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 CnH2n
• We also know that, the corresponding alkyne has two H atoms less. So the general formula for alkenes is CnH2n-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.
♦ 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.
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