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 CnH2n+2
• We also know that, the corresponding alkene has two H atoms less. So the general formula for alkenes is CnH2n
3. Chemists in the 19th 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 121.7o
• The bond angle between the two C-H single bonds in a CH2 group is 116.6o
• 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 |
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.
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