SOME BASIC PRNICIPLES OF ORGANIC CHEMISTRY

Transcription

1 CS 1 Syllabus : SOME BASIC PRNICIPLES OF ORGANIC CHEMISTRY Tetravalency of carbon; shapes of simple molecules-hybridisation (s and p); Classification of organic compounds based on functional groups :- C = C -, - C? C - and those containing halogens, oxygen, nitrogen and sulphur; Homologous series; Isomerism - structural and stereoisomerism. Nomenclature (Trivial and IUPAC) Covalent bond fission - Homolytic and heterolytic; free radicals, carbocations and carbanions; stability of carbocation and free radicals, electrophiles and nucleophiles. Electronic displacement in a covalent bond - Inductive effect, electrometric effect, resonance and hyperconjugation. Common types of organic reactions - Substitution, addition, elimination and rearrangement. Chemical Bonding and Molecular Structure Kossel Lewis approach to chemical bond formation, concept of ionic and covalent bonds. Ionic Bonding; Formation of ionic bonds, factors affecting the formation of ionic bonds; calculation of lattice enthalpy. Covalent Bonding : Concept of electronegativity, Fajan s rule, dipole moment; Valence shell Electron pair Repulsion (VSEPR) theory and shapes of simple molecules. Quantum methanical approach to covalent bonding : Valanece bond theory - Its important features, concept of hybridization involving s, p and d orbitals; Resonance. Molecular Orbital Theory : Its important features, LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configuration of homonuclear diatomic molecules, concept of bond order, bond length and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications. Purification and Characterisation of Organic Compounds : Purification Crystallization, sublimation, distillation, differential extraction and chromatography - principles and their applications. Qualitative analysis - Detection of nitrogen, sulphur, phosphorus and halogens. Quantitative analysis (basic principles only) - Estimation of carbon, hydrogen, nitrogen, halogens, sulphur, phosphorus. Calculations of empirical formulae of molecular formulae; Numerical problems in organic quantitative analysis. Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

2 C1A CS C O N C E P T S (Isomerism) Organic compounds having same molecular formulae but differing from each other at least in some physical or chemical properties or both are known as isomers. C1B There are two mains types of isomerism : (i) (ii) Structural isomerism Space or stereoisomerism Chain Isomerism : This type of isomerism is due to difference in the arrangement of carbon atom constituting the chain i.e., straight or branched chain of carbon atoms. For e.g., n-butane and Iso-butane are chain isomer. i.e., CH CH and Position Isomerism : This arises due to the difference in the positions occupied by particular atoms or groups (substituents) in the same carbon chain or due to different position of double or triple bonds in the alkenes and alkynes. For e.g., C 3 H 8 O Functional Isomerism : Compounds having the same molecular formulae but different functional groups show functional isomerism. As the functional group largely determines the properties of compound, such isomers differ in their physical and chemical properties. For e.g., (1) Alcohols and Ethers (C n H n+ O) () Aldehyde, Ketones & Unsaturated Alcohols (C n H n O) (3) Carboxylic Acids and Esters (C n H n O ) (4) Oximes and Amides (C n H n+1 NO) (5) 1 0, 0 and 3 0 amines (C n H n+3 N) (6) Alkynes and Alkadiene (C n H n ) (7) Cyanides and Isocyanides (8) Nitroalkanes and Alkylnitrites Ring and Chain Isomer : This type of isomerism is due to different mode of linking of carbon atoms i.e., the isomers possess either open or closed chain isomers. For e.g., (1) Alkenes and Cyclo Alkanes (C n H n ) () Alkynes and Cyclo Alkenes (C n H n ) Tautomerism : This is a special type of functional isomerism where the functional isomers exist in equilibrium with each other. The two forms which are in equilibrium in each other are called tautomers. For e.g., (1) Keto & Enol are tautomeric forms. Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

3 CS 3 () Nitro and Aci (3) Oximes and Nitroso Enolisation is in order as follows : CHO < CO < COCH CHO < COCH CO Metamerism : It is the isomerism in the same homologous series. It is due to presence of different allkyl groups attahced to the same polyvalent functional group or atom. For e.g., (1) C 4 H 10 O C OC C 3 H 7 O Diethylether Methylpropyl ether () C 5 H 10 O C COC C 3 H 7 CO Diethyl ketone Methylpropyl ketone (3) C 4 H 11 N C NHC C 3 H 7 NH Diethylamine Methylpropyl amine Practice Problems : 1. The number of isomers of C 6 H 14 is An isomer of ethanol is methanol acetone diethyl ether dimethyl ether 3. The compound C 4 H 10 O can show metamerism position isomerism functional isomerism all the three 4. C 7 H 8 O shows how many isomers (having onebenzene ring) Keto-enol tautomerism is observed in : ( ) 3 CCHO 6. Maximum enolisation takes place of CO COCH CHO COCH CO = O [Answers : (1) b () d (3) d (4) d (5) b (6) d] C Stereoisomerism : They are of two types : (1) Geometrical isomerism : Such isomers, which possess the same molecular and structural formula but differ int he arrangement of atoms or groups in space due to hindered rotation around the double bond i.e. ( ) The isomers which has similar group on same side are called -cis isomer and similar groups on opposite sides are called as -trans isomer. However in case of all the four groups are different, for e.g., ( ), it is not possible Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

4 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road to decide -cis and -trans isomers. In such cases E and Z system of nomenclature is used. CS 4 () Optical isomerism : The compounds having similar physical and chemical properties but differing only in the behaviour towards the plane polarised light are called as optical isomers. On the basis of study of optical activity, the various organic compounds are divided into three types : The optical isomer with rotate the plane of polarised light to the right are called as dextrorotatory or d-form or indicated by + sign. The optical isomer which rotate the plane or polarised right to the left are called as leavoratotory or l-form on indicated by ve sign. The optical powers of the above two isomers are same i.e. of same magnitude but of positive sign. The equimolar mixture of two forms, therefore will be optically inactive. This mixture is termed is Racemic Mixture. Cause of Optical Activity or Optical Isomerism : A plane which divides an object into two symmetrical halves is said to have a plane of symmetry. An object lacking a plane of symmetry is called dissymmetric or chiral. A symmetric object is called as achiral. A dissymmetrical or chiral object can be defined as the one which is not super imposable on its mirror image. In order to exhibit optical activity an object must be chiral. The compounds which are mirror images of each other but not super imposable are called as enantiomers. Generally compunds having carbon atom to which four different groups are when attached show optical isomerism. Molecule having a symmetric carbon atom lacks in plane of symmetry. Thus compound which consist of at least one asymmetric carbon atom is capable of showing the phenomenon of optical isomerism. Exception : Following compound with a cumulative system of double bond and having no chiral carbon atom shows optical isomerims. but not the following compound i.e.,. (i) Number of possible stereoisomers in compounds containing different numbers of asymmetric carbons : When the molecule has no symmetry : n number of asymmetric carbon atoms. The number of d- and l-forms a = n The number of meso forms m = 0 The total number of optical isomer = a + m = n (ii) When the molecule has symmetry : n number of asymmetric carbon atoms which is even. The number of d- and l-form, a = n 1 The number of meso forms, m = n 1 The total number of optical isomer = a + m = n 1 n 1.

5 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road (iii) When the molecule has symmetry and n no. of asymmetric carbon atom which is odd : CS 5 The number of d and l-forms, a = n 1 n1 n1 The number of meso forms m. Total optical isomer = a + m = n 1. Diastereomers : Stereoisomers which are not mirror images of each other are called diastereomers. Practice Problems : 1. The molecular formula of a saturated compound is C H 4 Cl. This formula permits the existance of two functional isomers position isomers optical isomers cis-trans isomers. Number of optically active isomers of tartaric acid CH(OH)COOH CH (OH)COOH is : [Answers : (1) b () a] C3A Reagents : Following types of reagents are used (i) (ii) Electrophiles : Electrophilic reagents (electron loving) are those reagents that are positively charged or contain an electron deficient atom. i.e., whose octet is not complete. They are also known as Lewis Acid (they are electron acceptor). For e.g., NO +, Cl +, H +, SO 3, BF 3 etc. Nucleophiles : Nucleophilic reagents (nucleus loving) or electron repelling reagents are negatively charged or having neutral atom with lone pair. They are also called as Lewis base (electron pair donor). For e.g., OH, CN, NH 3, H O etc. C3B Types of Intermediates Formed in the Organic Reaction : (i) Carbocation or Carbonium Ion : The carbon with a positive charge and six electron pair in valence shell is known as carbocation. This carbon is sp hybridised. (ii) (iii) The carbocation can undergo rearrangemets reaction. Order of Stability of Carbocation : Benzyl, Allyl > 3 0 > 0 > 1 0 > +. Carbanion : The carbon with a negative charge having three bond pair and one lone pair is called as carbanion. This carbon is sp 3 hybridised. Order of stability of carbanion is : Benzyl, Allyl > > 1 0 > 0 > 3 0. Free Radical : The carbon containing single unpaired electron is called as carbon free radical. It is sp hybridised Order of Stability of Free Radical : Benzyl, Allyl > 3 0 > 0 > 1 0. (iv) Carbene : There is another class of intermediate called carbene, (methylene). It exists in two forms : Singlet Carbene : In which the unshared electrons are paired. i.e.,. It is sp hybridised Triplet Carbene : In which unshared electrons are not paired. i.e.,. It is sp hybridised

6 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road (v) Nitrene (vi) Benzyne Practice Problems : 1. Most-stable carbonium ion is CS 6 p NO C6H4 C H C 6H5 C H p Cl C 6H 4 C H. Arrange the following in the decreasing order of the free radical p CH 3O C6H 4 C H II > I > III I > II > III III > II > I I > III > II [Answers : (1) b () a] C4 Different Effects for the Displacement of Shared Pair of Electron are as follows : (i) Inductive Effect (ii) Resonance Effect (iii) Electromeric Effect (iv) Hyperconjugation Effect (i) Inductiv Effect : Electrone withdrawing and electron releasing group is attached to the carbon chain permanent polarity is induced on the carbon atom. This displacement of electrons towards one of the atom because of difference in electronegativity is known as inductive effect. This effect is of two types : Positive Inductive Effect (+I) : This is due to electron releasing group. It develops a negative charge on the carbon chain. +I effect decreases as we move away from the electron releasing group. Greater the size of alkyl group greater the electron releasing effect. Due to electron releasing group electron density increases hence basic nature increased, thus acidic decreases Negative Inductive Effect ( I) : This is due to electron withdrawing groups. It developes positive charge on the chain. I effect also decreases as we move away from the electron withdrawing group. After third carbon, this effect is negligible. I effect is in the order as follows : (ii) NO > F > COOH >Cl > Br > I > OH > C 6 Due to I effect electron density decreases hence basic nature is decreased and acidic nature is increased. Resonance Effect : The phenomena by which two or more structures involving identical positions of atoms can be written for a particular compound is called Resonance. The various structures are called as resonating structures. Resonating structure can be written for the following : Presence of conjugated system (Alternate double-single bond or alternate single-multiple bonds) for e.g.,

7 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 7 Conjugated system attached to electron deficient atom with vacant p-orbital, for e.g., Conjugated system between electron rich atom i.e., containing lone pair and alternate double bond for e.g., (In this case Cl is having I as well as +R effect) Resonance effect is of two types : +R Effect : Groups which can released or donate the electron pairs through resonance are called as +R effect. for e.g., OH, O, NH etc. R Effect : Groups which attrach or withdraw the electron through resonance are of (iii) (iv) R effect for e.g., NO, CN,, COOH etc. Electromeric Effect : It is a temperory effect, it comes into play instantaneously at the demand of attacking reagent and as soon as the attaching reagent is removed the original conditions restored. The effect involving the complete transfer of shared pair of -electrons to one of the atoms joined by the multiple bond (double or triple) at the requirement of attacking reagent is known as electromeric effect. It is of two types : +E Effect E Effect Hyperconjugation Effect : It is the conjugation between and bond, for e.g., This type of electron released by alkyl group attached to the unsaturated group is called hyper conjugation. There should be atleast one H atom at carbon atom w.r.t. (C = C) bond. Practice Problems : 1. Delocalised molecular orbitals are found in H HS CH 4 CO 3. The correct order of the increasing C O bond length of CO, CO 3, CO is CO 3 < CO < CO CO < CO 3 < CO CO < CO 3 < CO CO < CO < CO 3 3. Among the following which one is most basic NH 3 NH CH NH C 6 NH 4. Most acidic compound is COOH C 6 COOH O NC 6 H 4 CO H C 6 OH 5. In the following compounds phenol (I), p-cresol (II), m-nitrophenol (III) and p-nitrophenol (IV), the order of acidity is III > IV > I > II I > IV > III > II II > I > III > IV IV > III > I > II

8 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road 6. The most unlikely representation of resonance structures of p-nitrophenoxide ion is CS 8 7. Increasing values of dissociation constant K a of COOH (I), HCOOH (II), COOH (III) : I < II < III III < II < I I < III < II II < III < I 8. Give the order of the stabilities of the followings : b > a > c > d > e a > b > c > d > e e > b > a > c > d d > a > b > c > e 9. The acylium ion, the structure R C O : is more stable than R C O :.. As in R C O : octet of every atom is complete. In R C O : structure there are more covalent bonds than R C O : Both are correct.. Given statement is wrong as R C O : is more stable than R C [Answers : (1) d () d (3) c (4) c (5) d (6) c (7) c (8) a (9) c].. O : C5 C6A Chemical Bonding INTRODUCTION A molecule will only be formed if it is more stable, and has a lower enrgy than the individual atoms. Normally only electrons in the outermost shell of an atom are involved in forming bonds. We divide elements into three classes : (A) (B) (C) Ionic Bonding Electropositive elements, whose atoms give up one or more electrons easily. They have low ionisation potentials. Electronegative elements, which can gain electrons. They have higher value of electronegativity. Elements which have little tendency to lose or gain electrons. An ionic bond is formed when a metal atom transfers one or more electrons to a nonmetal atom. Na Na F e As a result of this transfer the metal atom becomes cation and the nonmetal the anion. F e

9 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 9 Cations and anions attain noble gas configuration with outer shell octet of electrons. In some cases cations may not have octet but different outershell. Properties of Ionic solids They are good conductor of electricity in fused state and in aq. solution. They are soluble in polar solvents and insoluble in non polar solvents. They have high m.p. and b.p. They have strong force of attraction between cation and anion (Coulombic force) z1z e F Dr Practice Problems : 1. Which of following elements has the strongest tendency to form electrovalent compound? Li Na Be Mg. Which of the following is more ionic? NaCl KCl MgCl CaCl [Answers : (1) b () b] C6B Solubility of Ionic Solids : z By the Coulombic force of attraction 1z e F. We can study the solubility of ionic solids in different Dr solvents. Greater the force of attraction between ions, smaller the tendency of the ions to go into solution and thus smaller the solubility. For a given solute, greater the value of D then smaller the force of attraction between positive and negative ions, hence greater the solubility. D(H O) > D( CH OH) > D( O ) hence solubility of ionic solid (say NaCl) in H O > CH OH > O Practice Problems : 1. MgSO 4 is soluble while BaSO 4 in insoluble in H O. This is because lattice energy of BaSO 4 is greater than MgSO 4 BaSO 4 is more covalent than MgSO 4 hydration energy of Mg + is greater than Ba + lattice energy of MgSO 4 is greater than BaSO 4 [Answers : (1) c] C7A C7B C7C Covalent Bonding : If duplet () or octet (8) is completed by sharing of electrons between two electronegative elements, the bond formed is called covalent bond. The Octet Rule : For many light atoms a stable arrangement is attained when the atom is surrounded by eight electrons the octet rule. (In case of H, duplet is completed) Exceptions to the Octet Rule : In so many cases, the octet rule is violated : In BeF, octet of Be is not complete, in BF 3, octet of B is not complete. Other examples are PCl 5, SF 6, IF 7 where centre atom is having more than eight electrons. Coordinate Bonding A covalent bond results from the sharing of pair of electrons between two atoms where each atom contributes one electron to the bond. It is also possible to have an electron pair bond where both electrons originate from one atom and none from the other. Such bonds are called coordinate bonds or dative bonds. Since in coordinate bonds, two electrons are shared by two atoms, they differ from normal covalent bonds. It is represented as Atom/ion/molecule donating electron pair is called DONOR or LEWIS BASE. Atom/ion/molecule accepting electron pair is called ACCEPTOR or LEWIS ACID () points donor to acceptor. NH 4 + : NH 3 has three (N H) bonds and one lone pair. In NH 4 + formation this lone pair is donated to H + (having no electron).

10 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 10 C8A C8B Valence Shell Electron Pair Repulsion (VSEPR) Theory (Gillespie Theory) The shape of the molecule is determined by repulsions between all of the electron pairs represent in the valence shell. Repulsion is in order : (lp lp) > (lp bp) > (bp bp) The magnitude of repulsion between pairs of electrons depends on the electronegativity difference between the central atom and the other atom. Double bond cause more repulsion than single bonds, and triple bonds cause more repulsion than a double bond. The Effect of Bonding and Lone Pairs of Geometry and Bond Angles Example Orbitals on Theoretical Bond Angle Distorted Geometry Central Atom Shape due to Repulsion BeCl Linear Linear BF 3 3 Plane triangle 10 0 Plane triangle SO 3 Plane triangle Angular CH 4 4 Tetrahedral Tetrahedral NH 3 4 Tetrahedral Trigonal pyramidal NF 3 4 Tetrahedral Pyramidal H O 4 Tetrahedral Angular F O 4 Tetrahedral 10 0 Angular H S 4 Tetrahedral 90 0 Angular PCl 5 5 Trigonal bipyramidal 10 0 and 90 0 Trigonal bipyramidal SF 4 5 Trigonal bipyramidal and Irregular tetrahedral CIF 3 5 Trigonal bipyramidal T-shaped I 3 5 Trigonal bipyramidal Linear SF 6 6 Octahedral 90 0 Octahedral BrF 5 6 Octahedral Square pyramidal XeF 4 6 Octahedral 90 0 Square planar Practice Problems : 1. Molecular shapes of SF 4, CF 4 and XeF 4 are the same, with, 0 and 1 lone pair of electrons respectively. the same, with 1, 1 and 1 lone pair of electrons respectively. different, with 0, 1 and lone pairs of electrons respectively. different, with 1, 0 and lone pairs of electrons respectively. [Answers : (1) d] C9 Isoelectronic Principle : Isoelectronic species usually have the same structure. This may be extended to species with the same number of valence electrons. Species CH 4, NH 4+, BF 4 CO 3, NO 3, SO 3 CO, N 3, NO + Structures tetrahedral planar triangle linear

11 CS 11 C10A Valence Bond Theory : The covalent bond is a region of high electron charge density (high electron probability) that results from the overlap of atomic orbitals between two atoms. In general, the greater the amount of overlap between two orbitals, the stronger the bond. For each bond there is a condition of maximum atomic orbital overlap leading to maximum bond strength at a particular internuclear distance (bond length). This is called Valence Bond Approach. This is a localised electron model of bonding. Most of the electrons retain the same orbital locations as in a separated atoms, and the bonding electrons are localised (fixed) in the region of atomic orbital overlap. In H molecules, H H bond is by s s overlapping In F molecule : F F bond is by axial overlapping of two p-orbitals. In O molecule : One bond is by axial overlap of p-orbitals and bond is by lateral overlap of p-orbitals. In N molecule : One bond is by axial overlap of p-orbitals and two bonds are by lateral overlap of p-orbitals. C10B Hybridisation : Hybridisation is defined as the concept of intermixing of orbitals of same energy or of slightly different energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane. New orbitals formed are called hybrid orbitals. Only the orbitals of an isolated single atom can undergo hybridisation. The hybrid orbitals generated are equal in number to that of the pure atomic orbitals which mix up. A hybdrid orbitals, like the atomic orbitals, cannot have more than two electrons of opposite signs. C10C Sigma and PI Bond There can be following types of overlaping along the axes (end to end) : (s - s) overlapping when s-orbital overlaping with another s-orbital (s - p) overlapping, (p can be p x or p y or p z ) (p x p x ) overlapping any of the hybrid orbitals overlaps with another hybrid orbitals or s or p orbital. Bond formed in the manner is called sigma () bond in which electron density is concentrated in between the two atoms, and on a line joining the two atoms. Double or triple bonds occur by sideways overlap of orbitals (like (p y p y ) and (p z p z ) orbitals) giving pi () bonds in which electron density also concentrates between the atoms, but on either side of the line joining the atoms. The shape of the molecule is determined by the bonds (and lone pairs) but not by the bonds. Pi bonds merely shorten bond length. Thus Practice Problems : 1. In which of the following molecules would you expect the nitrogen to nitrogen bond to be the shortest? N H 4 N N O 4 N O Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

12 CS 1. Allyl cyanide has 9 bonds and 4 bonds 9 -bonds, 3 bonds and one lone pair 8 bonds and 5 bonds 8 -bonds, 3 bonds [Answers : (1) b () b] C10D Hybrid Orbitals and their Geometric Orientation Hybrid Orbitals Orientation Example Predict Bond Angle Actual Shape sp (two) linear BeCl, CO linear C H sp (three) trigonal planar BF 3,C H trigonal planar SO, SO 3 sp 3 (four) tetrahedral CH tetrahedral NH trigonal pyramidal (due to lp bp repulsion) H O angular (V-shaped) due to lp lp and lp bp repulsion sp 3 d(five) trigonalbi PCl and 90 0 trigonal pyramidal pyramidal SF , 36 and irregular tetrahedral I linear sp 3 d octahedral SF octahedral Practice Problems : 1. + The hybridisation of atomic orbitals of N in NO +, NO 3 and NH 4 are, respectively sp, sp, sp 3 sp, sp 3, sp sp, sp, sp 3 sp, sp 3, sp. If a molecule MX 3 has zero dipole moment, the sigma bonding orbitals used by M are pure p sp hybrids sp hybrids sp 3 hybrids 3. The shape of sulphate ion is hexagonal square planar trigonal bipyramidal tetrahedral 4. + The hybridisation of atomic orbitals of nitrogen in NO +, NO 3 and NH 4 are sp, sp 3 and sp respectively sp, sp and sp 3 respectively sp, sp and sp 3 respectively sp, sp 3 and sp respectively 5. Specify the coordination geometry around and hybridization of N and B atoms in a 1 : 1 complex of BF 3 and NH 3 N : tetrahedral, sp 3 ; B : tetrahedral, sp 3 N : pyramidal, sp 3 ; B : pyramidal, sp 3 N : pyramidal, sp 3 ; B : planar, sp N : pyramidal, sp 3 ; B : tetrahedral, sp 3 [Answers : (1) a () c (3) d (4) b (5) a] C11 Dipole Moment : In H, there is no displacement of the electric charge due to same electron affinity of both H-atoms and the bond is non-polar. In HCl, the Cl atom has a more electron affinity than does the H atom. Electronic charge distribution is shifted towards the Cl atom. The H-Cl bond is said to be polar. The magnitude of the charge displacement in a polar covalent bond is measured through a quantity called the dipole moment µ. It is the product of the magnitude of charges () and the distance separating them. (Here the symbol () suggests a small magnitude of charge, less than the charge on an electron) If µ = d = esu and d = 1Å = cm Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

13 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road then µ = = esu cm In S.I. unit, 1 D = coulomb meter (when charge = C and d = m). In diatomic molecule,µ = d CS 13 but in polyatomic molecule with angle, resultant dipole moment is the vector summation of the vector moments. Also µ resultant cos, µ 1. Symmetrical molecules without lone pair will have µ = 0. observed valueof µ % ionic character in a molecule = 100 theoretical valueof µ Practice Problems : 1. A molecule possessing dipole moment is CH 4 H O BF 3 CO. Correct order of dipole moment for the following molecule is (I) (II) (III) I = II = III I < II < III I > II > III II < III < I 3. Which of the following hydrocarbons has the lowest dipole moment? C = C CH C = CH CH CH C = CH 4. Dipole moment is exhibited by 1, 4-dichlorobenzene 1, -dichlorobenzene trans-1, dichloroethene trans 1, -dichloro--butene [Answers : (1) b () d (3) b (4) b] C1A Hydrogen Bonding : Hydrogen bonding is said to be formed when slightly acidic hydrogen attached to a strongly electronegative atom such as F, N and O, is held with weak electrostatic forces by the non-bonded pair of electrons of another atom. The coordination number of hydrogen in such cases is two. It acts as a bridge between two atoms, to one of which it is covalently bonded, and to the other attached through electrostatic forces, also called Hydrogen Bond. Of all the electronegative donor atoms, only F, N and O atoms enter into stable hydrogen bond formation. The weak electrostatic interaction leading to the hydrogen bond formation is shown by dott (...) lines. Thus X H...Y represent hydrogen bonding between hydrogen and Y atom. C1B Intramolecular H-Bonding : This type of H-bonding occurs when polar H and electronegative atom are present in the same molecule.

14 C1C Intermolecular H-Bonding : CS 14 This type of H-bonding takes place between H and electronegative element present in the different molecules of the same substance (as in between H O and H O) or different substances (as in between H O and NH 3 ) e.g. In water molecules : Due to polar nature of H O, there is association of water molecules giving a liquid state of abnormally high b.p. C13A Molecular Orbital Theory (MOT) : According to Valence Bond Theory only the half-filled orbitals of one atom overlaps with the half filled orbitals of other atom to form the covalent bond. According to MOT, on the other hand, the all atomic orbitals of one atom overlap with the all the atomic orbitals of the other atom provided the overlapping orbitals are of the same symmetry and of similar energy. The resulting polynuclear molecular orbitals contain the all electrons of the molecule. C13B The basic principles of the MOT : 1. When nuclei of two atoms come close to each other, their atomic orbitals interact leading to the formation of molecular orbital. The atomic orbitals of the atoms in a molecule completely lose their identity after the formation of molecular orbital.. Each molecular orbital is described by a wave function, known as molecular orbital wave function. 3. The molecular orbital wave function is such that represents the probability density or electron charge density. 4. Each molecular orbital wave function ( ) is associated with a set of quantum numbers which determine the energy and shape of the molecular orbital. 5. Each is associated with definite energy value. 6. Electrons fill the molecular orbitals in the same way as they fill the atomic orbital following the aufbau principle, Pauli s exclusion principle, and the Hund s rule of maximum multiplicity. 7. Each electron in a molecular orbital belongs to all the nuclei present in the molecule. 8. Each electron moving in a molecular orbital has a spin of 1 or 1. The basic difference between an atomic orbital and molecular orbital is that while an electron in an atomic orbital belongs to or influenced by one positive nucleus only, an electron in a molecular orbital is influenced by all nuclei of atoms contained in a molecule. Linear combination of Atomic orbitals (LCAO) in case of H + 1. A linear combination of two atomic orbitals A and B leads to the formation of two molecular orbitals and. The energy E + of molecular orbital is lowe than either of E A and E B (energies of isolated atoms). It is therefore designated as bonding molecular orbital (BMO). 3. The energy E of molecular orbital is higher than either of E A and E B. It is therefore designated as antibonding molecular orbital (ABMO) 4. The extent of lowering of energy of the bonding molecular orbital is equal to the extent of increase of energy of antibonding molecular orbital. Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

15 CS 15 Energies of bonding and antibonding molecular orbitals. The order of energy of molecular orbital for lighter elements like boron, carbon and nitrogen are as follows : (1s) < *(1s) < (s) < *(s) < py = pz < px < *py = *pz < *px The order of energy of molecular orbitals for heavier elements after nitrogen are : (1s) < *(1s) < (s) < *(s) < px < py = pz < *py = *pz < *px Practice Problems : 1. Which of the following molecular orbital in N has least energy? p y s p z * p z [Answers : (1) c] C14 C15 Heteronuclear Diatomic Molecule : The same principle apply when combining atomic orbitals from two different atoms as applied when atoms are identical that is only atomic orbitals of about same energy can combine effectively They should have maximum overlap. They must have the same symmetry Since the two atoms are different, the energies of their atomic orbitals are slightly different. Electronic Configuration of Molecules : While discussing the electronic configuration of molecules, we shall frequently make use of a term called bond order. B.O. = 1 [Nb N a ] N b number of bonding electrons N a number of anti-bonding electrons If B.O. = 0, 1,, 3 so on it means that no bond is formed, one bond is formed, two bonds or three bonds are formed between the atoms respectively. 1. B.O, Bond dissociation energy and B.O. 1 Bond Length. Electronic configuration helps to predict the magnetic character of the molecule. If all the electrons in a molecule are paired they are diomagnetic and if unpaired electron is present they are paramagnetic. Practice Problems : 1. Among KO, AlO, BaO and NO +, unpaired electron in present in KO only + NO and BaO KO and AlO BaO only. + + When N is formed from N, bond order... and when O is formed from O, bond order... increases decreases increases, decreases decreases, increases 3. Which of the following has longest bond length? O + O O O 4. Which of the following has identical bond order? (I) CN (II) O (III) NO + (IV) CN + I, III II, IV I, II, III I, IV [Answers : (1) a () d (3) d (4) a] Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

16 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 16 INITIAL STEP EXERCISE 1. The IUPAC name of picric acid is, 4, 6-trinitrophenol, 4, 6-trinitro-1-hydroxy hexane, 4, 6-trinitro-1-hydroxy benzene 1, 3, 5-trinitro-6-hydroxy benzene. Vinyl alcohol and acetaldehyde are geometrical isomers keto-enol tautomers chain isomers position isomers 3. Number of isomers of molecular formula C H Br is Which of the following compounds exhibits stereoisomerism? -methylbutene-1 3-methylbutyne-1 3-methylbutanoic acid -methylbutanoic acid 5. Which of the following hydrocarbons has the lowest dipole moment? C C CH C CH CH = CH C CH 6. Polarisation of electrons in acrolein may be written as C H CH CH O 8. Markownikoff s rule is applicable to which of the following reactions C H 4 + HBr C 3 H 6 + Cl C 3 H 6 + HBr C 3 H 6 + Br 9. Which of the following statements is correct about a carbonium ion? it reacts with nucleophile it can undergo rearrangement it can eliminate an H + to form an olefin all are correct 10. What is the decreasing order of strength of the bases OH (I), NH (II), H C C (III), CH (IV)? IV > II > III > I III > IV > II > I I > II > III > IV II > III > I > IV 11. The intermediate during the addition of HCl to propene in presence of peroxide is C H CH Cl C H CH C H CH C H 1. In the following compounds, anisole (I), benzene (II) and nitro benzene (III), the case of reaction with electrophiles is II > III > I III > II > I II > I > III I > II > III 13. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is MeCOCl MeCHO MeCOOMe MeCOOCOMe 14. The IUPAC name of the compound having formula C H CH C H O C H C H CH O C H C H CH O 7. Which of the following is not an electrophile AlBr 3 BF 3 SnCl 4 NH 3 3-amino-hydroxy propionic acid -amino-propan-3-oic acid Amino hydroxy propanoic acid -amino-3-hydroxy propanoic acid

17 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS The IUPAC name of the compound, is -Bromo-3-chloro-4-oxopentanoic acid 3-Chloro--bromo-4-oxopentanoic acid 4-Carboxybromo-3-chloro--butanone None 16. The IUPAC name of the compound, 0. Stability order of... in order : I C H, CH = CH H II 3 C H ( CH ) 3 C CH III IV IV < III < II < I IV < II < I < III I < II < III < IV IV < I < III < II 1. Which shows aromatic character? C 1, 3-Dibromo-3-methylbutane 3-Methyl-1, -bromobutane 3-Methyl-1, 3-bromopropane None g of the organic compound on heating with NaOH gave NH 3 which is neutralised by 00 ml of 1 N HCl. Percentage of nitrogen is 1% 60% % 6.67%. Degree of unsaturation in is 18. C H, N H 3,OH, F in increasing pk b values are : C H3 N H F OH F OH NH CH 3 OH NH CH3 none is correct F 19. B.E. of C H bonds designated by I, II, III, IV, V in increasing order is : V < IV < III < II < I III < II < IV < V < I III < II < V < IV < I all equal The electronegativity of cesium is 0.7 and that of fluorine is 4.0. The bond formed between the two is covalent electrovalent coordinate metallic 4. In which of the following molecules the bond angle is maximum? CH 4 H O NH 3 CO 5. Total number of line pair of electrons in XeF 4 is CO is isostructural with SO HgCl C H SnCl 7. Which of the following is the most polar bond? Cl Cl N F C F O F

18 CS (A) Tetracyanomethane (B) Carbon dioxide (C) Benzene (D) 1, 3 butadiene Ratio of and bonds is in order A = B < C < D A = B < D < C A = B = C = D C < D < A < B 9. In sp 3 d hybridisation, the d orbital that participates in hyubridisation is dx y dz dxy dxz 30. C C 3 sigma single bond in vinyl acetylene is due to the overlapping of sp sp sp sp sp sp sp sp 31. In the compound of the type ECl 3, where E = B, P, As or Bi the angles Cl E Cl for different E are in the order B > P = As = Bi B > P > As > Bi B < P = As = Bi B < P < As < Bi 3. A -bonds is formed by the overlap of s-s orbital s-p orbital p-p orbital in end to end fashion p-p orbital in sidewise manner Mg C 3 reacts with water and forms propyne C 3 which has two sigma and two pi - bond three sigma and one pi - bond two sigma and one pi - bond two sigma and three pi - bonds 34. Element X is strongly electropositive and Y is strongly electronegative. Both are univalent. The compound formed would be X + Y X Y X Y + X Y 35. The correct order of the O O bond length in O, H O and O 3 is O 3 > H O > O O > H O > O 3 O > O 3 > H O H O > O 3 > O 36. The molecule which has highest dipole moment amongts the following is Cl CH Cl CHCl 3 CCl 4 FINAL STEP EXERCISE 1. Pairs of related chemical species are given below. Which pair is not related through resonance? CO CH COOC, C(OH) = CH COOC 3. Give the decreasing order of acidic characters of the following : (i) (ii). Which is maximum basic in nature? (iii) (iv) I > II > III > IV III > IV > I > II II > III > IV > I I > III > IV > II Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

19 CS Arrange the following according to their stability 9. Acidic character of the following in the increasing order is : (i) (ii) II > IV > I > III > V > VI II > IV > I > III > V > VI I > II > III > IV > V > VI VI > V > IV >III > II > I 5. The 1, addition product of free radical addition of CBrCl 3 to 1, 3-butadiene (iii) None 6. Arrange in decreasing order of basicity. m - NO - C 6 H 4 - NH C 6 NH I II p - NO - C 6 H 4 NH o - NO - C 6 H 4 NH III IV II > I > III > IV I > II > III > IV IV > III > II > I IV > II > III > I 7. Among 3-Butenoic acid and 3-Butynoic acid the greater acidic character is 3-Butenoic acid 3-Butynoic acid both none 8. Basic character of following in the decreasing order is : iii < ii < i ii < i < iii i < ii < iii ii < iii < i 10. Arrange the following compounds in order of increasing b.pts. (i) COCl (ii) ( CO) O (iii) CONH (iv) COOH COCl < COOH < ( CO) O < CONH COCl > COOH = ( CO) O = CONH COCl = COOH > ( CO) O > CONH COCl < COOH = ( CO) O > CONH (i) ( ) 3 N (ii) (iii) C N iii > ii > i i > ii > iii ii > i > iii i > iii > ii Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

20 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 0 ANSWERS (INITIAL STEP EXERCISE) ANSWERS (FINAL STEP EXERCISE) 1. a. b 3. c 4. d 13. a 14. d 15. a 16. a 5. c 6. c 7. c 8. a 1. a. b 3. c 4. a 5. a 6. a 7. b 8. b 9. c 10. a 5. b 17. c 9. b 6. a 18. a 30. d 7. d 19. c 31. b 8. c 0. a 3. d 9. d 1. b 33. a 10. a. c 34. a 11. b 3. b 35. d 1. d 4. d 36. b AIEEE ANALYSIS [00] 1. Arrangement of ( ) 3 C, ( ) CH, CH When attached to benzyl or an unsaturated group in increasing order of inductive effect is ( ) 3, C < ( ) CH < CH CH < ( ) CH < ( ) 3 C ( ) CH < ( ) 3 C < CH ( ) 3 C < CH < ( ) CH. Mg Br is an orgno metallic compound due to C-H bond C-Mg bond Mg-Br bond C-Br bond 3. In which of the following species the inter atomic bond angle is ? NH 3 (BF 4 ) 1 (NH 4 ) +, BF 3 NH 3 BF 4 (NH ) 1 BF 3 4. In which of the following species is the underlined carbon having sp 3 hybridisation? C OOH C O CH = C H 5. Racemic mixture is formed by mixing two chiral compounds meso compounds isomeric compounds optical isomers 6. A similarity between optical and geometrical isomerism is that they have no similarity both are included in stereo isomerism if in a compound one is present then so is the other each form equal number of isomers for a given compound 7. Which of the following does not show geometrical isomerism? 1, 4-dichloro--pentene 1, 1-dichloro-1-pentene 1, 3-dichloro--pentene 1, -dichloro 1-pentene C H OH

21 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 1 8. Which of the following statements is true? In covalency transference of electron takes place Chemical bond formation takes place when forces of attraction overcome the forces of repulstion absolutely pure water does not contain any ions HF is less polar than HBr. 9. Which of the following compounds has wrong IUPAC name? CH CH COO CH ethyl butanoate -methyl 3-butanol methyl 3-pentanone 10. In XeF, XeF 4, XeF 6 the number of lone pairs on Xe are respectively 3,, 1 4, 1, 1,, 3, 3, 1 3 methyl 1-butanal 11. In the anion HCOO the two carbon-oxygen bonds are found to be of equal length. What is the reason for it? The anion is obtained by removal of a proton from the acid molecules Electronic orbitals of carbon atom are hybridised The C = O bond is weaker than C O bond The anion HCOO has two resonating structures 1. Among the following four structures I to IV, (I) (II) AIEEE ANALYSIS [003] 13. The general formula C n H n O could be for open chain dialdehydes diketones carboxylic acids diols 14. The IUPAC name of COCH ( ) is 3-methyl--butanone Isopropyl methyl ketone -methyl-3-butanone 4-methylisopropyl ketone 15. The correct order of increasing basic nature for the bases NH 3, NH and ( ) NH is NH < ( ) NH < NH 3 NH < NH 3 < ( ) NH ( ) NH < NH 3 < NH NH 3 < NH < ( ) NH 16. Which one of the following compounds has the smallest bond angle in its molecule? SO OH (III) SH NH Which one of the following pairs of molecules will have permanent dipole moments for both members? (IV) only II and IV are chiral compounds all four are chiral compounds only I and II are chiral compounds only III is a chiral compound SiF 4 and NO NO and CO NO and O 3 SiF 4 and CO 18. The pair of species having identical shapes for molecules of both species is CF 4, SF 4 XeF, CO BF 3, PCl 3 PF 5, IF 5

22 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS AIEEE ANALYSIS [004/005/006] 19. The correct order of bond angles (smallest first) in H S, NH 3, BF 3 and SiH 4 is 5. The IUPAC name of the is H S < SiH 4 < NH 3 < BF 3 NH 3 < H S < SiH 4 < BF 3 H S < NH 3 < SiH 4 < BF 3 H S < NH 3 < BF 3 < SiH 4 [004] 0. The bond order in NO is.5 while that in NO + is 3. Which of the following statements is true for these two species? bond length in NO + is greater than in NO bond length in NO is greater than in NO + bond length in NO + is equal to that in NO bond length is unpredictable [004] 1. Which one of the following has the regular tetrahedral structure? XeF 4 SF 4 BF 4 [Ni(CN) 4 ] (Atomic nos. : B = 5, S = 16, Ni = 8, Xe = 54) [004]. The maximum number of 90 0 angles between bond pair-bond pair of electrons is observed in dsp 3 hybridization sp 3 d hybridization dsp hybridization sp 3 d hybridization [004] 3. Which one of the following has largest number of isomers? [Ru(NH 3 ) 4 Cl ] + [Co(NH 3 ) 5 Cl] + [Ir(PR 3 ) H(CO)] + [Co(en) Cl ] + (R = alkyl group, en = ethylenediamine) [004] 4. The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is Fe 4 [Fe(CN) 6 ] 3 Na 3 [Fe(CN) 6 ] 3, 3-dimethyl-1-hydroxy cyclohexane 1, 1-dimethyl-3-hydroxy cyclohexane 3, 3-dimethyl-1-1-cyclohexanol 1, 1-dimethyl-3-cyclohexanol [004] 6. Which one of the following does not have sp hybridised carbon? acetone acetic acid acetonitrile acetamide [004] 7. Which of the following will have a mesoisomer also? -chlorobutane, 3-dichlorobutane, 3-dichloropentane -hydroxypropanoic acid [004] 8. Which of the following compounds is not chiral? 1-chloropentane -chloropentane 1-chloro--methyl pentane 3-chloro--methyl pentane [004] 9. Which one of the following species is diamagnetic in nature? He H H H [005] 30. Due to the presence of an unpaired electron, free radical are Chemically reactive Chemically inactive Anions Cations [005] Fe(CN) 3 Na 4 [Fe(CN) 5 NOS [004]

23 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS The molecular shapes of SF 4, CF 4 and XeF 4 are the same with, 0 and 1 lone pairs of electrons on the central atom, respectively the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively different with 0, 1 and lone pairs of electrons on the central atom, respectively different with 1, 0 and line pairs of electrons on the central atom, respectively [005] 3. The number and type of bonds between two carbon atoms in calcium carbide are One sigma, one pi One sigma, two pi Two sigma, one pi Two sigma, two pi [005] 33. Among the following acids which has the lowest pka value? COOH HCOOH ( ) CH COOH CH COOH [005] 34. Amongest the following the most basic compound is benzylamine aniline acetanilide p-nitroaniline [005] 35. Which types of isomerism is shown by, 3-dichlorobutane? Diastereo Optical Geometric Structural [005] 36. Which of the following molecules/ions does not contain unpaired electrons? O O B N + [006] 37. The IUPAC name of the compound is 1-bromo-3-chlorocyclohexene -bromo-6-chlorocyclohex-1-ene 6-bromo--chlorocyclohexene 3-bromo-1-chlorocyclohexene [006] 38. The correct order of increasing acid strength of the compounds (A) CO H, (B) MeOCH CO H, (C) CF 3 CO H, and (D) A < D < C < B B < D < A < C D < A < C < B D < A < B < C is [006] 39. The increasing order of stability of the following free radicals is ( CH3) C H (CH3) 3 C (C6H5) 3 C (C6H5) C H ( CH3) C H (CH3) 3 C (C6H5) C H (C6H5) 3 C ( C6H5) 3 C (C6H5) C H () 3 C () C H ( C6H5) C H (C6H5) 3 C (CH3) 3 C (CH3) C H [006]

24 CS 4 AIEEE ANALYSIS [007] 40. Which of the following hydrogen bonds is the strongest? O H... O O H... F O H... N F H... F 41. The IUPAC name of is 5, 5-diethyl-4, 4-dimethylpentane 3-ethyl-4, 4-dimethylheptane 1, 1-diethyl-, -dimethylpentane 4, 4-dimethyl-5, 5-diethylpentane 4. Which of the following species exhibits the diamagnetic behaviour? O NO O O Which of the following molecules is expected to rotate the plane of plane-polarised light? 44. In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed? O O + N N + C C + NO NO Which one of the following conformations of cyclohexane is chiral? Chair Boat Twist boat Rigid ANSWERS AIEEE ANALYSIS 1. b. b 3. a 4. b 5. d 6. b 7. a 8. b 9. c 10. a 11. d 1. c 13. c d 16. c 17. c 18. b 19. a 0 b 1. c. d 3. d 4. a 5. c 6. c 7. b 8. a 9. b 30. d 31. d 3. b 33. b 34. b 35. c 36. b 37. d 38. d 39. b 40. d 41. b 4. c 43. c c Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

25 Einstein Classes, Unit No. 10, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road CS 5 TEST YOURSELF 1. The IUPAC name of is 3-ethyl-4, 4-dimethylheptane 1, 1-diethyl-, -dimethylpentane 4, 4-dimethyl-5, 5-diethylpentane 5,5-diethyl-4, 4-dimethylpentane. The increasing order of stability of the following free radicals is ( CH3 ) CH (CH3 ) 3 C (C6H5 ) CH (C6H5 ) 3 C ( C6H5 ) 3 C (C6H5 ) CH (CH3 ) 3 C (CH3 ) CH ( C6H5 ) CH (C6H5 ) 3 C (CH3 ) 3 C (CH3 ) CH ( CH3 ) CH (CH3 )C (C6H5 ) 3 C (C6H5 ) CH 3. Increasing order of stability among the three main conformations of -fluoroethanol is eclipse, gauche, anti gauche, eclipse, anti eclipse, anti, gauche anti, gauche, eclipse 4. Which type of isomerism is shown by, 3-dichlorobutane? Geometrical Structural Optical Diastereo 5. Which one of the following does not have sp hybridized carbon? Acetone Acetic acid Acetonitrile Acetamide 6. Which of the following will have a meso-isomer also? -chlorobutane,3-dichlorobuane,3-dichloropentane -hydroxypropanoic acid 7. Which of the following pair of compounds are not matamers? OCH CH and CH OCH CH OCH and OCH( ) NHCH CH and CH NHCH NHCH CH and NHCH( ) 8. Which of the following molecule has the maximum number of stereoisomers? 9. H 3 CO show which type of effect. + R and + I + R and I R and + I R and I 10. Triplet carbene is hybridised of sp sp sp 3 dsp 1. a. a 3. a 4. c 5. c 6. b 7. d 8. d 9. b 10. a ANSWERS