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Nomenclature 
Badarla Sandeep
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
Contents 
1. Nomenclature 4 
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 --- 
Page No 
Trial Version only
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 IUPAC 
Suffix name 
(1) Saturated -ane Alkane 
( H3C CH3 
) 
(2) Unsaturated -ene Alkene 
With one double bond 
( H2C CH2 )
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 
) 
CH 
Unsaturated with -atriene Alkatriene 
three double bonds 
( ) 
CH 
CH 
CH 
CH 
CH2 
The compounds that are given above in the brackets are the 
H2C 
H2C 
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
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)
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 
1 
2 
3 
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 
4 
HO 
C 
O 
CH2 
CH 
Br 
CH3
Hence the IUPAC name of thhe compound is 
3-bromobutanoicacid 
Example 2 
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 
1 
2 
3 
4 
5 
CH2 
CH 
CH 
CH 
CH3 
CH3 
HO
3 2 1 
4-methylpent-2-ynoic acid 
Example 3 
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 
OH 
4-methylpent-2-ynoic acid 
Example 4 
CH3 
2-methylpropan-2-ol 
4 
5 
1 
3 2 
H3C 
CH C C C 
O 
H3C 
H3C C 
CH3 
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. 
= 
2. 
= 
3. 
= 
2 1 
Ethane 
H3C CH3 
2 
Propane 
1 
3 
1 
2 
3 
4 
2,3-dimethylbutane 
H3C CH2 
CH3 
CH3 
CH 
H3C CH 
CH3 
CH3
2,2-dimethylbutane 
2,2-dimethylpropane 
2-methylbut-2-ene 
HC 
but-1-yne 
HC 
3-methylbut-1-yne 
1 
2 
3 
4 
1 
2 
3 
1 
2 
3 
4 
1 
2 
3 
4 
1 
2 
3 
4 
4. 
5. 
6. 
7. 
8. 
9. 
H2C CH3 
CH3 
3-methylpent-1-ene 
1 
2 
3 
4 
5
OH 
OH 
OH 
OH 
O 
OH 
O 
O 
propan-1-ol 
propan-2-ol 
2,2-dimethylpropan-1-ol 
Ethanoic acid 
2-methylbutanoic acid 
pentan-3-one 
1 
2 
3 
1 
2 
3 
1 
2 
3 
1 
2 
1 
2 
3 
4 
1 
2 
3 
4 
5 
10. 
11. 
12. 
13. 
14. 
15.
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. 
1 
2 
3 
3-methylpentane 
CH2 
CH 
CH2 
CH3 
CH3 
In above example the longest chain is of 5 carbon 
4 
5 
H3C 
atoms.Hence it is a derivative of pentane.The substiuent is a methyl 
group. 
2. 
2,2,5,5-tetramethyloctane 
4 
C 
5 
CH3 
CH2 
6 
7 
CH3 
8 
CH2 
CH2 
CH2 
H3C 
Here the longest chain contains 8 carbons which is 
1 
2 
3 
CH3 
C 
H3C 
H3C 
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. 
H3C CH3 
In above example the longest chain contains 7 
CH 
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. 
CH2 CH3 
Correct numbering 
CH2 CH3 
Wrong numbering 
3 
4 
(a) (b) 
1 
2 
3 
4 
5 
6 
7 
5 
6 
4-ethyl-2-methylheptane 
1 
2 
3-ethyl-2-methylpentane 
5 
2 1 
3 
4 
5 
CH3 
CH2 
CH 
CH2 
CH2 
H2C 
CH3 
H3C 
CH HC 
CH2 CH3 
H3C 
H3C 
CH HC 
CH2 CH3 
H3C
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. 
CH CH2 
3 
4 5 1 
CH2 CH3 
2-methylpentane 
CH2 CH3 
2 
CH CH2 
3 
4 
5 
4-methylpentane 
H3C 
H3C 
Correct numbering Wrong numbering 
(a) (b) 
In this type of cases the numbering should 
1 
2 
H3C 
H3C 
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.
CH3 
H3C CH3 
CH3 
1 
2 
3 
4 
5 
1 
2 
3 
4 
5 
Correct numbering 
Wrong numbering 
2,3-dimethylpentane (Correct) 
3,4-dimethylpentane (Wrong) 
6. Lowest Sum Rule 
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 
2 
CH CH3 3 
3 
4 
5 2,2,4-trimethylpentane (Correct) 
1 
CH 
2 
CH2 
3 
4 
H3C 
5 
C 
CH3 
H3C 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
1 
2 
H3C 
3 
4 
5 
6 
7 
8 
1 
CH2 
2 
C 
3 
CH3 
CH2 
4 
CH2 
5 
CH2 
6 
C 
7 
CH3 
8 
2,2,6,6-tetramethyloctane (Correct) 
3,3,7,7-tetramethyloctane (Wrong) 
8. 
H3C 
CH3 
H3C 
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 
CH3 
6 
CH3 
2 
3 
5 
3-ethyl-2-methylhexane 
In above case they are two different alkyl substituents 
H3C 
1 
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.
10. 
It is same as the above studied case and the substituents 
are arranged in alphabetical order. 
11. 
1 2 
3 4 
5 6 
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 
H3C 
H3C 
CH3 
CH3 
CH3 
H3C 
CH3 
1 
2 
5 4 3 
7 6 
4-ethyl-2,2,3,4-tetramethylheptane 
H3C CH3 
H3C CH3 
3-ethyl-4-methylhexane
12. 
CH3 
2 3 4 5 
Same as the above case and the numbering is given 
1 
according to alphabetical order.Hence the name of the compound is 
3,3-diethyl-4,4-dimethylhexane 
15. Alkyl substituents : 
1 
(a) Methyl (CH3----) 
(b) Etyhl (CH3 CH2-----) 
H3C CH2 
(c) Propyl (CH3 CH2 CH2----) 
(d) Butyl (CH3 CH2 CH2 CH2----) 
"Q is any sustituent" 
H3C 
CH3 
H3C 
H3C CH3 
6 
3,3-diethyl-4,4-dimethylhexane 
H3C Q 
1 2 
1 
2 
3 
1 
2 
3 
Q 
H3C 
CH2 
CH2 
Q 
H3C 
CH2 
CH2 
CH2 
Q 
4
16.Trivial names for some alkyl substiuents : 
(a) 
Isopropyl 
(b) 
Isobutyl 
(c) 
Secondary butyl 
(d) 
Teritiary butyl 
(e) 
Teritiary pentyl 
2 
CH3 
" Q is any sustituent" 
H3C 
1 
2 
3 
1 
3 
4 
CH 
H3C 
Q 
H3C 
CH 
H3C 
CH2 
Q 
H3C 
CH2 
CH 
CH3 
Q 
H3C C 
CH3 
Q 
H3C 
CH2 
C 
CH3 
H3C 
Q
17. 
H3C 
CH3 
H3C 
CH3 
5-(2'-methylpropyl)nonane 
1 
2 
3 
4 
5 
6 
7 
8 
9 
1' 
2' 
3' 
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.
3' 
H3C CH3 
1' 
2' 
H3C H3C CH3 
CH3 
1 2 
3 4 
5 6 
7 8 
9 
5-(1'methylpropyl)-2,7-dimethylnonane 
18. 
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. 
2-methyl-4,4-di(propan-2'-yl)heptane 
(or) 
CH3 
In this case the same complex alkyl 
H3C 
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" 
H3C 
CH3 
H3C 
H3C 
CH3 
2-methyl-4,4-bis(propan-2'-yl)heptane 
1 
2 
3 
4 
5 
6 
7 
1' 
2' 
3'
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
2. 
halo = bromo 
alkane = ethane 
Hence the IUPAC name is bromoethane 
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. 
5. 
H3C 
Br 
1 
2 
bromoethane 
H3C 
Br 
1 
2 
3 
1-bromopropane 
I 
H3C CH3 
2-idopropane 
H3C CH3 
F 
2-fluorobutane 
1 
2 
3 
4 
2 3 
1
CH3 
CH3 
H3C 
Cl 
CH3 
5-chloro-2,3-dimethylhexane 
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. 
2 
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 
CH3 
CH3 
Cl 
H3C 
2-chloro-7-methyloctane 
1 
2 
3 
4 
5 
6 
1 
3 
4 
5 
6 
7 
8
(b) Trivial system of naming : 
1. 
2. 
3. 
4. 
5. 
H3C 
Cl 
Cl 
H3C CH3 
H3C CH3 
Cl 
CH3 
H3C 
Cl 
H3C 
CH3 
CH3 
Cl 
n-propylcholoride 
iso propylcholoride 
secondary butylchloride 
isobutylchloride 
teritiary butylchloride
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.
H3C 
Cl 
Cl 
H3C 
CH3 
Br 
Br 
H3C 
I 
I 
H3C 
H3C 
F 
F 
Ethylidene chloride 
Iso propylidene bromide 
Propylidene iodide 
Iso butylidene fluoride 
1. 
2. 
3. 
4. 
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. 
2. 
3. 
4. 
Cl 
I 
Now consider example 1.In this example the methane 
3 
molecule is replaced by two chlorine atoms as a result it is named as 
dichloromethane 
Cl 
Cl 
H3C 
Cl 
H3C 
CH3 
Br 
Br 
H3C 
H3C I 
dichloromethane 
1,1-dichloroethane 
2,2-dibromopropane 
1,1-diiodo-2-methylpropane 
1 
2 1 
1 2 
1 
2 
3
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. 
2. 
Cl 
Cl 
CH3 
Cl 
Cl 
1 
2 
1 
2 
3 
Ethylene chloride 
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. 
H3C 
Br 
Br 
1 
2 
3 
4 
5 
6 
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. 
2. 
Cl 
Cl 
H3C 
Br 
Br 
1,2-dichloroethane 
1,2-dibromobutane 
2 1 
1 
3 2 
4
CH3 
Cl 
Cl 
H3C 
CH3 
CH3 
1,2-dichloro-4-ethyl-5,7-dimethyloctane 
1 
2 
3 
4 
5 
6 
7 
8 
3. 
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. 
4 
Now let us consider example 1.In this structure 
1 
2 
3 
1 
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" 
Cl 
Cl 
Br Br 
2 
3 
4 
5 
6 
7 
tetramethylene chloride 
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. 
2. 
3. 
4. 
I I 
CH3 CH3 
Br Br 
H3C 
CH3 
Cl Cl 
CH3 
H3C 
Cl CH3 
1,3-diiodopropane 
1,7-dibromo-3,5-dimethylheptane 
1,7-dichloro-3,5-diethyl-4-methylheptane 
This is how one can name the terminal dihalides 
as shown above examples by proper numbering. 
Cl 
1,6-dichloro-2,3-dimethylhexane 
1 
2 
3 
1 
2 3 4 5 6 7 
1 2 3 4 5 6 7 
1 
2 
3 
4 5 
6
Dihalogen deivatives which contain different halogen atoms : 
1. 
Cl 
Br 
1 
2 
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 
CH3 
6 
CH3 
1 2 
3 
4 
5 
1-chloro-3-(iodomethyl)-4,5-dimethylhexane 
In example 2 the halogen atoms are chlorine and 
Cl 
I 
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"
1 
7 
CH3 CH3 
2 
6 5 4 3 
Cl Br 
CH3 
CH3 
2-bromo-6-chloro-3-ethyl-5-methylheptane 
3. 
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 
Cl 
6 
8 
5,5-bis(2-chloroethyl)nonane 
The above example is a complex molecule.The group that is 
1 
3 
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 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). 
IUPAC Trivial 
trichloromethane chloroform 
In above example the IUPAC name of the 
Cl 
Cl 
Cl 
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".
(b). IUPAC Trivial 
Br 
Br 
Br 
I 
I I 
(c). 
tribromomethane 
triiodomethane 
bromoform 
idoform 
8 
CH3 
I 
Br 
Cl 
3-bromo-7-chloro-1-iodooctane 
3. 
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. 
1 
2 3 4 5 6 
7 
Br 
Cl 
I 
CH3 
CH3 
H3C 
CH3 
2-bromo-4-chloro-5-ethyl-6-iodo-3-methylheptane 
1 
2 3 
4 
5 
6 
7
CH3 
2 1 
CH3 
Cl CH3 
H3C 
Br 
I 
4-(1'-bromoethyl)-4-(1"-chloroethyl)-3-iodoheptane 
3 
4 
5 
6 
1" 
2" 
1' 
2' 
7 
5. 
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
C tetrachloromethane 
Cl 
Cl 
Cl 
Cl 
C 
Cl 
Cl 
Br 
Br 
dibromo(dichloro)methane 
C 
Cl 
Cl 
Cl 
Br 
bromo(trichloro)methane 
C 
F 
Cl 
Br 
I 
bromo(chloro)fluoro(iodo)methane 
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. 
2. 
3. 
4. 
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. 
2. 
3. 
Cl Cl 
CH 
CH 
Cl Cl 
1,1,2,2-tetrachloroethane 
Cl 
Br C Br 
Br C 
Br 
Cl 
1,1,2,2-tetrabromo-1,2-dichloroethane 
1 
2 
1 
2 
CH2 
CH 
CH 
CH 
CH2 
CH2 
CH2 
Cl CH2 
Br 
Br 
I 
F 
2,8-dibromo-1-chloro-4-fluoro-3-iodooctane 
1 
2 3 4 5 6 7 8
1 
CH2 
2 
CH2 
3 
CH2 
4 5 6 
CH 
HC 
CH2 
CH2 
CH2 
HC 
CH2 
CH 
CH2 
Cl 
Br 
F 
I 
Cl 
Br 
1-bromo-4-(1'-bromo-2'-chloroethyl)-8-chloro-5-(1"-fluoro-2"-iodoethyl)octane 
4. 
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 
7 
8 
1' 
2' 
1" 
2"
(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. 
2" 
HC 
H3C 
3" 
1" 1' 2' 
1 2 
CH2 HC 
I F 
CH3 
4 
CH HC 
CH CH2 
3 
Cl Br 
6 
CH2 HC 
CH3 
7 
CH3 
5 
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 "badarlasandeep@gmail.com" which is of "free 
cost". 
Any feedbacks,suggestions and doubts are accepted. 
The full version book will be published online within 30 days. 
Hope this book will meet your expectations.

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Organic chemistry nomenclature

  • 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 1. Nomenclature 4 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 --- Page No Trial Version only
  • 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 ) CH Unsaturated with -atriene Alkatriene three double bonds ( ) CH CH CH CH CH2 The compounds that are given above in the brackets are the H2C H2C 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 1 2 3 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 4 HO C O CH2 CH Br CH3
  • 11. Hence the IUPAC name of thhe compound is 3-bromobutanoicacid Example 2 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 1 2 3 4 5 CH2 CH CH CH CH3 CH3 HO
  • 12. 3 2 1 4-methylpent-2-ynoic acid Example 3 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 OH 4-methylpent-2-ynoic acid Example 4 CH3 2-methylpropan-2-ol 4 5 1 3 2 H3C CH C C C O H3C H3C C CH3 OH
  • 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. = 3. = 2 1 Ethane H3C CH3 2 Propane 1 3 1 2 3 4 2,3-dimethylbutane H3C CH2 CH3 CH3 CH H3C CH CH3 CH3
  • 14. 2,2-dimethylbutane 2,2-dimethylpropane 2-methylbut-2-ene HC but-1-yne HC 3-methylbut-1-yne 1 2 3 4 1 2 3 1 2 3 4 1 2 3 4 1 2 3 4 4. 5. 6. 7. 8. 9. H2C CH3 CH3 3-methylpent-1-ene 1 2 3 4 5
  • 15. OH OH OH OH O OH O O propan-1-ol propan-2-ol 2,2-dimethylpropan-1-ol Ethanoic acid 2-methylbutanoic acid pentan-3-one 1 2 3 1 2 3 1 2 3 1 2 1 2 3 4 1 2 3 4 5 10. 11. 12. 13. 14. 15.
  • 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. 1 2 3 3-methylpentane CH2 CH CH2 CH3 CH3 In above example the longest chain is of 5 carbon 4 5 H3C atoms.Hence it is a derivative of pentane.The substiuent is a methyl group. 2. 2,2,5,5-tetramethyloctane 4 C 5 CH3 CH2 6 7 CH3 8 CH2 CH2 CH2 H3C Here the longest chain contains 8 carbons which is 1 2 3 CH3 C H3C H3C 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. H3C CH3 In above example the longest chain contains 7 CH 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. CH2 CH3 Correct numbering CH2 CH3 Wrong numbering 3 4 (a) (b) 1 2 3 4 5 6 7 5 6 4-ethyl-2-methylheptane 1 2 3-ethyl-2-methylpentane 5 2 1 3 4 5 CH3 CH2 CH CH2 CH2 H2C CH3 H3C CH HC CH2 CH3 H3C H3C CH HC CH2 CH3 H3C
  • 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. CH CH2 3 4 5 1 CH2 CH3 2-methylpentane CH2 CH3 2 CH CH2 3 4 5 4-methylpentane H3C H3C Correct numbering Wrong numbering (a) (b) In this type of cases the numbering should 1 2 H3C H3C 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. CH3 H3C CH3 CH3 1 2 3 4 5 1 2 3 4 5 Correct numbering Wrong numbering 2,3-dimethylpentane (Correct) 3,4-dimethylpentane (Wrong) 6. Lowest Sum Rule 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 2 CH CH3 3 3 4 5 2,2,4-trimethylpentane (Correct) 1 CH 2 CH2 3 4 H3C 5 C CH3 H3C 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. 1 2 H3C 3 4 5 6 7 8 1 CH2 2 C 3 CH3 CH2 4 CH2 5 CH2 6 C 7 CH3 8 2,2,6,6-tetramethyloctane (Correct) 3,3,7,7-tetramethyloctane (Wrong) 8. H3C CH3 H3C 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 CH3 6 CH3 2 3 5 3-ethyl-2-methylhexane In above case they are two different alkyl substituents H3C 1 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. It is same as the above studied case and the substituents are arranged in alphabetical order. 11. 1 2 3 4 5 6 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 H3C H3C CH3 CH3 CH3 H3C CH3 1 2 5 4 3 7 6 4-ethyl-2,2,3,4-tetramethylheptane H3C CH3 H3C CH3 3-ethyl-4-methylhexane
  • 22. 12. CH3 2 3 4 5 Same as the above case and the numbering is given 1 according to alphabetical order.Hence the name of the compound is 3,3-diethyl-4,4-dimethylhexane 15. Alkyl substituents : 1 (a) Methyl (CH3----) (b) Etyhl (CH3 CH2-----) H3C CH2 (c) Propyl (CH3 CH2 CH2----) (d) Butyl (CH3 CH2 CH2 CH2----) "Q is any sustituent" H3C CH3 H3C H3C CH3 6 3,3-diethyl-4,4-dimethylhexane H3C Q 1 2 1 2 3 1 2 3 Q H3C CH2 CH2 Q H3C CH2 CH2 CH2 Q 4
  • 23. 16.Trivial names for some alkyl substiuents : (a) Isopropyl (b) Isobutyl (c) Secondary butyl (d) Teritiary butyl (e) Teritiary pentyl 2 CH3 " Q is any sustituent" H3C 1 2 3 1 3 4 CH H3C Q H3C CH H3C CH2 Q H3C CH2 CH CH3 Q H3C C CH3 Q H3C CH2 C CH3 H3C Q
  • 24. 17. H3C CH3 H3C CH3 5-(2'-methylpropyl)nonane 1 2 3 4 5 6 7 8 9 1' 2' 3' 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. 3' H3C CH3 1' 2' H3C H3C CH3 CH3 1 2 3 4 5 6 7 8 9 5-(1'methylpropyl)-2,7-dimethylnonane 18. 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. 2-methyl-4,4-di(propan-2'-yl)heptane (or) CH3 In this case the same complex alkyl H3C 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" H3C CH3 H3C H3C CH3 2-methyl-4,4-bis(propan-2'-yl)heptane 1 2 3 4 5 6 7 1' 2' 3'
  • 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. halo = bromo alkane = ethane Hence the IUPAC name is bromoethane 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. 5. H3C Br 1 2 bromoethane H3C Br 1 2 3 1-bromopropane I H3C CH3 2-idopropane H3C CH3 F 2-fluorobutane 1 2 3 4 2 3 1
  • 28. CH3 CH3 H3C Cl CH3 5-chloro-2,3-dimethylhexane 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. 2 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 CH3 CH3 Cl H3C 2-chloro-7-methyloctane 1 2 3 4 5 6 1 3 4 5 6 7 8
  • 29. (b) Trivial system of naming : 1. 2. 3. 4. 5. H3C Cl Cl H3C CH3 H3C CH3 Cl CH3 H3C Cl H3C CH3 CH3 Cl n-propylcholoride iso propylcholoride secondary butylchloride isobutylchloride teritiary butylchloride
  • 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. H3C Cl Cl H3C CH3 Br Br H3C I I H3C H3C F F Ethylidene chloride Iso propylidene bromide Propylidene iodide Iso butylidene fluoride 1. 2. 3. 4. 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. 2. 3. 4. Cl I Now consider example 1.In this example the methane 3 molecule is replaced by two chlorine atoms as a result it is named as dichloromethane Cl Cl H3C Cl H3C CH3 Br Br H3C H3C I dichloromethane 1,1-dichloroethane 2,2-dibromopropane 1,1-diiodo-2-methylpropane 1 2 1 1 2 1 2 3
  • 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. 2. Cl Cl CH3 Cl Cl 1 2 1 2 3 Ethylene chloride Propylene chloride
  • 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. H3C Br Br 1 2 3 4 5 6 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. 2. Cl Cl H3C Br Br 1,2-dichloroethane 1,2-dibromobutane 2 1 1 3 2 4
  • 35. CH3 Cl Cl H3C CH3 CH3 1,2-dichloro-4-ethyl-5,7-dimethyloctane 1 2 3 4 5 6 7 8 3. 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 Now let us consider example 1.In this structure 1 2 3 1 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" Cl Cl Br Br 2 3 4 5 6 7 tetramethylene chloride 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. 2. 3. 4. I I CH3 CH3 Br Br H3C CH3 Cl Cl CH3 H3C Cl CH3 1,3-diiodopropane 1,7-dibromo-3,5-dimethylheptane 1,7-dichloro-3,5-diethyl-4-methylheptane This is how one can name the terminal dihalides as shown above examples by proper numbering. Cl 1,6-dichloro-2,3-dimethylhexane 1 2 3 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6
  • 38. Dihalogen deivatives which contain different halogen atoms : 1. Cl Br 1 2 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 CH3 6 CH3 1 2 3 4 5 1-chloro-3-(iodomethyl)-4,5-dimethylhexane In example 2 the halogen atoms are chlorine and Cl I 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. 1 7 CH3 CH3 2 6 5 4 3 Cl Br CH3 CH3 2-bromo-6-chloro-3-ethyl-5-methylheptane 3. 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 Cl 6 8 5,5-bis(2-chloroethyl)nonane The above example is a complex molecule.The group that is 1 3 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 4 I 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). IUPAC Trivial trichloromethane chloroform In above example the IUPAC name of the Cl Cl Cl 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 Br I I I (c). tribromomethane triiodomethane bromoform idoform 8 CH3 I Br Cl 3-bromo-7-chloro-1-iodooctane 3. 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. 1 2 3 4 5 6 7 Br Cl I CH3 CH3 H3C CH3 2-bromo-4-chloro-5-ethyl-6-iodo-3-methylheptane 1 2 3 4 5 6 7
  • 42. CH3 2 1 CH3 Cl CH3 H3C Br I 4-(1'-bromoethyl)-4-(1"-chloroethyl)-3-iodoheptane 3 4 5 6 1" 2" 1' 2' 7 5. 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. C tetrachloromethane Cl Cl Cl Cl C Cl Cl Br Br dibromo(dichloro)methane C Cl Cl Cl Br bromo(trichloro)methane C F Cl Br I bromo(chloro)fluoro(iodo)methane 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. 2. 3. 4. 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. 2. 3. Cl Cl CH CH Cl Cl 1,1,2,2-tetrachloroethane Cl Br C Br Br C Br Cl 1,1,2,2-tetrabromo-1,2-dichloroethane 1 2 1 2 CH2 CH CH CH CH2 CH2 CH2 Cl CH2 Br Br I F 2,8-dibromo-1-chloro-4-fluoro-3-iodooctane 1 2 3 4 5 6 7 8
  • 45. 1 CH2 2 CH2 3 CH2 4 5 6 CH HC CH2 CH2 CH2 HC CH2 CH CH2 Cl Br F I Cl Br 1-bromo-4-(1'-bromo-2'-chloroethyl)-8-chloro-5-(1"-fluoro-2"-iodoethyl)octane 4. 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 7 8 1' 2' 1" 2"
  • 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. 2" HC H3C 3" 1" 1' 2' 1 2 CH2 HC I F CH3 4 CH HC CH CH2 3 Cl Br 6 CH2 HC CH3 7 CH3 5 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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