In the previous section, we saw the details about Isomerism. In this section, we will see Isomerism in Alicyclic compounds.
• We have seen three types of structures for hydrocarbons:
(i) Straight chain (ii) Branched chain (iii) Cyclic
• We have seen some basic details about cyclic compounds in our earlier classes. (Details here)
• Now we will analyse them in greater detail. We will do the analysis by taking Cyclopentane as an example:
1. Consider the chain structure in fig.14.94(a) below:
• We know that it is the structure of pentane
♦ All the carbon-carbon bonds are single bonds
♦ 3 hydrogen atoms are attached to the carbon atom at the left end
♦ 3 hydrogen atoms are attached to the carbon atom at the right end also
♦ All interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.94(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.95 below
• Fig.14.95(a) is the same fig.14.94(b) that we saw above.
4. See the completed ring in fig.14.95(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
♦ The green dot which was in bond with that red dot is now free.
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.95(b) is Cyclopentane
5. Let us write the molecular formulae:
Pentane: C5H12
Cyclopentane: C5H10
• So the cyclopentane has two hydrogen atoms less than the pentane
■ This is true for all alkanes:
The number of hydrogen atoms in any cycloalkane will be 2 less than the corresponding alkane
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkene will be 2 less than the corresponding alkane
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkane is same as that in the corresponding alkene
This gives an interesting result:
■ An alkene and it's corresponding cycloalkane are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentane
♦ Corresponding alkane is: Pentane
♦ Corresponding alkene is: Pentene
♦ Cyclopentane and Pentene will be isomers
Another example:
• If the cyclic compound is: Cyclopropane
♦ Corresponding alkane is: Propane
♦ Corresponding alkene is: Propene
♦ Cyclopropane and Propene will be isomers
• Thus we saw the relation between a cycloalkane and it's corresponding alkene. For this result, we started out with a cycloalkane.
• Now we will start out another discussion with a cycloalkene. Let us see where we will reach:
1. Consider the chain structure in fig.14.96(a) below:
• We know that it is the structure of pentene
♦ One of the carbon-carbon bonds is a double bond
♦ 3 hydrogen atoms are attached to the carbon atom at the left end
♦ 2 hydrogen atoms are attached to the carbon atom at the right end
♦ 1 hydrogen atom is attached to the second carbon atom from the right end
♦ All other interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.96(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.97 below:
• Fig.14.97(a) is the same fig.14.96(b) that we saw above.
4. See the completed ring in fig.14.97(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
♦ The green dot which was in bond with that red dot is now free.
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.97(b) is Cyclopentene
5. Let us write the molecular formulae:
• Pentane: C5H12
♦ Cyclopentane: C5H10
• Pentene: C5H10
♦ Cyclopentene: C5H8
• So the cyclopentene has two hydrogen atoms less than the pentene
■ This is true for all alkenes:
The number of hydrogen atoms in any cycloalkene will be 2 less than the corresponding alkene
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkyne will be 2 less than the corresponding alkene
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkene is same as that in the corresponding alkyne
This gives an interesting result:
■ An alkyne and it's corresponding cycloalkene are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentene
♦ Corresponding alkane is: Pentane
♦ Corresponding alkene is: Pentene
♦ Corresponding alkyne is: Pentyne
♦ Cyclopentene and Pentyne will be isomers
Another example:
• If the cyclic compound is: Cyclopropene
♦ Corresponding alkane is: Propane
♦ Corresponding alkene is: Propene
♦ Corresponding alkyne is: Propyne
♦ Cyclopropene and Propyne will be isomers
• Thus we saw the relation between a cycloalkene and it's corresponding alkyne.
■ We can show the above findings in a simple fig:
• The green arrow indicates that Alkenes and Cycloalkanes can be isomers
• The yellow arrow indicates that Alkynes and Cycloalkenes can be isomers
■ In this section, we will be seeing problems related to the green arrow only
But before we see those problems, a more interesting result related to the green arrow is coming up:
1. We know that the pentene itself has isomers. This is based on the position of it's double bond. We saw details here.
Thus we have:
Pent-1-ene: CH3ㅡCH2 ㅡCH2 ㅡCH＝CH2
Pent-2-ene: CH3ㅡCH2 ㅡCH＝CHㅡCH3
2. The three compounds: cyclopentane, pent-1-ene and pent-2-ene have the same molecular formula C5H10.
• We can write:
The three compounds: cyclopentane, pent-1-ene and pent-2-ene are isomers
• pent-1-ene and pent-1-ene are chain isomers
3. For higher alkenes like octene, decene etc., more chain isomers are possible.
• All those isomers will be isomers of the corresponding cycloalkanes (cyclooctane, cyclodecane etc.,) also.
Solved example 14.21
Write the IUPAC names and structures of all possible isomers of the compound given below:
CH3ㅡCH2 ㅡCH＝CH2
Solution:
1. The given compound is But-1-ene
It has two possible isomers
(i) The chain isomer: But-2-ene
• It is produced by the change in the position of the double bond
• It's structure is: CH3ㅡCH＝CHㅡCH3.
(ii) The corresponding cycloalkane: Cyclobutane
• It's structure is shown in the fig.14.99(a) below:
2. So we have three compounds:
But-1-ene, But-2-ene and Cyclobutane
• All three of them have the same molecular formula: C4H8.
• All three are isomers.
Solved example 14.22
Write the IUPAC names and structures of all possible isomers of the compound shown in fig.14.99(b) above.
Solution:
1. The given compound is Cyclohexane
It has three possible isomers
(i) Hex-1-ene
• It's structure is: CH3ㅡCH2ㅡCH2ㅡCH2 ㅡCH＝CH2
(ii) Hex-2-ene
• It's structure is: CH3ㅡCH2ㅡCH2ㅡCH＝CHㅡCH3
(iii) Hex-3-ene
• It's structure is: CH3ㅡCH2ㅡCH＝CHㅡCH2ㅡCH3
2. So we have four compounds:
Hex-1-ene, Hex-2-ene, Hex-3-ene and Cyclohexane
• All four of them have the same molecular formula: C6H12
• All four are isomers
In the next section, we will see Reactions of Organic compounds
• We have seen three types of structures for hydrocarbons:
(i) Straight chain (ii) Branched chain (iii) Cyclic
• We have seen some basic details about cyclic compounds in our earlier classes. (Details here)
• Now we will analyse them in greater detail. We will do the analysis by taking Cyclopentane as an example:
1. Consider the chain structure in fig.14.94(a) below:
 |
| Fig.14.94 |
♦ All the carbon-carbon bonds are single bonds
♦ 3 hydrogen atoms are attached to the carbon atom at the left end
♦ 3 hydrogen atoms are attached to the carbon atom at the right end also
♦ All interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.94(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.95 below
 |
| Fig.14.95 |
4. See the completed ring in fig.14.95(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
♦ The green dot which was in bond with that red dot is now free.
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.95(b) is Cyclopentane
5. Let us write the molecular formulae:
Pentane: C5H12
Cyclopentane: C5H10
• So the cyclopentane has two hydrogen atoms less than the pentane
■ This is true for all alkanes:
The number of hydrogen atoms in any cycloalkane will be 2 less than the corresponding alkane
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkene will be 2 less than the corresponding alkane
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkane is same as that in the corresponding alkene
This gives an interesting result:
■ An alkene and it's corresponding cycloalkane are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentane
♦ Corresponding alkane is: Pentane
♦ Corresponding alkene is: Pentene
♦ Cyclopentane and Pentene will be isomers
Another example:
• If the cyclic compound is: Cyclopropane
♦ Corresponding alkane is: Propane
♦ Corresponding alkene is: Propene
♦ Cyclopropane and Propene will be isomers
• Thus we saw the relation between a cycloalkane and it's corresponding alkene. For this result, we started out with a cycloalkane.
• Now we will start out another discussion with a cycloalkene. Let us see where we will reach:
1. Consider the chain structure in fig.14.96(a) below:
 |
| Fig.14.96 |
♦ One of the carbon-carbon bonds is a double bond
♦ 3 hydrogen atoms are attached to the carbon atom at the left end
♦ 2 hydrogen atoms are attached to the carbon atom at the right end
♦ 1 hydrogen atom is attached to the second carbon atom from the right end
♦ All other interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.96(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.97 below:
 |
| Fig.14.97 |
4. See the completed ring in fig.14.97(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
♦ The green dot which was in bond with that red dot is now free.
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.97(b) is Cyclopentene
5. Let us write the molecular formulae:
• Pentane: C5H12
♦ Cyclopentane: C5H10
• Pentene: C5H10
♦ Cyclopentene: C5H8
• So the cyclopentene has two hydrogen atoms less than the pentene
■ This is true for all alkenes:
The number of hydrogen atoms in any cycloalkene will be 2 less than the corresponding alkene
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkyne will be 2 less than the corresponding alkene
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkene is same as that in the corresponding alkyne
This gives an interesting result:
■ An alkyne and it's corresponding cycloalkene are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentene
♦ Corresponding alkane is: Pentane
♦ Corresponding alkene is: Pentene
♦ Corresponding alkyne is: Pentyne
♦ Cyclopentene and Pentyne will be isomers
Another example:
• If the cyclic compound is: Cyclopropene
♦ Corresponding alkane is: Propane
♦ Corresponding alkene is: Propene
♦ Corresponding alkyne is: Propyne
♦ Cyclopropene and Propyne will be isomers
• Thus we saw the relation between a cycloalkene and it's corresponding alkyne.
■ We can show the above findings in a simple fig:
 |
| Fig.14.98 |
• The yellow arrow indicates that Alkynes and Cycloalkenes can be isomers
■ In this section, we will be seeing problems related to the green arrow only
But before we see those problems, a more interesting result related to the green arrow is coming up:
1. We know that the pentene itself has isomers. This is based on the position of it's double bond. We saw details here.
Thus we have:
Pent-1-ene: CH3ㅡCH2 ㅡCH2 ㅡCH＝CH2
Pent-2-ene: CH3ㅡCH2 ㅡCH＝CHㅡCH3
2. The three compounds: cyclopentane, pent-1-ene and pent-2-ene have the same molecular formula C5H10.
• We can write:
The three compounds: cyclopentane, pent-1-ene and pent-2-ene are isomers
• pent-1-ene and pent-1-ene are chain isomers
3. For higher alkenes like octene, decene etc., more chain isomers are possible.
• All those isomers will be isomers of the corresponding cycloalkanes (cyclooctane, cyclodecane etc.,) also.
Solved example 14.21
Write the IUPAC names and structures of all possible isomers of the compound given below:
CH3ㅡCH2 ㅡCH＝CH2
Solution:
1. The given compound is But-1-ene
It has two possible isomers
(i) The chain isomer: But-2-ene
• It is produced by the change in the position of the double bond
• It's structure is: CH3ㅡCH＝CHㅡCH3.
(ii) The corresponding cycloalkane: Cyclobutane
• It's structure is shown in the fig.14.99(a) below:
 |
| Fig.14.99 |
But-1-ene, But-2-ene and Cyclobutane
• All three of them have the same molecular formula: C4H8.
• All three are isomers.
Solved example 14.22
Write the IUPAC names and structures of all possible isomers of the compound shown in fig.14.99(b) above.
Solution:
1. The given compound is Cyclohexane
It has three possible isomers
(i) Hex-1-ene
• It's structure is: CH3ㅡCH2ㅡCH2ㅡCH2 ㅡCH＝CH2
(ii) Hex-2-ene
• It's structure is: CH3ㅡCH2ㅡCH2ㅡCH＝CHㅡCH3
• It's structure is: CH3ㅡCH2ㅡCH＝CHㅡCH2ㅡCH3
2. So we have four compounds:
Hex-1-ene, Hex-2-ene, Hex-3-ene and Cyclohexane
• All four of them have the same molecular formula: C6H12
• All four are isomers
In the next section, we will see Reactions of Organic compounds
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