Organic chemistry nomenclature


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Organic chemistry nomenclature

  1. 1. Nomenclature Badarla Sandeep
  2. 2. Preface Generally students have small doubtsregarding the nomenclature of the organic compounds. This bookis been written with the aim of needs and interests of students inorganic nomenclature. In this book every topic has been dealtprecisely and to the point in a sample and understandablelanguage.Things have been explained wherever possible. A goodnumber of illustration examples with answers is also provided inthis book.This helps the students in developing their reasoningskills in nomenclature. As every organic compound has itsown name.This book is helpful in determinining the IUPAC namesof the given organic compounds.This book aditionally provides asmall and interesting topic, "Molecular formula to an individualperson". Hope this book meets the needs ofthe students,chemists who want to learn organic chemistrynomenclature in detail. Badarla Sandeep Osmania University
  3. 3. Contents Page No1. Nomenclature 4 Trial Version only2. Saturated Aliphatic Compounds 163. Unsaturated Aliphatic Compounds ---4. Cyclic compounds ---5. Aromatic compounds ---6. Bicyclo Compounds ---7. Alcohols8. Ethers ---9. Aldehydes ---10.Ketones ---11.Carboxylic Acids ---12.Acid derivatives ---13.Nitrogen containing compounds ---14.Tricyclo Compounds ---15.Bio molecules ---
  4. 4. 1.NomenclatureWhat is organic chemistry? Organic chemistry is a branch ofchemistry which deals with carbon compounds.Organiccompounds mainly contains carbon and hydrogen as its mainconstituents.In addition to this it also contains N,S,P,Cl,Br and I.Why is nomenclature necessary? Organic chemistry is a vast branchas millions of organic compounds are already known andthousands of new compounds are beign added to this list everyyear.In order to facilitate the study of such large number ofcompounds.Therefore it is necessary to classify the organiccompounds.In order to classify this organic compounds the term"Nomenclature" comes into picture.What is nomenclature? Nomenclature means the assignmentof names to the organic componds.The naming of organiccompounds is an important aspect in the study of organicchemistry as their number is very large and variety of molecularstructures exist in their molecules.The field has become morecomplex on the phenomenon of the isomerism.They are two mainsystems of nomenclature of organic compounds.They are1.Trvial system2.IUPAC system
  5. 5. Trivial System In this system whenever a newcompound is discovered it is given an individual name.These namesare called Trivial names.These names are also called Commonnames.Examples(a) Acetic acid derives its name from vinegar of which it is chiefconstituent.(b) Formic acid was named as it was obtained from red ants.(c)The name oxalic acid, malic acid and citric acid is derived frombotanical sources.(d)Urea and Uric acid have been derived from animal sources.(e)The liqiud that is obtained by the destructive dstaillation of woodwas named as wood spirit.Later on it was called methyl alcohol.(f)Methane is called as marsh gas because it is produced in marshes.IUPAC system IUPAC stands for "International Union OfPure Applied Chemistry".By using this IUPAC system one can nameany complex organic compound easily.The name assigned to anorganic compound on the basis of latest IUPAC rules is known assystematic name.Features of IUPAC system(a) A given compound can be assigned only one name(b) This system is helpfull in naming the complex organic compounds
  6. 6. (c) This system is helpful in naming the multifunctional groups.(d) This is a simple,systematic and scientific method for thenomenclature of organic compounds.Rules for organic nomenclature For naming the organic compoundssytematically first we have to first study about the following threefeatures(a) Root word(b)Primary suffix(c)Secondary suffix(d)PrefixRoot word The basic unit in organic nomenclature is theroot word.Chains containing one to four carbon atoms are known byspecial root words while chains from C5 onwards are known by greeknumber roots.Chain length Root Word C1 Meth- C2 Eth- C3 Prop- C4 But- C5 Pent- C6 Hex- C7 Hept-
  7. 7. Chain length Root Word C8 Oct- C9 Non- C10 Dec- C11 Undec- C12 Dodec- C13 Tridec- C20 Eicos- C30 Triacont- C40 Tetracont- C50 Pentacont-Primary Suffix The primary suffixes are added to the root word toshow saturation or unsaturation in a carbon chain.Name of Carbon Chain Primary IUPAC Suffix name (1) Saturated -ane Alkane ( H3C CH3 ) (2) Unsaturated -ene Alkene With one double bond ( H2C CH2 )
  8. 8. Name of Carbon Chain Primary IUPAC Suffix name Unsaturated with -yne Alkyne one triple bond ( HC CH ) Unsaturated with -adiene Alkadiene two double bonds ( CH CH2 ) H2C CH Unsaturated with -atriene Alkatriene three double bonds ( H 2C CH CH ) CH CH CH 2 The compounds that are given above in the brackets are theexamples of the carbon chain mentioned above.Secondary Suffix Suffixes that are added after the primary suffixes toindicate the presence of a particular functianal group in carbon chain iscalled the secondary suffix. Functional Group Secondary Suffix Alcohol ( -OH ) -ol Aldehyde( -CHO ) -al Ketone( CO) -one Carboxylic acid ( -COOH) -oic acid Sulphonic acid( SO3H) -Sulphonic acid
  9. 9. Amine(-NH2) -amine Thioalcohol(-SH) -thiol Cyanide( -CN) -nitrile Esters(-COOR) -oate Amides(-CONH2 ) -amide Acid halide( -COX) -oyl halide Now let us see how the IUPAC names are formedfromed from the Root,Primary and the Secondary suffixesHomologus Root Primary Secondary IUPAC Series Word Suffix Suffix nameAlcohols Alk -ane -ol Alkanol(saturated)Alcohols Alk -ene -ol Alkenol(unsaturated)Alcohols Alk -yne -ol Alkynol(Unsaturated)One triple bondAldehydes Alk -ane -al Alkanal(saturated)Ketones Alk -ane -one Alkanone(saturated)
  10. 10. Prefix It is always be kept in mind that alkyl groupsforming branches of the parent chain are considered as sidechains.Atoms or groups of atoms such as fluoro(-F), Chloro(-Cl)bromo(-Br), Ido(-I), Nitro(-No2) and Alkoxy(-OR) are reffered to assubstituents.Root words are profixed with the name of the substituent orthe side chain.Arrangment of Prefixes,Root word and Suffixes These are arranged a follows while writing thename in such a manner.IUPAC name = Prefixes+Root word+Primary Suffix+ Secondary suffixExample1 Br HO 3 CH CH3 4 1C CH2 2 O 3-bromobutanoic acid If we consider the above example then 1.Prefix= Bromo 2.Root word = But 3.Primary suffix= ane 4.Secondary suffix= oic acid 5.No of Carbons = 4
  11. 11. Hence the IUPAC name of thhe compound is 3-bromobutanoicacidExample 2 1 CH2 CH CH3 3 5 HO CH 4CH 2 CH3 4-methylpent-2-en-1-olIn the above given example1 Prefix= Methyl2 Root word = Pent3 Primary suffix= ene4 Secondary suffix = ol5 Number of Carbons = 5Hence the name of the given compound is 4-methylpent-2-en-1-ol
  12. 12. Example 3 5 H3C OH 3 2 1 4 CH C C C H3C O 4-methylpent-2-ynoic acidIn the above given example1.Prefix = Methyl2.Root word = Pent3. Primary suffix = yne4. Secondary suffix = oic acid5. Number of Carbons = 5Hence the given compound name is 4-methylpent-2-ynoic acidExample 4 1 CH 3 3 2 H 3C C OH CH 3 2-methylpropan-2-ol
  13. 13. In the above example1.Prefix = Methyl2.Root Word = Prop3.Primary Suffix = ane4.Secondary Suffix = ol5.Number of Carbons = 3Hence the name of the Organic Compound is 2-methylpropan-2-olIUPAC Nomenclature of Organic compounds in Bond Line structures :1. 2 1 = H3C CH3 Ethane2. 1 CH3 = 2 H3C CH2 3 Propane 13. CH3 2 = CH CH3 3 H3C CH 4 CH3 2,3-dimethylbutane
  14. 14. 4. 2 4 2,2-dimethylbutane 1 35. 2 2,2-dimethylpropane 1 36. 2 4 2-methylbut-2-ene 1 37. 2 4 but-1-yne HC 3 18. 3 2 3-methylbut-1-yne HC 4 19. 2 4 1 3 H2C CH3 3-methylpent-1-ene 5 CH3
  15. 15. 10. 2 OH 1 propan-1-ol 3 OH11. 2 propan-2-ol 3 112. 2 OH 2,2-dimethylpropan-1-ol 3 1 O13. Ethanoic acid 1 OH 2 O14. 3 1 2-methylbutanoic acid OH 4 215. 4 2 3 pentan-3-one 5 1 O
  16. 16. 2. Saturated Aliphatic compounds Nomenclature In order to study about the nomenclature of the saturatedaliphatic compounds one nedd to know about the "Longest Chain Rule"Longest Chain Rule : This rule states that in naming an organiccompound we have to select the longest continuos chain of carbon atomswhich may or may not be horizontal.This continuos chain is called the"Parent Chain" or "Main Chain".Examples :1. 4 2 CH2 CH2 3-methylpentane 5 3 1 H3C CH CH3 CH3 In above example the longest chain is of 5 carbonatoms.Hence it is a derivative of pentane.The substiuent is a methylgroup. CH32. 1 H3C 2 4 CH3 C CH2 3 7 CH2 C CH2 2,2,5,5-tetramethyloctane H3C 5 CH2 CH3 H3C 6 8 Here the longest chain contains 8 carbons which isa continuos chain.At 2nd and 5th carbons they are two methyl derivativesie they are 4 methyl group substituents.Therefore the name of thecompound is 2,2,5,5-tetramethyloctane
  17. 17. 3. 3 7 H3C CH2 CH2 CH3 1 5 2 CH 4 CH CH2 6 4-ethyl-2-methylheptane CH3 H2C 5 6 CH3 In above example the longest chain contains 7carbons as indicated by blue numbers.The numbering should not bedone as indicated by red numbers because it is not the longest chain.Itcontain methyl at 2nd position as a substituent and ethyl group as asubstituent at 4 th position therefore the name of the compound is4-ethyl-2-methylheptane.4. 1 5 H3C CH2 CH3 H3C CH2 CH3 2 CH HC 4 CH HC 3 3 5 H3C 4 CH2 CH3 H3C CH2 CH3 Correct numbering 2 1 3-ethyl-2-methylpentane Wrong numbering (a) (b)
  18. 18. In above example the longest chain contains two possibilitiesas shown in (a) and (b). In such a case the longest chain is choosen insuch a way that it contains more number of substituents. If we cosider (a)it contains two substituents 2nd and 3rd position but if we consider (b) itcontains one substituent at 3rd position so the numbering as indicated bythe red numbers will be wrong in such that case and the name of thecompound is 3-ethyl-2-methylpentane.5. 1 4 5 5 1 H3C CH2 CH3 H3C CH2 CH3 2 2 CH CH2 4 CH CH2 3 3 H3C H3C 2-methylpentane 4-methylpentane Correct numbering Wrong numbering (a) (b) In this type of cases the numbering shouldbe done in such a way that the carbon atom carrying the first substituentget the lowest possible number.Hence in structure (a) the substituent isat 2nd carbon and in structure (b) the substituent is at 4th carbon as aresult structure (a) is correct numbering and structure (b) is wrongnumbering.Hence the name of the structure is 2-methylpetane.
  19. 19. 6. Lowest Sum Rule CH3 2 4 1 4 3 2 5 Correct numbering 2,3-dimethylpentane (Correct) H3C CH3 5 3 1 Wrong numbering 3,4-dimethylpentane (Wrong) CH3 In this case the numbering of carbon atomsshould be done in such a way that the sum of the positions of thesubstituent atoms attached should be minimum. In correct numbering the methyl groups are at 2nd and 3rdpositions. Sum of positions is 2+3=5. In wrong numbering the methyl groups are at 3rd and 4thpositions. Sum of positions is 3+4=7. In above 2 cases the sum of the positions of the substituentatoms is minimum in case 1. Hence the name of the structure is2,3-dimethylpentane.7. 1 CH3 CH3 H3C 5 2 4 C CH 2,2,4-trimethylpentane (Correct) 4 3 2 5 CH2 CH3 H3C 3 1 2,4,4-trimethylpentane (Wrong) Sum of positions in correct numbering = 2+2+4 = 8 Sum of positions in wrong numbering = 2+4+4 = 10 Hence the name of the compound is 2,2,4 -trimethylpentane
  20. 20. 8. 2,2,6,6-tetramethyloctane (Correct) CH3 H3C 1 7 CH2 CH3 2 6 8 3,3,7,7-tetramethyloctane (Wrong) 8 4 H3C C CH2 2C 1 3 5 5 3 7 CH2 CH2 H3C 4 6 CH3 Sum of positions in correct numbering = 2+2+6+6 = 16 Sum of positions in wrong numbering = 3+3+7+7 = 20 Hence the name of the compound is 2,2,6,6-tetramethyloctane9. CH3 4 6 2 CH3 3 5 H3C 1 3-ethyl-2-methylhexane CH3 In above case they are two different alkyl substituentsthen in that cases thier names are written in alphabetical orders.Hencethe name of the compound is 3-ethyl-2-methylhexane It should be kept in the mind that prefixes such asdi,tri, etc are not considered while arranging the substituentalphabetically.
  21. 21. 10. H3C 7 6 H3C CH3 5 4 3 CH3 2 CH3 H3C 1 CH3 4-ethyl-2,2,3,4-tetramethylheptane It is same as the above studied case and the substituentsare arranged in alphabetical order.11. 1 2 5 6 H3C CH3 3 4 H3C CH3 3-ethyl-4-methylhexane In above case the two different alkyl substituents(ethyl,methyl) are at equalent positions.Then in that case the numberingis done in such away that the alkyl group which comes first inalphabetical order gets the lowest number.Hence the name of thecompound is 3-ethyl-4-methylhexane
  22. 22. 12. H3C CH3 6 CH3 2 3 4 5 1 H3C CH3 H3C 3,3-diethyl-4,4-dimethylhexane Same as the above case and the numbering is givenaccording to alphabetical order.Hence the name of the compound is 3,3-diethyl-4,4-dimethylhexane15. Alkyl substituents : 1(a) H 3C Q Methyl (CH3----) Q(b) Etyhl (CH3 CH2-----) H3C CH2 1 2(c) 2 CH2 Q Propyl (CH3 CH2 CH2----) H3C CH2 1 3(d) 2 4 Butyl (CH3 CH2 CH2 CH2----) CH2 CH2 H3C CH2 Q 1 3 "Q is any sustituent"
  23. 23. 16.Trivial names for some alkyl substiuents : 3 H3C (a) Isopropyl 2 CH Q H3C 1 (b) H3C Q Isobutyl CH CH2 H3C (c) 3 CH2 Q 2 H3C CH Secondary butyl 4 CH3 1 CH3 (d) H3C C Q CH3 Teritiary butyl (e) CH2 CH3 H3C C Teritiary pentyl Q H3C " Q is any sustituent"
  24. 24. 17. 2 3 H3C H3C 1 4 2 CH3 3 5 1 6 7 8 9 CH3 5-(2-methylpropyl)nonane In this case first the longest chain is determinedthen look for the sustiuent groups. In above example the logenst chaincontains 9 carbon atoms and a sustituent is located at 5th position.Nowconsider the substituent and numbert it as 1,2, so on..........In aboveexample the substituent contains 3 carbon atoms with another methylgroupo at second position. Hence the compound can be named as 5-(2-methylpropyl)nonane Note that the sustituent group is always enclosedin brackets and then place in the IUPAC name.The above name can also be given in trivial system as follows 5-(isobutyl)nonaneAs we already studied that the substituent group at 5 th carbon is"iso butyl" group hence the compound can also be named as follows asstated above in trivial system.
  25. 25. 18. 3 H3C CH3 2 1 5 6 3 4 7 8 1 2 H3C H3C CH3 9 CH3 5-(1methylpropyl)-2,7-dimethylnonaneIUPAC name is 5-(1methylpropyl)-2,7-dimethylnonaneTrivial system name is 5-(secondarybutyl)-2,7-dimethylnonaneBecause in trivial system the substituent that is attached to 5th carbon inthe longest chain is secondarybutyl.19. 1 H3C 3 2 CH3 1 H3C 3 6 2 4 7 5 CH3 2-methyl-4,4-di(propan-2-yl)heptane H3C (or) H3C 2-methyl-4,4-bis(propan-2-yl)heptane CH3 In this case the same complex alkylgroup attached more than one time .In this case we use bis in place of diand tris in case of tri etc are use to indicate multiplicity of substitutedsubstituent.The above molecule can also be expressed in trivial naming "2-methyl-4,4-bis(isopropyl)heptane"
  26. 26. Halogenated Aliphatic Compounds : Halogens are the molecules that are situated in 17thgroup (VII A Group) elements in periodic table.They are1.Fluorine(F)2.Chlorine(Cl)3.Bromine(Br)4.Iodine(I)5.Astatine(At) In the above stated atoms astatine is radioactive elementhence it is not considered in our study.The halgen atoms are denoted byX in organic compounds. Compounds derived from alkanes by the replacement ofone or more hydrogen atoms by the coressponding number of halogenatoms are termed as "halo-alkanes".Generally depending upon the number of halogen molecules in thestructure they can be classified as mono,di,tri etc halgen derivatives.1.Mono halogen derivatives :(a)IUPAC names : In IUPAC naming of monohalogen derivatives one has to followthe following rule.The compounds should be named as "haloalkane"1. 1 H3C Cl chloromethane In the above example the halogen atom is chlorine and the alkylgroup is methyl as a result it is written as chloromethane halo = chloro alkane = methane
  27. 27. 2. H3C 1 bromoethane 2 Br halo = bromo alkane = ethane Hence the IUPAC name is bromoethane3. 2 Br 1-bromopropane H3C 3 1 In case of propane the Halogen atom can be placed inposition 1 or 2. Therefore in this case the position of the halogen atomshould be also shown.Hence the name of the structure is1-bromopropane.4. I 2-idopropane 2 H3C CH3 3 1 1 4 H3C CH35. 2-fluorobutane 2 3 F
  28. 28. CH36. 2 4 Cl 5 H3C 1 3 5-chloro-2,3-dimethylhexane CH3 CH3 6 In above example the longest chain is choosenand at the same time it should obey the lowest sum rule also.Hence thename of the compound is 5-chloro-2,3-dimethylhexane.7. CH3 3 Cl 7 5 8 6 4 H3C 2 1 CH3 2-chloro-7-methyloctane In the above example both the methyl groupand the chloro group are situauted at equal distances therefore in thiscase the chloro group should be given high preference and thenumbering should be done in such a way that halogen substituentcontanins least number. Hence the name of the compound is given bythe 2-choloro-7-methyloctane
  29. 29. (b) Trivial system of naming :1. Cl n-propylcholoride H3C2. Cl iso propylcholoride H3C CH33. H3C CH3 secondary butylchloride Cl4. CH3 isobutylchloride Cl H3C5. CH3 teritiary butylchloride H3C Cl CH3
  30. 30. This type of naming is based on the trivial system of the alkyl group orsubstituent that is attached to it.The trivial name added to the halogengives the trival naming of the mono halogen derivatives.For example consider structure 2In that the alkyl group that is present is iso propylThe halogen atom that is present is ChlorineHence the name of the compound is iso propyl chloride.Dihalogen derivatives : They are the halogen derivatives which contain two halo atoms inthe structure of the compound.This halogen derivatives can be classifiedinto 3 types.They are(a) Gem dihalides(b) Vic Dihalides(c) Terminal dihalidesGem dihalides : In these derivatives both the halogen atoms are attached tothe same carbon atom.These Gem dihalides can be named by 2 ways(a) Trivial naming(b) IUPAC namingTrivial naming : In trivial naming of this compounds, They are named as"alkylidene halides".This can be explained by the examples given below.
  31. 31. 1. H3C Cl Ethylidene chloride Cl2. CH3 H3C Br Iso propylidene bromide Br3. I H3C Propylidene iodide I4. F H3C Iso butylidene fluoride F H3C This is how one can express the the dihalogenderivatives in trivial naming. The alkyl substituents are named in thetrivial naming and they are added to the given halogen atom to form thetrivial name. For example consider structure 2 the alkyl groupis iso propyl group. The halogen atoms attached there is Bromine. As theresult the trivial name of the compund takes the form "alkylidene halide"as stated above.Therefore the name of the compound is "Iso propylidene bromide"
  32. 32. IUPAC naming : In IUPAC naming of these type of dihalogen derivativesthe position of the halogen atoms are noted and a suitable IUPAC nameis given for the respective compound.This can be shown by the givenexamples.1. Cl dichloromethane 1 Cl Cl2. 2 1 H3C 1,1-dichloroethane Cl3. 3 CH3 2,2-dibromopropane 1 2 H3C Br Br 34. H3C I 1 1,1-diiodo-2-methylpropane 2 H3C I Now consider example 1.In this example the methanemolecule is replaced by two chlorine atoms as a result it is named as dichloromethane
  33. 33. If we consider example 2 they are two chlorineatoms that are replaced in an ethane atom at 1st position as shown. As aresult the name of that structure takes the form 1,1-dichloroethane It should be noted that in this particular examplewe particularly specify "1,1-dichloroethane" because there is also achance of 1,2-dichloroethane.Therefore the numbering should be given inthis case. This is how one can name the gemdihalides.Vic Dihalides : In this dihalides the halogen atoms are attached to theadjacent carbon which is also called the vicinal carbon.These Vicdihalides can also be named in two ways(a) Trivial naming(b) IUPAC namingTrivial naming : In this trivial naming the alkyl group that is associated isgiven the trivial name and the halogen atom attached is also named andit takes the form "alkylene halides". This can be shown by the belowgiven examples.1. 1 Cl Ethylene chloride Cl 22. 3 CH3 2 Propylene chloride Cl Cl 1
  34. 34. If we consider the first example the alkyl group that isattached is ethyl group .The halogen atoms are attached to the adjacentcarbon atoms.The halogen atom in this example is chlorine.Thereforethe name of the compound should take the form "alkylene halide", as aresult the name of that compound is ethylene chloride3. 1 5 3 4 2 H3C Br 6 Br hexylene bromideIUPAC naming : In this IUPAC naming the positions of the halogenatoms that are attached should be given numbering and they areindicated in IUPAC name as shown in following examples.1. Cl 2 1 1,2-dichloroethane Cl2. 1 Br 1,2-dibromobutane 3 2 H3C Br 4
  35. 35. CH3 CH33. 1 7 5 3 8 6 4 2 H3C Cl Cl CH3 1,2-dichloro-4-ethyl-5,7-dimethyloctane In above example the longest chain contains 8carbon atoms which is continuos and the numbering is given accordingto the lowest sum rule. At 5th and 7th positions there exits methyl groupand at 4th position there exists ethyl group and at 1st and 2nd positionare attached by the chlorine atom as a result the name of the compoundis 1,2-dichloro-4-ethyl-5,7-dimethyloctane It should be noted that the substituent namesshould be in alphabetical order neglecting the di,tri suffixes.For clearpicture the alphabetical order for the above example ischloroetyhlmethyl As a result the name should be expressed inthese alphabetical order and added to the alkane ie octane in the aboveexample. This is how one should take care in naming such type ofcomplex molecules.
  36. 36. Terminal dihalides : These are the dihalogen derivatives in which thehalogen atoms are attached to the terminal carbon atoms.These terminaldihalides can be also named in two ways(a) Trivial system(b)IUPAC systemTrivial system : The dihalogen derivatives of this form can be expressedin the following way in trivial system.They are expressed in the form of"Polymethylene halides".This can be explained in the following examples.1. 2 4 Cl tetramethylene chloride Cl 1 32. 1 3 5 7 heptamethylene bromide Br Br 2 4 6 Now let us consider example 1.In this structurethey are 4 carbon atoms.1 and 4 are the terminal carbons in the aboveexample.The halogen atoms that are attached is chlorine atoms.Weknow that the dihalogen derivative of these form takes "Polymethylenehalide". Therefore the name of the compond is "tetramethylene chloride".Here we use the word tetra because they are 4 carbon atoms in thatexample.Simillarly they are 7 carbon atoms in example 2.The halogenatoms that are attached is bromine atoms.The name of the compound is "heptamethylene bromide"
  37. 37. IUPAC naming : In IUPAC naming the longest chain is chosen then thecarbon atoms are given numbering. Finally the IUPAC name is given.Thiscan be explained in the following way by the below examples.1. 1 3 1,3-diiodopropane I I 22. CH3 CH3 1,7-dibromo-3,5-dimethylheptane 1 2 3 4 5 6 7 Br Br H3C CH33. 1,7-dichloro-3,5-diethyl-4-methylheptane 1 2 3 4 5 6 7 Cl Cl CH3 6 Cl4. 4 5 H3C 3 2 1,6-dichloro-2,3-dimethylhexane 1 Cl CH3 This is how one can name the terminal dihalidesas shown above examples by proper numbering.
  38. 38. Dihalogen deivatives which contain different halogen atoms :1. 2 Br Cl 1 1-bromo-2-chloroethane Suppose if we consider the above example bothchlorine and bromine atoms are situated at equal distance from the alkylgroup. In such cases the numbering should be done such a way that thehalogen atom which will be first in alphabetical order will be given theleast number ie it is preffered first. In above example bromine is first inalphabetical order therfore this is numbered first.As a result the name ofthe compound is 1-bromo-2-chloroethane2. H3C 6 I 5 CH3 4 3 1 2 CH3 Cl 1-chloro-3-(iodomethyl)-4,5-dimethylhexane In example 2 the halogen atoms are chlorine andiodine.But these molecule is a complex molecule Cl atom is attached tothe 1st carbon.And at 3rd carbon the ido methyl group isattached.Therefore it is enclosed in the brackets.The longest chaincontains 6 carbon atoms as shown. Therefore the name of the compoundis " 1-chloro-3-(idomethyl)-4,4-dimethylhexane"
  39. 39. 7 13. CH3 CH3 2 6 5 4 3 Cl Br CH3 CH3 2-bromo-6-chloro-3-ethyl-5-methylheptane In the above case the longest chain contains 7carbon atoms but here the halogen atoms chlorine and bromine are atsame distance and they also contain methyl group and ethyl group at thesame distance in such a case the numbering should be done in such away that bromine should be given the first preference.The order of preference of halogen atoms is bromine>chlorine>fluorine>iodine In in the alphabetical order.4. Cl 2 4 5 7 9 H3C CH3 1 3 6 8 Cl 5,5-bis(2-chloroethyl)nonane The above example is a complex molecule.The group that isattached to the 5th carbon is chloro ethyl group.As they are attachedtwice to the same carbon "bis" comes into picture there.The abovemolecule can also be named as "5,5-di(2-chloroethyl)nonane".
  40. 40. Tri halogen derivatives : These tri halogen derivatives are derived by thereplacement of three hydrogen atoms from the alkanes with halogenatoms.This can be explained by the below given examples.1. 5 CH3 1 I 4 CH3 3 2 Cl Br 2-bromo-3-chloro-4-iodopentane In the above example we observe that thelongest chain contains 5 carbon atoms and the 2nd,3rd and 4th positionsare replaced by bromine,chlorine and iodine.We know that the halogenatoms should be in alphabetical order hence the name of the compoundis "2-bromo-3-chloro-4-idopentane"2.Haloforms : The trihalogen derivatives of first alkane(methane) aretermed as haloforms.They are named in trivial system as "Haloform".(a). Cl Cl IUPAC Trivial trichloromethane chloroform Cl In above example the IUPAC name of thecompound is trichloromethane.But in trivial system we know that it takesthe form "haloform".Here the halogen atom is chlorine. Therefore it isnamed"Chloroform".
  41. 41. (b). IUPAC Trivial Br Br tribromomethane bromoform Br(c). I triiodomethane idoform I I3. Br 8 CH3 1 7 3-bromo-7-chloro-1-iodooctane 2 3 4 5 6 I Cl In above example the numbering should be donein such a way that the lowest sum rule is applicable and the name isarranged in the alphabetical order.Hence the name of the compound is "3-bromo-7-chloro-1-idooctane"4. I 7 Cl 6 CH3 5 Br 4 2 3 CH3 H3C CH3 2-bromo-4-chloro-5-ethyl-6-iodo-3-methylheptane 1
  42. 42. 5. Cl CH3 2 2" H3C 1" I 1 4 3 2 1 Br 5 CH3 6 4-(1-bromoethyl)-4-(1"-chloroethyl)-3-iodoheptane CH3 7 The above example is a complex moleculestructure.This molecule contains 7 carbons as numbered in the longestchain.At 3rd position there is a halogen atom ie Iodine atom.Now let usfocus our view on fourth position.In fourth position they are two alkylsubstituents.They are(a) bromo ethyl(b) Chloro ethyl When comes to alphabetical orderbromine should be given first preference as a result the alkyl group ofbromine is given 1 and 2 as shown in the figure.Then the alkyl group ofchlorine should should be given 1" and 2" as shown in the figure. It should be noted that both the alkylgroups souldnt be given the same numbering.Coming to the halogenatoms bromine is attached to 1 positon as shown and chlorine isattached to 1" position as shown in the molecule.Therefore the alkylgroups with halogen substituents are arranged in brackets. Hence thename of the compound is 4-(1-bromoethyl)-4-(1"-chloroethyl)-3-idoheptane
  43. 43. Polyhalogen derivatives : These are the halogen derivatives whichcontain more than three halogen atoms in the structure of themolecule.This can be explained by the below given examples.1. Cl Cl C tetrachloromethane Cl Cl2. Br Cl dibromo(dichloro)methane C Br Cl3. Cl Cl bromo(trichloro)methane C Br Cl4. Cl F bromo(chloro)fluoro(iodo)methane C I Br This is how one can name the poly halogenderivatives of first alkane(methane).It should be noted that the names arealways arranged in alphabetical order only.
  44. 44. Let us now consider 1st example.In this example4 chlorine atoms are attached to carbon atom in place of hydrogen atoms.Hence it is called as "tetra chloromethane". This can be also named as"carbon tetrachloride" in trivial naming. Similarly in 3rd example they are 3 chlorine atomsand a bromine atom that is attached to the carbon atom In this casebromine should be given first preference because in alphabetiacl orderbromine comes first. The halogen atom that comes after bromine iechlorine should be enclosed in brackets.Hence the name of the structureis bromo(trichloro)methane. It should be noted that this rule is not applicablefor higher alkanes after methane.1. Cl 1 Cl CH 1,1,2,2-tetrachloroethane CH Cl 2 Cl2. Cl 1 Br C Br 1,1,2,2-tetrabromo-1,2-dichloroethane 2 Br C Br Cl I3. 1 CH2 CH CH2 CH2 Br 2 3 4 5 6 7 8 Cl CH CH CH2 CH2 2,8-dibromo-1-chloro-4-fluoro-3-iodooctane Br F
  45. 45. I4. 2" F CH2 CH 1" 1 3 7 CH2 CH2 HC CH2 Cl 2 4 5 6 8 Br CH2 CH CH2 CH2 HC 1 2 CH2 Br Cl 1-bromo-4-(1-bromo-2-chloroethyl)-8-chloro-5-(1"-fluoro-2"-iodoethyl)octane The above given structure is a complexstructure and it can be explained in the following way.The longest chaincontains 8 carbon atoms.Here the numbering can start from bromine orchlorine because they are at equal distance and the lowest sum rule forthe both types of numbering will be same.In such a case the firstpreference is given to the atom which comes first in the alphabeticalorder.Here bromine comes first in the alphabetical order hence it is giventhe first preference. At 4th and 5th positions they are alkylgroup sustituents.They are(a) 1-bromo-2-chloro ethyl group(b) 1-fluoro-2-ido ethyl group These sustituents are at 4th and 5th positions respectively.Byconsidering the alphabetical order(a) bromo(b) 1-bromo-2-chloro ethyl group(c) chloro
  46. 46. (d) 1-fluoro-2-idoethylgroup Therefore by considering this alphabetical order the name of thecompound is1-bromo-4-(1-bromo-2-chloroethyl)-8-chloro-5-(1"-fluoro-2"-iodoethyl)octane5. 3"CH3 I F 2" H3C HC CH CH2 1" 1 2 4 3 CH HC CH3 1 2 6 CH2 HC CH2 HC 5 7 Cl Br CH3 2-bromo-1-chloro-4-(2-fluoro-1-iodoethyl)-6-methyl-3-(propan-2"-yl)heptane In above example by choosing the proper numberingand following the lowest sum rule ,arranging the substituents inalphabetical oder one can generate the IUPAC name of the compoundeasily for any complex molecules.
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