a mixture of two miscible liquids ?
29. Explain with an example ‘fractional distillation’.
30. Explain with an example, the use of distillation under reduced pressure in the purification of organic compound.
31. Explain the use of ‘steam distillation’ in the purification of organic compounds.
32. How would you separate a mixture of two organic compounds having different solubilities ?
33. Explain with an example, the use of differential extraction in the purification of organic compounds.
34. Explain with an example the ‘extraction with a solvent’.
35. Suggest a scheme for the separation of a mixture containing water and benzene.
36. Explain the use of column chromatography in the purification of organic compounds.
37. Explain with an example ‘chromatography’.
38. Describe an experiment which is used to separate substances using the chromatographic technique. Draw a diagram of apparatus.
39. When is the process of fractional distillation employed ?
40. How does fractional distillation differ from ordinary distillation ?
41. Name the physical property of one of the constituent employed to separate it from other constituent employed to separate it from the other constituent in each of the following mixtures:
(i) Common salt + sand
(ii) Kerosene oil + water
(iii) Petrol + kerosene + diesel oil
(iv) Sulphur + powdered glass
(vi) Common salt + iodine
(v) Ammonium chloride + common salt
32. How can you detect the presence of carbon and hydrogen in an organic compound.
33. Describe the chemistry of Lassaigne’s test for detection of following in an organic compound (i) Nitrogen (ii) sulphur (iii) Chlorine
34. Will you get a precipitate , if you add silver nitrate solution to chloromethane ? If not , why ?
35. How can you estimate carbon and hydrogen in an organic compound ?
36. Describe Kjeldahl’s method for quantitative estimation of nitrogen in an organic compound.
37. Discuss Duma’s method for the estimation of nitrogen in an organic compound.
38. Describe Carius method for estimation of (i) Halogens (ii) Phosphorus
39. Describe Victor Meyer’s method for determining molecular mass of a volatile compound.
40. Describe volumetric method for determining the molecular mass of an acid (or a base)
41. 0.35 g of an organic compound Kjeldahlised and ammonia evolved was absorbed in 100 ml of N/5 sulphuric acid. The residual acid required 154 ml of N/10 NaOH for neutralisation. Calculate the percentage of nitrogen in the compound.
42. The reaction :
CH3CH2I + KOH(aq) CH3CH2OH + KI
is classified as :
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition
43. Which of the following is most stable ?
(a) (CH3)3C+CH2
(b) (CH3)C+
(c) CH3CH2+CH2
(d) CH3+CHCH2CH3
44. Will CCl4 give a white precipitate of AgCl on heating it with silver nitrate ? Give reason for your answer.
45. Exaplain why an organic liquid vapourises at a temperature below its boiling point in steam distillation ?
46. Name a suitable technique of separation of components from a mixture of calcium sulphate and camphor.
47. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen , sulphur and halogens.
48. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens ?
49. What is the difference between distillation, distillation under reduced pressure and steam distillation ?
50. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate ?