General Organic Chemistry Chapter E3 PDF

60 1014 General Organic Chemistry Chapter E3 23 General Organic Chemistry Bond ID Carbon is an essential element of organic compounds, it has four electrons in its outer most shell. According to the ground state electronic configuration of carbon, it is divalent. Tetravalency of carbon can be explai...

60 1014 General Organic Chemistry Chapter E3 23 General Organic Chemistry Bond ID Carbon is an essential element of organic compounds, it has four electrons in its outer most shell. According to the ground state electronic configuration of carbon, it is divalent. Tetravalency of carbon can be explained by promoting one of the 2s 2 electrons to the unocupied 2 p z 1 atomic orbital. Bond angle Geometry Four - 109.5 Tetrahedra l 25 or 1/4 U % s-character D YG The four valencies of carbon atom are similar and they are symmetrically arranged around the carbon atom. According to Le Bell and Van’t Hoff the four valencies of carbon do not lie in one plane. They are directed towards the corners of a regular tetrahedron with carbon atom at the centre and the angle between any two valencies is 109o28. Hybridisation in Organic Compounds (1) The process of mixing atomic orbitals to form a set of new equivalent orbitals is termed as hybridisation. There are three types of hybridisation, sp hybridisation (involved in saturated U (i) 3 organic compounds containing only single covalent bonds), sp 2 hybridisation ST (ii) (involved in organic compounds having carbon atoms linked by double bonds) and (iii) sp hybridisation (involved in organic compounds having carbon atoms linked by a triple bonds). Table : 23.1 Type of hybridisation Number of orbitals used Number of unused p-orbitals sp3 sp2 sp 1s and 3p 1s and 2p 1s and 1p Nil One Two Three - One - 120 Trigonal planar 33.33 or 1/3 Two - Two - 180 Linear 50 or 1/2 (2) Determination of hybridisation at different carbon atoms : It can be done by two methods, (i) First method : In this method hybridisation can be know by the number of   bonds present on that particular atom. Number of – bond/s Type of hybridisation 0 3 sp 1 2 2 sp sp Examples : O || (i) CH 3  CH  CH  C  CH 3      sp 2 sp 2 sp 2 sp 3 sp 3 (ii) CH 2  C  CH 2  sp 2 (iii)  sp  sp 2 CH 3  CH  CH  CH 2  C  N       2 2 3 3 sp sp sp sp sp sp (iv) HC  C  CH  CH 2     sp sp sp 2 sp 2  In diamond carbon is sp3 hybridised and in graphite carbon is sp 2 hybridised. General Organic Chemistry 1015 (ii) Second method : (Electron pair method) ep = bp + lp; where ep = electron pair present in hybrid orbitals , bp = bond pair present in hybrid orbitals Number of bp = Number of atoms attached to the central atom of the species Central atom H Orbital sp sp 2 sp 3 % s - character 50 33. 33 25  s-character in decreasing order and electroneg ativity in decreasing order C=C H Second atom H bp = 3 Thus sp-hybrid carbon is always electronegative Third atom 1 2 character. sp 2 -hybrid bp = 3 Example :  sp 2 5 sp 3 d 6 sp 3 d 2 (ii) CH 2  CH  bp  2 lp  1 ep  3, sp 2  bp  2 lp  0 ep  2, sp  (iv) CH  C ST U CH 2  C  CH 3 | CH 3 bp  3 lp  0 ep  3, sp 2   bp 1 lp  1 ep  2, sp  (v) CH 3  CH  CH 3  bp  3 lp  1 ep  4 , sp 3 (3) Applications of hybridisation (i) Size of the hybrid orbitals : Since s - orbitals are closer to the nucleus than p - orbitals, it is reasonable to expect that greater the s character of an orbital the smaller it is. Thus the decreasing order of the size of the three hybrid orbitals is opposite to that as well as sp Electronegative carbon having positive charge Electropositive carbon (c) Electronegativities of different hybrid and unhybrid orbitals in decreasing order is as follows U 4 sp 3  (i) CH 2  CH (iii) 3 sp 2 D YG 2 sp behave CH 2  CH ID Number of lp’s can be determined as follows, (a) If carbon has  - bonds or positive charge or odd electron, than lp on carbon will be zero. (b) If carbon has negative charge, then lp will be equal to one. Number of electron pairs (ep) tells us the type of hybridisation as follows, ep Type of hybridisation can   CH 3  CH 2 3 carbon electropositive (in carbocation) as electronegative (in carbanion) in character. 1 H 2 in character and sp 3 - hybrid carbon is electropositive in Central atom H  C  C H  bp  2 (ii) Electronegativity of different orbitals (a) Electronegativity of s-orbital is maximum. (b) Electronegativity of hybrid orbital  % scharacter in hybrid orbitals 60 H sp 3  sp 2  sp E3 First atom of the decreasing order of s orbital character in the three hybrid orbitals. s  sp  sp 2  sp 3  p  % s - character and electroneg ativity in decreasing order. (iii) Bond length variation in hydrocarbons % of s orbital character  1 1  C  C bond length C  H bond length Table : 23.2 Bond type (C – H) Bond length Bond type (C – C) Bond length sp 3  s 1.112Å sp 3  sp 3 (al 1.54 Å (alkanes) sp 2  s kanes) 1.103Å (alkenes) sp  s sp 2  sp 2 (al 1.34Å kenes) 1.08Å (alkynes) sp  sp (alk 1.20Å ynes) (iv) Bond strength in hydrocarbons : The shorter is the bond length, the greater is the compression between atomic nuclei and hence greater is the strength of that bond. Table : 23.3 Bond type (C – Bond energy Bond type (C – C) Bond energy 1016 General Organic Chemistry H) (kcal/mole) sp 3  s 104 sp 3  sp 3 (in (in alkanes) alkanes) sp  sp 106 2 (in alkenes) sp  s 80 – 90 the carbon having halogen. CH 122 – 164 2 | H 3 C  C  Cl (in alkenes) | CH sp  sp 121 (in alkynes) 3 123 – 199 Steric strain around this carbon  (More strained species) Steric strain is released (less strained H 3 C  C  CHspecies) 3  | CH 3 (2) Primary alkyl halide having quaternary  - (in alkynes) carbon does not form transition state because of the steric strain around  -carbon by the  -carbon. To (v) Acidity of hydrocarbons release the strain it converts into carbocation. Bulky group E3 (a) Hydrogen present on electronegative carbon is acidic in nature. CH 3 | CH 3  C | CH (b) Acidity of hydrogen is directly proportional to the electronegativity of the atom on which hydrogen is CH 2  Cl Strained carbon due to bulky group present around this carbon. 3 ID present. Thus H O  H  NH 3  CH  CH    (3) Steric strain inhibits the resonance. This phenomenon is known as steric inhibitions of Electronegativity of the atoms Acidity of compounds in decreasing order resonance. 25 pKa CH 2  CH 2 50 Electronic displacement in covalent bonds CH 3  CH 3 33.33 D YG % s-character U (c) Acidity of hydrocarbon  % of s-character CH  CH 25 44 50 s- character and acidity in decreasing order Acidity  Ka and Acidity  1 (pKa   log Ka) pKa Order of acidic nature of alkynes is, HC  CH  HC  C  CH 3 U The relative acidic character follows the order; H2O  ROH  HC  CH  NH 3  CH 2  CH 2  CH 3  CH 3 ST Obviously, the basic character of their conjugate bases follows the reverse order, i.e., CH 3 CH 2  CH 2  CH   NH 2  HC  C   RO   HO It is observed that most of the attacking reagents always possess either a positive or a negative charge, therefore for a reaction to take place on the covalent bond the latter must possess oppositely charged centres. This is made possible by displacement (partial or complete) of the bonding electrons. The electronic displacement in turn may be due to certain effects, some of which are permanent and others are temporary. The former effects are permanently operating in the molecule and are known as polarisation effects, while the latter are brought into play by the attacking reagent and as soon as the attacking reagent is removed, the electronic displacement disappears; such effects are known as the polarisability effects.  Electronic displacement Steric effect On account of the presence of bulkier groups at the reaction centre, they cause mechanical interference and with the result the attacking reagent finds it difficult to reach the reaction site and thus slows down the reaction. This phenomenon is called steric hinderance or steric effect. 3 60 sp  s 2 (1) Tertiary alkyl halides having bulky groups form tertiary carbocation readily when hydrolysed because of the presence of the three bulky groups on (kcal/mole) Polarisation effect (permanent) Inductiv e effect Mesomeri Hyperconjugati c effect ve effect Polarisability effect (temporary) Inductomer ic effect Electromeri c effect General Organic Chemistry 1017 Inductive effect or Transmission effect (1) When an electron withdrawing (X) or electron-releasing (Y) group is attached to a carbon chain, polarity is induced on the carbon atom and on the substituent attached to it. This permanent polarity is due to displacement of shared electron of a covalent bond towards a more electronegative atom. This is called inductive effect or simply as I – effect.   C  C    C C    C X   electron releasing power are said to have +I effect.  N H 3  NO 2  CN  SO 3 H  CHO  CO  COOH  COCl  COOR 2  F  Cl  Br  I  OH  OR  NH 2  C 6 H 5  H X  CH 3 CH 2 > X Secondary > Primary > Methyl (a) Any group or atom showing +I effect decreases the acid strength as it increases the negative charge on the carboxylate ion which holds the hydrogen firmly. Alkyl groups have + I effect. Thus, acidic nature is, HCOOH  CH 3 COOH  C 2 H 5 COOH  C3 H 7 COOH  C4 H 9 COOH +I effect increases, so acid strength decreases Formic acid, having no alkyl group, is the most acidic among these acids. (b) The group or atom having – I effect increases the acid strength as it decreases the negative charge on the carboxylate ion. Greater is the number of such atoms or groups (having – I effect), greater is the acid strength. U –I power of groups in decreasing order with respect to the reference H D YG ter. alkyl > sec. alkyl > pri. alkyl > CH 3  H + I power in decreasing order with respect to the reference H + I power  number of carbon in the same type of alkyl groups  CH 3 ID (2) Carbon-hydrogen bond is taken as a standard of inductive effect. Zero effect is assumed for this bond. Atoms or groups which have a greater electron withdrawing capacity than hydrogen are said to have–I effect whereas atoms or groups which have a greater > CONH CH 3 (iii) Relative strength of the acids :  Y X CH 60 C  C  X  H3C C H3C CH 3 E3  CH 3 Tertiary C  C  C  C Non polar C  (ii) Reactivity of alkyl halide : + I effect of methyl group enhances – I effect of the halogen atom by repelling the electron towards tertiary carbon atom. CH 3  CH 2  CH 2  CH 2   CH 3  CH 2  CH 2   CH 3  CH 2  U + I power in decreasing order in same type of alkyl groups Thus, acidic nature is, CCl 3 COOH  CHCl 2 COOH  CH 2 ClCOOH  CH 3 COOH Trichloro acetic acid Dichloro acetic acid Monochloro acetic acid Acetic acid (– Inductive effect increases, so acid strength increases) (c) Strength of aliphatic carboxylic acids and benzoic acid R COOH  C6 H 5  COOH  I group  I group (3) Applications of Inductive effect ST (i) Magnitude of positive and negative charges : Magnitude of +ve charge on cations and magnitude of –ve charge on anions can be compared by + I or – I groups present in it.   Magnitude of ve ve Magnitude of 1    I power of the group.  I power of the group HCOOH > C6 H 5 COOH > RCOOH Acid strength in decreasing order charge  Decreasing order of acids : 1   I power of the group.  I power of the group  Hence benzoic acid is stronger acid than aliphatic carboxylic acids but exception is formic acid. Thus, NO2CH 2COOH  FCH 2COOH  ClCH 2COOH  BrCH 2COOH. charge F3C  COOH  Cl3C  COOH  Br3C  COOH  I3C  COOH. CH 3 OH  CH 3 CH 2 OH  (CH 3 ) 2 CHOH  (CH 3 )3 COH Methyl alcohol Ethyl Alcohol Iso propyl alcohol Tert butyl alcohol 1018 General Organic Chemistry As compared to water, phenol is more acidic (–I effect) but methyl alcohol is less acidic (+I effect). OH  H  OH > CH 3 Water OH Methyl alcohol The difference in base strength in various amines can be explained on the basis of inductive effect. The +I effect increases the electron density while –I effect decreases it. The amines are stronger bases than NH 3 as the alkyl groups increase electron density on nitrogen due to + I effect while ClNH 2 is less basic due to –I effect. “So more is the tendency to donate electron pair for coordination with proton, the more is basic nature, i.e., more is the negative charge on nitrogen atom (due to +I effect of alkyl group), the more is basic nature”. (C 2 H 5 )2 NH  CH 3 CH 2 NH 2  CH 3 NH 2  NH 3  ClNH 2 Diethy l amine Ethy l amine Methy l amine Ammonia Chloro amine Alkyl groups (R– Relative base strength ) D YG R2 NH  RNH 2  R3 N  NH 3 CH 3 R2 NH  RNH 2  NH 3  R3 N C2 H 5 (CH 3 )2 CH (CH 3 )3 C CH 3  CH 3 CH 2  (CH 3 )2 CH   (CH 3 )3 C  Resonance effect or mesomeric effect (1) The effect in which  electrons are transferred from a multiple bond to an atom, or from a multiple bond to a single covalent bond or lone pair (s) of electrons from an atom to the adjacent single covalent bond is called mesomeric effect or simply as M-effect. In case of the compound with conjugated system of double bonds, the mesomeric effect is transmitted through whole of the conjugated system and thus the effect may better be known as conjugative effect. (2) Groups which have the capacity to increase the electron density of the rest of the molecule are said to have  M effect. Such groups possess lone pairs of electrons. Groups which decrease the electron density of the rest of the molecule by withdrawing electron pairs are said to have M effect, e.g., (a) The groups which donate electrons to the double bond or to a conjugated system are said to have  M effect or R effect. RNH 2  NH 3  R2 NH  R3 N  M effect groups : NH 3  RNH 2  R2 NH  R3 N....  Cl,  Br ,  I,  N H 2 ,  NR 2 ,OH ,OR,SH ,OCH 3 , S R..  The relative basic character of amines is not in total accordance with inductive effect (t  s  p) but it is in the following order: Secondary > Primary > Tertiary. U The reason is the steric hindrance existing in the tamines. of U The order of basicity is as given below; (ix) Stability of carbanion : Stability carbanion increases with increasing – I effect. ID Thus, the basic nature decreases in the order; (CH 3 )3 C   (CH 3 )2 CH   CH 3 CH 2  CH 3 60 (vi) Relative strength of the bases (Basic nature of  NH 2 ) E3 Phenol (viii) Stability of carbonium ion :+I effect tends to decrease the (+ve) charge and –I effect tends to increases the +ve charge on carbocation. ST  In gas phase or in aqueous solvents such as chlorobenzene etc, the solvation effect, i.e., the stabilization of the conjugate acid due to H -bonding are absent and hence in these media the basicity of amines depends only on the +I effect of the alkyl group thus the basicity of amines follows the order : (b) The groups which withdraw electrons from the double bond or from a conjugated system towards itself due to resonance are said to have M effect or R effect. M effect groups : O ||  NO 2 ,C  N ,  C ,CHO ,COOH ,SO 3 H (3) The inductive and mesomeric effects, when present together, may act in the same direction or (vii) Basicity of alcohols : The decreasing order of base strength in alcohols is due to +I effect of alkyl oppose each other. The mesomeric effect is more powerful than the former. For example, in vinyl chloride due to – I effect the chlorine atom should develop a negative charge but on account of mesomeric groups. effect it has positive charge. 3 o  2 o  1 o  NH 3. (CH 3 )3 COH  (CH 3 )2 CHOH  CH 3 CH 2 OH  CH 3 OH (3 o ) (2 o ) (1 o ) General Organic Chemistry 1019....  : Cl  CH  CH 2  : Cl  CH  CH 2.... (2) Structural hyperconjugation requirements hybrid carbon of either alkene alkyl carbocation or alkyl free radical. (1) Low reactivity of aryl and vinyl halides, (2) The acidic nature of carboxylic acids, (ii)  -carbon with respect to sp 2 hybrid carbon should have at least one hydrogen. (4) The stability of carbocations and carbanions. If both these conditions are fulfilled hyperconjugation will take place in the molecule. Difference between Resonance and Mesomerism : Although both resonance and mesomerism represent the same phenomenon, they differ in the following respect : Resonance involves all types of electron displacements while mesomerism is noticeable only in those cases where a multiple bond is in conjugation with a multiple bond or lone pair of Example : (i) .. :O.. :O:.. ||  (ii) R  C  O  H  R  C  O  H.... D YG ||  H H | H  C  CH  CH 2  H  C | | H H   CH  C H 2  Both (i) and (ii) are the examples of mesomerism and resonance effect. Let us consider the following..   H Cl :  H Cl. example Such an electron.. displacement is the example of resonance only (not the mesomerism). U Hyperconjugative effect ST (1) When a H  C bond is attached to an unsaturated system such as double bond or a benzene ring, the sigma () electrons of the H  C bond interact or enter into conjugation with the unsaturated system. The interactions between the electrons of  systems (multiple bonds) and the adjacent  bonds (single H  C bonds) of the substituent groups in organic compounds is called hyperconjugation. The concept of hyperconjugation was developed by Baker and Nathan and is also known as Baker and Nathan effect. In fact hyperconjugation effect is similar to resonance effect. Since there is no bond between the  -carbon atom and one of the hydrogen atoms, the hyperconjugation is also called no-bond resonance. H H   | | H C  CH  C H 2  H  C  CH  C H 2 | H H   U..  H 2 C  CH  CH  CH 2  H 2 C  CH  CH  C H 2 (iv) Resonating structures due to hyperconjugation may be written involving “no bond” between the alpha carbon and hydrogen atoms. ID electron. then (iii) Hyperconjugation is of three types E3 radicals, 60 (3) Basic character comparison of ethylamine and aniline, free sp 2 - (i) Compound should have at least one Application of mesomeric effect : It explains, some for (v) Number of resonating structures due to the hyperconjugation = Number of  -hydrogens + 1. Applications of hyperconjugation (1) Stability of alkenes : Hyperconjugation explains the stability of certain alkenes over other alkenes. Stability of alkenes  Number of alpha hydrogens  Number of resonating structures CH 3  CH  CH 2  CH 3  CH 2  CH  CH 2  CH 3  CH |  CH  CH 2 CH 3 Stability in decreasing order (2) Carbon-carbon double bond length in alkenes : As we know that the more is the number of resonating structures, the more will be single bond character in carbon-carbon double bond. (3) Stability of alkyl carbocations : Stability of alkyl carbocations  number of resonating structures  number of alpha hydrogens. (4) Stability of alkyl free radicals : Stability of alkyl free radicals can be explained by hyperconjugation. Stability depends on the number of resonating structures. (5) Electron releasing (or donating) power of R in alkyl benzene : CH 3  (or alkyl group) is R group, 1020 General Organic Chemistry ortho-para directing group and activating group for electrophilic aromatic substitution reaction because of the hyperconjugation. Cl | Cl  C  CH  CH 2 | Cl  Cl are as follows, CH 3   CH 3  CH 2   CH 3 CH 3 | CH   CH 3  C  | CH 3 effect  | Cl  CH  CH 2  Cl |  Cl  C  CH  CH 2 Cl  The meta directing influence and the deactivating of CX 3 group in electrophilic aromatic substitution reaction can be explained by this effect. X | X C || X | X  C X | Increasing inductive effect Electron donating power in decreasing order due to the hyperconjugation.  X X | X C || X | X C || +  X E3 CH 3 Cl | C | Cl   Cl  C  CH  CH 2  60 The electron donating power of alkyl group will depends on the number of resonating structures, this depends on the number of hydrogens present on carbon. The electron releasing power of some groups  Cl   X  Inductomeric effect (7) Dipole moment : Since hyperconjugation causes the development of charges, it also affects the dipole moment in the molecule. Inductomeric effect is the temporary effect which enhances the inductive effect and it accounts only in the presence of an attacking reagent. ID (6) Heat of hydrogenation : Hyperconjugation decreases the heat of hydrogenation. i.e., acetaldehyde (  2. 72 D) can be Example, U The increase in dipole moment, when hydrogen of formaldehyde (  2.27 D) is replaced by methyl group, referred to D YG hyperconjugation, which leads to development of charges. H H H | |  H  C  O , H  C  CH  O  H  C  CH  O (  2.27 D) | | H H H HO  H  C  Cl  HO............ C............ Cl H H H H  HO  C  H H Cl  (  2. 72 D) (8) Orienting influence of alkyl group in o, p - positions and of CCl 3 group in m -position : Ortho- U para directing property of methyl group in toluene is partly due to I effect and partly due to hyperconjugation. ST Reverse Hyperconjugation : The phenomenon 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. H In methyl chloride the –I effect of Cl group is further increased temporarily by the approach of hydroxyl ion. Electromeric effect (1) The phenomenon of movement of electrons from one atom to another in multibonded atoms at the demand of attacking reagent is called electromeric effect. It is denoted as E-effect and represented by a curved arrow ( ) showing the shifting of electron pair.   E A  B  A B : Reagent (2) (i)When the transfer of electrons take place towards the attacking reagent, the effect is called  E effect. The addition of acids to alkenes. General Organic Chemistry 1021  H   C  C | H.. A : B  A  B Free radical  CH  CH 2  H  CH 3  C H  CH 3 CH 3 Propene Since, CH 3 group is electron donating, the electrons are transferred in the direction shown. The attacking reagent is attached to that atom on which electrons have been transferred. (ii) When the transfer of electrons takes place away from the attacking reagent, the effect is called  E effect. Example, The addition of cyanide ion to carbonyl compounds. (ii) The factor which favours homolysis is that the difference in electronegativity between A and B is less or zero. (iii) Homolysis takes place in gaseous phase or in the presence of non polar solvents (CCl 4 , CS 2 ) , peroxide, UV light, heat ( 500 o C) , electricity and free radical. (iv) Mechanism of the reaction homolysis takes place is known as in which homolytic 60 C C mechanism or free radical mechanism. (2) Heterolytic bond fission or heterolysis (i) In heterolysis, the covalent bond is broken in such a way that one species (i.e., less electronegative) C  O | CN The attacking reagent is not attached to that atom on which electrons have been transferred. is deprived of its own electron, while the other species gains both the electrons. A : B  ID (3) Direction of the shift of electron pair : The direction of the shift of electron pair can be decided on the basis of following points. E3 C  O  CN   carbanion B carbocation 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. U (i) When the groups linked to a multiple bond are similar, the shift can occur in either direction.   A:  D YG (ii) When the dissimilar groups are linked on the two ends of the double bond, the shift is decided by the direction of inductive effect. In the case of carbonyl group, the shift is always towards oxygen, i.e., more electronegative atom. C  O    C O : In cases where inductive effect and electromeric effect simultaneously operate, usually electrometric effect predominates. U Cleavage (fission or breaking) of covalent bonds (iii) Mechanism of the reaction in which heterolysis takes place is known as heterolytic mechanism or ionic mechanism. (iv) The energy required for heterolysis is always greater than that for homolysis due to electrostatic forces of attraction between ions. Reaction Intermediates Breaking of covalent bond of the compound is known as bond fission. A bond can be broken by two ST ways, (ii) The factor which favours heterolysis is greater difference of electronegativities between A and B. (1) Homolytic bond fission or Homolysis (i) In homolysis, the covalent bond is broken in such a way that each resulting species gets its own Short lived fragments called reaction intermediates result from homolytic and heterolytic bond fission. The important reaction intermediates are free radicals, carbocations, carbanions, carbenes, benzyne and nitrenes. electron. This leads to the formation of odd electron species known as free radical. Table : 23.4 Characteristi c Free radical Carbocation Carbanion Carbene 1022 General Organic Chemistry Nature Neutral having odd electron Positive charge on C Negative charge on Neutral, C with 2 divalent unshared electrons Hybridisation sp2 sp2 sp3 (non- conjugated) (i) sp2 (singlet) (ii) sp (triplet) sp2 (Conjugated) Structure Planar Planar Pyramidal/Planar Magnetism Paramagnetic Diamagnetic Diamagnetic Stability... Ph3 C  Ph2 CH  Ph CH 2  Ph3 C  Ph2 CH  PhCH 2 . CH 2  CH  CH 2  3 o  2 o  CH 2  CH  CH 2 .. 1  CH 2  CH 2  CH 3 o  2 o  1o  CH 3 (i) Planar (singlet) (ii) Linear (triplet)  (1) 1, 2-Didehydrobenzene, PhCH 2  Allyl  Benzyne and C6 H 4 its  CH 2  1o  2 o  3 o (2) There is possibility of two spin states for nitrenes depending on whether the two non-bonding electrons (the normal nitrogen lone pair remains paired) have their spins paired or parallel. ID derivatives are called benzyne or arynes and the simplest member is benzyne... R –. N.  U (2) It is neutral reaction intermediate derived from benzene ring by removing two substituents, of ortho positions, one in the form of electrophile and other in the from of nucleophile leaving behind two electrons to be distributed between two orbitals. D YG Abnormal  bond  Ph3 C  Ph2 CH    o  60   E3 order (i) Diamagnetic (ii) Paramagnetic Triplet > singlet These two unpaired These two are lone pair of electrons may be paired or (3) In general nitrenes obey Hunds rule and the ground state triplet with two degenerate sp -orbitals containing a single electron each. R–N 2 Two sp -orbitals ouside the ring (3) Benzyne character. intermediate is aromatic in U (4) When halobenzene is heated with sodamide formation of benzyne takes place. Cl ST NaNH 2    (5) (i) It behaves as dienophile and gives DielsAlder reaction with diene.  (ii) It reacts with strong nucleophile like NH 2   N H 2  NH2  H sp-Triplet nitrene (4) Nitrenes can be generated, in situ, by the following methods, (i) By action of Br2 in presence of a base on a 1 o amide (Hofmann-bromamide reaction), O O O   || || ||..  Br2 / NaOH OH R  C  NH 2  R  C  NHBr  R  C  N  Br H 2O .. 1 o Amide   O ||.. Rearrangement  O  C  N  R   R  C  N    Br.. Isocy anate  NH2 KOH    R  NH 2  K 2 CO 3 .. Nitrenes (R – N : ) (1) The nitrogen analogous of carbenes are called nitrenes. (Hydrolysis) 1 o Amine (ii) By decomposition of azides in presence of heat or light.......   or h ν R  N  N  N :  R  N :  N  N Alkylazide Alkylnitrene General Organic Chemistry 1023 Unsubstituted nitrene.. (H  N :) can be obtained by photolysis of (or by passing electric discharge through) NH 3 , N 2 H4 or N 3 H. Attacking reagents (ii) Neutral nucleophiles : It can be classified into two categories : (a) Neutral covalent compound, in which central atom has complete octet, has at least one lone pair of electrons and all atoms present on central atom should not be electronegative, is neutral nucleophile. The fission of the substrate molecule to create centres of high or low electron density is influenced by attacking reagents. Most of the attacking reagents can be classified into two main groups. Electrophiles or electrophilic Nucleophiles or nucleophilic reagents. reagents............ N H 3, R  N H 2 , R2 N H , R3 N , N H 2  N H 2 (Nitrogen nucleophile)...... H  O H, R  O H, R  O R...... and (1) Electrophiles : Electron deficient species or electron acceptor is an electrophile. nucleophiles) E3 H , X , R, N O   , N  O, S O3 H O.......... P H 3 , R P H 2 , R2 P H , R3 P (Phosphorus nucleophiles) (b) Organic compound containing carbon, carbon multiple bond/ bonds behaves as nucleophile. ID  (Oxygen...... H  S  H , R  S  H , R  S  R (Sulphur nucleophiles)...... It can be classified into two categories : (i) Charged electrophiles : Positively charged species in which central atom has incomplete octet is called charged electrophile. 60 (iii) All cations are charged electrophiles except  cations of IA, IIA group elements, Al    and NH 4 Alkenes, Alkynes, Benzene, CH 2  CH  CH  CH 2 , CH 2  CH  C  CH (iii) Ambident nucleophiles : Species having two nucleophilic centres out of which, one is neutral (complete octet and has at least one lone pair of electrons) and the other is charged (negative charge) behaves as ambident nucleophile U (ii) Neutral electrophiles : It can be classified into three categories, D YG (a) Neutral covalent compound in which central atom has incomplete octet is neutral electrophile,..... BeCl 2 , BH 3 , ZnCl 2 , AlX 3, FeX 3 , CH 3 , CH 2 , CX 2 O.. ..   C  N , O  N  O, O  S  OH  (b) Neutral covalent compound in which central atom has complete or expended octet and central atom has unfilled –d-sub-shell is neutral electrophile,  O  Organometallic compounds are nucleophiles. SnCl 4 , SiCl 4 , PCl 5 , SF6 , IF7 U (c) Neutral covalent compound in which central atom is bonded only with two or more than two electronegative atoms is called neutral electrophile. BeCl 2 , BX 3 , AlX 3 , FeX 3 , SnCl 4 , PCl 3 ; ST.. PCl 5 , NF3 , C X 2 , CO 2 , SO 3 , CS 2 , Cl 2 , Br2 and I2 also behave as neutral electrophiles. Electrophiles are Lewis acids. (2) Nucleophiles : Electron rich species or electron donors are called nucleophiles. Nucleophiles can be classified into three categories : (i) Charged nucleophiles : Negatively charged species are called charged nucleophiles.       H , O H , R  O, C H 3 , X , S H , R  S  Nucleophiles are Lewis bases. Organic compounds which behave as an electrophile as well as a nucleophile : Organic compound in which carbon is bonded with electronegative atom (O, N, S) by multiple bond/bonds behaves as electrophile as well as nucleophile : O O O O || || || || R  C  H , R  C  R, R  C  OH , R  C  Cl , O O || ||   R  C  OR, R  C  NH 2 , R  C  N , R  N  C  During the course of chemical reaction electrophile reacts with nucleophile.  Strong Lewis acid is stronger electrophile   CO 2  N O 2  S O 3 H. Stronger is an acid, weaker is its conjugated base or weaker is the nucleophile. 1024 General Organic Chemistry Examples : HF  H 2 O  NH 3  CH 4  F   OH   NH 2  CH 3 Increasing order of nucleophilicity. Types of organic reactions It is convenient to classify the numerous reactions of the various classes of organic compound into four types,  Substitution reactions,  Addition reaction,  (iii) The leaving power of some nucleophilic groups are given below in decreasing order, O || O (Bromine atom is replaced by hydroxyl group) Types of substitution reactions : On the basis of the nature of attacking species substitution reactions D YG (2) Electrophilic substitution reactions (3) Free radical substitution reactions O  (i) Many substitution reactions, especially at the saturated carbon atom in aliphatic compounds such as alkyl halides, are brought about by nucleophilic reagents or nucleophiles. X U Leaving group Such substitution reactions are called nucleophilic substitution reactions, i.e., S N reactions (S ST stands for substitution and N for nucleophile). (ii) The weaker the basicity of a group of the substrate, the better is its leaving ability. Leaving power of the group  1 Basicity of the group HI  HBr  HCl  HF Example :          Decreasing acidity   ||  || || O O O   ||   I  Br  CF3  C  O  H O  Cl  F  CH 3  C  O (iv) In these reactions leaving group of the substrate is replaced by another nucleophile. If reagent is neutral then leaving group is replaced by negative part of the reagent. Negative part of the reagent is always nucleophilic in character.      E  Nu R  L    R  Nu  L ; R  L  Nu  R  Nu  L (v) In S N reactions basicity of leaving group should be less than the basicity of incoming nucleophilic group. Thus strongly basic nucleophilic group replaces weakly basic nucleophilic group of the substrate.   OH  R  OH  Cl.....(A) Example : R  Cl  ( NaOH ) (1) Nucleophilic substitution reactions Nucleophil e || U are classified into following three categories,  ID Ethyl alcohol E3 Substituting or attacking group R  X  OH   R  OH  O  C 6 H 5  S  O  CH 3  S  O CH 3  CH 2  Br  NaOH  CH 3  CH 2 OH  NaBr Substrate O || Replacement of an atom or group of the substrate by any other atom or group is known as substitution reactions. (1) Nucleophilic substitution reactions || || O O Examples : Leaving group SO  S  O  CH 3 ||  ||  || CF3  S  O  Br Rearrangement O O  Substitution reactions Ethyl bromide  60  Elimination reactions,  reactions,  I Br Cl F          Increasing basicity  Decreasing leaving ability   Basicity of OH is more than Cl hence OH replaces  Cl as Cl.   Cl R  OH   R  Cl  OH......(B) ( HCl )    Basicity of Cl is less than OH , hence Cl will not  replace OH as OH hence reaction (B) will not occur. (vi) Unlike aliphatic compounds having nucleophilic group as leaving group, aromatic compounds having same group bonded directly with aromatic ring do not undergo nucleophilic substitution reaction under ordinary conditions. The reason for this unusual reactivity is the presence of lone pair of electron or  bond on the key atom of the functional group. Another factor for the low reactivity is nucleophilic character of aromatic ring. (vii) The S N reactions are divided into two classes, S N 2 and S N 1 reactions. Table : 23.5 Distinction between SN2 and SN1 reactions General Organic Chemistry 1025 Factors SN2 Reactions  Number of steps Reaction rate and order SN1 Reactions  R : L  R   : L Slow One: R : L  : Nu R : Nu  : L Two: (i) Second order: Rate  [Substrate] [Nucleophile] or First order: Rate  [Substrate] or Rate = K1 [RL] Fast (ii) R   : Nu    R : Nu Rate = K 2 [RL ][: Nu ] Molecularity Bimolecular  TS of slow step  Unimolecular     : Nu    C    L    Nu : The nucleophile can attack the carbon of the substrate both from the back and front sides although the back side attack predominates. Stereochemistry Complete inversion of configuration takes place. Inversion and retention takes place. Reactivity order of alkyl halides Methyl>1°>2°>3°halides. (I  Br  Cl  F) 3°>2°>1° > methyl halides. (I  Br  Cl  F) Rearrangement No rearranged product is formed (except for allylic). Rearranged products can be formed. Nature of nucleophiles Favoured by strong and concentration of nucleophiles. Favoured by mild and low concentration of nucleophiles. Polarity Favoured by solvents of low polarity. Reaction rate determining factor By steric hindrance. Catalysis Not catalysed by any catalyst (phase transfer). (2) Electrophilic substitutions reactions : Electrophilic substitution involves the attack by an electrophile. It is represented as SE (S stands for substitution and E stands for elctrophile). If the order U   compound by hydrogen : R  M  H R  H  M   ST  MgBr H e.g., CH 3  CH 2  MgBr   CH 3  CH 2  CH 3  CH 3   H CH 3  CH 2  MgBr  H  Br CH 3  CH 2  CH 3  CH 3  MgBr2 CH 3  CH 2 Na  C6 H6 CH 3  CH 3  C6 H5 Na   Catalysed by Lewis and Bronsted acids, e.g.,  Ag , AlCl 3 , ZnCl 2 and strong HA. SE2 Reaction mechanism : Electrophilic substitution is very common in benzene nucleus (aromatic compounds) in which -electrons are highly 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.,   In halogenation; Cl  Cl  FeCl 3 Cl  Fe Cl 4 In (ii) Decarboxylation of silver salt of carboxylic acid by means of bromine:   By electronic factor (stability of R ). if the order is 2, it is S E 2 (Bimolecular). (i) Replacement of the metal atom in an organometallic  R3 C  C  OAg  Br  Br R3 C  C  O  Br  Br  Ag O R3C  Br  CO2  AgBr Favoured by solvents of high polarity. of reaction is 1, it is written as S E 1 (unimolecular)and SE1 Reaction mechanism : Electrophilic substitution in aliphatic compounds are very rare; some of the important examples are: || E3 ID high U D YG Reacting nucleophile 60 : Nu    C   : L The nucleophile attacks the carbon of the substrate exclusively from the back side. || O (iii) Isotopic exchange of hydrogen for deuterium or tritium: nitration;    HNO 3  2 H 2 SO 4 NO 2  2 HSO 4  H 3 O   In sulphonation; 2 H 2 SO 4 SO 3  HSO 4  H 3 O In Friedel-crafts  reaction;  R  Cl  AlCl 3 R  AlCl 4   RCOCl  AlCl 3 RCO  AlCl 4 H E H E theHaromatic E Step 2. The electrophile attacks    E + Benzene H E E 1026 General Organic Chemistry ring to form carbonium ion (or arenium ion) which is stabilized by resonance.   RH D⇋ RDH   R  H T ⇋ R T H 60 Step 3. Carbonium ion loses the proton to form substitution product. E3 The bromination of benzene in the presence of FeBr 3 is a example of electrophilic substitution reaction. ID Similarly, Nitration, sulphonation and Friedel-Crafts reaction…..etc., in benzene nucleus are the other examples of electrophilic substitution reactions. CH 4  Methane D YG (i) Chlorination of methane : The chlorination of methane in the presence of ultraviolet light is an examples of free radical substitution. UV Cl 2   CH 3 Cl  HCl light Methylchloride (ii) Arylation of aromatic compounds (Gomberg reaction) : The reaction of benzene diazonium halide with benzene gives substitution reaction. C6 H 5 N 2 X Benzene diazonium halide by a free radical Alkali   C6 H 5  C6 H 5  N 2  HX U C6 H 5  H  diphenyl Diphenyl ST (iii) Wurtz reaction : Ethyl bromide on treatment with metallic sodium forms butane, ethane and ethylene by involving free radical mechanism. (iv) Allylic bromination by NBS (NBromosuccinimide) : NBS is a selective brominating agent and it normally brominates the ethylenic compounds in the allylic (CH 2  CH  CH 2 ) position. This type of reaction involving substitution at the alpha carbon atom with respect to the double bond is termed Allylic substitution. It is also bromination. Some examples are: used for benzylic CH 2  CO CH 3  CH  CH 2  Propene U (3) Free radical substitution reactions : Free radical substitution reactions involves the attack by a free radical. These reactions occurs by free radical mechanism which involves Initiation, Propagation and Termination steps. Examples, CCl 4 N  Br   CH 2  CO NBS CH 2  CO Br  CH 2  CH  CH 2  Ally lbromide NH CH 2  CO Succinimide Addition reactions These reactions are given by those compounds which have at least one  bond, O || i.e., ( C  C ,C  C , C , C  N ). In such reaction there is loss of one  bond and gain of two  bonds. Thus product of the reaction is generally more stable than the reactant. The reaction is a spontaneous reaction. Types of addition reactions : Addition reactions can be classified into three categories on the basis of the nature of initiating species. (1) Electrophilic additions (2) Nucleophilic additions (3) Free radical additions (1) Electrophilic addition reactions (i) Such reactions are mainly given by alkenes and alkynes. (ii) Electrophilic addition reactions of alkenes and alkynes are generally two step reactions. General Organic Chemistry 1027 (iii) Alkenes and alkynes give electrophilic addition with those reagents which on dissociation gives electrophile as well as nucleophile. (iv) If the reagent is a weak acid then electrophilic addition is catalysed by strong acids (Generally H 2 SO 4 ). Cl | Anti- Markowniko ff's addition CH 2  CH  CHO  HCl    CH 2  CH 2  CHO (vii) Mechanism reactions is as follows, CC  Olefin This rule can be used only in those alkenes which fulfil the following conditions: (a) Alkene should be unsymmetrical. (b) Substituent/substituents present on doubly bonded carbon/(s) should only be +I group. For example, the following alkenes will give addition according to the Markownikoff’s rule. CH 3 D YG CH 3 C6 H 5 C  CH  C6 H 5, C6 H 5  CH  CH 2 C6 H 5 Following alkenes will not give addition reaction according to Markownikoff’s rule. R R C6 H 5 , R C6 H 5 U CC C6 H 5 C6 H 5 ST (vi) Unsymmetrical alkenes having the following general structure give addition according to anti Markownikoff’s rule. CH 2  CH  G , where G is a strong –I group such as | C  C E Fast   | | (2) 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 Example : The addition of HCN to acetone is an example of nucleophilic addition. CH 3 CH 3 OH C  O  HCN C CH 3 CH 3 CN Acetone Acetone cy anohy drin The mechanism of the reaction involves the following steps:  Step 1. HCN gives a proton (H ) and a nucleophile,  cyanide ion (CN ). HCN H   CN  Step 2. The nucleophile (CN ) attacks the positively charged carbon so as to form an anion [ H  does not initiate the negatively charged oxygen as anion is more stable than cation].  CH 3 O CH 3  CH 3  C CN C  O CN C  O or CH CH CN CH 3 3 3 Step 3. The proton (H  ) combines with anion to form the addition product. O ||  CX 3 , NO 2,  CN ,CHO ,COR ,COOH , C  Z (Z  Cl, OH, OR, NH 2 ) Example: | Carbonium ion Addition product R CC CH 2  CH 2 , R  CH  CH  R, Nucleophile | C C E X U C  CH 2 , CH 3  CH  CH 2 , | X ID (c) If phenyl group is present on doubly bonded carbon, then both doubly bonded carbons should be substituted by phenyl groups.  | C C E   Slow   addition 60 The negative part of the addendum adds on that doubly bonded carbon of the alkene which has least number of hydrogen atom.  Electrophile electrophilic E3 (v) Unsymmetrical alkenes and alkynes give addition reactions with unsymmetrical reagents according to Markownikoff’s rule.  E of CH 3 CN CH 3 CH 3 |  C  O  H NC  C  OH or | CH 3 CH 3 OH C CH 3 CN 1028 General Organic Chemistry gives cyanohydrin and the addendum is CN  ion and not HCN directly (addition is catalysed by bases or salts of weak acids and retarded by acids or unaffected by neutral compounds).   H CN C  O    ( A. R.)    CN C  O   O C  H   OH C CN CN Nucleophilic addition (AN) reactions on carbonyl compounds will be in order: H H 3C C O  H Product of the reaction is halocarbenes or dihalocarbenes. which are key intermediates in a wide variety of chemical and photochemical reactions. reactions: Consider the following reactions, C O H 3C Decreasing order of nucleophilic addition in some species. O O || ||  C 6 H 5  C  CH 3  C 6 H 5  C  C 6 H 5  CHO    CH 3  CH 2  CH 2  L CH 3  CH  CH 2  H  L   A reaction in which functional group (i.e., leaving group) is removed from - carbon and other group (Generally hydrogen atom) from the - carbon is called - elimination reaction. In this reaction there is loss of two  bonds and gain of one  bond. Product of the reaction is generally less stable than the reactant. ID C 6 H 5 CH 2 COCH 3  CH 3 COCH 3..  Alc. KOH / CHX 3   CX 2  X  H (II) -elimination reactions or 1, 2-elimination H 3C C O  H (I) -elimination reactions or 1,1-elimination reactions: A reaction in which both the groups or atoms are removed from the same carbon of the molecule is called - elimination reaction. This reaction is mainly given by gem dihalides and gem trihalides having at least one - hydrogen. 60 C  O compounds, the addition of liquid HCN E3 In COCH 3  COCl  COOCH 3  CONH 2  COOH (3) Free radical addition reactions : Those reactions which involve the initial attack by a free (1) Types of - elimination reactions : In analogy with substitution reactions, - elimination reactions are divided into three types: temperature, in presence of light or a free radical producing substance like O 2 and peroxides. (Elimination unimolecular conjugate base) reaction D YG U radical are known as free radical reactions. Addition of hydrogen bromide to alkenes (say, propylene) in the presence of peroxide (radical initiator) follows free radical mechanism. Free radical reactions generally take place in non-polar solvents such as CCl 4 , high Elimination reactions U Elimination reactions are formally the reverse of addition reactions and involve the removal of the two groups (Generally, one being a proton) from one or two carbon atoms of a molecule to form an unsaturated linkage or centre. ST Elimination reaction is given by those compounds which have a nucleophilic group as leaving group,    i.e., X, OH, OR, N 2 , N 3 , H 3 O, N R  R, S R R R Elimination reactions are generally endothermic and take place on heating. Elimination reactions are classified into two general types, (I) - elimination reactions or 1, 1-elimination reactions. (II) - elimination reaction or 1, 2-elimination reactions. (i) E1 (Elimination unimolecular) reaction, (ii) E2 (Elimination bimolecular) reaction and (iii) E1cb (i) E1 (Elimination unimolecular) reaction : Consider the following reaction, CH 3 | CH 3  C  Cl |  C H O/  2 5   CH CH 3 CH 3 2 C   C 2 H 5 OH  Cl CH 3 (a) Reaction velocity depends only on the concentration of the substrate; thus reaction is unimolecular reaction. Rate  [Substrate] (b) Product formation takes place by the formation of carbocation as reaction intermediate (RI). (c) Since reaction intermediate is carbocation, rearrangement is possible in E1 reaction. (d) Reaction is carried out in the presence of polar protic solvent. (e) The E1 reaction occurs in two steps, Step 1. General Organic Chemistry 1029 CH 3  C  Cl | CH 3    C - - - - - Cl    CH 3  C  CH 3  Cl |   Slow step | CH 3 | CH 3 CH 3 (TS 1 ) | ( RI ) Step 2.    ..    B  H  CH 2  C  CH 3  B - - - - H - - - -CH 2 - - - - C  CH 3  | |   CH 3 CH 3   TS 2   CH 3    B H  CH 2  C fast (ii) E2 (Elimination bimolecular) reaction : Consider the following reaction,   Base ( B) CH 3  CH 2  CH 2  Br   CH 3  CH  CH 2  H  Br   HCl 1 | CH 3 Saytzeff p roduct product is formed from - carbon which has maximum number of hydrogen. Product of the reaction in this case is known as Hofmann product. CH 3 Br | | CH 3 | Alc.KOH /  CH 3  C  CH 2  CH  CH 3  CH 3  C  CH 2  CH  CH 2  2 1 CH 3 | CH 3 Hofmann product  In E1 reactions, product formation always takes place by Saytzeff rule.  In E1cb reactions, product formation always takes place by Hofmann rule. ID (a) Reaction velocity depends only on the concentration of the substrate and the base used; thus reaction is bimolecular reaction. Rate [Substrate] [Base]  (ii) Hofmann rule : According to this rule, major product is always least substituted alkene i.e., major | CH 3 Alc.KOH /  CH 3  CH  CH  CH 3   CH 3  C  CH  CH 3 | CH 3 Cl 2 E3 | 60 CH 3 CH 3 U (b) Since the reaction is a bimolecular reaction, the product formation will take place by formation of transition state (TS).  In E2 reactions, product formation takes place by Saytzeff as well as Hofmann rule. In almost all E2 reactions product formation take place by Saytzeff rule. D YG (c) Rearrangement does not take place in E2 reaction but in case of allylic compound rearrangement is possible. (d) Reaction is carried out in the presence of polar aprotic solvent. (e).. B The E2 reaction occurs in  B  - - - - H     Slow step   CH 3  C CH 3  C  C  H  | | |  H  H Br TS  H | | U H one step,   C  H   Br   H | ST  fast   CH 3  CH  CH 2  BH  Br (2) Orientation in  - elimination reactions : If substrate is unsymmetrical, then this will give more than one product. Major product of the reaction can be known by two emperical rules. (i) Saytzeff rule : According to this rule, major product is the most substituted alkene i.e., major  product is obtained by elimination of H from that carbon which has the least number of hydrogen. Product of the reaction in this case is known as Saytzeff product. (3) Examples of  - elimination reactions (i) Dehydrohalogenation is removal of HX from alkyl halides with alcoholic KOH or KNH 2 or ter- BuOK (Potassium tertiary butoxide) and an example of - elimination, Alc.KOH CH 3  CH 2 X   H 2 C  CH 2 ; e.g., ( HX ) Ethene Alc.KOH CH 3  CH  CH 3   CH 3 CH  CH 2 | ( HX ) Propene X Alc.KOH CH 3  CH 2  CH  CH 3   CH 3  CH  CH  CH 3  | ( HX ) 2 - Butene (Major) X CH 3  CH 2  CH  CH 2 1- Butene (Minor) (ii) Dehydration of alcohol is another example of elimination reaction. When acids like conc. H 2 SO 4 or H 3 PO 4 are used as dehydrating agents, the mechanism is E1. The proton given by acid is taken up by alcohol. Dehydration is removal of H 2 O from alcohols, e.g., Conc. H 2 SO 4 , 170 C CH 3  CH 2  OH    H 2 C  CH 2 ( H 2 O ) Conc. H 2SO 4 , 170 C CH 3  CH 2  CH 2  OH    CH 3  CH  CH 2 Propan -1 - ol ( H 2 O ) Propene 1030 General Organic Chemistry  Dehydration of alcohols is in the order: Y B  | | | | Y  C  C    C  C   C  C  Tertiary  Secondary  Primary (3) (2) (1) are more easily dehydrated than the alcohols leading to non-conjugated alkenes. CH 2  CH  CH  CH 3 is easily | OH | OH OH > (3) Rearrangement or migration to free radical species (Free radical rearrangement) : Those OH E3 > rearrangement reactions in which the migrating group moves to a free radical centre. Free radical rearrangements are comparatively rare. (iii) Dehalogenation : It is removal of halogens, (4) Aromatic rearrangement : Those rearrangement reactions in which the migrating group moves to aromatic nucleus. Aromatic compounds of the type (I) undergo rearrangements in the manner X–H X–H X–Y mentioned below, Y + + in CH 3 OH , heat CH 2  CH 2  Zn dust   H 2 C  CH 2 (-ZnBr2 ) ID Ethy lene Ethy lene bromide Cu , 300 C   CH 3  C  CH 3 e.g., CH 3  CH  CH 3  ( H 2 ) OH || O Isopropy l alcohol Acetone U (iv) Dehydrogenation : It is removal of hydrogen, | (I) D YG Rearrangement reactions The reactions, 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. rearrangement U It is convenient to divide reactions into following types: (1) Rearrangement or migration to electron deficient atoms (Nucleophilic rearrangement) : Those ST rearrangement reactions in which migrating group is nucleophilic and thus migrates to electron deficient centre which may be carbon, nitrogen and oxygen. B: Position isomeris m Y The element X from which group Y migrates may be nitrogen or oxygen. Isomerism Organic compounds having same molecular formula but differing from each other at least in some physical or chemical properties or both are known as isomers (Berzelius) and the phenomenon is known as isomerism. The difference in properties of isomers is due to the difference in the relative arrangements of various atoms or groups present in their molecules. Isomerism can be classified as follows: Isomerism  X Constitutional or structural isomerism Without referring to space, the isomers differ in the arrangement of atoms within the molecule is called structural isomerism. Thus structural isomers have: Chain isomeris m | X electrophile and thus migrates to electron rich centre. OH | | 60 CH 3  CH 2  CH  CH 3 and so Br | (2) Rearrangement or migration to electron rich atoms (Electrophilic rearrangement) : Those rearrangement reactions in which migrating group is dehydrated than | | Bridged or non-classical carbocation X= Nucleophilic species, Y = Electronegative group, B = Another nucleophile. by E2 process. Alcohols leading to conjugated alkenes Br | X:  2° and 3° alcohol by E1 process and 1° alcohol e.g., | Ring chain isomeris m Functiona l isomeris m Configurational or stereo isomerism The isomerism arises due to different arrangement of atoms or groups in space. It deals with the structure of molecules in three dimensions. Thus stereoisomers have: Meta merism Tauto merism Geometric al isomerism Optical isomeris m Conformation al isomerism General Organic Chemistry 1031 (1) Chain, nuclear or skeleton isomerism : This type of isomerism arises due to the difference in the nature of the carbon chain (i.e., straight or branched) which forms the nucleus of the molecule. Examples : (i) : CH 3  CH 2  CH 2  CH 3 , C4H10 n Butane  Structural isomers which differ in the position of the functional group are called regiomers. For example, (i) CH 3  CH 2  CH 2  OH (ii) CH 3  CH  CH 3 | OH (3) Functional isomerism : This type of isomerism is due to difference in the nature of functional group present in the isomers. The following 60 Constitutional or structural isomerism pairs of compounds always form functional isomers with each other. CH 3  CH  CH 3 | E3 Examples : CH 3 (i) Alcohols and ethers (Cn H2n+2O) C2H6O : CH 3  CH 2  OH ; H 3 C  O  CH 3 Isobutane (ii) C5H12 : (Three) CH 3  CH 2  CH 2  CH 2  CH 3 , n  Pentane C3H8O : CH 3  CH 2  CH 2  OH ; C 2 H 5  O  CH 3 CH 3 n  propy l alcohol | C4H10O | | CH 3 Isopentane Ethy l methy lether CH 3  CH 2  CH 2  CH 2  OH ; ID CH 3  C H  CH 2  CH 3 , CH 3  C  CH 3 CH 3 Dimethy lether Ethy l alcohol : n Buty l alcohol C2 H 5  O  C2 H 5 Neopentane Diethy l ether (ii) Aldehydes, ketones alcohols …etc. (Cn H2nO) and U  Except alkynes chain isomerism is observed when the number of carbon atoms is four or more than four. D YG  Chain isomers differ in the nature of carbon chain, i.e., in the length of carbon chain.  The isomers showing chain isomerism belong to the same homologous series, i.e., functional group, class of the compound (Cyclic or open) remains unchanged.  Chain and position isomerism cannot be possible together between two isomeric compounds. If two compounds are chain isomers then these two will not be positional isomers. ST U (2) Position isomerism : It is due to the difference in the position of the substiuent atom or group or an unsaturated linkage in the same carbon chain. (ii) C3 H6 CH 3  CCl 2  CH 3 , CH 3  CH 2  CH  Cl 2 , 2, 2 Dichloro propane, (gemdihalide) Cl2 : 1,1Dichloro propane (gemdihalide) CH 3  C H  C H 2 , C H 2  CH 2  C H 2 | | | Cl Cl Cl 1, 2  Dichloro propane (Vicdihalide) | Cl 1, 3  Dichloro propane ( , - dihalide)  Aldehydes, carboxylic acids (and their derivatives) and cyanides do not show position isomerism.  Monosubstituted alicylic compounds and aromatic compounds do not show position isomerism. unsaturated O || C3H6O : CH 3  CH 2  CHO ; CH 3  C  CH 3 ; Propionald ehy de Acetone CH 2  CH  CH 3 ; CH 2  CH  CH 2 OH Ally lalcohol O 1, 2  Epoxy propane (iii) Acids, esters and hydroxy compounds …etc. (Cn H2nO2) C2H4O2 : CH 3 COOH ; HCOOCH 3 carbonyl Methy l formate Acetic acid C3H6O2 : CH 3  CH 2  COOH ; CH 3 COOCH 3 ; Methy l acetate Propionic acid O || CH 3 C HCHO ; CH 3  C  CH 2  OH | 1  Hy droxy propan - 2 - one OH 2  Hy droxy propanal (iv) Alkynes and alkadienes (Cn H2n-2) C4H6 : CH 3  CH 2  C  CH ; H 2C  CH  CH  CH 2 ; 1  Buty ne 1, 3  Butadiene CH 3  C  C  CH 3 ; H 2C  C  CH  CH 3 1, 2  Butadiene 2  Buty ne (v) Nitro alkanes and alkyl nitrites (  NO 2 and O  N  O ) O ; C2 H 5  O  N  O C2H5NO2 : C 2 H 5  N O Ethy l nitrite Nitro ethane (vi) Amines (Primary, secondary and tertiary) 1032 General Organic Chemistry C3H9N : CH 3  CH 2  CH 2  NH 2 ; isomerism is not possible, there will be metamerism e.g., Propan -1 - amine (a) CH 3  C  CH 2  CH 2  CH 3 ; H CH 3  CH 2  N || O CH 3 N  Methy l ethanamine (Pentan  2  one) CH 3 CH 2  C  CH 2 CH 3 are metamers and not position || CH 3 O (Pentan  3  one) CH 3  C H  CH 3 ; CH 3  CH 2  N isomers. | CH 3 NH 2 (b) CH 3  C  CH 2 CH 2 CH 3 ; || O 60 N , N  Dimethy l methanamin e Propan - 2- amine (vii) Alcohols and phenols (Pentan  2  one) CH2OH OH CH 3  C  C H  CH 3 || | O CH 3 CH3 C7 H8 O ; are metamers and not chain isomers. E3 (3- Methy lbuta n - 2 - one) Benzyl alcohol o-Cresol  Alkenes does not show metamerism. (6) Tautomerism (viii) Oximes and amides O || C2H5NO : CH 3  CH  NOH ; CH 3  C  NH 2 Acetaldoxime ID Acetamide (i) The type of isomerism in which a substance exist in two readily interconvertible different structures leading to dynamic equilibrium is known as tautomerism and the different forms are called tautomers (or tautomerides). H 2C | H 2C CH 2 | ; CH 2 D YG C6H12 : C H2 Cy clohexane CH 2  CH 2 | C H 2  CH 2 CHCH 3 ; CH 3 CH 2CH 2CH 2CH  CH 2 1 Hexene Methy l cy clopentane U  Ring – chain isomers are always functional isomers. ST (5) Metamerism : This type of isomerism is due to the difference in the nature of alkyl groups attached to the polyvalent atoms or functional group. Metamers always belong to the same homologous series. Compounds like ethers, thio-ethers ketones, secondary amines, etc. show metamerism. CH 3 ; C2 H 5  N CH 3 (i) C5H13N : C3 H7  N Dimethy l propy l amine (ii) C6H15N : The term tautomerism (Greek: tauto = same; meros = parts) was used by Laar in 1885 to describe the phenomenon of a substance reacting chemically according to two possible structures. U (4) Ring-chain isomerism : This type of isomerism is due to different modes of linking of carbon atoms, i.e., the isomers possess either open chain or closed chain sturctures. H2 C (ii) It is caused by the wandering of hydrogen atom between two polyvalent atoms. It is also known as Desmotropism (Desmos = bond and tropos = turn). If the hydrogen atom oscillates between two polyvalent atoms linked together, the system is a dyad and if the hydrogen atom travels from first to third atom in a chain, the system is a triad. (a) Dyad system : Hydrocyanic acid is an example of dyad system in which hydrogen atom oscillates between carbon  C  N H Examples : O H OH || | |  C C ⇌  C  C  | | (Keto) ; C 2 H 5  NH  C4 H 9 Buty l ethy l amine  If same polyvalent functional group is there in two or more organic compounds, then chain or position atoms. H  C  N ⇌ Keto-enol system : Polyvalent atoms are oxygen and two carbon atoms. Diethy l methy l amine Dipropy l amine nitrogen (b) Triad system C2 H 5 CH 3 C 3 H 7  NH  C 3 H 7 and (Enol) Acetoacetic ester (Ethyl acetoacetate) : O OH || | Keto form (92.1%) Enol form (7.9%) CH 3  C  CH 2 COOC 2 H 5 ⇌ CH 3  C  CHCOOC 2 H 5 Acetoacetic ester gives certain reactions showing the presence of keto group (Reactions with HCN , General Organic Chemistry 1033 H 2 NOH , H 2 NNHC 6 H5 , etc.) and certain reactions showing the presence of enolic group (Reactions with Na, CH 3COCl , NH 3 , PCl5 , Br2 water and colour with neutral FeCl 3 , etc.). Enolisation is in order CH 3 COCH 3  CH 3 COCH 2 COOC 2 H 5  C6 H 5 COCH 2 COOC 2 H 5  CH 3 COCH 2 COCH 3  CH 3 COCH 2 CHO (g) Tautomerism has no effect on bond length. (h) Tautomerism has no contribution in stabilising the molecule and does not lower its energy. (i) Tautomerism may occur in planar or nonplanar molecules.  Keto=enol tautomerism is exhibited only by such aldehydes and ketones which contain at least one  -hydrogen. For example O || H Acid catalysed conversion CH 3  C  CH 2 R 60 CH 3 CHO, CH 3 CH 2 CHO, CH 3 COCH 2 COCH 3 etc,. Keto OH OH | | H  CH 3  C  C H  R  CH 3  C  CH  R  | (Enol form) H Base catalysed conversion OH || CH 3  C  CH 2  R – H2 O Keto form Number of structural isomers O || Molecular formula CH 3  C  C H  R  OH | | H 2O  CH 3  C  CH  R   CH 3  C  CH  R OH  (Enol) HN nitrite form D YG Nitro form O O Nitro acinitro system O O nitro form (i) C4 H10 Two C5 H12 Three C6 H14 Five C7 H16 Nine C8 H18 Eighteen C9 H 20 Thirty five C10 H 22 Seventy five U (c) Triad system containing nitrogen : Examples Nitrous acid exists in 2 forms CH 3  CH 2  N Number of isomers Alkanes  O H O  N  O they do not have   H. ID O acetaldehyde, (CH 3 )3 C  CHO and chloral CCl 3  CHO as E3 (Intermediate)  Tautomerism is not possible in benzaldehyde (C6 H 5 CHO ) , benzophenone (C6 H 5 COC 6 H 5 ) , tri methyl CH 3  CH  N O OH Aciform (ii) ST U (iii) Characteristics of tautomerism (a) Tautomerism (cationotropy) is caused by the oscillation of hydrogen atom between two polyvalent atoms present in the molecule. The change is accompanied by the necessary rearrangement of single and double bonds. (b) It is a reversible intramolecular change. (c) The tautomeric forms remain in dynamic equilibrium. Hence, their separation is a bit difficult. Although their separation can be done by special methods, yet they form a separate series of stable derivatives. (d) The two tautomeric forms differ in their stability. The less stable form is called the labile form. The relative proportion of two forms varies from compound to compound and also with temperature, solvent etc. The change of one form into another is also catalysed by acids and bases. (e) Tautomers are in dynamic equilibrium with each other and interconvertible (⇌). (f) Two tautomers have different functional groups. Alkenes and cycloalkanes C3 H6 Two (One cycloalkane) C4 H8 Six (Four alkene cycloalkane) C5 H10 Nine (Five alkenes cycloalkanes) Alkynes C3 H4 Two C4 H 6 Six Monohalides C3 H 7 X Two C4 H 9 X Four C5 H11 X Eight Dihalides C2 H 4 X 2 Two C3 H 6 X 2 Four C4 H 8 X 2 Nine C5 H10 X 2 Twenty one alkene + + + one 2 - 4 – 1034 General Organic Chemistry Two (One alcohol and one ether) C3 H 8 O Three (Two alcohols and one ether) Seven (Four alcohols and three C4 H10 O ethers) Fourteen (Eight alcohols and six ethers) C5 H12 O Aldehydes and ketones Two (One aldehyde and one C3 H 6 O ketone) Three (Two aldehydes and one C4 H 8 O ketone) Seven (Four aldehydes and three C5 H10 O Two (One acid and one ester) C 3 H 6 O2 Three (One acid and two esters) C 4 H 8 O2 Six (Two acids and four esters) C5 H10 O2 Thirteen (Four acids and nine esters) (1) Conditions for geometrical isomerism : Compound will show geometrical isomerism if it fulfils the following two conditions (i) There should be frozen rotation about two adjacent atoms in the molecule. (a) frozen rotation about carbon, CC carbon double bond in alkenes. (b) frozen rotation about carbon, carbon single bond in cycloalkanes. (ii) Both substituents on each carbon should be different about which rotation is frozen. If these two conditions are fulfilled, then compound will show geometrical isomerism.  The compounds of the following type will not show geometrical isomerism. a  C a x  C a a  C a || || || x C y aC a x C  x Two (One 1°-amine and one 2°amine) Four (Two 1°-amines, one 2°amine and one 3°-amine) Eight (Four 1°-amines, three 2°amines and one 3°-amines) Four (2) Distinction between cis- and trans- isomers (i) By cyclization method : Generally, the cisisomer (e.g. maleic acid) cyclises on heating to form the corresponding anhydride while the trans-isomer does not form its anhydride. H  C  COOH H  C  CO Heat   || || H  C  COOH H  C  CO O U C8 H10 D YG Aliphatic amines Aromatic compounds Mesaconic acid (trans  isomer) Citraconic acid (cis- isomer) U C 2 H 4 O2 C4 H11 N H 3 C  C  COOH || HOOC  C  H H 3 C  C  COOH || H  C  COOH ID Monocarboxylic acids and esters C3 H 9 N Fumaric acid (trans) Maleic acid (cis) (c) C  N  frozen rotation about carbon, nitrogen double bond in oxime and imine. ketone) C2 H7 N H  C  COOH || HOOC  C  H 60 C2 H 6 O H  C  COOH || H  C  COOH E3 Alcohols and ethers  Note that the two reacting groups (–COOH) are near to each other. Geometrical or cis-trans isomerism H  C  COOH Heat   No anhydride || HOOC  C  H C9 H12 Five ST C7 H 8 O Nine The compounds which have same molecular formula but differ in the relative spatial arrangement of atoms or groups in space are known as geometrical isomers and the phenomenon is known as geometrical isomerism. The isomer in which same groups or atoms are on the same side of the double bond is known as cis form and the isomer in which same groups or atoms are on the opposite side is called trans-isomer. Examples : Maleic acid(cis) Maleic anhy dride Fumaric acid (trans)  Note that the two reacting groups (–COOH) are quite apart from each other, hence cyclisation is not possible. (ii) By hydroxylation (Oxidation by means of KMnO 4 ,OsO4 or H 2O2 in presence of OsO4 ) : Oxidation (Hydroxylation) of alkenes by means of these reagents proceeds in the cis-manner. Thus the two geometrical General Organic Chemistry 1035 isomers of an alkene leads to different products by these reagents. For example, H  C  COOH H KMnO 4 ||   H H  C  COOH OH OH COOH COOH meso  Tartaric acid Maleic acid (cis) H  C  COOH || HOOC  C  H OH H  HOOC KMnO 4 Fumaric acid (trans ) OH COOH H  HOOC H OH 60 () Tartaric acid H COOH OH cis Dichloroethy lene ( 1.9 D) trans  Dichloroethy lene ( 0.0 D) cis Butene  2 trans  Butene  2 (   0. 0 D) ID (iii) By studying their dipole moments : The cisisomer of a symmetrical alkene (Alkenes in which both the carbon atoms have similar groups) has a definite dipole moment, while the trans-isomer has either zero dipole moment or less dipole moment than the cisisomer. For example, 1,2-dichloroethylene and butene2. H  C  CH 3 H  C  CH 3 H  C  Cl H  C  Cl || || || || H C CH 3 H C Cl CH 3 C  H Cl C H E3 () Tartaric acid ST U D YG U In trans-isomer of the symmetrical alkenes, the effect produced in one half of the molecule is cancelled by that in the other half of the molecule. In case of unsymmetrical alkenes, the cis-isomer has higher dipole moment than the corresponding trans-isomer. General Organic Chemistry The symbol ‘E’ is assigned to an isomer in which the atoms or groups of higher preference are on the opposite side (E from German word Entgegen = across or opposite). The symbol ‘Z’ is assigned to an isomer in which the atoms or groups of higher preference are on the same side (Z from German word, Zusammen = together). For Example, CH 3  C  Cl || Cl  C  CH 2 CH 3 trans 2, 3  Dichloropentene  2 ( Less dipole moment ) cis  2, 3  Dichloropentene  2 (High dipole moment) Similar is the case with hexene-2. H3C H CH 2 CH 2  CH 3 H CC 1 cisHexene  2(more polar) C C CH 2 CH 2  CH 3 H trans  Hexene  2(Less polar) (iv) By studying other physical properties: (a) The cis-isomer of a compound has higher boiling point due to higher polarity, higher density and higher refractive index than the corresponding trans-isomer (Auwers-skita rule). CH 3  C  H CH 3  C  H || || H  C  CH 3 trans  2 Butene 1C 106 C cis 2 Butene 4 C 139 C H  C  Cl H  C  Cl H  C  Cl Cl  C  H || || cis 1, 2  Dichloroethene 60 C  80 C trans 1, 2  Dichloroethene 48 C  50 C Z  isom er 1 2  1 signifies higher preference and 2 signifies lower preference. Preference in most of the cases ‘Z’ corresponds to cis-form and ‘E’ to trans-form. However, there are many exceptions. The following rules are followed for deciding the precedence order of the atoms or groups; (i) Higher priority is assigned to the atoms of higher atomic number. For example, the order of preference in the following atoms, H , Cl , I, Br is : I (at. order of preference. For example, deuterium  D 2 1 is assigned higher priority in comparison to hydrogen D YG U ST CC CC no. 53)> Br (at. no. 35)>Cl (at. no. 17)>H (at. no. 1). (ii) If isotopes of the same element are attached, the isotope with higher mass number is given higher (b) The trans-isomer has higher melting point than the cis-isomer due to symmetrical nature and more close packing of the trans-isomer. (v) Stability : Trans-isomer is more stable than cis-isomer due to symmetrical structure.  Terminal alkenes such as propene, 1-butene and 2-methyl propene do not show geometrical isomerism.  Cis-trans isomers are configurational isomers but not mirror images, hence cis and trans isomers are always diastereomers.  Non-terminal alkenes with the same atoms or groups either on one or both the carbon atoms of the double bond such as 2-methyl-2-butene, 2,3-dimethyl – 2 – butene etc. do not show geometrical isomerism. (3) E and Z system of nomenclature : ‘Cis’ and ‘Trans’ designations cannot be used if four different atoms or groups are attached to the carbon atoms of a double bond. a b 2 ID b.p m.p. E  isom er 1 1  H . U CH 3  C  H 2 2 E3 H CH 3 CC 60 H 3 C  C  Cl || CH 3  CH 2  C  Cl 1033 d e In such cases, E and Z system of nomenclature is used. This system is based on a priority system developed by Cahn, Ingold and Prelog. In this system, the two atoms or groups attached to each of the doubly bonded carbon are put in order of preference on the basis of sequence rules. 1 1 (iii) In the groups, the order of preference is also decided on the basis of atomic number of first atom of the group. For example, in the following set, Cl,OH,COOH , NH  CH 3 ,SO 3 H. The order of the precedence is :  Cl   S O3 H   O H   N HCH 3   C OOH  ( at.no17)  (at.no.16)  (at.no.8)  (at.no.7)  (at.no. 6) When the order of preference of the groups cannot be settled on the first atom, the second atom or the subsequent atoms in the groups are considered. For example, in the set CH 2  CH 3 ,CH 3 ,COOH , the order cannot be decided on the basis of first atom as it is same in all the groups. However, in CH 2  CH 3 , the second atom is carbon, in CH 3 , the second atom is hydrogen while in COOH , the second atom is oxygen. Hence, the order of preference is : C O OH  CH 2  C H 3  C H 3  (at.no.8)  (at.no.6)  (at.no.1) (iv) A doubly or triply bonded atom is considered equivalent to two or three such atoms. For example, The group C  O is equal to C O and the group | O N | C  N is equal to  C N. | N 1034 General Organic Chemistry (4) Number of geometrical isomers in polyenes (i) When a compound has n double bonds and ends of a polyene are different, the number of Similarly consider the following structure C 2 H 5 C CH 3 || N OH geometrical isomers  2 n Sy n  m ethy lethy l ketoxim e or Anti  ethy l m ethy l ketoxim e C6 H5  CH  CH  CH  CH  CH  CH  CH  CH  Cl The given compound has four double bonds and the two ends are different (One is C6 H5 and other is number of geometrical (ii) When the ends of polyene are same. Case I : When number of double bonds (=n) is even then the number of geometrical isomers  2 n 1  2[(n / 2)1] Cl  CH  CH  CH  CH  CH  CH  CH  CH  Cl n  4 , even Number of geometrical isomers  2n 1  2(n / 2)1  2  2  8  2  10. Case II : When number of double bonds (=n) is odd. 1  n 1   2  1   Number of geometrical isomers  2 n 1  2 C6 H 6  CH  CH  CH  CH  CH  CH  C6 H 5 Number of geometrical Isomerism in  2 2  2 2 1 nitrogen C  N  bond. D YG  2 2  21  4  2  6. (5) Geometrical compounds isomers (i) Geometrical isomerism due to U The important class of compounds exhibiting geometrical isomerism due to C  N  bond are oximes, nitrones, hydrazones and semicarbazones. But the most common compound is oxime. Oximes : In aldoxime, when hydrogen and hydroxyl groups are on the same side, the isomer is known as syn. (analogous to cis) and when these groups are on the opposite side, the isomer is known as C6 H 5  C  H || anti (analogous to trans) N  OH Sy n - benzaldoxi me ST C6 H 5  C  H || HO  N Anti benzaldoxi m e In ketoximes the prefixes syn and anti indicate which group of ketoxime is syn or anti to hydroxyl group. For example: CH 3  C  C2 H 5 this compound will be named as; || N OH (a) Syn-ethyl methyl ketoxime  HO and C2 H5 are syn or (b) Anti-methyl ethyl ketoxime  HO and C 2 H 5 are anti. Anti azobenzene (6) Geometrical isomerism show by cumulatrienes : Cumulatrienes (Trienes with three adjacent double bonds) show only geometric isomerism. This is because their molecule is planar, as such the terminal CH 3 groups and H- atoms lie in the same plane. Therefore, in this case their planar structure can exist in two diastereoisomeric forms, cisand trans- but no enantiomeric forms are possible. H 3C H CH 3 H CCCC cis Hexa  2, 3 , 4  triene H 3C H CCCC trans  Hexa  2, 3 , 4  triene H CH 3 (7) Geometrical isomerism in cycloalkanes : Disubstituted cycloalkanes show geometrical isomerism. U (n  3 , odd ) 5 Sy n - azobenzene ID 3 6 isomers  2n  2 4  16. || N  C6 H 5 60 Therefore, || C H N E3 Cl). (ii) Geometrical isomerism due to N = N bond. C6 H 5  N C6 H 5  N CH3 H CH3 H OH H H OH HO H Cis-1,2Cis-1,2-dimethylcyc cyclohexanediol lopropane OH H Trans-1,2cyclopent anediol  Certain compounds show geometrical as well optical isomerism. Such type of isomerism is known as geometrical enantiomerism. Optical isomerism (1) Compounds having similar physical and chemical properties but they have the ability to rotate the plane of polarised light either to the right (Clockwise) or to the left (Anticlockwise) are termed as optically active or optical isomers and the property is called optical activity or optical isomerism. The optical activity was first observed in organic substances like quartz, rock-crystals and crystals of potassium chlorate (KClO 3 ) , potassium bromate (KBrO 3 ) and sodium periodate (NaIO4 ). (2) Measurement of optical activity : The measurement of optical activity is done in terms of specific rotation which is defined as the rotation produced by a solution of length of 10 centimetres (One General Organic Chemistry 1035 decimetre) and unit concentration (1 g/mL) for the given wavelength of the light at the given temperature.  obs C Specific rotation,  twavelength  lC (ii) Elements of symmetry : There are three elements of symmetry, length of the solution in decimeters and C is the number of grams in 1mL of solution. The specific rotation of the sucrose at 20°C using sodium light (Dline, =5893Å) is +66.5°C and is denoted as: (a) Plane of symmetry : It may be defined as a plane which divides a molecule in two equal parts that are related to each other as an object and mirror image. e.g.,  20D C  66.5C(C  0.02 g / mL water) + sign direction. indicates the rotation in COOH | H  C  OH | H  C  OH | COOH clockwise (b) Centre of symmetry : It may be defined as a point in the molecule through which if a line is drawn in one direction and extended to equal distance in opposite direction, it meets another similar or Centregroup of symmetry atom, eg. ID (i) The optical isomer which rotates the plane of the polarised light to the right (Clockwise) is known as dextrorotatory isomer (Latin: dextro = right) or d-form or indicated by +ve sign. CH 3 NH  CO | C CO  NH | H U (ii) The optical isomer which rotates the plane of the polarised light to the left (Anticlockwise) is known as laevorotatory isomer (Latin; laevo = left) or l-form or indicated by –ve sign. D YG (iii) The optical powers of the above two isomers are equal in magnitude but opposite in sign. An equimolar mixture of the two forms, therefore, will be optically inactive due to external compensation. This mixture is termed as racemic mixture or dl-form or () mixture. (iv) Optical isomer with a plane of symmetry is called meso form. It is optically inactive due to internal compensation, i.e., the rotation caused by upper half part of molecule is neutralised by lower half part of molecule. ST U (4) Chirality, (i) Definition : A molecule (or an object) is said to be chiral or dissymmetric, if it is does not possess any element of symmetry and not superimposable on its mirror image and this property of the molecule to show non-superimposability is called chirality. On the other hand, a molecule (or an object) which is superimposable on its mirror image is called achiral (non-dissymmetric or symmetric). To understand the term chiral and achiral let us consider the alphabet letters ‘P’ and ‘A’ whereas ‘P’ is chiral, ‘A’ is achiral as shown in fig. Mirror Non-superimposable (Chiral or dissymmetric) Mirror Superimposable (Achiral or nondissymmetric) Plane of symmetry E3 (3) On the basis of the study of optical activity, the various organic compounds were divided into four types : 60 is the rotation observed, l is the Where obs H CH 3 CH 3 NH  CO | | and C  C CO  NH | | H H cis  Dimethy ldiketo piperazine | C | CH 3 Trans  Dimethy ldiketo piperazine Since trans form contains a centre of symmetry, it is optically inactive. (c) Alternating axis of symmetry : A molecule is said to possess an alternating axis of symmetry if an oriention indistinguishable from the original is obtained when molecule is rotated Q degree around an axis passing through the molecule and the rotated molecule is reflected in a mirror that is perpendicular to the axis of rotation in step (I). Cl H H H F H F 180o H F Reflection perpendicular to axis rotation Cl rotation around axis H Cl F H Cl H Cl H F H F H H Cl (iii) Symmetric, Asymmetric and Dissymmetric molecules (a) Symmetric molecules : If any symmetry is present in the molecule then molecule will be symmetric molecule. (b) Dissymmetric molecules : Molecule will be a dissymmetric molecule if it has no plane of symmetry, 1036 General Organic Chemistry no centre of symmetry and no alternating axis of symmetry. Number of enantiomeric pair  a/2 Number of racemic mixture  a/2 (c) Asymmetric molecules : Dissymmetric molecule having at least one asymmetric carbon is known as asymmetric molecule. All asymmetric molecules are also dissymmetric molecules but the reverse is not necessarily true. Number of meso form Case I : When compound has even number of carbon atoms, i.e., n  2, 4, 8, 10, 12,..... : CHO | H — C  — OH | CH 3 No plane of symmetry  Dissymmetric molecule  Asymmetric molecule 60 (i) Number of optically by active forms  a  2 n 1 (ii) Number of enantiomeric pairs  a / 2 (iii) Number of racemic mixture  a / 2 (iv) Number of meso forms  m  2 (n / 2)1 (v) Total number of configurational isomers am Case II : When compound has odd number of carbon atoms, i.e., n  3, 5, 7, 9, 11,...... : | HO  C  H | COOH Lactic acid D YG  Carbons that can be chirality centres are sp 3 - and sp -hybridised carbons cannot be chiral carbons because they cannot have four group attached to them.  Isotopes of an atom behave as different group in stereoisomerism. H1 D | | Cl 35 — C * — Cl 37 H — C* — T | | H2 Br U of optically active forms (ii) Number of enantiomeric pairs  a / 2 (iii) Number of racemic mixutre  a / 2 (iv) Number of meso forms  m  2 (n 1) / 2 * hybridised carbons; sp 2 Number U d C b | c * (i)  a  2 n 1  2(n 1) / 2 ID (iv) Chiral or asymmetric carbon atom : A carbon bonded to four different groups is called a chiral carbon or a chirality centre. The chirality centre is indicated by asterisk. e.g., CH 3 a | (ii) If molecule is divisible into two identical halves, then the number of configurational isomers depends on the number of asymmetric carbon atoms. E3 COOH | H — C  — OH | H — C  — OH | C6 H 5 No plane of symmetry  Dissymmetric molecule  Asymmetric molecule 0 Lactic acid ST  Carbon of the following groups will not be a chiral carbon O ||  CH 3 ,  CH 2 OH ,  CHX 2 ,  CHO ,  C  Z  Maleic acid (HOOC  CH  CH  COOH ) show geometrical isomerism while malic acid (HOOC  CH 2  CHOH  COOH ) show optical isomerism. (5) Calculation of number of optical isomers (i) If molecule is not divisible into two identical halves and molecule has n asymmetric carbon atoms then Number of optically active forms  2 n  a (v) Total number of configurational isomers am (6) Optical activity of compounds containing one asymmetric carbon Examples :   CH 3  C HOH  COOH ; CH 3  C HOH  CHO Lactic acid  CH 2 OH  C HOH  CHO ;  C6 H 5  C HCl  CH 3 1 - chloro -1 - pheny letha ne Gly ceraldehy de Any molecule having one asymmetric carbon atom exists in two configurational isomers which are nonsuperimposible mirror images. COOH | H — C — OH | CH 3 (I) COOH | HO — C — H | CH 3 (II) (I) and (II) have the same molecular formula, the same structure but different configurations, hence (I) and (II) are known as configurational isomers. (I) and (II) are nonsuperimposable mirror images, hence (I) and (II) are optical isomers. Configurational isomers which are nonsuperimposable mirror images are known as enantiomers. Thus (I) and (II) are General Organic Chemistry enantiomers. Pair enantiomeric pair. of (I) and (II) is known 1037 as diastereomers have the same functional groups, their chemical properties are not very dissimilar. (i) Properties of Enantiomers : All chemical and physical properties of enantiomers are same except two Case II : When molecule is divisible into two physical properties. Mode of rotation : One enantiomer rotates light to the right and the other by an equal magnitude to the identical halves. Number of optical isomers  a  2 2 1  2 Number of meso forms  m  2 0  1 Total number of configurational isomers  3 opposite directions and cancel each other’s rotation. This phenomenon is called external compensation.  Racemic mixture can be separated into (+) and (–) forms. The separation is known as resolution. chemical reagents.  Racemic mixture is designated as being (  ) or D YG (7) Optical activity of compounds containing two asymmetric carbon a Case I : When molecule is not divisible into two identical halves. The number of optical isomers possible in this case is four (a  2  4 ). Further there will be two pairs 2 of enantiomers and two racemic modifications. In practice also it is found to be so. a CCC U ST b y (ii) Alkylidene cycloalkanes and spiro compounds : When one or both of the double bonds in allenes are replaced by one and two rings, the resulting systems are respectively known as alkylidene cycloalkanes and spirans. H 3C H C Properties of Diastereomers : Diastereomers Diastereomers have different chemical properties towards both chiral and achiral reagents. Neither any two diastereomers nor their transition states are mirror images of each other and so will not necessarily have the same energies. However, since the x C C C b Configurational isomers which are not mirror separated by chromatography. a or b images are known as diastereomers. have different physical properties, e.g., melting and boiling points, refractive indices, solubilities in different solvents, crystalline structures and specific rotations. Because of differences in solubility they often can be separated from each other by fractional crystallisation; because of slight differences in molecular shape and polarity, they often can be C C C Allenes exhibit optical isomerism provided the two groups attached to each terminal carbon atom are different, i.e., U (dl). (i) Allenes : Allenes are the organic compounds of the following general formulae. ID  The conversion of (+) or (–) form of the compound into a racemic mixture is called racemisation. It can be caused by heat, light or by (8) Optical activity in compounds containing no assymmetric carbon : Although the largest number of known optically active compounds are optically active due to the presence of chiral carbon atom, some compounds are also known which do not possess any chiral carbon atom, but on the whole their molecules are chiral (such molecules were earlierly called dissymmetric); hence they are optically active. Various types of compounds belonging to this group are allenes, alkylidene cycloalkanes, spiro compounds (spirans) and properly substituted biphenyls. 60 (ii) Racemic Mixture : An equimolar mixture of two enantiomers is called a racemic mixture (or racemate,  form, (dl) form or racemic modification). Such a mixture is optically inactive because the two enantiomers rotate the plane polarised light equally in E3 left direction. H a C b COOH 1-Methylcyclohexylidene-4acetic acid (Alkylidene CH 2 cycloalkane) CH 2 a C CH 2 C CH 2 b Spirans (iii) Biphenyls : Suitably substituted diphenyl compounds are also devoid of individual chiral carbon atom, but the molecules are chiral due to restricted rotation around the single bond between the two benzene nuclei and hence they must exist in two nonsuperimposable mirror images of each other. Such types of stereoisomerism which is due to restricted rotation about single bond, is known as atropisomerism 1038 General Organic Chemistry COOH O2N NO2 HOOC F HOOC nomenclature was given by Emil Fischer to designate the configurations of various sugars relative to the enantiomeric (+) and (–) glucose as reference. COOH F All sugars whose Fischer projection formula shows the OH group on the chiral carbon atom adjacent to the terminal CH 2 OH group on the right hand side The above discussion leads to the conclusion that the essential condition for optical isomerism is the molecular disymmetry or molecular chirality and not the mere presence of a chiral centre. However, it may be noted that the molecules having only one chiral centre belong to the D -series. Similarly if OH is on the left hand side, then the sugars belong to the L -series. | H — C — OH | CH 2 OH are always chiral and exhibit optical isomerism. (9) Fischer projection formulae D  series : The Examples : CHO | H — C — OH | CH 2 OH D(d ) gly ceraldehy de or D( ) gly ceraldehy de CHO | HO — C — H | CH 2 OH L (l ) gly ceraldehy de or L () gly ceraldehy de  It must be noted that there is no relation between the sign of rotation (+, – or d, l) and the configuration (D and L) of an enantiomer. ID Emil Fischer (1891) provided an easy method to represent the three dimensional formulae of various organic molecules on paper. Fischer projection is, thus, a planar representation of the three dimensional structure. L  series E3 arrangement of the atoms or groups in space that characterises a stereoisomer is called its configuration. | HO — C — H | CH 2 OH 60 and the stereoisomers are known and atropisomers. Examples By convention, the following points are followed in writing the Fischer formula.  Any compound that can be prepared from, or converted into D(+) glyceraldehyde will belong to Dseries and similarly any compound that can be prepared from, or converted into L(–) glyceraldehyde will belong to the L-series. U (i) The carbon chain of the compound is arranged vertically, with the most oxidised carbon at the top. D YG (ii) The asymmetric carbon atom is in the paper plane and is represented at the interaction of crossed lines. Asymmetric Carbon atom U C ST (iii) Vertical lines are used to represent bonds going away from the observer, i.e., groups attached to the vertical lines are understood to be present behind the plane of the paper. (iv) Horizontal lines represent bonds coming towards the observer, i.e., groups attached to the horizontal lines are understood to be present above the plane of the paper. (10) Name of optical isomers : Following three nomenclatures are used for optically active compounds, (i) D,L. System of nomenclature : This nomenclature is mainly used in sugar chemistry or optically active polyhydroxy carbonyl compounds. This  This nomenclature is also used in  -amino acids. (ii) Erythro and Threo system of nomenclature : This nomenclature is used only in those compounds which have (a) Only two chiral carbons and (b) The following structure, R  Cab  Cbc  R i.e., out of six substituents on two asymmetric carbons, at least two should be same. When two like groups in Fischer projection formula are drawn on the same side of the vertical line, the isomer is called erythro form; if these are placed on the opposite sides, the isomer is said to be threo form. General Organic Chemistry CH 3 | H — C — Cl | Br — C — H | C6 H 5 Erythro form Erythro form Threo form COOH 4 | |  3 —C —1 H 2 N — C — Br | | OH 2 C N O Br  Increasing priority with atomic number Rule 2 : If two or more than two isotopes of the same element is present, the isotope of higher (c) R,S Nomenclature (Absolute configuration) atomic mass receives the higher priority. 1 2 3 1 H 1H 1H Increasing priority 3 2 1 Rule 3 : If two or more of the atoms directly bonded to the chiral carbon are identical, the atomic number of the next atoms are used for priority assignment. If these atoms also have identical atoms attached to them, priority is determined at the first point of difference along the chain. The atom that has attached to it an atom of higher priority has the higher priority. ID The order of arrangement of four groups around a chiral carbon (stereocentre) atom is called the absolute configuration around that atom. System which indicates the absolute configuration was given by three chemists R.S. Cahn, C.K. Ingold and V. Prelog. This system is known as (R) and (S) system or the CahnIngold Prelog system. The letter (R) comes from the latin rectus (means right) while (S) comes from the latin sinister (means left). Any chiral carbon atom has either a (R) configuration or a (S) configuration. Therefore, one enantiomer is (R) and other is (S). A racemic mixture may be designated (R) (S), meaning a mixture of the two. (R) (S) nomenclature is assigned as 60 CH 3 | H — C — Cl | H — C — Br | C6 H 5 E3 R | a—C—b | c—C—b | R  U follows : 1039 D YG Step I : By a set of sequence rules given below the atoms or groups connected to the chiral carbon are assigned a priority sequence. I | 2 H and Br  I  H 2 C  H 2 C  C  C H 2  Br  2 H and C | CH 2  CH 2  CH 3  2 H and C Sequence Rules for Order of Priority In this example the atoms connected directly to the chiral carbon are iodine and three carbons. Iodine has the highest priority. Connected, to the three carbons are 2H and Br, 2H and C and 2H and C. Bromine has the highest atomic number among C,H and Br and thus CH 2 Br has highest priority among these number is given the next higher priority (3) and so on. three groups (i.e., priority no. 2).The remaining two carbons are still identical (C and 2H) connected to the second carbons of these groups are 2H and I and 2H and C. Iodine has highest priority among these atoms, so that CH 2  CH 2  I is next in the priority list and Cl 3 | | F — C — I  4 — C —1 | | Br 2 ST Thus, U Rule 1 : If all four atoms directly attached to the chiral carbon are different, priority depends on their atomic number. The atom having highest atomic number gets the highest priority, i.e., (1). The atom with the lowest atomic number is given the lowest priority, i.e., (2), the group with next higher atomic F Cl Br I  Increasing atomic number 4 3 2 1 Increasing priority CH 2  CH 2  CH 3 has the last priority. I | I  CH 2  CH 2  C  CH 2  Br | CH 2 CH 2 CH 3 1  2 3 4 Rule 4 : If a double or a triple bond is linked to chiral centre the involved atoms are duplicated or triplicated respectively. 1040 General Organic Chemistry O N O O | | || |  C  O   C  O ;  C  N   C  N ;  C  OH   C  OH | | | | N O By this rule, we obtained the following priority sequence CH:  CR 2 , CN , CH 2 OH , CHO, CO, COOH First interchange of two groups at the chiral centre inverts the configuration and this gives enantiomer of the original compound. Thus (A) and (B) are enantiomer. The second interchange involves the remaining two groups. 3 1 | 1 C  2 | Increasing priority 4 (B) For example: 2 2 | |  3 1 C | | 4 4 Clockwise arrangement 1 2 3of R and Let us apply the bromochlorofluoro methane. F | Br  C  H  | Cl , and S D YG , Anticlockwise arrangement 1 2 3of whole sequence to | 2 U ST 3 | 2 (A) 3 First interchange Between2 and 4 | 1 C  2 | 4 (B) 60 and 3 H CHO O O | H  C  OH  C | H  C  OH  | E3 | CH 2OH CH 2OH D-gly ceraldehy de 2 | 4 C 1 | 3 First interchange Between3 and4 CHO 2 | | HO  C  H   C  4 1 | | CH 2OH 3 L-Gly ceraldehy de 3 2 | C 1 | 4 1 Second interchange Between1 and2 2 3 4  Clockwise Rconfiguration (i) First interchange between 3 and 4 (ii) Sec. interchange between 1 and 2 1 3 2 2 4 Anticlockwise  S-configuration | 1 C 4 In such cases the Fischer projection formula of the compound is converted into another equivalent projection formula in such a manner that atom or group having the lowest priority is placed vertically downward. This may be done by two interchanges between four priority numbers. The first interchange involves the two priority numbers, one is the least priority number and other is the priority number which is present at the bottom of the plane. In the above case first interchange will takes place between 2 and 4. 1 C  4 | 4 3 In this Fischer projection the least priority number is not at the bottom of the plane. | Example : U 3 C 1 3 C  2 (A) Arrangement of1 2, are clockwise, hence configuration is R ID Step 2 : The molecule is then visualised so that the group of lowest priority (4) is directed away from the observes (At this position the lowest priority is at the bottom of the plane). The remaining three groups are in a plane facing the observer. If the eye travels clockwise as we look from the group of highest priority to the groups of second and third priority (i.e., 1 2 3 with respect to 4) the configuration is designated as R. If arrangement of groups is in anticlocwise direction, the configuration is designated as S. | Second interchange between remaining groups, i.e., 1 and3 Glyceraldehyde (For example) has one asymmetric carbon, hence it has two configurational isomers (I) and (II). CHO CHO | H  C  OH | | HO  C  H | CH 2 OH ( R ) gly ceraldehy de (I) CH 2 OH (S ) gly ceraldehy de (II) One can draw a number other configurations for glyceraldehyde but each of them will be a repetition of either (I) or (II). In this connection it is important to note that if two projection formulae differ by an odd number of interchanges (1, 3, 5, 7, …..) of positions of groups on the chiral carbon, they are different. But if the two differ by