Naming Organic Molecules 1j3i1t

Preface Generally students have small doubts regarding the nomenclature of the organic compounds. This book is been written with the aim of needs and interests of students in organic nomenclature. In this book every topic has been dealt precisely and to the point in a sample and understandable language.Things have been explained wherever possible. A good number of illustration examples with answers is also provided in this book.This helps the students in developing their reasoning skills in nomenclature. As every organic compound has its own name.This book is helpful in determinining the IUPAC names of the given organic compounds.This book aditionally provides a small and interesting topic, "Molecular formula to an individual person". Hope this book meets the needs of the students,chemists who want to learn organic chemistry nomenclature in detail.

B.Sandeep Chaitanya Bharathi Institute Of Technology (C.B.I.T)

Index 1. Nomenclature 2. Saturated Aliphatic Compounds

Page No Trial Version only

4 16

3. Unsaturated Aliphatic Compounds

---

4. Cyclic compounds

---

5. Aromatic compounds

---

6. Bicyclo Compounds

---

7. Alcohols 8. Ethers

---

9. Aldehydes

---

10.Ketones

---

11.Carboxylic Acids

---

12.Acid derivatives

---

13.Nitrogen containing compounds

---

14.Tricyclo Compounds

---

15.Bio molecules

---

1.Nomenclature What is organic chemistry? Organic chemistry is a branch of chemistry which deals with carbon compounds.Organic compounds mainly contains carbon and hydrogen as its main constituents.In addition to this it also contains N,S,P,Cl,Br and I. Why is nomenclature necessary? Organic chemistry is a vast branch as millions of organic compounds are already known and thousands of new compounds are beign added to this list every year.In order to facilitate the study of such large number of compounds.Therefore it is necessary to classify the organic compounds.In order to classify this organic compounds the term "Nomenclature" comes into picture. What is nomenclature? Nomenclature means the assignment of names to the organic componds.The naming of organic compounds is an important aspect in the study of organic chemistry as their number is very large and variety of molecular structures exist in their molecules.The field has become more complex on the phenomenon of the isomerism.They are two main systems of nomenclature of organic compounds. They are 1.Trvial system 2.IUPAC system

Trivial System In this system whenever a new compound is discovered it is given an individual name.These names are called Trivial names.These names are also called Common names. Examples (a) Acetic acid derives its name from vinegar of which it is chief constituent. (b) Formic acid was named as it was obtained from red ants. (c)The name oxalic acid, malic acid and citric acid is derived from botanical sources. (d)Urea and Uric acid have been derived from animal sources. (e)The liqiud that is obtained by the destructive dstaillation of wood was 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 Of Pure Applied Chemistry".By using this IUPAC system one can name any complex organic compound easily.The name assigned to an organic compound on the basis of latest IUPAC rules is known as systematic 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

(c) This system is helpful in naming the multifunctional groups. (d) This is a simple,systematic and scientific method for the nomenclature of organic compounds. Rules for organic nomenclature For naming the organic compounds sytematically first we have to first study about the following three features (a) Root word (b)Primary suffix (c)Secondary suffix (d)Prefix Root word The basic unit in organic nomenclature is the root word.Chains containing one to four carbon atoms are known by special root words while chains from C5 onwards are known by greek number roots. Chain length

Root Word

C1

Meth-

C2

Eth-

C3

Prop-

C4

But-

C5

Pent-

C6

Hex-

C7

Hept-

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 to show saturation or unsaturation in a carbon chain. Name of Carbon Chain

Primary Suffix

(1) Saturated ( H3C

CH3

name

-ane

Alkane

-ene

Alkene

)

(2) Unsaturated With one double bond ( H2C

IUPAC

CH2

)

Name of Carbon Chain Unsaturated with

Primary

IUPAC

Suffix

name

-yne

Alkyne

-adiene

Alkadiene

-atriene

Alkatriene

one triple bond (

HC

)

CH

Unsaturated with two double bonds (

CH H2C

)

CH2 CH

Unsaturated with three double bonds (

H 2C

CH CH

CH CH

CH 2

)

The compounds that are given above in the brackets are the examples of the carbon chain mentioned above. Secondary Suffix Suffixes that are added after the primary suffixes to indicate the presence of a particular functianal group in carbon chain is called the secondary suffix. Functional Group

Secondary Suffix

Alcohol ( -OH )

-ol

Aldehyde( -CHO )

-al

Ketone(

-one

CO)

Carboxylic acid ( -COOH) Sulphonic acid( SO3H)

-oic acid -Sulphonic acid

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 formed fromed from the Root,Primary and the Secondary suffixes Homologus Series Alcohols

Root

Primary

Secondary

IUPAC

Word

Suffix

Suffix

name

Alk

-ane

-ol

Alkanol

Alk

-ene

-ol

Alkenol

Alk

-yne

-ol

Alkynol

Alk

-ane

-al

Alkanal

Alk

-ane

-one

Alkanone

(saturated) Alcohols (unsaturated) Alcohols (Unsaturated) One triple bond Aldehydes (saturated) Ketones (saturated)

Prefix It is always be kept in mind that alkyl groups forming branches of the parent chain are considered as side chains.Atoms or groups of atoms such as fluoro(-F), Chloro(-Cl) bromo(-Br), Ido(-I), Nitro(-No2) and Alkoxy(-OR) are reffered to as substituents.Root words are profixed with the name of the substituent or the side chain. Arrangment of Prefixes,Root word and Suffixes These are arranged a follows while writing the

name in such a manner. IUPAC name =

Prefixes+Root word+Primary Suffix+ Secondary suffix

Example1

Br HO

3 CH

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

CH3 4

Hence the IUPAC name of thhe compound is

3-bromobutanoicacid

Example 2 1

CH2

CH

HO

CH 2

3

CH3 4CH

5

CH3

4-methylpent-2-en-1-ol

In the above given example 1 Prefix= Methyl 2 Root word = Pent 3 Primary suffix= ene 4 Secondary suffix = ol 5 Number of Carbons = 5 Hence the name of the given compound is 4-methylpent-2-en-1-ol

Example 3 5 H3C

OH 3 C

4 CH

2 C

1 C

H3C

O 4-methylpent-2-ynoic acid

In the above given example 1.Prefix = Methyl 2.Root word = Pent 3. Primary suffix = yne 4. Secondary suffix = oic acid 5. Number of Carbons = 5 Hence the given compound name is 4-methylpent-2-ynoic acid Example 4

1 CH 3 3 H 3C

2 C

CH 3 2-methylpropan-2-ol

OH

In the above example 1.Prefix

=

Methyl

2.Root Word = Prop 3.Primary Suffix = ane 4.Secondary Suffix = ol 5.Number of Carbons = 3 Hence the name of the Organic Compound is 2-methylpropan-2-ol IUPAC Nomenclature of Organic compounds in Bond Line structures : 1. 1

2

=

H3C

CH3

Ethane

2.

1

=

2

3

CH3

CH2

H3C

Propane

3.

1 CH3

2 3

4 2,3-dimethylbutane

=

CH H3C

CH3 CH

CH3

4. 2

4

2,2-dimethylbutane

1 3

5. 2 2,2-dimethylpropane

3

1

6. 2

4

1

2-methylbut-2-ene

3

7.

2

4

HC

but-1-yne

3

1

8. 3 2

3-methylbut-1-yne

HC

4

1

9.

2 1

H2C

4 3

CH3

5

CH3

3-methylpent-1-ene

10. 2

OH

propan-1-ol

1 3 OH

11.

propan-2-ol

2 1

3

12.

2

OH 2,2-dimethylpropan-1-ol

3

1

O

13.

Ethanoic acid

1

OH

2

O

14.

3

1

4

15.

OH

2

2-methylbutanoic acid

2

4 3 5 O

1

pentan-3-one

2. Saturated Aliphatic compounds Nomenclature In order to study about the nomenclature of the saturated aliphatic compounds one nedd to know about the "Longest Chain Rule" Longest Chain Rule : This rule states that in naming an organic compound we have to select the longest continuos chain of carbon atoms which may or may not be horizontal.This continuos chain is called the "Parent Chain" or "Main Chain". Examples : 1. 4 CH2

5

H3C

2

3

CH2

3-methylpentane

1

CH

CH3

CH3

In above example the longest chain is of 5 carbon atoms.Hence it is a derivative of pentane.The substiuent is a methyl group. 2.

CH3

1

H3C

2 C

CH3

4 3

CH2

CH2

7

C

5

H3C H3C

CH2 CH2

6

2,2,5,5-tetramethyloctane CH3

8

Here the longest chain contains 8 carbons which is a continuos chain.At 2nd and 5th carbons they are two methyl derivatives ie they are 4 methyl group substituents.Therefore the name of the compound is 2,2,5,5-tetramethyloctane

3. 3

H3C

CH2

1

2

CH

7

CH3

CH2

5

4 CH

CH3

H2C

CH2

6

4-ethyl-2-methylheptane

5 6

CH3

In above example the longest chain contains 7 carbons as indicated by blue numbers.The numbering should not be done as indicated by red numbers because it is not the longest chain.It contain methyl at 2nd position as a substituent and ethyl group as a substituent at 4 th position therefore the name of the compound is 4-ethyl-2-methylheptane. 4.

5

1

H3C

CH2

2 CH

CH3

H3C

4 CH

HC

3

5 4 CH2 CH3 Correct numbering

H3C

3-ethyl-2-methylpentane

(a)

CH2

CH3

CH2

CH3

HC

3 H3C

2

Wrong numbering

(b)

1

In above example the longest chain contains two possibilities as shown in (a) and (b). In such a case the longest chain is choosen in such 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) it contains one substituent at 3rd position so the numbering as indicated by the red numbers will be wrong in such that case and the name of the compound is 3-ethyl-2-methylpentane. 5. 1

4

H3C

CH2

2 CH

5

CH3

5

H3C

CH2

2

4 CH

CH2

3

1

CH3

CH2

3

H3C

H3C 2-methylpentane

Correct numbering (a)

4-methylpentane

Wrong numbering (b)

In this type of cases the numbering should be done in such a way that the carbon atom carrying the first substituent get the lowest possible number.Hence in structure (a) the substituent is at 2nd carbon and in structure (b) the substituent is at 4th carbon as a result structure (a) is correct numbering and structure (b) is wrong numbering.Hence the name of the structure is 2-methylpetane.

6. Lowest Sum Rule CH3

2 1

4

4

H3C

3

5

2

CH3

3

5

Correct numbering 2,3-dimethylpentane (Correct)

1

Wrong numbering 3,4-dimethylpentane (Wrong)

CH3

In this case the numbering of carbon atoms should be done in such a way that the sum of the positions of the substituent atoms attached should be minimum. In correct numbering the methyl groups are at 2nd and 3rd positions. Sum of positions is 2+3=5. In wrong numbering the methyl groups are at 3rd and 4th positions. Sum of positions is 3+4=7. In above 2 cases the sum of the positions of the substituent atoms is minimum in case 1. Hence the name of the structure is 2,3-dimethylpentane. 7.

CH3

1

H3C

2

5

4

C

4 3 CH2

H3C

CH3

3

CH

2

5

2,2,4-trimethylpentane (Correct)

CH3

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

8. CH3

7 8

CH2

2

H3C

H3C

6

1

CH3

4

C

3

CH2

5

5

4

H3C

3 CH2

CH2

2,2,6,6-tetramethyloctane (Correct)

1 8

2C 7

6

3,3,7,7-tetramethyloctane (Wrong)

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-tetramethyloctane 9.

CH3

H3C

6

4

2 3

CH3

5

1 3-ethyl-2-methylhexane

CH3

In above case they are two different alkyl substituents then in that cases thier names are written in alphabetical orders.Hence the name of the compound is 3-ethyl-2-methylhexane di,tri, etc alphabetically.

It should be kept in the mind that prefixes such as are not considered while arranging the substituent

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 substituents are arranged in alphabetical order. 11.

1 H3C

2

5

3

6 CH3

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 numbering is done in such away that the alkyl group which comes first in alphabetical order gets the lowest number.Hence the name of the compound is 3-ethyl-4-methylhexane

12. H3C

6

CH3

2

3

CH3

4

5

1 H3C

CH3 H3C

3,3-diethyl-4,4-dimethylhexane

Same as the above case and the numbering is given according to alphabetical order.Hence the name of the compound is 3,3-diethyl-4,4-dimethylhexane 15. Alkyl substituents : 1

(a)

H 3C

Methyl (CH3----)

Q

Q

(b) H3C

CH2

2

1

(c)

2

CH2 H3C

CH2

1

Propyl (CH3 CH2 CH2----)

4

Butyl (CH3 CH2 CH2 CH2----)

3 2

H3C

Q CH2

1

(d)

Etyhl (CH3 CH2-----)

CH2 CH2

3

Q

"Q is any sustituent"

16.Trivial names for some alkyl substiuents : 3

H3C

(a)

2 CH

Q

Isopropyl

H3C

1

(b)

H3C

Q

Isobutyl CH

CH2

H3C

(c)

3

CH2 H3C

Q

2

CH

4

Secondary butyl

CH3

1 CH3

(d)

H3C

C

Q

Teritiary butyl

CH3

(e)

CH3 CH2 H3C

C Q H3C

" Q is any sustituent"

Teritiary pentyl

17. H3C

3'

2

H3C

1 4

2'

3 5

CH3

1'

6 7

8 9 CH3 5-(2'-methylpropyl)nonane

In this case first the longest chain is determined then look for the sustiuent groups. In above example the logenst chain contains 9 carbon atoms and a sustituent is located at 5th position.Now consider the substituent and numbert it as 1',2', so on..........In above example the substituent contains 3 carbon atoms with another methyl groupo at second position. Hence the compound can be named as 5-(2'-methylpropyl)nonane Note that the sustituent group is always enclosed in brackets and then place in the IUPAC name. The above name can also be given in trivial system as follows 5-(isobutyl)nonane As we already studied that the substituent group at 5 th carbon is "iso butyl" group hence the compound can also be named as follows as stated above in trivial system.

18.

3' H3C

CH3

2'

1' 5

3 1

6

4

7

2

H3C

8

H3C

CH3

9

CH3 5-(1'methylpropyl)-2,7-dimethylnonane

IUPAC name is

5-(1'methylpropyl)-2,7-dimethylnonane

Trivial system name is

5-(secondarybutyl)-2,7-dimethylnonane

Because in trivial system the substituent that is attached to 5th carbon in the longest chain is secondarybutyl. 19.

1'

H3C

3'

2'

1 H3C

CH3

3 2

6

4 5

7 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 alkyl group attached more than one time .In this case we use bis in place of di and tris in case of tri etc are use to indicate multiplicity of substituted substituent.The above molecule can also be expressed in trivial naming "2-methyl-4,4-bis(isopropyl)heptane"

Halogenated Aliphatic Compounds : Halogens are the molecules that are situated in 17th group (VII A Group) elements in periodic table.They are 1.Fluorine(F) 2.Chlorine(Cl) 3.Bromine(Br) 4.Iodine(I) 5.Astatine(At) In the above stated atoms astatine is radioactive element hence it is not considered in our study.The halgen atoms are denoted by X in organic compounds. Compounds derived from alkanes by the replacement of one or more hydrogen atoms by the coressponding number of halogen atoms are termed as "halo-alkanes". Generally depending upon the number of halogen molecules in the structure 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 follow the 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 alkyl group is methyl as a result it is written as chloromethane halo

=

alkane =

chloro methane

2.

H3C 1 bromoethane

2

Br

halo

=

bromo

alkane

=

ethane

Hence the IUPAC name is bromoethane 3. 2

Br 1-bromopropane

H3C

1

3

In case of propane the Halogen atom can be placed in position 1 or 2. Therefore in this case the position of the halogen atom should be also shown.Hence the name of the structure is 1-bromopropane. 4.

I

2-idopropane

2 H3C

CH3

3

5.

H3C

1 4

1

CH3

2

F

3

2-fluorobutane

CH3

6.

2

4

Cl

5

H3C

3

1

5-chloro-2,3-dimethylhexane

CH3 6

CH3

In above example the longest chain is choosen and at the same time it should obey the lowest sum rule also.Hence the name of the compound is 5-chloro-2,3-dimethylhexane. 7.

CH3

3 8

H3C

7

6

5

Cl

4 2

1 CH3 2-chloro-7-methyloctane

In the above example both the methyl group and the chloro group are situauted at equal distances therefore in this case the chloro group should be given high preference and the numbering should be done in such a way that halogen substituent contanins least number. Hence the name of the compound is given by the 2-choloro-7-methyloctane

(b) Trivial system of naming : 1. Cl

n-propylcholoride

H3C

2.

Cl

iso propylcholoride

H3C

3.

CH3

H3C

CH3

secondary butylchloride

Cl

4.

CH3

isobutylchloride

Cl H3C

5.

CH3 teritiary butylchloride H3C

Cl

CH3

This type of naming is based on the trivial system of the alkyl group or substituent that is attached to it.The trivial name added to the halogen gives the trival naming of the mono halogen derivatives. For example consider structure 2 In that the alkyl group that is present is iso propyl The halogen atom that is present is Chlorine Hence the name of the compound is iso propyl chloride. Dihalogen derivatives : They are the halogen derivatives which contain two halo atoms in the structure of the compound.This halogen derivatives can be classified into 3 types.They are (a) Gem dihalides (b) Vic Dihalides (c) Terminal dihalides Gem dihalides : In these derivatives both the halogen atoms are attached to the same carbon atom.These Gem dihalides can be named by 2 ways (a) Trivial naming (b) IUPAC naming Trivial naming : In trivial naming of this compounds, They are named as "alkylidene halides".This can be explained by the examples given below.

1.

H3C

Cl Ethylidene chloride Cl

2.

CH3

H3C

Br

Iso propylidene bromide

Br

3.

I H3C

Propylidene iodide

I

4.

F

H3C

Iso butylidene fluoride F

H3C

This is how one can express the the dihalogen derivatives in trivial naming. The alkyl substituents are named in the trivial naming and they are added to the given halogen atom to form the trivial name. For example consider structure 2 the alkyl group is iso propyl group. The halogen atoms attached there is Bromine. As the result the trivial name of the compund takes the form "alkylidene halide" as stated above.Therefore the name of the compound is "Iso propylidene bromide"

IUPAC naming : In IUPAC naming of these type of dihalogen derivatives the position of the halogen atoms are noted and a suitable IUPAC name is given for the respective compound.This can be shown by the given examples. 1.

Cl dichloromethane

1 Cl Cl

2. 2

1

H3C

1,1-dichloroethane

Cl

3.

3

CH3

1

H3C

2,2-dibromopropane

2

Br

Br

4.

3

H3C

I

2

H3C

1

1,1-diiodo-2-methylpropane

I

Now consider example 1.In this example the methane molecule is replaced by two chlorine atoms as a result it is named as dichloromethane

If we consider example 2 they are two chlorine atoms that are replaced in an ethane atom at 1st position as shown. As a result the name of that structure takes the form 1,1-dichloroethane It should be noted that in this particular example we particularly specify "1,1-dichloroethane" because there is also a chance of 1,2-dichloroethane.Therefore the numbering should be given in this case. This is how one can name the gemdihalides. Vic Dihalides : In this dihalides the halogen atoms are attached to the adjacent carbon which is also called the vicinal carbon.These Vic dihalides can also be named in two ways (a) Trivial naming (b) IUPAC naming Trivial naming : In this trivial naming the alkyl group that is associated is given the trivial name and the halogen atom attached is also named and it takes the form "alkylene halides". This can be shown by the below given examples. 1.

1

Cl Ethylene chloride

Cl

2

2.

3

CH3

2 Cl

Cl

1

Propylene chloride

If we consider the first example the alkyl group that is attached is ethyl group .The halogen atoms are attached to the adjacent carbon atoms.The halogen atom in this example is chlorine.Therefore the name of the compound should take the form "alkylene halide", as a result the name of that compound is ethylene chloride 3.

1 5

3 4

H3C

2

6

Br

Br hexylene bromide

IUPAC naming : In this IUPAC naming the positions of the halogen atoms that are attached should be given numbering and they are indicated in IUPAC name as shown in following examples. 1.

Cl

2

1

1,2-dichloroethane

Cl

2.

1

3

H3C

4

Br 1,2-dibromobutane

2

Br

CH3

3.

CH3

1 8

H3C

7

6

5

3 4

2

Cl

Cl

CH3

1,2-dichloro-4-ethyl-5,7-dimethyloctane

In above example the longest chain contains 8 carbon atoms which is continuos and the numbering is given according to the lowest sum rule. At 5th and 7th positions there exits methyl group and at 4th position there exists ethyl group and at 1st and 2nd position are attached by the chlorine atom as a result the name of the compound is 1,2-dichloro-4-ethyl-5,7-dimethyloctane It should be noted that the substituent names should be in alphabetical order neglecting the di,tri suffixes.For clear picture the alphabetical order for the above example is chloro etyhl methyl As a result the name should be expressed in these alphabetical order and added to the alkane ie octane in the above example. This is how one should take care in naming such type of complex molecules.

Terminal dihalides : These are the dihalogen derivatives in which the halogen atoms are attached to the terminal carbon atoms.These terminal dihalides can be also named in two ways (a) Trivial system (b)IUPAC system Trivial system : The dihalogen derivatives of this form can be expressed in 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

Cl

4

1

tetramethylene chloride

Cl

3

2. 3

1 Br

5

7 heptamethylene bromide

2

4

6

Br

Now let us consider example 1.In this structure they are 4 carbon atoms.1 and 4 are the terminal carbons in the above example.The halogen atoms that are attached is chlorine atoms.We know that the dihalogen derivative of these form takes "Polymethylene halide". Therefore the name of the compond is "tetramethylene chloride". Here we use the word tetra because they are 4 carbon atoms in that example.Simillarly they are 7 carbon atoms in example 2.The halogen atoms that are attached is bromine atoms.The name of the compound is "heptamethylene bromide"

IUPAC naming : In IUPAC naming the longest chain is chosen then the carbon atoms are given numbering. Finally the IUPAC name is given.This can be explained in the following way by the below examples. 1.

1

3

I

1,3-diiodopropane

I

2

2. CH3

CH3 1,7-dibromo-3,5-dimethylheptane

1 3

2

Br

4

5

6

Br

CH3

H3C

3.

7

1,7-dichloro-3,5-diethyl-4-methylheptane

Cl

1

2

3

4

5

6

7

6

CH3

4.

4

H3C

2

Cl

1

Cl

Cl

5

3 1,6-dichloro-2,3-dimethylhexane

CH3

This is how one can name the terminal dihalides as shown above examples by proper numbering.

Dihalogen deivatives which contain different halogen atoms : 1.

2 Cl

Br

1

1-bromo-2-chloroethane

Suppose if we consider the above example both chlorine and bromine atoms are situated at equal distance from the alkyl group. In such cases the numbering should be done such a way that the halogen atom which will be first in alphabetical order will be given the least number ie it is preffered first. In above example bromine is first in alphabetical order therfore this is numbered first.As a result the name of the compound is 1-bromo-2-chloroethane 2.

H3C

3

1

Cl

6

5

I

2

CH3

4

CH3

1-chloro-3-(iodomethyl)-4,5-dimethylhexane

In example 2 the halogen atoms are chlorine and iodine.But these molecule is a complex molecule Cl atom is attached to the 1st carbon.And at 3rd carbon the ido methyl group is attached.Therefore it is enclosed in the brackets.The longest chain contains 6 carbon atoms as shown. Therefore the name of the compound is " 1-chloro-3-(idomethyl)-4,4-dimethylhexane"

7

3.

1

CH3

CH3

2 6

4

5

Cl

3

Br

CH3 CH3 2-bromo-6-chloro-3-ethyl-5-methylheptane

In the above case the longest chain contains 7 carbon atoms but here the halogen atoms chlorine and bromine are at same distance and they also contain methyl group and ethyl group at the same distance in such a case the numbering should be done in such a way 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

4

2 H3C 1

5

9 CH3

7

3 6

Cl

8

5,5-bis(2-chloroethyl)nonane

The above example is a complex molecule.The group that is attached to the 5th carbon is chloro ethyl group.As they are attached twice to the same carbon "bis" comes into picture there.The above molecule can also be named as "5,5-di(2-chloroethyl)nonane".

Tri halogen derivatives : These tri halogen derivatives are derived by the replacement of three hydrogen atoms from the alkanes with halogen atoms.This can be explained by the below given examples. 1.

5

CH3

1

4

I

CH3

3

Cl

2

Br

2-bromo-3-chloro-4-iodopentane

In the above example we observe that the longest chain contains 5 carbon atoms and the 2nd,3rd and 4th positions are replaced by bromine,chlorine and iodine.We know that the halogen atoms should be in alphabetical order hence the name of the compound is "2-bromo-3-chloro-4-idopentane" 2.Haloforms : The trihalogen derivatives of first alkane(methane) are termed as haloforms.They are named in trivial system as "Haloform". (a).

Cl

Cl

IUPAC trichloromethane

Trivial chloroform

In above example the IUPAC name of the compound is trichloromethane.But in trivial system we know that it takes the form "haloform".Here the halogen atom is chlorine. Therefore it is named"Chloroform". Cl

(b).

IUPAC

Br

Br

Trivial

tribromomethane

bromoform

triiodomethane

idoform

Br

(c). I

I

I

3.

8 Br

CH3

7

1 2

I

3

4

5

6

3-bromo-7-chloro-1-iodooctane Cl

In above example the numbering should be done in such a way that the lowest sum rule is applicable and the name is arranged in the alphabetical order.Hence the name of the compound is "3-bromo-7-chloro-1-idooctane" 4.

I

6

Cl

2

H3C

1

5

4

Br

7

CH3

3

CH3

CH3

2-bromo-4-chloro-5-ethyl-6-iodo-3-methylheptane

5.

CH3

Cl

2"

2' 1"

H3C

I

4

1'

5

Br

6

3 2

1 CH3

4-(1'-bromoethyl)-4-(1"-chloroethyl)-3-iodoheptane

CH3

7

The above example is a complex molecule structure.This molecule contains 7 carbons as numbered in the longest chain.At 3rd position there is a halogen atom ie Iodine atom.Now let us focus our view on fourth position.In fourth position they are two alkyl substituents.They are (a) bromo ethyl (b) Chloro ethyl When comes to alphabetical order bromine should be given first preference as a result the alkyl group of bromine is given 1' and 2' as shown in the figure.Then the alkyl group of chlorine should should be given 1" and 2" as shown in the figure. It should be noted that both the alkyl groups souldn't be given the same numbering.Coming to the halogen atoms bromine is attached to 1' positon as shown and chlorine is attached to 1" position as shown in the molecule.Therefore the alkyl groups with halogen substituents are arranged in brackets. Hence the name of the compound is 4-(1'-bromoethyl)-4-(1"-chloroethyl)-3-idoheptane

Polyhalogen derivatives : These are the halogen derivatives which contain more than three halogen atoms in the structure of the molecule.This can be explained by the below given examples. 1. Cl Cl C

tetrachloromethane

Cl Cl

2.

Br Cl dibromo(dichloro)methane

C Br Cl

3.

Cl Cl bromo(trichloro)methane C Br Cl

4. Cl F

bromo(chloro)fluoro(iodo)methane C I Br

This is how one can name the poly halogen derivatives of first alkane(methane).It should be noted that the names are always arranged in alphabetical order only.

Let us now consider 1st example.In this example 4 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 atoms and a bromine atom that is attached to the carbon atom In this case bromine should be given first preference because in alphabetiacl order bromine comes first. The halogen atom that comes after bromine ie chlorine should be enclosed in brackets.Hence the name of the structure is bromo(trichloro)methane. It should be noted that this rule is not applicable for higher alkanes after methane. 1.

Cl

Cl

1 CH

1,1,2,2-tetrachloroethane

CH

2

Cl

2.

Cl

Cl

1

Br

Br

C

1,1,2,2-tetrabromo-1,2-dichloroethane

2

C

Br

Br

Cl

I

3. 1

CH2 Cl

2

CH

CH

3

4

CH

CH2

5

6

CH2

CH2

7

8

Br

CH2 2,8-dibromo-1-chloro-4-fluoro-3-iodooctane

Br

F

I

4.

2" F

CH2 CH

1"

Br

1

3

CH2

CH2

2

7 HC

4

CH2

CH

HC

2' CH2

Cl

5

6 CH2

CH2

8

Cl

CH2

1' Br

1-bromo-4-(1'-bromo-2'-chloroethyl)-8-chloro-5-(1"-fluoro-2"-iodoethyl)octane

The above given structure is a complex structure and it can be explained in the following way.The longest chain contains 8 carbon atoms.Here the numbering can start from bromine or chlorine because they are at equal distance and the lowest sum rule for the both types of numbering will be same.In such a case the first preference is given to the atom which comes first in the alphabetical order.Here bromine comes first in the alphabetical order hence it is given the first preference. group sustituents.They are

At 4th and 5th positions they are alkyl

(a) 1-bromo-2-chloro ethyl group (b) 1-fluoro-2-ido ethyl group These sustituents are at 4th and 5th positions respectively.By considering the alphabetical order (a) bromo (b) 1-bromo-2-chloro ethyl group (c) chloro

(d) 1-fluoro-2-idoethylgroup Therefore by considering this alphabetical order the name of the compound is 1-bromo-4-(1'-bromo-2'-chloroethyl)-8-chloro-5-(1"-fluoro-2"-iodoethyl)octane

5.

3"CH3

H3C

I

F

2"

HC

CH

1"

1' 3 CH

1

CH2

2'

4 CH3

HC

2

HC

CH2

5

CH2

6

HC

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 numbering and following the lowest sum rule ,arranging the substituents in alphabetical oder one can generate the IUPAC name of the compound easily for any complex molecules.

This book is a trial version which contains only 2 topics. For full version book request, email me at "[email protected]" which is of "free cost". Any s,suggestions and doubts are accepted. The full version book will be published online within 30 days. Hope this book will meet your expectations.