This book provides a clear understanding about organic nomenclature in a clear and understandable way.

1. Nomenclature
Badarla Sandeep

2. 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.
Badarla Sandeep
Osmania University

3. Contents Page No
1. Nomenclature 4
Trial Version only
2. Saturated Aliphatic Compounds 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 ---

4. 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

5. 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

6. (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-

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 to
show 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. 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 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( CO) -one
Carboxylic acid ( -COOH) -oic acid
Sulphonic acid( SO3H) -Sulphonic acid

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 formed
fromed from the Root,Primary and the Secondary suffixes
Homologus Root Primary Secondary IUPAC
Series Word Suffix Suffix name
Alcohols Alk -ane -ol Alkanol
(saturated)
Alcohols Alk -ene -ol Alkenol
(unsaturated)
Alcohols Alk -yne -ol Alkynol
(Unsaturated)
One triple bond
Aldehydes Alk -ane -al Alkanal
(saturated)
Ketones Alk -ane -one Alkanone
(saturated)

10. 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 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. Hence the IUPAC name of thhe compound is
3-bromobutanoicacid
Example 2
1
CH2 CH CH3
3 5
HO CH 4CH
2
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

12. Example 3
5
H3C OH
3 2 1
4 CH C C 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 2
H 3C C OH
CH 3
2-methylpropan-2-ol

13. 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.
2 1
= H3C CH3
Ethane
2. 1 CH3
=
2 H3C CH2
3
Propane
1
3. CH3
2
= CH CH3
3 H3C CH
4 CH3
2,3-dimethylbutane

14. 4.
2 4 2,2-dimethylbutane
1
3
5.
2
2,2-dimethylpropane
1 3
6.
2 4
2-methylbut-2-ene
1 3
7. 2 4 but-1-yne
HC
3
1
8.
3
2 3-methylbut-1-yne
HC
4
1
9. 2 4
1 3
H2C CH3 3-methylpent-1-ene
5
CH3

15. 10.
2
OH
1 propan-1-ol
3
OH
11.
2 propan-2-ol
3 1
12. 2 OH
2,2-dimethylpropan-1-ol
3 1
O
13.
Ethanoic acid
1
OH
2
O
14. 3 1
2-methylbutanoic acid
OH
4 2
15. 4 2
3 pentan-3-one
5 1
O

16. 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 2
CH2 CH2 3-methylpentane
5 3 1
H3C 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.
CH3
2. 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 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

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 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.
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. 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 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 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.

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 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. 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. 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-tetramethyloctane
9. CH3
4 6
2 CH3
3 5
H3C
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
It should be kept in the mind that prefixes such as
di,tri, etc are not considered while arranging the substituent
alphabetically.

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 substituents
are 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 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

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 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 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. 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. 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 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.

25. 18. 3'
H3C CH3
2'
1'
5 6
3 4 7 8
1 2
H3C 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' 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 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"

26. 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 = chloro
alkane = methane

27. 2.
H3C 1
bromoethane
2 Br
halo = bromo
alkane = ethane
Hence the IUPAC name is bromoethane
3.
2 Br
1-bromopropane
H3C
3 1
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 1
1 4
H3C CH3
5.
2-fluorobutane
2 3
F

28. CH3
6.
2 4 Cl
5
H3C
1 3 5-chloro-2,3-dimethylhexane
CH3 CH3 6
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 Cl
7 5
8 6 4
H3C
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

29. (b) Trivial system of naming :
1.
Cl
n-propylcholoride
H3C
2. Cl
iso propylcholoride
H3C CH3
3.
H3C CH3
secondary butylchloride
Cl
4.
CH3
isobutylchloride
Cl
H3C
5. CH3
teritiary butylchloride
H3C Cl
CH3

30. 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.

31. 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"

32. 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
2,2-dibromopropane
1 2
H3C Br
Br
3
4. H3C I
1 1,1-diiodo-2-methylpropane
2
H3C 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

33. 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 Propylene chloride
Cl
Cl
1

34. 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 2
H3C Br
6
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 Br
1,2-dibromobutane
3 2
H3C Br
4

35. CH3 CH3
3.
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 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.

36. 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 4
Cl
tetramethylene chloride
Cl
1 3
2.
1 3 5 7
heptamethylene bromide
Br Br
2 4 6
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"

37. 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
1,3-diiodopropane
I I
2
2.
CH3 CH3
1,7-dibromo-3,5-dimethylheptane
1
2 3 4 5 6 7
Br Br
H3C CH3
3.
1,7-dichloro-3,5-diethyl-4-methylheptane
1 2 3 4 5 6 7
Cl Cl
CH3 6 Cl
4. 4 5
H3C
3
2
1,6-dichloro-2,3-dimethylhexane
1
Cl CH3
This is how one can name the terminal dihalides
as shown above examples by proper numbering.

38. Dihalogen deivatives which contain different halogen atoms :
1. 2 Br
Cl
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
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 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"

39. 7 1
3. 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 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
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 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".

40. 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
I 4 CH3
3 2
Cl 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 Trivial
trichloromethane chloroform
Cl
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".

41. (b). IUPAC Trivial
Br
Br tribromomethane bromoform
Br
(c).
I
triiodomethane idoform
I I
3. 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 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
7
Cl 6 CH3
5
Br 4
2 3
CH3
H3C CH3 2-bromo-4-chloro-5-ethyl-6-iodo-3-methylheptane
1

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 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

43. 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.

44. 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 1 Cl
CH 1,1,2,2-tetrachloroethane
CH
Cl 2 Cl
2. Cl
1
Br C Br
1,1,2,2-tetrabromo-1,2-dichloroethane
2
Br C Br
Cl
I
3.
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. I
4. 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 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.
At 4th and 5th positions they are alkyl
group 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.By
considering the alphabetical order
(a) bromo
(b) 1-bromo-2-chloro ethyl group
(c) chloro

46. (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 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 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.

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