Chapter 13.14 - Nomenclature and Isomerism in Alkynes

In the previous section, we completed a discussion on ozonolysis and polymerisation of alkenes. In this section, we will see alkynes.

Let us recall some properties of alkynes that we have seen in earlier chapters. They can be written in 3 steps:
1. Alkynes are unsaturated hydrocarbons. They contain at least one triple bond.
2. If there is one triple bond in an alkyne,
   ♦ it will contain four H atoms less than the corresponding alkane.
   ♦ it will contain two H atoms less than the corresponding alkene.
• For example:
    ♦ Molecular formula of butane is C₄H₁₀
    ♦ Molecular formula of butene is C₄H₈
    ♦ Molecular formula of butyne is C₄H₆
• We know that, the general formula for alkenes is C_nH_2n
• We also know that, the corresponding alkyne has two H atoms less. So the general formula for alkenes is C_nH_2n-2
3. Ethyne is the IUPAC name of the first member. It’s common name is acetylene.
• Acetylene is used for arc welding. In this process, a mixture of acetylene and oxygen is subjected to combustion. As a result, a flame with high temperature and heat is produced. This flame can be used for fusing metal parts together.

---

Structure of triple bond

Alkynes have at least one triple bond. We have seen the details about that triple bond in the previous chapters. [see fig.4.156 of section 4.27 ] Let us recall those details. They can be written in 6 steps:
1. The triple bond consists of:
    ♦ One sigma (σ) bond
    ♦ Two pi (𝜋) bonds.
2. The sigma bond is formed by the head-on overlapping of the sp hybridized orbitals.
    ♦ The pi bonds are formed by the sideways overlapping of the 2p orbitals.
3. The sigma bond is a strong bond.
    ♦ Bond enthalpy of a sigma bond is 397 kJ mol^-1
• Pi bond is a weak bond.
    ♦ Bond enthalpy of a pi bond is 284 kJ mol^-1
4. The triple bond is shorter in bond length.
    ♦ The C-C single bond in alkanes has a bond length of 154 pm.
    ♦ The C=C double bond in alkenes has a bond length of 134 pm.
    ♦ The C☰C triple bond in alkynes has a bond length of 120 pm.

5. The triple bond has greater strength.
    ♦ The C-C single bond has a bond enthalpy of 348 kJ mol^-1.
    ♦ The C=C double bond has a bond enthalpy of 681 kJ mol^-1.
    ♦ The C☰C triple bond has a bond enthalpy of 823 kJ mol^-1.
6. We have seen that, ethyne has a linear structure. The electron clouds between the two C atoms are cylindrically symmetrical around the internuclear axis.

---

Nomenclature of Alkynes

• We have seen the rules for writing the IUPAC names in an earlier chapter [see section 12.3]
• Let us recall some basic rules which are applicable to alkynes. It can be written in 3 steps:
1. The longest chain should be selected in such a way that, it contains the triple bonds.
2. Numbering must be done in such a way that, the triple bonds get the lowest possible numbers.
3. The suffix ‘yne’ is used instead of ‘ane’ of alkanes.
• Let us see some examples:

Example 1:
Write the IUPAC name of the structure: CH₃-C☰CH
Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: propane – ane + yne = propyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the triple bond.
    ♦ Numbering from left to right will give number ‘2’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Prop-1-yne.
7. Here we note an interesting point:
• In propyne, the triple bond has only two possible positions. Both those positions will give the same name: Prop-1-yne. So we can write the name simply as: Propyne.

Example 2:
Write the IUPAC name of the structure: CH₃-CH₂-C☰CH
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the triple bond.
    ♦ Numbering from left to right will give number ‘3’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-1-yne.

Example 3:
Write the IUPAC name of the structure: CH₃-CH☰CH-CH₃
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give number ‘2’ to the triple bond.
    ♦ Numbering from right to left will give number ‘2’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-2-yne.

Example 4:
Write the IUPAC name of the structure: CH☰C-C☰CH
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give numbers '1,3' to the triple bonds.
    ♦ Numbering from right to left will give number ‘1,3’ to the triple bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Buta-1,3-diyne.
• Note that, since more than one triple bond is present, we write 'buta' instead of 'but'
• Similarly, 'penta' is written instead of 'pent', 'hexa' is written instead of hex', so on . . .

---

Isomerism in Alkynes

Some basics can be written in 6 steps:
1. Consider the three structures in fig.13.86 below:

Fig.13.86

   ♦ Structure (a) is Pent-1-yne.
   ♦ Structure (b) is Pent-2-yne.
   ♦ Structure (c) is 3-Methylbut-1-yne.
2. Let us compare the number of atoms:
   ♦ In (a), there are five C atoms and eight H atoms.  
   ♦ In (b) also, there are five C atoms and eight H atoms.
   ♦ In (c) also, there are five C atoms and eight H atoms.
• So all the three structures have the same molecular formula C₅H₈
• It is clear that pent-1-yne, pent-2-yne and 3-Methylbut-1-yne are isomers.
3. We have discussed about isomers in a previous section [see section 12.9]. Also we have discussed about the various isomers among alkanes and alkenes.
• Based on those discussions, we can write:
   ♦ Structures (a) and (c) in fig.13.86, are chain isomers.     
   ♦ Structures (b) and (c) in fig.13.86, are chain isomers.
   ♦ Structures (a) and (b) in fig.13.86, are position isomers.     
(Recall that position isomers have the same carbon chain. But the positions of the functional groups will be different. In our present case, the functional group is the alkyne group)
4. In the case of alkanes, we have seen that, only those which have more than three C atoms will exhibit isomerism.
• In the same way, the first two alkynes (ethyne and propyne) do not exhibit isomerism.
• This is because, a minimum of four C atoms are required to obtain different structural arrangements.
5. We have seen that the first two members (ethyne and propyne) of the alkyne series do not exhibit isomerism.
• In examples 2 and 3 above, we have seen the isomers of the third member butyne.
• In fig.13.86 above, we have seen the isomers of the fourth member pentyne. Next let us see the isomers of the fifth member.
• It is explained as as solved example below:

Solved example 13.13
Write all the structures and IUPAC names of the structural isomers of alkynes corresponding to C₆H₁₀.
Solution:
Seven isomers are shown in fig.13.87 below:

Fig.13.87

---

In the next section we will see preparation of alkynes.

Previous

Contents

Next

Copyright©2022 Higher secondary chemistry.blogspot.com

Chapter 13.8 - Nomenclature of Alkenes

In the previous section, we completed a discussion on conformations of ethane. In this section, we will start a discussion on alkenes.

Let us recall some properties of alkenes that we have seen in earlier chapters. They can be written in 4 steps:
1. Alkenes are unsaturated hydrocarbons. They contain at least one double bond.
2. If there is one double bond in an alkene, it will contain two H atoms less than the corresponding alkane.
• For example:
    ♦ Molecular formula of butane is C₄H₁₀
    ♦ Molecular formula of butene is C₄H₈
• We know that, the general formula for alkanes is C_nH_2n+2
• We also know that, the corresponding alkene has two H atoms less. So the general formula for alkenes is C_nH_2n
3. Chemists in the 19^th century noticed that, the first member of the alkene series (ethene) formed a oily liquid on reaction with chlorine. So the alkenes were generally known as olefins. The word olefin means ‘oil forming’.
4. Ethene is the IUPAC name of the first member. It’s common name is ethylene.

---

Structure of double bond

Alkenes have at least one double bond. We have seen the details about that double bond in the previous chapters. [see fig.4.147 of section 4.26 ] Let us recall those details. They can be written in 8 steps:
1. The double bond consists of:
    ♦ One sigma (σ) bond
    ♦ One pi (𝜋) bond.
2. The sigma bond is formed by the head-on overlapping of the sp² hybridized orbitals.
    ♦ The pi bond is formed by the sideways overlapping of the two 2p orbitals.
3. The sigma bond is a strong bond.
    ♦ Bond enthalpy of a sigma bond is 397 kJ mol^-1
• Pi bond is a weak bond.
    ♦ Bond enthalpy of a pi bond is 284 kJ mol^-1
4. The double bond is shorter in bond length.
    ♦ The C-C single bond in alkanes has a bond length of 154 pm.
    ♦ The C=C double bond in alkenes has a bond length of 134 pm.
    ♦ The C-H single bond has a bond length of 110 pm.
• The bond angle between C-H single bond and C=C double bond is 121.7^o
• The bond angle between the two C-H single bonds in a CH₂ group is 116.6^o
• These details are shown in fig.13.46 below:

Fig.13.46

5. In a pi bond, there is sideways overlapping of the p-orbitals. The two p-orbitals combine together to form a 𝜋-cloud. We saw this in figs.4.148 and 4.149 of section 4.26.
6. The electrons in the 𝜋-cloud are loosely held.
• Those loosely held electrons are easily available for reagents which are in search for electrons (reagents in search for electrons are known as electrophilic reagents).
• So electrophilic reagents attack alkenes easily.
7. We can say that:
• Alkenes have lesser stability when compared to alkanes.
• Alkenes can be easily converted into alkanes by combining with electrophilic reagents.
8. In step (3), we wrote that:
    ♦ Bond enthalpy of the sigma bond in a double bond is 397 kJ mol^-1
    ♦ Bond enthalpy of the pi bond in a double bond is 284 kJ mol^-1
• So the bond enthalpy of a C=C double bond as a whole will be (397+284) = 681 kJ mol^-1
• The bond enthalpy of a C-C single bond is 348 kJ mol^-1

---

Nomenclature of Alkenes

• We have seen the rules for writing the IUPAC names in an earlier chapter [see section 12.3]
• Let us recall some basic rules which are applicable to alkenes. It can be written in 3 steps:
1. The longest chain should be selected in such a way that, it contains the double bonds.
2. Numbering must be done in such a way that, the double bonds get the lowest possible numbers.
3. The suffix ‘ene’ is used instead of ‘ane’ of alkanes.
• Let us see some examples:

Example 1:
Write the IUPAC name of the structure: CH₃-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: propane – ane + ene = propene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the double bond.
    ♦ Numbering from left to right will give number ‘2’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Prop-1-ene.
7. Here we note an interesting point:
• In propene, the double bond has only two possible positions. Both those positions will give the same name: Prop-1-ene. So we can write the name simply as: Propene.

Example 2:
Write the IUPAC name of the structure: CH₃-CH₂-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the double bond.
    ♦ Numbering from left to right will give number ‘3’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-1-ene.

Example 3:
Write the IUPAC name of the structure: CH₃-CH=CH-CH₃
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give number ‘2’ to the double bond.
    ♦ Numbering from right to left will give number ‘2’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-2-ene.

Example 4:
Write the IUPAC name of the structure: CH₂=CH-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give numbers '1,3' to the double bonds.
    ♦ Numbering from right to left will give number ‘1,3’ to the double bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Buta-1,3-diene.
• Note that, since more than one double bond is present, we write 'buta' instead of 'but'
• Similarly, 'penta' is written instead of 'pent', 'hexa' is written instead of hex', so on . . .

Example 5:
Write the IUPAC name of the structure shown in fig.13.47(a) below:

Fig.13.47

Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain containing double bond.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: propane – ane + ene = propene
4. Applying rule 4, we see that, there is one branch: methyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘2’ to the double bond.
6. Applying rule 6, we get: 2-methyl.
7. Rule 7 is not applicable here because, there is only one branch.
8. Applying rule 8, we get: 2-Methylprop-1-ene.

Example 6:
Write the IUPAC name of the structure shown in fig.13.48(a) below:

Fig.13.48

Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there is one branch: methyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘3’ to the double bond.
6. Applying rule 6, we get: 3-methyl.
7. Rule 7 is not applicable here because, there is only one branch.
8. Applying rule 8, we get: 3-Methylbut-1-ene.

---

Now we will see some solved examples:

Solved example 13.7
Write the IUPAC names of the three compounds shown in fig.13.49 below:

Fig.13.49

Solution:
Part (i):
We are given the condensed formula. It is shown in fig.13.50(a) below:

Fig.13.50

• The expanded form is shown in fig.b. Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are ten C atoms in the main chain containing both the double bonds.
2. Applying rule 2, we see that, ‘dec’ must be used.
3. Applying rule 3, we get decane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: decane – ane + ene = decene
4. Applying rule 4, we see that, there are two branches: two methyl branches.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give numbers '3,6' to the double bonds.
    ♦ Numbering from right to left will give numbers ‘4,7’ to the double bond.
6. Applying rule 6, we get: 2-methyl and 8-methyl.
7. Applying rule 7, we get: 2,8-dimethyl.
8. Applying rule 8, we get: 2,8-Dimethyldeca-3,6-diene.

Part (ii):
We are given the bond line formula. It is shown in fig.13.51(a) below:

Fig.13.51

• The expanded form is shown in fig.b. Based on the expanded form, we can write 6 steps: 
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.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: octane – ane + ene = octene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left. This is shown in fig.b.
    ♦ Numbering from left to right will give numbers '1,3,5,7' to the double bonds.
    ♦ Numbering from right to left will give the same numbers ‘1,3,5,7’ to the double bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Octa-1,3,5,7-tetraene.

Part (iii):
We are given the condensed formula. It is shown in fig.13.52(a) below:

Fig.13.52

• The expanded form is shown in fig.b. Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are five C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘pent’ must be used.
3. Applying rule 3, we get pentane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: pentane – ane + ene = pentene
4. Applying rule 4, we see that, there is one branch: propyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘4’ to the double bond.
6. Applying rule 6, we get: 2-propyl.
7. Rule 7 is not applicable because, there is only one branch.
8. Applying rule 8, we get: 2-Propylpent-1-ene.

Part (iv):
We are given the expanded form. It is shown in fig.13.53(a) below:

Fig.13.53

• Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are ten C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘dec’ must be used.
3. Applying rule 3, we get decane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: decane – ane + ene = decene
4. Applying rule 4, we see that, there are three branches: two methyl branches and one ethyl branch.
5. Applying rule 5, we see that numbering should be done from right to left. This is shown in fig.b.
    ♦ Numbering from left to right will give number '6' to the double bond.
    ♦ Numbering from right to left will give number ‘4’ to the double bond.
6. Applying rule 6, we get: 2-methyl, 4-ethyl and 6-methyl.
7. Applying rule 7, we get: 4-ethyl-2,6-dimethyl.
(Note that, ethyl gets preference over methyl when we consider the alphabetical order)
8. Applying rule 8, we get: 4-Ethyl-2,6-dimethyldec-4-ene.

Solved example 13.8
Calculate the number of sigma (𝜎) and pi (π) bonds in the four structures in the above solved example 13.7
Solution:
Part (a):
1. From the expanded form in fig.13.50(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 31 Nos.
        ✰ Number of C-H single bonds = 22 Nos. 
        ✰ Number of C-C single bonds = 9 Nos. 
   ♦ Counting the number of double bonds, we get: 2 Nos. 
        ✰ Number of C=C double bonds = 2 Nos.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 22
    ♦ Number of $\sigma_{C-C}$ = 9
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are two double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 2
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =22
    ♦ $\sigma_{C-C}$ = 9
    ♦ $\sigma_{C=C}$ = 2
3. Number of π bonds can be calculated in 2 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the two double bonds in this structure are between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the two double bonds = 2
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 2

Part (b):
1. From the expanded form in fig.13.51(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 13 Nos.
        ✰ Number of C-H single bonds = 10 Nos. 
        ✰ Number of C-C single bonds = 3 Nos. 
   ♦ Counting the number of double bonds, we get: 4 Nos. 
        ✰ Number of C=C double bonds = 4 Nos.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 10
    ♦ Number of $\sigma_{C-C}$ = 3
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are four double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 4
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =10
    ♦ $\sigma_{C-C}$ = 3
    ♦ $\sigma_{C=C}$ = 4
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the four double bonds in this structure are between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the four double bonds = 4
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 4

Part (c):
1. From the expanded form in fig.13.52(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 22 Nos.
        ✰ Number of C-H single bonds = 16 Nos. 
        ✰ Number of C-C single bonds = 6 Nos. 
   ♦ Counting the number of double bonds, we get: 1 Nos. 
        ✰ Number of C=C double bonds = 1 No.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 16
    ♦ Number of $\sigma_{C-C}$ = 6
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =16
    ♦ $\sigma_{C-C}$ = 6
    ♦ $\sigma_{C=C}$ = 1
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the one double bond in this structure is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bonds = 1
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1

Part (d):
1. From the expanded form in fig.13.53, we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 13 Nos.
        ✰ Number of C-H single bonds = 28 Nos. 
        ✰ Number of C-C single bonds = 12 Nos. 
   ♦ Counting the number of double bonds, we get: 1 No. 
        ✰ Number of C=C double bonds = 1 No.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 28
    ♦ Number of $\sigma_{C-C}$ = 12
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =28
    ♦ $\sigma_{C-C}$ = 12
    ♦ $\sigma_{C=C}$ = 1
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the one double bond in this structure is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bonds = 1
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1

---

In the next section we will see isomerism in alkenes.

Previous

Contents

Next

Copyright©2022 Higher secondary chemistry.blogspot.com

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.

---

• 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.

---

In the next section we will see preparation of alkanes.

Previous

Contents

Next

Copyright©2022 Higher secondary chemistry.blogspot.com