Recommended Content

Page 1 :
GENERAL ORGANIC, , CHEMISTRY, THEORY, NCERT SOLUTIONS, NCERT EXEMPLAR, WORK, SHEETS, FIND ALL YOUR QUESTIONS HERE, BEFORE GETTING THEM, IN YOUR EXAM ……...

Page 2 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , BASIC PRINCIPLES & TECHNIQUES IN, ORGANIC CHEMISTRY (GOC), , Introduction, ‘The branch of chemistry dealing with these compounds, which are widely distributed in, nature and play an important role in our daily lives’, is called organic chemistry., , In earlier period of development of chemistry, compounds were classified as follows, 1. Organic compounds derived from ‘living matter’ (plants and animals)., 2. Inorganic compounds prepared from ‘non-living matter’ (mineral sources)., Berzelius, a Swedish chemist proposed the mistaken notion that a ‘vital force’ present in, living matter was essential for the synthesis of organic compounds. However, the synthesis, of urea an organic compound present in urine, from ammonium cyanate, an inorganic, compound by Frederic Wohler in 1828 effectively destroyed the myth of organic compounds, being associated with a ‘vital force’., , Soon afterwards the pioneering work of Herman Kolbe who synthesized acetic acid and of, Berthelot who synthesized methane showed conclusively that organic compounds are, essentially the compounds formed by carbon with itself and other elements and that they, can be synthesized in a laboratory as easily as inorganic compounds., O], CH 3 − CHO ⎯[⎯→, CH 3 − COOH, , Acetaldehyde, , Acetic acid, , (First organic compound synthesized from its elements), , The chemistry of hydrocarbons and their derivatives constitutes organic chemistry., The number of organic compounds available today is more compare to total inorganic, compounds of all elements except carbon. This is due to unique catenation property of, Carbon., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 1 of 115

Page 3 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Some properties of Carbon, • Catenation: is the property of an element where a large, number of its own atoms join together through covalent, bonds. Due to which it forms single as well as multiple, covalent bonds with other carbon atoms. It is further, supplemented by the fact that it also forms covalent bonds, with atoms of other elements like hydrogen, oxygen,, nitrogen, Sulphur, phosphorus and halogens in a variety of, ways (i.e., single and multiple bonds)., This property gives a scope for the Chemists to synthesize new compounds., • Tetravalency of carbon atom, The atomic number of carbon is 6 and it has four electrons in its valence shell. In order to, acquire a stable noble gas configuration, it can share its 4 electrons with the electrons of its, atom or electrons of other atoms to form four covalent bonds. The bonds can be sigma(σ) or, pi(π)., , 2s2, , 2p2, , 2s1, , 2p3, , energy, Ground state, , Alkanes → CnH2n+2, , Alkenes→CnH2n, , Excited state, , Alkynes→CnH2n–2, , As per Lewis theory, Carbon can easily form tetra-bonds with other carbon atoms & other atoms. C-C,, C=C, C≡C exists in nature but not Carbon-Carbon tetrabonds., Carbon can form single, double or triple bond (covalent). When a single bond is formed between carbon, atoms (or any atoms), energy is released. Changing the molecule's electron arrangement to make a, double bond releases more energy, but not as much as when the first bond was made. Making a triple, bond again releases energy, but again to a lesser degree. However, when a fourth bond is attempted,, the overcrowded electrons between the atoms resist the change so strongly that it releases little, if, any, energy. This arrangement is therefore unstable, so a quadruple bond will be essentially impossible., , • Hybridization in carbon compounds, Hybridization is defined as intermixing of degenerate orbitals (orbitals at nearly same, energy) to produce entirely equivalent number of new orbitals of same energy, identical, shapes and symmetrically disposed in planes. The orbitals formed are called hybrid, orbitals.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 2 of 115

Page 4 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , ➢, ➢, ➢, ➢, , The orbitals of an isolated atom can undergo hybridization., Numbers of hybrid orbitals generated are equal to number of contributing atomic orbitals., The hybrid orbitals orient in the space providing definite geometry to molecule or ion., Like atomic orbital, a hybrid orbital cannot have more than two electrons of opposite spins., In diamond carbon is sp3 hybridized and in graphite carbon is sp 2 hybridized., There are three types of hybridization,, , (i) sp3 hybridization (contain saturated organic compounds with only single covalent bonds), (ii) sp2 hybridization (here organic compounds having carbon atoms linked by double bonds), (iii) sp hybridization (here organic compounds having carbon atoms linked by a triple bonds)., Type of hybridization, , sp3, , sp2, , sp, , Number of orbitals used, , 1s and 3p, , 1s and 2p, , 1s and 1p, , Number of unused p-orbitals, , Nil, , One, , Two, , Bond, , Four -, , Three -, One -, , Two -, Two -, , Bond angle, , 109.5, , 120, , 180, , Geometry, , Tetrahedral, , Trigonal planar, , Linear, , % s-character, , 25 or 1/4, , 33.33 or 1/3, , 50 or 1/2, , Prediction of hybridization- It can be done by two methods,, (i) First Method: In this method hybridization can be known by the number of  − bonds, present on that particular atom., Number of – bond/s, , 0, , Type of hybridization, , (i), , sp, , CH 3 − CH = CH − CH 2 − C  N, , ,  , , , 2, sp sp, sp 2 sp3, sp3 sp, , 2, 2, , sp, , (ii), , O, ||, CH3 − CH = CH −C − CH 3, ,   , , 2, sp 2 sp 2 sp3, sp3 sp, , (iii), , 1, 3, , (iv), , sp, , CH 2 = C = CH 2,  , , 2, sp 2 sp sp, , HC  C − CH = CH 2, ,  , , sp sp sp 2 sp 2, , (ii) Second Method (Electron pair method): The hybridized state of an atom of a molecule, or an ion or radical can be predicted by calculating number of orbitals or electron pairs, involved in hybridization (H) which is evaluated as follows., H = (number of  bonds formed with adjected atom/s + number of lone pairs of electrons, , 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 (do not include  electron pairs)., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 3 of 115

Page 5 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), Value of H, , 2, , 3, , 4, , Hybridization, , sp, , sp2, , sp3, , Structure, , linear, , trigonal planar, , tetrahedral, , Central atom, First atom, , H, , H, C=C, , H, , Second atom, , H, bp = 3, , Third atom, , Central atom, 1, , 2, , H − C  C− H, , bp = 2, , 1, H, , 2, 3, , bp = 3, , Number of lp’s can be determined as follows,, (a) If C has  - bonds or positive charge or odd electron, than lp on C will be zero., (b) If carbon has negative charge, then lp will be equal to one., Example:, , , CH 2 = CH, , bp = 2, lp = 0, ep = 2,sp, , , , CH  C, , bp=1, lp = 1, ep = 2,sp, , , , CH 2 = CH, , bp = 2, lp = 1, ep = 3,sp 2, , •, , CH 2 = C − CH 3, |, CH 3, bp = 3, lp = 0, ep = 3,sp 2, , , , CH3 − CH − CH3, , bp=3, lp = 1, ep = 4,sp3, , Pitfall - It may be noted that, the hybridized state of a heteroatom containing lone, pair electrons or carbanion or free radical bonded in conjugation with multiple bond/s is, sp2. For example, hybridized state of allylic or benzylic carbanions or free radicals or, atoms of aromatic ring is sp2., Problem 1: How many sigma () and pi () bonds are present in each of the following, molecules?, a) CH2 = C = CH2, b) CH3 – CH = CH – C  C – CH3, Solution: Every single bond is a  bond; every double bond contains one −bond and one, −bond while every triple bond consists of one − and two −bonds. Thus, a) C−C : 2, C−H : 4, C−C : 2, b) C−C : 5, C−H : 8, C−C : 3, Problem 2: Predict the shape of each of the following molecules., a) HC  CH, b) CH3Cl, c) H2C = O, 3, Solution: sp hybridization can always be correlated to tetrahedral shape, sp2 to trigonal, planar and sp to linear. Thus,, a) sp hybridized carbon atoms, linear, b) sp3 hybridized carbon, tetrahedral, c) sp2 hybridized carbon, trigonal planar, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 4 of 115

Page 7 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Structural representation of Organic Compounds, Structural formula (structure) is the sequence in which different atoms constituting the, molecule are bonded to one ano0ther. Structures of molecules of organic compounds can be, described in various ways. The most common types of representations are:, 1) Lewis structure (or electron dot structure) Here dots are used to represent all of the, valence electrons of all the bonded atoms in the molecule:, H H, , H, H C H, , H C, , C H, , Ethyne, , Ethene, , H, , H C C H, , Writing dot structure is tedious and time–consuming. The other representations are more, convenient and are, therefore, more often used., 2) Dash structural formula The Lewis structure can be simplified by representing a shared, electron pair by a ‘stick’ (dash, −) between the bonded atoms. When there is one dash, between two atoms, the atoms are said to be bonded by a single covalent bond. A double, covalent bond, in which two pairs of electrons are shared, is shown by two dashes between, the atoms. A triple bond is represented by three dashes between the atoms., The valence electrons that are not included in covalent bonds are called nonbonding, electrons (lone pairs). These are assigned to specific atoms and are represented by dots, drawn next to the symbols for these atoms., Lone pair of electrons on hetero atoms (e.g. oxygen, nitrogen, sulphur, phosphorus, halogens), may or may not be shown. Such structural formulae which focus only on the valence, electrons involved in bond formation are called complete structural formulae., H H, H H, | |, |, |, H − C− C− H, H − C = C− H, H−C  C −H, | |, H H, Ethane, Ethene, Ethyne, H, H, H, H, |, |, |, |, H − C− Cl or H − C− Cl, H − C− O− H or H − C− O − H, |, |, |, |, H, H, H, H, Methyl chloride, Methyl alcohol, chloromethane, methanol, , 3) Condensed structural formula Complete structural formulae can be shortened by leaving, out some or all of the covalent bonds and by denoting the number of identical groups, attached to an atom by a subscript. The resulting description of the molecule is called a, condensed structural formula. Thus,, CH3CH3, , H2C = CH2, , HC  CH, , CH3CH2Cl, , or, , or, , or, , or, , C2H6, , C2H4, , C2H2, , C2H5Cl, , Ethane, , Ethene, , Ethyne, , Ethyl Chloride, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 6 of 115

Page 9 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Ring or cyclic compounds - An organic compound in which carbon atoms are not bonded in, chain but are bonded in closed structures called rings are known as cyclic compounds. Such a, compound containing one or more rings is represented by drawing the suitable ring (polygon), without indicating the carbon and hydrogen atoms. The corner of the polygon denotes a, carbon atom and its sides represent a carbon – carbon bond. An atom or a group of atoms, (other than hydrogen) bonded to the carbon is however shown in the structure., Bond–line formulae of some cyclic compounds are:, CH2, , CH2, , H2C, , H2C, , CH2, , CH2, , H2C, , CH2, Cyclobutane, , Cyclopropane, , OH, , H2C, , CH2, , H2C, , CH2, , Cyclopentane, , CO2 H, , OH, , CH, , CO2 H, , CH, , H2C, , CH2, , H2 C, , CH2, , H2C, , CH2, , H2 C, , CH2, , CH2, Cyclohexanol, , CH2, Cyclohexane carboxylic acid, , Multiple bonds are also indicated in bond-line formulae., For example:, H3 C, , CH, C, , CH3, , H2 C, , CH3, , CH2, , ., , CH CH 2 OH, , OH, , Exp 1: Convert each of the following Lewis structures into complete structural formulae:, O, H, a), , H, , C, , O, , H, , b), , H, , C, , N, , c), , H C, , Cl, , H, Solution: Use the dash to represent the shared electron pair., a), , O, ||, H − C − O − H or, , O, ||, H − C− O − H, , b) H – C  N: or H – C  N, , H, |, c) H − C − Cl or, |, H, , H, |, H − C− Cl, |, H, , Exp2: Convert each of the complete structural formulae into condensed formulae., a), , H H, | |, H − O − C− C− N− H, | | |, H H H, H O H, c) H − C| − C|| − C| − H, |, |, H, H, , b), , H, H H H H, |, | | | |, H − C− C  C − C − C − C − C − H, |, | | | |, H, H H H H, H H H H, d) H − C| − C| − C| − C| − N = O, | | | | , H H H H O, , Solution: Omitting some or all of the dashes and indicating the number of identical groups by, a subscript we get, a) HO(CH2)2NH2, b) CH3C  C(CH2)3CH3, c) H3CCOCH3 d) CH3(CH2)3NO2,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 8 of 115

Page 11 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), ., ., , step 2, c) (CH3)2CHCH3CH(CH2)3NO2 ⎯⎯⎯, ⎯, →, , NO 2, , ., CH2, , NC, CH, , d) (CN)2CHCH2COCl ⎯⎯⎯→, step1, , CN, , Cl, C, O, , step 2, , Cl, , NC, , C, CN, , O, , Exp 6: Draw all possible bond-line formulae for a cyclic compound, C5H10., Solution: Start with the maximum number of carbon atoms in the ring and move towards the, ring of 3 carbon atoms. Explore all sorts of possibilities on this route:, , Three-dimensional representation of organic molecules, None of the formulae that we have described so far conveys any information about how the, atoms of a molecule are arranged in space. Shape or the three-dimensional (3-D) structure, of organic molecules can be described on a paper (two-dimensional) by exploring certain, conventions. For instance, by using solid, and dashed wedge, formula the 3–D, image of an organic molecule can be perceived on a two-dimensional paper., In solid-wedge and dashed-wedge description the solid-wedge denotes a bond projecting, out of the plane of the paper towards the viewer. The dashed-wedge depicts the bond, projecting behind the plane of the paper and going away from the viewer. Both the wedges, are drawn in such a way that the broad end of the wedge is near the viewer. The other two, bonds lying in the plane of the paper are shown by using a normal line ( ⎯ )., Let’s consider the wedge- and dashed-wedge representation of methane (CH4) molecule:, H, , H, , H, H, , H, , or, , H, H, , H, , The two carbon-hydrogen bonds represented by normal lines are in the plane of paper,, whereas the carbon-hydrogen bond represented with a solid wedge is aimed to be in front of, the plane of paper. The hydrogen bonded to carbon by dashed wedge is intended to be, behind the plane of paper. Note that the carbon atom is lying in the plane of paper., Wedge and dashed-wedge formulae are an important tool for clearly showing three, dimensions., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 10 of 115

Page 12 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Classification of organic compounds, , The compounds with C and H are called hydrocarbons. Organic compounds are considered as, the derivatives of hydrocarbons obtained by replacing H-atoms by other atoms or groups., Based on C - skeleton, organic compounds divided as below,, 1. Acyclic compounds - The open chain organic compounds are called acyclic compounds., The carbon chains may be linear or branched., The open chain compounds are classified into two groups., a. Saturated compounds: Open chain compounds in which carbon atoms of parent chain are, bonded with single bonds are called saturated acyclic compounds., Example:, O, CH3CH2CH2CH3, CH3, ||, |, CH, −, CH, −, CH, 3, 3, n–Butane, H − C− H, |, CH3, Isobutane, , Formaldehyde, , CH3 −C− OH, |, CH3, , tert-Butyl alcohol, , b. Unsaturated compounds: The open chain organic compounds in which parent chain contains, one or more carbon-carbon double bonds or triple bonds are called unsaturated acyclic, compounds., CH3, CH2 CH2, ethylene, , CH3CH CH2, propene, ,  ACTIVE SITE EDUTECH , , CH3CH CHCH3, 2-butene, , CH3C CHCH3, 2-methyl-2-butene, , CH3, CH3 CH C CH, 2-methyl-1-butyne, , CONTACT: , , Page 11 of 115

Page 16 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Homologous Series, A series of structurally related organic compounds with same functional group can be, represented by a general formula and differ by -CH2 group is called homologous series., The individual members of such series are called homologue and this phenomenon is known, as homology., , Characteristics of a homologous series are as follows., • Each homologues series can be represented by general molecular formula., Exp - General molecular formula of alkane is CnH2n+2. Molecular formula CnH2nO2, represents alkanoic acids (carboxylic acids) and alkyl-alkanoates (esters)., • All members of a given homologous series (homologues) possess the same functional group., • The successive members of a homologous series differ by a -CH2- group or by mass units., • Homologues of a homologous series can be prepared by general methods., • The physical properties such as boiling point, melting point, density, etc., of the members, of a homologues series show a regular gradation with increase in molecular mass., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 15 of 115

Page 17 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , •, , Chemical properties of the members of a homologous series are similar though the first, homologue may vary considerably from the rest of the homologues., , Nomenclature of Organic Compounds, The nomenclature deals with the naming of millions of organic compounds. The following, systems are employed., 1. Trivial system (common system), It is the oldest system of naming organic compounds. In the early stages of the development, of organic chemistry, organic compounds were named after the source from which they were, first isolated. Generally, the names chosen had Latin or Greek roots. The following illustrations, justify the statement., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 16 of 115

Page 19 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), Type of carbon chain, , root word, , primary suffix, , generic name, , Saturated (C-C single bonds), , alk, , ane, , alkane, , Unsaturated [C=C bond], , alk, , ene, , alkene, , Unsaturated [−CC−], , alk, , yne, , alkyne, , Note: A carbon-carbon double bond or triple bond must be included in parent chain even if it, contains less number of carbon atoms., • Secondary suffix added next to the primary suffix to indicate the presence of a functional, group in organic compounds which determines the class of organic compounds., Secondary suffix of a few functional groups are given., Functional group, , Structure, , Secondary suffix, , Class of organic compounds, , Hydroxyl, , –OH, , - ol, , Alcohols, , - al, , Aldehydes, , - one, , Ketones, , -oic acid, , Carboxylic acids, , alkylalkanote, , Esters, , - amine, , Amines, , Aldehydic, , O, C, , H, , Ketonic, , C, , O, , Carboxy, , O, C, , Ester, , OH, , O, C OR, , Amino, , –NH2, , Note - While adding the secondary suffix, the letter ‘e’ of the primary suffix (i.e., ane, ene, and yne) is dropped if the secondary suffix begins with a vowel (a, e, i, o or u). It is retained, if secondary suffix begins with a consonant., , c. Prefix, , (substituent), All the groups which are not names in parent chain and functional groups are called as, substituents. Its name placed before the root word., Exp – Alkyl group, halo atoms, nitro group, Substituents group, -CH3, -C2H5, -NO2, -Cl, -OR, , Prefix, , –C–C–, , Methyl, Ethyl, Nitro, Chloro, Alkoxy, Epoxy, , O, -NO, , Nitroso, , Alkyl groups – derived from an alkane, by removing a hydrogen atom bonded to carbon., These groups are named simply by dropping –ane from the name of the corresponding alkane, and replacing it by –yl. R is a general symbol, general formula for an alkyl group is CnH2n+1,, because it contains one less hydrogen atom than the parent alkane, CnH2n+2.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 18 of 115

Page 21 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , An alkyl group is described as,, ❖ primary if the carbon at the point of attachment is bonded to only one other carbon,, ❖ as secondary if bonded to two other carbons,, ❖ tertiary if bonded to three other carbons. Thus, if R is any hydrocarbon radical, the, different kinds of alkyl groups are, RCH2−, Primary, , R2CH−, Secondary, , R3C−, Tertiary, , Problem 1: Classify each of the following alkyl groups as primary, secondary or tertiary:, Butyl, Isopropyl, Isobutyl, sec-Butyl and tert-Butyl., Solution: Primary alkyl group is RCH2 −, CH3CH2CH2CH2−, Butyl, , CH3CHCH2−, |, CH3, Isobutyl, , Secondary alkyl group is, CH3CH2, , CH3, , R, , CH −, , CH −, , R, , CH −, , H3C, , CH3, , sec - Butyl, , Isopropyl, , R, R, , Tertiary alkyl group is, , C, CH3, |, CH3 −C−, |, CH3, , R, , (tert-Butyl), ,  Prefix n-(normal) is used for those alkanes in which all the carbon atoms form a, continuous chain with no branching., CH 3 CH 2CH 2CH 3 ;, CH 3 CH 2 CH 2 CH 2 CH 3, n − Butane, , n − Pentane, ,  Prefix iso is used for those alkanes in which one methyl group is attached to the next-toend carbon atom (second last) of the continuous chain., CH 3 − CH − CH 2 CH 3, , CH 3 − CHCH 3, , CH 3 − CHCH 2 CH 2 CH 3, , |, , |, , |, , CH 3, , CH 3, , CH 3, , Isopentane, , Isobutane, , Isohexane, ,  Prefix neo is used for those alkanes which have two methyl groups attached to the second, last carbon atom of the continuous chain., CH 3, |, , CH 3 − C − CH 3, |, , CH 3, Neopentane, ,  ACTIVE SITE EDUTECH , , CH 3, |, , CH 3 − C − CH 2 − CH 3, |, , CH 3, Neohexane, , CONTACT: , , Page 20 of 115

Page 23 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Note:, • If two different chains of equal length are possible, the chain with maximum number, of side chains or alkyl groups is selected., • Number used to specify the position of the substituents is called locant., Rule 2 - Lowest sum rule (rule of locant), Parent carbon chain is numbered using Arabic numerals 1, 2, 3, 4, 5..... in such a way that, functional group or substituents containing carbon receive least number., , CH3, |, CH3 − CH − CH 2 − CH 3, 1, , 2, , 3, , 4, , 3, , 2, , 1, 7, , 2, , 3, , 4, , 5, , 6, , 6, , 5, , 4, , 3, , 2, , 1, , 5, 6 4, 3, , 3, 6, , 4, 5, , 7, , 2, , 1 , , , , 8, , 5 CH 3 1, , 5, 3, , 4, , 7, 2, , 8, 1, , CH 3CH 2 CH 2 CH 2 CH CH 3, |, CH 3, , 1, , 4, , 3, 5, , 2, 6, , 4, , 1, , |, CH34 CH 2 2, |, |, CH3 − CH − CH − CH3, , CH 3CH 2CH 2CH 2− CH−CH 3, |, 2CH 2 6, |, 1CH 3 7, , 2, 4, , 1, 5, , 9, 8 7 6 5 4 3 2 1, 1 2 3 4 5 6 7 8 9, CH3CHCH 2CH2CH2CHCH 2CH2CH3, , 3, 3, , CH2CH3, , CH3, ,  When the parent chain contains two or more substituents, the numbering is done from the, end where the sum of the locants is least., 6, , CH3, 1, , 5, , CH3, , C2, , 2, , 4, , 3, , 2, , 1, , CH2, , CH2, , CH, , CH3, , 3, , 4, , 5, , (2 + 5 + 5 = 12, wrong), , 6, , (2 + 2 + 5 = 9, correct), , CH3, , CH3, ,  Parent chain with one substituent: Prefix the name of the substituent to the root word, of parent chain and indicate its position. The name of the substituent is separated from its, locant by a hyphen (-). Name of the organic compound is written as one word., 1, , 2, , CH3, , 1, , 3, , 2, , CH3, , CH CH3, , 3, , CH CH2, , 1, , 4, , 2, , 3, , CH3CH2, , CH3, , CH3, , CH3, 2-methylpropane, , 4, , 5, , 6, , CH CH2CH2CH3, CH2CH3, , 2-methylbutane, , 3-ethylhexane, ,  Naming of different substituents: When two or more different substituents are, present on the parent chain, they are named in alphabetical order along with their, appropriate locants., 1, , CH3, , 2, , 3, , 4, , 5, , 6, , 7, , CH CH2, , CH CH2CH2CH3, , CH3, , CH2CH3, , 4-ethyl-2-methylheptane, , 1, , CH3, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , CH CH2CH2CHCH2CH2CH3, Cl, , CH3, , 2-chloro-5-methyloctane, ,  Numbering of different substituents at equivalent positions: If two different alkyl or, halo groups are present at equivalent positions, the numbering of the parent chain is done in, such a way that the substituent which comes first in the alphabetical order gets the lower, number., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 22 of 115

Page 24 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), 6, , 5, , CH3, 1, , CH2 CH3, , 4, , CH2, 2, , 3, , 2, , correct numbering, , 1, , CH CH CH2CH3, 3, , 4, , 5, , 6, , incorrect numbering, , CH3, , 3-ethyl-4-methylhexane (NOT 3-methyl-4-ethylhexane), 1, , Br, , 2, , CH3, , 3, , CH, , 4, , 3, , correct, , 4, , CH, , CH3, , 2, , 1, , incorrect, , Cl, , 2-bromo-3-chlorobutane (NOT 3 - chloro-2-bromobutane), ,  Presence of the same substituent more than once: If the same substituent occurs, more than once, the prefixes di, tri, tetra are prefixed to the name of the substituent. It, may be noted that the position and name of the substituent are separated by a hyphen (-), whereas the numerals representing the positions of the substituents are separated by, commas., CH3, CH3, , CH3 CH CH CH3, , CH, , CH2CH3, , CH3, , CH, , C CH3, , CH, , CH3, , CH3, , CH3, 2, 3-dimethylbutane, , 4-ethyl-2,2,5-trimethylhexane, ,  Naming a complex substituent: In case the substituent on the parent chain is complex, (i.e., it has a branched chain), it is named as a substituted alkyl group by numbering the, carbon atom of this group attached to the parent chain as 1. Name of complex substituent is, enclosed in bracket to avoid confusion with the numbers of parent chain., CH3, 1, , 2, , CH3, , CH3 CH CH3, 2, , 3, , 4, , 5, , 6, , CH, , 1, , 2, , CH3, 3, , CH3CH2CH2CH2CHCH2CH2CH2CH3, , CH3 CH2 CH2 CH CH2 CH2 CH3, 1, , CH, , 1, , 7, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 5-(1, 2-dimethylpropyl)nonane, , 4-(1-methylethyl)heptane, , If same complex substituent occurs more than once on the parent chain, prefixes bis, tris,, tetrakis, etc., are used before the name of the complex substituent., The application of IUPAC rules to the structure of a molecule to arrive at IUPAC name, is illustrated as follows., CH2, CH3, 5, , CH, 4, , CH3, , CH, 3, , CH3, CH, 2, , CH3, 1, , Br, , • Parent chain has 5 carbon atoms, and it has more number of substituents. The root word is, •, •, •, •, •, , ‘pent’., Primary suffix is ‘ane’, Sum of locants is 9, the substituent which comes first in alphabetical order gets the lower, number., Prefixes to the rootword are bromo, ethyl and methyl. Hence, IUPAC name is, “2-bromo-3-ethyl-4-methylpentane”, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 23 of 115

Page 25 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , A few examples are given below:, CH3, , C, , 1, , CH2, , 2, , 3, , CH, 4, , CH3, , CH3, , CH3, 5, , 2 3, , 4, , 2, , 4, , CH2, , CH3, , 5, , 6, , CH3, , 2, , 1, , 3, , 4, , 6, , 5, , 7, , 8, , 9, , C H 3C H C H 2C H 2C H 2C H C H 2C H 2C H 3, |, |, CH 3, CH 2CH 3, , 2−Methylhexane, 3, , 4, , 2-bromo-3-ethyl-4-methylhexane, , CH3, , 2, , CH, , CH2, , CH3CHCH 2CH2CH2CH3, , 1, , CH, 3, , CH3, , 6, , 5, , CH, , 1, , 2,2,4-trimethylpentane, 1, , CH3, , Br, , CH3, , 6–Ethyl–2−methylnonane, 5, , 6, , CH3, 1, 2, 3, 4 5 6 7, CH3 CH 2 C CH 2CH2CH2CH3, , 7, , CH3CH2CHCH2CH2CH2CH3, |, CH3, , CH2, CH3, , 3−Methylheptane, 2, , 1, , 4, , 3, , 5, , 3–Ethyl–3–methylheptane, , 6, , C H 3C H −C H 2CHCH2C H 3, |, |, CH 2, CH 3, |, 4–Ethyl–2–methylhexane, CH 3, , not, 2–Methyl–4−ethylhexane because ethyl comes before methyl alphabetically, , 2, , 1, , CH 3, |3 4, 5, C H 3C H C HCH−CH 3, |, |, CH 3 CH 3, , 3, , C H 3− C H− C H − CH 3, |, |, CH 3 4 CH 2, |, 5, CH 3, , 1, , 2, 3 – Dimethylpentane, , 3 – Ethyl – 4, 4 – dimethylheptane, 8, 1, , 4, , 3, , 2, , 2, 2, 4, 4 – Tetramethylpentane, CH 3, 8, 4 3, 5, 2| 1, 7 6, CH 3CHC H 2CH 2− CHCH 2− C − CH 3, |, |, |, CH2CH3 CH 3, CH 3, 4 – Ethyl – 2, 2, 7− trimethyloctane, , 2, , 2, , 1, , 2, 3, 4 – Trimethylpentane, , CH 3CH 2 CH 3, | 3 |4 5, 6, 7, C H 3C H 2CH − C − C H 2C H 2CH 3, |, CH 3, 1, , CH3, CH3, | 3 4| 5, C H3−C − CH2− C− C H 3, |, |, CH3, CH3, , 2, , 6, , 5, , 7, , 10, , 7, , 9, , 3, , 5, , 2, , 4, , 6, , 1, , 8, , 4,5–Diethyl–3, 6-dimethyldecane, , 1, 7, 8, , 6, , 5, , 4, , 3, , 2, , 5-Ethyl–2, 2–dimethyloctane, (“di” not considered in alphabetical ordering), CH3, 1, , 2, , 3, , 4, , CH CH 3, 5, , 6, , 7, , 8, , 9, , 10, , CH3CH2CH2 CH CH CH 2CH2CH2CH2CH3, H3C CH CH 2 CH 3, , 5–(1, 1-Dimethylethyl)−3−ethyl-6-methyloctane, (“di” is being part of substituent name), ,  ACTIVE SITE EDUTECH , , 5–sec–Butyl– 4−isopropyldecane, , CONTACT: , , Page 24 of 115

Page 27 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , It may be noted that the groups -R, -OR, -NO2, -X, etc. are considered as substituents, and are indicated as prefixes., 1. Longest chain- The parent chain is the one which has functional group directly attached, to it, even if it violates the longest chain rule., 2. Numbering of parent chain-The numbering should be done in such a way that the carbon, linking to the functional group gets the lowest number even if it violates the lowest sum, rule or locant rule., When the functional group itself contains carbon atom, then that carbon atom is assigned as, number 1., 3. Naming of compounds with polyfunctional group- If there is more than one functional, group present in a compound, then one of the functional group is chosen as the principal, functional group (secondary suffix) and the remaining functional groups are treated as, (substituents) subordinate functional groups., The functional group with higher priority in the sequence given below is termed as principal, functional group., , 1. Write the correct IUPAC names of the following bond line formula:, Me, O, , C, , (i), , O, , (ii), , Cl, , (iii), H3CO, , Me, H, HO, , (v), , O, Me, , OH, H, Me, , (iv), , Me, Me, , (vi), ,  ACTIVE SITE EDUTECH , , OMe, OH Me, , Et, Me, , Me, Me, , (vi) H, , H, , (vii) Me, , CONTACT: , , COOH, , Page 26 of 115

Page 30 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Substituent + benzene → substituted benzene, CH3, , Br, , NO2, , methylbenzene, , bromobenzene, , nitrobenzene, , Example:, Special names of some monosubstituted benzene compounds, , Molecular formula, , Structural formula, , IUPAC name, , Special name, , methylbenzene, , Toluene, , benzenamine, , Aniline, , phenol, , Phenol, , CHO, , Benzene, Carbaldehyde, , Benzaldehyde, , COOH, , Benzene, Carboxylic acid, , Benzoic acid, , methoxybenzene, , Anisole, , CH3, , C6H5CH3, , NH2, , C6H5NH2, , OH, , C6H5OH, , C6H5CHO, , C6H5COOH, , C6H5OCH3, , OCH3, , If the functional group is attached to the carbon chain connected to benzene ring, then, benzene ring is considered as substituent and is prefixed before the root word as phenyl., Example:, CH, CH, CHO, CH, CH, Cl, 2, , (ii), , 2, , 3-phenylpropanal, , 2, , 2, , 2-phenylchloroethane, , Naming of disubstituted benzene, (a) If the substituents are same: In such case, the relative position of the substituents, must be indicated, by adding the symbols o − (1, 2) ; m − (1, 3) ; p − (1, 4)., X, 1, 2, , 6, 5, , 3, , o-ortho : positions 2, 6, m-meta : positions 3, 5, p-para : positions 1, 4, , 4, , In the trivial system of nomenclature, the terms ortho(o), meta(m) and para(p) are used as, prefixes to indicate the relative positions 1, 2-; 1, 3 and 1, 4-respectively., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 29 of 115

Page 31 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), SO3 H, , NO2, , Cl, Cl, , Example:, , NO2, 1,2-dichlorobenzene, (or) o-dichlorobenzene, , SO3 H, 1,4-benzenedisulphonic acid, (p-benzenedisulphonic acid), , 1,3-dinitrobenzene, (or) m-dinitrobenzene, , (b) If the two substituents are different, they are named in the alphabetical order., I, , Cl, , NO2, Br, 1-bromo-4-iodobenzene, (or) p-bromoiodobenzene, , 1-chloro-3-nitrobenzene, (or) m-chloronitrobenzene, , (c) If there are more than two substituents: The numbering is done in such a way that it, satisfies the lowest sum rule., Cl, , 1, , COOH, , 3, 2, , 1, , CH3, , Cl, , Br, , 2, 4, , 3, , NO2, , 1-bromo-3-chloro-2-methylbenzene 4-chloro-3-nitrobenzoic acid, , Nomenclature of Bicyclo compounds, (i) Bicyclo compounds contain two fused rings with the help of a bridge. We use the name of, the alkane corresponding to the total number of carbon atoms as the base name. The, carbon atoms common to both the rings are called bridge heads, and each bond or chain, of atoms connecting the bridgehead atoms called a bridge., (ii) While naming the bi-cycloalkane we write an expression between the word bicyclo and, alkane (in square bracket), that denotes the number of carbon atoms in each bridge. The, numerals are written in descending order and the numbers are separated by a point., (iii) If substituents are present, we number the bridged ring system beginning at one bridge, head, proceeding first along the longest bridge to the other bridge head, then along the, second next longest bridge back to first bridge head. The shortest bridge is numbered in, the last., Example:, Bridge head, One carbon bridge, , (1) Two carbon bridge, , CH2, , { CH, , 2, , CH2, CH2, , CH2, , 1, , }Two carbon bridge, CH, 2, , CH, , 6, , 2, 7, , 5, , 3, , Bicyclo [2, 2, 1] heptane, It's common name is norborane, , 4, , Bridge head, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 30 of 115

Page 33 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Isomerism, The existence of two or more compounds with the same molecular formula but different, physical and chemical properties is known as isomerism and the molecules themselves called, as isomers., The term was given by Berzelius. The difference, in properties of the two isomers is due to, difference in the arrangement of atoms within, their molecules., , Isomerism is mainly classified into structural isomerism and stereoisomerism., 1. Structural isomerism, It is due to the differences in structures of the isomers. Structural isomerism is further, classified into 5 types., (i) Chain isomerism (nuclear isomerism), Compounds with same molecular formula but differ in the arrangement and number of, carbon atoms within the molecule are called chain isomers and the phenomenon as chain, isomerism.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 32 of 115

Page 40 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , d) Secondary and tertiary amines exhibit metamerism due to the difference in the nature, |, , of the alkyl groups attached to the – NH – group and the − N− atom respectively. Thus, the molecular formula, C4H11N, represents the following metamers:, CH3CH2 − NH − CH2CH3, , CH3 − NH − CH2CH2CH3, , CH3 − NH − CH − CH3, |, CH3, , O, ||, e) Esters, R − C − O − R , exhibit metamerism due to the difference in the nature of the alkyl, O, ||, groups attached to the − C− O − group. Thus, the following esters are metamers:, O, O, ||, ||, CH3CH2 − C− O − CH3, CH3 − C− O − CH2CH3, , Note- If same polyvalent functional group is present in two or more organic compounds, then, instead of chain or position isomerism, treat the phenomenon as metamerism., (i) Pentan – 2- one and pentan – 3- one are metamers and not position isomers. They can be, included in position isomerism, if metamerism is not mentioned., (ii) Similarly, pentan – 2- one and 3 – methylbutan-2-one are metamers and not chain isomers., Metamers may be considered as position isomers. For instance, pentan – 2- one and penta-3one may be regarded as position isomers as well as metamers., (v) Tautomerism, Here a single compound exists in two readily interconvertible structures that differ in, position of hydrogen atom. Tautomer exhibits dynamic equilibrium with each other., A very common form of tautomerism is that between a carbonyl compound containing an  hydrogen and its enol form. This type of isomerism is also known as keto−enol isomerism., R', , R', , R − C − C − R'', H, , R − C = C − R", O−H, , O, , Keto form, , O, , Enol form, , H, H, , OH, O, CH3, , C CH3, , keto form, , O H, CH3, , C CH2, , enol form, , The percentage of enol form increases in the order simple aldehydes and ketones <, −keto esters < −diketones < −diketones having phenyl group < phenols. This, increase in the enol content is due to the fact that the enol form of the above type of, compounds is increasingly stabilized by resonance and hydrogen bonding than the, corresponding keto form., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 39 of 115

Page 41 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , (vi) Ring – chain isomerism, The phenomenon of existence of two or more compounds having the same molecular formula, but possessing open chain and closed chain (cyclic structure) is called ring – chain isomerism., This type of isomerism arises due to different modes of linking of carbon atoms. Thus ringchain isomers possess open chain or closed chain structures as illustrated by the following, examples:, i) Two ring-chain isomers are possible corresponding to the molecular formula C3H6:, CH3CH = CH2, , and, , Propene, Cyclopropane, ii) Six pairs of ring-chain isomers are possible for the molecular formula C4H8:, CH3CH2CH = CH2, , and, , CH3CH2CH = CH2, , and, , Me, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 40 of 115

Page 42 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , and, , CH3CH = CHCH3, CH3CH = CHCH3, , and, , CH3 − CH = CH2, |, CH3, , and, , CH3 − CH = CH2, |, CH3, , CH 3, , CH 3, , and, , iii) The molecular formula C3H4 represents the two ring – chain isomers:, and, , CH3 – C  CH, , Ring chain isomerism, can be included in functional isomerism, if not considered separately., , Concepts in organic reaction mechanism, The organic reactions involve the breaking of covalent bonds in the reacting molecules and, formation of new bonds to give product molecules, The organic molecule which reacts with attacking reagent is called substrate. In multistep, organic reactions, substrate react with reagent and leads to the formation of one or more, reaction intermediates. The general reaction path involving the formation of one reaction, intermediate is depicted as follows., Organic molecule, , + attacking reagent, , [reaction intermediate], , product(s) + byproduct (s), , (substrate), , Substrate + Reagent, , ⎯⎯→, , Intermediate, , ⎯⎯→ Products, , (Transition state), , Fission or Cleavage of covalent bond, The fission of covalent bond can take place in two ways depending on the nature of covalent, bond, nature of attacking reagent and conditions of the reaction., Like other chemical reactions, organic reactions are also a process of bond breaking and bond, making. A covalent bond between two atoms can be broken in essentially the following ways:, ., R., X, R:X, R : X, R :X1) Homolytic cleavage (Homolytic fission), “In this type of fission bond breaks symmetrically and each fragment formed gets an odd, electron from the shared bond pair electrons”., This cleavage results in the formation of, specie(s) with unshared electron called free, radicals. It is also known as free radical cleavage, or non-polar bond fission. The homolytic fission of a sigma () bond is shown as follows., A – B ⎯⎯→ A• + B•,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 41 of 115

Page 43 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), ., Cl, Cl, , Cl, Cl, , R, R, , X, X, , sunlight, sunlight, , heat or light, heat or light, , 2 Cl, . free radical, chlorine, 2 Cl, chlorine free radical, , ., , R, +, ., alkylRradical +, alkyl radical, , ., , X, ., X, halogen radical, halogen radical, , The free radicals contain unpaired electron (with odd number of electrons), electrically, neutral and paramagnetic., , The conditions favorable for homolytic cleavage are, The difference in electro negativity between A and B is less or zero., • Homolysis takes place in gaseous phase or in the presence of non polar, solvents(CCl4, CS4), peroxide, UV light, high temperature (≥5000C),, electricity and free radical., •, , 2) Heterolytic cleavage (Heterolytic fission), In this type of cleavage, covalent bond breaks asymmetrically & one of the two species, gets both the electrons and other loses resulting in the formation of ions. Heterolytic, cleavage results in formation of electron deficient and electron rich fragments., It is also known as ionic cleavage or polar bond or ionic fission., The electron deficient fragment is called electrophile, while electron rich fragment is known as nucleophile., 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. The, heterolytic fission is shown below, [Electronegativity of A is greater than B], , _, , A - B ⎯⎯, → A : + B+, , [Electronegativity of B is greater than A], , _, , A - B ⎯⎯, → A+ + B :, H, , CH3, , Br, , Br, , +, , H, +, electrophile, +, , .., , Br, nucleophile, , .., , CH3, +, Br, nucleophile, electrophile, , Both electrophile and nucelophile contain even number of electrons, influenced by strong, electrical field as they possess positive or negative charge and they are diamagnetic., •, •, •, •, •, , The factor which favors heterolysis is greater difference of electro, negativities between A and B., The energy required for heterolysis is always greater than that for, homolysis due to electrostatic forces of attraction between ions., Low temperature, Polar nature of substrate and attacking reagent, Presence of acid or base catalyst, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 42 of 115

Page 47 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Depending upon nature of carbon atom carrying the unpaired electron, free radicals are also, classified as primary (1), secondary (2) and tertiary (3) free radicals., , ., , CH3, methyl radical (1o ), , CH3, , ., , CH3, , ., , C, , CH2, , ethyl radical (1o ), , CH3, , CH2, , CH3, tertiary butyl radical (3o ), , ., , CH3, , CH, , iso propyl radical (2o ), , ., , ., , CH, , CH3, , CH2, , CH2, , allyl radical (1o ), , benzyl radical (1o ), , Structure of free radical: The free radicals are electron deficient since they contain seven, electrons on carbon atom. They are electrically neutral and paramagnetic. The structure of, alkyl radical is not known with certainty. For alkyl radicals two possible structures have been, proposed. The first is a planar sp2 hybridised radical similar to a carbocation. The second one, is a pyramidal sp3 hybridised radical similar to a carbanion., • Resonance stabilised free radicals such as allyl radicals and benzyl radicals are planar sp2, hybridised., •, , The bridge head free radicals are pyramidal (sp3-hybridised) because they cannot assume, planar geometry due to angle strain. Further, the free radicals in which carbon is bonded, to highly electronegative atoms are pyramidal., , ., , ., , CF3, , ., , trifluoromethyl radical, (sp3-hybridised), , bridge head radical (sp3-hybridised), , Stability of free radicals: The relative stability of alkyl free radicals is explained on the, basis of hyperconjugation and inductive effects. Greater the number of alkyl groups attached, to the carbon atom carrying unpaired electron, higher the delocalisation and hence more stable, is the alkyl radical. The order of stability of a few alkyl radicals is given below, CH3, , ., , C, , CH3, , CH3, , ., , CH, , CH3, , CH3, , ., , CH2, , ., , CH3, , CH3, t-butyl radical (3o), , iso propyl radical (2o), , ethyl radical (1o), , methyl radical (1o), , The order of stability of free radicals on the basis of resonance inductive effect is as follows:, (C 6 H 5 )3 C• > (C 6 H 5 )2 CH• > C 6 H 5• > CH 2 = CH CH •2 > 3° > 2° >1° > CH •3, , Like carbocations and carbanions, free radicals are highly reactive and short-lived, intermediates because of the strong tendency of the carbon atom carrying the unpaired, electron to acquire one more electron from an atom or a group to complete its octate. The, •, , reactivity of alkyl radical is reverse the order of stability CH3  1  2  3, The allyl and benzyl radicals are stabilised by resonance. The stability of allyl and benzyl, radicals is comparable., , ., , CH2, >, , ., , CH2, , CH, , CH2, , The allyl or benzyl radicals are more stable than alkyl radicals.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 46 of 115

Page 48 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , The reactions involving free radicals are, (i) catalysed by light, heat etc., (ii) proceed in vapour phase or in non polar solvents., (iii) autocatalytic., (iv) Carbenes, They are neutral and highly reactive species generally obtained by successive elimination of an, electrophile and a nucleophile from the same carbon atom (-elimination). The carbon atom of, carbene has six electrons in valence shell, out of which two constitute unshared electrons and, two bond pair electrons. So they are divalent carbon species containing two unshared electrons, and electrically neutral., Carbenes exist in two possible forms, (i) Singlet, (ii) Triplet, (i ) Sin glet, , (ii ) Tr iplet, , H, C, H, , H, sp 2 h ybr idized, Two elect r on s a r e, in sa m e or bit a l, (spin pa ir ed ), , H, , ., C., , carbene, , H, , R, , ., C., , H, , alkyl carbene, ,  ACTIVE SITE EDUTECH , , Cl, , ., C, ., , H, , C, , sp h ybridized, Two elect r on s a r e in, t wo differ en t or bit a ls, (spin fr ee), , O, H, , chloro carbene, , R, , C, , ., , C., , acyl carbene, , H, , ., , Cl C. Cl, dichloro carbene, , CONTACT: , , Page 47 of 115

Page 49 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Attacking Reagents, The organic reactions proceed by the attack of highly reactive reagents on the substrate molecule., These reagents are called attacking reagents which may be electron deficient or electron rich. They, are classified into two groups., , Electrophiles, The electron deficient molecules or positively charged ions which are capable of accepting an, electron pair from substrate molecule are called electrophiles., These species act as Lewis acids and attack the electron rich centre of the organic molecules., C.., , +, , +, , E, , E, , C, , electrophile, , substrate, , They are of following types,, (i) Positively charged: The species having a positive charge, e.g., H+ ,CH3+ ,NO2+ ,H3O + ,NO + ,Br + ,CH3CO + etc ., (ii) Neutral: The molecules containing electron deficient atom (i.e. Lewis acids) e.g. :CH2, AlCl3,, BF3, ZnCl2, FeCl3, SO3, etc., (ii) Ambident: Molecules with 2 electrons deficient centers,, e.g. , -unsaturated carbonyl compounds., It may be noted that all the positively charged species do not act as electrophiles. The positively, charged species which can accept an electron pair can act as electrophiles. The positively charged ions, +, , +, , +, , 2+, , such as H3 O, N H4 , Na, C a , etc. do not act as electrophile as they cannot accept electron pair, since all, the ions have an octet configuration., , Nucleophiles, The molecules or negatively charge ions which are capable of donating an electron pair, to electron deficient centre of the substrate are called nucleophiles., These species act as Lewis bases & attack on electron deficient centre of organic molecule., C+, , +, , substrate, , .., , Nu, , C, , Nu, , nucleophile, , The common examples of nucleophiles are given below:, (i) Negatively charged: The species having a negative charge, e.g. Cl−, Br−, OH−, CN−, NO2− etc., (ii) Neutral: The molecules having an unshared pair of electrons (i.e. Lewis base),, .., .., .., .., .., e.g. 𝑁𝐻3 , 𝑅 − 𝑁𝐻2 , 𝑅2 − 𝑁𝐻, 𝐻2 𝑂, 𝑅 − 𝑂 − 𝐻 𝑒𝑡𝑐., .., , .., , .., , .. .., , .. .., , (iii) Ambident: The molecule with two electron rich centres, e.g. − C  N,N, .. = O, O  N− O etc., Comparison between Nucleophiles and electrophiles, Nucleophiles, Electrophiles, • Electron rich, donate electron pair, generally • Electron deficient, accept electron pair, generally, anions., cations, • Act as Lewis bases, • Act as Lewis acids, • Attack an region of low electron density of • Attack on region of high electron density of, substrate, substrate, .., • They are represented by a general symbol ( Nu ), • They can be represented by a general symbol (E+), ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 48 of 115

Page 50 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Electron Displacement Effects in Covalent Bonds, The electron pair displacement in organic molecules takes place under the influence of a, hetero atom/group or by the attacking reagent. The displacement or shift of electron pair in, the organic molecule under the influence of substituent makes the molecule permanently, polar., , (i) Inductive effect (I effect), The inductive effect is defined as “the permanent displacement of sigma (σ) bond pair of, electrons towards more electronegative atom or group and as a result molecule becomes, permanently polar”., Larger the displacement of σ bond pair electrons greater the polarity. Consider the carbon, chain in which terminal carbon is bonded to a chlorine atom. Since chlorine is more, electronegative bond pair of electrons are displaced towards chlorine., 𝐂 − 𝐂 − 𝐂 − 𝐂 is non polar, +, , C , C , +3, , C, , +2, , C, , +1, , −, , C, , Cl, , −, , C , , +, , C , , −, , C , , +, , C , , C, −, , +, , C, , X, −, , −, , Y, , +, , order of intensity of I effect +3  +2   +, , There are two types of Inductive effect,, a. Positive inductive effect (+I effect), Here substituent (Y) releases electron pair away from itself. In other words,  bond pair, of electrons are displaced away from the substituent towards the C-chain., When less electronegative atom is attached to carbon atom (i.e. electron donating group),, the inductive effect is called + I effect., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 49 of 115

Page 51 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), electron releasing substituent, , , Y, , C, , +, , −, C3, , + I effect, , −−, C2, , −, C1, , +, Z, , Electronegativity of Y or Z is less than carbon atom, so the Y or Z group releases electron and, gets a partial positive charge (i.e. +) and induces a partial negative charge (i.e. −) on C1. As we, move along the chain from C1 to C3, the partial negative charge decreases., The examples of electron donating group are (CH3)3C, (CH3)2CH, CH3CH2, CH3 etc., The order of electron releasing or releasing ability of substituent is given below., O, , .., , .., , C6 H5 O >, C O > (CH3)3C, phenate ion carboxylate ion, 3oalkyl, , > (CH3)2CH, 2oalkyl, , > CH3, , > CH3, , CH2, , >, , o, , D >, , H, , 1oalkyl, , + I power in decreasing order with respect to the reference H, , b. Negative inductive effect (−I effect), Here, the sigma bond pair of electrons are displaced towards electron withdrawing, substituent (X) or away from the carbon chain., More electronegative atom is attached to carbon atom (i.e. electron withdrawing group), the, inductive effect is called – I effect., +, , The examples of electron withdrawing group 𝑁𝑅3 , 𝑁𝑂2 , 𝐹, 𝐶𝑙, 𝐼, 𝑂𝐻 𝑒𝑡𝑐., electron withdrawing substituent, +, , +, C3, , +, C2, , +, C1, , X, , C, , −, Z, , , , I effect, , Order of electron withdrawing ability (−I effect) of a few substituents is given below., ⊕, , 𝑵𝑯𝟑 > 𝑵𝑶𝟐 > 𝑪𝑵 > 𝑺𝑶𝟑 𝑯 > 𝑪𝑯𝑶 > 𝑪𝑶 > 𝑪𝑶𝑶𝑯 > 𝑪𝑶𝑪𝒍 > 𝑪𝑶𝑶𝑹 > 𝑪𝑶𝑵𝑯𝟐 > 𝑭 > 𝑪𝒍 > 𝑩𝒓 > 𝑰 > 𝑶𝑯 > 𝑶𝑹 > 𝑵𝑯𝟐 > 𝑪𝟔 𝑯𝟓 > 𝑯, –I power of groups in decreasing order with respect to the reference H, , Applications of inductive effect:, , (i) Magnitude of +ve and -ve charges: can be compared by + I or – I groups present in it., • Magnitude of, , +ve, , charge, , • Magnitude of, , −ve, , charge, , 1,  −I, + I power of the group, 1, ,  +I, − I power of the group, , , power of the group., power of the group., , (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, H 3C, , C, , CH 3, , X  H3C, , X  CH 3, , CH, , CH 2, , X  CH 3, , X, , CH 3, , Tertiary, , >, , Secondary, , >, , Primary, , >, , Methyl, , (iii) Relative strength of the acids:, (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  C2 H 5 COOH  C3 H7 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.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 50 of 115

Page 52 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , (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., Thus, acidic nature is, CCl3 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, , Hence benzoic acid is stronger acid than aliphatic carboxylic acids but exception is formic, RCOOH, acid. Thus, HCOOH, > C6 H 5 COOH, >, Acid strength in decreasing order, , As compared to water, phenol is more acidic (–I effect) but methyl alcohol is less acidic (+I, effect)., OH, OH  H − OH > CH 3, Water, , Phenol, , Methyl alcohol, , (vi) Relative strength of the bases (Basic nature of −𝑵𝑯𝟐 ), 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, , ClNH2, , 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”., Thus, the basic nature decreases in the order; (C 2 H 5 )2 NH  CH 3 CH 2 NH 2  CH 3 NH 2  NH 3  ClNH2, Diethyl, amine, , Ethyl, amine, , Methyl, amine, , Ammonia, , Chloro, amine, , The order of basicity is as given below;, Alkyl groups (R–), , Relative base strength, , CH3, , R 2 NH > RNH2 > R 3 N > NH3, , C2 H5, , R 2 NH > RNH2 > NH3 > R 3 N, , (CH3 )2 CH, , RNH2 > NH3 > R 2 NH > R 3 N, , (CH3 )3 C, , NH3 > RNH2 > R 2 NH > R 3 N, , (vii) Basicity of alcohols: The decreasing order of base strength in alcohols is due to +I effect, of alkyl groups. (CH 3 )3 COH  (CH 3 )2 CHOH  CH 3 CH 2 OH  CH 3 OH, (3 o ), , (2 o ), , (1 o ), , (viii) Stability of carbonium ion: +I effect tends to decrease the (+ve) charge and –I effect, tends to increases the +ve charge on carbocation., (CH 3 )3 C   (CH 3 )2 CH   CH 3 CH 2  CH 3, , (ix) Stability of carbanion: Stability of carbanion increases with increasing – I effect., CH 3−  CH 3 CH 2−  (CH 3 )2 CH −  (CH 3 )3 C −, , (x). Dipole moment: Inductive effect produces dipole moment in a molecule. As this effect, increases (electronegativity difference increases) the dipole moment also increases., CH3 − I  CH3 − Br  CH3 − Cl, 1.648 D, , 1.79 D, , 1.83 D, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 51 of 115

Page 53 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , (ii) Resonance effect (R−effect) or Mesomeric effect, “The permanent polarity produced in the molecule by the shift of pi() or lone pair, electrons in the conjugate system creating electron deficient and electron rich centres, called resonance effect (R−effect)., The conditions required for M or R effect:, 1. Molecule should be unsaturated with conjugated system (presence of alternate single and, double bond or heteroatom containing one or more lone pair electrons linked to multiple, bonded atom) of double bond., 1,3-butadiene, aniline, phenol, nitrobenzene, etc. In such systems,  or lone pair electrons are, delocalised and the molecule develops polarity., 2. Negative charge is in conjugation with double (or multiple) bond., 3. Lone pair of electrons in conjugation with double bond., The reactivity of compounds is affected by the presence of groups like, NO2 , C  N,, , O , Cl, NH 2 etc., , The movement of  electrons from one end to the other end of the chain through a conjugated, system of double bond is observed in resonance effect. It is a permanent effect., Depending upon the direction of shift of electron pair in conjugate system, R−effect is, classified into two types., a. Positive resonance effect (+R effect / +M effect): In this effect, of electron pair, ( or lone pair) moves away from the substituent or towards the conjugate system., The +R effect in aniline as shown below., , .., , .., , Substituents exert only +R effect as follows, −𝐂𝐥, −𝐁𝐫, −𝐈, −𝐍𝐇𝟐 , −𝐍𝐑𝟐 , −𝐎𝐇, −𝐎𝐑, −𝐒𝐇, −𝐎𝐂𝐇𝟑 , −𝐒𝐑, .., , The substituent which exerts +R effect are called electron releasing groups., The resonance effect in which resonance structure violates the octate rule should not be, considered. For example, structure (II) cannot be considered as resonance structure since it, violates the octate rule because oxygen has 10 electrons in the valence shell., CH2, , CH, I, , .., ..OCH3, , X, , +, , CH2, , .., ..OCH3, , CH, II, , b. Negative resonance effect (−R effect / −M effect): isthe shift of  or lone pair, electrons is towards the substituent attached to the conjugate system. The electron, displacement depicted in nitrobenzene represents −R effect., O, , +, , N, , O, , .., , O, , +, , N, , .., , O, , .., , .., , +, O, O N, , .., , .., , +, O, O N, , .., , +, O N, , .., , O, , +, , +, , +, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 52 of 115

Page 54 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), 𝑶, ||, , Substituents, which exert −R effect are given below. −𝑵𝑶𝟐 , −𝑪 ≡ 𝑵, −𝑪−, −𝑪𝑯𝑶, −𝑪𝑶𝑶𝑯, −𝑺𝑶𝟑𝑯, −𝑪𝑶𝑶𝑹, The substituent which exert −R effect is called electron withdrawn group. Resonance effect, provides explanation to least reactivity of haloalkenes and aryl halides towards nucleophilic, substitution reactions, acidic nature of phenols and carboxylic acids, mechanism of, electrophilic substitution reactions of benzene., , .., , (phenyl) etc. exert both +R and −R effects., Substituents like N O ,, ➢ Resonance, • Delocalization of p-electrons in conjugation is known as resonance., • When one structure is not sufficient to explain each and every property (chemical &, physical) then, a different structure has been drawn which is known as Resonating, Structure (canonical structure)., • All these structures contribute to the formation of a Real structure, known as, Resonating Hybrid., , or, (Actual Structure), (resonating structures), (Resonance hybrid), Condition for showing resonance:, (i) Molecule should be planar, nearly planar or a part of it is planar, (ii) Molecule should possess conjugated system., Note:, They are not resonating structures rather they are tautomers., ➢ Resonating Structure, • Hypothetical structure existing on paper., • The energy difference b/w different resonating structure is very small., • All R.S. contribute towards the formation of resonance hybrid (Their contribution, may be different)., • A single R.S. can't explain each & every property of that particular compound., Q. Draw the resonating structures:, Sol. (i), , (ii), ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 53 of 115

Page 55 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Resonance hybrid: It is a real structure that explains all the properties of a, compound formed by the contribution of different R.S., It has got maximum stability as compared to any R.S., • Resonance Energy: It is the difference b/w theoretical value of H.O.H &, experimental value. Or it can be defined as the difference b/w more stable R.S. &, R. H., The more resonance energy, the more stable will be the molecule., Resonance energy is an absolute term., Note: Cyclohexane is thermodynamically more stable than benzene, even though the, resonance energy of benzene is more., Contribution of different R. S. towards resonance hybrid:, • Non-polar R.S. contributes more than polar R.S., Example: (a) CH2 = CH - CH = CH2 (b) CH2-CH = CH - CH2 (c) CH2 - CH = CH - CH2, Order of stability of given R.S. ⇒ a > b = c, • Polar R. S. with complete octet will contribute more as compared with the one with an, incomplete octet, CH3 - CH - OCH3 (Incomplete octet), CH3 - CH = :O - CH3 (Complete octet), • In polar R. S. the -ve charge should be on more electro - ve atom & +ve charge should, be on more electro +ve atom., Example:, •, , (a), , •, •, •, , (more stable), , (b), (less stable), Compound with more covalent bonds will contribute more., Unlike charges should be closer to each other whereas like charges should be isolated., Extended conjugation contributes more than cross conjugation., Example:, , <, , (Cross conjugation < Extended conjugation), , ➢ Fries Rule:, • Compounds with more benzenoid structures are more stable., • As the Resonance energy is greater than those in which lesser no. of benzenoid, structures are present., •, •, , •, , R. E. is, <, If a double bond is participating in resonance then it will acquire a partial single, bond character as a result of which bond length increases & bond strength, decreases., If a single bond is involved in resonance then it will acquire partial double bond, character. As a result of which bond length decreases & bond strength increase., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 54 of 115

Page 56 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Q. Find the order of Stability in the following:, , (i), (ii) (a), , (b), , (c), , (d), , (iii), (iv) (a), (b) CH2 - CH = F, Sol. (i) a = e > b = d > c, (ii) a > b > c > d, c and are in complete, (iii) +CCl3 < +CF3, due to back bonding in +CF3, (iv) a > b (stability), , Note:, When lone pair, as well as a double bond, is present in some atom then only p bond will be, participating in resonance. Whereas lone pair remains sp2 hybridized orbital. When an atom, has two or more than two lone pairs then only one lone pair will participate in resonance and, the other one remains in sp 2 hybridized orbital., , (iii) Hyperconjugation (No bond resonance), The phenomenon of hyperconjugation is also known as Baker-Nathan effect as it was proposed, by Baker and Nathan. The hyperconjugation effect is much weaker compared to resonance, effect, yet it is quite useful is explaining relative stability, physical and chemical properties, of organic molecules., This effect is extension of resonance in which C–H sigma () bond pair electrons are involved, in delocalization. It is a permanent effect in which electron releasing alkyl group bonded, to unsaturated system in which delocalization of electrons takes place through overlap, between C–H sigma () orbital and Pi() bond orbital or vacant p-orbital is known as, hyperconjugation., The hyperconjugation is a stabilizing interaction. The delocalization of electrons by, hyperconjugation in propene molecule is depicted as in figure, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 55 of 115

Page 57 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022),  conjugation) hyperconjugation, , (, , H,  bond orbital, , H, C, , H, , C, , C, , H, , H, , H, ,  bond, Orbital diagram showing hyperconjugation in propene, , Propene molecule may be regarded as the resonance hybrid of the following hyperconjugative, structures (I – IV)., +, , H, H, , C, , H, CH, , CH2, , H, , C, , CH, , H, , .., , +, , H, , CH2, , C, , CH, , H, , .., , CH2, , H, , C, , CH, , .., , CH2, , +, , H, , H, , H, , H, , (I), , (II), , (III), , (IV), , Since there is no bond between carbon and hydrogen atoms in these structures (II –IV),, hyperconjugation is also called no bond resonance. It may be noted that although a free proton, (H+) has been shown in the above structures, it is still bound firmly to the - cloud and hence, is not free to move., Hyperconjugation effect in carbocations, let us take an example of ethyl carbocation, +, , (CH3 − CH 2 ) , in which the positively charged carbon atom has an empty p orbital. One of the C–, , H  bond orbital of methyl group align in the plane of empty p orbital and this bond pair, electrons delocalise into the empty p-orbital as shown in figure.,  conjugation) hyperconjugation, , (, H, sp3 s,  bond, C, , +, , C, , H, H, , H, H, empty p-orbital, , Orbital diagram showing hyperconjugation in ethylcarbocation, , The overlap of completely filled C–H  bond orbital with empty p-orbital of carbocation, causes dispersion of positive charge and stabilize the carbocation. The ethyl carbocation is, resonance hybrid of following contributing structures., +, , H, H, , C, , H, , H, +, , CH2, , H, , C, , CH2, , +, , H, , C, , H, CH2, , H, , C, , CH2, +, , H, , H, , H, , H, , (I), , (II), , (III), , (IV), , In general, larger the number of -hydrogen atoms of alkyl groups attached to a, positively charged carbon atom greater the stability of carbocations., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 56 of 115

Page 58 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Applications of hyperconjugation effect:, 1., Stability of alkenes: Larger the number of hyperconjugative structures higher the, delocalization of electron pairs and greater the stability of alkene. Number of, hyperconjugative structures is equal to number of -hydrogen atoms plus one., More number of methyl groups attached to double bonded carbon atom more would be the, stability of alkene., CH2 = CH2, <, CH3 – CH = CH2, <, (CH3)2 C = CH2, No hyperconjugation, 3 hyperconjugation, 6 hyperconjugation, structures, structures, structures, This order of stability is because of greater number of hyperconjugative contributing, structures causing larger delocalisation of -electrons solve and hence accounts for higher, stability of alkene., 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. Stability of carbonium ions: More, number of hyperconjugation structures of the, carbocation more will be its stability., tert–butyl, >, isopropyl, >, ethyl, 9 hyperconjugation, 6 hyperconjugative 3 hyperconjugation, structures, structures, structures, 3. Bond lengths: The bond length in a molecule change if there is hyperconjugation. In, C3H3 − C2H = C1H2, , •, , , the C1−C2 bond length is found to be more than 1.34 A (normal C = C bond, •, , length) while the C2−C3 bond distance is less than 1.54 A (normal C – C bond length)., 4. Directive influence of the group: +M effect of methyl group in toluene is due to, hyperconjugation., H, H, , H, , H, , +, , H, , H, H, , +, , H, , H, H, , +, , H, , H, , Due to hyperconjugation, there are nine different structures having negative charge at ortho, and para positions. Hence, + M effect of alkyl group attached to benzene ring follows the, order: 𝐦𝐞𝐭𝐡𝐲𝐥 > 𝐞𝐭𝐡𝐲𝐥 > 𝐢𝐬𝐨𝐩𝐫𝐨𝐩𝐲𝐥 > 𝐭𝐞𝐫𝐭 − 𝐛𝐮𝐭𝐲𝐥., In the same way, the meta directing influence and deactivating effect of –CCl3 group in, benzotrichloride can be explained on the basis of hyperconjugation as follows,, Cl, Cl, , Cl, Cl, , Cl, , Cl, Cl, , -, , Cl, , Cl, Cl, , Cl, , -, , Cl, , -, , Due to low electron density at ortho and para positions, the meta position becomes point, of high electron density, hence electrophilic substitution takes place in meta position., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 57 of 115

Page 59 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , (iv) Electromeric effect (E effect), The complete transfer or shift of  electron pair of a multiple bond to one of the bonded, atoms during the attack of positively or negativity attacking agents. As soon as the reagent is, removed, the molecule reverts back to original position., 𝐀=𝐁 →, , ⊕, , 𝐄, , Reagent, , −, , 𝐀 − 𝐁:, , Temporary polarization of the substrate molecule at the site of multiple bonds, by, complete shift of an  electron pair from one atom to the other under the influence of, attacking reagents., H, , H, , H, , H, , ⎯⎯⎯⎯⎯⎯⎯, →, ⎯⎯⎯⎯⎯⎯, ⎯, Re agent removed, Reagent added, , H, , +, , C, , H, , C, , H, , H, , Depending upon the direction of displacement, E effect is also of two types., • −E effect: An attacking reagent is said to have −E effect when the direction of , electron pair transfer of multiple bond is away from the attacking reagent., CH3, C, , O, , +, , H, , .., , CH3, , CN, , H, , C, , .., , O, , CN, , acetaldehyde; (-E effect), , The −E effect operates during nucleophilic addition reaction of aldehydes and ketones., • +E effect: An attacking reagent is said to have +E effect when the direction of , electron pair transfer of a multiple bond is towards the attacking reagent., CH3CH, , CH2, , +, , +, , +, , H, , CH3CHCH3, isopropyl carbocation, , Propene ; (+E effect), , The +E effect is observed during electrophilic addition reaction of alkenes and alkynes, CH3, , .., , HO +, , .., , .., , MgI, , H2N +, amide, , H, , H, , +, , CH3 + MgI, , .., , CH2 CHO, ethanal, , C, , CH, , CH2, , CHO +, , H2O, , ethanal carbanion, , .., , C CH + NH3, acetylide ion, , Steric Hindrance or Steric Strain, When two atoms are closer to each other than the sum of their van der Waal’s radii, they repel each other due to spatial crowding. Steric hindrance or steric strain refers, to non-bonded repulsive interactions between atoms which arise when the atoms come, too close due to the shape of the molecule. The repulsion arises due to nuclear-nuclear, and electron-electron repulsive forces which start dominating the attractive forces., Molecules with steric strain are relatively less stable than those having no strain or, less strain.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 58 of 115

Page 60 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , For example, in cis-but-2-ene steric hindrance is present because the two methyl, groups are quite close to each other wheras in trans-but-2-ene no steric hindrance is, there because the two methyl groups on the opposite sides., Steric hindrance, , NoSteric hindrance, H, , CH3, C=C, , C=C, H, , H, , CH3, , H, , cis-But-2-ene, (Sterically hindered, molecule, less stable), , trans-But-2-ene, (No Steric hindrance,, more stable), , Steric factors have great influence on orientation and reactivity of organic molecules, in various reactions., Types of Organic Reactions, I. Substitution reaction, The reactions in which an atom or a group in a molecule is replaced by another are called, substitution reactions. The incoming group gets attached to the same carbon atom to which, leaving group was attached. The substituting species may be a nucleophile, an electrophile or, a free radical., A – B + X – Y ⎯⎯⎯→ A – X + B – Y, , (i) Nucleophilic substitution (SN) reactions, The substitution reactions which are brought about by the attack of nucleophile (Nu).., are called nucleophilic substitution reactions., , .., Nu, , R//, + R/, , C, , R//, Nu, , L, , R, , C, , R/ + L, , .., , [ Nu = attacking reagent; L = leaving group], , R, , The common examples of S N reactions are as follows., aqueous, R−CH2−X + NaOH ⎯⎯⎯→, R−CH2−OH + X− (X = Cl, Br, I), alcohol, R−CH2−X + KCN ⎯⎯⎯→ R−CH2CN + X−, R−CH2OH + SOCl2 ⎯⎯, → R−CH2Cl + HCl + SO2, (ii) Electrophilic substitution (SE) reactions, The substitution reactions which are brought about by the attack of an electrophile are, called electrophilic substitution reactions. The substitution reactions of aromatic, compound such as chlorination, nitration of benzene are representatives of S E reactions., • Chlorination: Cl2 + FeCl3 → FeCl −4 + Cl+, H, , +, +, , + Cl, , slow, , H, Cl, , FeCl4, heat, , Cl, , + FeCl3 + HCl, , (iii) Free radical substitution, The substitution reactions which are brought about by the attack of free radical are, called free radical substitution reactions. The chlorination of aliphatic hydrocarbons in, presence of diffused sunlight is common example of free radical substitution.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 59 of 115

Page 62 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , III. Elimination reactions, The loss of atoms or group from adjacent carbon atoms (one in the form of a nucleophile and, other in the form of an electrophile) resulting in the formation of an unsaturated compound is, known as elimination reaction., CH 3 CH 3, , H 3C, , CH 3, , →, CH 3 ⎯⎯, , H 3C, H, , + H+ + X−, , H 3C, , X, , CH 3, , [Note: If these two groups or atoms are removed from adjacent carbon atoms, then it is known, as  elimination reaction.], R, , H, , OH, , con c., ⎯⎯⎯⎯, →, H SO, 2, , R, , 4, , CH 2, , + H 2O, , The elimination reactions are two types, -elimination reactions and -elimination reactions, (i) -elimination reactions, This type of reaction involves loss of two atoms or groups from vicinal (adjacent) carbon, atoms resulting in the formation of a  bond. Thus, it is the reverse of addition reactions., The most familiar example of, -elimination reactions are dehydrohalogenation reactions of alkylhalides, dehalogenation of, dihaloalkanes, dehydration of alcohols, pyrolysis of esters, Hofmann elimination of, quaternary ammonium hydroxide., Dehydrohalogenation, When alkylhalides are treated with alcoholic potassium hydroxide solution or sodamide the, corresponding alkenes are formed with the elimination of hydrogen halide., R, , , , , , CH, , CH, , H, , X, , + KOH, , R/, , alcohol, , R, , CH, , R/, , CH, , + KX + H2O, , alkene, , -elimination, , R, , H, , , , , , C, , C, , +, , NaNH2, , heat, , R, , H + NaX + NH3, , C, C, alkyne, , X, H, -elimination, , (i) Sodamide (NaNH2) is stronger dehydrohalogenating agent., (ii) An example of 1,4-elimination (-elimination) is, R, , CH, , CH, , CH CH2 + KOH, , R, , CH, , X, , H, , CH, , CH, , CH2 + KX + H2O, , 1,3-alkadiene, , 1,4-elimination, , Dehalogenation, The dehalogenation involves the removal of halogen molecule (X2) from vicinal dihalide by, heating with zinc dust in alcoholic medium., R, , CH, , CH, , Br, , Br, , R/ +, , Zn, , ethanol, , R, , CH, , CH, , R/ + ZnBr2, ,  -elimination, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 61 of 115

Page 64 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Purification and Characteristics of Organic Compounds, The study of organic compounds starts with the characterization of the compound and the, determination of its molecular structure. The procedure generally employed for this purpose, consists of the following steps:, (1) Purification of organic compounds, (2) Qualitative analysis of organic compounds, (3) Quantitative analysis of organic compounds, (4) Determination of molecular mass of organic compounds, (5) Calculation of Empirical formula and Molecular formula of organic compounds, (1) Purification of organic compounds: A large number of methods are available for the, purification of substances. The choice of method, however, depends upon the nature of, substance (whether solid or liquid) and the type of impurities present in it. Following methods, are commonly used for this purpose,, (i) Filtration, The process of filtration is used to separate insoluble, solid component of a mixture from the soluble, components in a given solvent., Exp- It is used to separate a mixture of Naphthalene, and Urea using a water as solvent. Urea dissolves in, water while Naphthalene remains insoluble. Upon, filtration, naphthalene remains on the filter paper, while urea is recovered from the filtrate by, evaporating water., Filtration is very slow and takes along time. In, such cases, filtration is carried out under reduced, pressure using a Buchner funnel and water suction pump., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 63 of 115

Page 65 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), (ii) Crystallization, , The method is based on the difference in, the solubilities of the compound and the, impurities in a suitable solvent. Pure, compound crystallises out from the, solution and highly soluble impurities, remain in the solution. Impure organic, compounds like glucose, urea, cinnamic, acid, etc are purified. Fractional, crystallisation is used for the separation, of a mixture of two compounds which are, soluble in the same solvent but to a, different extent., Ex: Separation of sugar and salt., (ii) Sublimation is the process of direct conversion of a solid into the gaseous state on, heating without passing through the intervening liquid state and vice versa on cooling., Heat, , Vapour, , Solid, Cool, , Only those substances whose vapour pressure becomes equal to the atmospheric pressure, much before their respective melting points are capable of undergoing sublimation., There exists an equilibrium between the solid and its vapours., , Camphor, naphthalene, anthracene, iodine, benzoic acid, salicylic acid, NH4Cl, HgCl2, etc are, purified by sublimation., (iii) Distillation- Distillation is a method used to separate constituents of a liquid mixture, which differ in their boiling points., Distillation is a process which involves two steps:, Vapourisation: Liquid is converted into vapours., Condensation: Vapours are condensed again into liquid., Depending upon the difference in the boiling points of the constituent liquids, different types, of distillation methods are employed.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 64 of 115

Page 66 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , •, , Simple distillation: Simple distillation is applied only for volatile liquids which boil without, decomposition at atmospheric pressure and contain non-volatile impurities., , This method can also be used for separating liquids having sufficient difference in their boiling, points. For example,, (i) benzene (boiling point 353 K) and aniline (boiling point 475 K), (ii) chloroform (boiling point 334 K) and aniline (boiling point 457 K), (iii) ether (boiling point 308 K) and toluene (boiling point 383 K), Nitrobenzene prepared in the laboratory can also be purified by distillation., • Fractional distillation: This method is used for the separation of two or more volatile, liquids from a liquid mixture which has boiling points close to each other. Liquids forming a, constant boiling mixture (azeotropic mixture) such as rectified spirit cannot be separated, by this method., , Fractional distillation is used these days in industries, especially, in the distillation of, petroleum, coal tar and crude alcohol. A mixture of methanol (boiling point 338 K) and, propanone (boiling point 330 K) or a mixture of benzene and toluene may be separated by, fractional distillation., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 65 of 115

Page 67 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , •, , Distillation under reduced pressure (Vacuum distillation): The compounds, which, decompose at a temperature below their normal boiling points, cannot be purified by, distillation under ordinary atmospheric pressure., , Glycerin is one such compound which decomposes at its boiling point. The pressure is reduced, by suction pump and the distillation is carried out at lower temperature as glycerine can be, distilled at 453 K (normal boiling point 563 K) under a pressure of 10-12 mm. Cane juice can, also be concentrated by this method. This technique can be used to separate glycerol from, spent lye in soap industry.,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 66 of 115

Page 68 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , •, , Steam distillation: This method is used to purify the organic compounds which, ❖ Are volatile in steam but are immiscible with water., ❖ Possess high vapour pressure at the boiling point of water., ❖ Contain nonvolatile impurities., , The compound to be purified is distilled with steam and impurities being nonvolatile remains in, mother liquor. For example, o- nitrophenol (volatile) and p-nitrophenol (non volatile) are, separated by this method., (iv) Differential extraction (or solvent extraction): The process, of separation of an organic compound (solid or liquid) from its, aqueous solution by shaking with a suitable, organic solvent is termed as solvent extraction. This method is, employed for non-volatile compounds. For example, benzoic acid is, extracted from its aqueous solution using benzene as solvent., (v) Chromatography, Chromatography is the technique of separating the components of mixture in which the, separation is achieved by the differential movement of individual components through a, stationary phase under the influence of a mobile phase., , jar adsorbent coated, on glass plate, sample dot, solvent, , base line, , Thin layer chrom atography, chromatogram being developed, , solvent, front, spot, x, , y, base line, , Developed chromatogram, , Dis tance moved by the substance formbaseline ( x ), , Rf = Dis tance moved by the solvent from base line ( y), , Depending upon the nature of the stationary phase (either a solid or a liquid tightly bound on, a solid support) and the nature of the mobile phase (either a liquid or a gas), different types, of chromatographic techniques are followed., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 67 of 115

Page 69 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Chromatography, , Adsorption Chrom., , Partition Chrom., , Column chromatography, , Paper Chrom., , High Performance Liquid Chrom. (HPLC), Ascending, , Thin Layer Chrom. (TLC), Gas-Liquid Chrom. (GLC), Type of chromatography, Stationary phase, , Column/adsorption, Thin layer chromatography (TLC, High performance liquid, chromatography (HPLC), , Solid, Solid, , Mobile phase, Liquid, Liquid, , Solid, , Liquid, , Gas liquid chromatography(GLC, , Liquid, , Gas, , Proper chromatography, , Liquid, , Liquid, , Decending, , Radial, , Application, Large scale separation, Qualitative analysis, Qualitative analysis and also, for separation, Qualitative analysis and also, for separation, Qualitative analysis and also, for separation, , The various components on the developed TLC plate are identified through their retardation, factor, i.e., Rf values., Rf =, , distance moved by the substance from base line, distance moved by the solvent from base line, , A component with highest value of Rf elute first. A component with greater tendency to adsorb, on solid has lesser the Rf value., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 68 of 115

Page 70 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , 2. Qualitative Analysis (Detection of Elements), The qualitative analysis of an organic compound involves the detection of all the elements, present in it. Carbon and hydrogen are generally present in all organic compounds. Other, elements which may be present in organic compounds are oxygen, nitrogen, sulphur, halogens,, phosphorus, etc. These elements are detected by the following tests., Detection of Carbon and Hydrogen by Copper oxide test, Principle Organic compounds undergo oxidation in the presence of a suitable oxidizing agent. In, this process, carbon is oxidized to CO2 and hydrogen is oxidized to water., , Procedure The compound is intimately mixed with dry cupric oxide. The mixture is strongly, heated in a hard glass test tube fitted with a cork and a delivery tube. The liberated gases, are passed into lime water., Organic compound + CuO, , CO2 + H2O +Cu, Carbon Water Copper, dioxide, , Passed into, lime water, Ca(OH) 2, , Passed into, anhydrous, CuSO4, , lime water, turns milky, , Copper sulphate, turns blue, , Carbon (C) present, in organic compound Hydrogen (H) present, in organic compound, , [𝐶], , + 2𝐶𝑢𝑂 →, , ℎ𝑒𝑎𝑡, , 𝑜𝑟𝑔𝑎𝑛𝑖𝑐𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑, , CO 2 +, , Ca(OH) 2 ⎯⎯, →, , lime water, , [2𝐻], , + 𝐶𝑢𝑂 →, , 𝑜𝑟𝑔𝑎𝑛𝑖𝑐𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑, , 𝐶𝑢𝑆𝑂4 + 5𝐻2 𝑂 →, (𝑤ℎ𝑖𝑡𝑒), ,  ACTIVE SITE EDUTECH , , 𝐶𝑂2 + 2𝐶𝑢, , CaCO3 + H 2O, (milky), , ℎ𝑒𝑎𝑡, , 𝐻2 𝑂 + 𝐶𝑢, , 𝐶𝑢𝑆𝑂4 . 5𝐻2 𝑂, (𝑏𝑙𝑢𝑒), , CONTACT: , , Page 69 of 115

Page 73 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , , dil. HNO3, , Organic compound, , Lassaignes extract, , Fused with sodium, (has covalent bonds, with carbon and halogens), , (has NaX, sodium halide), , and AgNO3, , White ppt of, silver chloride, (AgCl) soluble in, ammonia confirms, the presence of, chlorine in, organic compound, , Pale yellow ppt, of, silver bromide, (AgBr) sparingly, soluble in, ammonia, confirms the, persence of, bromine in, organic, compound, , Yellow ppt of, silver iodide, (Agl), insoluble, in ammonia, confirms the, presence of, iodine in, organic, compound, , Note: Beilstein’s test: Beilstein’s test is used to detect the halogen in an organic compound., A copper wire is heated in a Bunsen flame till no colour is imparted to the flame. The copper, wire is dipped in the given organic compound and exposed to the non-luminous zone of the, Bunsen flame. A bluish green coloured flame indicates the presence of halogen., Test for Phosphorus, Organic compound containing phosphrous is fused with sodium peroxide. The phosphorus of, the compound is oxidised to phosphate. The fused mass is extracted with water and filtered., The filtrate containing sodium phosphate is boiled with nitric acid and then treated with, ammonium molybdate. A yellow solution or precipitate indicates the presence of phosphorus., Na + P + O → Na3PO4, Na3PO4 + 3HNO3 → H3PO4 + 3NaNO3, H 3PO 4 + 12(NH 4 ) 2 MoO 4 + 2HNO 3 → (NH 4 ) 3 PO 4 12MoO 3 + 21NH 4 NO 3 + 12H 2O, ammonium molyblate, , , Organic compound, With Na2O, (sodium peroxide), , ammonium phosp hom olybdate, , Phosphorus present in organic compound, change to phosphate (Na, 3PO4), boiled with nitric acid, (HNO3) and added, ammonium molybdate, [(NH4)2MoO4], Yellow ppt of, ammonium, phosphomolybdate, [(NH4)3PO4.12MoO3], confirms the presence, of phosphorus in, organic compound, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 72 of 115

Page 74 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , (3) Quantitative analysis The quantitative analysis deals with the determination of, percentage of various elements. The following methods are employed for the determination, of percentage composition of elements present in the organic compounds., Estimation of carbon and hydrogen by Liebig’s combustion method, Principle: A known mass of organic compound is oxidized with cupric oxide in C02 free, atmosphere. C gets oxidized to C02, H to H2O. Water formed is absorbed in weighed anhydrous, CaCl2 U bulbs and carbon dioxide absorbed in weighed KOH bulbs. Increase in mass of CaCl2, and KOH bulbs are determined. Difference in mass corresponds to the mass of H2O and C02, respectively., Knowing the increase in mass, the percentage of C and H in the organic compound can be, calculated., , 𝜟, 𝒚, 𝒚, 𝑪𝒙 𝑯𝒚 + ቀ𝒙 + ቁ 𝑶𝟐 → 𝒙𝑪𝑶𝟐 + 𝑯𝟐 𝑶, 𝟒, 𝟐, , Calculations, • Let the mass of organic compound taken = m g, • Mass of water formed = m1 g, (increase in CaCl2 U-tube), , •, , Mass of carbon dioxide formed = m2 g, (increase in potash tubes), , (a) Percentage of Carbon, We know that 1 mole of carbondioxide (44 g) contains 1 gram atom of carbon (12g)., 44 g of CO2 contain C = 12g, 12, m2 g of CO2 contains C =, × m2 g, 44, 12m2, Hence Percentage of carbon =, × 100, 44×m, (b) Percentage of Hydrogen, We know that one mole of water (18g) contains 2gram atom of hydrogen (2g), H2O = 2H, 18g of H2O contain H = 2g, 2, m1 g of H2O contains H =, × m1 g, 18, 2m1, Percentage of hydrogen =, × 100, 18×m,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 73 of 115

Page 76 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , =, , 14(V − v)xN1, 1.4(V − v)x N1, 100, ×, =, 1000, w, w, , Percentage of nitrogen in the compoun =, , volume of the acid (cm3 ) × normality of the acid × 1.4, mass of organic compound in g, , Estimation of Nitrogen by Duma’s method, Principle: The organic compound containing nitrogen when heated with excess of copper oxide, in the atmosphere of carbon dioxide, yields nitrogen in addition to carbon dioxide and water., 𝑪 + 𝟐𝑯 + 𝟑𝑪𝒖𝑶 → 𝑪𝑶𝟐 + 𝑯𝟐 𝑶 + 𝟑𝑪𝒖, 𝟐𝑵 + 𝟐𝑪𝒖𝑶 → 𝑵𝟐 +oxide of nitrogen, , or, 𝑦, 𝑦, 𝑦, 𝑧, [𝐶𝑥 𝐻𝑦 𝑁𝑧 ], + ቀ2𝑥 + ቁ 𝐶𝑢𝑂 → ቀ2𝑥 + ቁ 𝐶𝑢 + 𝑥𝐶𝑂2 + 𝐻2 𝑂 + 𝑁2, 2, 2, 2, 2, 𝑂𝑟𝑔𝑎𝑛𝑖𝑐𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑, , Traces of nitrogen oxides formed during combustion of organic compound are reduced to, nitrogen by passing the gaseous mixture over a heated copper gauze. The percentage of, nitrogen present in a given organic compound is calculated from the volume of nitrogen, collected over potassium hydroxide solution from a known mass of organic compound., , Calculations, Let the mass of organic compound = wg, The volume of nitrogen collected = Vcm3, Atmospheric pressure from barometer) = P mm of Hg, 0, , Room temperature = t C, 0, , Aqueous tension at t C = a mm of Hg, Pressure of dry nitrogen = (P – a) mm of Hg, Let us first convert the volume of nitrogen to volume at S.T.P., Experimental conditions, S.T.P. conditions, P1 = (P – a)mmHg, P2 = 760 mm, T1 = (273 + t)K, T2 = 273, V1 = V cm3, V2 = ?, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 75 of 115

Page 78 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Percentage of halogen =, , atomic mass of X, mass of silver halide in grams, , 100, molecular mass of AgX mass of organic compound in grams, =, , 100, Xxa, ×, w, (108+ X), , Here X is the atomic mass of halogen, e.g. Cl = 35.5 , Br = 80 (79.9 exact), I = 127 (126.9 exact), Percentage of chlorine =, , 35.5, mass of silver chloride in grams, , 100, (108 + 35.5) mass of organic compound in grams, , Percentage of bromine =, , Percentage of iodine =, , 80.0, mass of silver bromide in grams, , 100, (108 + 80) mass of organic compound in grams, , 127, mass of silver iodide in grams, , 100, (108 +127) mass of organic compound in grams, , Estimation of sulphur by Carius method, Principle: An organic compound is digested with fuming nitric acid in a sealed tube. The sulphur, present in the compound is quantitatively oxidised into sulphuric acid. Sulphuric acid so formed, is precipitated as barium sulphate by adding excess of barium chloride., ⎯⎯3 → H2SO4, The main reactions are : S + H2O + 3O ⎯HNO, H2SO4 + BaCl2 ⎯→ BaSO4, ppt., , Calculations :, Let the mass of organic compound = w g, Mass of BaSO4 formed = ag, BaSO4, , 137 + 32 + 64 = 32 = 233, , 233 g of BaSO4 contain sulphur = 32 g, ag of BaSO4 will contain sulphur =, Percentage of sulphur =, , 32 x a, g, 233, , Mass ofsulphur, Massoforganiccompound, , 32𝑎, , × 100 =, , 233, , ×, , 100, 𝑤, , 32, mass of barium sulphate in grams, Percentage of sulphur =, , 100, 233 mass of organic compounds in grams, , Estimation of phosphrous by Carius method, Principle: The phosphorus present in the organic compound is oxidised to orthophosphoric acid, by heating with fuming nitric acid. The phosphoric acid so obtained is precipitated as, MgNH4PO4 which on ignition is converted into Mg2P2O7., [C, H, P], , fuming, + [O] ⎯⎯⎯, → H3PO4 + H 2O + CO2, nitric acid, , Organic compound, , H 3PO 4 + [NH 4Cl + NH 4OH + MgCl 2 ] → MgNH 4PO 4, magnesia mixture, ignite, 2MgNH 4 PO4 ⎯⎯⎯, →, , white precipitate, , Mg P O, , 2 2 7, magnesium pyrophoshate, , 62, , + H 2O + 2NH3, , mass of Mg 2 P2O 7 in grams, , Percentage of phosphorus = 222  mass of organic compound in grams 100, ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 77 of 115

Page 79 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Estimation of oxygen, There is no direct method for the estimation of oxygen present in the organic compound. The, percentage of oxygen in the compound is generally estimated by difference., Percentage of oxygen = 100 − [sum of percentage of all other elements present in it], The percentage of oxygen in an organic compound is usually found by difference between the, total percentage composition (100) and the sum of the percentages of all other elements., However, oxygen can also be estimated directly as follows:, A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas., The mixture of gaseous products containing oxygen is passed over red-hot coke when all the, oxygen is converted to carbon monoxide. This mixture is passed through warm Iodine, pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine., heat, ⎯, → O2 + other gaseous products, Compound ⎯⎯, 1373 K, 2C + O2 ⎯⎯⎯→ 2CO]  5, … (A), I2O5 + 5CO ⎯⎯, → I2 + 5CO2 ]  2, … (B), On making the amount of CO produced in equation (A) equal to the amount of CO used in, equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each, mole of oxygen liberated from the compound will produce two moles of carbon dioxide., Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated., Let the mass of organic compound taken be mg, Mass of carbon dioxide produced be m1g,  m1g carbon dioxide is obtained from,  Percentage of oxygen =, , 32  m1, g O2, 88, , 32  m1  100, %, 88  m, , The percentage of oxygen can be derived from the amount of iodine produced also., Presently, the estimation of elements in an organic compound is carried out by using, microquantities of substances and automatic experimental techniques. The elements, carbon,, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN, elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg), and displays the values on a screen within a short time. A detailed discussion of such methods, is beyond the scope of this book., ,  ACTIVE SITE EDUTECH , , CONTACT: , , Page 78 of 115

Page 82 :
GENERAL ORGANIC CHEMISTRY (FULLY SOLVED) FOR CBSE (IIT-JEE) EXAMS (2021, - 2022), , Do you want full versions of all Chemistry (Organic + Inorganic + Physical) notes?, Contact us through the following WhatsApp Number, 9844532971, Or click below link, https://wa.me/message/NZ7T434AHPHOD1, , HOW TO, FOCUS ON, STUDIES, DURING, EXAMS, , Turn Off, The Phone, , Make a List, , Shut Off, Everything, , Find a, Comfortable, Place, , Time Yourself, ?, , Clear Your Desk, ,  ACTIVE SITE EDUTECH , , Award Yourself, , Frame a Picture, Of Your Goal, , CONTACT: , 115, , Page 112 of