Chapter 13.5 - Mechanism of Halogenation

In the previous section, we completed a discussion on the physical properties of alkanes. We also saw the basics of a halogenation, which is a chemical property. In this section, we will see the mechanism of halogenation.

The mechanism involves three steps: Initiation, Propagation and Termination.
A. Initiation
This can be written in 3 steps:
1. Suppose that in the reaction mixture there is CH₄ and Cl₂.
• Then in the initial stage, there will be three types of bonds: C-C bonds, C-H bonds and Cl-Cl bonds.
2. When light or heat is applied, it is the Cl-Cl bond that breaks first. This is because, Cl-Cl bond is the weakest among the three.
3. The fission of Cl-Cl bond will be a homolysis. This is shown using Lewis structures in fig.13.32 below:

Fig.13.32

• So the result will be two Cl atoms with one unpaired electron. That means, we get two chlorine free radicals.
• This process can be written in a condensed form as shown below:
$\rm{Cl_2~ \color {green}{\xrightarrow[{homolysis}]{h \nu}} ~ \overset {•}{Cl}~+~\overset {•}{Cl}}$

B. Propagation
This can be written in 4 steps:
1. One of the chlorine free radicals obtained in the initiation step, attacks a CH₄ molecule.
• A C-H bond in the CH₄ molecule, undergoes homolysis. Thus we get a methyl free radical and a hydrogen free radical. This is shown using Lewis structures in fig.13.33(a) below:

Fig.13.33

• The hydrogen free radical combines with the chlorine free radical to give a molecule of HCl. This is shown in fig.b
• This process can be written in a condensed form as shown below:
$\rm{CH_4~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{h \nu}} ~ \overset {•}{C}H_3~+~H-Cl}$
2. The methyl free radical thus formed, attacks a second chlorine molecule.
• The Cl-Cl bond in that molecule undergoes homolysis and thus we get two chlorine free radicals. This is shown using Lewis structures in fig.13.34(a) below:

Fig.13.34

• One of those chlorine free radicals combine with the methyl free radical to form a chloromethane molecule. This is shown in fig.b
Thus we get one chloromethane molecule and one chlorine free radical.
• This process can be written in a condensed form as shown below:
$\rm{\overset {•}{C}H_3~+~Cl-Cl~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 - Cl~+~\overset {•}{Cl}}$
• Chloromethane is our required end product.
3. Remember that, the initiation step was possible due to the formation of chlorine free radical. But that radical was obtained with the help of sunlight.
• The chlorine free radical that we obtained just now was with the help of a methyl free radical.
• This new chlorine free radical will produce a new methyl free radical.
• The new methyl free radical will in turn produce a new chlorine free radical. So now a chain reaction will begin. Thus the reaction will propagate.
4. The condensed equations that we wrote in (1) and (2) represent the propagation steps. These steps give the principal products CH₃Cl and HCl.
• However, many other propagation steps are also possible. Two of them are written below:
(i) $\rm{CH_3 Cl~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{h \nu}} ~ \overset {•}{C}H_2 Cl~+~H-Cl}$
(ii) $\rm{\overset {•}{C}H_2 Cl~+~Cl-Cl~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_2 Cl_2~+~\overset {•}{Cl}}$
• The above propagation step (i) indicates that, the chlorine free radical, attacks a newly formed chloromethane instead of attacking a methane.
• Fig.13.35 below, shows the propagation step (i) using Lewis structures.

Fig.13.35

• Fig.13.36 below, shows the propagation step (ii) using Lewis structures.

Fig.13.36

• In these propagation steps, the principal products are: CH₂Cl₂ and HCl
• So we see that, highly halogenated products are also possible.

C. Termination
This can be written in 5 steps:
1. The reaction stops after some time due to the consumption of reactants.
2. The reaction may stop due to side reactions also.
3. One of the possible side reaction is shown below:
$\rm{\overset {•}{Cl}~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{{}}} ~ Cl-Cl}$
• If this reaction, occurs, there will be no more chlorine free radicals available for the propagation step.
• In such a situation, the reaction will stop even if the reactants are available.   
4. Another possible side reaction is shown below:
$\rm{H_3 \overset {•}{C}~+~\overset {•}{C}H_3~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CH_3}$
• If this reaction, occurs, there will be no more methyl free radicals available for the propagation step.
• In such a situation also, the reaction will stop even if the reactants are available.   
• This reaction helps us to explain the formation of ethane during the chlorination of methane.
5. Yet another possible side reaction is shown below:
$\rm{H_3 \overset {•}{C}~+~\overset {•}{Cl}~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - Cl}$
• If this reaction, occurs, there will be neither of chlorine free radicals or methyl free radicals available for the propagation step.
• In such a situation also, the reaction will stop even if the reactants are available.
• In this reaction, we get an useful product chloromethane. But not much chloromethane will be produced because, the reaction does not propagate.

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II. Combustion
• Combustion is the second chemical property of alkanes. It can be written 5 in steps:
1. Alkanes can be heated in the presence of air or dioxygen.
• Then a reaction will take place between the dioxygen and the alkane.
• The products are: CO₂, H₂O and large amounts of heat.
   ♦ Due to the intense heat, the H₂O will be in the gaseous state.
2. Let us see some examples:
(i) $\rm{CH_4 (g)~+~2O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ CO_2 (g)~+~2H_2 O (g);~\Delta_c H^\circleddash = -890~kJ~{mol}^{-1}}$
(ii) $\rm{C_4 H_{10} (g)~+~13/2 \; O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ 4\;CO_2 (g)~+~5\;H_2 O (g);~\Delta_c H^\circleddash = -2875.84~kJ~{mol}^{-1}}$
• These are thermochemical equations. We saw the details about such equations in section 6.6
3. The general combustion equation for any alkane is:
$\rm{C_n H_{2n+2} (g)~+~\left(\frac{3n+1}{2} \right) \; O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ n\;CO_2 (g)~+~(n+1)\;H_2 O (g)}$
• The validity of this general equation can be checked by putting n = 4. We will get the same equation in 2(ii).
4. Due to the evolution of large quantities of heat during combustion, alkanes are can be used as fuels.
• Recall that petrol is a mixture of alkanes. It is used as fuel in the internal combustion engines in automobiles. The large heat produced, causes expansion of the gaseous mixture inside the cylinder. The expanding gas pushes the piston of the cylinder, thereby rotating the wheel of the automobile.
• Kerosene is also a mixture of alkanes. The large heat produced is utilized to cook food.
5. We know that dioxygen is essential for the combustion of alkanes.
• If enough air or dioxygen is not available, all the carbon atoms in the alkane will not be oxidized to CO₂. Some C atoms will remain as such. This can be explained in 5 steps:
(i) During combustion of an alkane in the presence of insufficient dioxygen, all the H atoms will be oxidized to H₂O
• So the C atoms are now free.
(ii) There is not enough dioxygen for those C atoms.
• Some of them will be oxidized to CO₂
• The remaining C atoms form soot. Soot is also known as carbon black. Some images can be seen here.
(iii) This process is known as incomplete combustion.
• The equation for the incomplete combustion of methane is:
$\rm{CH_4 (g)~+~O_2 (g)~ \color {green}{\xrightarrow[{combustion}]{Incomplete}} ~ C (s)~+~2H_2 O (g)}$
(iv) Carbon black is used in the manufacture of ink, printer ink, pigments etc., It is also used in filters.
(v) We saw that, all H atoms will be oxidized even if there is insufficient dioxygen.
• This is because, oxidation of H atoms releases greater energy than the oxidation of C atoms. So the system can attain greater stability when H atoms are oxidized. That is the reason why the H atoms get preference.

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In the next section we will see the controlled oxidation.

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Chapter 13.2 - Problems in Nomenclature of Alkanes

In the previous section, we saw the nomenclature and isomerism in alkanes. In this section, we will see some advanced examples.

Example 1:
Write the IUPAC name of the structure shown in fig.13.13 below:

Fig.13.13

Solution:
• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are six C atoms in the main chain.
2. Applying rule 2, we see that, 'hex' must be used.
3. Applying rule 3, we get hexane.
4. Applying rule 4, we see that, there are two branches: methyl and ethyl
5. Applying rule 5, we see that, the correct way of numbering is indeed from left to right as shown in fig.13.13(a).
• If we number the C atoms from left to right as in fig.b, the branches will get the numbers 2,4. The sum is (2+4) = 6.
• If we number the C atoms from right to left, the branches will get the numbers 3,5. The sum is (3+5) = 8, which is larger and hence not acceptable.
6. Applying rule 6, we get: 2-methyl and 4-ethyl
7. Applying rule 7, we get: 4-ethyl-2-methyl
• This is because, 'e' comes before 'm' in the alphabetical listing.
8. Applying rule 8, we get: 4-Ethyl-2-methylhexane.

Example 2:
Write the IUPAC name of the structure shown in fig.13.14 below:

Fig.13.14

Solution:
In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also. Fortunately, there is a common name available for that branch. So we can first name the main carbon chain.
1. Applying rule 1, we see that, there are eight C atoms in the main chain.
2. Applying rule 2, we see that, 'oct' must be used.
3. Applying rule 3, we get octane.
4. Applying rule 4, we see that, there are four branches: two ethyl branches, one methyl branch and one isopropyl branch.
5. Applying rule 5, we see that, the correct way of numbering is from right to left as shown in fig.13.14(b).
• If we number the C atoms from right to left, the branches will get the numbers 3,3,4,5. The sum is (3+3+4+5) = 15.
• If we number the C atoms from left to right (fig.a), the branches will get the numbers 4,5,6,6. The sum is (4+5+6+6) = 21, which is larger and hence not acceptable.
6. Applying rule 6, we get: 3-ethyl, 3-ethyl, 4-methyl and 5-isopropyl
7. Applying rule 7, 7A and 7B, we get: 3,3-diethyl-5-isopropyl-4-methyl
• This can be explained in 2 steps:
(i) 'e' comes before 'i' and 'm' in the alphabetical listing.
'di' is not considered as part of the name. So 'd' should not be considered for alphabetical listing.
(ii) 'i' comes before 'm' in the alphabetical listing.
'iso' is considered as part of the name.
8. Applying rule 8, we get: 3,3-diethyl-5-isopropyl-4-methyloctane.

Example 3:
Write the IUPAC name of the structure shown in fig.13.15 below:

Fig.13.15

Solution:
In this problem, there are branches within a branches. So we must be ready to apply the rules that we saw in section 12.4 also. Fortunately, there are common names available for those branches. So we can first name the main carbon chain.
1. Applying rule 1, we see that, there are ten C atoms in the main chain.
2. Applying rule 2, we see that, 'dec' must be used.
3. Applying rule 3, we get decane.
4. Applying rule 4, we see that, there are two branches: one isopropyl branch and and one sec-butyl branch.
5. Applying rule 5, we see that, the correct way of numbering is from left to right as shown in fig.13.15(a).
• If we number the C atoms from left to right, the branches will get the numbers 4,5. The sum is (4+5) = 9.
• If we number the C atoms from right to left (fig.b), the branches will get the numbers 6,7. The sum is (6+7) = 13, which is larger and hence not acceptable.
6. Applying rule 6, we get: 4-isopropyl and 5-sec-butyl
7. Applying rule 7, 7A and 7B, we get: 5-sec-butyl-4-isopropyl
• This can be explained in 3 steps:
(i) 'b' comes before 'i' in the alphabetical listing.
(ii) 'sec' is not considered as part of the name. So 's' should not be considered for alphabetical listing.
(iii) 'iso' is considered as part of the name.
8. Applying rule 8, we get: 5-sec-butyl-4-isopropyl decane.

Example 4:
Write the IUPAC name of the structure shown in fig.13.16 below:

Fig.13.16

Solution:
In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also. The branch do not have any common name. So we first name the branch.
1. Applying rule 1, we see that, there are three C atoms.
2. Applying rule 2, we see that, 'prop' must be used.
3. Applying rule 3, we get propane.
4. Applying rule 4, we see that, there are two branches: two methyl groups.
5. Applying rule 5, we see that, the correct way of numbering is from the main branch as shown in fig.13.16(a).
6. Applying rule 6, we get: 2-methyl and 2-methyl
7. Applying rule 7, we get: 2,2-dimethyl
8. Applying rule 8, we get: 2,2-dimethylpropyl.
• This name must be written within parenthesis.

Now we can name the main chain.
1. Applying rule 1, we see that, there are nine C atoms.
2. Applying rule 2, we see that, 'non' must be used.
3. Applying rule 3, we get nonane.
4. Applying rule 4, we see that, there is only one branch.
5. Applying rule 5, we see that, numbering can be done in both ways. Both will give the number 5.
6. Applying rule 6 we get: 5-(2,2-dimethylpropyl)
7. Applying rule 7 is not required because there is only one branch.
8. Applying rule 8, we get: 5-(2,2-dimethylpropyl)nonane.

Example 5:
Write the IUPAC name of the structure shown in fig.13.17 below:

Fig.13.17

Solution:
• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are seven C atoms in the main chain.
2. Applying rule 2, we see that, 'hept' must be used.
3. Applying rule 3, we get heptane.
4. Applying rule 4, we see that, there are two branches: methyl and ethyl
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the numbers are 3 and 5.
• So we apply rule 5A:
If two branches are present in equivalent positions, then the lower number should be given to that branch which comes first in the alphabetical order.
   ♦ Here, ethyl comes first in the alphabetical order.
   ♦ Numbering in fig.a is correct.
6. Applying rule 6, we get: 3-ethyl and 5-methyl
7. Applying rule 7, we get: 3-ethyl-5-methyl
• This is because, 'e' comes before 'm' in the alphabetical listing.
8. Applying rule 8, we get: 3-Ethyl-5-methylheptane.

Solved example 13.3
Write the IUPAC names of the following compounds:
(i) (CH₃)₃CCH₂C(CH₃)₃
(ii) (CH₃)₂C(C₂H₅)₂
(iii) tetra-tert-butylmethane
Solution:
Part (i):
• We are given the condensed formula. Based on the condensed formula, we can draw the structural formula. It is shown in fig.13.18 below:

Fig.13.18

• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are four branches: four methyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the numbers are 2 and 4.
6. Applying rule 6, we get: 2-methy, 2-methyl, 4-methy, 4-methyl
7. Applying rule 7, we get: 2,2,4,4-tetramethyl
8. Applying rule 8, we get: 2,2,4,4-Tetramethylpentane.

Part (ii):
• We are given the condensed formula. Based on the condensed formula, we can draw the structural formula. It is shown in fig.13.19 below:

Fig.13.19

• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are two branches: two methyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the number is 3.
6. Applying rule 6, we get: 3-methy, 3-methyl
7. Applying rule 7, we get: 2,2-dimethyl
8. Applying rule 8, we get: 2,2-Dimethylpentane.

Part (iii):
• We have seen the 8 rules in section 12.3
• In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also.
• We are given the common name tetra-tert-butylmethane.
   ♦ This is similar to the common name: tetra-chloromethane.
• In tetrachloromethane, the four H atoms of methane are replaced by four Cl atoms.
• In the same way, in tetra-tert-butylmethane, the four H atoms of methane are replaced by four tert-butyl groups.
• We saw the structure of tert-butyl in an earlier section [see fig.12.33 in section section 12.4]. It is shown again in fig.13.20(a) below:

Fig.13.20

• Based on the structure of tert-butyl, the structure of tetra-tert-butylmethane will be as shown in fig.b. Now we can write the IUPAC name.
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are six branches: four methyl branches and two tert-butyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in two ways.
   ♦ In both fig. b and c, the numbers are 2, 3 and 4.
6. Applying rule 6, we get: 2-methy, 2-methyl, 4-methy, 4-methyl, 3-tert-butyl, 3-tert-butyl
7. Applying rule 7, we get: 3,3-di-tert-butyl-2,2,4,4-tetramethyl
• Remember that, 'tetra' is not considered as part of name. So 't' cannot be considered in the alphabetical listing.
8. Applying rule 8, we get: 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane.

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• In the above discussion, we were given the structures of hydrocarbons. We wrote the corresponding IUPAC names.
• We must be able to do the reverse also. That is., we will be given the IUPAC names. We must draw the corresponding structures.
• Let us see an example:
Draw the structure of 3-Ethyl-2,2-dimethylpentane
Solution:
1. In the given name, we have ‘pent’ as the root. Then there will be five C atoms in the longest chain. So we first draw a chain of five C atoms. This is shown in fig.13.21(a) below:

Fig.13.21

2. Next we number the C atoms from 1 to 5. This is shown in fig.b
3. Attaching the branches:
‘3-Ethyl’ indicates that, there is an ethyl group at the C atom number 3
So we draw an ethyl group at that C atom.
‘2,2-dimethyl’ indicates that, there are two methyl groups at the C atom number 2
So we draw two methyl groups at that C atom
This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
Thus we get the final structure shown in fig.d

Solved example 13.4
Write the structural formulas for the following compounds:
(i) 3,4,4,5-Tetramethylheptane
(ii) 2,5-dimethylhexane
Solution:
Part (i):
1. In the given name, we have ‘hept’ as the root. Then there will be seven C atoms in the longest chain. So we first draw a chain of seven C atoms. This is shown in fig.13.22(a) below:

Fig.13.22

2. Next we number the C atoms from 1 to 7. This is shown in fig.b
3. Attaching the branches:
‘3,4,4,5-tetramethyl’ indicates that, there are four methyl groups at the three C atoms as written below:
   ♦ One methyl group at C atom number 3
   ♦ Two methyl groups at C atom number 4
   ♦ One methyl group at C atom number 5
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d

Part (ii):
1. In the given name, we have ‘hex’ as the root. Then there will be six C atoms in the longest chain. So we first draw a chain of six C atoms. This is shown in fig.13.23(a) below.

Fig.13.23

2. Next we number the C atoms from 1 to 6. This is shown in fig.b
3. Attaching the branches:
‘2,5-dimethyl’ indicates that, there are two methyl groups at the two C atoms as written below:
   ♦ One methyl group at C atom number 2
   ♦ One methyl group at C atom number 5
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d

Solved example 13.5
Write structures for each of the following compounds. Why are the given names incorrect. Write the correct IUPAC names.
(i) 2-Ethylpentane
(ii) 5-Ethyl-3-methylheptane
Solution:
Part (i):
1. In the given name, we have ‘pent’ as the root. Then there will be five C atoms in the longest chain. So we first draw a chain of five C atoms. This is shown in fig.13.24(a) below:

Fig.13.24

2. Next we number the C atoms from 1 to 5. This is shown in fig.b
3. Attaching the branches:
‘2-ethyl indicates that, there is an ethyl group at the C atom number 2.
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d
5. For this structure, the name 3-Ethylpentane is wrong. The correct name can be obtained in 4 steps:
(i) For this structure, the C atoms should be numbered as shown in fig.e
(ii) Now we see that, there are six C atoms in the longest chain.
(iii) We also see that, the branch is methyl, not ethyl. The branch is at the third C atom.
(iv) So the correct IUPAC name is: 3-Methylhexane.

Part (ii):
1. In the given name, we have ‘hept’ as the root. Then there will be seven C atoms in the longest chain. So we first draw a chain of seven C atoms. This is shown in fig.13.25(a) below:

Fig.13.25

2. Next we number the C atoms from 1 to 7. This is shown in fig.b
3. Attaching the branches:
   ♦ ‘5-ethyl' indicates that, there is an ethyl group at the C atom number 5.
   ♦ ‘3-methyl' indicates that, there is a methyl group at the C atom number 3.
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d
5. For this structure, the name 5-Ethyl-3-methylheptane is wrong. The correct name can be obtained in 4 steps:
(i) For this structure, the C atoms can be numbered as shown in fig.e also
(ii) Both methods of numbering will give the same numbers 3 and 5.
• That means, the branches are at equivalent positions.
(iii) So we must apply rule 5A. [see section 12.3]
Then 'ethyl' gets the lower number because, it comes first in the alphabetical order.
(iv) So the numbering in fig.e is the correct method.
• Based on this numbering, the correct IUPAC name is: 3-Ethyl-5-methylheptane.

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In the next section we will see preparation of alkanes.

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Chapter 12.8 - Nomenclature of Substituted Benzene Compounds

In the previous section, we saw how the compounds with functional groups are named. In this section, we will see the nomenclature of substituted benzene compounds.

Let us first see some basics about benzene. It can be written in 3 steps:
1. Fig,12.52(a) below shows our familiar hexagonal ring.

Fig.12.52

• We know that, it is the bond-line formula of cyclohexane. We know how the C atoms are arranged in that ring. It is shown again in figs (b) and (c)
• Fig.b shows the condensed formula and fig.c shows the complete structural formula.
• From the complete structural formula, we can see how the valencies of C atoms are satisfied.
2. We know that, all bonds in the cyclohexane above are single bonds.
• Now, if alternate bonds are double bonds, there will be lesser number of H atoms. This is shown in fig.12.53 below:

Fig.12.53

• Fig.12.53(a) shows the bond line formula. Fig.b shows the condensed formula.
• Fig.c shows the complete structural formula. From the complete structural formula, we can see how the valencies of all the C atoms are satisfied.
◼ The compound in fig.12.53 is known as benzene.
3. Benzene has exactly half the number of H atoms that are present in cyclohexane.
    ♦ The molecular formula of cyclohexane is C₆H₁₂
    ♦ The molecular formula of benzene is C₆H₆

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A large number of organic compounds can be derived from benzene. So we must know how to name them systematically based on the IUPAC rules. We can learn those rules by analyzing some examples.

Example 1
This can be written in 4 steps:
1. Fig.12.54(a) below shows a compound in which, one H atom of benzene is replaced by a methyl group.

Fig.12.54

2. In such cases, the benzene is given importance and so it becomes the suffix in the name.
• We can write the name as: Methylbenzene.
3. Note that, the position number of the methyl group is not necessary. The methyl group can be anywhere along the ring. Whatever be the position, the compound will not change.
• This can be explained using the animation in fig.12.55 below.

Fig.12.55

• It can be written in 5 steps:
(i) Positions of the double bond:
• Consider the left side molecule in the animation. The top double bond is on the left side of the branch.
• Consider the middle molecule in the animation. The top double bond is on the right side of the branch.
(ii) So the two molecules appear to be different.
(iii) Now consider the right side molecule. Initially, it appears to be same as the left side molecule. But when it is rotated through 180^o, it becomes same as the middle molecule.
(iv) So we can write:
When there is only one branch, the position of the branch is not important.
4. Methylbenzene in fig.12.54(a) has many industrial applications. It’s common name is Toleune. This name is widely accepted.  

Example 2
This can be written in 5 steps:
1. Fig.12.54(b) above shows a compound in which, one H atom of benzene is replaced by a methoxy group.
2. As before, the benzene is given importance and so it becomes the suffix in the name.
• We can write the name as: Methoxybenzene.
• Note that, the position number of the methoxy group is not necessary. The group can be anywhere along the ring. Whatever be the position, the compound will not change. We saw this in the animation in fig.12.55 above.
3. Note how the valency of the O atom is satisfied. The two bonds around the O indicates that, it has obtained the required two new electrons.
4. The name 'methyl' can be abbreviated.
• IUPAC allows the following abbreviations:
    ♦ Methyl can be abbreviated as Me  
    ♦ Ethyl can be abbreviated as Et
    ♦ Propyl can be abbreviated as Pr
    ♦ Butyl can be abbreviated as Bu
• So our present Methoxybenzene can be shown as in fig.12.54(c) also
5. Methoxybenzene has many industrial applications. It’s common name is Anisole. This name is widely accepted.

Example 3
This can be written in 4 steps:
1. Fig.12.56(a) below shows a compound in which, one H atom of benzene is replaced by a ㅡNH₂ group.

Fig.12.56

2. We have seen that, the benzene is to be given importance and so it becomes the suffix in the name.
• We can write the name as: Aminobenzene.
• Recall that:
    ♦ When ㅡNH₂ group is the suffix, we use 'amine'
    ♦ When ㅡNH₂ group is the prefix, we use 'amino'
3. Here also, the position number of the ㅡNH₂ group is not necessary.
4. Aminobenzene has many industrial applications. It’s common name is Aniline. This name is widely accepted.

Example 4
This can be written in 4 steps:
1. Fig.12.56(b) above shows a compound in which, one H atom of benzene is replaced by a ㅡOH group.
2. We have seen that, the benzene is to be given importance and so it becomes the suffix in the name.
• We can write the name as: Hydroxybenzene.
• Recall that:
    ♦ When ㅡOH group is the suffix, we use 'ol'
    ♦ When ㅡOH group is the prefix, we use 'hydroxy'
3. Here also, the position number of the ㅡOH group is not necessary.
4. Hydroxybenzene has many industrial applications. It’s common name is Phenol. This name is widely accepted.

◼ Similar to the above four examples, we can explain Nitrobenzene and Bromobenzene also. They are shown in fig.12.56(c) and (d) above.

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• When there is more than one substituent, numbering must be done in such a way that, the substituents gets the lowest possible numbers.
• Let us see an example:
    ♦ In figs.12.57(a) and (b) below, the correct numbering is the one in fig.a.
        ✰ Based on this numbering, the name will be 1,2-dibromobenzene.
    ♦ Based on the numbering fig.b, the name will be 1,6-dibromobenzene.
        ✰ This name is not valid.

Fig.12.57

• Let us see another example:
    ♦ In figs.12.57(c) and (d) above, the correct numbering is the one in fig.c.
        ✰ Based on this numbering, the name will be 1,3-dibromobenzene.
    ♦ Based on the numbering fig.d, the name will be 1,5-dibromobenzene.
        ✰ This name is not valid.

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• When there are two ㅡBr functional groups in the benzene ring, only three arrangements are possible. So it is convenient to give three distinct names. This can be explained in 5 steps:
1. The three possible arrangements are shown in fig.12.58 below:

Fig.12.58

2. In fig.12.58(a), the functional groups are at positions 1,2
• The two functional groups are on adjacent C atoms.
• This arrangement is indicated by the word 'ortho'
• So the trivial name of 1,2-dibromobenzene is: ortho-dibromobenzene
    ♦ This can be further abbreviated as: o-dibromobenzene.
3. In fig.12.58(b), the functional groups are at positions 1,3
• There is one C atom between the two functional groups.
• This arrangement is indicated by the word 'meta'
• So the trivial name of 1,3-dibromobenzene is: meta-dibromobenzene
    ♦ This can be further abbreviated as: m-dibromobenzene.
4. In fig.12.58(c), the functional groups are at positions 1,4
• There are two C atoms between the two functional groups.
• This arrangement is indicated by the word 'para'
• So the trivial name of 1,4-dibromobenzene is: para-dibromobenzene
    ♦ This can be further abbreviated as: p-dibromobenzene.
5. The above trivial names are widely accepted.

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When there are more functional groups in the benzene ring, we will have to strictly follow the IUPAC rules. The steps can be explained with the help of some examples:

Example 1:
This can be written in 6 steps:
1. Fig.12.59(a) below shows a compound in which, one H atom of benzene is replaced by a ㅡOMe group.
• We know that, this compound is called anisole.

Fig.12.59

2. But besides OMe, two more groups are present: ㅡCl and ㅡCH₃.
• To write the IUPAC name, we have to assign proper locant numbers to these functional groups.
3. We take anisole as the base compound.
• The functional group which converts the benzene into anisole is given the number 1.
• So ㅡOMe gets the number 1
4. Once the number 1 is assigned, we can give numbers to the other functional groups. The direction of numbering is chosen in such a way that, the group closest to ㅡOMe gets the smallest number.
• In our present case, the group closest to ㅡOMe is ㅡCl.
5. So we number the atoms in a clockwise direction. This is shown in fig.a.
• The ㅡCl gets the number 2
• If the numbering is in the anti-clockwise direction, Cl will get number 6. This is not acceptable.
6. So we have the required locant numbers:
   ♦ ㅡOMe at 1
   ♦ ㅡCl at 2
   ♦ ㅡCH₃ at 4
• Number 1 for OMe need not be written because, it is the group which converts benzene into anisole. It is understood that, OMe will be at position 1 in anisole.
• We have to write the other two groups and it's numbers. They must be written in alphabetical order. So 'chloro' comes before 'methyl' 
• Thus the name of the compound is: 2-chloro-4-methylanisole.

Example 2:
This can be written in 6 steps:
1. Fig.12.59(b) above shows a compound in which, one H atom of benzene is replaced by a ㅡNH₂ group.
• We know that, this compound is called aniline.
2. But besides ㅡNH₂, two more groups are present: CH₃ and C₂H₅.
• To write the IUPAC name, we have to assign proper locant numbers to these functional groups.
3. We take aniline as the base compound.
• The functional group which changes the benzene into aniline is given the number 1.
• So ㅡNH₂ gets the number 1
4. Once the number 1 is assigned, we can give numbers to the other functional groups. The direction of numbering is chosen in such a way that, the group closest to NH₂ gets the smallest number.
• In our present case, the group closest to NH₂ is CH₃.
5. So we number the atoms in a clockwise direction. This is shown in fig.b. The CH₃ gets the number 2
• If the numbering is in the anti-clockwise direction, CH₃ will get number 6. This is not acceptable.
6. So we have the required locant numbers:
   ♦ ㅡNH₂ at 1
   ♦ ㅡCH₃ at 2
   ♦ ㅡC₂H₅ at 4
• Number 1 for NH₂ need not be written because, it is the group which converts benzene into aniline. It is understood that, NH₂ will be at position 1 in aniline.
• We have to write the other two groups and it's numbers. They must be written in alphabetical order. So 'ethyl' comes before 'methyl' 
• Thus the name of the compound is: 4-Ethyl-2-methylaniline

Example 3:
This can be written in 6 steps:
1. Fig.12.59(c) above shows a compound in which, one H atom of benzene is replaced by a ㅡOH group.
We know that, this compound is called phenol.
2. But besides ㅡOH, two more groups are present: CH₃ and CH₃.
• To write the IUPAC name, we have to assign proper locant numbers to these functional groups.
3. We take phenol as the base compound.
The functional group which converts the benzene into phenol is given the number 1.
So ㅡOH gets the number 1
4. Once the number 1 is assigned, we can give numbers to the other functional groups. The direction of numbering is chosen in such a way that, the group closest to OH gets the smallest number.
• In our present case, the group closest to OH is the right side CH₃.
5. So we number the atoms in a clockwise direction. This is shown in fig.c. The CH₃ gets the number 3
• If the numbering is in the anti-clockwise direction, this CH₃ will get number 5. This is not acceptable.
6. So we have the required locant numbers:
   ♦ ㅡOH at 1
   ♦ ㅡCH₃ at 3
   ♦ ㅡCH₃ at 4
• Number 1 for OH need not be written because, it is the group which converts benzene into phenol. It is understood that, OH will be at position 1 in phenol.
• We have to write the other two groups and it's numbers.
• Thus the name of the compound is: 3,4-Dimethylphenol.

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• In the above discussion, benzene was considered as the parent.
• Groups like , ㅡOH, ㅡCl, ㅡCH₃, ㅡCH₃CH₂ etc., were attached to the benzene ring.
   ♦ These groups are called substituents.
   ♦ An example is shown in fig.12.60(a) below.
   ♦ In this example, a propyl group is the substituent.

Fig.12.60

• In some cases, the benzene ring becomes the substituent.
   ♦ It will be attached to a parent.
• This can be explained in steps:
1. We know that, benzene is C₆H₆
• One H atom from the ring is removed so that, it becomes ㅡC₆H₅
   ♦ This is called phenyl group.
2. The phenyl group gets attached to another parent chain. An example is shown in fig.12.60(b) above.
• The compound in fig.b appears to be similar to the compound in fig.a.
• Then why is it that,
   ♦ Benzene ring in fig.a is the parent ?
   ♦ Benzene ring in fig.b is a substituent ?
3. The answer is:
If the branch attached to benzene has more than six C atoms, then that branch is considered as the parent and benzene is considered as the substituent.
4. There are other occasions also, where benzene becomes the substituent:
(i) If the branch of the benzene contains one or more double bonds.
(ii) If the branch of the benzene contains one or more triple bonds.
(iii) If the branch of the benzene contains one or more functional groups.
• In these three cases, the benzene will become the substituent even if the number of C atoms in the branch is less than 6. Figs (c) and (d) are examples.

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Let us see some solved examples:
Solved example 12.12
Write the structural formula of:
(a) o-Ethylanisole  (b) p-Nitro-aniline  (c) 2,3-Dibromo-1-phenylpentane (d) 4-Ethyl-1-fluoro-2-nitro-benzene.
Solution:
Part (a): o-Ethylanisole
1. 'anisole' indicates that, it is a benzene ring with OMe group as substituent.
2. Ethylanisole indicates that, an ethyl group is also present as a substituent
3. 'o' stands for 'ortho'. So the OMe and ethyl groups are in adjacent positions.
4. Thus we get the structure in fig.12.61(a) below:

Fig.12.61

Part (b): p-Nitro-aniline
1. 'aniline' indicates that, it is a benzene ring with NH₂ group as substituent.
2. Nitro-aniline indicates that, a NO₂ group is also present as a substituent
3. 'p' stands for 'para'. So the NH₂ and NO₂ groups have two C atoms in between them.
4. Thus we get the structure in fig.12.61(b) above.

Part (c): 2,3-Dibromo-1-phenylpentane
1. 'pentane' indicates that, a straight chain with five C atoms is the parent.
2. '1-phenylpentane' indicates that, a phenyl group is attached to the first C atom of the parent chain.
3. '2,3-Dibromo' indicates that, there are two Br groups, one at position 2 and the other at position 3
4. Thus we get the structure in fig.12.61(c) above.

Part (d): 4-Ethyl-1-fluoro-2-nitro-benzene
1. 'benzene' indicates that, it is the base.
2. '4-Ethyl-1-fluoro-2-nitro' indicates that, ethyl, fluoro and nitro groups are attached at positions 4, 1 and 2 respectively.
3. Thus we get the structure in fig.12.61(d) above.

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In the next section, we will see isomerism.

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Chapter 12.7 - Nomenclature When More Than One Functional Group Of The Same Type Are Present

In the previous section, we saw how the compounds with functional groups are named. In this section, we will see the nomenclature when more than one functional group of the same type are present.

Nomenclature of organic compounds having more than one functional group of the same type

We have seen that, when more than one branch of the same type is present, we use di, tri, tetra etc., We can apply the same method here also. Let us see some examples. We will learn some new rules also while analyzing those examples.

Example 1
Write the IUPAC name of CH₂(OH)CH₂(OH)
Solution:
1. We are given the condensed formula. Let us draw the complete structural formula. It is shown in fig.12.45(a) below:

Fig.12.45

2. We want to write the IUPAC name of the molecule in fig.12.45(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: ㅡOH and ㅡOH
All functional groups are of the same type. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the ㅡOH group.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
• Here, suffix is to be used because, the name is going to be based on ㅡOH
• So the main component of the final name will be: ethanol
(iii) Now we have to show that there are two ㅡOH groups.
So we write: ethan-diol
(iv) Also, we have to show the positions. So we write: ethan-1,2-diol
This is the IUPAC name.
(Note that in this example, the numbering from either sides will give the same result)

Example 2
Write the IUPAC name of CH₂=CH-CH=CH₂
Solution:
1. We are already given the structural formula. But we have to number the C atoms properly. This is shown in fig.12.45(b) above.
2. We want to write the IUPAC name of the molecule in fig.12.45(b). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and C=C
All functional groups are of the same type. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the C=C group.
• We know that, for this double bond, the 'ane' changes to 'ene'
• So the main component of the final name will be: butene
(iii) Now we have to show that there are two C=C groups.
So we write: but-diene
(iv) Also, we have to show the positions. So we write: buta-1,3-diene
This is the IUPAC name.
(The double bonds should get the lowest number possible. In this example, the numbering from either sides will give the same result)

Example 3
Write the IUPAC name of the compound shown in fig.12.46(a) below:

Fig.12.46

Solution:
• We want to write the IUPAC name of the molecule in fig.12.46(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C There is only one functional group. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the C=C group.
We know that, for this double bond, the 'ane' changes to 'ene'
• So the main component of the final name will be: hexene
• The numbering is done so as to include the double bond. If we do the number in an exact horizontal manner, we will get a longer chain. But such a numbering is not allowed. It is compulsory to select that chain which contains the double bond.
(iii) Now we have to show the position of the double bond. We get: hex-1-ene.
(iv) Also, we have to include the ethyl group at position 2.
So we write: 2-Ethylhex-1-ene
This is the IUPAC name.

Example 4
Write the IUPAC name of the compound shown in fig.12.46(b) above.
Solution:
• We want to write the IUPAC name of the molecule in fig.12.46(b). It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and C=C. There is only one type of functional group. So there is no need to assign any priority.
(ii) The IUPAC name of the molecule will be based on the C=C group.
We know that, for this double bond, the 'ane' changes to 'ene'
• So the main component of the final name will be: hexene
• The numbering is done so as to include both the double bonds. If we do the number in an exact horizontal manner, we will get a longer chain. But such a numbering is not allowed. It is compulsory to select that chain which contains all the double bonds.
(iii) Now we have to show that there are two C=C groups.
So we write: hexa-diene
(iv) Also, we have to show the positions. So we write: hexa-1,4-diene
(v) Now we have to include the ethyl group at position 2.
So we write: 2-Ethylhexa-1,4-diene
This is the IUPAC name.

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• Examples 3 and 4 are similar except that, there is an extra double bond in example 4.
• The names are also similar:
   ♦ 2-Ethylhex-1-ene
   ♦ 2-Ethylhexa-1,4-diene
• Note that:
   ♦ When there is only one double bond, we write 'hex'
   ♦ When there is more than one double bond, we write 'hexa'
• Another example is: 'pent' and 'penta'

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Now we will see some solved examples.

Solved example 12.10
Write the IUPAC names of the compounds shown in figs.12.47 (a) and (b) below:

Fig.12.47

Solution:
Part (a):
• We want to write the IUPAC name of the molecule in fig.12.47(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: >C=O and >C=O
All functional groups are of the same type. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the >C=O group.
    ♦ When the component obtained from >C=O is the suffix, we use ‘-one’  
    ♦ When the component obtained from >C=O is the prefix, we use ‘oxo-’
• Here, suffix is to be used because, the name is going to be based on >C=O
• So the main component of the final name will be: hexanone
(iii) Now we have to show that there are two >C=O groups.
So we write: hexane-dione
(iv) Also, we have to show the positions. So we write: Hexane-2,4-dione
This is the IUPAC name.
(Note that in this example, the numbering should be done from left to right. This is to give the functional groups, the lowest possible numbers) 

Part (b):
• We want to write the IUPAC name of the molecule in fig.12.47(b). It can be done in 7 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and C☰C
(ii) Next we compare the list with table 1. We see that, C=C gets top priority. So it is the principal functional group. The C☰C is the subordinate functional group.
(iii) Imagine that, the C☰C is not present. Then the IUPAC name of the molecule will be based on the C=C group.
• So the main component of the final name will be: hexene
(iii) Now we have to show that there are two C=C groups.
So we write: hexa-diene
(iv) Also, we have to show the positions. So we write: Hexa-1,3-diene
(v) Now we have to account for the C☰C group.
So we include 'yne' in the name.
(vi) The C☰C group is at position 5. So we write: Hexa-1,3-diene-5-yne
This is the IUPAC name.
(vii) Priority between double and triple bonds:
• Alphabetically, 'ene' comes before 'yne'. So 'ene' should be written first.
• While numbering, the lowest possible number should be given to the double bonds.

Solved example 12.11
Derive the structure of (i) 2-Chlorohexane, (ii) Pent-4-en-2-ol, (iii) 3-Nitrocyclohexene, (iv) cyclohex-2-en-1-ol, (v) 6-Hydroxy-heptanal.
Solution:
Part (i): 2-Chlorohexane
1. The suffix 'ane' indicates that the compound contains only single bonds. It is an alkane.
• 'hex' indicates that, there are six C atoms.
• So we can draw the skeletal structure as shown in fig.12.48(a) below:

Fig.12.48

2. 'Chloro' indicates the functional group: ㅡCl
• This is the only one functional group. The numbering must be in such a way that, the ㅡCl gets lowest number.
• So '2-Chloro' indicates that, the ㅡCl is at the second position.
• We can add this information to fig.a. We get the modified fig.b
3. Now we fill up the valencies of all C atoms. We get the final structure in fig.c

Part (ii): Pent-4-en-2-ol
1. The suffix '-ol' indicates that one ㅡOH group is present. It is an alcohol.
• It is the suffix. So it is the principal group.
• The numbering must be in such a way that, the ㅡOH gets lowest number
• So '-2-ol' indicates that, the ㅡOH is at the second position.
2. 'Pent' indicates that, there are five C atoms. So we can draw the skeletal structure as shown in fig.12.49(a) below:

Fig.12.49

3. 'en' indicates that there is one double bond.
'-4-en' indicates that, the double bond is between positions 4 and 5
We can add this information to fig.a. We get the modified fig.b
4. Now we fill up the valencies of all C atoms. We get the final structure in fig.c

Part (iii): 3-Nitrocyclohexene
1. The suffix 'ene' indicates that the compound contains one double bond. It is an alkene.
• 'hex' indicates that, there are six C atoms.
• 'cyclo' indicates that, it is a cyclic alkene.
• So we can draw the skeletal structure as shown in fig.12.50(a) below:

Fig.12.50

• Note that, the numbering is done in such a way that, the double bond gets the lowest numbers 1 and 2
2. 'Nitro' indicates that there is one ㅡNO₂ functional group.
• '3-Nitro' indicates that the ㅡNO₂ group is at position 3
• So we get the final structure shown in fig.b
3. From the tables-1 and 2 of the previous section, we know that, the double bond gets priority over the ㅡNO₂ group. We see that, ㅡNO₂ which is in table-2, will always be a subordinate group. That is why, 'ene' is the suffix.

Part (iv): cyclohex-2-en-1-ol
1. The suffix '-ol' indicates that one ㅡOH group is present. It is an alcohol.
• It is the suffix. So it is the principal group.
• The numbering must be in such a way that, the ㅡOH gets lowest number
• So '-1-ol' indicates that, the ㅡOH is at the first position
2. 'Cyclohex' indicates that, it is a cyclic structure with six C atoms.
3. 'en' indicates that, it contains a double bond.
'2-en' indicates that, the double bond is between 2 and 3.
4. Based on the above information, the structure will be as shown in fig.12.50(c) above.
5. From the table-1 of the previous section, we know that, the ㅡOH group gets priority over the double bond. That is why, it is given the lowest number.

Part (v): 6-Hydroxy-heptanal
1. The suffix 'al' indicates that, it is an aldehyde. It has a ㅡCHO group.
• This group is always at the end of the chain. So it do not require any number. (Details here)
2. 'hept' indicates that there are seven C atoms in the chain.
• The 'an' in 'heptan' indicates that, it is an alkane. All bonds are single bonds.
• So the skeletal structure will be as shown in fig.12.51(a) below:

Fig.12.51

• Note that, the seven C atoms include the C atom coming from the ㅡCHO group.   
3. '6-Hydroxy' indicates that, there is a ㅡOH group at the sixth C atom.
• So the skeletal structure can be modified as shown in fig.12.51(b) above.
4. Now we fill up the valencies of all C atoms. We get the final structure in fig.c

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In the next section, we will see nomenclature of substituted Benzene compounds.

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Chapter 12.6 - Nomenclature When Functional Groups are Present

In the previous section, we saw how the cyclic compounds are named. In this section, we will see the nomenclature when functional groups are present.

Nomenclature of organic compounds having functional groups

Some basics about functional groups can be written in 5 steps:
1. We know that, functional group is an atom or a group of atoms. It gives special characteristic properties to the organic compound to which it is attached.
• We have seen some basics about this topic in our earlier chemistry classes (Details here).
2. Consider a molecule of an organic compound containing a functional group.
• That molecule will be more reactive at the region where the functional group is attached.
3. Different organic compounds may be having the same functional group.
• For example, the three compounds in fig.12.39 below, have the same functional group ㅡOH

Fig.12.39

4. The above three compounds have similar chemical properties.
• For instance, they all liberate hydrogen on reaction with sodium metal.
• That means, different compounds become similar due to the presence of common functional groups.
5. So compounds carrying common  functional groups can be put together into various classes. Let us see some examples:
• All organic compounds carrying the functional group ㅡOH come under the class: Alcohols.
• All organic compounds carrying the functional group ㅡCHO come under the class: Aldehydes.
• All organic compounds carrying the functional group >C=O come under the class: Ketones.
(What does the '>' of the keto group indicate? Answer can be seen here)
• We will see more such classes as we continue our discussion.

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We have seen the basics of naming functional groups in our earlier chemistry classes. (Details here) Now we will see some advanced details. They can be written in 5 steps:
1. Identify the functional group present in the given molecule.
• Based on the functional group, the class of the molecule can be decided.
• Based on the class, the appropriate suffix can be decided.
• For example, if the molecule falls in the class of alcohols, the suffix is ‘ol’
2. Identify the parent chain (the chain with the largest number of C atoms)
• Based on the number of C atoms in the parent chain, the root name can be fixed.
• For example, if the parent chain has three C atoms, and if the hydrocarbon is an alkane, the root name will be propane.
3. Number the C atoms in the parent chain in such a way that, the C atom carrying the functional group gets the lowest possible number.
4. Now remove ‘e’ from the root name and put the following format in it’s place: -[no space]number of C atom[no space]-[no space]suffix denoting the functional group.
• For example, the IUPAC name of the molecule in fig.12.40(a) below is: Butan-2-ol
• Note that, 'Butan' is obtained by removing 'e' from Butane.

Fig.12.40

5. In some cases, in addition to the functional group, the parent chain may contain branches also. In such cases, the names of the branches should be written as prefixes.
• Let us see an example. It can be written in steps:
(i) Consider the molecule in fig.12.40(b) above.
• Imagine that, there is no methyl branch. Then the name of the molecule would be: octan-2-ol
(ii) Now, we must give appropriate name for the methyl branch.
• We know that, it’s name is: 6-methyl.
• We must use this as the prefix to the result in (i).
(iii) So the IUPAC name will be: 6-Methyloctan-2-ol

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• We have seen that in IUPAC system, the suffix is based on the functional group.
• For example, in 6-Methyloctan-2-ol, the suffix is octan-2-ol , which is based on the ㅡOH functional group.
• If there are different functional groups, which one shall we choose?
• We can write the answer in 7 steps:
1. If more than one functional group is present in a molecule, we must select one of them as the principal functional group. The remaining functional groups in the molecule will be called subordinate functional groups.
2. How do we know which one is the principal functional group?
• For that, first we make a list of all the functional groups in our molecule.
• Then we compare the list with the table 1 given below:

Table 1: Order of priority

ㅡCOOH, ㅡSO3H, ㅡCOOR, ㅡCOCl, ㅡCONH2, ㅡC☰N, ㅡHC=O, >C=O, ㅡOH, ㅡNH2, >C=C<, ㅡC☰Cㅡ • R represents an alkyl group like CH2ㅡ, CH3CH2ㅡ etc.,

• When compared with the above table 1, the functional group which comes in the top most position in our list will be the principal functional group.
(Note that, the above table-1 does not contain all the groups specified by the IUPAC. It contains only those groups which we encounter frequently in our present discussions)
3. There are some groups which are always considered as subordinate groups. They are given in table 2 below:

Table 2: Groups which are always subordinate

RㅡX, RㅡOㅡR, RㅡSㅡR, NO2 etc., • R represents an alkyl group like CH2ㅡ, CH3CH2ㅡ etc., • X represents a halogen like Cl, Br etc.,

(Note that, the above table-2 does not contain all the groups specified by the IUPAC. It contains only those groups which we encounter frequently in our present discussions)
4. Let us apply the above table 1 to an example.
• Consider the molecule in fig.12.41(a) below.

Fig.12.41

• We want to write the IUPAC name of the molecule. It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: ㅡCOOH and ㅡOH
(Recall how the COOH group is attached to a molecule. We have seen it in our earlier chemistry classes. Details here)
(ii) Next we compare the list with table 1 above. We see that, ㅡCOOH gets top priority. So it is the principal functional group. The ㅡOH is the subordinate functional group.
(iii) Imagine that, the ㅡOH is not present. Then the IUPAC name of the molecule will be based on the ㅡCOOH group.
    ♦ When the component obtained from ㅡCOOH is the suffix, we use ‘-oic acid’  
    ♦ When the component obtained from ㅡCOOH is the prefix, we use ‘carboxy’
• Here, suffix is to be used because, ㅡCOOH has priority and the final name is going to be based on it.
• So the main component of the final name will be: hexanoic acid
(Main component will be always the suffix. Recall that, in methane, the main component is 'ane', which is the suffix)
(iv) Now we have to account for the ㅡOH group.
• The component of the name obtained from the ㅡOH will be the prefix.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
(v) Now we give the position number. The ㅡOH is at position 4. So the component will be 4-hydroxy
• Thus the IUPAC name of the molecule will be: 4-Hydroxyhexanoic acid.
5. Let us apply the above table 1 to another example.
• Consider the molecule in fig.12.41(b) above. We want to write the IUPAC name of the molecule. It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: ㅡC☰N and ㅡOH
(ii) Next we compare the list with table 1 above. We see that, ㅡC☰N gets top priority. So it is the principal functional group. The ㅡOH is the subordinate functional group.
(iii) Imagine that, the ㅡOH is not present. Then the IUPAC name of the molecule will be based on the ㅡC☰N group.
    ♦ When the component obtained from ㅡC☰N is the suffix, we use ‘nitrile’  
    ♦ When the component obtained from ㅡC☰N is the prefix, we use ‘cyano’
• Here, suffix is to be used because, ㅡC☰N has priority and the final name is going to be based on it.
• So the main component of the final name will be: hexanenitrile
(Main component will be always the suffix. Recall that, in methane, the main component is 'ane', which is the suffix)
(iv) Now we have to account for the ㅡOH group.
• The component of the name obtained from the ㅡOH will be the prefix.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
(v) Now we give the position number. The ㅡOH is at position 4. So the component will be 4-hydroxy
• Thus the IUPAC name of the molecule will be: 4-Hydroxyhexanenitrile
6. Let us apply the above table 1 to yet another example.
• Consider the molecule in fig.12.42(a) below.

Fig.12.42

• We want to write the IUPAC name of the molecule. It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: ㅡNH₂ and ㅡOH
(Recall how the ㅡNH₂ group is attached to a molecule. We have seen it in our earlier chemistry classes. Details here)
(ii) Next we compare the list with table 1 above. We see that, ㅡOH gets top priority. So it is the principal functional group. The ㅡNH₂ is the subordinate functional group.
(iii) Imagine that, the ㅡNH₂ is not present. Then the IUPAC name of the molecule will be based on the ㅡOH group.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
• Here, suffix is to be used because, ㅡOH has priority and the final name is going to be based on it.
• So the main component of the final name will be: hexan-2-ol
(Main component will be always the suffix. Recall that, in methane, the main component is 'ane', which is the suffix)
(Note that, the numbering of C atoms is done in such a way as to give the lowest number to the priority group which is ㅡOH)
(iv) Now we have to account for the ㅡNH₂ group.
• The component of the name obtained from the ㅡNH₂ will be the prefix.
    ♦ When the component obtained from ㅡNH₂ is the suffix, we use ‘-amine’  
    ♦ When the component obtained from ㅡNH₂ is the prefix, we use ‘amino-’
(v) Now we give the position number. The ㅡNH₂ is at position 4. So the component will be 4-amino
• Thus the IUPAC name of the molecule will be: 4-Aminohexan-2-ol
7. Let us apply the above table 1 to one more example.
• Consider the molecule in fig.12.42(b) above. We want to write the IUPAC name of the molecule. It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and ㅡOH
(ii) Next we compare the list with table 1 above. We see that, ㅡOH gets top priority. So it is the principal functional group. The C=C is the subordinate functional group.
(iii) Imagine that, the C=C is not present. Then the IUPAC name of the molecule will be based on the ㅡOH group.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
• Here, suffix is to be used because, ㅡOH has priority and the final name is going to be based on it.
• So the main component of the final name will be: hexan-2-ol
(Main component will be always the suffix. Recall that, in methane, the main component is 'ane', which is the suffix)
(Note that, the numbering of C atoms is done in such a way as to give the lowest number to the priority group which is ㅡOH)
(iv) Now we have to account for the C=C group.
• The component of the name obtained from the C=C will be the prefix.
• But for double bonds and triple bonds, we need not worry about prefix or suffix. • Both prefix and suffix can be written using 'ene' or 'yne'.
• In our present case, hexane becomes hexene.
• Since ㅡOH is present, we remove 'e' from hexene and add 'ol'. Thus we get: hexenol
To indicate the position of the ㅡOH, we write: hexen-2-ol
(v) Now we give the position number of the C=C. It is at position 4.
• Thus the IUPAC name of the molecule will be: 4-Hexen-2-ol

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Now we know how to use the table 1. Let us see a solved example.

Solved example 12.8
Write the IUPAC names of (i) HOCH₂(CH₂)₃CH₂COCH₃ (ii) BrCH₂CH=CH₂
Solution:
Part (i):
1. We are given the condensed formula. Let us draw the complete structural formula. It is shown in fig.12.43(a) below:

Fig.12.43

2. We want to write the IUPAC name of the molecule in fig.12.42(a). It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: >C=O and ㅡOH
(Recall how the >C=O group is attached to a molecule. We have seen it in our earlier chemistry classes. Details here)
(ii) Next we compare the list with table 1 above. We see that, >C=O gets top priority. So it is the principal functional group. The ㅡOH is the subordinate functional group.
(iii) Imagine that, the ㅡOH is not present. Then the IUPAC name of the molecule will be based on the >C=O group.
    ♦ When the component obtained from >C=O is the suffix, we use ‘-one’  
    ♦ When the component obtained from >C=O is the prefix, we use ‘oxo-’
• Here, suffix is to be used because, >C=O has priority and the final name is going to be based on it.
• So the main component of the final name will be: heptan-2-one
(Note that, the numbering is done in such a way as to give the lowest possible number to the principal functional group)
(iv) Now we have to account for the ㅡOH group.
• The component of the name obtained from the ㅡOH will be the prefix.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
(v) Now we give the position number. The ㅡOH is at position 7. So the component will be 7-hydroxy
• Thus the IUPAC name of the molecule will be: 7-Hydroxyheptan-2-one.

Part (ii):
1. We are given the condensed formula. Let us draw the complete structural formula. It is shown in fig.12.43(b) above.
2. We want to write the IUPAC name of the molecule in fig.12.42(b). It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and ㅡBr
(ii) Next we compare the list with table 1 above. We see that, C=C gets top priority.
• Also from table 2, we know that X (a halogen) will always be a subordinate group.
• So C=C is the principal functional group. The ㅡBr is the subordinate functional group.
(iii) Imagine that, the ㅡBr is not present. Then the IUPAC name of the molecule will be based on the C=C group.
• The component of the name obtained from the C=C will be the suffix.
• For double bonds and triple bonds, we need not worry about prefix or suffix. Both prefix and suffix can be written using 'ene' or 'yne'.
• In our present case, since there are three C atoms, we get propene.
• To show the position of the double bond, we modify 'propene' as: prop-1-ene
(Note that, the numbering is done in such a way as to give the lowest possible number to the principal functional group)
(iv) Now we have to account for the ㅡBr group.
• The component of the name obtained from the ㅡBr will be the prefix.
    ♦ When the component obtained from ㅡX is the prefix, we use ‘halo’
         ✰ ㅡF becomes fluro
         ✰ ㅡCl becomes chloro
         ✰ ㅡBr becomes bromo
         ✰ ㅡI becomes iodo
(v) Now we give the position number. The ㅡBr is at position 3. So the component will be 3-bromo
• Thus the IUPAC name of the molecule will be: 3-bromoprop-1-ene.

Solved example 12.9
Write the IUPAC names of the compounds shown in figs.12.44 (a) and (b) below:

Fig.12.44

Solution:
Part (a):
• We want to write the IUPAC name of the molecule in fig.12.43(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We see that there is only one functional group, which is: ㅡOH
• So the name will be based on this group.
(ii) Imagine that, the ㅡCH₃ is not present. Then the IUPAC name of the molecule will be based on the ㅡOH group.
• The component of the name obtained from the ㅡOH will be the suffix.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
• Here, suffix is to be used because, ㅡOH has priority and the final name is going to be based on it.
• So the main component of the final name will be: octan-3-ol
(Note that, the numbering is done in such a way as to give the lowest possible number to the principal functional group)
(iii) Now we have to account for the ㅡCH₃ group.
We know that, it is a methyl group.
(iv) Now we give the position number. The ㅡCH₃ is at position 6. So the component will be 6-methyl
• Thus the IUPAC name of the molecule will be: 6-Methyloctan-3-ol.

Part (b):
• We want to write the IUPAC name of the molecule in fig.12.43(b). It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: ㅡCOOH and >C=O
(ii) Next we compare the list with table 1 above. We see that, ㅡCOOH gets top priority. So it is the principal functional group. The >C=O is the subordinate functional group.
(iii) Imagine that, the >C=O is not present. Then the IUPAC name of the molecule will be based on the ㅡCOOH group.
    ♦ When the component obtained from ㅡCOOH is the suffix, we use ‘-oic acid’  
    ♦ When the component obtained from ㅡCOOH is the prefix, we use ‘carboxy’
• Here, suffix is to be used because, ㅡCOOH has priority and the final name is going to be based on it.
• So the main component of the final name will be: hexanoic acid
(Note that, the numbering is done in such a way as to give the lowest possible number to the principal functional group)
(iv) Now we have to account for the >C=O group.
• The component of the name obtained from the >C=O will be the prefix.
    ♦ When the component obtained from >C=O is the suffix, we use ‘-one’  
    ♦ When the component obtained from >C=O is the prefix, we use ‘oxo-’
(v) Now we give the position number. The >C=O is at position 5. So the component will be 5-oxo
• Thus the IUPAC name of the molecule will be: 5-Oxohexanoic acid.

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In the next section, we will see nomenclature when more than one functional group of the same type are present.


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