Chapter 12.1 - Structural Representations of Organic Compounds

In the previous section, we saw some basic details about sigma and pi bonds in organic compounds. In this section we will see structural representations of organic compounds.

Readers are advised to refresh their knowledge acquired about organic compounds from earlier classes. Those notes are available here.

Given below are the different methods by which we can represent organic compounds:
A. Complete structural formula
B. Condensed structural formula
C. Bond-line formula

We will now see each of them in detail
A. Complete structural formula
• This can be explained in 5 steps:
1. We have seen Lewis structures in an earlier chapter. It is very easy to draw Lewis structures of organic compounds. It will give all the details about how the electrons are shared between various atoms in the molecule.
2. We know that in Lewis structures, electrons are represented using dots.
• However, when we use Lewis structures for organic molecules, dots can be avoided.
   ♦ If one pair of electrons are shared, we use a ‘-’   
   ♦ If two pairs of electrons are shared, we use a ‘＝’   
   ♦ If three pairs of electrons are shared, we use a ‘☰’
3. Recall that, electrons that do not take part in bonding are called lone pairs. In Lewis structures, we show them using pairs of dots.
• However, in the case of organic molecules, it is not compulsory to show them. They can be avoided if they do not have any role in the problem under consideration.
4. Based on the above steps, ethane, ethene, ethyne and methanol can be represented as shown in figs. 12.5 (a), (b), (c) and (d) respectively.

FIG.12.5

• Recall that O has 6 electrons in the valence shell.
• In methanol, two of those electrons will be used for forming the single bonds.
• So 4 electrons remain. They do not take part in bonding. They are shown as two lone pairs.
5. Such structural representations are called complete structural formulas.

B. Condensed structural formula
This can be explained in steps:
1. The complete structural formula can be abbreviated. For that, we make three modifications.
(i) Bonds to be retained:
   ♦ We retain the ‘-’ (single bonds) between C atoms
   ♦ We retain the ‘=’ (double bonds) between C atoms
   ♦ We retain the ‘☰’ (triple bonds) between C atoms
(ii) Bonds to be discarded:
We discard the ‘-’ (single bonds) between C and H
(iii) Grouping of identical atoms
   ♦ For each C atom, we group together identical atoms attached to that C.
   ♦ Number of atoms in such groups are indicated by subscripts
2. The expression obtained in this way is called condensed structural formula.
3. The condensed structural formula of ethane, ethene, ethyne and methanol are given below:
Ethane: H₃C-CH₃
Ethene: H₂C=CH₂
Ethyne: HC☰CH
Methanol: CH₃OH
4. Another example:
• Consider the molecule:
H₃C-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₃
• It is already in the condensed form. But it can be further condensed into the form:
CH₃(CH₂)₆CH₃

C. Bond-line formula
This method is used to further abbreviate the condensed formula. In this method, we see a large number of lines. Letters C and H are not written. Letters like O, N, OH, Br etc., will be seen rarely. It can be explained using some examples.
Example 1:
This can be written in 4 steps:
1. Consider the complete structural formula shown in fig.12.6(a) below:

Fig.12.6

• It can be converted into a condensed structural formula. This is shown in fig.b
• Now, the first step to write the bond-line formula, is to write the condensed formula in a 'zigzag manner' as shown in fig.c
2. The next step is to remove all C and H. We will get fig.d
• Fig. d is the bond-line formula of the molecule in fig.a
3. But how can we be sure that fig.d and fig.a represent the same molecule?
• To realize both to be the same, we must remember certain rules:
Rule (i): This rule is related to the junction between any two zigzag lines.
• At any junction between any two zigzag lines, there will be both C and H
   ♦ Number of C will be 1
   ♦ Number of H will be that required to satisfy the valency of that C.
• Consider the junction number 6 in fig.e
   ♦ As per rule, there will be one C at that junction.
   ♦ Two lines meet at that junction.
         ✰ That means, two valencies of the C at that junction are already satisfied.
   ♦ For the remaining two valencies, we provide two H atoms.
   ♦ So there will be one CH₂ at that junction.
• Consider the junction number 3 in fig.e
   ♦ As per rule, there will be one C at that junction.
   ♦ Three lines meet at that junction.
         ✰ That means, three valencies of the C at that junction are already satisfied.
   ♦ For the remaining one valency, we provide one H atom.
   ♦ So there will be one CH at that junction.
Rule (ii): This rule is related to the terminal points.
• At the terminal points, there will be a CH₃ (methyl group).
• For example, consider the terminal point 1 in fig.d
   ♦ As per rule, there will be one C at that point.
   ♦ One line start from that point.
         ✰ That means, one valency of the C at that junction is already satisfied.
   ♦ For the remaining three valencies, we provide three H atoms.
• So there will be one CH₃ at that point.
• The same is applicable to the terminal point 8.
   ♦ That means, there will be a CH₃ at the terminal point 8.
• The same is applicable to the terminal point 3'.
   ♦ That means, there will be a CH₃ at the terminal point 3'.
4. So it is clear that, fig.d and fig.a represent the same molecule.

Example 2:
This can be written in 3 steps:
1. Consider the complete structural formula shown in fig.12.7(a) below:

Fig.12.7

• It can be converted into a condensed structural formula. This is shown in fig.b
• Now, the first step to write the bond-line formula, is to write the condensed formula in a 'zigzag manner' as shown in fig.c
2. The next step is to remove all C and H. We will get fig.d
• Fig. d is the bond-line formula of the molecule in fig.a
3. But how can we be sure that fig.d and fig.a represent the same molecule?
• To realize both to be the same, we must remember certain rules:
Rule (i): This rule is related to the junction between any two zigzag lines.
• We have already seen the rule in example 1. We can readily apply it for our present example.
• We will get CH₂ at junction 3.
    ♦ Because, two lines meet at this junction.
• We will get CH at junction 2.
    ♦ Because, three lines meet at this junction.
Rule (ii): This rule is related to the terminal points.
• We have already seen the rule in example 1. We can readily apply it for our present example. We will get CH₃ at the terminal points 1 and 4.
Rule (iii): This rule is applicable when functional groups like Br, Cl, OH, COOH etc., are present at the terminal points.
• In our example, Br is present at the terminal point 2'
• We would expect a CH₃ at a terminal point. But since it is written specifically as 'Br', we must avoid CH₃ and use Br in it's place.
• Fig.a shows how the Br satisfies the valency of the C at junction 2

◼ In general we can write: (i) At all junctions in a bond-line formula, • There will be one C atom • H atoms should be attached to that C based on the number of lines meeting at that junction. (ii) At all terminal points in a bond line formula, • There will be a CH3 (unless there is a functional group).

Example 3:
In fig.12.8(a) below, a bond line formula is given.

Fig.12.8

• We want to convert it into a complete structural formula. It can be done in 3 steps:
1. Consider the top most junction
• There will be one C at this junction.
• Two lines meet at this junction. So two valencies of that C are satisfied.
• To satisfy the remaining two valencies, we provide two H atoms.
• So we have CH₂ at the top most junction.
2. The remaining two junctions in the given bond-line formula are similar to the top most junction. Because, two line meet at both those junctions.
• So we can write:
This is a molecule with CH₂ at all the junctions.
• Based on the above information, we can draw the condensed formula as shown in fig.b
3. Based on the condensed formula, we can draw the complete structural formula as shown in fig.c

Example 4:
In fig.12.9(a) below, a bond line formula is given.

Fig.12.9

• We want to convert it into a complete structural formula. It can be done in 3 steps:
1. Consider the top most junction
• There will be one C at this junction.
• Two lines meet at this junction. So two valencies of that C are satisfied.
• To satisfy the remaining two valencies, we provide two H atoms.
• So we have CH₂ at the top most junction.
2. The remaining four junctions in the given bond-line formula are similar to the top most junction. Because, two line meet at each of those junctions.
• So we can write:
This is a molecule with CH₂ at all the junctions.
• Based on the above information, we can draw the condensed formula as shown in fig.b
3. Based on the condensed formula, we can draw the complete structural formula as shown in fig.c

Example 5:
In fig.12.10(a) below, a bond line formula is given.

Fig.12.10

• We want to convert it into a complete structural formula. It can be done in 5 steps:
1. There is a total of six junctions.
• From one of those six junctions, a line is projecting out. We will consider this junction in a later step (4).
2. First consider any one of the remaining 5 junctions
• There will be one C at that junction.
• Two lines meet at that junction. So two valencies of that C are satisfied.
• To satisfy the remaining two valencies, we provide two H atoms.
• So we have CH₂ at that junction.
3. The five junctions are similar. Because, two line meet at each of those junctions.
• So we can write:
This is a molecule with CH₂ at five junctions.
4. Now we consider the sixth junction.
• There will be one C at that junction.
• Three lines meet at that junction.
    ♦ One of those three lines carry a Cl.
    ♦ That means, one of those three valencies are satisfied by one Cl
    ♦ The other two are satisfied by the adjacent C atoms
    ♦ Thus three valencies of that C are satisfied.
• To satisfy the remaining one valency, we provide an H atom.
• So we have CHCl at that junction.
• Based on the above information, we can draw the condensed formula as shown in fig.b
5. Based on the condensed formula, we can draw the complete structural formula as shown in fig.c

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Now we have a basic idea about Complete structural formula, Condensed structural formula and Bond-line formula. Let us see some solved examples.

Solved example 12.4
Expand each of the following condensed formulas into their complete structural formulas.
(a) CH₃CH₂COCH₂CH₃
(b) CH₃CH=CH(CH₂)₃CH₃
Solution:
The required complete structural formulas are shown in figs.12.11 (a) and (b) respectively.

Fig.12.11

 

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In the next section, we will see a few more solved examples.


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