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.

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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.

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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 . . .

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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

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In the next section we will see preparation of alkynes.

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Chapter 13 - Hydrocarbons

In the previous section, we completed a discussion on some basic principles and techniques in organic chemistry. In this chapter, we will see hydrocarbons.

Some basics about hydrocarbons can be written in 7 steps:
1. Consider LPG that we use commonly for cooking purposes. LPG is the abbreviated form of Liquefied Petroleum Gas.
• Let us see some basic details about LPG. It can be written in 4 steps:
(i) Crude oil is a thick black liquid that occurs naturally at many places below the earth’s crust. This oil is refined to obtain useful products like petrol, kerosene etc.,
(ii) Petroleum gas is obtained during this refining process. This gas is liquefied by applying pressure and lowering temperature.
(iii) In the liquid state, this fuel can be easily stored and transported in cylinders and tanks. Thus it becomes available to us for cooking purposes.
(iv) When LPG burns, pollution occurs only to a very small extent.
2. CNG is another fuel. It is used in automobiles. CNG is the abbreviated form of Compressed Natural Gas.
• Let us see some basic details about CNG. It can be written in 3 steps:
(i) Natural gas is a gas which forms naturally above crude oil deposits.
(ii) This gas is collected and compressed to reduce volume. It can then be transported in cylinders, tanks or through pipe lines.
(iii) Like LPG, the CNG also causes pollution only to a very small extent.
3. Other fuels like petrol, diesel and kerosene are already familiar to us.
4. All these fuels are mixtures of various hydrocarbons.
◼ Hydrocarbons are compounds containing only hydrogen and carbon.
5. We know that hydrogen can undergo combustion in the presence of oxygen.
• The balanced equation is:
2H₂ (g) + O₂ (g) ⟶ 2H₂O(g)
6. Similarly, carbon can also undergo combustion in the presence of oxygen.
• The balanced equation is:
C (s) + O₂ (g) ⟶ CO₂ (g)
7. Hydrocarbons are compounds containing both hydrogen and carbon. So they are good fuels.
• For example, methane can undergo combustion in the presence of oxygen. The balanced equation is:
CH₄ (g) + 2O₂ (g) ⟶ CO₂ (g) + 2H₂O (g)
• The main components of LPG are propane and butane.       
• The main component of CNG is methane.
• Petrol, diesel and kerosene contain mixtures of various hydrocarbons.

Classification of hydrocarbons

• Some basics about classification can be written in 6 steps:
1. Since there are a very large number of hydrocarbons, it is essential to classify them into various categories.
• The three main categories are:
(i) Saturated hydrocarbons
(ii) Unsaturated hydrocarbons
(iii) Aromatic hydrocarbons.
2. We know that, carbon is tetravalant. That is., it needs four electrons to attain stability.
• Those four electrons can be obtained through single bonds, double bonds or triple bonds.
3. In saturated hydrocarbons, all bonds will be single bonds.
• Saturated hydrocarbons can be further classified as alkanes and cycloalkanes.
• In alkanes, the first and last C atoms in the chain will not be bonded together. So they do not form closed chains. They are open chain hydrocarbons.   
• In cycloalkanes, the first and last C atoms in the chain will be bonded together. So they form closed chains.   
4. In unsaturated hydrocarbons, one or more bonds will be double or triple bonds.
• If one or more bonds are double bonds, it is classified as an alkene.
• If one or more bonds are triple bonds, it is classified as an alkyne.
• Like cycloalkanes, cycloalkenes and cycloalkynes are also possible.
5. So the classification of hydrocarbons can be represented as in fig.13.1 below:

Fig.13.1

6. Aromatic hydrocarbons are a special type of cyclic compounds. The name is derived from the fact that, they have a pleasant odour.
• They are also known as arenes.

Alkanes

• Some basics about alkanes can be written in 13 steps:
1. All bonds in alkanes are single bonds.
2. The first member of the alkane family is methane (CH₄).
3. If we remove one H atom from CH₄, and put a C atom in it’s place, we get the second member.
• But the valencies of the new C atom should also be satisfied with H atoms. So the second member is C₂H₆
4. If we remove one H atom from C₂H₆, and put a C atom in it’s place, we get the third member.
• But the valencies of the new C atom should also be satisfied with H atoms. So the third member is C₃H₈
5. In this way we can obtain a large number of members. We can write:
• Each member is obtained by two steps:
(i) Remove an H atom from the preceding member.
(ii) Put a -CH₃ group in the place of that H atom.
6. We see that:
    ♦ Each member has one C atom more than it’s preceding member.
    ♦ Also each member has two H atoms more than it’s preceding member.
7. The general formula of this homologous series is: C_nH_2n+2
• n is the number of C atoms.
• If we know the value of n, we can easily calculate the number of H atoms.
8. The first member methane has the simplest structure among all alkanes. It is a tetrahedral structure. We saw the details in an earlier chapter. [see section 4.25]
• Fig.13.2 (a) below shows the ball and stick model of methane.

Fig.13.2

Some features of this model can be written in 8 steps:
(i) The central pink sphere represents the C atom.
(ii) The four orange spheres represent the four H atoms.
(iii) The sticks are painted with two different colors.
    ♦ The pink color indicates that, one electron in the bond belongs to C.
    ♦ The orange color indicates that, the other electron in the bond belongs to H.
(iv) The red arrow represents x-axis.
(v) The green arrow represents y-axis.
(vi) The C atom is situated at the origin.
(vii) Consider the stick between the top H atom and the C atom. This stick is perpendicular to the xy-plane.
(viii) If we rotate this CH₄ molecule through 90^o about the y-axis, the top H atom and it's stick will become aligned with the x-axis.
• This rotation is indicated by the yellow curved arrow.
• The resulting orientation obtained after rotation, is shown in fig.b. It is a rotated tetrahedral structure.
9. Now we can write about the structure of the second member ethane (C₂H₆). It can be written in 4 steps:
(i) Consider the orientation of tetrahedral CH₄ molecule in fig.13.2(b) above.
• Two such tetrahedra are lying along the x-axis and are facing each other in fig.13.3(a) below:

Fig.13.3

(ii) Consider the tetrahedron on the left side. One of it's H atoms lies on the x-axis. This H atom is removed. It then becomes -CH₃
(iii) Consider the tetrahedron on the right side. One of it's H atoms lies on the x-axis. This H atom is removed. It then becomes -CH₃
(iv) Removal of H atoms will make the molecules unstable. But the two -CH₃ can combine together by making a direct bond between the two C atoms. This is shown in fig.18.3(b). It is a C₂H₆ molecule.
• We can write:
The ethane molecule contains two tetrahedral structures.
10. Now we can write about the structure of the third member propane (C₃H₈). It can be written in 5 steps:
(i) Consider the orientation of tetrahedral CH₄ molecule in fig.13.2(b) above.
• Two such tetrahedra are lying along the x-axis and are facing each other in fig.13.4(a) below. Also a third tetrahedron is approaching from the +ve side of the y-axis.

Fig.13.4

(ii) Consider the tetrahedron on the left side. One of it's H atoms lies on the x-axis. This H atom is removed. It then becomes -CH₃
(iii) Consider the tetrahedron on the right side. One of it's H atoms lies on the x-axis. This H atom is removed. It then becomes -CH₃
(iv) Consider the third tetrahedron which approaches from the +ve side of the y-axis. Two of it's H atoms are removed. It then becomes -CH₂
(iv) Removal of H atoms will make the molecules unstable. But the two -CH₃ can combine with the -CH₂ by making direct bonds between the three C atoms. This is shown in fig.18.4(b). It is a C₃H₈ molecule.
• We can write:
The propane molecule contains three tetrahedral structures.
(v) It may be noted that, the three C atoms do not fall along a line.
11. In this way, tetrahedra are joined together to form members of the alkane series.
• The 3D model of butane is shown in fig.13.5 below.
• We can see that:
    ♦ One -CH₃ tetrahedron is present at each end of the chain.
    ♦ Two -CH₂ tetrahedra are present in the middle.

Fig.13.5

By Ben Mills and Jynto

The original file can be seen here.

12. Scientists have determined the bond lengths in alkanes,
    ♦ The C-C bond length is 154 pm
    ♦ The C-H bond length is 112 pm
13. It is important to remember that, in alkanes, there are no π bonds.
    ♦ All C-C bonds are 𝜎 bonds.
    ♦ All C-H bonds are 𝜎 bonds.
• We saw those details in the previous chapter.

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In the next section we will see nomenclature of alkanes.

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