Fundamental Concepts in Organic Reaction Mechanism PDF

Fundamental Concepts in Organic Reaction Mechanism In an organic reaction, the organic molecule (also referred as a substrate) reacts with an appropriate attacking reagent and leads to the formation of one or more intermediate(s) and finally product(s) The general reaction is depicted as follows...

Fundamental Concepts in Organic Reaction Mechanism In an organic reaction, the organic molecule (also referred as a substrate) reacts with an appropriate attacking reagent and leads to the formation of one or more intermediate(s) and finally product(s) The general reaction is depicted as follows : ✓ Substrate is that reactant which supplies carbon to the new bond and the other reactant is called reagent. ✓If both the reactants supply carbon to the new bond then choice is arbitrary and in that case the molecule on which attention is focused is called substrate. Reaction mechanism: A sequential account of each steps as follows: describing details of electron movement, energetic during bond cleavage and bond formation, and the rates of transformation of reactants into products (kinetics) is referred to as reaction mechanism.  Nucleophiles and Electrophiles Nucleophile (Nu) : A reagent that brings an electron pair i.e., nucleus seeking and the reaction is then called nucleophilic. Electrophile (E+): A reagent that takes away an electron pair i.e., electron seeking and the reaction is called electrophilic. ✓ During a polar organic reaction, a nucleophile attacks an electrophilic centre of the substrate which is that specific atom or part of the electrophile that is electron deficient. ✓Similarly, the electrophiles attack at nucleophilic centre, which is the electron rich centre of the substrate.  Electron Movement in Organic Reactions The movement of electrons in organic reactions can be shown by curved-arrow notation. Presentation of shifting of electron pair is given below :  Physical Effects (electron movement) 1. Resonance or Mesomeric effect 2. Inductive effect 3. Electromeric effect 4. Hydrogen bond 5. Hyper conjugation 6. Steric effect  Resonance or Mesomerism 1. All the properties of a compound cannot be explained by single structure. 2. Canonical structures or resonance contributing structures–differ in positionof electrons. 3. Delocalization of electrons leads to decrease in potential energy of molecule. canonical structures of Resonance hybrid benezene structure of benezene  Resonance or Mesomerism 4. Resonance hybrid is more stable than canonical structures. 5. Resonance structures are imaginary. 6. Resonance energy = Actual energy of hybrid–energy of most stable contributing structure. 7. Resonance is measure of stability. Rules for Drawing Resonance Structure 1. The molecule should be planar. 2. It contains an alternating system of single and double bonds (a conjugated system). 3. The relative positions of nuclei should remain unchanged (e.g. tautomerism). 4. The negative charge must preferably lie on the most electronegative atom. 5. The charge needs to be preserved in all the resonating structures. 6. The electrons always move away from a negative charge. 7. Arrows should be drawn to indicate the direction of the movement of electrons. Types of Resonance +R or +M effect – + C C OH —C—C OH (+M or +R effect of –OH group) –R or –M effect + – C C CH O —C—C CH — O (–R effect of –CHO group) Types of Resonance For substituted benzene + + NH2 NH2 NH2 – etc. – +R effect of –NH2 group. – – – – – O O O O O O + + + N N N + etc. + –R effect of –NO2 group.  Benzene Bond Lengths 1.33 Å 1.52 Å 1.42 Å 1.42 Å  Resonance Structures of Benzene The 6 -electrons are able to flow (or resonate) continulally around the -molecular orbital formed from the six p atomic orbitals on each of the 6 carbon atoms on the ring structure. This is represented by the two resonance structures below (which are identical or degenerate). This flow of electrons leads to a very stable electronic structure, which accounts for benzenes low reactivity relative to alkenes. This stabilty is referred to as the Resonance Stablisation Energy. Canonical structurtes The -electrons are referred to as being conjugated.  Resonance Imparts Stability to Anionic Structures (and Cationic Structures H H No adjacent double H3C bond to the oxygen O lone pair CH2 Relatively difficult to form Replaced by C=O O O H3C H3C O O Relatively easy to form  VERY IMPORTANT POINT The ability to be able to delocalise (spread out) charge via resonance allows an assessment of (i) the degree of ease of formation of the charged species, and (ii) the stability of the charged species  The Resonance Arrow and its Physical Meaning The resonance arrow is not an equilibrium arrow O O H3C H3C O O The resonance arrow shows only the distribution of electrons. Thus, for the two degenerate structures above, the implication is that there is an even distribution of the two electrons between the two oxygen atoms, at all times. O H3C O Experimentally it is found that both C-O bonds are the same length and are intermediate in length between the C-O single and double bond, as are the C-C bonds in benzene. General Structure that will Display Resonance of Charges and Lone Pairs of Electrons R 1 R3 2 R R4 R1 R3 R2 R4  Some Important Aromatic Resonance Structures Nitro Group: An Electron Withdrawing Group O O O O O O O O N N N N Canonical structures O O N O O N Repels E Attracts E  Methoxy Group: An Electron Donating Group Canonical structures OMe OMe OMe OMe …Note in a reaction mechanism we would not show the lone pairs on the carbons carrying the –ve charge… OMe OMe OMe OMe Canonical structures These resonance structures allow us to rationalise (and predict) reactivity OMe OMe OMe Attracts E E Repels E  N N O O N OMe Inductive Effect 1. Permanent effect in saturated carbon chain compounds. 2. Group attached to carbon chain should have tendency to release or withdraw electrons. Types of inductive effect + I effect effect –electron donating groups e.g., CH3 , C2H5 – I effect effect –electron withdrawing groups e.g., - NO2 , –CN Features of Inductive Effect 1. Chloroacetic acid is a stronger acid than acetic acid because O O Cl - CH2 C Cl-CH2 - C + + H OH -I O Ka = 1.4 × 10-3 O O + CH3-C CH3-C + H OH +I O Ka = 1.75 × 10-5 Features of Inductive Effect 2. The larger is the electron-withdrawing effect of a group, the greater is the –I (inductive) effect. F CH2 COOH Br CH2 COOH Ka 2.5×10-3 1.3×10-3 3. Inductive effect is additive Cl3CCOOH Cl2CHCOOH Ka 2.3×10-1 5.4×10-2 Features of Inductive Effect 4. Since this effect is transmitted through a chain it becomes less effective with distance ClCH2CH2COOH ClCH2CH2CH2COOH Ka 8.32×10-4 3.02×10-5 Electromeric Effect Temporary effect which is observed in presence of reagents involving transfer of electrons in an unsaturated system. in presence of reagent + – X Y X Y in absence of reagent Electromeric Effect Addition of HBr to an alkene R 2 R 1 H H HBr C C C—C H H + – H H R – + + Br H R — CH — CH3 C — CH3 H Br Hyperconjugation or no bond resonance (a) Involves  and  bond orbitals (b) More the number of hyperconjugative structures, more will be the stability of ion or molecule + H H H H H H H + H – C – C+ H–C=C–H H C=C H–C=C + H H H H H H H H Structure of ethyl carbonium ion Hyperconjugation or no bond resonance (c) The number of hyperconjugative structures in an alkene is obtained by the number of C — H bonds attached to the carbon bonded directly to the double bonded carbon atoms. H H+ H – + – H C CH CH2 H C CH CH2 H C CH CH2 H H H H – H C CH CH2 H + Significance of Hyperconjugation H3C — CH — CH CH2 H3C — CH CH — CH2 + H 1–butene (2 hyperconjugative structures) – H3C — CH CH — CH3 H3C — CH — CH CH2 + H (6 hyperconjugative structures) 2–butene More stable Relative strength of organic acids O O R-C O R-C + + H R-C OH O O Resonance structures  Bond Cleavage and reaction Intermediate 1. Fission of a covalent bond: homolysis and heterolysis 2. Reaction intermediates: carbocations, carbanions and free radicals 3. Different reagents: electrophiles, nucleophiles and carbene 4. Brief idea about type of organic reactions: substitution, addition, elimination,condensation, rearrangement, isomerisation Reaction Mechanism Detailed description of sequence of steps involved in group from reactants to products. Reactant intermediate product Bond Cleavage Heterolytic Cleavage A : B ⎯⎯→ A + : B − A : B A – : B+ +vely charged ion – carbocation -vely charged ion – carbaanion Homolytic Cleavage A : B ⎯⎯→ A  + B  Free radicals. Carbonium ion Planar – sp2 hybridised bond angle 120o Has six electrons Stabilized by resonance or Empty unhybridised inductive effort p-orbital or hyperconjugation + C sp2 Hybridisation of carbon Planar Strucutre of carbnion Examples of Carbonium ion + CH2 CH2 CH2 etc + + Benzyl cation + + CH 2 CH CH 2 CH 2 CH CH 2 Allyl cation Stabilised through resonance H2 C CH+ no resonance hence unstable. Vinyl cation Stability of Cabocation The resonance effect is always more predominant than the inductive effect in stabilizing an ion. (i) By inductive effect CH3 CH3 CH3 CH3 > C + > CH3 > C + > H C+ CH3 H H 3° 2° 1° Stability of Cabocation (ii) By hyperconjugation CH3 CH3 + H3C — C H3C — C + CH2 — H CH2H + CH3 CH2H etc. + H3C — C H CH2 C CH3 CH3 Thus, tertiary carbocation is more stable than secondary and so on. Carbanion Pyramidal - sp3 hybridised bond angle 109.28 Has eight electrons Stabilized by resonance or by inductive effect... sp3 hybrid orbital containing lone pair Tetrahedral structure of carboanion Stability of Carbanion (i) By resonance - - H H H Cyclopentadienyl carbanion Stability of Carbanion (ii) By inductive CH3 CH3 CH3 CH3 C CH3 C H C CH3 H H 3° 2° 1° Stability of Carbanion (iii) Electron-donating groups destabilize a carbanion while electron-withdrawing groups stabilize it. CH2− CH −2 > NO2 OCH3 Free Radical Planer or Pyramidal Has seven electrons Stabilized by resonance or by inductive effect. Order of stability of free radical 3o >2o> 1o Unhybridised orbital containing odd electron + 120oC ( C sp2 hybridised carbon Planar Sturcutre Classification of Reagents Nucleophilic Reagents (Nucleophiles) Attacks the positive end of a polar bond or nucleus- loving is known as nucleophile. Generally, negatively charged or electron rich species are nucleophilic.  − e.g. OH , OCH3− , − − − −CN , −I , CH3 COO , NH2 , CH3− + H2 O, NH3 , NH3 — NH2........ H 2O, CH 3 — O.. — CH 3, C 2H 5 — OH,.. NH 3, N.. All nucleophiles are in general Lewis bases. Classification of Reagents Electrophilic Reagents (Electrophiles) Attacks a region of high electron density or electron-loving is known as electrophile. All positively charged or electron deficient species are electrophilic. H+ , CH3+ , NO+2 , Cl+ , Br + , Ag+ Classification of Reagents Neutral reagents which contain an electron-deficient atom are also electrophiles. AlCl3, SO3, BF3, SOCl2, POCl3, FeCl3, ZnCl2 All electrophiles are in general Lewis acids. Carbenes Divalent carbon compound. Carbon atom is linked to two adjacent groups by covalent bonding. A carbene is neutral and possesses two free electrons, i.e. a total of six electrons. Electron deficient. Carbenes Carbene is of two types (i) Singlet carbene: CH2 hybridisation sp2 it is v-shaped (ii) Triplet carbene: CH2 hybridisation sp it is linear shaped Triplet carbene is more stable than single carbene. Types of Organic Reaction Substitution Addition Elimination Rearrangement Condensation Isomerisation Types of Organic Reaction Substitution Reaction Replacement of an atom or group by other atom of group Nucleophilic substitution: R − X + OH− ⎯⎯⎯ → R OH + X− SN1 Reaction: Unimolecular nucleophilic substitution reaction. Types of Organic Reaction - SN1 Reaction (1) CH3 CH3 SN1 + CH3 — C — CH2Cl – CH3 — C — CH2 OH CH3 slow CH3 (2) CH3 CH3 + 1, 2-Methyl anion CH3 — C — CH2 shift CH3 — C — CH2 – CH3 + CH3 CH3 – Fast OH CH3 — C — CH2CH3 OH Types of Organic Reaction - SN2 Reaction SN2 Reaction: This is called bimolecular nucleophilic substitution and it is one-step process. CH3 CH3 CH3 – H – –. Fast H — C — Br + OH OH C Br HO — C — H slow CH2CH3 CH2CH3 CH2CH3 Transition state unstable Addition The reagent often adds to  C = C , −C  C−,  C = O or −C  N bond and the  bond is  converted into bond. Can be electrophilic addition or nucleophilic addition. 2 Cl CH2 = CH2 ⎯⎯⎯→ Cl CH2 − CH2Cl CCl4 OH + H2 O H+ (Hydration) Elimination Reaction Two groups on adjacent atoms are lost as a double bond is formed. Conc. H2SO4 CH3 – CH – CH – CH3 CH3 — CH CH – CH3 – H2 O OH H We divide elimination reactions into three classes. (1) E1 (2) E1 CB (3) E2 Rearrangement Migration of a group takes place within the same molecule. O C6H5 H+ C=N ⎯⎯⎯→ C6H5 — C — N — C6H5 ether C6H5 OH H (Beckmann rearrangement) OH H+ (Dehydration and rearrangement) Condensation Two molecules of same or different reactants combine to give a new product with the elimination of simple byproducts like H2O, NH3, etc, O O CH3 O dil. H3C — C — CH3 + H3C — C — CH3 H3C — C — CH — C — CH3 NaOH,  The Criteria for Aromaticity—Hückel’s Rule Four structural criteria must be satisfied for a compound to be aromatic. A molecule must be cyclic. To be aromatic, each p orbital must overlap with p orbitals on adjacent atoms.  A molecule must be planar. All adjacent p orbitals must be aligned so that the  electron density can be delocalized. Since cyclooctatetraene is non-planar, it is not aromatic, and it undergoes addition reactions just like those of other alkenes.  A molecule must be completely conjugated. Aromatic compounds must have a p orbital on every atom.  A molecule must satisfy Hückel’s rule, and contain a particular number of  electrons. Hückel's rule: Benzene is aromatic and especially stable because it contains 6  electrons. Cyclobutadiene is antiaromatic and especially unstable because it contains 4  electrons. Note that Hückel’s rule refers to the number of  electrons, not the number of atoms in a particular ring. Considering aromaticity, a compound can be classified in one of three ways: 1. Aromatic—A cyclic, planar, completely conjugated compound with 4n + 2  electrons. 2. Antiaromatic—A cyclic, planar, completely conjugated compound with 4n  electrons. 3. Not aromatic (nonaromatic)—A compound that lacks one (or more) of the following requirements for aromaticity: being cyclic, planar, and completely conjugated. Note the relationship between each compound type and a similar open-chained molecule having the same number of  electrons. 62 Examples of Aromatic Rings Completely conjugated rings larger than benzene are also aromatic if they are planar and have 4n + 2  electrons. Hydrocarbons containing a single ring with alternating double and single bonds are called annulenes. To name an annulene, indicate the number of atoms in the ring in brackets and add the word annulene. 63  -Annulene has 10  electrons, which satisfies Hückel's rule, but a planar molecule would place the two H atoms inside the ring too close to each other. Thus, the ring puckers to relieve this strain. Since -annulene is not planar, the 10  electrons can’t delocalize over the entire ring and it is not aromatic. 64  Two or more six-membered rings with alternating double and single bonds can be fused together to form polycyclic aromatic hydrocarbons (PAHs). There are two different ways to join three rings together, forming anthracene and phenanthrene. As the number of fused rings increases, the number of resonance structures increases. Naphthalene is a hybrid of three resonance structures whereas benzene is a hybrid of two. 65  Stereoisomerism Isomers same molecular formula Constitutional Isomers Stereoisomers Different nature/sequence Different arrangement of of bonds groups in space Conformational Isomers Configurational Isomers Differ by rotation about a Interconversion requires breaking single bond bonds Enantiomers Diastereoisomers Non-superposable mirror Not mirror images images  Structural representations Bond line structure (Pentane) O Ethylmethyl ether 1. Shows three dimensional structure Wedge and dash representation Wedged bonds Above the plane Dashed bonds Below the plane plain lines On the plane of the paper. H e.g., Methane C H H H PROJECTION FORMULAS OF CHIRAL MOLECULES Configuration of a chiral molecule is three dimensional structure and it is not very easy to depict it on a paper having only two dimensions. To overcome this problem the following four two dimensional structures known as projections have been evolved. 1. Fischer Projection 2. Newman Projection 3. Sawhorse Formula 4. Flying Wedge Formula 1. Fischer Projection COOH COOH Away H C OH or H OH from the veiwer CH2OH CH2OH (I) Towards (I) the veiwer Characteristic features of Fischer projection: Rotation of a Fischer projection by an angle of 1800 about the axis which is perpendicular to the plane of the paper gives identical structure. However, similar rotation by an angle of 900 produces non - identical structure.  2. Newman Projection In Newman projection the carbon atom away from the viewer is called 'rear' carbon and is represented by a circle. The carbon atom facing the viewer is called 'front' carbon and is represented as the centre of the above circle which is shown by dot. The remaining bonds on each carbon are shown by small straight lines at angles of 120o as follows: i) Bonds joined to 'front' carbon intersect at the central dot. ii) Bonds joined to 'rear' carbon are shown as emanating from the circumfrance of the circle.  The concept of Newman projection for n-butane can be understood by the following drawings: Out of sight three groups emanating from Visible circumfrance Rear Carbon Newman Projection Front Carbon CH3 CH3 H H H3 C H H H H H H CH3 Staggered Eclipsed  3. Sawhorse projection The bond between two carbon atoms is shown by a longer diagonal line.The bonds linking other substituents to these carbons are shown projecting above or below this line. H CH3 H 3 1 H H3C 4 CH H 3 H3 C H 2 H H H Alkanes and Cycloalkanes: Conformations and cis-trans Stereoisomers Stereochemistry: three-dimensional aspects of molecules Conformation: different spatial arrangements of atoms that result from rotations about single (σ) bonds Conformer: a specific conformation of a molecule 3.1: Conformational Analysis of Ethane H H H H H H H H C C H H H H Sawhorse 74 There are two conformations of ethane: Back carbon H H H H H H Front carbon Newman projection Staggered Eclipsed Dihedral (torsion) angle: angle between an atom (group) on the front atom of a Newman Projection and an atom (group) on the back atom Dihedral angles of ethanes: Staggered conformation: 60° (gauche), 180° (anti), and 300° (-60°, gauche) Eclipsed conformation: 0°, 120°, and 240° (-120°) 75 Energy vs. dihedral angle for ethane http://www2.chem.ucalgary.ca/Flash/ethane.html The barrier (Eact) for a 120° rotation of ethane (from one staggered conformer to another) is 12 KJ/mol. The eclipsed conformer is the barrier to the rotation. An H-H eclipsing interaction = 4 KJ/mol Torsional Strain: strain (increase in energy) due to eclipsing 76 interactions Conformations of Propane H3C H CH3 H H H H H C C H H H H staggered The barrier to C-C rotation for 5.0 KJ/mol propane is 13 KJ/mol = 1 (CH3-H) + 2 (H-H) eclipsing Interactions. A CH3-H eclipsing interaction is 5 KJ/mol eclipsed 77 Conformational Analysis of Butane Two different staggered and eclipsed conformations Staggered: anti H3C H CH3 H H H H H C C H CH 3 H3C H Staggered: gauche 3 KJ/mol H3C H CH3 H H H 3C H CH 3 C C H H H H 78 Steric Strain: repulsive interaction that occurs when two groups are closer than their atomic radii allow 3 KJ/mol Eclipsed conformations of butane: rotational barrier of butane is 25 KJ/mol. A CH3-CH3 eclipsing interaction is 17 KJ/mol. CH3 - H CH3 - CH3 79 Energy diagram for the rotation of butane Summary: H - H eclipsed 4.0 KJ/mol torsional strain H - CH3 eclipsed 5.0 KJ/mol mostly torsional strain CH3 - CH3 eclipsed 17 KJ/mol torsional + steric strain 80 CH3 - CH3 gauche 3.0 KJ/mol steric strain Conformations of Cyclohexane - ΔHcomb suggests that cyclohexane is strain-free; favored conformation is a chair. Axial and Equatorial Bonds in Cyclohexane Chair cyclohexane has two types of hydrogens: axial: C-H axis is “perpendicular” to the “plane of the ring” equatorial: C-H axis is “parallel” to the “plane of the ring” Chair cyclohexane has two faces; each face has alternating axial and equatorial -H’s a e a a e axial e e e equatorial a e a a 81 top face bottom face All H-H interactions are staggered - no torsional strain; minimal angle strain (~111°) Other conformations of cyclohexane: half chair; twist boat, and boat Conformational Inversion (Ring-Flipping) in Cyclohexane Ring flip interchanges the axial and equatorial positions. The barrier to a chair-chair interconversion is 45 KJ/mol. 45 KJ/mol 82 83 Chair-Chair Interconversion of Cyclohexane axial equatorial Half-chair Chair (+ 45 KJ/mol) Twist-boat Twist-boat Boat (+23 KJ/mol) (+ 23 KJ/mol) (+ 32 KJ/mol) axial equatorial Half-chair Chair 84 (+ 45 KJ/mol)  Conversion of Fischer Projection into Sawhorse Projection. Fischer projection of a compound can be converted into sawhorse projection first in the eclipsed form by holding the model in horizontal plane in such a way that the groups on the vertical line point above and the last numbered chiral carbon faces the viewer. Then one of the two carbons is rotated by an angle of 180o to get staggered form (more stable or relaxed form). Rear carbon 1COOH 1COOH 2 2 H OH HO H H OH 3 4 Rotation by HO H COOH HOOC COOH 4 180o along C2 - C3 axis COOH 3 Front HO H H HO carbon Eclipsed Staggered Fischer Projection Sawhorse Projection  Conversion of Sawhorse projection into Fischer projection First the staggered sawhorse projection is converted in eclipsed projection. It is then held in the vertical plane in such a manner that the two groups pointing upwords are away from the viewer i.e. both these groups are shown on the vertical line. Thus, for 2,3-dibromobutane. CH3 H Br Br H CH3 CH3 CH3 Rear carbon Rotation by H Br CH3 H Br 180o Br H Front carbon Staggered Sawhorse H Br Projection CH3 Eclipsed Sawhorse Projection Fischer Projection  Conversion of Sawhorse to Newman to Fischer Projection Rear H 2N Cl Ph carbon H 2N Cl Ph CH3 View through Front carbon t h e f ro n t c a r b o n Br OH Br OH CH3 Staggered Staggered S a w h o r s e P ro j e c t i o n N e w m a n P ro j e c t i o n Rotate the front carbon along the central bond by 180o CH3 HO NH 2 Br H 2N Cl H o l d i n vertical p l a n e Cl keeping front carbon as the HO Br lowest F ro n t c a r b o n H3C Ph Ph Eclipsed F i s c h e r P ro j e c t i o n N e w m a n P ro j e c t i o n  Conversion of Fischer to Newman to Sawhorse Projection Rear carbon CHO H OH Cl H Rotate front carbon H OH by 180o H Cl CHO CH2OH Front carbonCH2OH Eclipsed Newman Projection Fischer Projection CH2OH H OH H OH CHO View through HOH2C central bond Cl H CHO Cl H Staggered Staggered Sawhorse Newman Projection Projection Conversion of Fischer Projection into Flying Wedge The vertical bonds in the Fischer projection are drawn in the plane of the paper using simple lines consequently horizontal bonds will project above and below the plane. CO O H H W h e n Br above the plane CH 3 CO O H Br H Br W h e n H above the plane CH 3 COOH Br H3C H  OH H C HO 1. Plane polarized light H3C F C 2. Stereoisomerisms F H 3. Chirality H3 C 4. Naming stereocenters - R/S configuration 5. Molecules with 2 or more stereocenters 6. Optical activity 7. Separation of Enantiomers, Resolution  Plane-Polarized Light Light vibrating in all planes ⊥ to direction of propagation Plane-polarized light: light vibrating only in parallel planes Plane-polarized light the vector sum of left and right circularly polarized light  Optically Activity Enantiomers (chiral) interact with circularly polarized light rotating the plane one way with R center and opposite way with S. Rotation of plane-polarized light clockwise (+) or counterclockwise (-).  Plane-Polarized Light (polarimeter) achiral sample no change in the plane of polarised light. Plane-Polarized Light (polarimeter)  H A R C L I rotates the plane  Specific rotation t C obs D = C where  → Specific rotation, obs → Observed angle of rotation, → Length of the solution in decimetre, C → Concentration of the active compound in grams per millilitre. The observed rotation of the plane of polarised light produced by a solution depends on the following. 1. The amount of substance in tube. 2. Length of solution examined. 3. Temperature of the experiment. 4. Wavelength of light used.  Chiral Center Common source of chirality - tetrahedral (sp3) carbon (atom) –bonded to 4 different groups. Chiral center - carbon (atom) with 4 different groups, e.g., 2-Butanol - 1 chiral center. OH C H H3 C CH2 CH3 (1) Enantiomers: stereoisomers are nonsuperposable mirror images. All chiral centers are stereocenters. Not all stereocenters are chiral centers.  Elements of Symmetry Conformations of 2,3-butanediol* syn - plane of symmetry anti - point of symmetry H OH CH3 C C.H CH HC. CH3 HO 3 3 C HO H C HO H If symmetry is present, the substance is achiral.  Elements of Symmetry mirror plane (a) Plane of symmetry HO OH H H HO CH2 OH H CH2 H H H H H H H H H OH OH H H H H  Fischer projection formula Convenient way to represent the three dimensional structures in two dimensions. Molecule is drawn in the form of a cross with the chiral carbon at the intersection of the horizontal and vertical lines. First we orient the molecule in such a way so that the carbon chain is vertical. The horizontal lines represent the bonds directed towards the viewer, and the vertical lines away from the viewer.  Drawing of Fischer formula H H3C H3C CH2OH HOH2C C H C CH2CH3 CH2CH3 H3C HOH2C C H CH2CH3 In Fischer projection formula rotation by 180o in plane of paper is allowed.  R,S Convention Each atom bonded to the chiral center assigned a priority by atomic number; higher atomic number, higher the priority. (1) (6) (7) (8) (16) (17) (35) (53) -H -CH3 -NH2 -OH -SH -Cl -Br -I increasing priority Same atoms bonded to the chiral center look to the next set of atoms priority assigned to 1st point of difference. H ( 1) H H (6) H ( 7) H (8) C H C C H C N H C O H H H H H H H increasing priority The isotope with higher mass number is given higher order of precedence.  R,S Convention Double (triple) bond atoms viewed as bonded to an equivalent number of atoms by single bonds C C H is treated as H C C C C H H H H O C O is treated as C C O H H C C is treated as C C H C C H C C  Naming Chiral Centers 1. Locate the chiral center, prioritize four substituents 1 (highest) to 4 (lowest) 2. Orient molecule so that lowest priority (4) group is directed away ( behind ) 3. Read three groups toward you (in front) (1) to (3) Clockwise R configuration; counterclockwise S (4) (1) H Cl (3) (2)  Naming Chiral Centers ( R ) -3-Chlorocyclohexene (3) (4) H R Cl(1) (2) (R)-mevalonic acid 1 4 HO CH3 O HO OH 3 2  Illustrative Example What is the relation between the molecules represented by the following two structures? OH H CH3 C 2H 5 H 3C C 2H 5 H Br H Br HO H  Solution Make two interchanges at each of the two chiral carbon atoms in second structure in such a way that CH3 — group is held vertically upward and C2H5 — group vertically downward. H CH3 HO CH3 Make two interchanges H OH at both the chiral carbon atoms H C2H5 Br H Br C 2H5 By doing so we get first structure. Thus, the two structures are identical.  Illustrative Example Write the R, S configuration of the following compound. Cl H H 3C CH 3 H Cl  Solution 1CH CH3 3 H Cl H 2 Cl C 3 Cl H C Cl H CH 4 3 CH 3 1 2 3 CHClCH3 Cl CH3 C 2 C 3 3 Cl CHClCH3 CH3 2 1 So configuration is 2S, 3S.  Illustrative Example Two isomeric alkenes A and B having molecular formula C5H9Cl on adding H2, A gives optically inactive compound, while B gives a chiral compound. What are the two isomers? Solution : Cl Cl H2 CH 3 — CH 2 — C CH — CH 3 CH 3 — CH 2 — CH CH2 — CH 3 3-chloro-2-pentene (A) Optically inactive Cl Cl H2 CH 3 — CH 2 — CH C — CH 3 CH 3 — CH 2 — CH 2 CH — CH 3 2-chloro-2-pentene (B) Optically active Therefore A is 3-chloro-2-pentene, B is 2-chloro-2-pentene.  D and L system This system has been used to specify the configuration at the asymmetric carbon atom. In this system, the configuration of an enantiomer is related to the two forms of glyceraldehyde were arbitrarily assigned the absolute configuration. CHO CHO H C OH OH C H CH2OH CH2OH (+) - glyceralodehyde (–) - glyceralodehyde D Configuration L Configuration  D and L system CHO CHO H OH OH H HO H H HO H OH OH H H OH OH H CH2OH CH2OH D - (+) - glucose L - (–) - glucose  Diastereomers Mirror plane CH3 CH3 H Cl Cl H c c c c H OH HO H CH3 CH3 Not mirror images. Not mirror images. Not mirror images. CH3 CH 3 Cl H H Cl c c c c H OH HO H CH3 CH3 Diastereomers: different physical properties  Meso-compounds Mirror plane CH3 CH3 H Cl Cl H c c c c Cl H Cl H CH 3 CH 3 Plane of CH3 CH3 symmetry H Cl Cl H c c c c Same molecule, H Cl Cl H Optically inactive. CH3 CH3  Illustrative Example What is the relationship between the molecule given below? HO COOH and HO COOH Solution : The two compounds are identical since they have a plane of symmetry.  Racemic mixture equal amounts of (+) and (-) enantiomers - rotation is 0 COOH COOH C H H C 21 21 [ ] D = +2.6° H3 C OH CH3 [ ] = -2.6° HO D Net rotation is zero.  Composition of a mixture of Enantiomers []sample optical purity % = x 100 []pure enantiomer enantiomeric excess (ee): Difference between the percent of 2 enantiomers in a mixture. [R] − [S] Enantiomeric excess =  100 [R] + [S]  Illustrative Example (+)-Mandelic acid has a specific rotation of +158°. What would be the observed specific rotation of a mixture of 25% (–)-mandelic acid and 75% (+)-mandelic acid? Solution : Specific rotation of the mixture  75   25  =  ( +158) +   ( −158)  100   100  = +79°  Resolution of racemic mixture (a) One strategy: convert enantiomeric pair into 2 diastereomers. diastereomers - different compounds, different physical properties. separate diastereomers remove reagent leaves pure enantiomers  Resolution by acid-base reactions H H3C H O H CH3 C Ph H N C H Pure-Sb H3C C C O H H N H F3C Ph H H H3C O C Ph O C H O H N H H3C C CH3 C C H CF CF3 O + 3 H + O O ----resolved---- C H C H3C H C Ph CH3 O CH3 CF3 CF3 C C H N H H3C O H O H CH3 H racemic O O F3C H N C H C C C O mix H H Ph H3C C F3C H  Illustrative Example How (+) and (–) lactic acid can be separated? Solution : (+) and (–) lactic acid on treatment with an optically active base such as (+) or (–) brucine to form diastereomers which have different-melting /boiling point and solubilities and hence can be separated. Class Exercise – 1 Select the most stable carbocation among the following. CH3 + + (a) CH (b) (C H ) 6 5 3 C CH3 + (c) CH3 CH2 C H2 (d) (CH3 )3 C + Solution C6H5 + C — C6H5 C+ C6H5 This carbocation is highly stabilised through resonance with three benzene rings. Hence answer is (b). Class Exercise - 2 Which of the following is an addition reaction? (a) CH CH CH OH− 3 3 ⎯⎯⎯⎯ → CH2 = CH CH3 Alcohol | Br h (b) CH3 CH3 + Cl2 ⎯⎯⎯ → CH3 CH2 Cl (c) CH3 CH2 Br + CN − ⎯⎯→ CH3 CH2 CN + Br − (d) CH3CH = CH2 + H Br ⎯⎯⎯ → CH3 CH CH3 | Br Solution – OH H3C — CH — CH3 H2C CH — CH3 Alcohol (Elimination) Br h H3C — CH3 + Cl2 CH3CH2Cl + HCl Substitution – H3C — CH2Br + CN H3C — CH2CN + Br– (Substitution) H3C CH CH2 + HBr H3C CH CH3 Br Addition Hence answer is (d). Class Exercise - 3 Which of the following is the most effective group in stabilizing a free radical inductively? (a) F (b) I (c) Br (d) Cl Solution Since free radical is electron deficient, any substituent with more electron releasing and less electron withdrawing ability will stabilize the radical inductively. The decreasing order of electronegativity of halogens is: F > Cl > Br > I Hence answer is (b). Class Exercise - 4 Which of the following is not a nucleophile? (a) CN– (b) BF3 (c) RNH2 (d) OH– Solution Among the following, BF3 is only electron deficient. Hence, it will not act as a nucleophile. Hence answer is (b). Class Exercise - 5 Which of the following is the correct order regarding –I effect of the substituents? (a) –NR2 > –OR > –F (b) –NR2 > –OR < –F (c) –NR2 < –OR < –F (d) –OR > –NR2 > –F Solution –I effect increases with electronegativity of atom. The decreasing order of electronegativity is F>O>N  The correct order for –I effect is –NR2 < –OR < –F Hence answer is (c). Class Exercise - 6 The least stable carbonium ion is (a) (b) + + H3 C C H2 C6 H5 — CH2 — C H2 (c) (d) + + C6 H5 — C H2 C6 H5 — C H — C6 H5 Solution Among the following, (a) is stabilised through +I effect and (b) is destabilized through –I effect of phenyl ring. Other two are stabilised through resonance. Hence answer is (c). Class Exercise - 7 Arrange the following ions in the decreasing order of stability. + C H2 CH3 CH3 CH3 + + + (a) (b) (c) (d) Solution +. It is a primary cation. CH2 Hence, minimum stability. CH3 CH3 + and are secondary cations. + (c) (b) Hence, stabilised through +I effect of –CH3 group which decreases with distance. (c) is more stable as compared to (b). (d) is most stable as it is tertiary cation and stabilised through +I effect of –CH3 group and hyperconjugation.  The order is (d) > (c) > (b) > (a) Class Exercise - 8 Arrange the following radicals in order of their decreasing stability CH3 C H2 , (CH3 )3 C, C6 H5 C H2 , CH2 = CH C H2 Solution Radicals are stabilized through electron releasing resonance and inductive effect. CH2 CH2 etc. More resonating structure H2C CH — CH2 H2C — CH CH2 Solution One resonating structure, although both are primary radicals. Among H3C — CH2 and (CH3)3C , later is a tertiary radical. Hence, more stable. The decreasing order of stability is C6H5CH2 > H2C CH — CH2 > (CH3)3C > H3CCH2 Class exercise 1 The hybridization of carbon atoms C — C single bond in vinylacetylene is (a) sp3 - sp3 (b) sp - sp2 (c) sp2 - sp (d) sp3 - sp Solution : 1 2 3 4 H2C CH — C CH Hence answer is (c). Vinylacetylene Class exercise 2 Allyl isocyanide has (a) 9  bonds and 4  bonds (b) 8  bonds and 5  bonds (c) 8  bonds, 5  bonds and 4 non-bonding electrons (d) 9  bonds, 2  bonds and 2 non-bonding electrons Solution: + – H2 C CH — CH2 — N C Allyl isocyanide The compound has 3  bonds and one lone pair, i.e. two non-bonding electrons. It also contains 9 - bonds. Hence answer is (d). Class exercise 3 Among the following which has the most acidic  -hydrogen? O O O (a) CH3CCH2CHO (b) CH3CCH2CCH3 O (c) CH CCH COOCH (d) CH3CHO 3 2 3 Solution O O  H3C — C — CH2 — CH one keto and one aldehydic carbonyl group. O O  H3C — C — CH2 C — CH3 two e-withdrawing keto groups. O O  H3C — C — CH2 C — OCH3 keto and ester group. O  H3C CH one aldehyde group. Solution Since e-withdrawing nature of C O gas varies as aldehyde > keto > ester Then most acidic a-H atom is present in O O H3C C — CH2 — CH Hence answer is (a).  Class exercise 4 The decreasing order of acidity among phenol, p- methylphenol, m-nitrophenol and p-nitrophenol is (a) m-nitrophenol, p-nitrophenol, phenol, p-methylphenol (b) p-nitrophenol, m-nitrophenol, phenol, p-methylphenol (c) p-methylphenol, phenol, m-nitrophenol, p-nitrophenol (d) phenol, p-methyl phenol, p-nitrophenol, m-nitrophenol Solution OH OH OH OH NO2 CH3 NO2 +I –I –I, –R Electron withdrawing groups increase acidic strength while electron donating group decreases the same. So the proper decreasing order of acidic strength is OH OH OH OH > > > NO2 Hence answer is (b). NO2 CH3 Class exercise 5 In the following compounds, the order of basicity is O N N N N H H H (I) (II) (III) (IV) (a) I > IV > II > I (b) II > I > IV > III (c) III > I > IV > II (d) IV > I > III > II Solution O N N N N sp3 sp2 sp3 H sp2 H H I II IV III Between I and IV, IV is less basic because of the –I effect of oxygen atom. II is more basic than III as the lone pair on N-atom in III is not available for protonation as it is involved in resonance. Therefore, the correct order is I > IV > II > III Hence answer is (a). Class exercise 6 Account for the order acidity in the following compounds. O (i) CH3CCH2COOH CH3CH2COOH (ii) HC CCH2 COOH H2 C = CHCH2 COOH Solution O   (i) H3C C CH2 COOH > CH3 CH2 COOH (a) (b) In compound (a), electron-withdrawing keto group increases the acidic strength by decreasing the O — H bond strength, while no such effect is there in compound (b). (ii) HC C CH2 COOH > CH2 CH CH2 COOH Carbon atoms attached to triple bond is sp hybridised and more electron-withdrawing than sp2 hybridised carbon atom. Hence, such order in acidic strength is observed. Class exercise 7 Which of the following two amines is more basic and why? CCl3CH2CH2CH2NH2 or CCl3CH2CH2NH2 Solution Electron-withdrawing groups decrease the charge density on N-atom of organic amines and hence decrease the basic strength. In Cl3C CH2 CH2 CH2 NH2, the electron withdrawing — CCl3 is far apart from — NH2 group as compared to Cl3C CH2 CH2 NH2. Hence, the former is more basic in nature. Class exercise 8 Do you think CH3CH CH2 OH can show tautomerism? Solution: Yes. The tautomeric form is CH3COCH3. Class exercise 9 Which one will be more acidic ? OH OH CH3 CH3 CH3 CH3 NO2 C N I II Solution Because of steric inhibition of resonance conjugate base of I will not be stabilised by resonance. But for II there is no such steric inhibition of resonance. Class exercise 10 Which hydrogen is maximum acidic in the following compound? H—C C NO2 O—H COOH Solution: Carboxylic hydrogen is maximum acidic.