LECTURE NOTESSession - 2009-10 ORGANIC CHEMISTRY TOPIC : GOC - I CONTENTS : 1 IUPAC Nomenclature Refer sheet GOC- I JEE Syllabus [2009] Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centers, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections); Page No.1 IUPAC NOMENCLATURE 1.1 INTRODUCTION TO ORGANIC COMPOUNDS:Organic compounds are compounds of carbon and hydrogen and the following elements may also be present: (Halogens, N, S, P, O). There are large no. of organic compounds available and large no of organic compounds are synthesized every year.The most important reason for large no of organic compounds is the property of catenation (self–linkage) in carbon. Element Bond Energy C C–C (strongest bond) Si Si–Si Ge Ge–Ge Sn Sn–Sn Pb Pb–Pb Decreasing order Bond energy depends on (i) Size of atom (Inversely proportional) Size : C–C>Si–Si (ii) E.N difference along period (Directly) E.N Diff : C – H < N – H < O – H < H – F (iii) Bond order (no. of covalet bonds b/w two atoms) (Directly) Bond order : C – C < C = C < C C Catenation in carbon: The element carbon has strongest tendency to show catenation or self–linkage due to – (a) its tetravalency so that it can form bonds with many elements as well as carbon itself. (b) It can form multiple bonds (C = C, C C). Due to its small size, there is efficient colateral overlapping b/w two P–orbitals. Does not exist (c) High bond energy of C – C bond so that it can form strong bonds in long chains and in cylic compounds. 1.2 CLASSIFICATION OF ORGANIC COMPOUNDS : Page No.2 Ex. Compounds Classification Unsaturated Heterocyclic , saturated Unsaturated Saturated Saturated, Alicylic 1.3 IMPORTANT TERMS: Saturated compounds: When all the valencies of an element are satisfied by covalent bonds. Unsaturated compound: When a compound contains one or more bonds (C = C, C = N, N = N, C = O or C C, C N, N N) Molecular Formula (M.F.) : The molecular formula of a compound indicates the actual number of atoms of each element present in one molecule. Structural Formula (S.F.): It indicates the linkage due to covalent bond between different atoms in a molecule. (i) Expanded Structural Formula (E.S.F.) (ii) Condensed Structural Formula (C.S.F.) (iii) Bondline Structural Formula (B.S.F.) Ex. M.F. C3H8 Ex. H Ex. H M.F. C4H10 H H H H H E.S.F. H – C – C – C – H H H H C.S.F. CH3 – CH2 – CH3 E.S.F. H – C – C – C – C – H H H H H C.S.F. CH3 – CH2 – CH2 – CH3 B.S.F. B.S.F. or B.S.F. C.S.F. M.F. Ex. CH 3 – CH – CH 2 – CH 3 CH3 B.S.F. M.F. C6H12 C5H12 Homologous series : Homologous series may be defined as a series of similarly constituted compounds in which the members possess the same functional group, have similar chemical characteristics and have a regular gradation in their physical properties. The two consecutive members differ in their molecular formula by CH2. Calculation of Degree of Unsaturation (DU):(a) It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE) Page No.3 –2H (b) H3 C – H2C – CH3 (DU O) –2H CH3 – C CH or CH2 = C = CH2 or That means Deficieny of 2H is equivalent to 1 DU (c) (i) 1DU = Presence of 1 Double Bond or Presence of 1 Ring closure (ii) 2DU = Presence of 2 Double bond or 1 Triple bond or two ring closure or 1 double bond + 1 ring. (d) G.F. D.U. (i) CxHy y (x + 1) – 2 (ii) CxHyOz yo (x + 1) – 2 (iii) CxHyXs ys (x + 1) – 2 (iv) CxHyNw y–w (x + 1) – 2 (v) CxHyOzXsNw ys– w (x + 1) – 2 Ex Calculate DU of following compounds (a) C6H6O DU = 4 (b) C6H5Cl DU = 4 (c) C6Br6 DU = 4 (d) C5H11OCl DU = 0 (e) C9H12N2 DU = 5 (f) C6N6 DU = 10 (g) C10H8SO5 N4Cl2 e.g. M.F. DU = 8 DU S.F. H 1. C4H6 2 H C–C=C=C H H H H C–C–CC C–CC–C C=C–C=C C CH2 C Page No.4 H | Primary (1°) carbon : | H 1° Secondary (2°) carbon : C | Tertiary (3°) carbon : | 3° H C | Quaternary (4°) carbon : | C Superprimary : 4° (1°) Ex. That means at least 1 Benzene ring is present.2. C7H8 (Aromatic) =4 Note : In case of the aromatic Compounds minimum DU = ‘4’. Degree of carbon : It is defined as the number of carbon atoms attached to a carbon atom. 10 – 5 carbon CH3 CH3 1 2 | 3| 4 5 CH3 – C – CH – CH2 – CH3 | CH3 0 2 – 1 carbon 0 3 – 1 carbon 40 – 1 carbon Page No.5 . C7H6O (Aromatic) =5 4. C2H2Cl2 =1 Total isomers = 3 3. yne one bond diyne two bond (ii) Secondary Suffix:.It is used for the principal functional group. bromo.It indicates saturation or unsaturation existing in the main chain. viz. of carbon atoms present in the main chain. Type of C – C bonds & type of replaceable H-atoms in saturated hydrocarbon. Note : Total no. chloro.4 IUPAC NOMENCLATURE OF ORGANIC COMPOUNDS: General Scheme of Naming:Secondary Prefix + Primary Prefix _ _Word Root_ _Primary Suffix_ _Secondary suffix The organic compound is always named according to the general scheme as given by IUPAC.5 NAMING OF SATURATED HYDROCARBONS Rules :(Branched and substituted Alkanes) (1) Selection of parent chain (a) Chain with maximum number of ‘C’ atoms (longest chain). The following substituent groups are always cited in the prefix. Prefix : It is the first part of the name. It is represented as Alk. (i) Primary Prefix : ‘Cyclo’ (ii) Secondary Prefix : Normal substitutents and junior functional groups are treated as a substituent than their name is treated as secondary prefix. word root and primary suffix. In every compound.6 . of monosubstituted product does not depend upon type of C-atoms but depends upon type of replaceable H-atoms. iodo) (iv) –NO2 Nitro (v) –N = O Nitroso (vi) Junior functional group The prefixes are always written in alphabetical order The position of substituent group in the main carbon chain is mentioned by writing the number just before the name of substituent by writing a small dash (–). *Degree of hydrogen is same as the degree of carbon to which it is attached.Ex. always exist. Page No. Suffix : (i) Primary Suffix:. two parts. 1. (i) R – Alkyl (ii) R – O – Alkoxy (iii) X halo (fluoro. Word Root: It indicates the no. 1. ane single bond (saturated) ene one = bond diene If two double bonds Polyene if plenty of double bond are present. 7 .(b) If number of carbon atoms are same in more than one longest chain then more substituted longest chain will be the parent chain. Methylpropane CH3 CH3 CH3 – CH – CH – CH3 2. * If alphabets are also same then numbering is done from that side of the parent chain having its substituent of substituent at lower number. (c) If number of side chain are also same then that will be parent chain having its substituent at lower number. 3-Dimethyl butane – CH3 CH3 – C – CH3 CH3 – – Ex. then numbering is done from alphbetical order. – – CH3 CH3 – CH – CH3 Dimethyl propane 2-Methylbutane Page No. C C C–C–C–C–C–C–C–C C–C–C C (2) Numbering (a) Numbering is done from that side of the parent chain having it substituent at lower number (lowest set of locant) (b) If position of substituent are same from both the end of the parent chain. 4-Diethyl-2.(a) –H CnH2n + 2 CnH2n + 1 Alkyl Radical –H CnH2n CnH2n – 1 Alkenyl Radical –H CnH2n – 2 CnH2n – 3 Alkynyl Radical R (structural formula) Name CH3 – Methyl CH3 – CH2 – Ethyl Page No.8 . 3-dichloro hexane – – – – – – Cl C–C–C–C–C Br Cl Cl Cl Br (3) Rules for writing Alkyl Radicals:Radicals:. 5-Dimethyl heptane – – – – – – C–C–C C–C–C–C–C C–C–C 3-Ethyl hexane CH3 – CH2 – CH2 – CH2Cl 1-Chlorobutane C–C–C–C–C Cl Br 3-Bromo-2-chloropentane C–C–C–C–C Br Cl 2-Bromo-4-chloro pentane – – – – – C–C–C–C–C C–C–C 4-Bromo-2.– C C–C–C–C–C – 3-Ethyl-2-methyl hexane C–C–C 4. 2-dichloro pentane C–C–C–C–C–C 5-Bromo-2. 4-Diethyl heptane C C–C–C C–C–C–C–C C–C–C C 4. 2 – Dimethylpropyl) nonane 5 – (1.1 – Dimethylpropyl) nonane 5 – (1 – Ethylpropyl) nonane 5 – (Dimethylethyl) nonane Alkene / Alkyne radicals Alkene Alkenyl Alkyne CH2 = CH2 CH3 – CH = CH2 Alkynyl CH2 = CH – ethenyl Prop – 1 – enyl Methylethenyl Prop – 2 – enyl Page No.9 .CH3 | (b) CH3 – C – CH3 | H CH3 3 2| 1 –H CH3 – C – CH2 – Isobutyl | H –H CH3 | CH3 – C – CH3 Tert butyl (Tertiary butyl) | (4) Systematic Names of Radicals : Complex Radical : Ex 5 – ( 1. Tris Page No. Benzyl 3. Tert are not considered for alphabetization (c) Bis . -naphthyl (6) Numeral Prefixes (1) Following prefixes are considered for alphabetization : (a) Iso (b) neo (c) Di. Tri. Benzo 5. (b) Sec. Alkylidene 8. Phenyl 2. Tetra of complex radical are considered for alphabetization (2) Following are not considered for alphabetization (a) Di. m. Tolyl CH3 (o. Tri.10 .CH3 – C CH CH3 – C C – Prop-1-ynyl CH3 – C CH – CH2 – C C – H Prop-2-ynyl (5) Benzene Radical 1. p) Methylene 7. Tetra for simple radicals.naphthyl 9. . Benzal 4. 2. 2-dimethyl No other substituent.(7) Retained Names of Alkanes: (1) Normal (n) : Radical or hydrocarbon which has straight chain and if it has free valency it must be present at either of ends. C C– C–C– C–C C C (3) Neo : – Iso-octane (commercial name) in petroleum industry – – Q. 4-trimethylpentane There should be 5 to 6 carbon atoms 2. IUPAC name is 2. – – C C– C–C C Neopentane C C – C – C – C Neohexane C – – Ex. Page No. Ans.11 . C – C – C – C (n butane) (2) Iso : Two methyl group at the end of linear chain (unbranched chain) Isobutane or methyl propane (ii) C C– C –C–C Isopentane or methylbutane C C– C –C–C–C Isohexane or 2-methylpentane – – – (i) C C– C –C (iii) Conditions :- There should be 4 to 6 carbon atoms only There should not be any other alkyl group present in the chain. If first and second factor are common then chain with lowest locant (multiple bond) is selected as main chain. (Multiple bond prior to substituent). (C) Rules for Numbering: Lowest locant Rule till first point of difference (irrespective of double bond or triple bond). Alphabetization. 5-triene 3 5 Page No.1 (5) Tertiary : First member NAMING OF UNSATURATED COMPOUND (ALKENES AND ALKYNES) : (A) General formula:.12 . naming and longest chain selection. 3.CnH2n and CnH2n – 2 respectively (B) Rules for selection of main chain: Longest carbon chain with a multiple bond.3-diene (b) C C – C C Butadiyne (c) C = C – C C Butenyne – – 1. Longest carbon chain with maximum number of multiple bonds. Ex.(4) Secondary : It is applicable only for radical. CH3 | CH3 – CH – CH = CH – CH3 4-Methylpent-2-ene Cl 5 4 3 2 1 CH3 – C – CH2 – CH2 – CH = CH2 Cl 5. 5-Dichlorohex-1-ene (a) C = C – C = C Buta-1. Then double bond is prefer over triple bond in numbering.6 6 CC–C–C=C–C 1 2 3 5 4 2 1 4 6 Hex-4-en-1-yne 6 Hexa-1. Lowest locant Rule is followed till first point of difference. If all the other factors are common CH3 – CH2 – CH = CH2 But-1-ene CH3 – CH = CH – CH3 But-2-ene CH3 CH3 – CH – CH2 – CH = CH2 4-Methylpent-1-ene – Ex. of substituents > Nearest locant > Alphabetization. (b) If all factors are similar in cyclic and acyclic part.1-Dimethylpropyl)-4-ethenylhepta-1.4 4-Ethenylhept-2-en-5-yne Ex. If cyclic part constitutes the side-chain (substituent) then prefix cyclo is considered for alphabetization:- Ex:- Cyclopropane Cyclobutane Page No.3-Dimethylhex-1-en-4-yne 1.2 4 -(1.13 .. 2-Dimethylbutyl) hept-2-en-5-yne Ex. of carbon atoms > Maximum no.Ex.7 NAMING OF CYCLIC HYDROCARBON (ALICYCLIC COMPOUNDS) : (A) Main chain selection: (a) Multiple Bond > No.6 2. 4-Dimethylpenta-1. then Cyclic > Acyclic (B) Numbering: (a) Lowest Locant (b) Alphabetization (C) Naming: Prefix ‘cyclo’ is used just before the word root if it constitutes the main chain. 3-diene Ex. 5-diene Ex.5-diene Ex.1 3-(2-Methylpropyl) hept-1-ene Ex.6 2. If cyclic part is the main chain then the prefix ‘cyclo’ is not considered for alphabetical order.5 4-(1.3 3-Ethynylhexa-1. 14 .Cyclohexane Cycloheptane Cyclooctane Methylcyclopropane Ethylcyclopropane Propylcyclopropane 1-Cyclopropylbutane – Cyclopentane – C – C – Cl 2-Chloroethylcyclopropane – Cl – C – C Chloroethylcyclopropane – Cl C–C–C – 3 2 1 – Cl 3 C–C–C 1-Chloro-3-cyclopropylpropane 1 2 – – – Cl C–C–C 1 2 3 2-Chloro-1-cyclopropylpropane 1-Chloro-1-cyclopropylpropane – – C–C–C 1-Chloro-2-propylcyclopropane Cl – 1 2 3 F – 4 Br – – Cl 2-Bromo-1-chloro-3-fluoro-4-iodocyclohexane I Methylethylcyclopropane or isopropylcyclopropane – Cl – Br 4 5 I – 2 3 1 6 1-Bromo-2-chloro-4-iodocyclohexane 6 5 1 2 3 4 3-Chlorocyclohex-1-ene Cl Br 6 1 5 2 3 4 5-Bromo-3-chlorocyclohex-1-ene Cl Page No. 8 FUNCTIONAL GROUP TABLE (Seniority order): Class Name Suffix Prefix 1. R–C–O–C–R || || O O Alkanoic anhydride – anhydride ------------ 4. Page No. R–C–H || O Alkanal – al (carbaldehyde) formyl / Oxo Alkanone – one Oxo / Keto – ol hydroxy 9. + Sub.G. > substituent > Alphab) (C) Naming:. > Max.G. R–C–X || O Alkanoyl halide – oyl halide (carbonyl halide) halo carbonyl 6. R – SO3H Alkane sulphonic acid – sulphonic acid sulpho 3. R – COOH Alkanoic acid – oic acid (carboxylic acid) Carboxy 2. R – NH2 Alkanamine – amine amino 1.15 . (Similar group) > Multiple bond > Max.General scheme : The senior most functional group constitutes secondary suffix. no. R – COOR Alkyl alkanoate – alkanoate (carboxylate) Alkoxy carbonyl 5. > M. Other junior F.G. (i) Lowest locant (F.9 NAMING OF FUNCTIONAL GROUP CONTAINING COMPOUNDS : (A) Selection of Main Chain: Senior F.): The carbon atom bearing functional group (or C – atom of terminal functional group) is given lowest possible number. no.B. of F. R – OH Alkanol 11. no.’s are written as prefix in alphabetical order. R – C – NH2 || O Alkanamide – amide (carboxamide) Carbamoyl 7. R – SH Alkanethiol – thiol mercapto 12.1.G.G. of locants > lowest locant > alphabetization (B) Numbering (See F. R–C–R || O 10. R–CN Alkanenitrile – nitrile (carbonitrile) cyano 8. of ‘C’ atoms > Max. 9.1.1 Carboxylic acid F.'C' of COOH is not considerd in parent chain Alkane carboxylic acid Carboxy Rule : If first alphabet of sec. Prefix – SO3H Sulpho Ex. – COOH Prefix Suffix IUPAC name Oic acid . u. i. y then ‘e’ of primary suffix will be dropped. Suffix Sulphonic acid IUPAC name Alkanes sulphonic acid (1) CH3 – SO3H Methanesulphonic acid (2) C – C – C – C – C SO3H Pentane-2-sulphonic acid – 1.G.16 .2 (9) HOOC – COOH Page No.G.9. 3-tricarboxylic acid Sulphonic acid F. 2. Ex. suffix is begin from a.'C' of COOH considered in the parent chain Alkanoic acid Carboxylic acid . o. (1) HCOOH Methanoic Acid (2) CH3 – COOH Ethanoic Acid (3) C – C – COOH Propanoic Acid (4) C – C – C – COOH Butanoic Acid 2 3 4 5 6 (5) C – C – C – C – C – C – C – C 1 COOH 2-propylhexanoic acid – COOH (6) Cyclopropanecarboxylic acid 2 1 (7) – CH2 – COOH Cyclopropylethanoic acid – COOH 3' C–C–C 1' 2' 4-(2-carboxypropyl) cyclohexane-1-carboxylic acid – (8) COOH Ethanedioic acid (10) HOOC – CH2 – COOH Propanedioic acid (11) HOOC – CH2 – CH2 – CH2 – CH2 – COOH Hexanedioic acid (12) Cycloprop-1-ene-1. 9. (1) H– C –O– C –H O (2) O C C C–C–C–O–C–C–C O Methanoic Anhydride (unstable ) 2-Methylpropanoic anhydride O (3) Butanedioic anhydride (4) C – C – C – C – O – C – C – C – C || || O O Butanoic anhydride O || C O (5) Cyclohexane-1.3 5-Sulphopentanoic acid Anhydride (a) R – C – O – H + H – O – C – R Heat –H2O O O R– C –O– C –R O O Alkanoic anhydride (b) Mixed Anhydride (according to alphabet) CH3 – COOH + Ethanoic acid CH3 – CH2 – COOH Pr opanoic Acid CH3 – C – O – C – C2H5 O O Ethanoic propanoic anhydride Ex.SO H – 3 (3) Cyclobutanesulphonic acid 1 2 3 4 5 (4) HOOC – C – C – C – C – SO 3H 1. 2-dicarboxylic anhydride C || O (6) –C–O–C– O (7) Cyclopropane carboxylic anhydride O Benzene dicarboxylic anhydride (Phthalic anhydride) Page No.17 . alkanoate or Alkanoyloxy or carboxylate O (Ester) or Alkyl.Group Prefix –C–X Halo Carbonyl O Suffix Name Alkanoyl halide Oyl-halide or or Carbonylhalide Alkane carbonyl halide – O = C – Cl Example:- (1) Benzene carbonyl chloride Page No..5 Acid halide:F.9..group Prefix Suffix Name – C – OR Alkoxy Carbonyl Oate Alkyl....Alkanecarboxylate –H2 O R – C – O – R' R – C – O – H + H – O – R' O O Example:(1) H – C – O – CH3 Methyl methanoate O (2) C – C – C – O – C – C – C Propyl propanoate O (3) C – C – O – C – C – C – C Ethyl butanoate O – C (4) C – C – C – O – C – H Isobutyl methanoate O 3 2 1 (5) C – O – C – C – C – COOH 3-(Methoxycarbonyl)propanoic acid O (6) C – C – O – C – C – COOH 3-(Ethanoyloxy)propanoic acid O (7) 2-(Methoxycarbonyl)-4-(Propanoyloxy) cyclohexan-1-carboxylic acid O – – C O– (8) Cyclopropyl benzene carboxylate O (9) – O – C – CH2 – CH3 Methyl-3-propanoyloxy cyclohexane carboxylate – – O=C OCH3 1.9..4 Ester : F.18 .1.. 9. 7-dioyl Chloride – – C – CH2 – CH – CH = CH – CH 2 – COCl Cl CH 3 (8) 4-Bromocarbonyl-2-chlorocarbonyl-5-propanoyloxycycloexane-1-carboxylic acid 1.Group Prefix Suffix Name – C – NH2 Carbamoyl amide or Carboxamide Alkanamide or Alkane Carboxamide O Example:- (1) H – C – NH2 Methanamide (Formamide) O (2) CH3 – C – NH2 Ethanamide (Acetamide) O (3) CH3 – C – NH – CH3 N-methylethanamide O – (4) CH3 – C – N – C – C – C O N-butyl-N-propylethanamide C C C C Page No. 5-Dichlorocarbonylpentanoic acid – COOH COOH – – (6) 3-Chlorocarbonyl cyclopropane-1.6 Amide:F.19 . 2-dicarboxlic acid C – Cl O O (7) 5-Methylhept-3-ene-1.(2) CH3 – C – Cl Ethanoyl chloride O (3) Prop-2-enoyl bromide (4) 6-Methyl hept-4-en-2-ynoyl chloride O – Cl C (5) CH2 – CH2 – CH2 – CH – C – Cl COOH O – – 5. – O = C – NH2 (5) Benzene carboxamide – O = C – NH2 Benzene-1. 2. N’-Dimethyl propane-1.2. 3-dicarboxamide – (6) C – NH2 O – COOH (7) 3-Carbamoyl cyclohexane-1.carboxylic acid – C – NH2 O O C – NH – CH3 (8) CH2 N.7 (5) Benzene carbonitrile (7) CH2 – CH – CH2 CN CN CN Propane-1.9. 3-tricarbonitrile – – CN CN CN – 1.20 .3-tricarbonitrile – – 3-Cyano propanoic acid – CH2 – CH2 – COOH CN – (6) Page No. 3-diamide C – NH – CH3 O Nitrile:- (1) H – CN Methanenitrile (2) CH3 – CN Ethanenitrile (3) CH3 – CH – CH2 – CN CH3 – Example:- 3-Methyl butanenitrile – CN (4) Cyclopropane-1. 5.Group Name Formyl or oxo al or carbaldehyde Alkanal or Alkane Carbaldehyde – CH = O (1) HCHO Methanal (2) CH3 – CH2 – CH = O Propanal (3) CCl3 – CH2 – CH2 – CH2 – CH = O 5.8 Isocyanide :- F. 3-dicarbaldehyde – CH = O (8) HOOC – CH2 – CH2 – CH2 – CH = O – (9) HOOC – CH2 – CH – CH2 – COOH CH2 – CH = O 5-Oxopentanoic acid 3-(2-Oxoethyl) pentane-1. 3-diphenyl propanal CH2Cl – CH = O (7) Benzene-1.carboxylate CH2– CH2 – CH = O Page No.Group Prefix Suffix Name –NC Isocyano Isocyanide Alkyl Isocyanide Example:- (1) CH3 – N C Methyl isocyanide (2) CH3 – CH2 – N C Ethyl isocyanide – NC (3) Aldehyde:Prefix F.1. 5-dioic acid – COOC2H5 (10) – Example:- Suffix – 1.9 Phenyl isocyanide ` Ethyl 3-(3-oxopropyl) cyclohexane-1.9.21 . 5-Trichloro pentanal – CH = O (4) Cyclohexane carbaldehyde – CH = O (5) Benzaldehyde/Benzene carbaldehyde Ph (6) – CH – CH Ph – CH = O – 2-Chloromethyl-3.9. 3.11 Alcohol:F.1.10 Ketone:Prefix F. 5-Dimethylhexane-1. 2. 5-trione O O O – CH2 – C – CH3 (4) 2-(2-Oxopropyl) cyclohexanone O – COOH 3-(3-Oxobutyl) cyclohexane -1-carboxylic acid – (5) CH2 – CH2 – C – CH3 O (6) OHC – CH2 – C – CH3 3-Oxobutanal O 1.9. 4. 6-diol – – – (1) CH3 – CH – CH2 – OH CH3 – HO – – OH – (3) – OH HO – Cyclohexane-1. 3. 6-hexaol OH OH (Glyceraldehyde) 2.Group Prefix Suffix Name –OH Hydroxy ol Alkanol 2-Methyl Propanol (2) CH 3 – CH – CH 2 – CH 2 – CH – CH 3 CH 2 – OH CH 2 – OH 2.9.Group Name one Alkanone – Keto or oxo C=O – Example:- Suffix (1) CH3 – C – CH3 Propanone O O (2) Cyclohexanone O (3) Cyclohexane-1. 5. 3-Dihydroxy propanal – – (4) CHO CH – OH CH2 – OH – OH Cl – CH2 – CH – CH2 – CH – CH2 – CH2 – CH2 – CH2 – OH – (5) – CH3 – CH2 – CH2 OH Cl – Example:- 7-Chloro-5-(3-Chloro-2-hydroxy propyl) octane-1. 6-diol Page No.22 . 5-trione – O OH O (Glycerol) Propane-1. 6-trihydroxy cyclohexane-1.23 .9. N-Dialkylalkanamine R' (d) R – N – CH3 N-Ethyl-N-methylalkanamine CH2 – CH3 (1) CH3 – NH2 Methanamine (2) C – C – C – C – C NH2 Pentan-3-amine (3) C – C – C – NH – C – C N-Ethylpropan-1-amine (4) C – C – C – N – C N-Ethyl-N-methylpropan-1-amine – – Ex:- C C N-Ethylpropan-2-amine – (5) C – C – C NH – C – C – OH (6) CH2 – CH = CH – CH – CH3 NH – CH2 – CH3 – – (7) OHC – CH2 – CH – CH2 – CH3 4-(N-Ethyl amino) pent-2-en-1-ol 3-(N-Phenyl amino) pentanal NH – Ph If hetero atom’s are count as a ‘C’ atom in the parent chain then they are written as (1) –NH– Aza (2) –O– Oxa (3) –S– Thia (4) –Se Selena (1) HOOC – CH2 – NH – CH2 – CH3 3-Aza pentanoic acid H N – Example:- (2) Aza Cyclo pentane Page No. 3-triol – – – (7) CH2 – CH – CH2 OH OH OH 1.O OH – (6) – HO 2. 3.12 Amines:F.Group Prefix Suffix Name –NH2 amino amine Alkan amine (a) R – NH2 1º = amine Alkanamine (b) R – N – R' 2º = amine H N Alkylalkan a min e (R' ) (R ) (c) R – N – R' 3º = amine N. 4. 2. Group Prefix Suffix Name R – O – R' alkoxy – Alkoxy alkane Alkoxy + Alkane less no.24 .– H N (3) 3-Methyl aza cyclo hexane – CH3 1. – OH – Br 1 2 6 (9) – 3 5 4 6-Bromo-5-methoxycyclohex-2-en-1-ol O – CH3 – OH 1 Br – 2 6 (10) 3 – 5 4 6-Bromo-5-ethoxycyclohex-3-en-1-ol O–C–C Page No. numbering always starts from senior most functional group. of ' C' of ' C' Methoxymethane (2) C – O – C – C Methoxyethane (3) C – O – C – C – C 1-Methoxypropane (4) C – C – C O C 2-Methoxypropane – (i) Acyclic Ethers:(1) C – O – C – – C (5) C – C – C – O – C – C 1-Isopropoxypropane 1-(Methylethoxy) propane – O–C–C (6) Ethoxycyclohexane – O– (7) Cyclopropoxycyclohexane – Br O–C–C–C – 3 (8) 4 2 5 3-Bromo-6-chloro-2-propoxycyclohexan-1-ol 1 – – 6 OH Cl In cyclic system.10 ETHERS (R – O – R’):F. more no. 10-Tetraoxacyclododecane 1.25 .11 AROMATIC COMPOUNDS: Classification: H Cation 2 e– Cyclopropenium ion Cation 2 e– Cyclobutenium dication – H – Non-Benzenoid: – + H – – + + – H H – H H Page No. 6. 6. 13-Tetraoxapentadecane 2 3 1 12 4 O O O O7 5 (4) 11 10 6 8 9 1. 4. 3-Epoxypropane Polyethers:(1) C – C – O – C – C – O – C – C 1 2 1. 3-Epoxypropane (3) C – C – C O 1. 2-Epoxypropane – Cl (4) C – C – C O 1-Chloro-2. 9. 2-Diethoxyethane 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 10 11 (2) C – C – O – C – C – O – C – C – O – C – C 3. 9-Trioxaundecane 12 13 14 15 (3) C – O – C – C – C – O – C – C – O – C – C – C – O – C – C 2.(ii) Cyclic Ether: (3-membered Ring) Hetero cyclic compounds (1) C – C O Oxirane or Epoxyethane (2) C – C – C O 1. 7. 3.. 5-Trimethyl benzene (Symmetrical Trimethyl benzene) Page No. 2.26 . then only 3 trisubstituted benzene derivative are possible. 3-Dimethylbenzene (Meta-xylene) – CH3 – CH3 (4) (Para) 1. 4-Trimethyl benzene (Unsymmetrical Trimethyl benzene) C – C C – Q. 4-Dimethylbenzene (Para-xylene) – CH3 Write structures of – C–C (i) Metadiethylbenzene – C–C – C–C (ii) 1-ethyl-4-isopropylbenzene – C–C–C (iii) Orthodiethenylbenzene – – C=C C=C Trialkyl Substituted Benzene: If all the three substituents are similar. 3-Trimethyl benzene (Vicinal Trimethyl benzene) C – – – C – C 1. 2. Naming of Aromatic Hydrocarbons (Arenes): – CH3 (1) Methylbenzene (Toluene) – CH3 – CH3 (2) (Ortho) 1.– Anion – – . 2-Dimethylbenzene (Ortho-xylene) – CH3 (3) (Meta) 1. – – C 1. H H – H – H – + 6 e– Cycloheptatrienyl cation (Tropylium cation) H – – – H Cyclopentadienyl anion H H – –H – H H 6 e– H Benzenoid Aromatic Compounds: All organic compounds which contain atleast one benzene ring are known as benzenoid aromatic compounds. – C C 1. O Phenyl ethanoate O – C – Cl 11. Benzene carboxylic anhydride (Benzoic anhydride) – SO3H 8. Benzene carbonyl chloride (Benzoyl chloride) O – NH – C – CH3 12. C–C C – C 1-Chloro-4-methyl-2-nitrobenzene – – 2. Benzene carboxylic acid (Benzoic acid) O O – – C–O–C 7. NO2 Cl C=C 3.Examples : Isopropylbenzene (Cumene) – – 1. Benzene sulphonic acid O – C – O – C2H5 9. 1-Phenylprop-1-ene – COOH 6.27 . Phenylethene (Double bond > Phenyl) (Common name : Styrene) – C–C=C 4. N-Methylbenzene carboxamide Page No. Ethyl benzene carboxylate (Ethyl Benzoate) – O – C – CH3 10. N-Phenylethanamide O – C – NH – CH3 13. 3-Phenylprop-1-ene (Allylbenzene) – CH = CH – CH3 5. – CN 14. Benzene carbonitrile (Benzonitrile – popular) – CH = O 15. Benzene carbaldehyde (Benzaldehyde – popular) O – – C 16. Diphenyl ketone (Benzophenone) O – – C 17. CH3 Methyl phenyl ketone (Acetophenone) – OH 18. Phenol – OH 19. ––– CH3 Methylphenol [o-Cresol, m-cresol, p-cresol] – OH OH – 20. o-Hydroxyphenol (Catechol) – OH m-Hydroxyphenol (Resorcinol) – 21. OH – OH 22. p-Hydroxyphenol (Quinol)(Hydroquinone) – OH – NH2 23. Benzenamine (Aniline) – SH 24. Benzenethiol O O – C – CH3 O 25. –C–O–H 2-Ethanoyloxy (Acetoxy) benzene -1-carboxylic acid (Aspirin) Page No.28 LECTURE NOTES Session - 2009-10 ORGANIC CHEMISTRY TOPIC : GOC - I CONTENTS : 1 Isomerism * Structure Isomerism * Stereo Isomerism Geomertical Optical Conformational Refer sheet GOC- I JEE Syllabus [2009] Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centers, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections); Page-29 Isomerism Isomers: Compounds with same general formula or molecular formula but different physical and chemical property. Ex:- CH3 – CH2 – OH and CH3 – O – CH3 Ex:- CH3–COOH and HCOOCH3 Homologs: Compounds with same general formula differing by same structural unit – CH2 – or molecular weight by 14 unit. Ex:- CH3 – OH and CH3 – CH2 – OH Difference between isomers and homologs:- Page-30 Structural isomers: When two or more number of organic compounds have same molecular formula but different structural formula these are called structural isomers. Stereo isomers: When two or more compounds have same Molecular Formual (M.F.) and same Structural Formula (S.F.) but have different stereochemical formula (S.C.F.), these are called stereoisomers. Stereo Chemical Formula (S.C.F) : It indicates different arrangements of atoms or groups in space around a stereo centre or it indicates different spatial orientations of atoms or groups around a stereo centre. C4H8 CH3 (iii) CH 3 – C = CH 2 – M.F. S.F. (i) CH3 – CH2 – CH = CH2 These are structural isomers. For (ii), S.C.F. are H3C H CH3 H H3C and C=C (ii) CH3 – CH = CH – CH3 C=C H CH3 Stereocentre These are stereoisomers. H Structural isomers are also known as:(i) Skeletal isomers (ii) Linkage isomers (iii) Constitutional isomers Chain isomers : They have different size of main carbon chain and / or side alkyl chain (i) The two chain isomers should have same nature of F.G./multiple bonds/substituents (except –R group) (ii) The position of F.G./M.B./substituent (locants) is not. considered here. Ex:- Alkanes:(a) C1 – C3 (b) C4H10 chain isomers not possible H3C – CH2 – CH2 – CH3 and CH3 – (c) C5H12 CH3 – CH2 – CH2 – CH2 – CH3 , CH3 – CH2 – (d) C6H14 (i) CH3 – CH2 – CH2 – CH2 – CH2 – CH3, (ii) CH3 – CH – CH2 – (iii) CH3 – CH2 – (v) CH3 – CH2 – – CH3 – CH2 – CH3, (iv) CH3 – – CH3 , – CH3 – – CH3 – CH3, – CH3, – CH3 (i) and (ii) – Chain isomers (i) and (iii) – Chain isomers (i), (iv) – Chain isomers (i), (v) – Chain isomers (ii), (v) – Chain isomers (ii), (iii) – Position isomers (iv), (v) – Position isomers Ex. Ans. and shows which types of isomerism ? Chain Isomers Page-31 Cl Write all Positional isomers of Dichlorocyclobutane ? Cl Cl– .) or substituents in the same skeleton of C-atoms. (ii) – Chain (ii). C – C – C – C – NH2. – (i) C – C – C – C – C – C – Ex. Cl – Cl– Cl – Ans. (iii) – Position (i). – Cl Q. C–C=C–C (3) C C – C – C – C . – C C – C – C – NH2 Positional isomers: They have different position of locants Functional group (F. C – C – C – C – OH OH Write all Positional isomers of Dichlorobenzene ? – Ex. or M. . – (1) – Ex. (4) . . – – Ans.G. C–CC–C–C (4) C – C – C – C . – . – . The skeleton of C-atom should not change. Page-32 . Cl Cl – – Cl – Q. should not change. – Cl– Ans. Nature of F. . – Cl C – C (iii) C – C – C – C – C – C – C–C – (ii) C – C – C – C – C – C. (3) – . – Cl– Cl – Cl Write all Positional isomers of Dichlorocyclopentane ? Cl Cl Cl – – – Cl – Q. .B. Write chain isomers of N-alkanamine (1º) containing 4 carbon atoms ? Ans. (1) C – C – C – C – C . Cl Write all Positional isomers of Dichlorocyclopropane ? Cl Q. C–C– C –C–C C C – – (2) C = C – C – C .Q.G. – Cl Ans. (iii) – Chain – C6H12 (Cycloalkanes):(2) .) or Multiple Bond (M. C C (i). .B. 2º.– – – (10) – – – Ans. The chain and positional isomerism is ignored (not considered). (a) C – C – NH – C 1º amine 2º amine C–C–CN Cyanide (Propanenitrile) C – C – NC Isocyanide (Ethane isocyanide) (4) (5) (6) .). (11) .Following compounds don’t exist at room temperature therefore not consider as a structure isomer (ii) – C C – OH (iii) – C – OH – – – (i) – C = C – OH – (iv) – C – OH OR – (v) – C – O – C = C – – OH (vi) Any peroxy compound OH Page-33 . (b) are not functional isomers Alkyne (b) Cycloalkene among themselve (c) Bicyclo (2) C – C – C – OH Alcohol C–C–O–C Ethers (3) C – C – CH = O C– C –C Aldehydes O Ketones C – C – COOH C– C –O–C Carboxylic acids O Esters C – C – C – NH2 .. Compound Functional isomer C–C–C–C Nil C–C–C=C Compound C–C–CC Isomer Functional (a) C = C – C = C Alkadiene Remarks (a). 3º amines are functional isomers – Important point :. C–C (9) C–C–C – – (5) C C–C . (7) . (1) (Ring-chain isomers are Functional isomers) 1º. (b) C – N – C C 3º amine – Ex. . – – – – (6) – . C–C – (3) and (4) – Positional isomer (3) and (6) – Positional isomer (4) and (6) – Positional isomer (5) and (10) – Positional isomer (All are chain isomers of 1) Remaining are chain isomers Functional isomers: They have different nature of functional group (F. (8) ..G. C C– C –O–C–C (6 ethers). C2H3N Ans. C C C – C – C – O – C – C.F. C – – – – C – C – C – O – C. C3H9N 2º Amine (R – NH – R) 3º Amine R – N – R R G. CnH2n + 1N 2 n (1) 2 M. CnH2n – 1N n 2 M. C – C – C – C – C. (i) C – C – C – C – C – OH. – C – C – C – O – C. N. G. OH C C – C – C – OH C – – – – – C C – C – C – C – C. (c) >C = O (keto) group does not show metamerism Following functional groups show metamerism Ethers (All isomeric ethers are metamers) (b) R’ – NH – R 2º Amines (c) R’ – N – R R'' (d) R – S – R’ 3º Amines – (a) R – O – R’ Thioethers Page-34 .F. 1º Amine (R – NH2) Q – – C – H2C– – CH2 Azine N H – Q. OH C C – C – C – OH. G.F. C C – C – C – C – OH. then these are called metamers. Conditions: (a) Same functional group (b) Chain or position isomerism is not considered. – Q. S) but have different nature of alkyl or aryl (aromatic radical) group attached to hetero atom. (7 alcohols) C C – C – O – C. – C – C – C – C – O – C. CnH2n + 3N n (1) 3 M. C – C – C – C – OH. Cyanides and isocyanides H3 C – C N CH2 = C = NH HC C – NH2 H3 C – N C Metamers: When two isomers have same functional group (containing a hetero atom –O.F.Write acyclic isomers of C5H12O Ans.F.F. C2H5N 1º Amine (1) C = C – NH2 Ethenamine 2º Amine 3º Amine C–C=N C–N=C Ethanimine N-Methylene methanamine (Unstable at room temperature) Q. Ans.(e) R – C – O – R’ Esters O (f) R – C – O – C – R’ Anhydrides O O O (g) R – S – O – R’ Sulphonate esters O Q. Cyclic compound (P) contains 18 1º H atoms. – C C CH3 Page-35 .F. – Ans. Write it Structural formula ? C Q. 3º Amines of M. – C – C and Show which type of isomerism ? – C Q. C–C–C– N–C C – C–C– N–C–C C – Ans. two types of ‘C’ atoms. All H are identical. It has 12 1º H atoms. – – Metamerism – – Ans. It has 2 types of H-atoms and 3 types of C-atoms. C C O C H and C C O C are metamers || || O O Q. – – or – CH3 H3C – O – C – CH3 – – Ans. Identify (A). C – C C Compound X is an ether. Identify (P). Acyclic compound (A) contains 18 1º H atoms. two types of ‘C’ atoms. How many esters are possible for C3H6O2 ? Ans. All H are identical. C C C – C – C – C C8H18 C C Q. C5H13N (All metamers) C– C– N–C O–C – – O–C–C Q. a ring structure. asymmetric carbon atom (*Cabcd). Geometrical Isomerism 1.g. E. eclipsed ethane and staggered ethane These isomers change into each other at room temperature and can never be isolated. So these are not considered as true isomers. a b a (I) b Page-36 . CH3 C=C H CH3 CH3 H H (I) C=C (II) H CH3 CH3–CH2–CH=CH2 (III) I.Stereoisomers (Classification) Stereoisomers : The stereoisomers has different orientation of groups along a stereo centre. Cause of Geometrical isomerism : Geometrical isomerism arises due to the presence of a double bond or a ring structure C = C. structural formula and molecular formula but different stereochemical formula. Such isomers are called conformational isomers. or Ring structure (Stereo centres) Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecule exist in two or more orientations. III I. a a C=C b (I) b The root form of geometrical isomers lie in restricted rotation.g.g. This rigidity to rotation is described as restricted rotation/hindered rotation/no rotation. III II. C = N. Cis–2–Butene Trans-2-Butene Conformational isomers : When different orientations arise due to the free-rotation along a sigma covalent bond. E.g. A stereocentre can be C = C (any double bond). E. They can separated by physical and chemical method. II Positional Positional Stereoisomer Configurational isomers : Configurational isomerism arises due to different orientations along a stereocentre and these isomers can be seperated and these isomers do not convert into one-another at room temperature. These isomers has same general formula. Therefore. N = N. E. they are true isomers. 5. Geometrical isomerism across – Nil and By E / Z 2. Geometrical isomerism across (a) Imine ( ) Imine compounds are produced from carbonyl compounds on reaction with ammonia.Condition (i) Restricted rotation (ii) The two groups at each end of restricted bond must be different. 2-Dideuteroethene H2C = CH (7) Phenylethene Ans. (1) Ethene (2) Propene (3) 2-Methylbut-2-ene CH3 C CH CH3 | CH3 (4) But-2-ene CH3 – CH = CH – CH3 (5) Penta-1. (8) Buta-1. Q.I. 3-diene C = C – C = C – C (6) 1. Cbd Cab Cde (iii) In two geometrical isomer the distance between two particular groups at the ends of the restricted bond must be changed. 6 1. 1 3 C=C 2 4 Caa Caa X Caa Cbd X Cae Cbb X Cab Cab Cab . 3-diene C = C – C = C 4. (Syn and anti) Imines prepared from unsymmetrical addehydes and ketones. always show geometrical isomerism. Which of the following compounds show G. Page-37 . O || (b) H C H O || (c) H C D (e) Ph C CH3 || O (f) Ph C Ph || O (h) (i) c. Eg. 7 Q. The lowest molecular weight of acyclic ketone and its next homologue are mixed with excess of NH2 – OH to react. * Unsymmetrical ketoes form two oximes. 2.Q. O || (a) CH3 C CH3 O || (d) CH3 C H (g) Ans. g. How many oximes are formed after the reaction ? 3 Ans. 4. i (b) Oximes C = N – OH : These are prepared by reacting carbonyl compuond with hydroxyl amine (NH2 – OH) R C=O H H2O + H2 N–OH R R C = N–OH H (Aldoxime) OH R C=N H (I) (syn) and OH C=N H (II) (anti) Syn and anti in aldehyde only not for ketones. d. Which of the following ketones will form two oximes. Which of the following compounds show geometrical isomerism after reaction with NH3. (1) Propanone O || CH3 C CH3 (2) Butanone O || CH3 CH2 – C CH3 (3) 3-Pentanone C–C–C–C–C (4) Acetophenone C 6H 5–C–CH 3 O O O (5) Benzophenone C 6H 5–C–C 6H 5 (6) Cyclohexanone O O Me (7) Methyl Cyclohexanone Ans. * Except formaldehyde (CH2O) All other aldehyde form two oximes. Page-38 . e. N=N Ph Ph syn ... diastereomers) Q.(c) Hydrazones C = N – NH2 : R R H2O C = O + H2N... Ph . .. O + + H2N – NH2 + + NH2 CH 3 CH3 O CH 3 Number of isomer = 3 (3) Geometrical isomerism across azo compounds (– N = N – ) (i) H – N = N – H (H2N2) (ii) Ph2 N2 (Azobenzene) . . Two chain isomer of a cycloalkanone which are next higher homologue of lowest molecular weight.. Ans.. N=N anti Ph (4) Geometrical isomerism across ring structure (i) Restricted rotation (ii) (iii) Page-39 .NH2 H hydrazine C = N–NH2 H R NH2 R + C=N H (I) NH2 C=N H (II) (Geo. cycloalkanone reacted with hydrazine. Identify the structure and number of isomer of hydrazones prepared ? . N NH2 NH 2 N N . Thus – F > – OH > – NH2 > – CH3 (ii) If the first atom is identical. or optical isomerism) Those stereoisomers which are not mirror images of each other are called diastereomers.(5) Geometrical isomerism in cycloalkenes across double bonds : In cycloalkenes. exists across double bonds with ring size equal to or greater then 8 carbon atoms (due to ring strain) Stereocentre:An atom or bond across which stereoisomerism exists (either G.I. (a) (b) < (c) < (d) > Page-40 . G. Those compounds which are non-superimposable mirror-images of each other are called enantiomers are enantiomers E/Z Nomenclature : Z (Zussamen = together) a > b and e > d E (Entegegen = opposite) a > b and e > d Rules :(i) The group with the first atom having higher atomic number is senior.I. then second atom is observed for deciding the seniority of the group. that is isotopes. (iv) If the group has unsaturation.Ex. Then heavier isotope has higher seniority. (1) HC C C(CH3)3 C == C E CH2 – CH = CH2 (2) Z N (3) CH2 – C(CH3)3 C H2C = C = CH C == C E C CH (4) Z (5) E (6) E Page-41 . (1) (2) – CH = CH2 < – C CH (Hypothetical) (v) Bond pair is always senior to lone pair. E (iii) If the first atom has same atomic number but different atomic mass. Z Ex. CH 2 = CH Ex. then a hypothetical hypothetical equivalent in drawn for it and it is compared with other group for seniority. of Geometrical isomerism = 2n–1 + 2 Ex.1 Ans. = 2n where n = number of stereocentre Ex. Stereocentre = 2 Geometrical isomerism = 4 (a) (b) (c) (d) Case II : Compounds with similar ends & even no. (i) > (ii) Cis > Trans < COOH COOH C=C (iii) H Trans > Cis H COOH H C=C > COOH Cis > Trans H Page-42 . of stereocentre.F. n Ex. No.Number of geometrical isomers Case I : Compounds having dissimilar ends No.C.I. No. of Geometrical isomerism = 2n–1 + 2 2 CH3 – CH = CH – CH = CH – CH3 Stereocentre = 2 Geometrical isomerism = 21 + 20 = 3 1 Case III : Compounds with similar ends but odd number of stereocentre. of G. Ans. n 1 2 CH3 – CH = CH – CH = CH – CH = CH – CH3 Stereocentre = 3 Geometrical isomerism = 6 Physical properties of Geometrical isomers : The physical properties of organic compounds can be compared through the knowledge of their molecular formula. structural formula and stereochemical formula (S.1 Ans.) More Polar geometrical isomer is more soluble in water. Page-43 . It is represented as follows:- t 25 º C [ ] 580 nm where c = observed angle of rotation c = concentration in g/mL (or density) = Path length of Sample tube in dm Q. = (1) (5) (3) (6) Recorder (8) (1) Source of light (2) Polychromatic Non-polarised light (3) Slit (Monochromator) (4) Monochromatic Non-polarised light (5) Polariser (Prism-setting) (6) Monochromatic plane-polarised light (7) Sample tube ( = path length) (8) Recorder for measurement of optical rotation Observations:. Optically active. (I) (II) With symmetrical With asymmetric molecule molecule [] = 0 [] 0 Optically inactive compounds:For symmetrical molecule optical rotation observed after interaction of light is zero. If concentration is reduced to half. Specific Rotation:.The specific rotation of a compound indicates the optical rotation of unit concentration (1 g/mL) present in a sample tube of 1 dm of path-length at given temperature and given wavelength of light.When plane polarised light is passed through sample tube. the observed rotation() will be 140º but = 35º If = +70º. laevorotatory or l or (–). then following changes can be observed in the recorder:Angle of rotation () (1) = 0º (2) = +xº (clockwise rotation) (3) = –xº (anticlockwise rotation) Inference Compound is optically inactive. dextrorotatory or d or (+). The optically active compound can show optical isomerism. Measurement of optical activity It is measured by an instrument called polarimeter. Page-44 .5º to distinguish. Calculate ? Ans.Optical Isomerism Some organic compounds can rotate the plane of plane-polarised light. d will have +35º and will have –145º (not distinguish) still halfed. it will be a (i) d compound or (ii) compound of = –(360º – 70º) = –290º It can be decided by changing the concentration or by changing the length of the tube (c or ). Such compounds are called optically active compounds.5º and –72. Optically active. Compound ‘X’ has = +70º for 2 g/mL solution in sample tube of = 1 dm. it will give +17. = (1) (2) 70 = +35º 2 Its concentration is made twice. The molecules are asymmetric in nature and show non-zero optical rotation. The molecule is said to have n-fold axis of symmetry Ex.S.But in case of optically active compounds.) (2) Plane of symmetry (P.S. then its original or identical or superimposable.The imaginary plane which divides a molecule into two equal halves which are related as mirror image is known as plane of symmetry Centre of symmetry : . meets the identical atom or group. C 2 (180 ºrot ) Ex.A centre of symmetry in a molecule is said to exist if a line is drawn from any atom or group to this points and then extended to an equal distance beyond this point. (axis) There is no relation whatsoever with chirality and axis of symmetry. Page-45 . Meso molecule does not have axis of symmetry no axis of symmetry . Cl H R H C=C H R Centre of symmetry H Cl Centre of symmetry Axis of symmetry : It is defined as Cn axis of symmetry that means of the molecule is rotated by 360 º n angle.) (3) Axis of symmetry (4) Alternating axis of symmetry Plane of symmetry : . Symmetry of elements :(1) Centre of symmetry (C. (C4 axis) Ex. When all the elements of symmetry (23 including Centre of symmetry and Plane of symmetry are absent) the molecule is said to be asymmetry. Page-46 . enantiomers. The two enantiomers can rotate the plane-polarised light with equal magnitude and opposite signs. So. optical isomers. Dissymmetry.The human hand does not have Centre of symmetry and Plane of symmetry . They have similar physical properties except the sign of optical rotation. It is represented by Sn Ex. Two types:(a) Dextrorotatory (b) Laevorotatory Enantiomers are always mirror image isomers. so it is dissymmetry.Dissymmetry/Chirality is the minimum and sufficient condition for a molecule to be optically active. but the reverse is not true.The molecules or objects in which minimum two elements of symmetry (Centre of symmetry and Plane of symmetry) are absent are called dissymetric molecules/objects. Asymmetry:. These can be distinguished only by polarimeter. All asymmetric molecules are also optically active. Chirality (Dissymmetry) and Optical Isomers:(a) (b) (c) (d) (e) The dissymmetric molecules have two orientations in space. and P. a C.The dissymmetric molecules are always non-superimposable on their mirror-image orientations. are absent. Note : All dissymmetric compounds are chiral. So.S. if in a molecule. H H Me Me H Me 180º rotation (I) H Me Me H H Reflextion (II) Me (III) I and III are equivatent. Dissymmetry:. the molecule will be optically active. Every enantiomer is optically active. Because of dissymmetry. These are mirror images (enantiomers). Chirality and Optical Activity:. So. structural formula but have different orientation in space. Optical Isomers:Enantiomers:. so it can be called asymmetry. They have same molecular formula. The non-superimposable orientations are non-identical orientations. these two isomers are capable of rotating plane-polarised light. so are called isomers.S.All dissymmetric molecules are chiral and are optically active. Any molecule which has this property is called chiral.The mirror-image stereoisomers are called enantiomers. Dissymmetry and chirality:. These are the isomers which have maximum resemblance with each other. all asymmetric molecules are always dissymmetric. then its original molecule is obtained. Non-superimposability of enantiomers:.Alternating Axis of Symmetry : When a molecule is rotated by 360º angle and its mirror image is taken in the perpendicular plane of n rotation. It does not have any element of symmetry. Chirality (Hand-like property or Handedness):. These two orientations are called stereoisomers. They have dissymmetry/chirality. these are called optical isomers. (ii) Groups at Vertical line are away from observer. (v) High priority group lies at the top of vertical line (IUPAC Numbering starts from top). VII VIII Projection Formula of Chiral Molecules:(i) Wedge-Dash Projection formulae down up Ex:- (i) Butan-2-ol CH3 C H C2H5 OH (ii) Fisher Projection formula (i) Butan-2-ol (CH3 – CH2 – CH – CH3) – Ex:- CH3 H OH CH2CH3 /////////////////////// OH Rules of writing Fisher Projection formula :(i) It is represented by a cross (+).1:. (iv) Central ‘C’ atom of the cross is chiral. b Centre of symmetry Plane of symmetry Optical active H C (1) H H H Absent Yes No Absent Yes No Absent Yes No Absent No Yes H C (2) H H Cl H C (3) Br H Cl H C (4) Br Cl F Ex.Mark the chiral objects:(i) Cup (ii) Plate (vii) Shoe (viii) Glove Ans. (iii) Letter A (iv) Letter G (v) Fan (vi) Door IV. a C* e d If a molecule contains only one asymmetric carbon atom. then the molecule as a whole becomes chiral and optically active and show optical isomers.Optically Active Carbon Compounds:If a carbon-atom is attached with four different group. then if does not have any element of symmetry. Page-47 . which is represented as *Cabde. It is known as asymmetric carbon atom. (iii) Groups at Horizontal line are towards the observer. Draw Fisher Projection formula of following molecules:- (1) 2-chlorobutane (2) Pentan-2-ol – * (3) CH2 – CH – CH OH OH O – Glyceraldehyde 1 2 3 4 5 – * * (4) CH3 – CH – CH – CH2 – CH3 OH OH – Ex:- . They are true isomers (not superimposable) CHO CHOH CHOH CHOH CHOH CH2OH – – – – – (5) Glucose Mark:. of C* 4 (ii) Draw one Fisher Projection Formula CH = O H OH H OH H OH H OH Page-48 . (I. IV) Enantiomers. . II). (III. All other diastereomers. .(i) No. one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is designated as an L-sugar. b to c. even though we may not know the absolute configuration of either.Configuration nomenclature in Optical Isomers :Relative configuration : The experimentally determined relationship between the configurations of two molecules. Relative configuration is expressed by D-L system. it is (R). its configuration is (S). (I) R/S Configuration : R Rectus S Sinister Examples : a b c d (1) Seniority order a > b > c > d (2) Put junior most group at dotted line b (3) Q. & those with (-) glyceraldehyde are assigned as L. Absolute configuration : The detailed stereochemical picture of a molecule. A sugar whose highest numbered chiral centre (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designated as a Dsugar. The configuration of (+)-glyceraldenyde has been assigned as D and the compounds with the same relative configuration are also assigned as D. Br Br Cl I F (R) Cl F (S) I Page-49 . (i) If it follows clockwise route.g. (I) D . Examples : Sugars have several asymmetric carbons. (ii) If it follows anticlockwise route. e.L System (Relative configuration) : Application on correct Fisher Projection Formula This method is used to relate the configuration of sugars and amino acids to the enantiomers of glyceraldehyde. including how the atoms are arranged in space. Alternatively the (R) or (S) configuration at each chirality centre. c a d Now go from a to b. (1) Ans. Properties:(i) Dipole moment (ii) Boiling Point (iii) Melting Point (iv) Solubility (v) Specific rotation [] same same same same different Molecules with more than one chiral carbons:(I) Calculation of No. [] = 0 due to external compensation of optical rotation.When all Chiral carbon atoms are differently substituted (all are dissimilar C*) optical isomer = (2)n Case II:. then clockwise R & anticlockwise S Q. R (3) Ans.R/S Configuration in Fisher Projection Formula :(c) Me(c) Ex:- H(d) Et(b) (d) Pr (a) (b) abc Clock wise R (a) If the junior most group is at horizontal line clockwise S & anticlockwise R If juniormost at vertical line. of optical isomer :Case I:. S Properties of enantiomers :One chiral carbon : (i) Number of optical isomer = 2 (d or ) (ii) Number of Racemic Mixture Equimolar mixture of d and . R (2) Ans. n (i) n = even x = 2n – 1 + 2 2 –1 (ii) n = odd x = 2n – 1 Page-50 .When all the Chiral carbon atoms are similar at ends. of Racemic Mixture – Two (I + II. Cl H H Cl C2H5 C2H5 I [] = +xº C2H5 II –xº C2H5 III +yº Analysis:(a) I. n = 3. Thus. Meso isomer In (III) Specific rotation [] = +xº In (IV) Specific rotation [] = –xº (III) (IV) Page-51 . III – Diasteromers II.CH3 CH3 – – Molecules with two Asymmetric Carbon Atoms of Dissimilar Nature:(i) Structural formula CH3 – CH – CH – CH2 – CH3 Cl Cl n 2 (ii) Optical isomer = 2 = 2 = 4 (iii) Stereochemical Formula:- CH3 CH3 H Cl Cl H H Cl Cl H H Cl Cl H. III – Diasteromers IV –yº III. [] = 0. III + IV). (i) (ii) (ii) (iii) (iv) and their mirror images. IV – Diasteromers II. (iii) Stereo chemical formula :- Me H OH H OH Me (I) (II) Plane of symmetry present. pairs) – Ans. II – Enantiomers I. I and II are identical. x = 23 = 8 (4 d. – Q. optically inactive. – CH3 – CH – CH – CH – CH2 – CH3 Cl Cl Cl Calculate total number of optical isomer. IV – Diasteromers (b) No. 4 Racemic Mixtures (4 different fractions on fractional distillation) (i) n = 2 – – Compounds with 2 Asymmetric Carbons of Similar Nature:Ex:CH3 – CH – CH – CH3 OH OH n (ii) x = 2n – 1 + 2 2 –1 = 3 (Optical stereoisomers). IV – Enantiomers I. In (I and II) Superimposable on its mirror image. P. They are non resolvable. Q. it has different physical properties than d – -pair.Optical Diastereomers have different physical properties. Presence of more than one asymmetric ‘C’ atoms. diff. They are achiral (optical rotation = 0).. of d – pairs = 1 (III + IV) = Racemic mixtures In (I) and (II) : [] = 0 due to internal compensation and Non-Resolvable In (III) and (IV) : Resolvable [can be separated into two isomers (enantiomers)] Meso isomers: The optical stereoisomers which have more than one asymmetric carbon atoms but have a plane of symmetry are called meso compound.F..) (ii) By fractional crystallization (different solubility) (iii) Chromatography (different solubility) (iv) Differential Melting (M. Stereo Chemical Formula :- \\\\\\\\\\\\\\\\\\\\ (I) 2S 3S 4R (II) 2S 3R 4R (III) 2S 3 achiral 4S (IV) 2R 3 achiral 4R Optically active (i) & (ii) Achiral due to Plane of symmetry. Mark meso compound among following H H COOH OH (1) (2) H Ans. Cl Cl Cl n=3 x = 2n – 1 = 4 stereoisomers. Conclusion:.No. They are diastereomer of d – pair. They have [] = 0 due to internal compensation of optical rotation.) Page-52 . so these can be separated by normal physical methods of separation.P. Properties of Optical Diastereomers:The optical isomer which are neither mirror image nor superimpossible to each other are called optical diastereomers. OH COOH Cl Me Cl (3) (4) Me OH H (5) * OH (1) (2) (4) and (5) – – – Molecules with three similar chiral carbon: Ex:CH3 – CH – CH – CH – CH3 M. (i) By fractional distillation (different B. (Optically inactive). So. so these cannot be separated by normal physical methods of separation.R – C – O – H + H – O – R ' R* – C – O – R' (H2 SO 4 ) O O – H2 O Me H COOH (d) Et Et H Me + OH COOH ( ) H D (d') H H + H (dd') C–O O (d') O D ( d ') (dd') Me H (d' ) C–O Me D Et Me Et Me Me Pair of Diastereomers Ester upon still hydrolysis will give back the carboxylic acid and alcohol.The enantiomers in a d – pair have identical physical properties. General scheme of resolution:- d+ + d' Racemic Optically mixture active Reagent Hydrolysis Hydrolysis d + d' (I) Resolution (by using Esterification) of RCOOH and R'OH Example:- Esterifica tion General Reaction:. The special method is used for separation of d – pair known as optical resolution.Geometrical isomer are always diastereomers – True Chemical method of separation (resolution) by using optically active reagent:Separation of Racemic Mixture:.All diastereomers have different polarities (Both Geometrical and Optical) – True Assertion:. Assertion:. (II) Separation of (d ) pair of alcohol by using optically active acid:- D Ph + H H (dd') O–C O Me Me Ph Me + H H C–O O D (d' ) Me Page-53 . %E.):Ex:Compound X has [] = 70º We have mixture of enantiomers in different percentage. (obs) 100% []specific Page-54 .P.(III) By salt formation (RCOOH + R’NH2) + + Diastereomers (Separable) Optical Purity or Enantiomeric Excess (O. or E.E.E. The molecule will exist in two enantiomeric forms. then inversion does not take place and such molecules are optically active [ ] 0 . the orbital diagram of this structure will be - The groups at the end of allene structure lie in same plane (ZX plane). The molecules lacks molecular dissymmetry & it will not show optical activity hence optical isomerism. – – – So.Optically active compounds without a chiral carbon:Rapid R R Inversion – Asymmetric N:R' – N N – R' R'' R'' A nitrogen atom attached with 3 different groups (sp3) is chiral. having trigonal pyramidal shape undergoes rapid inversion of shape at room temperature. If asymmetric Nitrogen atom is present in ring structure. it will not show geometrical isomerism. Nitrogen converts into its enantiomer.g. asymmetric. Overall the structure has molecular dissymmetry which is the sufficient condition for optical activity. So at room temperature. everytime a racemic mixture of ‘d’ and ‘ ’ forms exists. the orbital diagram of this structure will be Since the groups at the end of allene are in perpendicular plane. Page-55 . In case of acyclic compounds. the Nitrogen.. Therefore it will have a plane of symmetry (ZX plane). (b) Allenes with odd bonds : e. The molecule lacks centre of symmetry as well as plane of symmetry. This racemic mixture can never be resolved at room temperature. acyclic molecules with chiral nitrogen are chiral but optically inactive ([] = 0) and are non-resolvable.g. (otherwise ring will break) Ex:- Me N [ ] 0 Optical activity without asymmetric carbon : CCC–C–C (Cumulated double bond) One after another CC–CC–C (Conjugated double bond) Alternate C C–C–C C (Isolated double bond) Separated (I) Case of allene : (a) Allenes with even bonds : e. But the compound will exist in two geometrical diastereomeric forms. (III) Case of cycloalkylidene : (IV) Case of ortho-ortho-tetrasubstituted biphenyls : becomes non-planar at room temperature in order to have minimum electronic repulsion among the substituent.g. (a) spiranes with even rings : shows optical isomerism. e. In this orientation (phenyl planes perpendicular to each other) the free rotation of C – C single bond is restricted and molecule shows optical activity due to molecular disymmetry. while the spiranes with odd rings shows geometrical isomerism. The spiranes with even rings and different groups at terminal carbons show optical activity & optical isomerism.(II) Case of spiranes : A similar case like allenes is observed in spiranes. Page-56 . (b) spiranes with odd rings : shows geometrical isomerism. C–N. (C–C.E.5 kJ / mol Energy . 5. difference between conformational at potential energy minima and maxima. Conformational Energy : The rotational energy barrier is known as conformational energy.CONFORMATIONAL ISOMERISM : 1. It is the P. CH3–CH2–CH3 Page-57 . N–N.)/Torsion Angle : The interfacial angle between the groups attach at two -bonded atoms is defined as dihedral angle. C –O. 4. Conformational Isomers : Those conformations which are most stable and have minimum P. These are not true isomers and can never be isolated. Free Rotation : A sigma covalent bond undergoes free rotation at room temperature. Angle of rotation/Dihedral Angle (D.Level Diagram of ethane Ex. CH3 – CH3 H(a) H(c) rotation H (b) Eclipsed H (c) H(b) H(a) Staggered (b) Newmann Projection formula 1 12. Conformers / Rotamers or conformations : The infinite number of spatial oreintation of molecule arises due to free rotation along a sigma covalent bond 3. O–O) 2.E.A. are defined as conformational isomers. Projection formula (a) Saw horse Projection formula Ex.H. C3 CH3 CH3 CH3 60 º CH3 CH3 H 120 º H H H (I) H H H (II) 180 º H H H 240 º H CH3 (IV) 300º CH3 H H H CH3 H H (V) 360º Page-58 . So draw the newman between C2 . Ethyl-hydrogen repulsion is less than methyl-methyl repulsion.Saw horse Projection formula Eclipsed Staggered Newman Projection Formula CH3 H CH3 60 º H H H H 120 º H II Staggered Eclipsed HH H III Eclipsed CH3 H H H H H H H IV Staggered Propane : Butane : Ex. 300º) (ii) Vander Waal strain : It is the repulsion between the group attached at adjacent bonds. The conformational isomers with different energies are called non-degenerate isomers. minima in the P. It is active at torsional angles 0º. there is no distortion of bond-angle due to free rotation. are defined as conformational isomers. Strains : (i) Torsional Stran (eclipsing strain) : It is defined as the electronic repulsion between the bond-pairs electrons of two adjacent eclipsed bonds.A. The vander waal strain is maximum in eclipsed conformation. all conformations have zero angle strain in acyclic compound Page-59 . Order : IV < II < III < I Potential Energy Diagram of butane : Conformational Isomers : Due to free rotation from 0º to 360º those conformations which are most stable and have minimum P. = 60º.E. It is almost zero for H-atom since the sum of vander waal radii is less than internuclear distance between two H atom. 240º in which the molecule has eclipsed conformation. diagram with respect to rotational angle.I/VII = Fully eclipsed II/IV = Gauche form III/V = Partially eclipsed IV = Anti form Stability Order : IV > II > III > I Anti > Gauche > Partially eclipsed > Fully eclipsed P. The conformational isomes lie at the P. 180º.H. while intermediate in gauche conformations. (D. (iii) Angle Strain : It arises due to distortion in normal bond-angles.E. So. 120º. It is considered almost zero in the staggered conformation.If these conformers have same energy then there isomers are called degenerate isomers or equienergic isomers.E. In acyclic compounds. and minimum in anti conformation.E. Draw conformation isomers of following compound (a) CH3 – CH – CH3 (b) CH3 – (CH2)3 – CH3 CH3 Ques. H H H 60 º H3C 180 º CH3 CH3 CH3 CH3 CH3 120 º CH CH3 CH3 3 Eclipsed H3C H CH3 H H H3CCH 3 H CH3 H3C CH3 H Staggered Page-60 .Ques. Draw conformation isomers of following compound CH3 – CH –CH – CH3 with respect to C2 and C3 carbon CH3 CH3 atoms. CH2OH OH H OH H H H (II) Gauche Hydrogen Bonding : (II) Gauche form most stable due to hydrogen bonding. O2N – CH2 – CH2 – OH Interconversion of projection formulae : Example Tartaric Acid : COOH – CHOH – CHOH – COOH no. where G = – OH. – CHO the Gauche form is more stable than the anti form due to intramolecular hydrogen bonding i. – NR2. – COOH. – NH2 . – NO2.Case of intramolecular hydrogen bonding : In case of G – CH2 – CH2 – OH. of Sterioisomers : 3 COOH 1. stability : Gauche form > anti form. Stability Order : II > IV > III > I Example 2. CH2–OH (Glycol). Meso H OH H OH COOH Page-61 . Example 1. – F.e. 2. d/ COOH COOH H Meso COOH H OH H OH COOH OH OH H H COOH OH COOH COOH COOH H H COOH OH H OH COOH OH H OH H OH d/ COOH COOH COOH OH H OH COOH H OH H H OH COOH COOH OH HH COOH H OH COOH COOH H OH OH OH COOH H H COOH OH H OH H COOH OH Conformation isomers of cyclohexane : Chair form and Boat form Page-62 . amino and nitro.I JEE Syllabus [2009] Practical organic chemistry: Detection of elements (N. carboxyl.I CONTENTS : 1 Structure Identification * Monochlorination * Catalytic Hydrogenation * Ozonolysis * Elements detection * Identification of Functional Group by Lab.2009-10 ORGANIC CHEMISTRY TOPIC : GOC . carbonyl (aldehyde and ketone). S. halogens). Test Refer sheet GOC. 63 .LECTURE NOTES Session . Detection and identification of the following functional groups: hydroxyl (alcoholic and phenolic). Chemical methods of separation of mono-functional organic compounds from binary mixtures. Ex. of position isomers (structural) of monochloro compounds is equal to the number of different types of H-atoms present in the reactant. it can be concluded that the total no. CH3 Ex. Application:.. it forms a mixture of monochloroisomers. X = Neopentane Q.If a molecule has more than one type of H-atom. If one H-atom is substituted by one halogen atom.Hence. This is known as monohalogenation reaction. All these isomers are position isomers. The different type of H-atoms are also known as non-identical Hydrogens or non-equivalent Hydrogens or chemically different Hydrogens. Cl / h Two monochloro P(C6H14) 2 How many isomers of P will give two monochloro compounds ? Ans. X(C5H12) 2 Ans. Br2. Cl / h only one monochloro isomer. – – CH2Cl Cl / h 2 64 . I2). then on monochlorination. Monochlori nation 2 Products (structure isomers) (b) C – C – C – C Monochlori nation 2 Products (structure isomers) (c) C – C – C – C – C Monochlori nation 3 Products (structure isomers) (d) C – C – C – C Monochlori nation 4 Products (structure isomers) – (a) C – C – C – C CH3 Monochlori nation 5 Products (structure isomers) (e) Q. C – – C C – C – C – C only one isomers Remark : In aromatic hydrocarbons.When an alkane or a cycloalkane is treated with halogen (Cl2. F2.STRUCTURE INDENTIFICATION Monochlorination:(a) Cl / h CH3Cl + HCl (i) CH4 2 Cl / Sunlight (ii) CH3 – CH3 2 CH3 – CH2Cl + HCl – Cl Cl / h 2 (iii) + HCl – – – – C C Cl2 / h C – C – C – Cl + HCl (iv) C – C – C C C Cl – Cl / h 2 (v) + HCl Remarks:. but H-atoms of Benzene ring are stable. a photochemical reaction takes place and a C – H bond cleaves and a C – Cl bond is formed. the hydrogen atoms of the side-chain are chlorinated. Conclusion:. Exceptions:Aromatic bonds which are stable at room temperature but can be hydrogenated at high temperature. of moles of H2 consumed by 1 mole of compounds is equal to the no. All positional isomers of alkenes or alkynes (due to multiple bond) always give same product on hydrogenation. (X) (Y) Catalytic Hydrogenation of C = C. 65 . C C) are hydrogenated.Q. Alkynes. Cl / h Two mohochloro X(C8H10) (Aromatic) 2 Cl / h One monochloro Y(C8H10) (Aromatic) 2 Ans. polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room temperature. The reaction can’t be stopped at any intermediate stage. The no. 2 CH – CH3 – 2 [Reaction cannot be stopped at any intermediate stage] Remarks:(a) Alkenes. It can be concluded that the hydrogenation product of an alkene or alkyne or any unsaturated compound is always a saturated compound. C C General reaction:Ni R – CH2 – CH2 – R (a) R – CH = CH – R + H2 Ni / Pt / Pd R – CH2 – CH2 – R (b) R – C C – R + 2H2 H2 H2 R – CH CH – R R – CH2 – CH2 – R (Not isolated) 2H2 / Ni CH3 – CH2 – CH2 – CH3 (c) CH2 = CH – CH = CH2 (d) (e) 3H2 / Ni CH = CH2 – H2 / Ni room temperature CH – CH3 – 2 H 2/Ni (100 – 150ºC) (f) H / Ni. (b) All C – C bonds(C = C. of bonds presents. C Q. – Ans. Ans. of monochloro products of a fully saturated isomer of C4H6. C–C–C–C–C–C 2 monochloro product Find the structural isomers of product? fully saturated cycloalkane of M. Cl / h 2 A C 8H17 Cl (C8H18 ) (Only one type ) Identify A ? 66 .During catalytic hydrogenation and monochlorination. Ex:. . DU = 2 – – Q. Ans. C– C–C C – Cl – 2H2 / Ni (1) Cl Y = (X) Cl / h 2 Y + Z Z = Cl CH3 CH3 H / Ni 2 (2) CH3 CH3 H / Ni 2 (3) Q. carbon skeleton remain unchanged. Y. Z Identify the lowest molecular weight alkane which gives four structural isomeric monochloro products ? C– C –C–C – Q. C6H12 which gives two monochloro – Q. Z Ans. – Q. Cl / h 2 5 Monochloro product 5 Monochloro product H / Ni Cl / h Z(only one monochloro product) X(C4H6) 2 Y 2 Identify X.C = C – C = C + Cl2 X Ex. C C – C – C – C. Cl – C C5H12 = 72g Identify the structure of hexane which gives 3 monochloro products ? Ans. Find the no.F. DU = 2 X Y . C = C = C or C – C C Q. (3) The products are carbonyl compounds (aldehydes or ketones).W. Ans. C–C (i) C C – C – C – C (b) Ans. – – – C (ii) C C – C – C – C C CH2 Ans. This type of ozonolysis is known as reductive ozonolysis. Remarks:(1) Alkene and polyalkene on ozonolysis undergo oxidative cleavage.G. Determine the M. C C C– C– C–C C C Write all isomeric alkynes which produce an isomer of heptane which on further monochlorination gives (a) three monochloro products.s. (2) (a) The reagent of reductive ozonolysis is (i) O3 (ozone) (ii) Zn and H2O or Zn and CH3COOH or (CH3)2 S (b) The reagent of oxidative ozonolysis is O3 and H2O2. General Reaction:.R – CH = CH – R Ex:- (1) CH2 = CH2 (1) O 3 R – CH = O + O = CH – R + ZnO + H2O (2) Zn/H2O (1) O 3 CH2 = O + CH2 = O (2) Zn/H2O (2) CH3 – CH2 – CH = CH2 (1) O 3 CH3 – CH2 – CH = O + O = CH2 (2) Zn/H2O (3) CH2 = CH – CH2 – CH = CH – CH3 O3 / Zn CH2 = O + O = CH – CH2 – CH = O + O = CH – CH3 67 . CH2 CH2 C6H6 = 78g Ozonolysis: It tells about position of unsaturation.– – Q. two monochloro Nil Q. Find the structure of lowest molecular weight hydrocarbon and maximum unsaturation which on hydrogenation produce such an alkane which gives two monochloro products ? Ans. of maximum unsaturated hydrocarbon which on hydrogenation gives C6H12 which on further chlorination gives two monochloro. (4) Ozonolysis does not interfere with other F. – – Ans. Ans. 5 monochloro product 68 . . 1 mol CO2. 1 mol Find the structure of the hydrocarbon and the no. At higher temperature. the aromatic double bonds can also undergo ozonolysis. An unsaturated hydrocarbon on ozonolysis produces 1 mole of . P= Q. If the products are rejoined. All C = C (except aromatic ones) undergo oxidative cleavage under normal conditions. the position of C = C can be determined in the reactant molecule. of monochloro products formed followed by hydrogenation. C O3 O = C – C – C – C – C = O + O = CH2 (1) Zn O C O () 3 (2) (4) Q. CH = CH – CH3 – low temperatur e CH = O – (3) Zn O3 / Zn + O = CH – CH3 O3 C6H5 – CH = O – CH = CH – Zn H / Ni Cl / h 2 CnH2n 2 2 CnH2n1Cl (P) CnH2n – 2 (Q ) (m products) Single Compound m3 (no isomer ) – – CH3 HCOOH + CH3 – C – COOH CH3 Identify P ? Ans.O / Zn 3 + OHC – CH2 – CHO (Propandial) (4) Applications: The process is used to determine the position of C = C in a molecule. Q.C. X is C – C = C – C = C – C 69 . H2 C–C–C–C–C–C X O3 (Zn) CH3CHO + CHO – CHO Identify structure of X ? Ans. Identify X ? C X – Ans. X is Q.) Cl / h 2 m products (m 7 ) O 3 (Zn/H2O) HCHO + 4(1-oxoethyl) Cyclohexan-1-one. X = H / Ni 2 X (Unsat. H. Identify structure of X ? Ans. C=C–C Q. hydrocarbon) H / Ni 2 – Q.– X (Unsat. Cl2 m products h (m 3 ) O 3 (Zn/H2O) O identify structure of X ? O Ans. O O Q.1 Identify A.3 Identify A & B Sol. O3 Zn. A= or Ex. A Methyl glyoxal + Formaldehyde Ex. A = CH3 C CH CH CH2 | CH3 B = CH3 CH CH2 CH2 – CH3 | CH3 70 . H2O – CH3 CH – 3 O3 Zn. Sol. -Terpinene Ex.B & C with the help of following reactions.Q.2 C C | | B = CCCC | | C C C = Identify -Terpinene and P-Menthane. H2O – CH3 2 CH3 + CH3 C – C– +2 – CH – 3 Sol. Cl2 / h (A) (C 9H18 ) Single monochloro produc ( Saturated Hydrocarbon ) t Cl2 / h (B) (C8H18) Single monochloro product Cl / h (C) (C 7H14 ) 2 s Two monochloro products ( Saturated Hydrocarbon) Sol. m=3 Ex.4 Identify A & m A = Ph C C Ph | | CH3 CH3 Sol.5 H 2 / Ni A Cl / h n-products 2 O3 Zn/H 2O + H– C –H + H – C – C – CH2 – CH2 – C – C – H Sol. A= n=7 71 .Ex. Halogens of presence of sulphur Lassaigne’s test A white ppt. 3Na4[Fe(CN)6] + 4FeCl3 Fe4[Fe(CN)6]3 + 12NaCl 2. soluble in NH4OH solution indicates iodine A white ppt. completely in- .1 Identification of Elements in Organic Compounds Element 1.1. of magnesium 4. Nitrogen Test / Reaction Remark Lassaigne’s test The appearance of green or Na + C + N NaCN prussian blue colour confirms FeSO4 + 6NaCN Na4 [Fe(CN)6] + Na2SO4 the persence of nitrogen. partly soluble in NH4OH solution indicates bromine. Nitrogen and Sulphur Blood red colouration confirms Lassaigne’s test FeCl Na + C + N + S NaSCN 3 Fe(SCN)3 presence of both nitrogen & sulphur 72 .S] 3. Sulphur Formation of a white ppt. soluble in NH4OH X + Na NaX solution indicates chlorine. (a) Oxidation test indicates presence of sulphur 3KNO3 3KNO2 + 3[O] Na2CO3 + S + 3[O] Na2SO4 + CO2 BaCl2(aq) + Na2SO4(aq) BaSO4 + 2NaCl(aq) Appearance (b) Lassaigne’s test purple colouration confirms the 2Na + S Na2S Na2S + Na2[Fe(CN)5NO] Na4[Fe(CN)5NO. A yellow ppt. NaX + AgNO3 NaNO3 + Ag X A dull yellow ppt. Phosphorus pyrophosphate indicates phosphorus H3PO4 + Magnesia mixture MgP2O7 + H2O 2MgNH4PO4 Mg2P2O7 + 2NH3 + H2O 5. 2 Identification of Functional Groups by Laboratory Tests Lucas Test I. R – OH + HCl white ppt + H2 O cloudiness R – C C Ag (white) R – C C Cu (red) RCOOH + RCOOH 2HCHO 2ROH + Na 2RONa + H2 R – C CH + Ag+ R – C CH + CuCl Red ppt. = O Compounds O3(ozone) O3 Red colour decolourises H2C = CH2 + O3 -------------- NR NR NR NR Pink colour Disappears Reaction Observation Br2 / H2O [Bayer’s reagent] alk. Na (b) AgNO3 + NH4OH (a) Cuprous chloride + NH4OH R – C C – R Acid formed.alcohol II. HCl + anhyd.73 Bubbles of H2 come out (3)° Cloudiness appears immidiately (2°) Cloudiness appears within 5 min. sec. cold KMNO4 conc.alcohol Presence of active ‘H’ Ozonolysis Ozonolysis Bromination Hydroxylation Inert paraffins Remarks . ZnCl2] ROH 3° 2° 1° White ppt. alcohol III. H2SO4 conc. pri. Lucas Reagent [Conc. NaOH KMnO4 LiAlH4 Reagent (R – OH) R – C CH CC CC C=C/ CC C–C Functional Groups 1. dil. (1°) Cloudiness appear after 30 min. ter. NaHCO3 solution NaOH. Iodoform reaction Tollen’s test Fehling’s test DNP-test Test of enols / phenols Remarks Litmus change to red. The pink colour is resumed by RCHO. H2O + CO2 RCHO + Ag+ RCOOH + 2Ag (Silver mirror) RCHO + Cu+2 RCOOH + Cu2O + 2H2O Fehling soln. RCONH2 + NaOH R COOH + R’ OH (Colourless solution) (pink) R COOR’ + NaOH + Phenophthalein Litmus test. or silver mirror Tollen’s reagent Red ppt. blue. Yellow orange ppt. Red H + H2 Reaction Blue litmus Yellow ppt of CHI3 (iodoform) Pink colour resume Schift’s Reagent * I2 / NaOH Black ppt. green buff) Observation FeCl3 (Neutral) Reagent . NaOH. 4-Dinitrophenyl hydrazine (2. 4-DNP) solution Fehling solution A&B Coloured ppt. Smell of NH3 Conc. Conc.74 Amides Ester CH3CHO or ArCOCH3 or R – COCH3 R – CHO Ar – OH Enols Functional Groups (yellow orange ppt. (violet. RCOONa + NH3 Schiff’s reagent : p-Rosiniline hydrochloride saturated with SO2 so it is colourless. 2. phenophthalein. Pink colour disappear on heating.) Sodium bicarbonate test Effervescence evolve. (i) NaNO2 + H2SO4 (ii) Phenol R2NH Sec. Molisch’s reagent (10% -naphthol in alcohol). Amines red colouration Liebermann test Orange red dye is formed Effervescence of N2 Nauseating odour (Carbylamine) black ppt Observation Carbohydrate HNO2 (NaNO2 + HCl) + -Naphthol HNO2 (NaNO2 + HCl) CHCl3.) RNH2 Nitro Compounds (RCH2NO2) or ArNO2 Functional Groups ArNHOH C C=N – C (Blue colour) CO CO C OH CO + H2N.2 % sol.HCl ROH + N2 + H2O NaCl + HNO2 + HNO2 Ag Ag RNC + 3KCl + 3H2O NaNO2 + HCl R NH2 + HONO R NH2 + CHCl3 + 3KOH Reaction Ninhydrin test Dye test Carbylamine Reaction Remarks .COOH (Amino acid) OH OH (Ninhydrin) CO CO Benzenediazonium chloride – Naphthol N=N-Cl + OH N=N OH + CO2+ RCHO + H2O orange-red dye + 2H2O N2Cl NH2. ArNH2 Amines (pri.CHR.n) Amino acids Reddish violet colour. KOH Mulliken’s test Reagent Ar.75 Violet colour Blue colour Ninhydrin reagent (0. amines. 1 Ex.p. / b. test for Nitrogen. 2. Identification of Organic Compounds on the bacis of Chemical Properties Ex. 3 Ex. : Structure Determination Identification of Organic Compounds on the bacis of Physical Properties (a) Physical state (b) Odour (c) Water solubility (d) m.5 Which of the following will not give (+ve) L. 4. 2 Ex. (A) CH3–CH2–NH2 (B) (C) (D) NH4NO3 ( Here is no carbon for the formation of CN–) 76 .S. (e) relative Ex.Examples 1.p. 1 Ans.3 Ans. Ex.Instant turbidity with Lucas reagent H2C = CH – CH2 – OH CH2 = CH – CH+ CH2+ – CH = CH2 Allyl C+ . 77 . slow r×n with HCl/ZnCl2 Other Examples Ex. 1° Benzyl C+ . but has angle strain .2 Ans. Ex.4 - CH3 – CH2 – C C – CH2 – CH3 Ans. 1° Angle strain / less stable / unstable C+ Although 3° . Ex. 6 C4H8 Ans. CH3 – CH2 – CHO Ex. O || CH3 C O CH3 Ans.10 A (C7H8O) FeCl3 (neutral) –ve Lucas –ve Reagent Ex.5 Ans. 78 .11 R gives following tests.7 (A) Ans. CH3 – CH = CH – CH3 Ex. A= Ans.9 Na metal –ve Ex. CH3 – CH2 – COOH Ans.Ex. A= Ex.8 Ex. Ans. (–) (V) – – CH3 (a) CH3 – C – OH CH3 Lucas Reagent instant turbidity Na metal (+) Ceric Ammonium (+) In 5 min. Na metal (+) (+) In 30 min. A C D F G H I J (B) NaHCO3 (C) 2. Tick mark the reagents which will give positive response with the following compound. 79 . How will you distinguish the following pair of compounds. T = . . Write S.1. Na metal (+) (+) (VI) (a) CH3 – CH2 – CH2 – CH2 – NH2 NaNO2/HCl (N2 gas ) Na metal (+) (b) CH3 – CH2 – CH2 – NH – CH3 N2 gas not liberated Na metal (+) – (b) CH3 – CH – CH2 – CH3 OH (c) CH3 – CH2 – CH2 – CH2 – OH 2. (J) Dil KMnO4 (cold) T (C7H6O2) is an aromatic white solid which liberates a colourless odourless gas on heating with NaHCO3. (A) Na metal O O H2N O Ans.F.. of X and its all possible functional isomers (all aromatic) Sol. 4-DNP (D) AgNO3 + NH4OH OH (E) Fehling solution (F) NaNO2/HCl (cold) (G) ZnCl2 (Anhyahous)/HCl (H) Cu2Cl2 + NH4OH (I) Br2/H2O 3. Compounds Isomers Reagents (I) (a) CH3 – CH2 – COOH (b) CH3 – C – O – CH3 1 2 3 NaHCO3 (+) NaHCO3 (–) Acidic odour Fruity/Sweet odour Na metal (+) Na metal (–) O (II) (a) Ph – CH2 – C – OH Same O (b) Ph – C – O – CH3 Same Same O (III) (a) Ph – CH2 – CH = CH2 Br2/H2 O (+) Dil KMnO4 (+) (b) Ph – Br2/H2O (–) Dil KMnO4 (–) (IV) (a) CH3 – CH2 – CH2 – OH Na metal (+) (b) CH3 – CH2 – O – CH3 Na metal (–) Ceric Ammonium Nitrate (+) . Aliphatic compoud with atleast two functional group ( which can form H-bonding) are water soluble. So they can be separated by use of appropriate solvents. sucrose. Ex. diamines. Physical methods : (i) Crystallisation (ii) Sublimation (iii) Distillation (iv) Solvent extraction (v) Chromatography 2. Hence these can be applied to solid as well as liquid compounds.e to obtain the organic compound in pure state. A chemical method can be applied only when one of the components of the mixture is soluble in a particular solvent while the other is insoluble in the same solvent . glycerol. The usual systematic scheme for separating a solid binary mixture is discussed below. Most of the aromatic compounds are water insoluble due to large hydrophobic group of six carbon atom Aromatic acids are insoluble in water but soluble in aqueous NaHCO3 solution or NaOH solution. 2º. Purification means the removal of undesirable impurities associated with a particular organic compound.POC . –COOH) . Purification of organic compounds : The organic compounds derived from natural sources or prepared in the laboratory are seldom pure. due to salt formation. oxalic acid . Various methods have been developed to purify organic compounds 1. They are usually contaminated with other substances. Amino acids (– NH2 . Aromatic amine (Aniline 1º.II Separation of binary mixture of organic compounds Theory of separation : Organic compounds have different solubilities in different solvents. Diacids. Aromatic hydroxy compounds are water insoluble but are soluble in aqueous NaOH solution due to salt formation. glycol. It is the most important step in the sense that if separation is incomplete the result will not be correct because the impure compound will give tests of different functional group and its melting point will also be very much different from that of the pure compound obtained from complete separation. fumaric acid. 3º) are organic base and water insoluble but are soluble in aqueous HCl solution due to salt formation. Separation with water Separation with sodium bicarbonate Separation with sodium hydroxide Separation with hydrochloric acid Solubility of two components. 80 . maleic acid. Separation Scheme Solvent H O 2 (1) Some important point : The mixture of organic compounds can be separated by using appropriate solvent. glucose. diols. (i) (ii) (iii) (iv) Separation of Binary mixtures of organic compounds. i. malonic acid. hydroxy acids (– OH. Separation is the first step during the actual analysis of organic mixture. Chemical methods:Chemical methods of separation depend upon the nature of the functional group present in the component. –COOH). 1 Binary mixtures .(Two components) Compound A Compound B Appropriate Solvent (1) + H2O (2) + H2O 81 .Ex. Q.3 + Identify I & II Sol. P Ex. NaHCO3 (5) + aq. HCl Ex.2 Sol. Q. NaOH (6) + aq.(3) Fructose + H2O (4) + aq. I A.4 + + P+Q+R Identify P. R Sol. II B Ex. + Identify I & II. PI Q II R III 82 . Ex. Mixture separation (1) (2) (3) Succinic Acid + Phthalic Acid + ortho–Cresol + + III 83 . 5.