Organic chemistry-Some basic principles and techniques
CHAPTER ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES 12 LEARNING OBJECTIVES (i) Understand reasons for tetravalence of carbon and shapes of organic molecules. In organic compounds can be described in terms of orbitals hybridisation concept, according to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. (ii) Write structures of organic molecules in various ways. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula. (iii) Classify the organic compounds. Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds. (iv) Name the compounds according to IUPAC system of nomenclature and also derive their structures from the given names. The naming of the organic compounds is carried out by following a set of rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature, the names are correlated with the structure in such a way that the reader can deduce the structure from the name. (v) Understand the concept of organic reaction mechanism. Organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. (vi) Explain the influence of electronic displacements on structure and reactivity of organic compounds. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain carbon atom or other atom positions as places of low or high electron densities. (vii) Recognise the types of organic reactions. Organic reactions can be broadly classified into following types; substitution, addition, elimination and rearrangement reactions (viii) Learn the techniques of purification of organic compounds. Understand the principles involved in quantitative analysis of organic compounds. Purification, qualitative and quantitative analysis of organic compounds are carried out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties. Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography. Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in a pure form, its qualitative analysis is carried out for detection of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of percentages of all other elements present.
INTRODUCTION Organic chemistry is the study of the preparation, properties, identification, and reactions of those compounds of carbon not classified as inorganic. The latter include the oxide of carbon, the bicarbonates and carbonates of metal ions, the metal cyanides, and a handful of other compounds. There are several million known carbon compounds, and all but a very few are organic. Virtually all plastics, synthetic and natural fibers, dyes and drugs, insecticides and herbicides, ingredients in perfumes and flavouring agents and all petroleum products are organic compounds. All the foods you eat consist chiefly of organic compounds in the families of the carbohydrates, fats and oils, proteins, and vitamins. The substances that make up furs and feathers, hides and skins, and all cell membranes are also organic. Originally, the distinction between inorganic and organic substances was based on whether or not they were produced by living systems. For example, until the early nineteenth century, it was believed that organic compounds had some sort of “life force” and could only be synthesized by living organisms. The view was dispelled in 1828 when German chemist Friedrich Wohler prepared urea from the inorganic salt ammonium cyanate by simple heating : Gyaan Sankalp 1 Organic chemistry-Some basic principles and techniques
Heat NH4 OCN H 2 N C NH 2 Ammonium cyanate || O Urea Since urea is a component of urine, it is clearly an organic material, yet here was clear evidence that it could be produced in the laboratory as well as by living things.
STRUCTURE AND SHAPES OF ORGANIC MOLECULES : 1. Tetracovalency of carbon : The atomic number of carbon is 6 and it has four electrons in its valence shell. In order to acquire stable inert gas configuration, it can share its electrons with the electrons of other to form four covalent bonds. Thus, carbon has a covalency of four or is tetracovalent. In 1874, Vant Hoff and Le Bel predicted that the four bonds of carbon in methane and other saturated compounds do not lie in a plane but are directed towards the four corners of a regular tetrahedral.
109º28' C C
A Regular Space Model of The Normal Direction Tetrahedron Carbon Atom of Valencies with 4similar Faces
Figure : Tetrahedral representation of carbon valencies 2. Formation of organic molecules : We know that organic compounds are carbon compounds and carbon atom does not take part as such in bond formation but its 2s and 2p orbital (s) mixed together to form new orbitals known as hybrid orbitals and the process in turn is known as hybridization. The types of hybridizations encountered in organic compounds are sp3 (involved in saturated organic compounds containing only single covalent bonds, eg. methane), sp2 (involved in organic compounds having carbon linked by double bonds, e.g. ethylene) and sp (involved in organic compounds having carbon linked by a triple bond, e.g acetylenes). The bond angles and geometry associated with the three types of hybridization are summarized below :
Hybridisation sp3 sp 2 sp
Angle 109º28' 120º 180º
Geometry Tetrahedral Trigonal Linear
Example Alkane, Cycloalkane Alkenes Alkynes and all other and in saturated part and other compounds compounds containing of all organic molecules containing C=C, C=O, C C and C N triple bonds, C=N and C=S double bonds
Bond Four- Three- Two-
One- Two-
s% 25 33.3 50
p% 75 66.7 50
Electronegativity 2.48 2.75 3.25
2 Gyaan Sankalp Organic chemistry-Some basic principles and techniques 3. Types of bonds : Organic compounds normally contain only two types of carbon–carbon bonds, i.e, –(sigma) and –(pi) (i) Sigma bond : (a) A carbon–carbon, –bond can be formed by overlap of two sp3, sp2 or sp–hybridized orbitals of carbon atoms. Alkanes and cycloalkanes contain only sp3–sp3 C—C, –bonds, alkenes contains sp2–sp2, C—C –bonds while alkynes contain sp—sp, C—C, –bonds. (b) In a similar way, carbon can form sigma bonds with hydrogen atoms, while alkanes contain only sp3–s, C – H, –bonds, alkenes contain sp2–s, C–H, –bonds and alkynes contain sp–s, C–H, –bonds. (ii) –bond : (a) A pi–bond is formed by sideways or lateral overlap of two p–orbitals, Thus, alkenes contain one sp2 — sp2, C – C, –bond and one –bond. (b) In alkynes, the carbon atoms are sp–hybridized. Therefore, alkynes contain one sp–sp, C – C, –bond and two –bonds which are mutually perpendicular to each other. (c) The two carbon atoms and two hydrogen atoms of acetylene molecule lie along a line with C – C bond angle of 180º. Thus, acetylene is a linear molecule. 4. Effect of hybridization on bond length and bond strengths : The bond length and bond strength of any bond depends upon the size of the hybrid orbitals involved. (i) Bond lengths : Since a p–orbital is much bigger in size than a s–orbital of the same shell, therefore, as we go from sp3 sp2 sp, the percentage of p–character decreases from 75 66.7 50%. Accordingly, the size of the orbital decreases in the same order : sp3 > sp2 > sp. Since a bigger orbital forms a longer bond, therefore, C–C single bond lengths decrease in the order: C(sp3) – C(sp3) > C (sp2) – C(sp2) > C(sp) – C(sp) 1.54 Å 1.34Å 1.20 Å (ii) Bond strengths : Shorter the bond, greater is its strength. Thus, the -bond formed by sp–hybridized carbon is the strongest (i.e. maximum bond energy) while that formed by sp³–hybridized carbon is the weakest (i.e. minimum bond dissociation energy). For example, C(sp) – H > C(sp2)–H > C(sp3)–H 121 kcal mol–1 106 kcal mol–1 98.6 kcal mol–1 C(sp) – C(sp) > C(sp2) – C(sp2) > C(sp3) – C(sp3) 200 kcal mol–1 142 kcal mol–1 80-85 kcal mol–1 Since the extent of overlap in sideways overlap is low, a carbon–carbon –bond is always weaker than a carbon–carbon –bond. A carbon–carbon double bond is, however, stronger than a carbon–carbon single bond since it consists –bond and a weak – bond. In a similar way, a carbon–carbon triple bond is still stronger than carbon–carbon double bond. 5. Types of carbon and hydrogen atoms : There are four types of carbon atoms : (i) A primary (1°) carbon atom is bonded to either one more carbon atom or to more. (ii) A secondary (2°) carbon atom is bonded to two other carbon atoms. (iii) A tertiary (3°) carbon atom is bonded to three other carbon atoms. (iv) A quaternary (4°) carbon atom is bonded to four other carbon atoms. The 1°, 2°, 3° and 4° carbon and 1°, 2° and 3° hydrogen atoms are illustrated below :
1º CH3 1º| 2º 3º 1º 4º Ex : CHCCHCHCH3 2 3 | | CHCH3 3 1º 1º STRUCTURAL REPRESENTATION OF ORGANIC' COMPOUNDS : The structures of organic compounds can be represented in several ways. Connnonly used methods to describe the structure of any organic compound are : (a) Lewis formula (or Electronic formula) of a compound : The formula showing the mode of electron-sharing between different atoms in the molecule of a compound is called its electronic formula or Lewis formula. Writing of the Lewis formula is primarily based on the octet rule (with certain exceptions). To construct the Lewis formula for a molecule proceed as follows: (i) Determine the total number of available outer electrons in the atoms that make up that molecule, (ii) Arrange these electrons around the atoms so that each atom (except hydrogen) is surrounded by eight electrons (leave aside a few exceptions). The shared electrons are common to the two atoms. Hydrogen atom in any compound will have only two electrons.
Gyaan Sankalp 3 Organic chemistry-Some basic principles and techniques
CH4 : 6C contributes four electrons to bonding, each 1H contributes one electron H. H . H...... H C. H H : .C . : H H C H . H H H In acetylene (HCCH), for each carbon, two of the four bonding electrons are used in forming single bonds to one hydrogen and the other carbon. Two electrons are left over on each carbon, and these are shared to create a triple bond between the carbons.
6C Four electrons each A triple bond 1H One electron each
H...... H C. C. H H C. C. H H C C H . Acetylene H (b) Structural formula : The formula showing the spatial arrangement of various atoms present in a molecule of the compound is called its structural formula. In other words, the formula showing the way various atoms are linked to each other in a molecule of any compound is called its structural formula. In structural formula, single bonds are shown by single lines (–), double bonds are shown by double lines (=), and triple bonds by three lines (). Such structural formulae are also called complete structural formulae. Thus, structural formula of a compound is its electronic formula in which one pair of shared electrons ( or ) is . . . × replaced by a single bond, two pairs of the shared electrons (. . or . ×) by a double bond, and three pairs of the shared electrons . . . × ( . . or . × ) by a triple bond. . . . × (c) Condensed structural formula : Structures can be abbreviated by omitting some or all of the covalent bonds and indicating the number of identical groups attached to an atom by a subscript. Such a compact formula is called condensed structural formula or simply as condensed formula. Like methane can be represented as CH4, ethane as CH3 – CH3. (d) Bond line structural formula : Writing structural formulae for big complex organic molecule is both time and space consuming. Bond-line notation is a simple and convenient method of describing organic molecules. In this method. (i) Bonds are represented by lines. Thus, single line ( – ) represents a single bond, two parallel lines (=) represent a double bond, and three parallel lines () represent a triple bond. (ii) Carbon atoms are represented by the line-ends and line-intersections. The termini denote a methyl group (– CH3) unless indicated, otherwise by a functional group. (iii) Functional group or any substituent may be shown at the appropriate place by using its symbol or formula. (iv) Hydrogen atoms are assumed to be present in the required number so as to satisfy the tetravalency of carbon. The bond - line structure for common compounds : Methane Ethane Propane Butane Pentane Decane CH4 C2H6 C3H8 C4H10 C5H12 C10H22 – – Methylbutane Dimethylpropane Ethanoic acid Prpanoic acid C4H9.CH3 C3H6.(CH3)2 CH3 – COOH CH3 – CH2 – COOH O O
OH OH Cyclopropane Cyclobutane Cyclopentane Cyclohexane
Methylcyclobutane Cyclobutene Cyclobutyne CH3
4 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (e) Three-dimensional representation : Covalent bonds are directional. Thus, bonds in an organic compound are oriented at certain angles to each other. The three-dimensional (3-D) structures of organic molecules cannot be visualized from two-dimensional structures written on paper. However, structures of the three-dimensional organic molecules can be shown on paper by using solid and dashed wedge method of describing organic molecules. The following conventions are followed while writing solid and dashed wedge formulae. (i) The solid wedge ( ) is used to describe a bond projecting from the plane of the paper towards the observer.. (ii) The dashed wedge ( ) is used to describe a bond below the plane of the paper away from the observer.. (iii) The bond lying in the plane of the paper is shown by a normal line bond. The solid and dashed wedge formula of methane is : H solid wedge C H C H dashed wedge CLASSIFICATION OF ORGANIC COMPOUNDS :
Organic Compounds
Open chain Closed chain or or Acyclic Cyclic or or Aliphatic compound Ring compound
Saturated Unsaturated Homocyclic Heterocyclic
Alicyclic Aromatic
Saturated Unsaturated Benzenide Nonbenzenide Aliphatic or open chain compounds : Those compounds in which first and last carbon atoms are not connected with each other. Branched or unbranched chains are possible in these compounds. For example : C | C – C – C – C C – C – C – C C – C – C | | C C
(unbranched) (branched) There are two varieties in these compounds- Saturated hydrocarbons : (a) In such type, adjacent carbons are attached with single bonds. Ex. CH3 – CH2 – CH3 (b) General formula of these compounds is CnH2n+2 (alkane) (c) These are also called as paraffins (Parum + Affinis i.e. little reactivity) because these are less reactive due to absence of – bonds.
Gyaan Sankalp 5 Organic chemistry-Some basic principles and techniques Unsaturated Hydrocarbons : (a) There will be a double bond or a triple bond between any two carbons atoms. CH2 = CH – CH3 , CH C – CH3 Propene Propyne (b) General formula is CnH2n and CnH2n–2. (c) Alkenes are also called as olefins because they reacts with halogens to form oily substances olefins (Oleum + fines i.e. Oil forming). (d) Due to presence of bonds these are more reactive. Cyclic or closed compound : In these compounds first and last carbons are attached with each other.
Ex. cyclopropane. These are of two types Homocyclic compounds : These are the compounds in which the complete ring is formed by carbon atoms only. These are also of two types. (a) Alicyclic compounds : These are the compounds having the properties like aliphatic compounds. These may be saturated or unsaturated like aliphatic compounds.
Cyclopropane, Cyclopropene Cyclobutene (b) Aromatic compounds : These compounds consist of at least one benzene ring i.e. a six-membered carbocyclic ring having alternate single and double bonds. These compounds have some fragrant odour and hence, named as aromatic (greek word aroma means sweet smell) CH3 CH CH2 Ex.
Benzene Toluene Styrene Heterocyclic compounds : These are cyclic compounds having ring or rings built up of more than one kind of atoms. (i) Aromatic hetero cyclic compound :
O S N | H Furane Thiophene Pyrrole (ii)Aliphatic hetero cyclic compound : O
CH2 CH2 HC C O O HC C Oxirane or Epoxyethane or Ethylene oxide & or dimethylene oxide or Oxo cyclo propane O Maleic anhydride Groups : Atom or a group of atoms which possess any 'charge' on it or any 'free valency' are called as groups. Normal group : (a) It is represented by 'n'. (b) Straight chain of carbon atom is known as normal group. (c) Free bond will come either on Ist carbon atom or on last carbon atom. n – butyl C – C – C – C – n – propyl C – C – C – Iso group :
(a) It is represented by following structure : H3 C CH | CH3 6 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (b) When two methyl groups are attached to the same carbon atom, group is named as iso ;
C C Iso propyl ; C C C Isobutyl ; C C C C Isopentyl C C C Secondary group : (a) It is represented by following structure – C C C C (b) When ethyl and methyl groups attached to the terminal carbon atom, group is named as secondary- Ex. C C C secondary butyl C Tertiary group : C (a) It is represented by following structure – C C C (b) When three alkyl groups (similar or dissimilar) are attached to the same carbon atom, group is named as tertiary.
C C
C – C – Tertiary butyl and C – C – C – Tertiary pentyl
C C Neo group : (a) When a carbon atom is attached to other four carbon atom group is named as neo group. C C | | (b) It is represented by following structure : CCC e.g. CCC Neo pentyl | | C C Alkyl group : When a hydrogen is removed from saturated hydrocarbon then alkyl group is formed. It is represented by R and its general formula is CnH2n+1. A bond is vacant on alkyl group on which any functional group may come. CH4 H CH3 – Methyl CH3 – CH3 H CH3 – CH2 – Ethyl (a) C3H7 can be represented as : (i) Normal propyl CH3 – CH2 – CH2 – (ii) Isopropyl (1-methyl ethyl) CH3 – CH | CH3
(b) C4H9 can be represented as : (i) n-Butyl H3C – CH2 – CH2 – CH2 –
(ii) Iso butyl (2-methyl propyl) CH3 – CH CH2– | CH3
(iii) Secondary butyl (1-methyl propyl) CH3 – CH2 – CH | CH3
CH3 (iv) Tertiary butyl (1, 1-dimethyl ethyl) CH3 C CH3
Gyaan Sankalp 7 Organic chemistry-Some basic principles and techniques
(c) C5H11 can be represented as : (i) n-Pentyl H3C – CH2 – CH2 – CH2 – CH2 – H3C CH CH2 CH2 (ii) Isopentyl (3-methyl butyl) H3C (iii)Active amyl (2-methyl butyl) H C – CH CH CH 3 2| 2 HC3
HC 3 | (iv) Tertiary pentyl (1, 1-dimethyl propyl) H C – CH C 3 2 | H3C
HC3 | (v) Neo pentyl (2, 2-dimethyl propyl) H C C CH 3| 2 HC3
(vi) Active secondary amyl (1-methyl butyl) H C – CH – CH CH 3 2 2 | HC3
(vii) Secondary amyl (1-ethyl propyl) H C – CH CH 3 2 | HC2 | HC3
H C CH CH (viii)Active isosecondary amyl 3 | | HCHC3 3 (1, 2-dimethyl propyl) Alkenyl group : CH2 = CH – Vinyl (ethenyl) ; CH2 = CH – CH2 – Allyl (2-Propenyl) CH3 – CH = CH – Propenyl (1-propenyl) ; CH3 – C = CH2 Isopropenyl | Alkynyl group : CH C – Acetynyl (Ethynyl) CH C – CH2 – Propargyl (2-propynyl) & CH3 – C C – Propynyl (1-propynyl) Functional group : Organic compounds madeup of only carbon and hydrogen are termed as hydrocarbons and they are parent organic compounds. All other compounds are considered to have been derived from them by replacing one or more of they hydrogen some other atom or group of atoms. These atoms or group are termed as functional group and they largely determines the chemical properties of the compound. The functional group show their individual identity in their chemical reactions. Generally compound having same functional group show same chemical reactions. Some important functional groups are shown in table. Name and formula of functional groups :
Name of functional group Formula of functional group Example | Alkane C CH – CH | 3 3
Alkene C = C CH2 = CH2 Alkyne – C C – CH CH Halide – X CH3 – CH2Cl
8 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
Alcohol – OH CH3 – CH2 – OH Thio alcohol – SH CH3 – CH2 – SH Ether – O – CH3 – O – CH3 Aldehyde – CHO CH3 – CHO
C CH3 C CH 3 Ketone || || O O
Carboxylic acid – COOH CH3 – COOH Ester – COOR CH3COOCH3 Acid chloride – COCl CH3COCl
COC CH3 C O C CH 3 Acid anhydride || || || || OO OO
Acid amide – CONH2 CH3CONH2 Cyanide – CN CH3CN Amine – NH2 CH3NH2 Nitro – NO2 CH3NO2 Sulphonic acid – SO3H CH3CH2SO3H Isocyanide – NC CH3NC
HOMOLOGOUS SERIES : When structurally similar organic compounds are arranged in the order of increasing molecular weight, then the series of compounds so obtained is called a homologous series. The members of such a series are known as homologous of one another and this property is called homology. Characteristics of Homologous Series : 1. All the members of a homologous series can be represented by only one general formula. 2. The members of a homologous series differ in their molecular weight by 14 or its multiple. 3. The members of a homologous series differ in their molecular formulae by CH2 or its multiple. 4. The members of a homologous series can be synthesised by some general methods of preparation. 5. Physical properties of the members of a homologous series normally exhibit a regular gradual change with increase in molecular weight. 6. The members of a homologous series normally exhibit similar chemical properties (reactions). 7. Homologous cannot be isomers due to difference in their molecular formulae. Therefore, two or more than two isomers can never be included in the same homologous series.
Ex. : Series Member General formula Alkane – CH4, C2H6 etc. CnH2n+2 Alkene – C2H4, C3H6 etc. CnH2n Alkyne – C2H2, C3H4 etc. CnH2n–2 Alkanol – CH3OH, C2H5OH etc. CnH2n+1OH Alkanol – HCHO, CH3CHO etc. CnH2nO Alkanoic acid – HCOOH, CH3COOH etc. CnH2nO2 Alkylamines (1º or primary amines), Dialkylamines (2º or secondary amines) and Trialkylamines (3º or tertiary amines) constitute different homologous series.
NOMENCLATURE : Mainly three systems are adopted for naming of organic compound – (i) Common Names or Trivial System (ii) Derived System (iii) IUPAC system
Gyaan Sankalp 9 Organic chemistry-Some basic principles and techniques COMMON OR TRIVIAL SYSTEM : Initially organic compounds are named on the basis of source from which they were obtained.
S. No. Organic Compound Trivial Name Source
1. CH3OH Wood spirit or Methyl spirit Obtained by destructive distillation of wood. 2. C2H5OH Grain Alcohol Obtained by fermentation of Barley 3. NH2CONH2 Urea Obtained from urine 4. CH4 Marsh gas (fire damp) It was produced in marsh places. 5. CH3COOH Vinegar Obtained from acetum i.e. Vinegar 6. COOH Oxalic acid Obtained from oxalis plant | COOH 7. HCOOH Formic acid Obtained from formicus [Red ant] 8. CH CH COOH Lactic acid Lactum = milk 3 | OH 9. CH COOH Malic acid Apple (Malum) | 2 CH(OH)COOH
10. CH3CH2CH2COOH Butyric acid Obtained from butter
Some typical compounds in which common and trivial names are also differ
S. No. Compound Trivial Name Common Name
1. CH4 Marsh gas Methane 2. CH3OH Woodspirit Methyl alcohol 3. CH3COOH Vinegar Acetic acid 4. CH C CH Acetone Dimethyl ketone 3|| 3 O O || 5. CH2 = CH – CH Acrolein Acryl Aldehyde CH O |3 || 6. CH3 C C H Pyvaldehyde Tertiary valeraldehyde | CH3
(Common – Names R is termed as alkyl -) :
S. No. Compound Name
1. R – X Alkyl halide 2. R – OH Alkyl alcohol 3. R – SH Alkyl thio alcohol 4. R – NH2 Alkyl amine 5. R – O – R Dialkyl ether 6. R – S – R Dialkyl thioether 7. RCR Dialkyl ketone || O 8. R – NH – R Dialkyl amine
10 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
9. RNR Trialkyl amine | R 10. R – O – R' Alkyl alkyl' ether
11. RCR Alkyl alkyl' ketone || O 12. R – S – R' Alkyl alkyl' thio ether 13. R – NH – R' Alkyl alkyl' amine
14. RCR Alkyl alkyl' alkyl" amine || R Position of double bond : In an unsaturated hydrocarbon if the position of double is on 1st or last carbon then it's prefix will be (alpha) if it is on 2nd carbon it is termed as (Beta) and the (gamma) and (delta) and so on. Ex. H2C = CH – CH2 – CH3 – butylene H3C – CH = CH – CH3 – butylene H3C – CH2 – CH = CH2 – butylene H2C = CH – CH3 or H3C – CH = CH2 (Both are same positions, propylene)
H3 C C CH 2 Isobutylene | CH3
CH3 – CH2 – CH = CH – CH2 – CH3 – hexylene CH3–CH2–CH2–CH=CH–CH2–CH2–CH3 – octylene Common – Naming of di-hydroxy compounds : (a) When two – OH groups are attached to adjacent carbon's they are termed as alkylene glycol. OH CHCH2 2 CH3 CH 2 CH CH 2 | Ex. | | | | & CH CH C CH OH OH OH OH OH 3 2| 2 CH3 Ethylene glycol Butylene glycol Active amylene glycol (b) When two – OH group are attached at the two ends of a carbon chain, these compounds are named as polymethylene glycol. Poly Number of CH2 groups.
CH2 CH 2 CH 2 CH 2 Ex. | | Tetra methylene glycol OH OH
CH2 CH 2 CH 2 CH 2 CH 2 CH 2 Hexamethylene glycol | | OH OH
CH OH dimethylene glycol (wrong) Exception : | 2 CH2 OH ethylene glycol (right) Common – Naming of dihalides : (a) When two same halogen atoms are attached to the same carbon such compounds are called Gemdihalides. (b) Common names of such compounds are alkylidene halides. Cl I Ex. CH3 CH Ethylidene chloride ; CH3 CH CH Isobutylidene Iodide Cl I CH3 X Methylidene halide (wrong) Exception : CH2 Methylene halide (right) X
Gyaan Sankalp 11 Organic chemistry-Some basic principles and techniques (c) When two same halogen atoms are attached to adjacent carbon, these are called as vicinal dihalides common names of such compounds are alkylene halide. Cl | Ex. CH CH CH CH CH 3 2 H3C C CH2 Cl 2 2 | | | I I Cl Cl CH3 Propylene Iodide Isobutylene chloride Ethylene chloride (d) When two same halogen atoms are attached at the two ends of a carbon chain its common naming will be polymethylene halide. 'poly' word indicates the number of –CH2– groups. –CH2– 2 3 4 5 6 Poly di tri tetra penta hexa CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 Ex. l l I I Br Br Trimethylene Iodide Pentamethylene Bromide dimethylene halide (wrong) CH2 X Exception : CH2 X ethylene halide (right) Example 1 : Make the structure of following organic compounds – 1. Isopropylidene Bromide 2. Active amylene Iodide 3. Isobutylene glycol 4. Isobutylene 5. Trimethylene glycol
I OH Br CH C CH I CH3 C CH2 OH Sol. 1. CH3 C 2. 3 2 3. Br CH CH CH3 CH3 2 3 4. H3C C = CH2 5. CH2 CH2 CH2 CH3 OH OH Common – Naming of the functional group having carbon :
Functional group Suffix Functional group Suffix O O || -aldehyde || -ic Acid CH C OH O O || -yl halide || -amide CX C NH2 – C N -o-nitrile – N C -oisonitrile O O C O || -ate C -ic anhydride COR O
No. of Carbon Prefix No. of Carbon Prefix 1 Carbon Form- 2 Carbon Acet- Normal 3 Carbon Propion- 4 Carbon Butyr Iso Normal – 3 C + (=) double bond Acryl- 5 Carbon Valer Iso – Secondary – 4 C + double bond Croton- Tertiary – 12 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
O O O O O || || Ex. || || || CHCOH CH CH C Cl CH3 CH C NH 2 HCH 3 3 2 | CHCH3 CH3 Formaldehyde Acetic Acid Propionyl chloride Isobutyr amide Acetaldehyde O Nomenclature of Ester : C O R The group which is attached to the oxygen is written as alkyl and the remaining structure is named according to previous discussions. O O O O || || || || Ex. CH3 C O CH 3 CH3 CH 2 C O CH 2 CH 3 CH3 C O CH 2 CH 3 H C O CH3 Methyl acetate Ethyl propionate Ethyl acetate Methyl formate O O O || || CH3 C O H CHCHCOCHCH2 2 3 CH3 CH CHC O CH3 Acetic acid Ethyl acrylate Methyl crotonate O R C O Nomenclature of anhydride : R C O Rule : Add the total number of carbon atoms and divide it by 2, the substract will give you the number of C - atom. O
Total CH3 CH2 C 6 = Substract = Number of C atom. O ; 3 (Propionic anhydride) 2 CH3 CH2 C 2 O
O O
R – C CH3 – C In O If R R', You need not to find out substract. Eg. O R' – C CH3 – CH2 – C
O O Acetic propionic anhydride (right). Propionic acetic anhydride (wrong) Divide it in two parts as above and name it by suffixing 'ic anhydride' (alphabatically)
O || CH3 CH2 C Ex. O Butyric propionic anhydride CH CH CH C 3 2 2 || O O || CH3 CH C | O CH3 O || CH2 CH C Ex. CH3 CH2 CH C Isobutyric secondary valeric anhydride ; Ex. OAcrylic anhydride | || CH2 CH C CH O || 3 O
Gyaan Sankalp 13 Organic chemistry-Some basic principles and techniques DERIVED SYSTEM : According to this system name to any compound is given according to the parent name of the homologous series. This system is reserved for the following nine homologous series.
S. No. Name of Homologous Derived Name Structure of group | 1. Alkane Methane C | 2. Alkene Ethylene > C = C < 3. Alkyne Acetylene – C C – | 4. Alkanol Carbinol C OH |
| 5. Alkanal Acetalddehyde C CHO |
| 6. Alkanoic acid Acetic acid C COOH |
| 7. Alkanoyl halide Acetyl halide C COX |
| 8. Alkanamide Acetamide C CONH2 |
| | CCC 9. Alkanone Acetone | || | O
CH3 CH3 CH3 CH3
Ex. CH3 C CH3 , CH3–C C – CH3 , CH3 C OH , CH3 C CHO , CH3 C COOH
CH3 CH3 CH3 CH3 Tetra methyl Dimethyl Trimethyl Trimethyl Trimethyl Methane Acetylene Carbinol Acetalddehyde Acetic acid
Types of Ethylene : (Symmetrical and Unsymmetrical) (a) Symmetrical : In the given two alkyl groups one group is attached to the one carbon of ethylene and next on the next carbon. (b) Unsymmetrical : When both the given groups are attached on the same carbon. Note :– Symmetrical and Unsymmetrical terms are used only when two alkyl groups are given. Ex.Symmetrical dimethyl ethylene Unsymmetrical dimethyl ethylene
H CH3 CH3 H C = C C = C CH 3 H CH3 H H H
C = C Symmetrical ethyl methyl ethylene CH3CH2 CH3
14 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
CH CH3 CH3 CH3 3
C = C Tri methyl ethylene C = C Tetra methyl ethylene CH3 H CH3 CH3
CH3 – C C – CH3 Dimethyl acetylene
IUPAC SYSTEM : IUPAC system of nomenclature of Aliphatic compounds According to IUPAC system, the name of an organic compound consists of three parts (i) Word root (ii) Suffix (iii) Prefix (i) Word root : Word root denotes the number of carbon atoms present in the principal chain which is the longest possible chain of carbon atoms. According to chain length word root name up C12 are given below.
Chain Length Word root Chain Length Word Root C1 Meth C7 Hept (a) C2 Eth C8 Oct(a) C3 Prop C9 Non(a) C4 But(a) C10 Dec(a) C5 Pent(a) C11 Undec (a) C6 Hex(a) C12 Dodec(a)
Extra ‘a’ given in parenthesis is used only if the primary suffix to be added to the word root starts with a consonant. Consonant – di, tri, tetra etc are not started with vowel, then extra ‘a’ has been added to the word root. (ii) Suffix – Suffixes are of two types. Primary and secondary suffixes. (a) Primary suffix – A primary suffix indicates the type of linkage in the carbon atoms. If the carbon atoms are linked by single covalent bonds, the primary suffix is ‘ane’. If these are linked by double and triple bonds, the primary suffixes ‘ene’ and ‘yne’ are respectively used to represent them. Thus, ane : primary suffix for C – C bond ene : primary suffix for C = C bond yne : primary suffix for C C bond If the parent chain of carbon atoms contains more than one double or triple bonds, numerical prefixes like di (for two), tri (for three) tetra (for four) etc. are added to primary suffix. Example – Hydrocarbon Word root Primary suffix IUPAC Name CH3–CH2 – CH2 – CH3 But ane Butane CH2 = CH – CH = CH2 Buta* diene Butadiene CH3 –CCH Prop yne Propyne (b) Secondary suffix: A secondary suffix is used to represent the functional group if present in organic molecule and is attached to the primary suffix while writing its IUPAC name. Secondary suffixes of some of the functional groups are :
Functional group Symbol Suffix Prefix
Sulphonic –SO3H sulponicacid sulpho Carboxylic –COOH –oic acid carboxy O Ester Alkyl - Alkanonate alkyl carboxylate or R – C – OR alkoxy carbonyl O Acid halide - oylhalide Halo formyl – C – Cl O Acid Amide amide carbamoyl – C – NH2
Gyaan Sankalp 15 Organic chemistry-Some basic principles and techniques O Aldehyde –al Formyl/ Aldo – C – H cynide – C N –nitrile Cyano isocynide N C –isocynide Isocyano ketone > C = O – one keto, oxo hydroxy –OH –ol hydroxy Thioalcohol – SH –Thiol mercapto Amine –NH2 –Amine amino R Secondary amine – N N-Alkyl-amine N-alkyl-amine H R Tertiaryamine – N N-alkyl-N-alkyl-amine N-alkyl-N-alkyl-amine R O + Nitro – N nitrite nitro O-
Groups given below does not have any Suffix Ether –OR ...... alkoxy Epoxide – O – ...... epoxy Azo – N = N – ...... azo Nitroso – NO ...... nitroso Halogen – X (F, Cl, Br, I) ...... halo While adding a secondary suffix to the primary suffix, the terminal ‘e’ of the primary suffix (ane, ene, or yne) is dropped if the secondary suffix begins with ‘a’, ‘i’ ‘o’ ‘u’ or ‘y’. In case, it begins with consonant, then the terminal ‘e’ of the primary suffix is retained. Ex. Organic compounds Word root Primary suffix Secondary suffix IUPAC Name
CH3CH2OH Eth an(e) ol Ethanol CH3CH2CHO prop ane(e) al Propanal CH3CH2CH2NH2 prop ane(e) amine Propanamine CH3CH2CN prop ane nitrile Propanenitrile
(iii) Prefix : Prefix is a part of IUPAC name which appear before the word root. Prefixes are of two types. (a) Primary prefix – A primary prefix “cyclo” is used in order to differentiate a cyclic compound from an acyclic compound. For example HC CH CH2 2 2
CHCHCH3 2 3 CHCHCHCH3 2 2 3 Propane Butane HC2 CH2 HC2 CH2 Cyclopropane Cyclobutane Propane and butane are the IUPAC names of acyclic compounds while cyclopropane and cyclobutane are for cyclic compounds. (b) Secondary prefix – In the IUPAC system of nomenclature, certain characteristic groups are not considered as or secondary suffixes. These are regarded as substituents and are denoted by secondary prefixes. The secondary prefixes of a few substituents are given – Substituent Group Secondary prefix Substituent group Secondary prefix –F Fluoro –NO2 Nitro –Cl Chloro –CH3 Methyl –Br Bromo –C2H5 Ethyl –I Iodo –OCH3 Methoxy – NO Nitroso –OC2H5 Ethoxy
16 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Besides these, some other functional groups are also treated as substituents in case of poly functional organic compounds. These are listed as
Functional group Name Secondary prefix SO3H sulphonyl sulpho – COOH carboxylic acid carboxylic O || Ester alkyl carboxylate or alkoxy carboyl R C O OR ' – COX Acid halides Halo formyl
– CONH2 Acid amide Carbamoyl O
R – C O Acid unhydride Oic anhydride R – C O – CHO Aldehyde Formyl – CN Cynide Cyano NC Isocynide Iso Cyano R C = O Ketone Keto, oxo R' – OH Alcohol Hydroxy –SH Thioalcohol Mercapto – NH2 Amine Amino – O – Epoxy Epoxide
General rules for naming long chain aliphatic compounds : Many organic compounds contain long chains which may include substituents, multiple bonds and functional groups. The IUPAC names of such compounds are based on certain general rules.
Rules for IUPAC nomenclature of straight and Branched chain (Alkanes) : In writing the IUPAC name of an aliphatic compounds the following sequence is used. Secondary prefix + primary prefix + word root + primary suffix + secondary suffix 1. Longest chain rule – Select the longest continuous chain of carbon atoms in a given molecule of alkane which need not be straight. It is known as parent chain or principal chain.
If two chains of equal lengths are possible, then the one with more number of side chains (or substituents) represents the parent chain. For example,
H3 C – CH – CH – CH 2 – CH2 – CH 3
CH3 CH 2 CH 3 Parent chain of six carbon atoms has two substitute (Right)
Gyaan Sankalp 17 Organic chemistry-Some basic principles and techniques 2. Lowest number rule – For molecules with only one substituent, number the carbon atoms as 1, 2, 3, 4 ...... etc. in the parent chain starting from one end in such a way that the carbon atom carrying the substituent gets the lowest number.
1 2 3 4 5 6 7 7 6 5 4 3 2 1 HCCHCHCHCHCHCH3 2 2 2 2 3 HC3 CH 2 CH CH 2 CH 2 CH 2 CH 3 | | CH3 CH3 The numbering of parent chain is correct because The numbering of parent chain is incorrect CH3 group is attached to C-3 atom The number which indicates the position of the substituent (or side chain) in the parent chain is called its position number or locant. The name of the substituents is separated from its locant by hyphen (-) and the final name of the alkane is always written as one word i.e. 3-methylheptane. In case, same alkyl group occurs more than once at different positions in the parent chain, the positional number of each alkyl group is separated by commas and suitable prefixes like di (for two), tri (for three) tetra (for four) etc. are attached to the name of the alkyl groups. 3. Lowest set of locants rule – In case, there are two or more substituents, then the parent or principal chain is numbered from the end which gives the lowest set of locants. This rule (IUPAC 1993) replaces the lowest sum rule used earlier. When two or more substituents are present in a molecule, then the end of the parent chain which gives the lowest term at the first point of difference in the set of locants is considered for numbering the carbon atoms in the parent chain.
10 9 8 7 6 5 4 3 2 1 CHCHCHCHCHCHCHCHCHCH3 2 2 2 2 2 3 Ex. | | | CHCHCH3 3 3 Set of locants : 2, 4, 9 locants Correct
1 2 3 4 5 6 7 8 9 10 CH3 CH CH 2 CH 2 CH 2 CH 2 CH CH 2 CH CH 3 | | | CH3 CH 3 CH 3 Set of locants : 2, 7, 9 locants Wrong Alphabetical order for the side chain (or substituents) – When two or more different alkyl groups (side chains or substituents) are present on the parent chain. Such groups prefixed by their locants (or positional numbers) are arranged in alphabetical order irrespective of their positional number, before the name of the parent alkane. For example,
It may be remember that the prefixes di, tri, tetra are not considered in case the same alkyl group occurs more than on the parent chain at different positions while arranging them in an alphabetical order. For example
CH2 –CH 3 5 4 3 2 1
CH3 – CH 2 – C – – CH – CH 3
CH3 CH 3 3-Ethyl-2,3-dimethylpentane Naming the different substituents at equivalent positions – If two different substituents are present at equivalent positions from the two ends in the parent chain, then the numbering of the chain is done in such a way that the substituent which comes first in the alphabetical order gets the lower number. For example
18 Gyaan Sankalp Organic chemistry-Some basic principles and techniques 4. Naming the complex substituents or alkyl groups : (a) An alkyl group is said to be complex in case one or more carbon atoms in it are further substituted i.e., it is a substituted substituent (or alkyl group). It is named as a substituted alkyl group by numbering from the carbon atom of this group attached to the parent chain as “1”. The name of the complex substituent is generally enclosed in brackets in order to avoid any confusion with either numbering the parent chain or naming the other substituents attached to the chain. For example -
1 2 3 4 5 6 7 CH3 – CH 2 – CH2 – CH – CH 2 – CH 2 – CH 3 1 H3 C – C – CH 3 2 CH3 4-(1,1-dimethylethyl) heptane
(b) It may be noted that while deciding the alphabetical order of the various substituents, the name of the complex substituent is considered to begin with the first letter of its complete name. For example-
CH3 1 2 3 CH3 – CH – CH – CH 3 1 2 3 4 5 6 7 8 9 10 CH3 – CH2 – CH – CH 2 – CH – CH 2 – CH 2 – CH 2 – CH 2 – CH 3
CH2 – CH 3 5-(1,2-Dimethylpropyl)-3-ethyldecane (c) When the names of the complex substituents are composed of identical words, the priority is decided by comparing the locant at the first cited point within the complex substituents. This means that the complex substituent gets lower priority which has the lowest locant. For example
CH3 CH3 3 2 1 1 2 3 CH3 – CH 2 – CH CH2 – CH– CH 3 1 2 3 4 5 6 7 8 9 10 CH3 – CH 2 – CH2 – CH 2 – CH – CH– CH 2 – CH 2 – CH 2 – CH 3 5-(1-methylpropyl)–6–(2-methyl propyl) decane In this case, 1-methylpropyl gets priority over 2-methylpropyl since the locant for the methyl substituent in the first case is less. Note : (a) Initially write the number of that carbon on which the side chain is attached, then write it in alkyl group. (b) If there are more than one side chains (same) initially write the numbers altogether using symbol (,) in between them, after last number use the symbol hyphen (-). Write the numbers in increasing order. CCCCC Ex. | | | 2, 3, 4-trimethyl pentane CCC (c) IUPAC name is written according to english alphabates (strictly). C | C 1 2 3 4 |5 6 7 — Correctnumbering CCCCCCC Ex. 7 |6 5 |43 2 1— Incorrectnumbering CC | C 2-methyl-4-Isopropyl-5-ethyl heptane (wrong) 5-ethyl-4-Isopropyl-2-methyl heptane (right) Gyaan Sankalp 19 Organic chemistry-Some basic principles and techniques (d) Capital word is used for first substitute. 1 2 3 4 5 6 H3 C CH CH CH CH2 CH3 Ex. H3C NH2 C2H5 3-Amino-4-ethyl-2-methyl hexane. (more correct)
Rule in IUPAC nomenclature of unsaturated hydrocarbons (alkenes and alkynes) In writing the IUPAC name of a particular member of the alkene or alkyne family, the primary suffix, ‘ane’ of corresponding in alkane is replacing by the correct ‘ene’ or ‘yne’ respectively.
1. The parent chain must include the multiple bond regardless of the fact whether it is longest continuous chain or not. For example,
CH3 – CH 2 – CH – CH2 – CH 2 – CH 3
CH = CH2 2. The numbering of the parent chain must be done in such a way that the first C-atom involved in the multiple bond gets the lowest number. According to the latest IUPAC convention, the locants for the multiple bonds (double or triple bonds) are placed immediately before the primary suffixes (ene or yne) which they represent. The earlier practice was to place the locant either before the word root or after the name of the primary suffix but it is not regarded as proper. For example, 4 3 2 1 CH3 – CH 2 – CH = CH2 But-2-ene 3. If the selected principal or parent chain contains two (or three) double or triple bonds, then the primary suffix -diyne (or triene) or -diene (or triyne) are used to represent these. In all these cases, terminal ‘a’ is also added to the word root. For example, 6 5 4 3 2 1 CH3 – C C – CH 2 – C CH Hexa-1, 4-diyne 4. If the parent chain includes both double and triple bonds, then the following points must be kept in mind. (a) the unsaturated compound is always named as a derivative of alkyne i.e., primary suffix ‘ene’ always comes before ‘yne’. In all such cases, the terminal ‘e’ of ‘ene’ is dropped if it is followed by the suffix starting with a, i, o, u or y. For example, 6 5 4 3 2 1 CH3 – C C – CH 2 – C CH Hexa-1, 4-diyne (b) In case, the numbering from two different ends gives two different sets of locants, then the lower set of locants is preferred. For example,
5 4 3 2 1 CH3 – CH= CH – C CH Pent-3-en-1-yne 5. If the unsaturated hydrocarbon contains a side chain (or substituent) along with the multiple bonds, then the numbering of the parent or principal chain is done in such a way that the multiple bonds gets the lowest set of locants. However, if the numbering from both ends gives the same set of locants to the multiple bonds, then the locant for the substituent must be minimum. For example, CH 3 CH 4 3 2 1 3 CH = CH – C = CH 1 2 3 | 4 5 2 2 HC C CH C CH 2-Methylbuta-1,3-diene 3-methylpenta-1,4-diyne (Minimum locant for the CH3 substituent) In some cases, more than two double bonds are present in the hydrocarbon and it is not possible to include all of them in the parent chain. In such cases, the following prefixes are used for the double bonded groups not involved in the chain. CH2 = CH3 – CH = CH2 = CH2 – CH2 = CH – CH2 – (Methylidene) (Ethylidene) (Vinyl or ethenyl) (Allyl)
20 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Nomenclature of monofunctional compounds : Compounds having one functional group are called monofunctional compounds. The various rules are : 1. That longest chain is selected which contain functional group, even if other chain contains more number of carbon atoms. If the functional group contains a carbon atom, that atom is also counted as a part of the chain.
CH3 – CH2 – CH – CH 2 – CH 2 – CH 3 CHO 2. When more than one same functional groups are present, the carbon chain which contains maximum number of functional group is selected.
CH3 – CH2 – CH2 – CH – CH 2 – COOH COOH 3. When the functional group does not have carbon atom then the carbon atom to which the functional group is attached should be included in the main chain.
CH3 – CH2 – CH – CH2 – CH 3
CH2
CH2 – OH 4. The selected carbon chain is numbered from that end to which the functional group is nearer. 5 4 3 2 1 CH3 – CH – CH2 – CH – CH 3
CH3 OH 5. If the functional group is at the same distance from both the ends, then numbering is done from the end in which the substituent is nearer. 1 2 3 4 5 6 7 CH3 – CH – CH2 – CH – CH 2 – CH2 – CH 3
CH3 OH 6. While writing name of the compound, the name of functional group along with its position is represented by suitable suffix and all the substituents are represented by suitable prefixes in their alphabetical order. 1 2 3 4 5 CH3 – CH – CH2 – CH – CH 3
OH CH3 4-methyl pentane-2-ol Classification of name : (a) – OH group is present as functional group so it is represented by suffix – ol (b) Longest chain consist of five carbon atoms all having single bond between them so parent hydrocarbon is pentane. (c) – OH group is present on carbon number 2 and substituent methyl group is present on carbon number 4. Above name is derived as : 4-methyl + Pent + ane + 2-ol Prefix for Parent Saturated Suffix for the substituents hydrocarbon chain functional group 7. The chain terminal functional group, viz., – COOH, – CHO, – COOR, – CONH2, – COCl, – C N etc., when present in the parent chain is always given number 1. Generally, the number 1 is omitted while writing the final names of the compound, For example,
2 3 4 4 CH3 CH 2 CH CH 2 CH 2 CH 3 is 2-ethylbutanoic acid | 1COOH
Gyaan Sankalp 21 Organic chemistry-Some basic principles and techniques 8. If an organic compound contains a functional group, multiple bond, and a sidechain (substituent) then the following order of preference must be followed. Functional group > Double bond > Triple bond > Sidechain
4 3 2 1 5 4 3 2 1 Ex. CH3 – CH – CH – COOH CH3 – CH2 – C – CH – CH 3
CH3 CH3 O CH3 2-methyl pentane-3-one CH3 2-ethyl-3-methyl butanoic acid
6 5 4 3 2 1 CH3 – CH – CH2 – CH – CH 2 – CN
Cl CH3 5-chloro-3-methyl hexane nitrile
Nomenclature of ether : The small alkyl group attached with oxygen is written as alkoxy in which oxygen is included and the longest chain of remaining carbons is selected. — O —, no. — alkoxy alkane Ex. Write the IUPAC name 3 2 1 1. 1-Methoxy propane CH3 CH 2 CH 2 O CH 3
1 2 3 2. CH3 – O – CHCHCH2 2 3 1-Methyoxy propane
3 2 3. CHCHOCH3 3 2-Methoxy propane 1 | CH3
Ex. Write the correct IUPAC name of the compound 3-ethoxy butane.
CHCHCHCH 3 2| 3 O | CH 2-Ethoxy butane | CH3 Nomenclature of keto group : The carbon of a ketone is included in longest chain it never comes on first or last positions, whereas – CHO (Aldehyde group) comes at first and last position only.
5 4 3 Ex. CH3 C H 2 CH CH 3 2| CO 1| CH3
Ex. Modify the following IUPAC name – 3-methyl 2-butanone 3 2 1 CH3 CH2 C CH3 3-Methyl-2-pentanone 4 CH2 O 5 CH3
22 Gyaan Sankalp Organic chemistry-Some basic principles and techniques IUPAC Nomenclature for compounds containing more than one different functional groups (Poly functional compounds) When an organic compound contains two or more different functional groups, then one of these is selected as principal functional group while the others are called secondary functional groups and are treated as substituents. The choice of the principal functional group is made on the basis of the of preferences.
S. No. Name Formula Functional group Prefix Suffix
1. Sulphonic acid – SO3H Sulpho – Sulphonic acid 2. Carboxylic acid – (C)OOH Carboxy – oic acid – COOH – Carboxylic acid (no count) (C) O 3. Anhydride O — – oic anhydride (C) O
4. Ester – (C)OOR — – oate – COOR Alkoxy carbonyl carboxylate 5. Acid halide – (C)OX — – oyl halide – COX Haloformyl – carbonyl halide 6. Acid amide – (C)ONH2 — – amide – CONH2 Carbamoyl carboxyamide 7. Aldehyde – (C)HO Oxo – al – CHO formyl carbaldehyde 8. Cyanide – (C) N — – nitrile – C N Cyano carbonitrile 9. Iso cyanide – N (C) — – iso nitrile – N C Carbyl amine — O || 10. Ketone –(C)– Oxo – one 11. Alcohol – OH Hydroxy- – ol 12. Thio alcohal – SH Mercapto- – thiol 13. Phenols – OH hydroxy — 13. Amine – NH2 Amino- — Amine 14. Ethers – OR alkoxy – 15. Alkenes > C = C < – ene 16. Alkynes – C C – – yne 17. Halogens X (F, Cl, Br, I) halogen (halo) – 18. Nitroso compounds – NO nitroso – 19. Nitro compounds – NO2 nitro – 20. Alkoxy – OR alkoxy – 21. Alkyl – R alkyl –
In writing the IUPAC name of a poly functional compound, the functional group with the maximum preference (principal functional group) is represented by the secondary suffix which is added to the word root along with the primary suffix. All other secondary functional groups if present, are indicated by suitable prefixes which are added to the root word. (In alphabatic order) Thus, a functional group can act either as a secondary suffix or a prefix depending upon its priority in the priority sequence. A functional group, in fact, has two names; one while acting as secondary suffix and the other while acting as a prefix. Suppose an organic compound contains both –CHO and –COOH groups attached on both sides of the carbon atom chain. Then –COOH group is regarded as the principal functional group and its secondary suffix is oic acid. At the same time, –CHO group is the substituent and its prefix is formyl as shown below :
The IUPAC name of the compound is : 4-Formylbutanoic acid. Gyaan Sankalp 23 Organic chemistry-Some basic principles and techniques If the organic compound more than one similar complex substitute than the numerical prefix di, tri, tetra, extra replaced by bis, tris, tetrakis, etc. respectively.
If two substituents of the same priority occupy identical positions from either end of the parent chain, lower number must be given to the substituent whose prefix comes first in the alphabetical order. This, when Br and Cl atoms are present at the equivalent positions in a chain, then the carbon atom having Br is numbered lower,
1 2 3 4 4 3 2 1 Br CH2 CH 2 CHCl 2 BrCH 2 CH 2 CH 2 CHCl 2 Correct Wrong (1-bromo-4-chlorobutane) (4-bromo-1-chlorobutane) NOMENCLATURE OF AROMATIC COMPOUNDS Aromatic is regarded as aroma means pleasant smell. Aromatic compounds may be classified into two main types. These are benzenoids and non-benzenoids. They may be nuclear substituted and side-chain substituted (i) Nuclear substituted : Those in which the functional group is directly attached to the benzene ring. In the IUPAC system, they are named as derivatives of benzene. The positions of the substituents in disubstituted benzenes are indicated either by prefixes or by arabic numerals such as o-(ortho) for 1, 2 ; m-(meta) for 1, 3 and p-(para) for 1, 4. For tri and higher substituted benzene derivatives, the positions of the substituents are indicated only by arabic numerals. (ii) Side-chain substituted : Those in which the functional is present in the side chain of the benzene ring. Both in the common and IUPAC systems, these are usually named as phenyl derivatives of the corresponding aliphatic compounds (except arenes which are named as derivatives of benzene in the IUPAC system). The positions of the substituents on the side chain including the benzene ring are indicated by Greek letters i.e., , , .... etc, in the common system, and by arabic numerals, i.e. 1, 2, 3 ...... etc. in the IUPAC system. However, many of these compounds are better known by their common names.
BENZENOIDS : These are cyclic compounds which contains in them either one or more benzene rings. A few examples benzenoids are as follows.
Benzene Napthalene Anthrancene It may be noted that the presence of double bonds in the alternate positions is note the sole criteria for a compound to be aromatic. For example, cyclooctatetraene. (that contain alternative single double bond but non aromatic)
Cyclooctatetraene Non-benzenoics : Certain compounds do not contain a benzene ring but they fulfil the characteristics of aromatic compounds. These are called non-benzenoids. Some heterocyclic compounds belongs to this class of aromatic compounds. For example,
Pyrrole Furan Thiophene Pyridine
24 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Certain ions can also behave as aromatic compound. For example
H Cyclopentadiene Cyclopropenecation Anion 2 -electron (6 -electrons)
Nomenclature of Benzenoids : Benzene (C6H6) is a hexagonal ring of carbon atoms with single and double bonds in the alternate positions. This ring is also called nucleus. In case any one or more hydrogen atoms attached to the carbon atoms of the ring are substituted by some alkyl (R) groups or other functional group then such groups form the side chains.
R Side chain
Ring
Monosubstituted : Since all the carbon atoms in the ring are identical, by replacing a hydrogen atom attached to any of them only one mono substituted derivative will be formed – Cl A
ex. Chlorobenzene Disubstituted : A disubstituted derivative can be formed by substituting any of the five available hydrogen atoms in the monosubstituted derivative by suitable substituents (x).
A A A B B meta(or 1, 3) (II) B
It may be noted that the arabic numerals are used in the IUPAC names while the specific prefixes ortho (o–), meta (m–), para(p) are used for the common names. The aromatic compounds are generally known by their common names or by commercial names in some cases. Tri and higher substituted – These are generally represented by the arabic numerals as shown below
A A A A 1 A A 2 A A A 5 3 A A 4 A 1, 3, 5-trisubstituted A
The IUPAC and common names (given in brackets) of a few important members of each family are given below. 1. Aromatic hydrocarbons (Arenes) : Hydrocarbons which contain both aliphatic and aromatic units are called arenes. These are of three types :
Gyaan Sankalp 25 Organic chemistry-Some basic principles and techniques (i) Benzene and alkylbenzenes
CH3 CH3 CH3 CH3 1 1 1 CH3 2
3 CH Benzene Methylbenzene 1,2-Dimethylbenzene 3 4 1,3-Dimethylbenzene (Toluene) (o-Xylene) CH (m-xylene) 3 1,4-Dimethylbenzene (p-xylene)
1 2 3 CH3 CH2 CH 3 CH3 –CH–CH 3 1
HC3 5 3 CH3 1, 3, 5-Trimethylbenzene Ethylbenzene 2–Propylbenzene (Mesitylene) (Isopropylbenzene or Cumene)
(ii) Alkenylbenzenes
1 2 3 2 1 CH = CH2 CH = CHCH3 CH2 CH = CH 2
Ethenylbenzene 1-Propenylbenzene 3–Propenylbenzene (Styrene, vinylbenzene ( - Methylstyrene) (Allylbenzene) or phenylethylene) (iii) Alkynylbenzene 1 2 3 3 2 1 C CH C C – CH3 CH2 – C CH
Ethynylbenzene 1–Propynylbenzene 3–Propynylbenzene (Phenylacetylene) (b-Methylphenylacetylene) (Propargylbenzene or Benzylacetylene) (iv) Hydrocarbons containing condensed or fused rings :
3 4 2 8 1 8 9 1 5 1 7 2 7 2 6
6 3 6 3 7 10 5 4 5 10 4 8 9 Naphthalene Anthracene Phenathrene
CH2 – CH CH
Phenyl Benzyl Benzal Benzo
26 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
CH3 CH3 CH3 2 3 4 1
1 1 2–Tolyl or o–Tolyl 3–Tolyl or m–Tolyl 4–Tolyl or p – Tolyl 2. Hydroxy derivatives – The nuclear substituted hydroxy derivatives are called phenols while the side substituted are known as aromatic alcohols.
OH OH OH 1 OH 1 1 2 2 OH 2 3 3 OH 4 OH 1,2-Dihydroxybenzene 1,3-Dihydroxybenzene 1,4-Dihydroxybenzene Phenol (Catechol) (Resorcinol) (p-Quinol or hydroquinone)
OH OH OH 1 1 1 2 OH 2 2 OH 3 3 HO 5 OH 3 OH 4 4 OH 1, 2, 3- Trihydroxybenzene 1,2, 4- Trihydroxybenzene 1,2, 5- Trihydroxybenzene (Pyrogallol) (Hydroquinol) (Phloroglucinol)
OH OH 1 1 2 6 2 ON2 NO2 3 5 3 4 4
CH3 NO2 4-Methylphenol 2, 4, 6-Trinitrophenol (p-Cresol) (Picric acid)
3. Halogen derivatives : Cl Cl Cl 1 1 Cl 2
3 Cl Chlorobenzene 1, 2– Dichlorobenzene or 1, 3-Dichlorobenzene or o-Dichlorobenzene m - Dichlorobenzene
Cl 1 CH2 Cl CHCl2
4 Cl 1, 4-Dichlorobenzene or Phenylchloromethane Phenyldichloromethane p-Dichlorobenzene (Benzyl chloride) (Benzal chloride)
Gyaan Sankalp 27 Organic chemistry-Some basic principles and techniques
2 1 3 2 1 CCl3 CH2 CH2Cl CH2 CH2 CH2Br
Phenyltrichloromethane 1-Chloro-2-phenylethane 1-Bromo-3-phenylpropane (Benzotrichloride) ( - Phenylethyl chloride) ( - Phenylpropyl bromide)
2 1 1 2 3 CH CH–Cl CH CH–CH2 Br
1-Chloro-2-phenylethene 3-Bromo-1-phenylpropene ( -Chlorostyrene) (Cinnamyl bromide)
3 2 1 CH CH CHO CHO CH2 CHO 2 2 CHO OH
2-Hydroxybenzaldehyde Phenylethanal 3-Phenylpropanal Benzaldehyde (Salicylaldehyde) (Phenylacetaldehyde) ( -Phenylpropionaldehyde)
1 2 COC6 H 5 C O CH3 1 2 3 C O CH2CH3
1-Phenylethanone Diphenylmethanone (Acetophenone or (Benzophenone or 1-Phenylpropanone Methyl phenyl ketone) Diphenyl ketone) (Propiophenone) 5. Ethers :
OCH3 O–C 6 H 5
Methoxybenzene Phenoxybenzene (Anisole or Methyl phenyl ether) (Diphenyl ether) 6. Nitro compounds :
NO2 CH3 NO 1 2 1 NO2
NO2 2 3 2-Nitrotoluene or Nitrobenzene 1, 3-Dinitrobenzene or m-Dinitrobenzene o-Nitrotoluene
28 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
OH OH CH3 1 1 6 1 6 O 2 N 2 NO2 O N 2 2 NO2
4 4 4 NO NO2 2 NO2 4-Nitrophenol or 2,4,6-Trinitrophenol p-Nitrophenol (Picric acid) 2,4,6-Trinitrotoluene (T.N.T.)
7. Carboxylic acids :
COOH COOH COOH 1 CH3 1 OH 2 2
Benzoic acid 2-Methylbenzoic acid 2-Hydroxybenzoic acid or Benzene carboxylic acid (o-Toluic acid) (Salicylic acid)
COOH COOH COOH 1 1 COOH 1 2 3 4 COOH COOH Benzene-1, 2-dicarboxylic acid Benzene-1, 3-dicarboxylic acid Benzene-1, 4-dicarboxylic acid (Phthalic acid) (Isophthalic acid) (Terephthalic acid)
8. Sulphonic acid : SO 3 H SO 3 H SO3 H
NH2 CH3 Benzene sulphonic acid Pera amino benzene Pera Toluene sulphonic acid (PTS) sulphonic acid (Sulphanilic acid) 9. Amines : (i) Arylamines NH2 1 NH2 N(CH ) NHC H NH2 3 2 6 5 1 NH2 2 4 CH3 Benzenamine N,N-Dimethylbenzenamine N-Phenylbenzenamine 4-Methylbenzenamine Benzene-1, 2-diamine (Aniline) (N, N-Dimethylaniline) (Diphenylamine) (p-Toluidine) (o-Phenylenediamine) (ii) Aralkylamines
2 1 CH NH 2 2 CH2CH2 NH2
Phenylmethanamine 2-Phenylethanamine (Benzylamine) ( -Phenylethylamine)
Gyaan Sankalp 29 Organic chemistry-Some basic principles and techniques 10. Arenediazonium salts + + N = N C l¯ N = NHS O 4 Benzenediazonium chloride Benzenediazonium hydrogen sulphate
11. Cyanides and Isocyanides :
C N CH 2 C N N C
Benzonitrile or Benzyl cyanide Phenyl isocyanide or Phenyl cyanide Phenylacetonitrile Phenyl carbylamine 12. Grignard’s reagent – Mg – Br
Phenyl magnesium bromide Note : Bond–line notation of organic compounds : In this notation, bonds are represented by lines and carbon atoms by line ends and intersections. It is assumed that required number of H–atoms are present wherever they are necessary to satisfy tetracovalency of carbon. For example,
2 CH3 4 | 1 CH2 = C CH= CH2 is represented as 3 2- Methylbuta-1,3-diene 2-Methylbuta-1, 3-diene or Isoprene Some other examples are :
2 3 4 OH 3 2 4 1 2 4
5 1 1 6 2-Ethenyl-3-methyl- O 5 4-Ethylhex-4-en-2-ol 1, 3-cyclohexadiene 3-Ethyl-4-methylhex
NAMING ALICYCLIC COMPOUNDS : Monocyclic compounds : The name of alicyclic compounds are based on the following rules. 1. The names of the alicyclic compound are obtained by adding there primary prefix ‘cyclo’ to the word root that corresponds to the number of carbon atoms in the ring. For the cyclic compounds containing all single bonds in the ring, primary suffix ‘ane’ is added to the word root. For those containing one double or triple bond, the primary suffix ene or yen is added.
2. If only one substituent is attached to the ring, its position is note mentioned. If two or more substituents are present, their positions are indicated by arabic numerals i.e., 1,2,3,4 ...... etc. which are used for numbering the carbon atoms in the ring. The numbering is done in such a way (clockwise or anticlockwise) that the substituents get the lowest set of locants. All other rules 30 Gyaan Sankalp Organic chemistry-Some basic principles and techniques relating to aliphatic or acyclic compounds are then followed. For example.
CH3 CH3 CH2 CH3
3. If a multiple bonds and some other substituents are present in the ring, the numbering is done in such a way so as to assign lowest number to the multiple bond. For example
CH3 NO2 3 2 C2 H5 1 3-Nitrocyclohex-1-ene 4. In case, some functional group along with some substituents are present in the ring, the numbering of the carbon atoms should be done in such a way so that the functional group gets lowest number. NH OH 2 O 1 CH3 2 3 CH3 CH3 CH3 In case, the functional group directly attached to the ring contains carbon atom, suitable suffixes are used to represent such a group. Functional group Suffix – COOH Carboxylic acid – CHO Carbaldehyde – CN Carbonitrile – COCl Carbonylchloride – CONH2 Carboxamide – COOR R—Carboxylate
COOH CONH 1 2 2
3 CH2 5 Ex. NO2 3-Nitrocyclohex-2-ene-1- carboxylic acid
5. If the ring contains lesser carbon atoms than the alkyl group attached to it, the compound is named as the derivative of alkane and the ring is treated as cycloalkyl substituent. Otherwise, it is named as the derivative of cycloalkane. For example.
1 2 3 4 5 6 H C – H C– CH – CH H3 C – CH – CH 2 – CH2 – CH 2 – CH 3 3 2 3
2-cyclopentylhexane Sec-Butylcyclohexane
Gyaan Sankalp 31 Organic chemistry-Some basic principles and techniques In case, the side chain contains a multiple bond or a functional group, the alicyclic ring is treated as the substituent irrespective of its size. For example,
3 2 1 3 2 1 H3 C – CH – CHO CH3 – CH = CH2
2-Cyclohexylpropane 3-Cyclopropylprop-1-ene 6. If the ring contains fewer carbon atoms than the alkyl group attached to it or when more than one ring system is attached to a single chain the compound is named as a derivative of alkane and the ring is treated as cycloalkyl substituent. 7. When both ring as well as side chain contains the same functional group, then the parent hydrocarbon is decided on the basis of the number of carbon atom. 8. If a compound contains an a cyclic ring as well as a benzene ring, it is named as a derivative of benzene. Nomenclature of Bicyclic Compounds : (a) Many hydrocarbons and their derivatives contain two fused or bridged rings. (b) The carbon atoms common to both rings are called bridge head atoms and each bond or chain of carbon atoms connecting both the bridge head atoms is called as bridge. (c) The bridge may contain 0, 1, 2... etc. carbon atoms. For example,
Bridge head atom Bridge head atom Zero carbon One carbon bridge CH bridge CH
Two CH2 CH2 Two One CH2 CH One 2 carbon carbon Carbon Carbon CH2 bridge bridge bridge CH bridge CH2 CH2 Bridge head atom CH
Bridge head atom
III III
(d) These bicyclic compounds are named by attaching the prefix 'bicyclo' to the name of the hydrocarbon having the same total number of carbon atoms as in the two rings. (e) The number of carbon atoms in each of the three bridges connecting the two bridge head carbon atoms is indicated by arabic numerals, i.e. 0, 1, 2,.... etc. These arabic numerals are arranged in descending order ; separated from one another by full stops and then enclosed in square brackets. (f) The complete IUPAC name of the hydrocarbon is then obtained by placing these square brackets containing the arabic numerals between the prefix bicyclo and the name of the alkane. For example,
Bicyclo [2, 2, 1] heptane Bicyclo [3, 1, 1] heptane Bicyclo [2, 2, 2] octane
Bicyclo [4, 4, 0] decane Bicyclo [1, 1, 0] butane (also called decalin) (Simplest bicyclo compound)
32 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (g) If a substituent is present, the bicyclic ring system is numbered. The numbering begins with one of the bridge head atoms, proceeds first along the longest bridge to the second bridge head atom, continues along the next longest bridge to the first bridge head atom and is finally completed along the shortest path. For example, 1 2 8 Cl 1 9 2 3 2 7 CH3 1 3 3 5 8 6 6 4 6 4 5 7 7 4 5 2, 6, 6–Trimethylbicyclo [3, 1, 1] hept-2-ene 8–Chlorobicyclo [3, 2, 1] octane 8– methylbicyclo [4, 3, 0] nonane
NAMES OF SPIRO HYDROCARBONS (a) Spiro bicyclic hydrocarbons contain two rings consisting of carbon atoms only the two rings are linked by a common carbon. These compounds are named by placing prefix ‘spiro’ before the name of the acyclic parent hydrocarbon with same number of skeleton carbon atoms. The numbers of skeleton atoms linked to the spiro atom are indicated by arabic numbers, separated by a full stop. The numbers are written in ascending order and enclosed in square brackets. Numbering of a spiro bicyclic hydrocarbon starts with a ring carbon next to the spiro atoms and proceeds first through the small ring and then through the spiro atom and around the second ring. Some examples are given below :
6 9 7 8 1 8 9 1 7 5 6 1 2 Cl 7 2 4 3 7 4 8 1 3 2 10 3 spiro[2, 2] pentane 6 5 5 4 4 6 5 Simplest spiro compound 3 2 spiro [3.4]octane Spiro [3.5] nonane Spiro [2.4] heptane 2-Chlorospiro [3, 5] decane (b) IUPAC nomenclature of Unbranched Assemblies consisting of two or more identical hydrocarbon units joined by a Single Bond. These systems are named by placing a suitable numerical prefix such as bi for two, ter for three, quarter for four, quinque for five etc. before the name of the repetitive hydrocarbon unit. Starting from either end, the carbon atoms of each repetitively hydrocarbon unit are numbered with unprimed and primed arabic numerals such as 1, 2, 3...., 1' , 2', 3' ...., 1'' , 2'', 3'' ... etc. The points of attachment of the repetitive hydrocarbon units are indicated by placing the appropriate locants before the name,
2 2' 3 4 4" 3" 1 1' Ex. 1' 2' 3 3' 2 1 1" 2" 1,1'-Bicyclopropane 4' 3' 1, 1', 2', 1" - Tercyclobutane (c) As an exception, unbranched assemblies consisting of benzene rings are named by using appropriate prefix with the name phenyl instead of benzene. For example,
3 2 2' 3' 2'' 3'' 1 4' 4 4'' 1' 1'' 5 6 6' 5' 6'' 5'' 1,1',4',1''-Terphenyl Example 2 : Give IUPAC name of the compound : C2H5 – CH(CH3) – CH2 – CH(CH3)2 Sol. The compound, C2H5 – CH(CH3) – CH2 – CH(CH3)2 may be written as H H 6 5 4 3 2 1 CH3 – CH2 – C – CH2 – C – CH 3
CH3 CH3 The main carbon chain consists of six carbon atoms and hence the parent alkane is hexane. The numbering is done from the right end of the chain. The two methyl groups are attached to the main chain at carbon number 2 and 4. Thus, the name of the compound is, 2, 4 dimethylhexane.
Gyaan Sankalp 33 Organic chemistry-Some basic principles and techniques Example 3 : Write down the the IUPAC name of the compound : O
CH3 – C – CH2 – CH 2 – CH 3 O
Sol. The compound, CH3 – C – CH2 – CH 2 – CH 3 contains a carbonyl group, and a chain consisting of five carbon atoms. Therefore, the given compound is named after pentane (5 carbon alkane) and the numbering is done from the carbon nearer to the carbonyl group. Thus, the correct name of the compound is O 1 2 3 4 5 CH3 – C – CH2 – CH2 – CH 3 pentan-2-one Example 4 : Give the name of the compound :
CHCHCHCHCH3 2 2 3 | OH Sol. The given compound has a chain of five carbon atoms and a hydroxy group as the functional group. The numbering is done from the end closer to the carbon having – OH group. Therefore, the compound is 1 2 3 4 5 CH3 – CH – CH2 – CH2 – CH 3
OH pentan-2-ol Example 5 : Give IUPAC name of alkane compound : CH3 – CH = CH2 – C C – CH2 – CH3 Sol. The longest chain contains 8 carbons. So, the parent alkane is octane. Numbering of the chain is done from the end nearer to the carbon having double bond. Thus, 1 2 3 4 5 6 7 8 CH3 – CH = CH – CH2 – C C – CH 2 – CH 3 The compound contains one double and one triple bond. So, the suffix would contain both the suffixes ene and yne. Indicating their positions in the chain, the suffix is 2-en-5-yne. So, the molecule is named as, oct-2-en-5-yne. Example 6 :
Name the compound : C(CH)3 3
Sol. The molecule has a terty-butyl group bounded to a nine-membered cycloalkane. It is tert-butylcyclononane. Alternatively, the tert-butyl group could be named systematically as a 1,1-dimethylethyl group, and the compound would then be named (1, 1-dimethylethyl) cyclononane. (Parentheses are used when necessary to avoid ambiguity. In this case the parentheses alert the reader that the locants 1,1 refer to substitutents on the alkyl group and not to ring positions.)
TRY IT YOURSELF Q.1 Write IUPAC name of the following compounds : Me CH3 | Me Me (a) CH3 – CH2 – CH – CH – CH2 – CH2 – CH3 (b) CH2 = CH – CH(CH3)2 (c) | Me Me CH – CH2 – CH3 | CH3
34 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
CH3 O | || (d) (CH3)3COH (e) CH3 – CH – CH2 – CH2 – C– CH2Cl O CH3 CH6 5 | (f) CH2 = CH – C = CH – CH = CH2 (g) CH3 – C – CH2 – CHO (h) CH3 C CH C CH3 | | Br CH3 CHCl
CH3 (i) Glycerol (j) Acetone (k) Formic acid (l) Oxalic acid
(m) Isopropyl alcohol (n) (o)
OH (p) (q)
Q.2 What is the IUPAC Name of t-butyl alcohol. (A) Butanol–2 (B) 2–Methyl&Propanol–2 (C) Butanol–1 (D) Propanol&2
Q.3 The compound is known by which the following name – (A) Bicyclo [2, 2, 2] octane (B) Bicyclo [2, 2, 1] octane (C) Bicyclo [1, 2, 1] octane (D) Bicyclo [1, 1, 1] octane
Q.4 The IUPAC name of is -
(A) 3-methyl cyclohexene (B) 1-methyl cyclohex-2-ene (C) 6-methyl cyclohexene (D) 1-methyl cyclohex-5-ene Q.5 The general formula CnH2nO2 could be for open chain - (1) diketones (B) carboxylic acids (C) diols (D) dialdehydes
C C CH CH CH Q.6 Consider the following compound : 2 3 OH The IUPAC name for the same is – (A) 5–Phenyl–4–pentyne–2–ol (B) 1–Phenyl–1–pentyne–1–ol (C) 1–(2–hydroxypropyl)–2–phenyllacetylene (D) None of these Q.7 The IUPAC name of CH3COCH (CH3)2 is - (A) isopropyl methyl ketone (B) 2-methyl-3-butanone (C) 4-methylisopropyl ketone (D) 3-methyl-2-butanone H
CH3 C CH2 CH 3 Q.8 IUPAC name of the following compound :
(A1) 2–Cyclohexylbutane (B) 2–Phenylbutane (C) 3–Cyclohexylbutane (D) 3–Phenylbutane ANSWERS (1) (a) 3, 5–Dimethyl-4-propyl heptane; (b) 3–Methyl–1–butene; (c) 5, 6–Diethyl–3–methyl-dec-4-ene; (d) 2-Methylpropanol-2; (e) 1-Chloro-5-methyl-hexan-2-one ; (f) 3-Bromo-hex-1, 3, 5-triene ; (g) 3–cyclopropyl-3- methylbutanal; (h) 3–(1’-Chloroethyl)–4–methyl-4-phenyl-pent-2-one; (i) propane- 1, 2, 3-triol ; (j) Propanone; (k) Methanoic acid; (l) Ethanedioic acid; (m) Propan-2-ol ; (n) 3-Methyl hexane ; (o) 3, 3, 4-Trimethyl octane; (p) 2-Methyl but-1-ene-3-yne (q) 1, 2-Ethanedioic acid; (2) (B) (3) (A) (4) (A) (5) (B) (6) (A) (7) (D) (8) (A)
Gyaan Sankalp 35 Organic chemistry-Some basic principles and techniques ISOMERISM Different organic compounds with the same molecular formula but different physical and chemical properties are called isomers and the general phenomenon is known as isomerism. It is classified as follows. Classification or types of isomerism : ISOMERISM
Structural isomerism Stereoisomerism
Confirmational Chain Position Functional Metamerism Configrational Tautomerism isomerism isomerism isomerism isomerism Isomerism
Geometary Optical Structural Isomerism : Compounds which possess the same molecular formula but differ in bonding arrangement of atoms (or) groups within the molecule i.e. differ in structure [A structural formula for a compound conveys which atom is directly linked to which other atom. It, however, tells nothing about the shape of the molecule but indicates the groups of elements present which provide clues to the properties of the substance], are structural isomers and this phenomenon is known as structural isomerism. (A) Chain or Nuclear Isomerism: In this type the isomers differ in the variation of the carbon chain (or) skeleton of the molecule. The same molecular formula may represent a straight chain of carbons as well as a branched chain. The molecular formula C4H10 stands for two isomers namely n-butane and isobutane.
H3C CH3 CH and CH3 CH 2 CH 2 CH 3 n Bu tane CH3 isobutane (2 methyl propane)
Ex. C5H12 stands for three chain isomers
CH3
CH CH3 (i) (ii) H3C CH2 CHCHCHCHCH3 2 2 3 3 (n-pentane) isopentane
CH3
(iii) H3C C CH3
CH3 neopentane (2,2-dimethyl propane)
Ex. Cyclohexane and methyl cyclopentane are nuclear isomerism
CH3 CH2 H2C CH2 CH H2C CH2 H2C CH2 H2C CH2 CH2
36 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
CH3 OH and H C Ex. C4H10O : H3C OH 3 butan-1-ol 2-methylpropan-1-ol
Ring chain isomerism :– They can also be considered as functional isomers.
Ex. (i) C4H8 , 2-butene cyclobutane
(ii) C4H6 or , , , 1-butyne 3-methy 1,3-butadiene bicyclo (1,1, 0) butane cyclobutene cyclopropene (B) Position Isomerism : Position isomerism is shown by the compounds in which there is difference in the position of attachment of functional group, multiple bond or substituent along the same chain length of carbon atoms - Ex. (i) Molecular formula : C3H7X (X = halogen, NH2, OH, or OR)
CH3 – CH2 – CH2 CH3 – CH – CH3 | | X X (i) (ii) In these structure three carbon atoms form a chain, and X is joined at the end in (i), while at the middle carbon in (ii). To be specific.
(a) CH3 – CH2 – CH2Cl and CH3 – CH – CH3 : Position isomers | Cl 1-Chloropropane 2-Chloropropane
(b) CH3 – CH2 – CH2OH and CH3 – CH – CH3 : Position isomers | OH 1-Propanol 2-Propanol
(c) CH3 – CH2 – CH2 – NH2 and CH3 – CH CH3 : Position isomers | n-Propylamine NH2 Isopropylamine (ii) Molecular formula : C4H8 CH3 – CH2 – CH = CH2 and CH3 – CH = CH – CH3 : Position isomers 1-Butene 2-Butene In the disubstiuted benzene derivatives position isomerism also exists because of the relative position occupied by the substituents on the benzene ring. Thus, Chlorotoulene, C6H4(CH3)Cl exist in three isomeric forms - ortho, meta and para.
CH3 CH3 CH3 Cl
Cl Cl o-Chlorotoluene m-Chlorotoluene p-Chlorotoluene
Gyaan Sankalp 37 Organic chemistry-Some basic principles and techniques
(C) Functional Isomerism : If the molecules have the same molecular formula but differ in the type of the functional group , then it is known as functional group isomerism. The following pairs of families show this isomerism. (i) Monohydric alcohol and ether (ii) Aldehyde and ketone (iii) Acid and ester (iv) Cyanide and isocyanide (v) nitroalkane and alkyl nitrite (vi) Oxime, amide and many more (vii) Alkene and Cycloalkane (viii) Alkyne, Alkadiene and Cycloalkene Ex. (i) Molecular formula : C2H6O CH3 – CH2 – OH and CH3 – O – CH3 : Functional isomers. Ethyl alcohol Dimethyl ether (Alcohol) (Ether) (ii) Molecular formula : C3H6O O O || || CH3 – CH2 – C – H and CH3 – C – CH3 : Functional isomers Propanal Propanone (Aldehyde) (Ketone) (iii) Molecular formula : C3H6O2 O || CH3 – CH2 – COOH and CH3 – C – O – CH3 : Functional isomers Propanoic acid Methyl acetate (Acid) (Ester) (iv) Molecular formula : CH3NO2 O || CH3 – N O CH3 – O – N = O : Functional isomers Nitromethane Methyl nitrite (Nitro) (Nitrite) (iv) Metamerism : It arises when different alkyl radicals are joined with the same divalent functional group present in the molecule, e.g. ethers, thioethers, secondary amines, ketones, esters etc.
H3C O O CH a) H3C O CH3 H3C CH3 3
diethyl ether 1-methoxypropane H3C 2-methoxypropane
H C S S 3 H3C CH3 CH3 b) H3C S CH3 1-(methylthio)propane (ethylthio)ethane H3C 2-(methylthio)propane
(v) Tautomerism : It is a special type of functional isomerism in which the isomers are readily interchangeable and maintain a dynamic equilibrium with each other. Hence, the name dynamic isomerism. Both the isomers represent one and a single sub- stance. Reversibility of change is due to mobility of a group or atom which can move from one position o another in the molecule, often with rearrangement of a double bond. These two forms are called tautomers and tautomerides of each other. Cause of tautomerism : Migration of acidic hydrogen from -carbon to electronegative atom which is bonded with multiple bond is the cause of tautomerism and this phenomenon is known as tautomerism.
OH
CH3 H C H 2 CH3 O (I) (II)
38 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (i) Both (I) and (II) are isomers known as tautomers. (ii) Tautomers are always in equilibrium state. (iii) Number of sigma, pi and lone pair of electrons in both tautomers are always same. (iv) Tautomerism is a chemical phenomenon which takes place only in liquid state. (v) Tautomerism is catalysed by acid as well as base. (vi)Tautomers are always reversible functional groups, functional isomers. Tautomerism is of several types the two important types are dyad and triad systems : (a) Dyad System: It involves the oscillation of H atom between two polyvalent atoms, e.g.., hydrogen cyanide (I) and hydrogen isocyanide (II). H C N H N C (I) (II) In the above example the H atom vibrates between carbon and nitrogen atoms. The alkyl derivatives of (I) are called cyanides while those of (II) isocyanides. (b) Triad System: In it one H atom oscillates between three polyvalent atoms. Keto-Enol Tautomerism : When the tautomers exist in the two forms keto & enol then, such type of tautomerism is called keto- Enol tautomerism’. It was discovered by the scientist ‘Knorr’ in 1911 in acetoacetic ester. The Keto means the compound has a Keto group > C = O, and the enol form has both double bond and OH (hydroxy) group joined to the same carbon. H O OH | || | – C – C – – C = C – | | Keto forms Enol forms O H || |
Ex. (i) CHC–H3 – CH2 = C – OH Acetaldehyde Vinyl alcohol O OH || | – = (ii) CHC–CH3 3 CHC–CH2 3 Acetone Isopropenyl alcohol O H || | OH | (iii) CH3 – C – CH – COOC2 H 5 CH3 – C = CH – COOC2 H 5 Aceto acetic ester (AAE) -Hydroxy crotonic ester Keto form Enol form
= (iv) CH3 – C – CH2 – CH3 CH2 C – CH2 – CH3 CH2 – C = CH – CH3 || | | O OH OH Butanone (Keto) (Enol) (Enol) OH OH | || | (v) CH3 –C= CH–COCH 3 CH3 –C–CH–COCH 3 Acetyl acetone In order for conversion of a keto form to its enol form it must have an -hydrogen (i.e., hydrogen attached to the carbon adjacent to the carbonyl group). Thus benzaldehyde, m-chlorobenzaldehyde (in general, aromatic aldehydes) formaldehyde, trimethylacetaldehyde do not exist as their enol forms. O O || C – H CH3 || | C – H H – C – H CH3 – C – C – H || | || O CH3 O Cl Benzaldehyde m-Chlorobenzaldehyde Methanal Trimethyl-acetaldehyde
Gyaan Sankalp 39 Organic chemistry-Some basic principles and techniques Cyanide-iso cyanide tautomerism H – C N C N – H Nitro-acinitro-tautomerism O O CH – N CH = N 3 O 2 OH
Structural requirement for tautomerism : (i) Compound should have electronegative atom bonded with multiple bond (i.e. N and O). (ii) Compound should have atleast one acidic hydrogen present on -carbon of the molecule. It the compound favour these two conditions it will show tautomerism. For example.
O O Electronegative atom Electronegative atom H C H3C 3 H CH3
Electronegative atom N Electronegative atom H3C N O H3C
CH3 O These compounds fulfill both the conditions, hence these will show tautomerism.
C H CH3 O C6H5 O 6 5
H3C H3C N H5C6 H O N CH3 C6H5 CH3 In these compounds electronegative atom is bonded with multiple bond but compound has no hydrogen on -carbon hence these compounds will not show tautomerism. If compound has active methylene group (or) active methyne group then -carbon for tautomerism is always carbon of active methylene (or) methyne group. For example
O H3C O CN O CH2 O CH2 HC NC CH2 H5C2OOC CH CH H C CH 3 3 3 3 CN active methylene group active methylene group active methylene group These compounds will also show tautomerism because these compounds fulfill both structural requirements. Enolisation : The conversion of a keto form into an enol form is known as enolisation and the enolisation of a compound has been found to depend upon various factors such as structural factor, temperature and nature of solvent. However, the most important is the structural factor like resonance and hydrogen bonding. Quantity of enol µ no. of –CO groups µ no. of a–H O O O || || || Ex. CH – C– CH – C– CH CH – C– CH – C– H CH – C– CH CH – C– H 3 2 || 3 3 2 || 3 3 3 || O O O
40 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Enol content of some compounds are :
Compound Enol Percentage Compound Enol Percentage
O O O H C 3 0.00025 H3C O 31.00 CH 3 CH3 CH3
O O O CH3 H3C 0.0056 80.4 H3C CH3 O
O O O O
H3C O 4.8 89.0 H5C6 CH3 CH3
O O O
H3C O 7.7 99.99
C2H5
Note : (a) Ketonic form predominates in simple monocarbonyl compounds like acetaldehyde, acetone and cyclohexanone. This is due to the greater bond strength of C = O {>C = O, 365 kJ/mole) present in keto form than the carbon-carbon double bond (C = C, 250 kJ/ mole) present in enolic form. (b) Enolic form predominates in -di ketones due to intramolecular hydrogen bonding and resonance. Intramolecular hydrogen bonding stabilizes enol form by 7 kcal/mole and resonance stabilizes enol form by 15 kcal/mole. Thus enol form is more stable than keto form by 22 kcal/mole in 1,3-diketones. Mechanism of tautomerism : Acid catalysed tautomerism is a two step process. (i) Proton transfer from the acid catalyst, H—A, to the carbonyl oxygen forms the conjugate acid of the aldehyde (or) ketone.
O O H
fast H C C CH + :A H3C C CH3 H—A 3 3 (conjugate acid of ketone)
(ii) Proton transfer from the -carbon to the base, A gives the enol and generates a new molecule of the acid catalyst, H—A.
O H O H
slow H—A H3C C CH2 H H 3C C CH 2 Enol form with unsymmetrical ketones, enolization may occur in either of two directions. OH O OH H2C C CH2 CH3 H3C C CH2 CH3 H3C C CH2 CH3 but-1-en-2-ol 2-Butanone but-2-en-2-ol
Gyaan Sankalp 41 Organic chemistry-Some basic principles and techniques the ketone is by far the most abundant species present at equilibrium. Both enols are also present, but in very small concentra- tions. The enol with the more highly substituted double bond is the more stable of the two enols and is present in higher concentration than the other.
O O H H O O O O
H3C CH3 H3C CH3 H3C CH3
H H H H O O O O O O O O
H C CH 3 3 H3C CH3 H3C CH3 H3C CH3
STEREO ISOMERISM : Two or more compound which have the same molecular formula and the same connectivity of atoms but different three dimensional arrangement of their constituent atom or groups are said to be stereoisomers and the phenomenon is termed as stereoisomerism. The branch of chemistry which deals with the study of molecular structures in three dimensional space is called stereo chemistry. Thus stereoisomers have : (a) same molecular formula (b) same molecular weight (c) same empirical formula and same structures (d) different properties and orientation of atom or molecules in space. It is of two types : (i) Geometrical (ii) Optical. Stereoisomers can also be classified as (a) conformational isomers that include rotation about single bonds and amine inversion. (b) configurational isomers that include cis-trans isomers and isomers that contains chirality centers. Conformational isomers can interconvert rapidly at room temperature. Because they interconvert, conformational isomers cannot be separated. Conformational isomers are also called conformers. Congiurational isomers cannot interconvert (unless a covalent bond breaks, which can happen only under-vigorous conditions. Because they cannot interconvert, configurational isomers can be separated.
GEOMETRICAL ISOMERISM : Isomerism where the isomers possess the same structural formula containing a double bond and differ only in respect of the arrangement of atoms or groups about the double bond. It is shown by alkenes or their derivatives in which two different atoms or groups are attached to each carbon containing the double bond. Double bond create restricted rotation.
N Ex. > C = C < , > C = N – , – N = N – , N N or N
The compound having the formula abc = cab occur in two forms and exhibits geometrical isomerism. a c b a c b
a c b b c a (cis form) (trans form) (Same groups lie on same side) (Same groups lie on opposite side) If the two atoms or groups linked to same doubly linked carbon are similar as in the molecule aac = cab, the compound does not show geometrical isomers. Here the two possible configurations are, in fact the same. a – c – a a – c – a || || b – c – a b – c – b (I) (II) Cyclic compounds can also have cis and trans isomers because the cyclic system prevents free rotation about the single bonds. Nomenclature of geometrical isomerism : (i) Cis and trans Cab b – c – a a – c – b || || || Cab b – c – a b – c – a cis trans 42 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
Ex. CH3 – CH = CH – CH3 [2-Butene] CH H CH3 CH3 3 C = C C = C
H H H CH3 cis-2-Butene trans-2-Butene CH (COOH) = CH (COOH) [Butenedioic acid] H H HOOC H
C = C C = C
HOOC COOH H COOH Maleic acid (cis) fumaric acid (trans) CH3 – CH = CH COOH [2-Butenoic acid] H H C CH3 3 COOH C C C C H COOH H H crotonic acid (trans) isocrotonic acid (cis) CHCl = CHCl [1,2-Dichloroethene] H H H Cl
C C
Cl Cl Cl H (cis) (trans) (ii) E-Z system of designating configuration : Cis and trans designations for geometrical Isomers are suitable when structurally similar or Identical groups are joined to the both carbon atoms. [cis-trans] Method does not work when there are no similar groups. Ex. [Br ClC = CFI] The configuration can be specified by using the notations Z and E. The method of E, Z-system : Find out the groups of highest priority on each carbon :– If two high priority groups are on the same side the configuration is Z. (zusammen German word for together). If the two groups of high priority are on apposite side, the configuration is E. (entgegen = opposite) (a) Naming of compound is based on priority order (atomic number of the atoms bounded directly to a particular sp2 carbon) atomic number priority atomic number priority Ex. I > Br > Cl > O > C > H (b) In the case of a tie in the atomic numbers, the atoms that are attached to the “tied” atoms must be considered. If an atom is doubly bonded to another atom, the priority system treats it as if it were bonded to two of those atoms. In the case of isotopes (atoms with the same atomic number but different mass numbers), the mass number is used to determine their relative priorities. Deuterium and hydrogen have the same atomic number, but deuterium has a greater mass number, so it has the higher priority.
H3C H C H H3C 3 C C C C Ex. (i) Br H Br H3C ; priority is : Br > CH3 , CH3 > H E(trans) Z(cis)
Br H Cl F C C C C (ii) Cl I Br I E(trans) Z(cis)
Gyaan Sankalp 43 Organic chemistry-Some basic principles and techniques
Higher Higher Higher Cl H CH3 CH2CH2CH3
(iii) C C C C CH CH H 2 3 H Br Higher (Z)-3-ethyl-2-hexene (E) -1-bromo-2-chloroethene (iii) Geometrical isomerism due to – N = N– bond : Ex. (a) H2N2O2 (Hyponitrous acid) : N – OH N – OH || || N – OH HO – N Syn isomer Anti Isomer (b) Azobenzene (C6H5 – N = N – C6H5) : C H C H H CH6 5 6 5 C6H5 H 6 5 N N C C
N N (c) N C H C6H5 OH 6 5 OH Syn Anti Syn Anti Properties related to geometrical isomers. (i) Geometrical isomerism arises due to restricted rotation about a double bond (ii) Stability : trans > cis The trans isomers of alkenes are more stable than their corresponding cis isomer. In the cis isomer, the two bulkyl groups remain very close to each other. The repulsion is more pronounced hence less stable. In trans isomer the bulky groups lie on opposite side, far apart hence less repulsion more stable.
H R R R C C C C Ex. H R H H trans cis (iii) Physical properties : (a) Dipolemoment() : cis > trans
H C Cl H C Cl
Ex. H C Cl Cl C H cis trans
H – C – CH3 H – C – CH3 || || H – C – CH2 – CH3 CH3CH2 – C – H Imp. Exception : 1-chloro propene
H – C – CH H – C – CH3 3 || || H – C – Cl Cl – C – H
cis trans (b) Polarity :cis > trans polarity (c) Solubility : cis > trans Polar naturesolubility 44 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (d) Boiling point : cis > trans (e) Melting point : trans > cis Melting point is not depends on polarity but depends on packing capacity of molecule. Packing capacity Melting point
Ex.
(f) Separation : cis and trans isomers can be separated by fractional distillation or fractional cystillisation. (g) cis and trans form can be interchange : The interconversion cis trans or trans cis is possible. On heating -bond is broken and reforms on cooling 373 K Ex. Maleic acid (cis) Maleic anhydride HO2 513 K HO Fumaric acid (trans) Maleic acid 2 maleic anhydride. Anhydride formation in maleic acid is easier than fumaric acid The number of geometrical isomers for alkenes : (a) If alkene has one double bond : Then geometrical isomers is two Ex. 2-butene CH CH3 CH3 H 3 C C , C C H H CH3 H cis trans (b) If alkene has two or more double bond : (i) In structure if the terminal groups are nonsimilar. Then we can use formula. Then number of isomers = 2n [Where n = no. of double bond (n = odd/even)] Ex. CH3 – CH = CH – CH = CH – CH = CH – Br No. of isomers = 23 = 8 (ii) In structure if the terminal groups are similar. Then we can use formula. Then number of isomers = 2n–1 + 2p–1 (n = double bond) p = n/2 when n is even number n 1 p = when n is odd number 2
Ex. (i) H3C – CH = CH – CH = CH – CH = CH – CH3 2, 4, 6–octatriene 3 1 n = 3, p = ; Number of isomers = 22 + 21= 4 + 2 = 6 2
(ii) H3C – CH = CH – CH = CH – CH = CH – CH = CH – CH3 2, 4, 6-decatetraene n = 4, p = 21 = 2 Number of isomers = 23 + 21= 4 + 2 = 6
OPTICAL ISOMERISM : (i) The geometrical isomers differ in physical properties such as melting point, boiling point, density etc. and also in certain chemical properties. (ii) The optical isomers will have the same chemical reactions and will be alike in all physical properties mentioned above. (iii)They can only be distinguished by their action on plane polarized lights this property which is often referred to as the optical activity. Optical activity : (i) Plane polarised light can be obtained by passing ordinary light through Nicol prism.
Gyaan Sankalp 45 Organic chemistry-Some basic principles and techniques
ordinary light Nicol prism Plane polarised waves vibrating in all directions waves vibrating in one direction.
Plane rotated to the left Plane rotated to the right (ii) When certain organic compounds in their solutions are placed in the path of a plane polarized lights have the remarkable property of rotating its plane through a certain angle which may be either to the left or to the right if the polarized light has its vibrations in the plane AB before entering such a solution, the direction on leaving it will be changed to say A'B', the plane have been rotated through the angle as in a (fig.). (iii)This property of a substance of rotating the plane polarized light is called optical activity and the substance possessing it is said to be optically active. A A' A'
solution or B B' B' (A) The observed rotation of the plane of polarised light produced by a solution depends on. (i) Nature of the substance (ii) Solvent used (iii) Concentration of solution (iv) Length of polarimeter tube (v) the temperature of the experiment and (t). (vi) the wavelength of the incident light used (). (B) For the measurement of optical rotations, a term specific rotation is introduced. (i) This is a physical constant characteristic of a substance as much as the melting point, boiling point, density or its refractive index. (ii) It is defined as the number of degrees of rotation observed when light is passed through 1 decimeter of its solution having concentration 1 gram per milliliter. (iii)The specific rotation of a given substance can be calculated by using the following expression. [] obs c
where [] = specific rotation determined at tºC, obs – the observed angle of rotation, = length of the polarimeter tube c = concentration of the active compound in gm/ml. The sign attached with the angle of rotation signifies the direction of rotation. Negative sign(–) indicates that the rotation is toward the left, while positive (+) sign means that the direction of rotation is towards right. The simple organic compounds which show optical activity are : CH3 * | Lactic acid CH – CH (OH) – COOH ; 2-Methyl butanoic acid 3 C2H5 – CH – COOH *
CH3 | 2-Methyl butanol C2H5 – CH – CH2OH * All these substances are known to exist in two forms. (1) One rotating the plane of polarised light to the left. This form is named as laevorotatory. (Latin, laevous = left) or direction (–) form (2) one rotating the plane of polarized light exactly to the same extent but to the right this form is named dextrorotatory (Latin Dexter - right) or direction (+) – form.
46 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (3) An inactive form which does not rotate the plane polarized light at all. This mixture of equal amounts of (+) and (–) — forms and hence it is optically inactive. It is named (±) - mixture or Racemic mixture. (Latin, Racemic - mixture of equal compounds) Note : Three lactic acids are known they are : (a) (+) - lactic acid, (b) (–) - lactic acid and (c) (±) mixture since the (±) acid is only a mixture of (+) - and (–) forms, in reality lactic acid exists in two forms. (a) These two acids are exactly identical in physical and chemical properties but differ in their action on the plane polarized light as they have different sign of specific rotation.
CHIRALITY CENTER (Stereogenic centre, stereo centre or assymetric carbon) A carbon bonded to four different groups is called a chirality center. The chirality center in each of the following compounds is indicated by an asterisk. For example, the fourth carbon in 4-octanol is a chirality center because it is bonded to four different groups (H, OH, CH2CH2CH3, and CH2CH2CH2CH3). Notice that the difference in the groups bonded to the chirality center is not necessarily right next to the chirality center the propyl and butyl groups are different groups. The starred carbon in 2,4-dimethylhexane is a chirality center because it is bonded to four different groups – methyl, ethyl, isobutyl, and hydrogen.
CH3 * * CH CH CH CHCH CH CH CH CH CHCH CH | * 3 2 2 2 2 2 3 3 2 3 CH CHCH CHCH CH | | 3 2 2 3 OH Br | 4-octanol 2-bromobutane CH3 2,4-dimethylhexane Notice that the only carbons that can be chirality centers are sp3 hybridized carbons, sp2 and sp hyrbidized carbons cannot be chirality centers because they cannot have four groups attached to them. All organic compounds containing on asymmetric carbon atom (lactic acid, amyl alcohol etc.) are optically active. OH CH3 | CH3 COOH H5 C2 – C* – CH2 OH | H H lactic acid amyl alcohol (Optically active) (optically active) Example 7 : Examine the following for stereogenic centers : (a) 2-Bromopentane (b) 3-Bromopentane Sol. A stereogenic carbon as four different substitutents. (a) In 2-bromopentane, C-2 satisfies this requirements. (b) None of the carbons in 3-bromopentane have four different substitutents, and so none of its atoms are stereogenic centers. H H | | CH3 C CH 2 CH 2 CH 3 CH3 CH 2 C CH 2 CH 3 | | Br Br 2-Bromopentane 3-Bromopentane Elements of symmetry : A molecule as a whole is asymmetric if it does not possess any element of symmetry such as (i) Plane of Symmetry (ii) Centre of symmetry (i) Plane of symmetry or mirror plane : A imaginary plane which bisect a molecule such a way that one half of the molecule is exactly the mirror image of the other half.
(ii) Centre of symmetry : A imaginary point in a molecule which divides the structures of molecule in to same parts. Centre of symmetry COOH
HCOO H trans-cyclohexane-1, 4-dicarboxylic acid Gyaan Sankalp 47 Organic chemistry-Some basic principles and techniques Symmetrical molecule is always optically inactive. Chirality : This term has been recently used to describe such molecules which have no elements of symmetry, thus asymmetrical molecules are also called chiral molecules and optical activity is attributed to certain chiral centres in them. An asymmetrical carbon is a chiral centre. An asymmetrical object cannot be superimposed on its mirror image. Chirality is lost when the two atoms bonded to an asymmetric carbon become similar thus while lactic acid is optically active, propionic acid is not. Chirality or molecular dissymmetry cause of optical Isomerism : (i) The necessary condition for a molecule to exhibit optical isomerism is dissymmetry or chirality. (ii) Thus all organic compound which contain asymmetric carbon (c* abde) are chiral and exist in two tetrahedral forms. (iii) Although the two forms (I and II) have the same structure, they have different arrangements of groups a, b, d, e about the asymmetric carbon in fact, they represent asymmetric molecules they do not have a plane of symmetry, they are related to each other as an object to its mirror image and are non superimposable. (iv) The two models or structures (I and II) actually stand for dextro or (+) and laevo or (–) isomers. Since they are related to each other as mirror images, they are commonly called Enantiomers (Greek word, enantio = opposite, morph - form) or enantiomers thus optical isomerism is now often referred to as an enatiomers. (v) Optical isomers or enantiomers due to the presence of an asymmetric carbon atom in a compound differ only in the arrangement or configuration of groups of tetrahedral perspective. (vi) Examples of compounds which exist as (+) and (–) enantiomers. Mirror
COOH COOH
C C H OH CH HO 3 CH3 (I) (+) and (–) lactic acid (II) Mirror
H H
C C C6H5 C6H5 CH CH Cl 3 3 Cl (I) (+) and (–) -1-chloro-1-phenyl ethane (II) Criterion of enantiomerism : (i) The compound which contain one or more asymmetric carbon atoms show enantiomerism. (ii) But there are some known compound which have asymmetric carbon but due to presence of plane of symmetry do not show enantiomerism. (iii) Meso tartaric acid has two asymmetric carbons but is optically inactive.
COOH Plane of | symmetry H – C *– OH
H – C *– OH | COOH Meso tarteric acid Drawing Enantiomers : Once you have determined that a compound has a chirality center, there are several ways to draw the enantiomers. Perspective formulas show two of the bonds to the chirality center in the plane of the paper, one bond as a solid wedge coming out of the paper, and the fourth bond as a dashed line projecting back from the paper. You can draw the first enantiomer by putting the four groups on the four bonds to the chirality center in any order. Draw the second enantiomer by making a mirror image of the first enantiomer.
48 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
Br Br
C C H C CH3 3 H H CH2CH3 CH2CH3 Perceptive formulas of the enantiomers of 2-bromobutane You can also draw enantiomers using wedge-and-dash structures. Draw the first enantiomer by arranging the four groups bonded to the chirality center in any order. Draw the second enantiomer by interchanging two of the groups. It does not make any difference which two groups you interchange. (It is a good idea to make models to convince yourself that this is true.) Many organic chemists prefer to interchange the horizontal groups, because the two enantiomers then look like mirror images on paper. Br Br
CH3 C CH2CH3 CH2CH3 C CH3 H H Wedge and dash representations of the enantiomers of 2-bromobutane A shortcut for showing the three-dimensional arrangement of groups bonded to a chirality center was devised in the late 1800s by Emil Fischer. Fischer used a single dot to represent a bond that was slanted away from the viewer and a solid line for a bond that was directed toward the viewer. Several years later, Viktor Meyer modified Fischer's method. He used the point of intersection of two perpendicular lines to represent the chirality center; horizontal lines represent the bonds that project out of the plane of the paper toward the viewer, and vertical lines represent the bonds that project back from the plane of the paper away from the viewer. This method of representing the arrangement of groups bonded to a chirality center is known as a Fischer projection formula or, more simply, a Fischer projection. chirality center Br Br Br Br CH3 C CH2CH3 CH3CH2 C CH3 CH3 CH2CH3 CH3CH2 CH3 H H H H original Fischer projections Fischer projections used now
DL Configuration : In this system the standard chosen for assigning relative configurations was glyceraldehyde. The absolute configurations of (+) and (–) glyceraldehyde are as. CHO CHO
H OH HO H
CH OH 2 CH 2 OH D-Glyceraldehyde L-Glyceraldehyde – OH on right side – OH on left side The other stereoisomers can be assigned D or L notation by comparison with the structural arrangements of D- and L- glyceraldehyde. For the purpose of comparison the following steps may be followed. (i) The most oxidised carbon attached to the chiral center is placed on the top of vertical line (ii) The group with carbon atom forming a part of the chain is kept at the bottom of vertical line. (iii)The remaining groups are assumed to be projected towards the viewer and are placed along horizontal line. Now comparison can be made with the configurations of glyceraldehyde. For example, the two enantiomeric forms of lactic acid can be compared as follow : CHO CHO HO H Glyceraldehyde H OH CH 2 OH CH 2 OH D (+)-Glyceraldehyde L (–)-Glyceraldehyde
Gyaan Sankalp 49 Organic chemistry-Some basic principles and techniques
COOH COOH
Lactic acid H OH HO H
CH CH3 3 D(+)-Lactic acid L(–)-Lactic acid Note : (a) It may be noted that the capital letters D and L do not pertains to dextro or laevo rotatory. Specific rotation of a stereoisomer is an experimentally determined property whereas D, L notations are based on comparison. (b) D & L notation are applicable only for those stereoisomers which can be converted to or prepared from D & L glyceraldehyde.
R – S configuration (absolute configuration) (a) This is a newer and more systematic method of specifying absolute configuration to optically active compounds. This is totally depend on sequence rule or CIP-system (sequence rule proposed by R.S. Cahn, C.K. Ingold and V.Prelog) (b) This system of designating configuration has been used increasingly since the early 1960 and may eventually replace the DL- system. (c) In this system, the configuration of a stereiosomer is designated by using prefix R derived from Greek word rectus (means-right) and S derived from Greek word sinister (means-left). The procedure involves two main steps as described below. Step 1. Rank the groups (or atoms) which are bonded to the chirality centre in order of priority. The criterion of priority is based on certain set of rules known as sequence rules or Cahn. Ingold-Prelog (CIP) priority rules given as under. The sequence Rules or CIP-Priority rule : (i) The atoms or group directly bonded to the asymmetric carbon are arranged in order of decreasing atomic number and assigned priority 1, 2, 3, 4, accordingly. Thus in chlorobromofluoromethane (CHClBrF), the substituent Br (at no = 35), Cl (at no = 17), F (at no = 9) and H (at no 1) given the order of priorities. Br > Cl > F > H (1) (2) (3) (4) (ii) When two or more groups have identical first atoms attached to asymmetric carbon, the priority order is determined by considering the atomic numbers of the second atoms; and if the second atoms are also identical the third atoms along the chain are examined. Let us consider the three groups 1 2 1 2 3 1 2 3 methyl ; – CH – CH – H ethyl ; – CH – CH – CH n-propyl – CH 2 – H 2 2 2 2 3 In methyl and ethyl the first atom is carbon and therefore, atomic numbers of the second atoms II (at no 1) and (at no 6) decide the priority order, ethyl > methyl. While considering ethyl and n-propyl the second atom is also identical (carbon) and hence the third atoms (H, C) give the priority order n-propyl > ethyl. (iii) If the first atoms of the two groups have same substituents of higher atomic number, the one with more substituents takes priority. Thus — CHCl2 has a higher priority than — CH2Cl (iv) A doubly or triply bonded atom ‘A’ present in a group appended to asymmetric carbon, is considered equivalent to two or three singly bonded ‘A’ s, respectively. A A Thus, = A A equals (A, A) ; A A equals A (A, A, A) A A H | Hence between groups – C = O(O,O,H) and – CH2OH (O, H, H), the former will have higher priority. A phenyl group is handled as if it had one of the Kekule structures. A C A A It may be noted R and S system is merely a normeclature device and has nothing to do with the sign and magnitude of optical activity. Thus the complete description of an optically active compound must include both the direction of rotation (+ or –) and the configuration of the compound (R or S). A racemic modification is an equimolecular mixture of two enantiomers and is given the prefix (RS).
50 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Step 2. After deciding the priorities of atoms or groups around the chirality centre, the molecule is visualised in such a way that atom or group with lowest priority is directed away from the eye. Now, observe the remaining atoms or groups in decreasing order of priorities, i.e., from 1 to 2 to 3. In doing so, if movement of eye occurs in clockwise direction the configuration is specified as R. On the other hand, if eye moves in anticlockwise direction, the configuration is specified on S. If you forget which is which, you can imagine driving a car and turning the steering wheel to make a right or left turn. Let us now, use step 1 and step 2 to specify the configurations of enantiomers of (i) 2-butanol and (ii) lactic acid. * (i) 2-Butanol (CH3–CH– C2H5). The sequence of priority of groups around the chirality centre is OH(1), C2H5(2), CH3(3), H(4). | OH Visualising the enantiomeric forms with H directed away from the eye.
1 1 2 Eye moves 2 OH Eye moves HO C H C H 2 5 clockwise 2 5 H anticlockw ise H C from highest from highest to lowest to lowest 3 priority 3CH priority CH3 3 R-2-Butanol S-2-Butanol
* (ii) Lactic acid (CH3–CH – COOH). The sequence of priority of various groups around the chirality centre is OH(1), COOH(2), | OH CH3(3), H(4). Visualising the formulae with H directed away from eye.
1 2 1 HO 2 HO COOH HOOC H H Eye moves Eye moves
3 clockwise anticlockw ise CH3 CH3 3 R-Lactic acid S-Lactic acid
R and S Notations for the optical isomers represented by fischer projections (by planar formulae). Case 1 : When the atom or group of lowest priority is at the bottom : Simply rotate the eye in decreasing order of priority and find the configuration of chiral carbon. For example, (1) (2) I COOH
(2) (3) (1) (3) Br Cl Br CH3 (4) (4) H H (S)-Bromochloroidomethane (R)-2-Bromopropanoic acid Case 2 : When the atom or group of lowest priority is at the top. Rotate the entire molecule through an angle of 180º so that the atom or group of lowest priority is at the bottom. For example,
(4) (1) H NH2 (3) (2) (3) (2) H5C2 CH3 H3 C C2H5 ((R)-Butan-2-amine) (1) NH (4) 2 H
Gyaan Sankalp 51 Organic chemistry-Some basic principles and techniques Case 3 : When the atom or group of lowest priority is on the left hand side. Without changing the position at the top, rotate the molecule in the anticlockwise direction so that the atom or group of lowest priority comes to the bottom. For example,
(1) (1) NH2 NH2 (3) (2) (3) (4) H3 C COOH H CH3 (4) (2) H COOH (R)-2-Aminopropanoic acid Case 4 : When the atom or group of lowest priority is on the right hand side of the horizontal line. Without changing the position at the top, rotate the molecule in the clockwise direction so that the atom or group of minimum priority comes to the bottom. For example,
(1) (1) OH OH (2) (3) (2) (4) HOOC H H3 C COOH (3) (4) H CH3 (R)-2-Hydroxypropanoic acid (Lactic acid) When comparing two Fischer projections to see if they are the same or different, never rotate one 90° or turn one over, because this is a quick way to get a wrong answer. A Fischer projection can be rotated 180° in the plane of the paper, but this is the only way to move it without risking an incorrect answer. Example 8 : Do the following structures represent identical molecules or a pair of enantiomers ? CH3 OH
C C H and CH3 HO CH CH CH CH2CH2CH3 3 2 2 H Sol. The easiest way to determine whether tow structures represent identical molecules or a pair of enantiomers is to determine their configurations. If they have the same configuration, they are identical, if they have opposite configurations, they are enantiomers. The first structure shown has the S configuration and the second structure has the R configuration. Therefore, the structures represent a pair of enantiomers. Optical Isomerism In compounds with more than one Asymmetric carbon Atom : (i) An asymmetric carbon can produce molecular asymmetry. (ii) Thus the molecules containing an asymmetric carbon exist in two optically active forms, (+) - isomer and (–) - isomer, and an equimolar mixture of the two, (±) - mixture, which is optically inactive. (iii) When there are two or more asymmetric carbon atoms in a molecule, the problem is complicated considerably. (iv) An organic compound which contains two dissimilar asymmetric carbons, can give four possible stereoisomeric form. Thus 2- bromo-3-chlorobutane may be written as * * CH3 CH CH CH3 1 2 3 Br Cl (v) The two asymmetric carbons in its molecule are dissimilar in the sense that the groups attached to each of these are different. C2 has CH3, H, Br, CHClCH3 C3 has CH3, H, Cl, CHBrCH3 Such a substance can be represented in four configurational forms.
CH3 CH2 CH3 CH3 | | | | H C Br Br C H Br C H H C Br | | | | H C Cl Cl C H H C Cl Cl C H | | | | CH3 CH3 CH3 CH3 (I) (II) (III) (IV)
52 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (vi) The forms I and II are optical enatniomers (related as object and mirror image) and so are forms III and IV. These two pairs of enantiomers will give rise to two possible racemic modifications. (vii) It may be noted that forms I (2S, 3S) and III (2S, 3R) are not mirror images or enantiomers, and yet they are optically active isomers. Similarly, the other two forms i.e., II (2R, 3R) and IV (2R, 3S) are also not enantiomers but optically active isomers. (viii) Such stereoisomers which are optically active isomers but not mirror images, are called Diastereoismers or Diastereomers. (ix) Diastereoisomers have different physical properties. Thus they have different melting points, solubilities in a given solvent, densities, and refractive indices. They also differ in specifivrotations ; they may have the same or opposite signs of rotations. Like geometrical isomers, the diastereoisomers may be separated from each other (i) by fractional distillation due to their difference in boiling points. (ii) by fractional crystallisation due to their difference in solubility. (iii)by chromatography due to their different molecular shapes and polarity.
OPTICAL PURITY : Whether a particular compound consists of a single enantiomer or a mixture of enantiomers can be determined by its observed specific rotation. For example, an enantiomerically pure sample of (S)-( + )-2-bromobutane will have an observed specific rotation of +23.1° because the specific rotation of (S)-( + )-2-bromobutane is +23.1°. If, however, the sample of 2-bromobutane has an observed specific rotation of 0°, we will know that the compound is a racemic mixture. If the observed specific rotation is positive but less than +23.1°, we will know that we have a mixture of enantiomers and the mixture contains more of the enantiomer with the S configuration than the enantiomer with the R configuration. From the observed specific rotation, we can calculate the amount of each enantiomer present in the mixture. First we must calculate the optical purity using the following formula. observed specific rotation optical purity = specific rotation of the pure enantiomer For example, if a sample of 2-bromobutane has an observed specific rotation of +9.2°, its optical purity is 0.40. In other words, it is 40% optically pure. 9.2 optical purity = 0.40 or 40% 23.1 Because the observed specific rotation is positive, we know that the solution contains excess (S)-(+)-2-bromobutane. The optical purity tells us how much excess (S)-( + )-2-bromobutane is in the mixture. The mixture is 40% optically pure, which means that 40% of the mixture is excess S enantiomer and 60% is a racemic mixture. Half of the racemic mixture plus the amount of excess S enantiomer equals the amount of the S enantiomer present in the mixture. Thus, 70% of the mixture is the S enantiomer (1/2 × 60 + 40) and 30% is the R enantiomer. ) Number of optical isomers : Case : 1. When the molecule is unsymmetrical. (It cannot be divided into two halves) Number of d and isomers(a) = 2n, Number of meso form(m) = 0 Total number of optical isomers (a + m) = 2n ; where n is the number of chiral carbon atoms. Ex. 2, 3-Pentane diol : CH3 | H C* OH | H C* OH | CH2 5 d and isomers = 22 = 4 meso form = 0 Case : 2 When the molecule is symmetrical. (number of chiral carbon = even number) Number of d and forms(a) = 2(n–1) , Number of meso form(m) = 2(n/2–1), Total number (a + m) = 2(n–1) = 2(n/2–1) Ex. Tartaric acid
COOH | H C* OH | H C* OH | COOH
Number of d and forms = 2(2 – 1) = 2, Number of meso form = 2(2/2–1) = 2º = 1, Total optical isomers = 3 Gyaan Sankalp 53 Organic chemistry-Some basic principles and techniques Case : 3. When the molecule is symmetrical. (Number of chiral carbon = odd number) Number of d and forms(a) = 2(n–1) – 2(n/2–½) Number of meso form (m) = 2(n/2–½) Total number of isomers (a + m) = 2n – 1 Ex. HOOC – CH(CH3) – CHOH – CHBr – CHOH – CH(CH3) – COOH Number of d and form = 24 – 22 = 12 Number of meso form = 22 = 4 Total optical isomers = 12 + 4 = 16 Isomerism of tartaric acid : Let us consider the optical isomerism of tartaric acid which contains two similar asymmetric carbon atoms, in detail. The two asymmetric carbon atoms in tartaric acid, *CH(OH)COOH | *CH(OH)COOH are attached to the groups H, OH, COOH and CH(OH)COOH. Its molecule can be represented by space models of two tetrahedra joined at corner but for the sake of convenience we will use the planed formulas. The end groups being identical, in all four arrangements are possible according as one or both H groups and OH groups are on the left or on the right
COOH COOH COOH COOH | | | | H C OH HO C OH H C OH HO C OH | | | | HO C OH H C OH H C OH HO C OH | | | | COOH COOH COOH COOH (I) (II) (III) (IV)
Of these, formula IV when rotated through 180º in the plane of the paper becomes identical with formula III. Therefore, for tartaric acid we can have only three different arrangements.
COOH COOH COOH | H – C – OH HO – C – H H – C – OH | HO – C – H H – C – OH H – C – OH | COOH COOH COOH
(I) (II) (III) Three forms of tartaric acid (i) structure I will rotate the plane of polarised light to the right and will represent (+) - tartaric acid. (ii) structure II will rotate the plane of polarised light to the left and will represent (–) - tartaric acid, and (iii)structure III will represent optically inactive tartaric acid, since the rotatory power of the upper half of the molecules is balanced by that of the lower half. It may also be noted that formulas I and II are mirror images of each other and hence represent (+)-and (–)-isomers ‘Formula III, however, has a plane of symmetry (dotted line) and hence represent an inactive isomer of tartaric acid. In actual practice, four tartaric acids are known : (i) (+) - Tartaric acid ; (ii) (–) - Tartaric acid ; (iii) Inactive Tartaric acid ; this is also known as meso - Tartaric acid or m-tartaric acid ; and (iv) (±) Tartaric acid ; this form of tartaric acid being a mixture of equal amounts of (+) - and (–) isomers. This is optically inactive due to internal compensation. The three tartaric acids, (+), (–) -, m-, are all space isomers but m-tartaric acid is not a mirror image of either of the active forms. Hence it differs from them in melting point, density and other physical properties. External and internal compensation : (i) If equimolecular amounts of d-and -isomers are mixed in a solvent, the solution is inactive. (ii) The rotation of each isomer is balanced or compensated by the equal but opposite rotation of the other. (iii) Optical inactivity having this origin is described as due to External Compensation. Such mixtures of (+) - (–)-s isomers (Racemic mixture) can be separated into the active components.
54 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (iv) In meso tartaric acid the inactivity is due to effects within the molecule and not external. The force of rotation due to one half of the molecule is balanced by the opposite and equal force due to the other half. (v) The optical inactivity so produced is said to be due to Internal Compensation. (vi) It occurs whenever a compound containing two or more asymmetric carbon atoms has plane or point of symmetry. (vii) Since the optical inactivity of such a compound arises with in the molecule, the question of separating into active components does not arise.
COOH
H – C – OH
H – C – OH
COOH Inactivity of Meso Tartaric acid by Internal Compensation
RACEMISATION : It is a process of conversion of an optically active compound in to the optically inactive racemic mixture by the application of heat/light/acid/base. a a Ex. b C* x C + (sp2 Hybrid & Trigonal plane geometry) c b c
Nu–
a a b C*— Nu + Nu —C* b (Nu = Nucleophilic) c c (d mixture)
ASYMMETRIC SYNTHESIS : The synthesis of an optically active compound (Asymmetric compound) from an optically inactive compound (Symmetric compound). Ex. (a) Nucleophilic addition of HCN on Benzaldehyde & acetaldehyde)
CH3 CH OH CH OH 3 HO 3 C = O + HCN C* 2 C* H H CN H COOH
(Acetaldehyde) (Acetaldehydecyanohydrine) (Lactic acid)
¼Prochiral centre½
CH 6 5 CH6 5 OH CH OH HO2 6 5 C = O + HCN C C* H H CN H COOH (Benzaldehyde) (Benzaldehyde (Mendaleic acid) cyanohydrine)
Gyaan Sankalp 55 Organic chemistry-Some basic principles and techniques (b) Reduction of Pyruvic acid
Prochiral centre Chiral centre sp2 c-atom Red P * CH3 — C — COOH + 2H CH3 —CH — COOH || | O OH (Lactic acid) (c) HVZ reaction [Hell volhard zelinsky reaction] Red P CH3 — CH 2 — COOH + Br2 CH3 — CH — COOH + HBr | Br (Propanoic acid) ¼2–Bromo Propanoic acid½
CONFORMATION : (i) Molecules which differ from one another only by rotations about single– single bond. (ii) It one CH3 group of ethane is kept constant & other methyl group is permitted to rotate through C – C bond axis, an infinite no. of atomic arrangements are possible, which are called conformations. (iii) Structure of conformers including bond length & bond angle in same configuration is also same. (iv) Configuration is also same. (v) Conformers can not be separated so they are not isomers, they are only different forms of same molecule. Ethane is the best example of demonstrate conformational isomerism. In ethane molecule there is free rotation around the carbon- carbon single bond. The two orientations are : (a) Eclipsed confirmation : In this orientation hydrogen atoms on one carbon are exactly eclipsing the hydrogen atoms on the other. There is minimum distance between the various hydrogen atoms. The eclipsed form has the greater energy content due to maximum repulsive interactions (less stable). (b) Staggered confirmation : In this orientation the hydrogen atoms of one carbon are at maximum possible distance from that of the other. The staggered form has the less energy content due to minimum repulsive interaction. (more stable) Eec. – Est. = 3 kcal/mole The two forms of ethane are rapidly interconvertible at room temperature and are thus not separable. HH | | Canformers of ethane : HCCH | | HH (i) Sawharse projection formulae : (3D representation)
H H H C H H H H H 60º H H C
H H Staggered Conformation Eclipsed Conformation
56 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (ii) Newman projection formulae : (2 D representation) H H H H H H H
H H H H H H H H H H H staggered Skew Eclisped It many be noted that one conformation of ethane can be converted in to other by the rotation of 60º about the bond the infinite other conformations of ethane lying between the two extranes are called skew conformation. Stability order & energy diagram : Staggered > Eclipsed Eclipsed (Less stable)
y –1 g
r 12.5 kJ mol e n e
l a i t n e t o
P Staggered Staggered (Stable)
Rotation Changes in energy during rotation about C – C bond in ethane. Conformations of n-Butane : Butane molecule can be represented as derivative of ethane as : H H 1 2 3 4 H 3 C CH3 H H Considering the rotation around single bond between C-2 and C-3, we get many conformations of which the main conformations are. CH 3 CH3 CH3 H H H HC3 H
H H H H H H H CH3 CH3 H I. Staggered (Anti) II. Eclipsed III. Staggered (Gauche)
CH3 CH3 CH3 HC3 H CH3 H
H H H CH3 H H H H H H H IV. Fully Eclipsed V. Staggered (Gauche) VI. Eclipsed Gyaan Sankalp 57 Organic chemistry-Some basic principles and techniques Staggered conformation (I), in which methyl groups are as far apart as possible, is most stable due to minimum repulsion between methyl groups. This conformation is also called anti conformation. This on rotation through 60º gives eclipsed conformation (II), in which methyl group on one carbon is overlapped by the hydrogen atom on the other carbon. Further rotation through 60º gives another staggered conformation (III) in which methyl groups or two carbons are 60º apart. This conformation, is also called Gauche conformation. Gauche conformation on further rotation through 60º gives fully eclipsed conformation (IV) in which methyl groups on two carbons are just opposite to each other. In this conformation steric strain in maximum hence this conformation is most unstable. Further rotation through 60º gives again gauche conformation (V) which is mirror image of gauche conformation (III). Conformation (V), on rotation through 60º gives conformation (VI) which is again eclipsed conformation. The relative energies of various conformations of n-butane is The order of relative stabilities various conformers of n-Butane is Anti staggered > Gauche > Eclipsed > Fully Eclipsed
Fully Eclipsed
Eclipsed Eclipsed
18.4
y kJ/mole g r e
n 14.6
e 14.6 kJ/mole kJ/mole g n i s a e r c 3.76 kJ/mole 3.76 kJ/mole n I Anti Gauche Gauche Anti
0° 60° 120° 180° 240° 300° 360°
Angle of rotation Energy difference between various conformations of n-butane Conformations of Cyclohexane : If cyclohexane molecule were to be planar, the C-C-C angle would have been 120º. Therefore, ring would be highly strained. Cyclohexane avoids this strain by assuming conformations in which all bond angles between carbon atoms are close to tetrahedral angle, 109º 28'. The two most important conformations of cyclohexane are the chair form and the boat form as shown in fig. Of the two, the chair form is more stable than the boat form, the energy difference being about 30 kJ mol–1. However, the energy barrier between the two conformations is of about 44 kJ mol–1). This is because in the boat form, may hydrogens on adjacent carbons correspond to the unfavourable eclipsed conformation of ethane. Also, the two hydrogens marked H also called flagpole hydrogen in the boat form are quite close and repel one another. On the other hand, the chair form does not have these unfavourable interactions and all the hydrogens correspond to the more stable staggered conformation of ethane.
H Flagpole hydrogen H H H H H H H H H H H H H H H H H H H H H H H Chair conformer Boat conformer
58 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
H H H H
H H H H H H
H H H H H H H H
Chair Boat Conformation Conformation Chair and boat conformations of cyclohexane. Example 9 : Give the functional isomer for acetaldoxime.
O Sol. H3C H3C N OH NH (1E)-acetaldehyde oxime 2 acetamide
Oximes and amides are functional isomers. Example 10 : Give the possible chain isomers for propyl benzene.
H3C CH3 CH2CH2CH3
Sol.
propylbenzene isopropylbenzene
Example 11 : Give the possible positional isomerism for di chlorobenzene.
Cl Cl Cl Cl Sol.
Cl 1,2-dichlorobenzene 1,3-dichlorobenzene Cl 1,4-dichlorobenzene
Example 12 : Give the metamer for ethyl propionate. O
O CH3 CH3 H3C H3C O Sol. propyl acetate O ethyl propionate Gyaan Sankalp 59 Organic chemistry-Some basic principles and techniques Example 13 : Give the possible chain isomers for C5H13N
NH2 H3C CH3 H C NH 3 2 H3C Sol. NH CH3 2 CH3 Example 14 : A solution prepared by mixing 10 mL of a 0.10M solution of the R enantiomer and 30mL of a 0.10M solution of the S enantiomer was found to have an observed specific rotation of +4.8°. What is the specific rotation of each of the enantiomers ? Sol. One millimole (mmol; 10mL × 0.10 M) of the R enantiomer is mixed with 3 mmol of the S enantiomer; 1 mmol of the R enantiomer plus 1 mmol of the S enantiomer will form 2 mmol of a racemic mixture. There will be 2 mmol of S enantiomer left over. Therefore, 2 mmol out of 4 mmol is excess S enantiomer (2/4 = 0.50). The solution is 50% optically pure. observed specific rotation Optically pure = 0.50 = specific rotation of the pure enantiomer 4.8 0.50 = x = +9.6° x The S enantiomer has a specific rotation of +9.6°, the R enantiomer has a specific rotation of –9.6°.
TRY IT YOURSELF Q.1 The type of isomerism shown by CH3CH(OH) COOH is - (A) Position isomerism (B) Stereo isomerism (C) Optical isomerism (D) Cis-trans isomerism
CH3 H Q.2 The compound C=C COOH shows – C CH3 H CH3 (A) Optical isomerism (B) Geometrical isomerism (C) Tautomerism (D) Optical & Geometrical isomerism Q.3 Which are isomers – (A) Ethanol and ethoxy ethane (B) Methanol and Methoxy methane (C) Propionic acid and ethyl acetate (D) Propionaldehyde and acetone Q.4 A compound whose molecule is superimposable on its mirror image inspite of the presence of chiral centres is called :– (A) Diastereomer (B) Meso compound (C) An enantiomer (D)Athreo enantiomer Q.5 The two isomers given below are –
CH3 CH3 H OH H OH OH H H OH COOH COOH (A) Enantiomers (B) Diastereomers (C) Mesomers (D) Position isomers Q.6 Which type of isomerism is mutually shown by the following structures-
Cl | CH3 – CH – CH2 – CH2 – CH3 CH3 – CH2 – C CH3 | | Cl CH3 (A) Only functional group isomerism (B) Only chain isomerism (C) Positional and chain isomerism (D) Only positional isomerism
60 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Q.7 Which of the following is not a functional group isomer of ethyl acetate - (A) Isobutyric acid (B) 2, 3-Butanediol (C) 3-Hydroxybutanal (D) 4-Hydroxybutanone
Q.8 Among the following structure which one is called erythro isomer :–
B B B A X A X A B (A) (B) (C) A X X A X A Y Y Y (A) (A) and (B) (B) Only (B) (C) Only (A) (D) Only (C) Q.9 The number of asymmetric carbon atoms and the number of optical isomers in CH3(CHOH)2COOH, are respectively – (A) 3 and 4 (B) 1 and 3 (C) 2 and 4 (D) 2 and 3 Q.10 Which of the following represent E isomer -
Cl Br Cl CH2 5 (A) C = C (B) C = C HC3 CH2 CH 3 CH3 CHO H CHCl HC3 CH = CH2 2 (C) C (D) C = C H Cl CH3 CHCl2
Q.11 The pair of structure given below represent Me H H H H H
and H H H H H Me (A) Enantiomers (B) Conformers (C) Position isomers (D) homologous Q.12 For which of the following parameters, the structural isomer C2H5OH and CH3OCH3 would be expected to have the same values ? (Assume ideal behaviour) (A) Boiling points (B) Vapour pressure at the same temperature (C) Heat of vaporization (D) Gaseous densities at the same temperature and pressure Q.13 An organic compound exhibits optical isomerism when – (A) Four groups linked to carbon atom are different (B) Three groups linked to carbon atom are different (C) Two groups linked to carbon atom are different (D) All the groups linked to carbon atom are same Q.14 Tautomerism is exhibited by – (A) (CH3)3CHO (B) (CH3)2NH (C) R3CNO2 (D) RCH2NO2 Q.15 Total number of isomers of C6H14 are – (A) 4 (B) 5 (C) 6 (D) 7
ANSWERS (1) (C) (2) (A) (3) (D) (4) (B) (5) (B) (6) (B) (7) (B) (8) (C) (9) (B) (10) (B) (11) (B) (12) (D) (13) (A) (14) (D) (15) (B)
Gyaan Sankalp 61 Organic chemistry-Some basic principles and techniques REACTION MECHANISM : A chemical equation is a symbolic representation of a chemical change. It indicates the initial reactants and final products involved in a chemical change. Reactants generally consist of two species: (i) One which is being attacked; it is called a substrate. (ii) Other which attacks the substrate; it is referred to as a reagent. These two interact to form products. Substrate + Reagent Products It is important for us to know not only what happens in a chemical change but also known how it happens. Most of the reactions are complexes and take place via intermediates which may or may not be isolated. The intermediates are generally very reactive. They react readily with other species present in the environment to form the final products. The detailed step by step description of a chemical reaction is called its mechanism. Substrate intermediates Products Mechanism is only a hypothesis which explains various facts regarding a chemical change. The reactions of organic compounds essentially involve changes in the existing covalent bonds present in their molecules. These changes may involve electronic displacements in the bonds, breaking of the bonds, energy changes accompanying the cleavage and formation of new bonds. To understand clearly the mechanism of various organic reactions, it is thus essential to have knowledge about the following well established concepts: (a) Cleavage (fission or breaking) of covalent bonds, (b) Nature of attacking reagents. (c) Electronic displacements in bonds,
TYPES OF BOND FISSION : (a) Homolytic (symmetrical) fission or Homolysis : (i) If a covalent bond breaks in such a way that each atom takes away one electron of the shared pair, it is called homolytic or symmetrical fission or homolysis. Homolytic fission A B A + B Free radicals
H3 C – OH H3 C + OH, H° = 92 kcal/mole methanol (ii) The neutral chemical species (such as A and B) which contain an odd or unpaired electron and which are produced by homolytic fission of covalent bonds are called free radicals. (iii)Favourable conditions : (a) High temperature (b) Light of suitable wave length (c) Nonpolar solvent (d) Presence of peroxide or oxygen (e) Ability of substrate and attacking reagent to produce free-radicals. (b) Heterolytic Bond Fission or Heterolysis In heterolysis, the covalent bond is broken in such a way that one species (i.e., less electronegative) is deprived of its own electron, while the other species gains both the electrons.
A : B A + B + – – + H C OH HC + OH 3 3 or H3 C OH HC3 + OH methyl cation Hydroxide ion methyl anion Hydroxyl cation Thus formation of opposite charged species takes place. In case of organic compounds, if positive charge is present on the carbon then cation is termed as carbocation. If negative charge is present on the carbon then anion is termed as carbanion. Carbocation and carbanion are the reaction intermediates. The factor which favours heterolysis is a greater difference of electronegativity between A and B. Thus (1) Heterolytic bond fission gives carbocation or carbanion as reaction intermediate. (2) Mechanism of the reaction in which heterolytic bond fission takes place is known as heterolytic mechanism or ionic mechanism. (3) The energy required for heterolytic bond fission is always greater than that for homolytic bond fission due to electrostatic force of attraction between ions. Favourable conditions : (a) Low temperature (b) Polar solvent (c) Presence of acid or base catalyst (d) Polar nature of the substrate and attacking reagent.
CLASSIFICATION OF REAGENTS : Electrophiles or electrophilic reagents (E+) : (i) They are electron deficient species. (ii) They are always in search of electrons. (iii) They have a tendency to accept electron pair from another molecule. (hence electro = electron, phile = love). 62 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (iv) In a reaction, an electrophile attacks the substrate at the point of maximum electron density. Electrophiles may be neutral or positively charged. In neutral electrophiles, centre atom has deficiency of electrons like in BF3, BCl3, AlCl3, BeCl2, FeCl3, SO3, etc. (v) All the positive ions behave like electrophiles. + + + + + + + + Cl , Br , I , NO2 , NO , H , H3O , NH4 , R , R — C = O , etc. (vi) Electrophiles are generated by heterolysis of a covalent bond. (vii) Transitional metal cations are electrophiles Fe3+, Fe2+, Ag+, Hg2+, Cd2+, etc. (viii) All Lewis acids are electrophiles. Nucleophilic reagents or Nucleophiles (Nu–) : (i) They are electron rich species they have a tendency to donate electron pair. (ii) They attacks the centre of minimum electron density in a chemical reaction. They are two types : (a) Charged Nucleophiles : All the negative ions qualify as nucleophiles. F–, Cl–, Br–, I–, OH–, CN–, RCOO– – – – – – RO , R , R—C C , NH2 , SH , etc. (b) Neutral Nucleophiles : Centre atom has electron pair. NH3 , R — NH2 , R – NHR , RN3 , HO:2 , R – ..O – H, R – ..O – R, H2 S.., R R – S.. – R, R – C = O , C = O | R H O || , NH OH , NH – NH , CH = CH , R C O H 2 2 2 2 2 R MgX
ELECTRON DISPLACEMENT IN ORGANIC MOLECULES: In organic molecules different reactions depend on the electron density in their molecules. Since majority of attacking reagents are polar, i.e. nucleophilic (or) electrophilic, hence in organic compounds permanent (or) temporary polarity is developed by temporary (or) permanent electron displacements. The different electronic effects which operates in covalent bonds are (a) Inductive effect (b) Mesomeric and resonance effect (c) Electromeric effect and (d) Hyper conjugation (a) Inductive Effect : In a covalent single bond between unlike atoms, electron pair forming the bond is never shared absolutely equally between the two atoms ; it tends to be attracted a little more towards the more electronegative atom of the two. Thus in an alkyl chloride, electron density tends to be greater nearer chlorine than carbon, as the former is the more electronegative; this is generally represented as in
Cl Cl
If the carbon atom bonded to chlorine is itself attached to further carbon atoms, the effect can be transmitted further .
For example C2H5 – Cl.
Cl C H3 CH2 2 1 a-I group, with draw electrons.
a slightly positive charge a polarization of this bond on the C 2 atom. Electron displacement along the chain
In ethyl chloride the effect of the chlorine atom, partial appropriation of the electrons of the carbon chlorine bond is to leave C1 slightly electron deficient this it seeks to rectify by in turn, appropriating slightly more than its share of the electrons of the bond joining it to C2, and so on down the chain. The effect of C1 on C2 is less than the effect of Cl on C1. However it decreases as the distance from source (more electronegative atom) increases. From practical point of view, it may be neglected after third Gyaan Sankalp 63 Organic chemistry-Some basic principles and techniques carbon atom. Inductive effect is denoted by the symbol ‘I’ and represented by a straight arrow (), the arrow head pointing towards the source (more electronegative element). The inductive effect causes certain degree of polarity in the bond which in term renders the bond much more liable to be attached by other charged atoms (or) group. The characteristic of inductive effect are (i) A permanent effect (ii) The electron never leave their original atomic orbital (iii)Operates through bond (iv) Polarisation of electrons is always in single direction. (v) Its magnitude (i.e. electron with drawing or donating power) decreases with increase in distance. (i) – I effect : Atoms or group of atoms which attract the bonded electrons more strongly than hydrogen atom, are said to have – I effect and are termed as electron attracting (–I groups). Such groups when linked with a carbon chain make it electron poor. NR Ex. 3 >– NO2> – CN > – COOH > –F > – Cl > – Br > – I > – OH > – OR > –C6H5 > – H (ii) + I Effect : Those atoms or group of atoms which attract the shared electron pair (bond pair) less strongly then hydrogen atom are said to have +I (electron repelling) effect. Such groups when attached with a carbon chain displace the shared electrons towards the chain and make it electron rich. (CH3)3C – > (CH3)2CH – > CH3 – CH2 – > CH3 – Inductive effect of hydrogen is zero, i.e. it neither attracts nor repels the bond pair between carbon and hydrogen. Applications of inductive effect : (i) Dipole moment: Since, inductive effect leads to a dipolar character in the molecule, it develops some dipole moment in the molecule, which increases with the increase in the inductive effect. CH3 I CH3 Br CH3 Cl 1.648D 1.79D 1.83D Increasing dipole moment (ii) In Bond Length: With increase in inductive effect, the bond length usually decreases because of increased ionic character, CH3 F CH3 Cl CH3 Br CH3 I 1.38Å 1.78Å 1.94Å 2.14Å (iii) Reactivity of Alkyl Halides: The carbon-halogen bond in tertiary alkyl halides is most reactive while it is least reactive in primary alkyl halides. This can be explained on the basis of greater + I effect in t-alkyl halides which pushes the – I effect; hence the bond becomes highly polar and weak. CH 3 H3C
H C H3C Cl H3C Cl Cl 3 Cl H3C CH3 (iii) Strength of Fatty Acids: As the number of alkyl groups attached to –COOH group increases, the acid strength decreases. Thus formic acid is stronger acid than acetic acid which is stronger than propionic acid and so on, due to increasing +I effect of alkyl groups.
O O H3C
H H3C O O H O H O H (Formic acid) (Acetic acid) (Propanoic acid)
(iv) Strength of Substituted Acids: Chlorinated acetic acids are stronger than acetic acid due to the – I effect of chlorine atom. O O H3C Cl H -I +I O O H (mono chloro acetic acid) (Acetic acid)
Larger the number of chlorine atoms, the greater will be –I and the stronger will be the acid.
64 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Cl Cl
Cl COOH COOH Cl H3C COOH Cl Cl COOH The relative strength of the different halogen substituted acids : FCH2COOH ClCH2COOH BrCH2COOH ICH2COOH (v) Basicity of aliphatic primary amines : In aliphatic amines, R – NH2, the +I alkyl group increases the electron density on nitrogen atom of –NH2 group. As a result, its tendency to donate the electron pair to a proton increases. Increasing order of basic strength of aliphatic primary amines is : NH3 < (CH3)3N < CH3NH2 < CH3CH2NH2 < (CH3)2NH Base strength of aliphatic and aromatic amines varies as follows : R – NH2 > NH3 > Ar – NH2
ELECTROMERIC EFFECT (E-EFFECT) : The electromeric effect may be defined as : The temporary effect which operates in the organic compounds having multiple bonds i.e. double or triple bonds under the influence of an outside attacking species. As a result, one -electron pair of the multiple bond gets completely transferred to one of the bonded atoms which is usually more electronegative. The electromeric effect is shown by a curved arrow ( ) representing the electron transfer originating from the centre of the multiple bond and pointing towards one of the atoms which is more electronegative. The effect can be illustrated by the attack of H+ ion on the molecule of alkene.
C C H C C
H Carbocation Types of electromeric effect : The electromeric effect is of two types i.e. + E-effect and – E-effect. (A) +E-effect :– If the -electron pair of the multiple bond is transferred to the atom to which the attacking reagent gets attached, the effect is called +E effect. For example,
CH H CH3 CH 2 CH3 CH CH3 Propene Isopropyl carbocation
In this case, CH3 group has +I (inductive) effect and the -electron shift can take place only to the right. (B) –E-effect :– In case the -electron pair of the multiple bond is transferred away from the atom which gets linked to the attacking reagent, the effect is known as –E-effect. For example, attack of CN– ion on formaldehyde, O– H – H O CN C H H CN Formaldehyde Anion
MESOMERIC EFFECT OR RESONANCE : Mesomeric effect (M effect) operates only in the molecules having a conjugated system, i.e., only in the molecules having alternate single and double bonds. As a result of mesomeric effect, the -electrons get delocalized, giving rise to a number of resonance structures of the molecule. The atoms or groups which draw electrons away from carbon atom are said to have a – M effect. – M effect atoms or groups : NO2, CN, > C = O The atoms or groups which release electrons towards carbon atom are said to have a +M effect. +M effect atoms or groups : Cl, Br, I, NH2, NR2, OH, OCH3 The +M effect of Cl atom is shown :
. . . .
.
......
. . . . .
. +
. . .
+
. . + . Cl Cl Cl Cl . Cl .– . . . –
.– .
Gyaan Sankalp 65 Organic chemistry-Some basic principles and techniques The – M effect of nitro group is shown : . . – O + O + + – C = C – C = C – N – C – C = C – C = N | – | | | O – O
RESONANCE There are many organic compounds which can be represented by two or more such structures which have same arrangement of atoms but different arrangements of electrons . All the properties of these compounds cannot be explained by anyone of these structure because actual structure of these compounds is combined form of all possible structures .This phenomenon is called resonance. All possible structures are known as resonating structures or canonical forms and the combined form of all these resonating structures which can explain all the properties of that compound is termed as resonance hybrid.
An example is that of benzene which has three alternate double bonds.
As per this structure of benzene, it should exhibit two different bond lengths due to three C – C bond three C = C double bond. However , as determined experimentally all the six C --- C bond length is 1.39 Å. This value is in between C – C & C = C bond length. Thus, the structure of benzene cannot be represented adequately by the above structure but it is resonance hyhrid of following two resonating structure (I & II).
I II Resonance hybrid The resonance structure are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion of their stability. Energy of actual molecule (Resonance Hybrid) is less than that of any of the resonating structure. The difference of energy between the actual structure and most stable resonance structures (structures with lowest energy) is called the resonance energy or resonance stabilization energy. The more the number of resonating structures the more is the resonance energy. Resonance is particularly important when contributing structures have equivalent energy . Some important points regarding resonance are : 1. The resonating structure have same position of atoms or nuclei . 2. All the resonating structures have same number of unpaired electrons. 3 Energy of different resonating structures is same or almost similar. 4. Among various resonating structures of a compound the resonating structure will be stable where (a) Number of covalent bond is more. (b) There is less separation of opposite charges. (c) Negative charge is present on more electronegative atom. (d) Positive charge is present on less electronegative atom. 5. Atoms which take part in resonance should lie in same plane. Rules for drawing resonance contributors : In order to draw resonance contributors, the electrons in one resonance contribution are moved to generate the next resonance contributor. As you draw contributing resonance structures, note that only electrons move. The nuclei of the atoms never move. Further, the only electrons that can move are electrons and nonbonding electrons. Also notice that the total number of electrons in the molecule does not change, nor do the numbers of paired and unpaired electrons. The electrons can be moved in one of the following ways. 1. Move electrons toward a positive charge or toward a bond. 2. Move a nonbonding pair of electrons toward a bond. 3. Move a single non bonding electron toward a bond. Notice that, in all cases, the electrons are moved toward an sp2 hybridized atom. (Remember that an sp2 hybridized carbon is either a double-bonded carbon or a carbon that has a positive charge or an unpaired electron.) Electrons cannot be moved toward an sp3 hybridized carbon because an sp3 hybridized carbon has four bonds and cannot accommodate any more electrons.
66 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
+ + Ex. (1) CH CH CH CHCH CH CH CH CHCH CH3 CH CH CHCH 3 3 3 3 3 resonance hybrid resonance contributors
+ + CH2 CH2 CH2 CH2 CH2 CH2
+ + (2)
+ resonance contributors resonance hybrid
. . . . .
– . . . O . O . O (3) R – C R – C R – C . . . . NH2 NH NH2 + 2
resonance hybrid ......
resonance contributors . . .
. – . –
. . .
O . . O O O C C C C
. . . . .
– . . . . – . .
– . . .
. – . . (4) O . O O O ...... O . . . .O . . . O O resonance hybrid Conditions for Resonance : (i) The atomic arrangement is the same in all the canonical forms. (ii) Same number of paired or unpaired electrons must be present in each canonical form. (iii) The canonical forms must possess same or nearly same energy. (iv) The molecule must have a planar structure. (v) All the canonical forms do not contribute equally towards the structure of a resonance hybrid. A more stable canonical structure contributes more to the resonance hybrid. The contribution depends upon the following factors. (a) Presence of isolated charges or increase in charge separation decreases the stability of a canonical form and hence its contribution in the hybrid structure. (b) When all the structures have formal charge, that structure having negative charge on electronegative and positive charge on electro + ve element (atom) will be more stable. For example, carbonyl group is a resonance hybrid of the following structures. –– COCOCO I II III The polar structure (II) is more stable than (III) because in this structure negative charge resides on oxygen (more electronegative atom) and +ve charge is present on more electropositive carbon atom. (c) Resonating structure having electron deficient positively charged atoms possess very high energy and hence unstable. (d) Resonating structures with a greater number of covalent bonds is more stable. Contribution of Resonating structures: The contribution of an individual factors: (i) Neutral species is more stable than the charged (or dipolar spices). (ii) Species having complete octet is more stable than the species having incomplete octet
R—C==O R—C O e=6 e=8 e=8 e=8 (I) (II) (I) and (II) are resonating structure of acyl cation. (II) will be more sable than (I).
Gyaan Sankalp 67 Organic chemistry-Some basic principles and techniques (iii) If all structure have formal charge, the most stable one is that in which the positive and negative charges reside on the most electropositive and most electronegative atoms of the species respectively. O O H—C—OH H—C—OH I II (iv) Resonating structure with a greater number of covalent bonds is more stable. (v) Increase in charge separation decreases the stability of a resonating structure.
OH O—H O—H O—H
(I) (II) (III) (IV) Hence stability of II and IV will be the same and both will be more stable than III. The order of stability of resonating structures in decreasing order will be as follows : I > II IV > III All the resonating structure do not contribute equally to the real molecule. Their contribution is a direct function of their stability. Characteristics of resonance and resonance hybrid : (i) A resonance hybrid is more stable than any of the canonical forms. Greater the number of resonating structures which can be written for a molecule, greater will be its stability. Thus, resonance gives extra stability to a molecule and hence decreases its reactivity. Resonance hybrid possesses lesser energy than any of the canonical forms. Resonance energy : It is a measure of the stability of a resonance hybrid. It may be defined as, “the difference in the energy of most stable (maximum contributing) canonical form and the resonance hybrid". Resonance Energy = Theoretical heat of formation – Experimental heat of formation. Higher the resonance energy of a molecule greater is its stability. (ii) Bond lengths between atoms in a resonance hybrid are different from those in canonical forms / For example in actual benzene molecule the carbon–carbon bond distance between adjacent atoms is equal (1.397 Å) while in the Kekule structure, it is 1.54 Å and 1.34 Å alternately. Resonance and Bond order : Bond order in conjugated compound or bond order in compounds which exhibit resonance Total number of bonds on central atom = Number of resonating structures
Two bods Total bonds on this carbon For example : One bond
2 1 Bond order of carbon in benzene = = 1.5 2 2– Examples: (i) Bond order between carbon and oxygen in carbonate (CO3 ) ion : All the C – O bond distances are equal in this ion due to resonance 2 1 1 4 O O O ; Bond order = = 1.33 C = O C – O C – O 3 3 O O O – (ii) N – O bond order in nitrate (NO3 ) ion : O O O 2 1 1 4 ; Bond order between O and N = = = 1.33 O – N O = N O – N 3 3 O O O (iii) C – C bond order in benzene :
2 1 3 ; C — C bond order = = = 1.5 2 2 68 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
(iv) Bond order in SO3 : O O O O = S O – S O – S O O O 2 1 1 4 Bond order between S and O = = = 1.33 3 3 Steric Inhibition of Resonance: The most important condition for resonance to occur is that the involved atoms in resonating structure must be coplanar or nearly coplanar for maximum resonance. If this condition does not fulfill, the involved orbitals cannot be parallel to each other and as a consequence delocalisation of electrons or positive charge cannot occur. The planarity of orbitals are inhibited by the bulky groups present on adjacent atoms. This phenomenon is known as steric inhibition of resonance. For example, in dimethyl aniline (I) the orbital having lone pair of electrons present on nitrogen atom is in the plane of the benzene hence lone pair of electrons present on nitrogen atom is in the plane of the benzene ring hence lone pair takes part in the delocalisation.
H3C CH3 H3C CH3 N N O2N NO2
(I) (II)
In N, N-dimethyl-2, 6-dinitroaniline (II) the N (CH3)2 groups is out of the plane of the benzene ring owing to the presence of the two bulky nitro groups and consequently the lone pair of electrons on the nitrogen atoms of N (CH3)2 group cannot get delocalised through lone pair, conjugation. Thus bulky groups present at ortho position inhibit delocalisation of lone pair of electrons or positive charge present on key atom of the molecule. Steric inhibition of resonance has profound effect on: (i) Physical properties (ii) Acidity and Basicity and (iii) Reactivity of organic compounds In nitro-benzene (I) bond length between carbon-nitrogen (bond -a) is in between single and double due to the resonance but in compound (II) bond length between carbon-nitrogen is only of single bond due to the inhibition of resonance.
O O O O N N a a
(I)
O O O O H3C CH3 N CH3 N CH3 CH —C C—CH C C 3 3 H3C CH3 CH3 CH3 H3C CH3 (II) Bond strength is dependent on the extent of the overlap of the combining atomic orbitals, so that in these conjugated systems the more nearly equal in size the p-orbitals are, the more effective is the -orbital overlap. Hence fluorine is more effective than chlorine in conjugating with carbon, and oxygen is more effective than sulphur . Hence the order of + M effects is : —NR2 — OR — F
+ M effect possessing groups are: —OH,—OR,—NH2 ,—NHR,—NR 2 ,—SR,—X etc.
– M effect possessing groups are: —CHO, CO, — CN, — NO2 , — SO 3 H etc.
Gyaan Sankalp 69 Organic chemistry-Some basic principles and techniques APPLICATION OF MESOMERIC OR RESONANCE (i) Effect of resonance on the structure : Resonance decreases bond length of single bond and increases bond length of double bond in real structure. Consider structure of benzene: 1 1 2 2
I II
I and II are resonating structure of C6H6. (a) According to resonating structure I, C–C bond length between C1 and C2 will be 1.33 Å (b) According to resonating structure II, C–C bond length between C1 and C2 will be 1.54 Å (c) According to resonance, bond length between C1 and C2 will neither be 1.33 nor 1.54 Å but will be in between 1.33Å & 1.54Å , i.e., bond length between C1 & C2 is > 1.33Å & < 1.54 Å. (d) experimental value is 1.40 Å, this result coincides with the result obtained by resonance (ii) The low reactivity of halogens : The low reactivity of halogens bonded to unsaturated carbon is due to the +M effect of the halogen. The C—Br bond in vinyl bromide has a partial double-bond character due to the +M effect of bromine with consequent low reactivity of bromine.
CH2 CH—Br CH—CH 2 Br
(iii) Acidity of phenol : The acidity of phenol is due to the + M effect of OH group. The mesomeric transfer of the lone pair on the oxygen atom of phenol to the electrons of the benzene ring results in several resonance structures with positive charge on the oxygen atom. This aids the hydrogen atom of OH group will leave as proton.
OH OH OH OH O
H 2 O H 3O
The ionization is specially aided due to the formation of the relatively more stable phenoxide ion. The charge delocalization in phenoxide ion affords greater stability over phenol in which charge separation occurs in the canonical forms.
O O O O
Hence , phenol prefers to ionize, i.e. it is acidic . (iv) Aromatic character of compounds According to the Hückel rule, a compound will be aromatic if it fulfils the following four conditions: (a) Compound should be cyclic (b) Compound should be planar (c) Compound should be conjugated : (1) Double bond, Single bond, Double bond (2) Double bond, Single bond, electron pair (3) Double bond, single bond, positive charge (iv) Compound should have (4n + 2) conjugated or delocalised electrons where n is a whole number n = 0 , 1 , 2 , 3 , ...... 4n
Structure N
No. of conjugated electrons 6 14 6 2 6
70 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Anti-Aromatic compounds : According to Huckel rule, compound will be anti aromatic if it will fulfil the following four conditions: (i) Compound should be cyclic (ii) Compound should be planar (iii) compound should be conjugated and (iv) Compound should have (4n) conjugated or delocalised electrons where n is whole number. n = 1, 2, 3, 4, 5, 6
Structure
(4n) 4 4 4 Exception : Although cyclooctatetraene has (4n) electrons even then it is not an anti aromatic. Geometry of this compound is non planar
Cyclooctatetraene Aromatic compounds are diamagnetic in character whereas anti-aromatic compounds are paramagnetic in character: (v) Stability of conjugated species : (1) Stability of a conjugated compound is more than the corresponding non-conjugated compound. For example CH2 = CH – CH = CH – CH2 CH2 = CH – CH2 – CH = CH2 (I) (II) Conjugated compound Non-conjugated compound Hence (I) will be more stable than (II) (2) Stability of an aromatic compound is more than the corresponding non-aromatic conjugated compound. For example (III) is more stable than (IV)
CH2 = CH – CH = CH – CH = CH2
(III) (IV) Aromatic compound Conjugated non-aromatic compound Thus stability series of different compounds in decreasing order is as follows : Aromatic compound > conjugated non-aromatic compound > non-conjugated compound > anti aromatic
CH2 = CH – CH 2 C6 H 5 – CH 2 Number of resonating structure = 2 Number of resonating structure = 4 Hence the benzyl carbocation is more stable than the allyl carbocation. (vi) Stability of substituted benzyl carbocations: Stability of substituted benzyl carbocation depends on the nature of group present in the benzene ring. The group may be + I, –I, +R or –R. Case I : When ring has a group which is –I and –M group. – I group withdraws electrons, increases magnitude of positive charge, decreases stability.
– M group (When present at o- or p-position) withdraws electrons, increases magnitude of positive charge, decreases stability.
CH CH2 2 CH2 NO2
NO2
(I) (II) (III)NO2 (i) Increase in the magnitude of (i) Increase in positive charge only (i) Increase in positive charge by – I positive charge by – I and by – I effects and – M effect – M effect (ii) – I and – M power is maximum (ii) – I and – M power is minimum Hence (II) is more stable than (III) which is more stable than (I). Thus meta derivative is more stable than p-derivative which is more stable than o-derivative. Gyaan Sankalp 71 Organic chemistry-Some basic principles and techniques Case II : When ring has a group which is +I and +R group.
CH CH2 2 CH2 CH3
CH3 CH3 (I) (II) (III) Positive charge is decreased by + I Stabilised by + I group only Stabilised by + I and + M effect and + M group or stability by + I and + M group. Hence (I) is more stable than (III) which is more stable than (II). Thus o-derivative is more stable than p-derivative which is more stable than m-derivative. Case-III : When group has + M effect and – I effect.
CH2 CH2 Stabilised by +M effect OCH destabilised by – I effect 3 Destabilised by –I effect –I power is maximum (due to distance) OCH 3 OCH3 Hence p-derivative is more stable than o-derivative which is more stable than m-derivative In case of halo derivatives, result depends only on –I power of the group.
CH2 CH2 CH2 X X
X stability in decreasing order (vii) Stability of substituted Benzyl carbanions Case I : When group has + M and + I effect.
CH2 CH2 +M effect decreases stability CH3 +I effect also decreases stability Destabilised only by +I effect +I power is maximum CH3
CH2
destabilised by +M effect and +I effect, +I power is minimum
CH3 Hence m-derivative is more stable than p-derivative which is more stable than o-derivative Case-II : When group has –R and –I effect.
CH CH 2 CH 2 2 Stabilised by –M and –I effect NO Stabilised by –M and –I effect 2 and – I power is minimum (– I power is maximum) Stabilised by – I effect NO 2 NO2 Thus o-derivative is more stable than p-derivative which is more stable than m-derivative. 72 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
CH2 CH2 OCH3 Destabilised by + M effect and stabilised by – I effect Stabilised by –I effect –I power is maximum OCH3
CH2 Destabilised by +M effect stabilised by –I effect and –I power is minimum
OCH3 (viii) Acidity of aromatic acids : Acidity of carboxylic acids depends on the stability of acid anion thus. Electron withdrawing group stabilises anion, hence increases acidity. O EWG C—O
Similarly , electron donating group destabilises anion, hence decreases acidity. O EDG C—O
The First member of aromatic acid is a benzoic acid which dissociates as follows
HO O O O C C
+ H
Thus acidity of benzoic acid will depends on the stability of benzoate anion which is only stabilised by inductive effect and not by resonance because carboxylic group and carboxylate ion are not in the plane of the ring. These two groups will not take
part in the delocalisation with the benzene ring . In substituted benzoic acid there is no direct interaction between COO and substituent . Acidity of substituted Acids: Ortho substituted benzoic acid is always a strong acid than m- and p-derivative due to the ortho effect. Case I : When group is –M and –I group.
O O O O Carbon becomes electron C C deficient due to resonance O O effect of NO2 group. This N N electron deficient carbon O O withdraws electrons from the carboxylate group by (I) inductive effect.
Thus , anion is stabilised by – M and – I effect and – I power is maximum O C O
Anion is stabilised only by –I effect of NO2 group NO2 (II)
Gyaan Sankalp 73 Organic chemistry-Some basic principles and techniques
O O O O O C C C O
N O N O (IV) O O (III)
Anion is stabilised by –M and –I effect of NO2 group
Thus decreasing order of the stability of these anions is follows: I > III > II > IV We know that ortho derivative is the most acidic therefore decreasing order of acidity of these acids is as follows : o-derivative > p-derivative > m-derivative > benzoic acid Acidity in decreasing order
Case II : When group has +M and +I effect. O OH OH C O C O C O
CH 3 CH Ortho derivative will be most Anion is destabilised due Anion is destabilised3 by +M and +I effect acidic due to ortho effect. to +I effect and +I power is minimum benzoic acid > o-derivative > m-derivative > p-derivative Thus Acidity in decreasing order Case III : When group has +M and –I effect O O O O C C
Stabilised by –I effect ; Destabilised by +M effect stabilised by –I effect
OCH3 OCH3 o-derivative > p-derivative > m-derivative > benzoic acid Thus Acidity in decreasing order
(ix) Basicity of Aromatic amines : (a) Basicity of nitrogen containing compounds electrons on nitrogen In aromatic amines, lone pair of electrons present on nitrogen is delocalised. hence, electron density decreases due to resonance. 1 Thus Basicity Number of resonating structures Structure Number of resonating structure
CHNH6 5 2 4
C6 H 5– NH–C 6 H 5 7
11 C6 H 5– N–C 6 H 5
CH6 5
74 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (b) Basicity of other Nitrogen containing compounds: Basicity depends n the following factors: 1 (1) Electronegativity of nitrogen : Basicity Electronegativity of nitrogen
(2) Inductive effect : Basicity + I power of the group on nitrogen 1 Basicity I power of the group on nitrogen
R—NH2 NH2—OH +I group (I) (II) –I group (3) Resonance : Delocalisation of lone pair of electrons present on nitrogen decreases basicity.
N N Lone pair H Lone pair is not delocalised delocalised Hence (II) is more basic than (I)
(c) Basicity of Substituted Anilines: Para substituted aniline is more basic than ortho substituted aniline and the effect is known as para effect. Case I : When group has – R and – I effect.
NH2 Decrease in electron density nitrogen by NH2 NO2 – R and –I effect and Decrease in electron density on nitrogen by –I effect is maximum ; – effect only (minimum electron density on nitrogen) NO2
NH NH2 2 Decrease in electron density on nitrogen by –R and –I effect and ; no group – I power is minimum
NO2 aniline > m-derivative > p-derivative > o-derivative Thus order of basicity is as follows : Basicity in decreasing order
Case II : When group is + R and + I group . NH NH2 2 Increase in electron density on nitrogen by No group, hence electron density does + R and +I group and CH3 not change due to group. ; +I power is maximum
NH2 NH2 Increase in electron density on nitrogen by Increase in electron density on nitrogen +I group ; by +I group and +I group and +I power is minimum. CH3 CH3 Thus aniline will be least basic and p-derivative will be most basic hence order of basicity is as follows p-derivative > o-derivative > m-derivative > aniline Decreasing order of Basicity p-derivative is more basic than ortho derivative due to a para effect.
Gyaan Sankalp 75 Organic chemistry-Some basic principles and techniques
Case III : When group + R and – I group.
NH2 OCH Increase in electron density on nitrogen by +R effect 3 Decrease in electron density on nitrogen by – I effect – I power is maximum.
NH NH 2 2 Increase in electron density on nitrogen by + R effect Decrease in electron density on nitrogen by decrease in electron density on nitrogen by – I effect – I effect OCH3 – I power is minimum OCH3 p-derivative is more basic than ortho derivative due to para effect. p-derivative > aniline > o-derivative > m-derivative Thus , Decreasing order of Basicity Effect of cross conjugation on basicity : O O
CH3 —C—NH 2 CH3 —C = NH 2 Due to delocalisation and – I effect of CO group, amides are less basic than amines. O
C6 H 5— C—NH 2 In this amide there is cross conjugation which increase basicity; thus C6H5CONH2 is more basic than CH3CONH2 . bond of C=O group is in conjugation to benzene ring as well as lp of NH2 groups. (x) Acidity of -Hydrogens : -Hydrogen of carbonyl compounds , nitrites , acids , nitro compounds are acidic in character . In other words we can say that (i) -Hydrogens are acidic in character when –I group is present on the -carbon. (ii)Acidity of -hydrogens depends on the stability of carbanion which is obtained by the ionisation of the compound. Thus acidity depends on the stability of CHG2 Stability of anion depends on two factors.: (a) Stability of carbanion –I power of the group present on -carbon For example; CH3 – NO 2 CH2 – NO 2 NO–2 CH – NO2
NO2 NO2 – I power in increasing order; acidity in increasing order (b) Stability of carbanion number of resonating structures.
O O CH – N CH2 = N O O Thus -hydrogen of nitro methane is acidic due to inductive effect as well as delocalisation of negative charge.
HYPERCONJUGATION, BAKER - NATHAN EFFECT (NO BOND RESONANCE): It involves the delocalization of C—H sigma () electrons with –electrons of a multiple bond when they are in conjugation. Therefore, it is also called conjugation. In the system H – C –C = C, the electrons of C – H bond are released in the unsaturated carbon chain. This phenomenon is called hypercojugation.
76 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
H H
– C – C = C – C = C – C
The hydrogen atom of the H – C bond acquires a +ve charge but remains very close to () carbon atom although there is not any bond between the two. Hence, the name No–bond resonance for hyperconjugation. Greater the number of H–C bonds attached to an unsaturated system, more will be the electron release in the unsaturated carbon chain, greater will be the number of resonating structure. For example : In propylene, there are three H—C sigma bonds in conjugation with the bond. So, only three resonating forms can be written.
H H H H+ H – C = CH – CH + H – C – CH = CH2 2 H C = CH – CH2 H – C = CH – CH2 H+ H H H Propylene contains 3H – C hyperconjugated bond Propylene contains [3H–C hyperconjugated bond] In 1–Butene, there are two C – H bonds in conjugation with double bond (–electrons), So, only two canonical forms can be written. H + H H – | – CH CH3 CH 2 CH3 C = CH CH2 CH3 C CH CH2 | H H H 1-butene [Two C – H hyperconjugated bonds] When an isopropyl group is attached with an olefinic system only one C – H bond is hyperconjugated. H H – CH – C – CH = CH 2 CH3 C CH CH2 3 | CH3 CH3 [Only one H - C hyperconjugated bond] [Only one H–C hyperconjugated bond] In case a tertiary butyl group is attached with an olefinic system, there is not any C – H hyperconjugated bond, hence no hyperconjugation. CH3
CH CH3 C CH 2
CH3 [No hyperconjugation, no s-electron release because there is no C – H bond]
When the alkyl groups 3º–, 2º–, 1º– and CH3– are attached with an olefinic bond, their electron releasing tendencies are reversed. CH3 | CH3 C CH3 CH CH3 CH2 CH3 | | CH3 CH3 [t-Butyl group No electron release by hyperconjgation]
Gyaan Sankalp 77 Organic chemistry-Some basic principles and techniques Effects of Hyperconjugation : 1. Stability of Carbonium ions: The order of stability of carbonium ions is as follows + + + (CH3)3C (CH3)2CH CH3CH2 above order of stability can be explained by hyper conjugation. In general greater the number of hydrogen atoms attached to - carbon atoms, the more hyper conjugative forms can be written and thus greater will be the stability of carbonium ions.
H CH3 + H C H H3C C 3 carboniumion H3C C CH3 CH (9 equivalent forms) 3
H CH3 + H C H H C C 3 2 carbonium ion H C C H 3 (6 equivalent forms) H
H CH + 3 H C H H C H C 1° carbonium H H (3 equivalent forms)
2. Stability of free radicals: Stability of free radicals can also be explained as that of carbonium ions. (CH3 ) 3 C (CH 3 ) 2 CH CH 3 CH 2 CH 3
3 2 1 3. Bond lengths : The bond length in a molecule changes if there is hyper conjugation, for example, in propene , CH—CHCH3 2
the C—C1 2 bond length is found to be more than 1.34Å while the C—C2 3 bond distance is less than 1.54 Å (C—C single bond length). 4. Dipole moment: since hyper conjugation causes the development of charges, it also affects the dipole moment of the molecule. 5. Orientation influence of methyl group : The o, p directing influence of the methyl group in methyl benzenes is attributed partly to inductive and partly of hyper conjugation effect.
CH3
[ Orientation influence of the methyl group due to + I effect ] The role of hyper conjugation in O , P directing influence of methyl group is evidence by the fact that nitration of P-isopropyl toluene and p-tert-butyl toluene from the product in which – NO2 group is introduced in the ortho position with respect to methyl group and not to isopropyl or t-butyl group although the latter groups are more electron donating than methyl groups i.e. the substitution takes place contrary to inductive effect. Actually this constitutes an example where hyper conjugation overpowers inductive effect.
78 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
H H H C H H C H
H C CH3 H3C C CH3 CH3 CH3 6. Stability of alkenes : (a) Alkylated alkenes are more stable than others. Greater the alkylation of an alkene [greater the number of alkyl groups in an alkene] greater is its stability. Thus, the stability order is : R C C R R 2C CHR . | | R R [ Tetraalkyl ethylene ]
RCH = CHR > R2C = CH2 > R – CH = CH2 > CH2 = CH2 Least stable (b) Greater the number of C – H bonds in an alkene, greater will be the number of hyperconjugated structures, hence greater will be the stability conjugated structures, hence greater will be the stability. For instance, propene is more stable than ethene. CH3 – CH = CH2 > CH2 = CH2 Propene Ethene (3C – H hyperconjugated bonds) (No hyperconjugation) 7. Stability of alkyl benzene : [Stability -H atoms]
CH3 | CH3 CH2 – CH3 CH3 – CH – CH 3 CH3 – C – CH 3
Methyl Ethyl Isopropyl tartery Butyl benzene benzene benzene benzene -H % 3 2 1 0 Reverse Hyperconjugation : The phenomenon a of hyperconjugation is also observed in the system given below: X
– C – C = C ; where X = halogen
In such system the effect operates in the reverse direction . Hence the hyperconjugation in such system is known as reverse hyperconjugation.
Cl Cl Cl
Cl – C – CH = CH2 Cl – C = CH – CH2 Cl C = CH – CH2 Cl Cl Cl
Cl
Cl – C = CH – CH2 Cl
Gyaan Sankalp 79 Organic chemistry-Some basic principles and techniques
The meta directing influence and deactivating effect of CX3 group for electrophilic aromatic substitution reaction can be explained by this effect. X X X X – C – X X – C X X – C X
REACTIVE INTERMEDIATES A great majority of organic reactions are multistep processes involving intermediates whose life time are usually very short. An understanding of the structures and properties of these intermediates which are normal covalent compounds, ions or radical’s is of paramount importance to an understanding of organic reaction mechanisms. Intermediates possess sufficiently low energy to be formed under the reaction conditions, but in most of the cases are not stable enough to permit isolation. This is especially true of those intermediates in which carbon has fewer than four covalent bonds. Common intermediates : Free Radicals : | | Homolytic fission • • CX – C X | | If EN of C –~ EN of X (a) An atom or group of atoms possessing an odd or unpaired electron is called free radicals It is electrically neutral and shows paramagnetism. Free radicals are electrophiles. (b) Types of Free Radicals : CH3 , CH3 CH2 , (CH3)2 CH , (CH3)3 C , CH2 = CH – CH2 , C6H5 CH2 1º 2º 3º (c) Formation of free radicals : h (i) Cl2 Cl· + Cl· [Homolytic fission] h (ii) CH COCH CH + CH C O CO + CH 3 3 140 C 3 3 3 (iii)(C2H5)4Pb Pb + 4 C2 H5 (iv) CH3 – N = N – CH3 N2 + 2 CH3 O O || || (v) C6H5 – C – O – O – C – C6H5 2C6H5COO· 2C6H5· + 2CO2 (d) Salient features : (i) Free radical reactions proceed in vapour phase or in nonpolar solvents. (ii) Free radical reactions are frequently autocatalytic. (e) Reactions : (i) Chlorination of alkanes. (ii) Pyrolysis of alkanes. (iii)Wurtz reaction. (iv) Anti-markownikoff rule. (v) Kolbe electrolytic synthesis. (vi) Polymerisation initiated by free radical. (f) Stability of free radicals, Factor : (i) Resonance (ii) Hyperconjugation Order of stability of free radicals : t-butyl > isopropyl > ethyl > methyl (g) Stability on the basis of resonance : Stability Number of resonating structures Triphenyl methyl free radical has the maximum number of resonating structures. Hence it is the most stable free radical. The order is : ()3C· > ()2CH· > CH2· > CH2 = CH – CH2· [where = C6H5]
80 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (h) Resonance in allyl free radical :
CH2 CH CH2 CH2 CH CH2 Resonance in benzyl free radical :
CH2 CH2 CH2 CH2
etc.
Carbocations : Carbocations are the key intermediates in several reactions and particularly in nucleophilic substitution reac- tions. Structure : Generally, in the carbocations the positively charged carbon atom is bonded to three atoms and has no lone pairs on carbon. Carbon has six valence electrons. It is sp2 hybridized with a planar structure and bond angles of about 120°. R
C R R Orbital picture of carbocation 2 There is a vacant unhybridized p orbital which e.g., in the case of CH3 lies perpendicular to the sp hybridised plane.. Formation : Carbocations can be generated in a variety of way’s some of the reactions in which carbocations are formed are summarized below. 1. Solvolyis of C—X bond (X = halogens, etc.] R—XRX – 2. Deamination of amines by nitrous acid.
HNO2 R—NHRNRN2 2 2 3. Protonation of alcohols followed by dehydration
H R — OH R OH2 R H 2 O Stability: There is strong evidence, both physical and chemical, that alkyl groups are more electron-donating than hydrogen. It therefore, follows that when these groups are attached to the electron-deficient carbon of the carbocation, they tend to release electrons and partially compensate for the electron-deficiency of the positive carbon. The positive charge thus gets dispersed over all the alkyl groups and this dispersal of charge increases the stability of the whole system. Accordingly, tertiary carbocations are more stable than secondary ones which in turn are stabler than primary carbocations.
R H H H
R C R C R C H C
R R H H The greater stability of alkyl substituted carbocations is sometimes partly ascribed to the phenomenon of hyperconjugation. According to this, the electrons of an C—H bond can be delocalized into the unfilled p orbital of the positive carbon atom, thus spreading the charge over all such bonds. For an alkyl-substituted carbocation, several hyperconjugative resonance forms can be written each having the same number of covalent bonds as the first structure.
H H H H
H C CH H C CH2 H C CH2 H C CH2 2
H H H H In case of a secondary carbocation, more equivalent structures can be written than for a primary carbocation whereas still greater number of such structures can be written for a tertiary carbocation.
Gyaan Sankalp 81 Organic chemistry-Some basic principles and techniques
H H
H C H H C H
H C C Five more eqivalent H3C C 3 structures H H
H H
H C H H C H H C C Eight more eqivalent H3C C 3 structures
CH3 CH3 Resonance is an important factor that enhances the stability of a carbocation by delocalization of its charge in systems like allyl and benzyl carbocations. CH2 = CH — CH2 CH2 — CH = CH2
CH2 CH2 CH2 CH2
Triphenyl carbocation is so stable that when triphenylmethyl bromide is placed in liquid sulphur dioxide (a solvent with which carbocation does not react) it is possible to determine it quantitatively by measuring the electrical conductivity of the solution.
C6H5 C6H5
liquid SO2 H5C6 C Br H5C6 C Br
C6H5 C6H5
CPh2 CPh2 CPh2
etc. CPh2
Steric effects also play key role in the ease of formation and stability of carbocations derived from highly substituted substrates. The stability of such carbocations is attributed to the steric relief. In substrates such as tri-isopropyl chloride, the three bulky isopropyl groups are pushed together due to sp3 angle of 109.5°. This pushing together results in a strain called B strain (or back strain). When this ionizes, the angle expands from 109.5° (sp3) to 120° (sp2) resulting in the relief of this strain due to increase in space between the alkyl groups.
CH 3 H3C CH3 CH H3C HC 109.5° Cl H3C C CH CH H3C CH 109.5° 120° CH3 CH H C 109.5° 3 H3C CH3 H3C CH3 Such a carbocation would resist addition of a nucleophile as it would result in the crowding of bulky groups together.
82 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
| Carbanion : C: : |
| | Heterolytic fission CX C:X | | If EN of C > EN of X An organic species containing negatively charged carbon atom is called carbanion. There are eight electrons in the valence shell of negatively charged carbon. Two electrons remain as unshared pair. Carbanions are nucleophile. Organic metallic compound acts as carbanion : R....Li R.... Mg Br
(a) Types of carbanion ion : C H3, CH3 – C H2,(CH3)2 C H, (CH3)3 C 1º 2º 3º (b) Formation of Carbanion :
(i) OH + CH2 CHO CH2 - CHO + H 2 O
H
C H O + (ii) 2 5 CH2 COOC 2 H 5 CH2 COOC 2 H 5 + C 2 H 5 OH
H (c) Reactions : (i) Aldol condensation (ii) Claisen condensation (iii) Decarboxylation sodalim e Ex. RCOOH RH O || R –– C –– O R + CO2 (d) Stability of Carbanion : Factor : (i) Inductive effect (ii) Resonance (iii) s-character in Hybridisation 1 Stability of Carbanion –I , I
Ex. (i) CH3–CH2 < CH3 < NO2–CH2 (ii) –, CH2 > CH3 > , – CH2 (iii) C H3 > CH3 C H2 > (CH3)2 C H > (CH3)3 C Explaination : Stability of alkyl carbanions can be explained by inductive effect. Greater the number of alkyl groups [+ I effect] attached to the carbon atom bearing negative charge, lesser is the stability. 3º < 2º < 1º < C H3
H H CH3 CH 3 C H C CH3 C CH CH3
CH3 H H CH3
Gyaan Sankalp 83 Organic chemistry-Some basic principles and techniques
(iv) C H2 — NO2 > C H2 — CHO (v) C H2 — CHO > C H3 (vi) C H2 — CH2 — NO2 > CH3 — C H — NO2 > (CH3)2 C H (vii) F — C H2 > Cl — C H2 (viii) Resonance exists in C H2 — C N, hence it is stable carbanion. CH = C = N CH2 ––– C N 2 Carbanion in which negative charge and double bond are in conjugation. (f) Resonance in acetonyl carbanion
CH3 ––– C ––– CH2 CH3 ––– C = CH 2 || | O O Benzyl carbanion CH CH2 CH2 2
3 sp
Electron pair is present in sp³ C (g) Geometry : H H Hybrid orbitals character, 25% H Pyramidal
Carbenes : Carbenes can be defined as neutral, divalent carbon intermediates in which a carbon is covalently bonded to two atoms and has two non-bonding orbitals containing two electrons between them. Theoretical considerations suggest that there are two possible kinds of carbenes, singlet and triplet carbenes. In the singlet state, a carbon atom is presumed to approximate sp2 hybridization. Two of the three sp2 hybrid orbitals are utilized in forming two covalent bonds whereas the third hybrid orbital contains the unshared pair of electrons. The unhybridized p orbital remains vacant. Thus singlet carbene (A) resembles a carbocation very closely. On the other hand, carbon atom of a triplet carbene (B) is sp hybridized and it is a linear or near-linear species. These two hybrid orbitals are involved in the bond formation with two groups and the two electrons are placed one each, in the equivalent, mutually perpendicular py and pz orbitals . Since these electrons have parallel spins, a carbene with this structure is said to be in a triplet state.
R R C C (A) (B) R' R'
At first sight, it appears that a singlet carbene has lower energy as the unshared electron pair is in a sp2 hybrid orbital ; but the considerations of the electron repulsion energy that must be overcome to pair two electrons in a single orbital places it at the higher energy level than a triplet structure. It is, therefore, reasonable to believe that the triplet state of a carbene is more stable than the singlet state and should be expected to be the ground state. Formation : Photochemical or thermal cleavage of cyclopropanes and oxiranes is a common method for the generation of carbenes.
84 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
H5C6 CH2 hv C CH2 or HCCHCHCH2 6 5 2 H
H5C6 O O hv C CH C6 H 5 HC HC — C 6 H 5 H C6H5
Stability: Carbenes in which the carbene carbon is attached to two atoms, each bearing an unshared pair of electrons, are somewhat more stable due to resonance.
R2N R2N R2N C C C
R2N R2N R2N
Nitrene : It is a neutral intermediate species containing nitrogen atom having one bond pair and two lone pair. It is an electron deficient nitrogen. Formation : Amide + bromine + caustic potash 2Br CH3 –– C –– NH 2 + Br2 + 2OH CH3 –– C –– N: + 2H2O + || || O O (Acylnitrene) Two form of nitrene :
N (i) Singlet : N (ii) Triplet :
sp2, Linear sp, Linear Stability :– Triplet > Singlet (Reaction : Hoffmann bromamide reaction) Arynes : Arynes may be defined as aromatic compounds containing a formal carbon-carbon triple bond. The best known aryne is benzyne which may be regarded as the aromatic counterpart of acetylene or in other words, it is benzene minus two ortho hydrogens and can also be called as dehydrobenzene.
The benzyne bond is not like the triple bond of acetylene where the two carbons form a bond using sp orbitals and the remaining p orbitals are used to form bonds. Such a structure is not possible in benzyne because of the hexagonal geometry associated with the benzene ring. Most probably the new bond of benzyne is formed by the overlap of sp2 orbitals belonging 2 to two neighbouring carbon atoms. These sp orbitals are orthogonal to the molecular orbital of the benzene ring.
Formation: Benzyne has been shown to be intermediate in several important organic reactions. For example, in the presence of a strong base, aryl halides eliminate HX from 1, 2-positions to produce benzyne which then rapidly reacts with the available nucleophile to regenerate the aromatic system.
Gyaan Sankalp 85 Organic chemistry-Some basic principles and techniques
Br NH Br NH2 3 H
NH 3 NH2
NH3
Stability : The new bond of benzyne, formed by the overlap of sp2 orbitals belonging to two neighbouring carbon atoms in unstable, and therefore benzynes are extremely reactive chemical species.
TYPE OF ORGANIC REACTIONS The reactions of organic compounds can be classified into four main types: (i) Substitution or displacement reactions (ii) Addition reactions (iii) Elimination reactions (iv) Rearrangement reactions (1) Substitution or displacement reactions : Substitution or displacement reactions are those reactions in which an atom or group of atoms attached to a carbon atom in a substrate molecule is replaced by another atom or group of atoms. During the reaction no change occurs in the carbon skeleton, i.e., no change in the saturation or unsaturation of the initial organic compound. Depending on the mechanism, the substitution reactions are further classified into three types: (a) Free radical substituting reactions (b) Electrophilic substituting reactions (c) Nucleophilic substituting reactions Some of the examples of substitution reactions are : uv (i) CH + Cl CH Cl + HCl 4 2 light 3 Methane (Hydrogen atom is replaced by chlorine) (ii) CH3CH2Br + aq.KOH CH3CH2OH + KBr Ethyl bromide Ethyl alcohol (Bromine atom is replaced by hydroxyl group) (iii) CH3OH + HBr CH3Br + H2O Methyl alcohol Methyl bromide (Hydroxyl group is replaced by bromine)
H2 SO 4 (conc.) (iv) C6H6 + HNO3 C6H5NO2 + H2O Benzene (conc.) Nitrobenzene (Hydrogen is replaced by NO2 group) 500 C (v) CH3CH = CH2 + Cl2 ClCH2CH = CH2 + HCl Propylene Allyl chloride (Hydrogen is replaced by chlorine)
FeCl3 (vi) C6H6 + Cl2 C6H5Cl + HCl Benzene Chlorobenzene (Hydrogen is replaced by chlorine) Mechanism of substitution reactions : These reactions may follow free radical, nucleophilic or electrophilic mechanism. Some typical examples are considered to explain the three types of mechanism. (a) Free radical substitution reactions : Chlorination of methane : The chlorination of methane in the presence of ultraviolet light is an example of free radical substitu- tion (Homolysis). uv CH + Cl CH Cl + HCl 4 2 light 3 Methane Methyl chloride The reaction does not stop with the formation of methyl chloride (CH3Cl) but the remaining hydrogen atoms are replaced one by one with chlorine atoms to give rise chain reaction.
86 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
CH3Cl + Cl2 CH2Cl2 + HCl CH2Cl2 + Cl2 CHCl3 + HCl CHCl3 + Cl2 CCl4 + HCl Mechanism : The reaction is initiated by the breaking of chlorine molecule into chlorine–free radicals in presence of UV light. uv I Step : Chain initiation : Cl Cl + Cl 2 light II Step : Chain propagation : The chlorine–free radicals attack methane molecule. CH4 + Cl CH3 + HCl Cl CH3 + Cl2 CH3Cl + Each of the methyl–free radicals, in turn, reacts with chlorine molecule to form methyl chloride and at the same time chlorine–free radical is produced. III Step : Chain termination : The chain of reactions initiated and propagated as shown above may be terminated if free radicals combine amongst themselves without giving rise to any new radicals. C l + C l Cl2 CH3 + C l CH3Cl CH3 + CH3 CH3 – CH3 Reactivity of the halogen for free radical substitution is in the order : F2 > Cl2 > Br2 > I2 (b) Nucleophilic substitution : Many substitution reactions, especially at the saturated carbon atom in aliphatic compounds such as alkyl halides, are brought about by nucleophilic reagent or nucleophiles. R — X + OH– R — OH + X– Substrate Nucleophile Leaving group Such substitution reactions are called nucleophilic substitution reactions, i.e., SN reactions (S stands for substitution and N for nucleophile). The nucleophilic substitution reactions are divided into two classes: (i) S Reactions : These are bimolecular reactions. When the rate of reaction depends on the concentration of both substrate and N2 the nucleophile, the reaction is said to be S , i.e., 2nd order change. N2 Rate [Sustrate] [Nucleophile] Hydrolysis of methyl chloride is an example of S reaction N2 When the methyl chloride is attacked by OH– strong nucleophile from the opposite side of the chlorine atom, a transition state (TS) results in which both OH and Cl are partially bonded to carbon atom. H H H H HO C Cl HO C Cl HO C H + Cl H H H H Transition state Methyl Alcohol When (—) 2–bromo ethane is allowed to react with sodium hydroxide under conditions where second order kinetics are followed, there is obtained (+) 2–ethanol.
H H H H H C Br HO C Br HO C H + Br HO CH3 CH3 (–) 2-Bromo ethane CH3 (+) 2-ethanol Transition state
Gyaan Sankalp 87 Organic chemistry-Some basic principles and techniques Configuration : Hence, an S reaction proceeds with complete stereo–chemical inversion. This is also known waldon Inver- N2 sion. S order : Methyl > ethyl > isopropyl > tertiary butyl > allyl > benzyl halides N2 (ii) S Reactions : S stands for unimolecular reaction. When the rate of nucleophilic substitution reaction depends only on the N1 N1 concentration of the substrate, the reaction is of first order change and is represented as S . N1 Rate [Substrate] The hydrolysis of tert–butyl bromide is an example of S reaction. The reaction consists of two steps : N1 Step 1. The substrate undergoes heterolytic fission forming a carbonium ion. This is the slow process and rate determining step.
CH3 CH3 | Slow step + CH3 C Br CH3 C + Br | CH 3 CH3 The carbonium ion is planar as the central positively charged carbon atom is sp2 – hybridized. Step 2. The nucleophile (OH–) can attack the planar carbonium ion form either side to form tert–butyl alcohol and the low concentration of OH– favours S reaction. N1
CH3 | CH3 CH3 | | CH3 C OH Fast step | CH3 C OH or HO C CH3 | CH3 | CH3 CH3 In another example, the carbonium ion formed (in Step I) can undergo rearrangement to give more stable carbonium ion.
CH3 CH3 CH3 | | | 1 2methyl Slow CHCCHCH Step 1. CHCCHI3 2 CH3 C C H 2 3 2 3 | | shift 3º CH3 CH3 (1º )
CH3 CH3 | | — Fast Step 2. CHCCHCH + OH CHCCHCH3 2 3 3 2 3 | 3º carbonium ion OH
As a result of S reaction, there can be racemization and inversion. The racemization is due to inverting nucleophilic displace- N1 ment of halogen atom from the alkyl halide by the halide in solution. For example, when (–) 2–bromo butane having chiral centre is treated with low concentration of nucleophile (OH—), it forms (+) 2–butanol. There is also loss in optical activity due to formation of d–and –isomer (racemic) because of S reaction. N1 In another example of 1–Bromo–1–phenyl ethane [C H CH(Br)CH ], the S reaction involves racemization plus inversion. 6 5 3 N1
88 Gyaan Sankalp Organic chemistry-Some basic principles and techniques OH¯
CH6 5 (a) CH H (b) Slow 6 5 H C Br –(Br ) + C CH3 CH3 (a) (b)
CH6 5 CH6 5
HO C H H C OH
CH3 CH3 Enantiomers (a) Inversion (b) Inversion The nucleophilic reagent attacks both (a) the back side and (b) the front side of the carbonium ion. Back side attack (a) predominates. The two enantiomers constitute the racemic modification.
Thus, in S 1 reaction, racemization as well as inversion is observed, while in case of S complete inversion takes place (where N N2 chiral carbon exist). S reaction is favoured by heavy (bulky) groups on the carbon atom attached to halogens; i.e., N1
R3C — X > R2CH — X > RCH2 — X > CH3 — X Tertiary Secondary Primary Methyl (c) Electrophilic substitution : Electrophilic substitution involves the attack by an electrophile. It is represented as SE (S stands for substitution and E stands for electrophile). If the order of reaction is 1, it is written as S (unimolecular) and if the order is 2, it E1 is S (Bimolecular). E2 (I) SE1 Reaction Mechanism : Electrophilic substitution in aliphatic compounds ( S ) are very rare : some of the important examples are : E1 (a) Replacement of the metal atom in an organometallic compound by hydrogen R — M + H+ R — H + M+ CH3CH2Na + C6H6 CH3 — CH3 + C6H5Na (b) Isotopic exchange of hydrogen for deuterium or tritium R — H + D+ R — D + H+ R — H + T+ R — T + H+ (II) SE2 Reaction Mechanism : Electrophilic substitution ( S ) is very common in benzene nucleus (aromatic compounds) in which –electrons are highly E2 delocalized and an electrophile can attack this region of high electron density. In all electrophilic aromatic substitution reactions, it involves : Step 1. The formation of an electrophile, E+, i.e., E – Nu E+ Nu
E = Cl , NO2 , SO3 , R , RCO Step 2. The electrophile attacks the aromatic ring to form carbonium ion (or arenium ion) which is stabilized by resonance. H E H E H E + + + E +
Benzene + Resonance form
Gyaan Sankalp 89 Organic chemistry-Some basic principles and techniques Step 3. Carbonium ion loses the proton to form substitution product. H E E + + H+
Ex. The bromination of benzene in the presence of FeBr3 is an example of electrophilic substitution. Br FeBr3 + Br2 + HBr
Mechanism : Step 1. Formation of electrophile takes place. + – Br — Br + FeBr3 Br + FeBr4 Step 2. The electrophile (Br+) attacks the benzene ring to form a resonance stabilized carboniumion. + H H H + Br Br Br Br + +
Resonance forms H + Slow +CH C6H5H + Br 6 5 Br Step 3. Elimination of proton occurs and the substitution product is formed. H Br Br + H+ + + – (H + FeBr4 FeBr3 + HBr) Similarly, Nitration, Sulphonation and Friedel – Crafts reaction .... etc., in benzene nucleus are the other examples of electrophilic substitution. (ii) Addition reactions : Addition reactions are those in which the attacking reagent adds upto the substrate molecule without elimination. Such reactions are given by those compounds which possess double or triple bonds. In the process a triple bond may be converted into double bond or single bonds and a double bond is converted into single bond. For each -bond of the molecule two sigma bonds are formed and the hybridisation state of carbon atoms changes from sp and sp2 and sp2 to sp3 . Like substitution reactions, addition reaction are also of three type: (a) Free radical addition reactions. (b) Nucleophilic addition reactions (c) Electrophilic addition reactions. Some of the examples of addition reactions are:
Ni H2 /Ni (i) CH CH + H2 CH2 =CH 2 CH3 –CH 3 Acetylene 140ºC Ethylene 200ºC Ethane
(ii)CH2=CH2 + Br2 CH2Br–CH2Br Ethylene 1,2-Dibromoethane (Ethylene bromide) H OH (iii) CH3 –C=O + HCN CH3 – C Acetaldehyde CN Cyanohydrin O Acid (iv)CH3CN + H2O CH3–C–NH2 (HO)2 2 Methyl cyanide Acetamide HBr (v) CHCH + HBr CH2=CHBr CH3CHBr2
90 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Acetylene Vinyl bromide 1,1-Dibromoethane (Ethylidine bromide) Mechanism of addition reaction : The addition reactions are the reaction of the double or triple bonds. These reactions may be initiated by electrophiles, nucleophiles or free radicals. The molecules having > C = C < or — C C — are readily attacked by electrophilic reagents while molecules having > C = O or – C N are readily attacked by nucleophilic reagents. (a) Electrophilic addition reactions : In electrophilic addition reactions, an electrophile approaches the double or triple bond and in the first step forms a covalent bond with one of the carbon atoms resulting in the formation of carbonium ion which then takes up a nucleophile to result in addition product. E
+ Slow C = C + E C – C –
Olefin Electrophile Carbonium ion E E + C – C – + X¯ Fast C – C – Nucleophile X Trans-addition Addition product Polar molecule attacks the double bond of alkene, then electron of double bond shift to one of carbon atom due to electromeric effect. The addition of HBr on ethylene is an example of electrophilic addition. Ethylene is a symmetrical olefin. Br
CH2 = CH2 + HBr CH2 – CH 2 Ethylene H Ethyl bromid Mechanism : Step 1. Hydrogen bromide gives a proton and bromide ion. + HBr H + • Br¯ Electrophile Nucleophile Step 2. The electrophile attacks the double bond to form a carbonium ion.
+ + CH2 = CH2 + H H2 C – CH 3 Carbonium ion Step 3. The nucleophile (Br¯ ion) now attack the carbonium ion to form the addition product.
+ • CH3 – CH2 + • Br¯ CH3 –CH 2 Br Identical + • CH2 – CH 3 +• Br¯ BrCH2 –CH 3
This clearly indicates that the reaction proceeds in two steps : (a) Formation of carbonium ion (more stable) and (b) Attack of nucleophile on the carbonium ion. In case both alkene and the adding reagent are unsymmetrical, two products are expected.
CH3 CH 2 CH 2 Br (n-Propyl bromide)
CH3 – CH = CH2 + HBr¯ Propene CH3 – CH – CH 3 Br (Isopropyl bromide) Gyaan Sankalp 91 Organic chemistry-Some basic principles and techniques Experimentally, it is observed that isopropyl bromide is the major product. This can be explained on the basis of following mechanism. Consider the addition of HBr to propene which is unsymmetrical in nature. Step 1. Hydrogen bromide gives a proton (H+) and a bromide ion (Br¯) HBr H+ + :Br¯ Step 2. The proton (H+) attacks the -bond to give a stable carbonium ion. + CH3 – CH–CH 3 (More stable) + CH3 CH = CH2 + H + CH3 – CH 2 –CH 2 (Less stable) Step 3. The nucleophile bromide ion attack the more stable carbonium ion to give isopropyl bromide (major product). Br
+ · Fast CH3 – CH – CH3 + · Br CH3 – CH – CH 3
CH3
CH3 – CH – CH – CH 3 (2-Bromo-3-methyl butane) Br (Minor product) CH3 +H – Br Ex. CH – CH – CH = CH 3 2 (A.R.) 3-Methyl-1-butene CH3
CH3 – CH – CH – CH 3 (2-Bromo-2-methyl butane) Br (Major product)
It can be explained by 1, 2–hydride shift. Addition reactions in alkadienes : Conjugated alkadienes (1, 3–butadiene) reacts with halogens, halogen acids hydrogen and water, etc. to yield a mixture of 1, 2, and 1, 4–addition products. Some of the important examples are : (i) Addition of halogens :
CH2 Br – CHBrCH = CH 2 3, 4-Dibromo-1-butene (1, 2-addition) CH2 = CH - CH = CH2 + Br 2 1, 3-Butadiene
BrCH2 – CH = CH – CH 2 Br 1, 4–Dibromo-2-butene (1, 4–addition)
(ii) Addition of halogen acids (HBr or HCl) :
CH3 CHBr – CH = CH 2 3-Bromo-1-butene (1, 2-addition) CH2 = CH – CH = CH2 + HBr 1, 3-Butadiene
CH3 – CH = CH – CH 2 Br 1-Bromo-2-butene (1, 4-addition)
92 Gyaan Sankalp Organic chemistry-Some basic principles and techniques (b) Nucleophilic addition reactions : When the addition reaction occurs on account of the initial attack of nucleophile, the reaction is said to be a nucleophilic addition reaction. Due to presence of strongly electronegative oxygen atom, the –electrons of the carbon–oxygen double bond in carbonyl group (>C = O) get shifted towards the oxygen atom and thereby such bond is highly polarised. This makes carbon atom of the carbonyl group electron deficient. + C = O C — O C = O
C O E
Nu C O E Nu
C O Nu
E The addition of HCN to acetone (>C = O compounds) is an example of nucleophilic addition. CH3 CH3 OH C = O + HCN C CH3 CH3 CN Acetone Acetone cyanohydrin The Mechanism of the reaction involves the following steps : Step 1. HCN gives a proton (H+) and nucleophile, a cyanide ion (CN—). HCN H+ + CN– Step 2. The nucleophile (CN—) attacks the positively charged carbon as to form an anion [H+ does not attack the negatively charged oxygen as anion is more stable than cation]. CH3 CH3 CH3 O CN C = O CN C O or C CH3 CH3 CH3 CN Step 3. The proton (H+) combines with anion to form the addition product. CH3 CH CH OH 3 + 3 CN C O + H NC C OH or C CH3 CH3 CN CH3 Other examples of nucleophilic addition reaction are : Addition of Grignard reagent to >C = O compounds : OH OMgX H 2 O C = O + X – Mg – R C C Grignard R R reagent Addition Alcohol product (c) Free radical addition reaction : (i) This type of addition reaction takes place in vapour phase or in non–polar solvents (i.e., Cl2, Br2, H2, CO2 and CH4 .... etc.) in presence of sunlight. (ii) The additives are free radicals and the rate determining step suggests for addition of free radicals. (iii)For example the photochemically catalysed addition of chlorine to ethylene may be shown as follows : h or heat H C = CH + Cl H C – CH 2 2 2 (light) 2 2 Ethylene (gas) | | Cl Cl Ethylene (di) chloride (iv) Addition of HBr on unsymmetric alkene in presence of peroxide is. Gyaan Sankalp 93 Organic chemistry-Some basic principles and techniques Peroxide effects Peroxide CH3 – CH = CH2 + HBr CH3 – CH2 – CH2 – Br Mechanism : O O Homolysis 2RO• R R (i) Chain initiation : RO• + HBr ROH + Br• (ii) Chain propagation : (two steps) • CH3 – CH = CH2 + Br• CH3 – CH – CH2Br 2º free radical • CH3 – CH – CH2Br + HBr CH3 – CH2 – CH2Br + Br• (iii) Chain termination : • Br• + Br• or 2CH3 – CH – CH2Br • or CH3 – CH – CH2Br + Br• (iii) Elimination reactions The reverse of addition reactions are termed as elimination reactions. In these reactions generally atoms or groups from two adjacent carbon atoms in the substrate molecule are removed and multiple bond is formed. In the process two sigma bonds are lost and a new -bond is formed, i.e., state of hybridization of carbon atom changes from sp3 and sp2 to sp. Some example are : Br Br Heat (i) H—C—C—H + Zn H—C==C—H + ZnBr2 (–Br2 ) H H H H 1,2-Dibromoethane Ethylene (Dehalogenation)
H O P2 O 5 , Heat (ii) H—— C C—NH2 CHCN3 (–H2 O) Methyl cyanide H Acetamide (Dehydration)
H H OH Conc. H2 SO 4, 170ºC (iii) H—C—C —C—H CH3 CH=CH 2 (–H2 O) Propylene H H H (Dehydration) Mechanism of elimination reactions : An elimination reaction, generally, involves loss of atoms or groups from adjacent carbon atoms resulting in the formation of a –bond between these carbon atoms, so these are reverse to addition reactions. The elimination reactions are divided into two classes : (I) E2 reactions. E2 stands for bimolecular elimination. (II) E1 reactions. E1 stands for unimolecular elimination. (I) E2 reactions : The dehydrohalogenation of the alkyl halides with alcoholic alkali is an example of this type. It occurs in one step. – CH3CH2CH2Br + C2H5 O CH3CH = CH2 + Br + C2H5OH 1–Bromopropane Propene
CH3H CH3H Slow CH3 H C 2 H 5 O+ H C C Br C 2 H 5 O H C C Br C C + C H OH + Br H H H H 2 5 Transition state H H Rate equation : v = k [halide] [:Br] ; Molecularity –2, Order –2
94 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
(II) E1 reactions : These occur in two steps . Step 1. The alkyl halide ionise to give the carbonium ion.
CH3 CH3 Slow CH C X C + X 3 CH3
CH3 CH3 Carbonium ion Step 2. Carbocation loss a proton from the –carbon atom by the base (nucleophile) to form alkene. CH CH3 3 + Fast C CH2 H + OH C CH2 + H 2 O CH3 CH3 Rate equation : v = k [RX] Molecularity –1 Order –1 Dehydrohalogenation is removal of HX from alkyl halides with alcoholic KOH or KNH2 or KO–ter–Bu (potassium tertiary butoxide) and an example of elimination, Alc.KOH Ex. CH3 — CH2X H2C = CH2 (HX) Ethene Alc.KOH CH – CH2 CH CH 3 CH CH = CHCH +CH CH CH = CH 3 | (HX) 3 3 3 2 2 X 2 - Halobutane 2 - Butene (major) 1-Butene (minor)
The more substituted alkene is more stable (according to Saytzeff rule), hence the formation of 2–butene is preferred to 1–butene.
Slow, E1 (i) CH3 CHCH 2 CH 3 CH3 CH CH2CH3 | Br Br 2–Bromo butane Carbonium ion (2º) Fast (ii) CH CH CH CH CH CH = CHCH 3 2 3 H (from C3 ) 3 3 2º Carbonium ion 2–Butene (more substituted) Dehydration of alcohol is another example of elimination reaction. When acids like conc. H2SO4 or H3PO4 are used as dehydrat- ing agents, the mechanism is E1. The proton given by acid is taken up by alcohol. H | O – H CH3 CH2 O H H CH3 – CH2 – H | Slow CH CH + H O CH3 CH2 O H 3 2 2 Carbonium ion The carbonium ion loses a proton to form alkene.
+ + H CH2 CH2 H + CH2 CH2 Dehydration is removal of H2O from alcohols, e.g., Conc. H2 SO 4 ,170ºC CH3CH2OH H2C = CH2 HO2
Conc. H2 SO 4 ,170ºC CH3CH2CH2OH CH3CH = CH2 HO2 Gyaan Sankalp 95 Organic chemistry-Some basic principles and techniques Dehalogenation : It is removal of halogens, e.g., in CH OH,heat CHCH2 2 + Zn dust 3 H C = CH | | 2 2 ZnBr2 Br Br Ethylene bromide Ethylene Dehydrogenation : It is removal of hydrogen, e.g., Cu, 200ºC CH3CH2OH CH3CHO
Cu, 300º C CH3 CH CH 3 CH3 C CH 3 | H2 || OH O Isopropyl alcohol Acetone (iv) Rearrangement reactions: The reaction which involve the migration of an atom or group from one site to another within the molecule (nothing is added from outside and nothing is eliminated) resulting in a new molecular structure are known as rearrangement reactions. The new compound is actually the structural isomer of the original one. Some of the examples are : anhydrous AlCl 3 CH —CH—CH (i) CH3 CH 2 CH 2 CH 3 3 3 n-Butane CH3 Isobutane
AlCl3 or Al 2 (SO 4 ) 3 (ii) CH3 CH 2 CH= CH 2 CH3 CH=CHCH 3 + CH3 –C=CH 2 n-Butane 2-Butene CH3 2-Methylpropene Heat (iii) NH4CNO NH2–CO–NH2 Ammonium cyanate Urea O
OCCH3 OH OH anhy.AlCl (iv) 3 COCH3 [Fries rearrangement] Phenylacetate COCH3 (ester) o– and p -hydroxy acetophenone
O AlBr (v) CH CH CH Br 3 CH NH—C—CH 3 2 2 Heat 3 3 n-Propyl bromide N-Methyl propionamide
TRY IT YOURSELF Q.1 Which is electrophilic reagent in the following – – (A) RO (B) BF3 (C) NH3 (D) R–OH Q.2 Which of the following is the fundamental cause of electron donor effect of alkyl radicals (A) +I Effect (B) +R Effect (C) Hyperconjugation (D) Electromeric Effect Q.3 Electrophiles are always – (A) Positively charged (B) Negatively charged or neutral (C) Neutral (D) Neutral or positively charged Q.4 Ethylene gives fastly – (A) Addition reaction (2) Substitution reaction (3) Elimination reaction (4) Rearrangement reaction
96 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
Q.5 Decreasing order of – I effect of the triad [– NO2, NH3 , – CN] is -
(A) – NH3 > – NO2 > – CN (B) – NH3 > – CN > – NO2 (C) – CN > – NO2 > NH3 (D) – NO2 > – CN > – NH3 Q.6 Which of the following properties of ethylamine makes it a stronger base than aniline – (A) Positive inductive effect (B) High steric hindrance (C) Mesomeric effect (D) Presence of lone pair Q.7 Which statements is false in favour of resonance, it – (A) Increase stability of molecule (B) It create same type of bond (C) Increase reactivity of molecule (D) Decrease reactivity of molecule Q.8 Which of the following species can act as an electron deficient reagent – (A) Acetate ion (B) Methyl carbanion (C) Methoxide ion (D) Dichlorocarbene Q.9 Toluene is more reactive than benzene towards electrophilic reagents due to - (A) Inductive effect only (B) Hyperconjugative effect only (C) Both inductive as well as hyperconjugative effects (D) Strong mesomeric effect Q.10 Which of the following is an example of a nucleophile – (A) Fluoride ion (B) Hydronium ion (C) Chlorine atom (D) Aniline hydrochloride Q.11 CH3 is less stable than - (A) CH3 – CH2 (B) CH3 – CH – CH3 (C) CH3 – NO2 (D) CH3 – CH – C2H5 Q.12 Which of the following species has highest number of electron pairs on the central carbon atoms & (A) Free radical (B) Carbanion (C) Carbocation (D) Carbene Q.13 Addition of Hl on double bond of propene yields isopropyl iodide as major product, it is because addition proceeds through - (A) More stable carbanion (B) More stable carbocation (C) More stable free radical (D) Homolysis Q.14 The nitration of benzene by HNO3 + H2SO4 is – (A) Electrophilic substitution (B) Electrophilic addition (C) Nucleophilic substitution (D) Free radical substitution Q.15 Which of the following is an example of a carbocation – (A) (CH3)3 C (B) (CH3)3 C: (C) (CH3)3 C (D) (CH3)3C– Q.16 The formation cynohydrine from ketone is an example of – (A) Electrophilic addition (B) Nucleophilic addition (C) Nucleophilic substitution (D) Electrophilic substitution Q.17 Lowest boiling point will be shown by -
NH2 NO 2 NO2 HO (A) NO2 (B) (C) (D) HO
Q.18 Which of the following has +R (resonance) effect HO NO2 (A) – CN (B) – CHO (C) – NH2 (D) – NO2 Q.19 Most acidic compound is -
(A) CH3 – COOH (B) C6H5 – COOH (C) C6H5 – OH (D) O 2 N CO 2 H Q.20 The stabilization due to resonance is maximum in -
(A) (B) (C) (D) ANSWERS (1) (B) (2) (C) (3) (D) (4) (A) (5) (A) (6) (A) (7) (C) (8) (D) (9) (C) (10) (A) (11) (C) (12) (B) (13) (B) (14) (A) (15) (C) (16) (C) (17) (B) (18) (C) (19) (D) (20) (C)
Gyaan Sankalp 97 Organic chemistry-Some basic principles and techniques PURIFICATION OF ORGANIC COMPOUNDS : For characterization and determination of the structural formulae of an organic compound it is essential that it should be in the purest form . Methods usually adopted for the purification of organic compounds are :
1. Crystallization : The technique used for obtaining a compound in pure solid form often of well-defined geometrically shaped crystals. The method is based on the difference in the solubility of the organic compound and the impurities present in it. A saturated solution of the impure compound in a suitable solvent is prepared. Of course the temperature has to be increased for more dissolution of the organic compound. Subsequently the solution is filtered and cooled, pure crystals get separated. Solvent chosen must have the following properties : (i) Not react chemically with the compound to be purified (ii) Not dissolve the impurities at all or dissolve to a limited extent so that they remain in the mother liquor (iii) The solubility of the compound in the solvent at room temperature should be low, whereas at elevated temperature the solubility should be high. Solvents : Water, ethanol, dioxane, acetone, benzene, chloroform, petroleum ether etc. Ex.Compound and impurity Solvent chosen Benzoic acid + naphthalene Water Sugar + sand Water Sugar + Common salt Ethanol Iodoform + impurity Ethanol Fractional Crystallization : Separation of different compounds of a mixture by repeated crystallization is called fractional crystallization. A hot saturated solution containing two substances is cooled, the component with lower solubility crystalizes first and these crystals can then be separated out by filtration. The more soluble component crystallizes out on further cooling. Ex. [KClO3 + KCl] mixture can be separated by fractional crystallisation. KClO3 crystallized first being less soluble. After the separation of KClO3, KCl will be crystallized subsequently being more soluble.
2. Sublimation : In the sublimation process, solid directly transforms into vapour state on heating and the vapours on cooling transform into solid state. This method is used for the purification of solids which sublime from the non–volatile impurities. heating Solid Vapour Cooling Ex. Camphor, naphthalene, ammonium chloride, salicyclic acid, iodine, AlCl3, HgCl2 etc. are purified by sublimation.
3. Distillation : Distillation is used for separating the constituents of a liquid mixture which differ in their boiling points. Simple Distillation : Distillation implies conversion of liquid into vapour state by heating followed by condensation of the vapours by cooling. (The non–volatile impurities are left in the flask itself) The method is suited for liquids which are stable at their boiling point. Ex. Ethanol, benzene, chloroform, carbon tetrachloride, toluene, xylene, nitrobenzene, chlorobenzene etc. A mixture of two liquids can be separated if their boiling points differ by a wide range. (i) Mixture of hexane (b.p. 342 K) and toluene (b.p. 384 K) (ii) Mixture of benzene (b.p. 353 K) and aniline (b.p. 457 K) Fractional Distillation : Fractional distillation is a suitable process for separating two or more miscible liquids which have boiling points quite close to each other. Ex. (i) Mixture of acetone (b.p. 330K) and methyl alcohol (b.p. 338K) (ii) Separation of petroleum into gasoline, kerosene oil, diesel oil etc. Distillation assembly is provided with fractionating column. The fractionating column is a long tube provided with obstructions to the passage of the vapours moving upwards and liquids moving downwards. It should be kept in mind that fractionating column increase the cooling surface. On heating the liquid in the distillation flask, vapours of more volatile liquid rise up in the fractionating column and condense while passing through the condenser [cooled by circulating water] and is collected in the receiver. The vapours of less volatile liquid simultaneously rise in the column but due to several obstructions they get condensed in the column itself and flow back to the distillation flask. Obviously the receiver contains the fraction rich in more volatile liquid. On repeating the process separation of mixture of two liquids can be accomplished. Figure shows fractional distillation.
98 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
Distillation under reduced pressure (vacuum distillation) : Vacuum distillation is used for the distillation of those liquids which decompose before their boiling point in reached. Under reduced pressure the liquid will be boiled at lower temperature (without decomposition) and so its distillation can be carried out. Ex. Fatty acids, glycerol etc. Steam Distillation : Steam distillation is used to purify the substances (liquids or solids) (a) Which are volatile in steam, (b) Immiscible with water (c) Contaminated with non–volatile impurities
Steam distillation Ex. (i) Purification of aniline (ii) Separation of ortho and para nitrophenol. The proportion of the organic substance that distills over with steam is related to the vapour pressures as well as molecular
w1 p 1 M 1 masses by the following equation : w2 p 2 M 2
where w1 and w2 are the masses of water and the compound (which is steam distilled) respectively, p1 and p2 are the vapour pressures of water and the compound respectively at the temperature of distillation. M1 and M2 are the molar masses of water and the compound respectively. 4. Differential Extraction : In order to recover organic compounds from their aqueous solution differential extraction method is prominently used. An aqueous solution of organic compound is taken in the separating funnel and is shaken with organic solvent which is immiscible with water but in which the organic compound is highly soluble. Organic compound is dissolved in the organic solvent which is extracted by the separating funnel. Organic compound is then recovered by evaporating the solvent in a water bath. The process is repeated to make further recovery of organic compound. It should however be noted that with a given amount of the solvent, if more number of extractions are carried out with small amount of the solvent greater will be amount of the extracted substance. Solvents : the solvents Usually used in the extraction process are benzene, ether, chloroform, dioxane, acetone etc. Ex. To remove benzoic acid from aqueous solution of benzoic acid, benzene is used as the extracting solvent.
Gyaan Sankalp 99 Organic chemistry-Some basic principles and techniques 5. Chromatography : Chromatography is the most versatile analytical technique used for the separation, purification, isolation and identification of the constituents present in a mixture (coloured or colourless). It was discovered by Tswett (1906) for the separation of plant pigments. Chromatography may be defined as the technique of separating the components of a mixture in which separation is affected by movement of individual components through a stationary phase (fixed) under the influence of mobile phase. Different types of important chromatography are :
Nature of phase Type of chromatography Uses Mobil Stationary Liquid Solid Column chromatography Preparative scale separations Liquid Solid High performance liquid (HPLC) Qualitative and quantitative analysis Liquid Solid Thin layer chromatography (TLC) Qualitative analysis Gas Liquid Gas-liquid chromatography (GLC) Qualitative and quantitative analysis Liquid Liquid Paper chromatography Qualitative quantitative analysis of polar organic and inorganic compounds
Adsorption or column chromatography : Sample containing various constituents is dissolved in a suitable solvent and the solution is allowed to percolate through a column (glass tube or burette containing silica gel.) The solution moves down slowly and the various constituents of the mixture gets adsorbed to different extent over the adsorbent. The constituent which is adsorbed most strongly will be retained at top. The constituent which is adsorbed less strongly will be retained at the lower level. The constituent which is adsorbed still less will be retained at the still lower level and so on. Supposing there are three organic compounds A, B and C present in the mixture. The decreasing order of their adsorbing tendency is : A > B > C Obviously A will be adsorbed near the top of the column while C having least adsorbing tendency will be adsorbed in the last. The column will appear as coloured bands or zones of various constituents. Further by the process of elution i.e. by percolating suitable solvents through the column. The least adsorbed constituent is eluted first by the solvent while the strongly adsorbed constituent will be eluted afterwards by a solvent. The pure compound is recovered by evaporation of the solvent. Ex. Chlorophyll from plants, pigments or dyes etc.
Mixture (A + B + C) + A Solvent Adsorbent (Alumina or B Bands Megnesia or Silica zel) C
Glass wool
(a) Adsorbent Column (b) Separation of components
(c) Elution Column chromatography
100 Gyaan Sankalp Organic chemistry-Some basic principles and techniques QUALITATIVE ANALYSIS The qualitative analysis of an organic compound involves the detection of all the elements present in it. Detection of Carbon and Hydrogen : A small amount of dry and powdered substance is mixed with about double amount of pure and dry copper oxide. The mixture is heated in a dried hard glass tube delivery tube and fitted with a carrying a bulb in the horizontal length The bulb of delivery tube contains anhydrous copper sulphate (white). When the mixture is heated, if the carbon present in the compound is oxidised to carbon dioxide which turns lime water milky and if the hydrogen present in the organic compound is oxidised to water which turns anhydrous copper sulphate in the bulb to blue. Reactions : Heat C + 2CuO CO2 2Cu In organic compound
CO2 + Ca(OH)2 CaCO3 + H2O lime water milky Heat 2H + CuO H2 O Cu In organic compound
CuSO4 + 5H2O CuSO4 . 5H2O anhydrous hydrated copper copper sulphate sulphate (white) (blue) This method is known as copper oxide test. Detection of Nitrogen : Nitrogen in an organic compound is detected by the following tests : (a) Soda lime test. A pinch of an organic compound is heated strongly with soda lime (NaOH + CaO) in a test tube. If ammonia gas evolves, it indicates nitrogen. CaO CH3 CONH 2 NaOH CH 3 COONa NH 3 acetamide Limitation : This method has a limitation. A large number of organic compounds such as nitro and diazo compounds do not liberate ammonia on heating with sodalime. (b) Lassaigne’s method : A small piece of a dry sodium is heated gently in an ignition tube till it melts to a shining globule. Then, a small amount of organic substance is plunged in to distilled water contained in a porcelein dish. The contents are heated for 5 min. added and the tube is heated strongly till it becomes red hot. The red hot tube is then cooled and filtered. The filtered liquid is known as sodium extract or Lassaigne’s extract. The Lassaigne’s extract is usually alkaline. If not, it may be made alkaline by adding a few drops of a dilute solution of sodium hydroxide. To a part of sodium 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 added. 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 In organic compound FeSO4 + 2NaCN Fe(CN)2 + Na2SO4 Fe(CN)2+ 4NaCN Na4[Fe(CN)6] sodium ferrocyanide 3Na4[Fe(CN)6]+4FeCl3 Fe4[Fe(CN)6]3 + 12NaCl ferric ferrocyanide (blue colour) When nitrogen and sulphur both are present in any organic compound, sodium thiocyanate is formed during fusion. When extracted with water sodium thiocyanate goes into the sodium extract and gives blood red colouration with ferric ions due to the formation of ferric thiocyanate
Na C N S NaCNS |— from organic — | sodium thiocyanate compound
3 3NaCNS Fe Fe(CNS)3 3Na ferric thiocyanate (blood red)
Gyaan Sankalp 101 Organic chemistry-Some basic principles and techniques Detection of Sulphur : Lassaigne's Test : The sulphur in the compound reacts with sodium metal to form sodium sulphide. 2 Na + S Na2S The Lassaigne’s extract is divided into three parts and following tests are performed. (i) Lead acetate test : The second part of the Lassaigne’s extract is acidified with acetic acid and then lead acetate solution is added. Formation of black precipitate confirm the presence of sulphur. Na2S + Pb(CH3COO)2 PbS + 2CH3COONa lead acetate black (ii) Silver nitrate test : The third part of Lassaigne's extract is acidified with acetic acid and then AgNO3 solution is added. Formation of black ppt. confirms the presence of sulphur.– Na2S + 2AgNO3 Ag2S + 2NaNO3 Black (iii) Sodium nitroprusside test. To one portion of the extract few drops of sodium nitroprusside are added. The appearance of violet colouration indicates sulphur. Na2S + Na2[Fe(CN)5NO] Na4[Fe(CN)5NO.S] sod. nitroprusside violet colouration Detection of Halogens : Beilstein test : A stout copper wire is heated in non–luminous flame till it ceases to impart green colour to the flame. The heated end of the wire is dipped into the organic compound and again taken to flame. Appearance of green or bluish green colour of the flame indicates the presence of halogens. Urea, thiourea etc. also give this test due to the formation of volatile Cu(CN)2 Lassaigne’s test. Sodium extract is prepared. During fusion, sodium will combine with the halogen (from the organic compound) to form sodium halide. Na + X Fusion NaX(X = Cl, Br, ) Now, we add AgNO3 solution. (i) If white precipitate appears then chlorine is confirmed. NaCl + AgNO3 AgCl + NaNO3 white ppt. (ii)If light yellow precipitate appears then bromine is confirmed. NaBr + AgNO3 AgBr + NaNO3 light yellow ppt. (iii) If dark yellow precipitate, appears then iodine is confirmed. Na + AgNO3 Ag + NaNO3 dark yellow ppt. Special test for bromine and iodine. (NaBr & Na come from Lassaigne's extract. Acidified lassaigne's extract + Chlorine water Br2 or 2 evolved 2NaBr + Cl2 2NaCl + Br2 (turns CS2 layer orange) 2Na + Cl2 2NaCI + 2 (turns CS2 layer violet) Detection of Phosphorus : Organic compound is fused with sodium peroxide (Na2O2) or fusion mixture (Na2CO3 + KNO3) and the mass is extracted with water. The extract is boiled with conc. HNO3 and then ammonium molybdate solution is added. The appearance of yellow precipitate or colouration (due to formation of Ammonium phosphomolybdate), shows the presence of phosphorus. 2P + 5 Na2O2 + 4H2O 2Na2HPO4 + 6NaOH (NH ) M O Na2HPO4 + HNO3 4 204 (NH4)3PO4.12 M0O3 yellow QUANTITATIVE ANALYSIS : It involves the estimation of following elements
[C and H] [N] [X] [S] [P] [O] Estimation of C and H [Liebig's method] : A known weight of O.C. is heated with cupric oxide in an atmosphere of air or oxygen (free from CO2). C and H of O.C. are oxidised to CO2 and H2O respectively. The former is collected over KOH and latter is absorbed by CaCl2. From the amounts of CO2 and H2O vapours the percentage of C and H can be calculated.
102 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
C + 2CuO CO2 + 2Cu; 2H + CuO H2O + Cu In the compound In the compound 12 mass of CO formed 2 100 % of C = 44 mass of organic compound taken
2 mass of H O formed 2 100 % of H = 18 mass of organic compound taken Estimation of N : There are two methods for the estimation of nitrogen. (a) Duma's Method (b) Kjeldahls Method (a) Duma's Method : A known weight of organic compound is heated with cupric oxide in an atmosphere of CO2. C and H are oxidised to CO2 and H2O respectively. N is converted to N2. It is anticipated that some oxides of nitrogen may also be formed. It is reduced to N2 by passing over heated copper gauze. C + 2CuO CO2 + 2Cu 2H + CuO H2O + Cu 2N + CuO N2 + oxide of nitrogen Oxide of nitrogen + Cu CuO + N2 The gaseous nitrogen is collected over conc. KOH solution (all other gases are absorbed except nitrogen) and its volume is measured at room temperature and atmospheric pressure. The volume of the nitrogen gas so obtained is converted in to NTP and its mass is calculated. 28 Volume of N at NTP % of nitrogen = 2 100 22400 Mass of organic compound taken
Volume of N2 at NTP can be calculated as : PVPV 1 1 2 2 (NTP) TT1 2
PVT1 1 2 (P a) V 1 273 V2 PT2 1 760 (273 t)
Where : V1 = Volume of N2 at room temperature, a = Aqueous tension at t°C (mm of Hg) t°C = Room temperature, V2 = Volume of nitrogen at NTP, P = Atmospheric pressure (b) Kjeldahl's Method : Kjeldhal’s method is a faster method than Dumas method. However, this method is used only for organic compounds which on heating strongly with concentrated sulphuric acid are converted quantitatively into ammonium sulphates. Kjeldahl’s method cannot be used for the organic compounds, (a) containing nitrogen in the ring, e.g., pyridine, quinoline etc. (b) containing nitro (– NO2) and diazo (– N = N –) groups. A known weight of organic compound is heated strongly with conc. H2SO4[CuSO4 catalyst]. As a result the nitrogen in the compound is converted quantitatively into ammonium sulphate. The solution is then treated with excess of NaOH. The ammonia gas evolved is passed into excess volume of HCl or H2SO4 solution of known strength. The acid left back is treated with standard alkali. From the acid consumed by ammonia. percentage of nitrogen can be calculated. conc. H SO C + H + N 2 4 (NH ) SO + CO + H O 4 2 4 2 2
(NH4)2SO4 + 2NaOH Na2SO4 + 2NH3 + 2H2O 2NH3 + H2SO4 (NH4)2 SO4 1.4 Normality of the acid Volumeof used acid 1.4 N V % of nitrogen = Mass of organic compound taken W Estimation of X (Carius method) : The organic compound containing halogen is oxidised with fuming nitric acid in presence of AgNO3. The halogens are converted in to silver halides The silver halides are washed and separated and dried. The percentage of halogen can be calculate.
XF.HNO3 AgX AgNO3 35.5 Mass of AgCl 100 80 Mass of AgBr 100 % of chlorine 143.5 Mass of organiccompound taken ; % of bromine 188 Mass of organic compound taken
Gyaan Sankalp 103 Organic chemistry-Some basic principles and techniques
127 Mass of AgI 100 % of iodine 235 Mass of organiccompound taken
Estimation of S : The O.C. containing sulphur is heated with fuming HNO3 sulphur is oxidised to H2SO4. Precipitation of sulphate is carried out with barium chloride solution in the form of BaSO4. The precipitate is separated washed and weighed. F.HNO3 SHSO 2 4
H2SO4 + BaCl2 BaSO4 + 2HCl 32 Mass of BaSO 100 4 % of sulphur 233 Mass of organiccompound taken
Estimation of P : Phosphorus is estimated by Carius method. The organic compound is heated with F.HNO3 which oxidises phosphorus to phosphoric acid. It is treated with solution of magnesia mixture(MgSO4 + NH4Cl + NH4OH). A precipitate of MgNH4PO4 is obtained. On ignition it gives magnesium pyrophosphate Mg2P2O7. Its weight is determined.
P + F.HNO3 H3PO4 H3PO4 + Magnesia mixture MgNH4PO4 Magnesium ammonium phosphate 2MgNH4PO4 Mg2P2O7 + 2NH3 + H2O Magnesium pyrophosphate 62 Mass of Mg P O 100 % of phosphorus 2 2 7 222 Mass of O.C. taken Estimation of O : It is calculated by difference. It is estimated by adding the percentage of all other elements and subtracting this sum from 100. Example 15 : 0.244g of an organic compound on combustion with dry oxygen produced 0.616g of and 0.108g of H2O. Determine the percentage composition of the compound. Sol. Mass of the organic compound taken = 0.244g. Mass of CO2 formed = 0.616g Mass of H2O formed = 0.108g We know that CO2 C and H2O 2H 44g 12g 18g 2g 0.616 12 (i) Mass of C in the CO formed = g 2 44 0.616 12 100 Percentage of C in the compound = 68.85 44 0.244 0.108 2 (ii) Mass of H in the H O formed = g 2 18 0.108 2 100 Percentage of H in the compound = 4.92 18 0.244 (iii) Percentage of O in the compound = 100 – (68.85 – 4.92) = 26.23 The percentage composition of the compound is, C = 68.85, H = 4.92 and O = 26.23. Example 16 : 0.196g of an organic substance on combustion gave 0.22g of CO2 and 0.0675g of H2O. In Carius determination, 0.3925g of the substance gave 0.717g of dry AgCl. Find the percentage composition of the substance. Sol. (a) Percentage of C and H Mass of the substance taken for combustion = 0.196g Mass of CO2 formed = 0.22g Mass of H2O formed = 0.0675g
104 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
0.22 12 Percentage of C in the compound = 100 30.6 44 0.196 0.0675 2 100 Percentage of H in the compound = 3.8 18 0.196 (b) Percentage of chlorine Mass of substance taken for Carius determination = 0.3925g Mass of AgCl formed = 0.717g AgCl = Cl (Molar mass of AgCl = 108 + 35.5 = 143.5) 143.5g 35.5g 35.5 So, Mass of chlorine in 0.717g of AgCl = 0.717g × 0.174g 143.5 0.1774 100 Then, percentage of Cl in the simple = 45.2 0.3925 Therefore, the percentage composition of the compound is, C = 30.6%, H = 3.8%, Cl = 45.2%, O (by difference) 20.4%
CHEMICAL METHODS OF DETERMINATION OF MOLECULAR MASS There are two methods for determination of molecular mass – (i) Silver Salt Method (for organic acids) (ii) Chloroplatinate salt method (for organic bases) (i) Silver Salt Method : A known mass of the acid is dissolved in water followed by the subsequent addition of silver nitrate solution till the ppt of silver salt is complete. The ppt is separated, dried, weighed and ignited till decomposition is complete. The residue of pure silver left behind is weighed. AgNO Organic acid 3 Silver salt combustion Ag Let the mass of silver salt formed = W gm Mass of silver left behind = w gm Mass of silver salt Equivalent massof silver salt Mass of silver left behind = Equivalent mass of silver
W E = w 108 W E = × 108 w Hence , Equivalent mass of organic acid = Equivalent mass of silver salt – Equivalent mass of silver + Equivalent mass of hydrogen = E – 108 + 1 = E – 107 Molecular mass of organic acid = Equivalent mass of acid × Basicity of acid Example 17 : 0.41 g of silver salt of a dibasic organic acid left 0.216 g residue of silver on ignition. Calculate the molecular mass of the acid. (equivalent weight of silver = 108) Sol. Mass of the silver salt taken = 0.41 g Mass of silver left behind = 0.216 g Mass of silversalt Equivalent mass of silver salt Mass of silver residue = Equivalent mass of silver
0.41 108 Equivalent mass of silver salt = = 205 0.216 Equivalent mass of acid = 205 – 108 + 1 = 98 Molar mass of acid = Equivalent mass of acid × basicity of acid = 98 × 2= 196
Gyaan Sankalp 105 Organic chemistry-Some basic principles and techniques (ii) Chloroplatinate salt method : This chemical method is used to find the molecular mass of organic bases. A known mass of organic base is allowed to react with chloroplatinic acid (H2 Pt Cl6) to form insoluble chloroplatinate salt. The ppt of chloroplatinate salt is separated, dried, weighed and is subsequently ignited till decomposition is complete. The residue left is platinum which is again weighed. The molecular mass is then calculated by knowing the mass of chloro platinate salt and that of platinum left. If B represents the molecule of mono acid organic base, then the formula of chloro platinate salt is B2H2 Pt Cl6 . H2 Pt Cl 6 Ignite organic base B2H2 Pt Cl6 Pt (chloroplatinate salt) Mass of chlorplatinate salt = W g Mass of platinum residue left = w g If eq. weight of organic base =E Thus molecular mass of chloroplatinate salt = 2E + (2 + 195 + 35.5 × 6) = 2E + 410 1 mole of chloroplatinate salt (2E + 410) on heating gives one atom of platinum (195). Massof chloro platinate Molar massof chloroplatinatesalt Massof platinum residue = Atomic weight of platinum
W 2E 410 w 195
W 1 W 2E = 195 – 410 ;E = x195 410 w 2 w Molar mass of organic base = equivalent weight of base × acidity of the base Example 18 : 0.532 g of the platinic chloride of a monoacid organic base left 0.195 g of platinum as residue on ignition . Calculate the equivalent & molecular mass of the base. (At mass of Pt = 195) mass of chloroplatinatesalt molar massof chloroplatinatesalt Sol. mass of Pt residue = atomic weight of platinum
0.532 2E 410 0.195 195
0.532x195 2E = 410 0.195
1 0.532 195 E = 410 2 0.195 122 Equivalent weight of base = = 61 2 The acidity of organic base is one thus molecular mass is equal to equivalent mass.
TRY IT YOURSELF Q.1 The most suitable method of separation of 1 : 1 mixture of ortho and para nitrophenols is – (A) Distillation (B) Crystallisation (C) Sublimation (D) Chromatography Q.2 Duma's method involves the determination of content of nitrogen in the organic compound in the form of – (A) Gaseous NH3 (B) Gaseous N2 (C) NaCN (D) (NH4)2SO4 Q.3 A compound has C, H and N in the following percentage C = 40%, H = 13.335 and N = 46.67%. What is the empirical formula – (A) CH2N (B) CH4N (C) CH2N2 (D) CH5N Q.4 Kjeldahl's method is used in the estimation of – (A) Nitrogen (B) Halogens (C) Sulphur (D) Oxygen Q.5 Which of the following is not used for the purification of solid impurities – (A) Distillation (B) Sublimation (C) Crystallisation (D) All of these are used for the purification of solid impurities
106 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Q.5 Chromatography is used for the purification of – (A) Solids (B) Gases (C) Liquids (D) All Q.7 A compound with empirical formula CH2O has a vapour density of 30. Its molecular formula is – (A) CH2O (B) C2H4O2 (C) C3H6O3 (D) CH5N Q.8 In a Lassaignes's test for sulphur in the organic compound with sodium nitrosulphide solution the purple colour formed is due to – (A) Na4[Fe(CN)5NOS] (B) Na3[Fe(CN)5S] (C) Na2[Fe(CN)5NOS] (D) Na3[Fe(CN)6] Q.9 The organic compound which does not give blue colour in Lassaigne test is – (A) Aniline (B) Glycine (C) Hydrazine (D) Urea Q.10 0.2 g of an organic compound on complete combustion produces 0.44 g of CO2, then the percentage of carbon in it is – (A) 50 (B) 60 (C) 70 (D) 80 ANSWERS (1) (A) (2) (B) (3) (B) (4) (A) (5) (D) (6) (D) (7) (B) (8) (A) (9) (C) (10) (A)
USEFUL TIPS 1. The order of decreasing electronegativity of hybrid orbitals is sp > sp2 > sp3. 2. Conformational isomers are those isomers which arise due to rotation around a single bond. 3. A meso compound is optically inactive, even though it has asymmetric centres (due to internal compensation of rotation of plane polarised light) 4. An equimolar mixture of enantiomers is called racemic mixture, which is optically inactive. 5. Tautomerism is the type of isomerism arising by the migration of hydrogen. 6. Reaction intermediates and reagents : Homolytic fission Free radicals Heterolytic fission ions (Carbonium, ion carbonium, etc.) 7. Nucleophiles – electron rich Two types : 1. Anions 2. Neutral molecules with lone pair of electrons (Lewis bases) Electrophiles : electron deficient. Two types : 1. Cations 2. Neutral molecules with vacant orbitals (Lewis acids). 8. Inductive effect is due to electron displacement along a chain and is permanent effect. 9. +I (inductive effect) increases basicity, – I effect increases acidity of compounds. 10. Resonance is a phenomenon in which two or more structures can be written for the same compound but none of them actually exists. 11. Tautomers are in dynamic equilibrium interconvertible and no such equilibrium in resonance. 12. Tautomerism has no contribution on stabilization of a molecule resonance has its effect on stabilization. 13. Tautomerism may occur in planar or non-planar while resonance only in planar. 14. Enolization order : CH3COCH3 CH3COCH2COOC2H5 C6H5COCH2COOC2H5 CH3COCH2CHO CH3COCH2COCH3 15. Percent enol content conjugation / Temperature / Hydrogen bonding / Base strength 16. The low reactivity of halogens bonded to unsaturated carbon is due to the +M effect of the halogen. 17. The acidity of phenol is due to the + M effect of OH group. 18. Aromatic compounds are diamagnetic in character whereas anti-aromatic compounds are paramagnetic in character: 19. The benzyl carbocation is more stable than the allyl carbocation. 20 Characteristic reactions : Homologous series Type of reactions (a) Alkanes Substitution (Mostly free radical) (b) Alkenes and alkynes Electrophillic addition (c) Arenes Electrophillic substitution (d) Alkyl halides Nucleophillic substitution (e) Aldehyde and ketones Nucleophillic addition
Gyaan Sankalp 107 Organic chemistry-Some basic principles and techniques MISCELLANEOUS SOLVED EXAMPLES
Example 1 : Give hybridisation state of each carbon in following compounds : CH2 = C = O, CH3 – CH = CH2, (CH3)2CO, CH2 = CHCN, – + CH3CH2 , CH3CH2 , CH3CH2. Sol. The hybridisation state of each carbon in the given compounds, starting from the left is given below. 2 (i) CH2 = C = O sp , sp 3 2 2 (ii) CH3 – CH = CH2 sp , sp , sp 3 2 (iii)(CH3)2CO sp , sp 2 2 (iv) CH2 = CHCN sp , sp , sp – 3 3 (v) CH3CH2 sp , sp + 3 2 (vi) CH3CH2 sp , sp • 3 2 (vii) CH3CH2 sp , sp
Example 2 : Write the IUPAC names of following compounds - COOH O O (i) (ii) (iii) (iv) (v)
O
CH2 —CHCH 3 (vi) (vii) (viii) (ix) O
5 4 CH3 CH — CH 3 3 | 2 1 CH — C — CH — C — CH Sol. (i) CH3 —CH 2 —CH—CH—CH 3 (ii) 3 2 3 (iii) 1, 3, 5-cyclohexatriene | || || OO CH3 3-Ethyl-2, 4-dimethyl pentane 2, 4-pentane dione
(iv) CH3 — CH 2 CH 2 (v) Cyclopropane carboxylic acid (vi) 1, 3-cyclobutadiene | || C C — CH 1-Hexene-3-yne 1 2 3 4 5 CH2 C—CH 2 —CH—CH 3 (vii) 1, 2-epoxy propane (viii) | | (ix) Cyclopentanone CH3 — CH — CH 3 CH 3 2-isopropyl-4-methyl-1-pentene
Example 3 : Indicate the and bonds in the following molecules : C6H6, C6H12, CH2Cl2, CH2 = C = CH2, CH3NO2, HCONHCH3 Sol. C6H6 12 and 6 bonds C6H12 18 bonds CH2Cl2 4 bonds CH2 = C = CH2 6 bonds and 2 bonds CH3NO2 6 bonds and 1 bond HCONHCH3 8 bonds and 1 bond
108 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Example 4 : Write bondline formulae for : tert-Butylcyclopentane, Isopropyl alcohol, 2, 3-Dimethyl butanal, Heptan-4-one, Cyclohexanone. Sol. Name Structural formula Bond line formula
CH3
CH3 – C – CH3 (i) Tert-butylcyclopentane
CH3 CHCH 3 2 (ii) Isopropyl alcohol | OH OH 3 1
CH3 | CH CH CCHO 2 (iii)2, 3-dimethylbutanal 3 2 1 H | 4 3 CH3 O
O 1 2 3 4 5 6 7 (iv) Heptan-4-one || CH3 CH 2 CH 2 CCH 2 CH 2 CH 3 O
O C O HC2 CH2
(v) Cyclohexanone HC2 CH2 C
H2 Example 5 : (a) Explain compounds like fulvene (I) and tropone (II) have high dipolemoments ?
CH 2 O (I) (II) (b) Although ethers are generally inert to bases, 2, 4-dinitroanisole is readily converted to 2, 4-dinitrophenol and methanol on refluxing with NaOH aqueous. Account for this observation. Sol. (a) In the resonance forms shown, there is a considerable contribution from structure containing aromatic ions. Hence high dipole moments are exhibited. Tropone II Fulvene (I)
– ; +
CH CH O 2 2 CH2 O- (b) 2, 4-dinitroanisole has in its ring structure two nitro groups which are strongly electron withdrawing. The presence of these nitro groups enables the nucleophilic substitution of the –OCH3 group by the –OH group very easy because the electrons that constitute the C–OCH3 bond are drawn towards the nitro groups and the bond is easily broken.
Gyaan Sankalp 109 Organic chemistry-Some basic principles and techniques
OCH3 OH HO OHC3
NO2 NO2 NO2 OCH 3 + OH + CH3OH
NO NO2 2 NO2 Example 6 : (a) Explain why CHCl3 is more acidic than CHF3? (b) C2H5SH is a stronger acid than C2H5OH. Why ? • • Sol. (a) (i) : C Cl3 is less basic than CF3. (ii) In addition the inductive effect, Cl uses its empty 3d orbitals F can disperse the charge only inductive effect. to disperse the charge by p-d bonding. See Figure. O O F Cl •
F C • Cl C
F Cl
While F is a second period element with no d orbital whereas Cl is a third period element with a 3d orbital. There is no 2d orbital for F. Hence CHCl3 is more acidic than CHF3. – (b) Since C2H5SH is a stronger acid, C2H5S its conjugate base is a weaker base than the corresponding C2H5O . – Although both O and S belong to the same group in the periodic table, as S is larger in size than O, the negative charge on C2H5S can disperse over a larger base and hence is weaker. Example 7 : (a) The elimination shown below fails to take place. Why ?
base/ Br (HBr) H
H H (b) What stereosiomer of tartaric acid is formed when erythro , ’-dichloro succinic acid is treated with moist Ag2O ? Sol. (a) In the elimination shown by the bicyclic compound, the halogen viz., Br is attached to a bridgehead carbon and if the loss of HBr takes place, it leads to double bond involving the bridge head carbon which is a violation of Bredt’s rule. (Bredt noticed that there were no examples of bicyclic molecules with double bonds at the bridgehead position). (b) The reaction given is COOH COOH H – C – Cl H – C – OH moist H – C – Cl Ag2 O H – C – OH COOH COOH erythro , ’-dichloro meso tataric acid succinic acid In erythro , ’-dichlorosuccinic acid, the two halogen atoms are on the same side of the compound the two chiral centres. This on hydrolysis leads to meso tartaric acid, which is optically inactive.
110 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Example 8 : Give the IUPAC names of the following compounds
CN (a) (b) (c) O
(d) Cl Br (e) Cl H
1 2 3 5 1CN 6 4 3 3 4 5 6 2 1 6 2 7 Sol. (a) 7 5 3 (b) 4 2 (c) 1 4-propylhept-2-yne 3-methylpentanenitrile 2, 5-dimethylheptane 2 4 6 3 5 O 1 7 3 1 (d) Cl Br (e) Cl 2 H 3-bromo-3-chloroheptane 3-chloropropanal Example 9 : Give suitable reason for the following statements. (a) Although both benzamide (C6H5CONH2) and aniline (C6H5NH2) contain the –NH2 group, only aniline is soluble in dil. HCl, (b) Although both C6H5CH2OH (I) and C6H5OH (II) contain the –OH group, only (II) is soluble in aqueous NaOH. Sol. (a) When we consider the structure of benzamide C6H5CONH2 one observes that the two groups –CO– and –NH2 is an electron donor. The pair of electrons of nitrogen are drawn towards the carbonyl group and –NH2 group loses its basicity considerably and becomes neutral. On the other hand, the amino group in aniline is directly attached to the phenyl ring and it is basic and hence soluble in dil. HCl.
• • O HN O HN2 2 C C
Nitrogen loses the electrons and acquires a positive charge. (b) C6H5CH2OH (I) is an alcohol it is known as benzyl alcohol. In this compound the –OH is attached to the phenyl ring not directly but through a bridging group viz, –CH2. Here the H of the OH group is not acidic. C6H5OH(II) is a phenol. In this compound, the –OH is attached directly to the phenyl ring and the hydrogen of the OH is acidic – and once the hydrogen is lost, the remaining C6H5O cas stabilize easily through resonance. OH O
+ H
O O O O O –
– Such a resonance stabilization is not possible in C6H5CH2O when a hydrogen is lost from C6H5CH2OH. The solubility of C6H5OH (II) in aqueous NaOH is due its acidic nature. Gyaan Sankalp 111 Organic chemistry-Some basic principles and techniques Example 10 : (a) Write the electron dot structure for phosgene (COCl2) (b) Are the following structures given for phosgene correct ? If not, explain why ?
...... (i) : Cl : :C: :O: Cl (ii) : Cl : :C: :O: Cl : ...... O ×× Sol. (a) Cl ×C×Cl (b) Both are incorrect In structure (i), oxygen has 10 electrons which is more than the octet. The number of valence electrons that must appear in the structure should be 24, that is 4 for C, 6 for O, and 2 × 7 = 14 for Cl. In the structure (i), there are only 18 electrons in all. One of the chlorine atoms in structure (i) has a double bond. In structure (ii), the total number of valence electrons are 22 as against the required 24. C and O do not have their normal covalencies. Example 11 : (a) Account for the fact SO2 has a dipole moment ( = 1.63 D), whereas CO2 has no dipole moment ( = 0) (b) Account for the observation that ethanol C2H5OH (M. wt. = 46) boils at 78ºC whereas ethylamine C2H5NH2 (Mol. wt. = 45), boils at 17ºC ? Sol. (a) CO2 is a linear molecule and the C – O bond dipoles mutually cancel each other. On the other hand SO2 molecule is angular and not linear and hence SO2 has a dipole moment. (b) Oxygen being more electronegative than nitrogen, strong hydrogen bonds are formed in ethanol compared to the hydrogen bonds formed in ethylamine. Though the molecular weights are nearly the same for the two compounds, the boiling points are different. Example 12 : Consider the following pairs of structures. Mark the pairs as identical (1) cis isomers (2) trans compounds (3) enantiomers and (4) structural isomers. H CH H 3 CH C 3 (a) and (b) and (c) and C Br Br
Cl CH3 H Cl
(d) H OH and H3C OH (e) and Cl COOH COOH Cl Sol. (a) Trans compounds (b) Structural isomers (c) Identical compounds (d) Enantiomers (e) cis isomers
Example 13 : O OH The compound shown : readily loses both its protons to form a dianion. In the dianion, all the C – C bonds O OH and C – O bonds are of the same length. Provide an explanation. Sol. The explanation is given in terms of resonance. A number of resonance structures can be drawn as shown below for the dianion, where the electrons located on the oxygen anions travel throughout the molecule, thereby making the C – C and C – O bonds similar in length.
O O O O O O O O O O O O O O O O
112 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Example 14 : Which of the following represents the correct IUPAC name for the compound concerned : (a) 2, 2-dimethylpentane or 2-Dimethylpentane, (b) 2, 3-dimethylpentane or 3, 4-Dimethylpentane (c) 2, 4, 7-Trimethyloctane or 2, 5, 7-Trimethyloctane (d) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (e) But-3-yn-1-ol or But-4-ol-1-yne Sol. The correct IUPAC names from each pair is given below : (a) 2, 2-dimethylpentane (b) 2, 3-dimethylpentane (c) 2, 4, 7-trimethyloctane (d) 2-chloro-4-methylpentane (e) But-3-yn-1-ol Example 15 : Give the functional isomer for acetic acid
O O H Sol. H3C
OH O CH3 acetic acid methyl formate
Monocarboxylic acids and Ester’s are functional isomers. Example 16 : Give the functional isomer for ethyl thioalcohol. H C S H3C 3
S H CH3 ethanethiol (methylthio)methane Thioalcohols and thio ethers are functional isomers. Example 17 : Draw formula for the first five members of each homologous series beginning with the following compounds. (a) H – COOH (b) CH3COCH3 (c) H – CH = CH2 O O O Sol. (a) H – C CH3 – C CH 3 CH 2 – C O – H O – H O – H O O
CH3CH2 CH2– C CH3CH2 CH2 CH2 – C O – H O – H
O O O || (b) || || CHCCH3 3 CHCHCCH3 2 3 CHCHCCHCH3 2 2 3 O O || || CHCHCHCCHCH3 2 2 2 3 CHCHCH3 2 2 C CHCHCH 2 2 3
(c) H – CH = CH2 CH3 – CH = CH2 CH3 – CH = CH – CH3 CH3 – CH2 – CH = CH – CH2 – CH3 CH3 – CH = CH – CH2 – CH3 Example 18 : Rationalize the following observations 1 (a) SN reactions always lead to racemic products (b) Tautomers of a compound are in equilibrium with each other. (c) Optically active 2-iodobutane on treatment with Nal in acetone gives a product which does not show optical activity. (d) 2,6- dimethylbenzoic acid fails to undergo esterification with alcohol under normal esterification conditions. 1 Sol. (a) In an SN reaction. two steps are involved. The first step is a rate determining slow step which involves the ionization of the substrate to give a carbocation. For example slow + – (H3C)3 CCl (CH3)3C + Cl (step I)
Gyaan Sankalp 113 Organic chemistry-Some basic principles and techniques The second step is a rapid step involving the attack of the incoming. nucleophile to form the product. + – fast (CH3)3C + OH (CH3)3COH (Step II) Since a planar carbocation is formed as an intermediate and the attack by the entering nucleophile can occur from the front and the back (being a planar structure), the product is racemic in nature. (b) Tautomerism is a type of isomerism known as keto - enol tautomerism. The two forms viz, the keto and the enol can coexist in equilibrium Depending on the structure, conditions and the medium, either one of the forms dominate. For example, acetone exists as a keto form predominantly phenol exists in the enol form. However, the keto or enol can mutually be interconverted. They are in dynamic equilibrium. For example OH | H3C – COCH3 CH—CCH3 2 (keto) (enol) 2 (c) To start with 2- iodobutane is optically active. When NaI in acetone medium attacks, an SN process occurs which should lead to inversion of configuration. However at equilibrium there are fairly equal quantities of both (+) and (–) forms of the product which is also 2- iodobutane. Hence there is loss of optical activity due to racemization. (d) The two bulky methyl groups in the 2 and 6 positions of the ring- vicinal to the main functional group which is COOH group prevents the attack on carboxyfunction. Hence either the esterification is slow or does not take place at all. O OH C
CH3 H3C
Example 19 : (a) What is a carbanion? How is it formed ? Give two examples of reactions involving a carbanion. (b) What are electrophilic reagents ? Give a few examples. Sol. (a) Carbonaion is a reactive intermediate in the area of organic chemistry. Carbanion an ion in which carbon carries a negative charge. The hybridisation in a carbanion is sp3 and the structure is pyramidal. Carbanions are formed as follows : (i) By the heterolytic cleavage of C–C bond or C–H bond
.. C H C C H C: a carbanion ; a cabanion
(ii) Carbanions are also formed when an electron withdrawing group is adjacent to a carbon. For example. H O O | H—C— C base H—C— C | a carbanion H | H H H Examples of reactions involving carbanion. COOC H COOC2 H5 COOC 2 H5 2 5 NaOC H – C H CH X (1) H 2 C 2 5 HC 6 5 2 C6H5CH2 CH COOC H COOC H 2 5 COOC 2 H5 2 5 ( malonic ester) (a carbanion) O O – (2) H3 C–C + base (NaOH) H2 C–C H H a carbanion
114 Gyaan Sankalp Organic chemistry-Some basic principles and techniques
O O – OH O One more + H 3C–C H 2C–C | molecule of HC—CH—CH –C CH3 CHO H H 3 2 H (an aldol) (b) electrophilic reagents : A species which can accept a pair of electron is called an electrophilic reagent. An electrophilic reagent may be positively charged or an electron deficient molecule. Few examples of electrophilic reagent are Positively charged ions. HO3 ,NO 2 ,NO,CHN 6 5 N Electron- deficient molecules BF3, AlCl3 SO3, SnCl2, FeCl3 ZnCl2. Example 20 : What are carbenes and nitrenes ? Sol. Carbenes are neutral, electron deficient intermediates containing carbon. Carbon is covalently bonded to two hydrogen atoms and have two non- bonded electrons. H Cl Example of a simple carbene :C (carbene) , Example of a dichlorocarbene ; :C (dichlorocarbene) H Cl The hybridisation in a carbene is sp2 and the structure is planar. Nitrenes are highly reactive neutral species containing nitrogen having six electrons around nitrogen. N– H. The hybridisation of nitrogen in a nitrene is sp2 and the structure is planar.. Example 21 : Draw valid lewis structures for the following molecules/ Species. Indicate the presence of lone pair of electrons in every structure. – (a) Dimethylketene (b) An azide ion (N3 ) (c) Cyanamide (d) Acrolein (e)Methyl-N-cyclohexylcarbamate. :O: HC 3 O NH–C–OCH .. – + + 3 Sol. (a) C=C = O.. (b) N = N = N (c) H2 N – C = N (d) H2C = CH–C (e) H H 3 C
Example 22 : Draw the structures of the conformational isomers of 2,3- dimethylbutane. Indicate the order of their stabilities. HH | | Sol. CH—C—C—CH3 3 | | CH3 CH 3 The conformational isomers are
H H H H CH3 H CH H CH3 CH3 3
H H3C CH H3C CH3 H3C 3 H C H C 3 CH CH3 3 CH CH3 H 3 3 CH3 Staggered I Eclipsed II Eclipsed III Gauche IV Gyaan Sankalp 115 Organic chemistry-Some basic principles and techniques The staggered (anti) conformation does not have torsional strain since the groups are staggered. This is the most stable form The eclipsed conformation II has only CH3–CH3 eclipse while III has two; II is more stable than III. The gauche form IV will have stability intermediate between I and II.
Example 23 : Explain why alkyl groups act as donors when attached to a system. Sol. Carbon is slightly more electronegative than hydrogen. Thus, carbon atom in an alkyl (say methyl) group has higher electron density around it as compared with a H atom. As a result, alkyl groups are able to donate electrons inductively when attached to a system.
Example 24 : In the estimation of sulphur by carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find the percentage of sulphur in the given compound. Massof BaSO 32 100 0.668 32 100 Sol. Percentage of sulphur in the compound = 4 × = 233 Mass of compound 233 0.468 Percentage of sulphur in the compound = 19.6%
Example 25 : 0.243 g of a monoacidic base containing C, H and N only, on combustion gives 330 ml of CO2 at STP. 0.667 g of the same base on kjeldahl's digestion liberates ammonia equivalent to 26.7 ml of 0.25 NH2SO4 solution 1g of the salt of the base with chloroplaitinic acid gives on ignition 0.321 g of platinum metal. Deduce the molecular weight and probable structure of the base. 0.33 Sol. Number of moles of CO produced = = 0.01473 2 22.4 0.01473x12 Percentage of C in the compound = × 100 = 72.74 0.243 Number of mili equivalents of H2SO4 reacted = 26.7 × 0.25 = 6.675 Number of milli equivalents of NH3 produced = 6.675 6.675 103 14 100 Percentage of N in the compound = = 14.01 0.667 Since the compound contains only C, H & N. Percentage of H in the compound = 13.16 72.75 Empirical formula C ; = 6.0625 : 6.0 12 14.09 13.16 N : = 1.0 : 1.0 H ; = 13.16 : 13.0 14 1 Empirical formula C6H13N 1 g of the chloroplatinate produces 0.321 g of pt. 195 195 g of pt will be produced from = 608 g of chloroplatinate. 0.321 Chloroplatinate is B2H2PtCl6 where B is the base. Molecular weight of H2PtCl6 = 410. 608 410 Molecular weight of Base = = 99 5 The molecular formula = C6H13N It molecular formula = C6H13N
NH2 It is a cyclic compound ; cyclohexylamine.
116 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Example 26 : Give condensed and bondline structural formulae and identify the functional group(s) present, if any, for (a) 2, 2, 4-Trimethylpentane (b) 2-Hydroxy-1, 2, 3-propanetricarboxolic acid, (c) Cycloocta-1, 5-diene (d) Hexanedial, (e) 2-(4-Isobutylphenyl) propanoic acid (f) 2-Hydroxy-1, 2-diphenylethan-1-one. Sol. Name of the compound Condensed formula Bondline structure formula
(a) 2, 2, 4-trimethylpentane (CH3)2CHCH2C(CH3)2CH3
HO COOH OH 3 1 (b) 2-hydroxy-1, 2, 3-propane HOOCCH2C(OH)COOHCH2COOH O OH O tricarboxylic acid
H H2 C C 5 6 4 H2C 7CH2
8CH (c) Cyclocta-1, 5-diene H2C 3 2 2 1 C C H2 H
O H (d) Hexanedial OHC(CH2)4CHO H O
O 3 2 1 3 CH3 CHCOOH 2 1 1 OH 6 2 (e) 2-(4-isobutylphenyl) propanoic acid 5 3 4 CH3 CHCH 2CH3
2 CH 2 1 1 6 5 CH OH (f) 2-hydroxy-1, 2-diphenylethan-1-one C6H5 CH(OH) C OC6H5 6 5 O Example 27 : 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in carius estimation. Calculate the percentage of chlorine present in the compound. 35.5 0.5740 Sol. Percentage of chlorine = × 100 = 37.6% 143.5 0.3780 Example 28 : Identify the functional groups in the following compounds.
CHO NH2 O CNHCH3
(a) (b) (c) CH3 CH2 CH(CH2 ) 2 COCl (d) OMe OH CH = CHNO O OCH2 CH2 N(C2 H5 )2 2
Gyaan Sankalp 117 Organic chemistry-Some basic principles and techniques
Sol. (a) – CHO (Aldehydic group) (b) – NH2 (Amino group) O – OMe (Methoxy group) || (Ester group) CO
– OH (Phenolic group) – N(C2H5)2 (Substituted amino group) (diethylamino group) (c) – CONHR (Substituted amide group) (d) C = C (Double bond) – COCl (Chlorocarbonyl) – NO2 (Nitro group) Example 29 : A mixture consisting of bromobenzene, p- nitroaniline and o- cresol is give. Indicate the steps to be followed for the separation of the mixture. Sol. The given mixture is dissolved completely in solvent ether , so that it forms a clear and homogenous solution . The either solution containing the mixture is shaken with 10% NaOH solution in a separating funnel and the two layers are allowed to separate. The lower aqueous layer is drained to a beaker and neutralised with dil. HCl, when o- cresol comes out. Phenols are extracted by aqueous NaOH. The phenolic component is further extracted with ether and purified by distillation. The remaining ether solution in the separating funnel is now shaken with 6N (1 :1) HCl and the two layer are separated. Ether layer is always on the top being lighter. The heavier aqueous acid layer containing the amine. p- nitroaniline is neutralized and the free amine is obtained as an yellow solid. The remaining ether solution, which contains bromobenzene is evaporated carefully (BP. 35°C) to get bromobenzene. Example 30 : – – Which is expected to be more stable, O2NCH2CH2O or CH3CH2O and why ? HH HH | | | | Sol. ONCCO2 HCCO | | | | HH HH
–NO2 group is electron-withdrawing group. So, it favours delocalisation of the negative charge and therefore stabilizes the – – anion. Therefore, O2N-CH2CH2O is more stable than CH3–CH2O . Example 31 : Draw two enantiomers of 1, 1-chlorobromoethane.
Cl Cl C C H Sol. H CH3 Br Br CH3 R S The four atoms directly bonded to asymmetric carbon are Br (at no = 35), Cl (at no = 17), C(at no = 6) and H (at no = 1). Hence the priority order is Br > Cl > CH3 > H (1) (2) (3) (4) Now viewing the groups Br, Cl, CH3 in tetrahedral perspective, the configuration R and S are assigned as in fig. Example 32 : Write two enantiomers of lactic acid.
COOH COOH
C C H H OH Sol. CH3 OH CH3
R S According to the sequence rules, the priorities of the four groups are OH > CO2H > CH3 > H 1 2 3 4 and the configuration R and S as shown in diagram.
118 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Example 33 : Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. + (a) C6H5OH, (b) C6H5NO2 (c) C6H5 CH2 (d) CH3CH = CHCHO, + (e) CH3CH = CH CH2 (f) C6H5CHO (g) CH2 = CHOCH3 Sol. (a) C6H5OH ...... + +OH .. .. :OH OH +OH :OH :OH :– : .. – resonance structures of phenol resonance hybrid (b) C6H5NO2
–1/2 –1/2 .. – – – – – – O + O O + O O + O O + O – O + O N N O + O N N N N + +
+ resonance structure of nitrobenzene resonance hybrid + (c) C6H5 CH2. + CH + CH 2 CH CH2 CH2 2 CH2 2 + +
+
(d) CH3CH = CHCHO ...... – O: O..: O..: + + CH3CH = CH – C CH3CH = CH – C CH3 CH – CH = C H H H + (e) CH3CH = CH CH2
+ + CH3CH = CH – CH 2 CH3 CH – CH = CH2 (f) C6H5CHO .. – – .. ..– H O: .. – .. H O H + O: H O: C .. H O: H O: C C C C .. C + +
+ resonance structures of benzaldehyde resonance hybrid (g) CH2 = CHOCH3 .. .. + –CH = CH – O – CH CH2 = CH – O.. – CH 3 2.. 3 This structure is relatively unstable because it has separated charges and its most electronegative atom is the one with positive charge. So, it makes only a small contribution to the resonance hybrid. Example 34 : State whether the following statements are True of False. Justify your answers with a brief explanation. (a) NH2 is a weaker base than OH (b) The specific rotation of one enantiomer was found to be +15º. This isomer has been obtained from a reaction in which there was 20% racemization and 80% retention of configuration. The angle of rotation of the mixture will be +12.0º (c) C – Cl bond is polar, but CCl4 is non-polar. (d) CH3(CHOH)2 COOH exhibits optical isomerism and the number of optical isomers it can have will be 2. Gyaan Sankalp 119 Organic chemistry-Some basic principles and techniques Sol. (a) False. Between N and O the atom having the larger number of lone pair of electrons around it will have a greater spread of the electronic charge around it. In this case, OH has three lone pairs as shown.
whereas NH2 has only two pairs as lone pair of electrons, . Greater spreading makes the base weaker. So NH2 is a stronger base then OH. (b) True. The mixture contains 90% of the pure enantiomer (10% form the racemic mixture plus 80% from the retention of configuration). The remaining 10% is due to inversion. So, the angle of rotation of the mixture observed will be = 0.90 (+ 15) + 0.10 (– 15) = + 12.0º (c) True. The bond C – Cl is polar in nature because of the difference in electronegativities. However, the molecule as a whole CCl4 is non-polar, and the net dipole moment is zero because of the symmetry of the molecule and the mutual bond moments or bond dipoles cancel each other. (d) False. Only the first part of the statement is true as the compound can exhibit optical isomerism as there are two chiral centres in the molecules. However, the latter part of the statement is not correct as there can be 4 optical isomers for the compound based on the rule, number of optical isomers = 2n, where n is the number of chiral centres in the molecule. Example 35 : Classify the following reactions in one of the reaction type. – – (a) CH3CH2Br + HS CH3CH2SH + Br (b) (CH3)2C = CH2 + HCl (CH3)2ClC – CH3 (c) (CH3)3C – CH2OH + HBr (CH3)2CBrCH2CH3 – (d) CH3CH2Br + HO CH2 = CH2 Sol. (a) Substitution reaction (nucleophilic) (b) Addition reaction (c) Rearrangement reaction (d) Elimination reaction Example 36 : Rearrange the compounds/species given in the order specified. (a) CH3 , HC C , NH2 , OH , OCH3 in the decreasing order of base strength. (b) p-cresol,p-dihydroxybenzene, p-methoxyphenol, p-nitrophenol, p-chlorophenol, in the increasing order of reactivity towards bromination. (c) CH3COCH2CHO, CH3COCH3, CH3CHO, CH3COCH2 COCH3, C6H5OH in the increasing order of enol content (d) Decreasing order of reactivity towards E2 elimination. CH3CH2CH2CH2CH2Br, (H3C)2 C(Br) – CH2 CH3, (CH3 CH2)2 CHBr, CH3CH(Br)CH(CH3)2 and (CH3)2 CH – CH2 CH2 Br. Sol. (a) CH3 > NH2 > HC C > OCH3 > OH Stronger the base means weaker the conjugate acid and vice versa. The corresponding conjugate acids are (CH4, NH3, C2H2, CH3OH, H2O) arranged in the increasing order acid strength. CH4 is the weakest acid and H2O is the strongest in the group given.
(b)
The substituents in the para position viz. –NO2 and –Cl groups are electron withdrawing and hence exert a deactivating influence on bromination. The other two groups namely –CH3 and –OCH3 in the para position are activating and hence the reaction is a little faster. The –OH group in the para position is the most activating group. (c) CH3CHO < CH3COCH3 < CH3COCH2CHO < CH3COCH2COCH3 < C6H5OH In acetaldehyde and acetone, there is only one carbonyl function so that enol content is low. There is only one CH3 group in acetaldehyde, but there are two CH3 groups in acetone, so the enol content in acetone is slightly higher. In the case of CH3COCH2CHO, there are two carbonyl groups and an active methylene centre. In the case of acetyl acetone, there is an active methylene group, two carbonyl groups and two methyl groups. So the enol forms can be OH OH | | H3 C C CH COCH 3 CH2 C CH 2 COCH 3
120 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Hence comparatively greater enol content is expected. In the case of phenol, the compound exists exclusively in the enol form, as it is completely aromatic and the keto form is least favourable as the aromaticity is destroyed. (d) The deciding factor towards the order of reactivity in E2 elimination is that a more branched alkyl halide leads to a more branched olefin which is more stable. So the stability of alkenes formed decides the rate of dehydrohalogenation. So the reactivity order is 3º > 2º > 1º. On this basis, the alkyl halides can be arranged in the decreasing order as follows.
CH3 Br | | HCCCHCH3 2 3 > H3C . CHCHCH 3 | | Br CH3
> H3C – CH2 CH – CH2CH3 > H3 C CH CH 2 CH 2 Br > H3C CH2 CH2 CH2 CH2 Br | | Br CH3 Example 37 : Explain why (a) an allyl cation is more stable than n-propyl cation although both the species contain the same number of carbons ? (b) meso tartaric acid is optically inactive, although it contains 2 chiral centres ? (c) SN2 reaction of an optically active substrate always proceeds with inversion of configuration ? Sol. (a) An allyl cation is represented as H2C = CH – CH2 which can exist in two resonance forms as
H2 C – CH = CH2 The positive charge can migrate from carbon 1 to carbon 3 and hence the cation is well stabilised. On the other hand, in n-propyl cation CH3CH2 CH2 , the positive charge on the terminal carbon is highly localized and cannot shift to other positions in the molecule, through bonds. Hence the species is less stable and more energetic, although both the species have three carbon atoms each. (b) Meso-tartaric acid is represented as
COOH COOH | | H C OH HO C OH | | H C OH HO C OH | | COOH COOH
Although it contains two chiral centres, the optical activity exhibited by one centre is exactly equal and opposite to that produced by the other centre and hence they mutually cancel one against the other internally and the net optical activity is zero in the meso form. (c) In an SN2 mechanism, the reaction always proceeds by the rearside attack by the entering nucleophile on the central carbon, while the leaving nucleophile leaves from the opposite side. Further, the process is concerted and hence always proceeds with the inversion of configuration. This is applicable only to optically active substrates. Example 38 : (a) A chlorohydrocarbon contains 64% carbon and 4.4% hydrogen. Find out the empirical formula and molecular formula if its molecular weight is 112.5. The compound does not give a white precipitate with alc. AgNO3 solution. The compound burns with a smoky flame. Suggest a structure for the compound. (b) (i) An organic compound gave on analysis C = 43%, H = 7.2% and the rest nitrogen. 200 ml of the compound at NTP weighs 1 gm. What is the molecular formula of the compound ? (ii) Benzene gets burnt in oxygen completely according to the equation
2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O(g) How many litres of oxygen at NTP will be needed to complete the reaction when 39 grams of benzene are used ? Sol. (a) The compound is a chloro hydrocarbon. So the compound contains C, H and Cl. Percentage of C = 64, Percentage of H = 4.4 The rest should be chlorine. Percentage of Cl = 31.6 Gyaan Sankalp 121 Organic chemistry-Some basic principles and techniques Elemental composition from the Percentage : 64 4.4 31.6 C = = 5.3 ; H = = 4.4 ; Cl = = 0.9 12 1 35 So, the empirical formula is 5.3 4.4 0.9 C = ; H = ; Cl = 0.9 9 0.9
Empirical formula is C6H5Cl ; Molecular weight is 112.5 Hence, molecular formula and empirical formula are the same. So, the molecular formula is C6H5Cl. The compound burns with a smoky flame shows that the compound is aromatic. Since the compound contains Cl, but still does not give a precipitate with alcoholic AgNO3 shows that the chlorine is attached to the aromatic nucleus. Based on the information available, one structure that can be proposed for the compound is
Chlorobenzene (b) (i) Carbon percentage = 43, Hydrogen percentage = 7.2, Nitrogen percentage = 50 43.0 7.2 50 Empirical formula is C = = 3.5; H = = 7.2; N = = 3.6 12 1 14
Dividing by the smallest number, CH2N. 1 g of the compound occupies 200 ml. Weight of the compound occupying 22.4 litres = 112 g. Molecular weight of the compound = 112 Empirical formula weight = 28 Molecular formula = C4H8N4 (ii) 2 × C6H6 = 2 × 78 g of benzene requires 15 × O2 = 15 × 32 g of oxygen of complete burning. 156 g of benzene requires 480 g of oxygen. 39 g of benzene requires 120 g of oxygen. 22.4 litres of oxygen at NTP weighs 32 g oxygen. 22.4 120 g of oxygen will occupy a volume, × 120 = 84 litres 32 Example 39 : Classify the reagents shown in bold in the following equations as nucleophiles or electrophiles. Use curved arrow notation to show the electron movement. – – – (a) CH3COOH + HO CH3COO + H2O (b) CH3COCH3 + NC CH3C(CN)OHCH3 + (c) C6H6 + CH3C O C6H5COCH3 .. ..– .. O: :O: O:
.. – – Sol. (a) CH3 – C – OH + : OH.. CH3 – C – O – H CH3 – C – O + H 2 O OH OH– is a negative nucleophile ...... – : O: :O: OH – H – C N – CH3 – C – CN CH3 – C – CN + CN (b) CH3 – C – CH 3 + : CN
CH3 CH3 CN– is a negative nucleophile O + O C – CH3 + .. 3 : C – CH3 + (c) + CH3 – C = O H + : B + BH
+ CHCO is a positive electrophile. 122 Gyaan Sankalp Organic chemistry-Some basic principles and techniques Example 40 : What is the relationship between the members of the following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ? O O Cl H H CH Cl C = C 2 (a) (b) H C = C CH2 Cl Cl H
OH OH || | D H D D (c) H C OH H C OH (d) C = C C = C H D H H Sol. (a) Structural isomers (b) Identical (c) Resonance contributors (d) Geometrical isomers Example 41 : For the following bond cleavages, use curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediate products as free radical, carbocation and carbanion. . . (a) CH O – OCH (b) O + OH– O + H O 3 3 CH3 O + OCH 3 2 E (c) + + Br (d) + E+ + Br . . Sol. (a) CH3 O – OCH 3 CH3 O + OCH 3 Homolysis Free radicals – O + H (b) O + OH 2O Heterolysis Carbanion CH 3 – C = O + OH CH3 C = O + H2 O H–CH2 – :CH2
E (c) + + Br– (d) + E+ Heterolysis + Br Heterolysis Carbocation Electrophile Carbocation Example 42 : Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. Sol. Sublimation. Camphor sublimes on heating and can be collected as the sublimate. Example 43 : Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test ? Sol. This is because if sulphuric acid is used, it will decompose sodium sulphide (Na2S) formed during fusion. As a result, the solution will give a negative test for the presence of sulphur in the sample. Example 44 : Will CCl4 give white precipitate of AgCl on heating it with silver nitrate ? Give reason for you answer. Sol. No. Because CCl4 is a covalent compound. Example 45 : An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. Sol. Element Percentage Atomic mass Relative number Simplest Simplest whole of atoms atomic ratio no. ratio C 69 12 69/12 = 5.75 5/75/1.64 = 3.5 7 H 4.8 1 4.8/1 = 4.8 4.8/1.64 = 2.93 6 O 26.2 16 26.2/16 = 1.64 1.64/1.64 = 1 2 Empirical formula = C7H6O2 Then, C7H6O2 7CO2 + 3H2O Stoichiometry : 122g 308g 34g
Gyaan Sankalp 123 Organic chemistry-Some basic principles and techniques
308g 0.2g Mass of carbon dioxide formed = = 0.50 g 122g
54g 0.2g Mass of water formed = = 0.089 g 122g
Example 46 : Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound. Sol. Carbon dioxide is weakly acidic. Potassium hydroxide is strongly alkaline. It reacts with CO2 to form KHCO3 and K2CO3
Example 47 : A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. Sol. Volume of 0.5M H2SO4 solution = 50 mL V Using molarity equation, 60 mL × 0.5 M = 2 × HSO2 4 × 0.5 M V HSO2 4 = 30 mL Thus, Volume of 0.5 M H2SO4 used for neutralising ammonia = (50 mL – 30 mL) = 20 mL
1.4 Normality of acid Volume of acid used for neutrali sin g NH3 Percentage of nitrogen in the sample = Mass of the compound taken
1.4 1.0 20mL Percentage of nitrogen in the sample = = 56% 0.5
124 Gyaan Sankalp