Chapter 12 - Basic Principles in Organic Chemistry

In this chapter, we will see some basic principles and techniques in organic chemistry.

Catenation

• Carbon exhibits the property of catenation.
• It is the property by which an element can bind with itself through covalent bonds to form long chains, branched chains or rings.
   ♦ Due to this property, carbon can form a large number of compounds.
• Since the number of such compounds is large, they are studied under a separate branch of chemistry called, organic chemistry.

Relation between structure of an organic molecule and it’s chemical properties

This can be explained in 5 steps:
1. We have seen that, the C atom in CH₄ is sp³ hybridized. (Fig.4.136 of section 4.25)
• There will be four orbitals. Those four orbitals will be directed towards the four corners of a tetrahedron.
2. In C₂H₄, the C atoms are sp² hybridized. (Fig.4.149 of section 4.26)
• The sp² hybrid orbitals have greater s character than sp³ hybrid orbitals.
3. Similarly, the sp hybrid orbitals of C₂H₂ have still more s character. (Fig.4.157 of section 4.27)
4. This greater or lesser s character will influence the chemical and physical properties of molecules. We will see more details in later sections.
5. Another aspect is the presence of π bonds.
• We have seen that in C₂H₄, all atoms lie on the same plane. But the p orbitals are oriented out of plane in the perpendicular direction to the plane. (Fig.4.150 of section 4.26)
• These perpendicular electron clouds are easily available for the attacking reagents. This will influence the general chemical properties of C₂H₄.

As we proceed with our discussion, we will see more examples.

Let us see a solved example

Solved example 12.1
How many σ and π bonds are present in (a) HC☰CCH＝CH₃ (b) CH₂＝C＝CHCH₃
Solution:
Part (a):
1. We are given the condensed formula. Based on that, we will draw the structural formula. In the structural formula, all the single, double and triple bonds will become visible. It is shown in fig.12.1(a) below. Note that, we will learn the details about how to draw those structural formulae, in later sections of this chapter.

Fig.12.1

2. Number of bonds:
   ♦ Counting the number of single bonds, we get: 8 Nos.
        ✰ Number of C-H single bonds = 6 Nos. 
        ✰ Number of C-C single bonds = 2 Nos. 
   ♦ Counting the number of double bonds, we get: 1 No. 
        ✰ Number of C=C double bonds = 1 No. 
   ♦ Counting the number of triple bonds, we get: 1 No. 
        ✰ Number of C☰C triple bonds = 1 No. 
3. Number of σ bonds can be calculated in 4 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 6
    ♦ Number of $\sigma_{C-C}$ = 2
(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) We know that, all triple bonds will be have one σ bond each.
• The triple bond is between two C atoms. In our present case, there is one triple bond.
• So we can write:
    ♦ Number of $\sigma_{C☰C}$ = 1
(iv) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ = 6
    ♦ $\sigma_{C-C}$ = 2
    ♦ $\sigma_{C=C}$ = 1
    ♦ $\sigma_{C☰C}$ = 1
4. 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 only one double bond in this molecule is between two C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bond = 1
(ii) We know that, all triple bonds will be have two π bonds each.
• We see that the only one triple bond in this molecule is between two C atoms.
    ♦ So the number of $\pi_{C☰C}$ from the one double bond = 2
(iii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1
    ♦ $\pi_{C☰C}$ = 2

Part (b):
1. We are given the condensed formula. Based on that, we will draw the structural formula. In the structural formula, all the single, double and triple bonds will become visible. It is shown in fig.12.1(b) above.
2. Number of bonds:
   ♦ Counting the number of single bonds, we get: 7 Nos.
        ✰ Number of C-H single bonds = 6 Nos. 
        ✰ Number of C-C single bonds = 1 Nos. 
   ♦ Counting the number of double bonds, we get: 2 No. 
        ✰ Number of C=C double bonds = 2 No. 
3. Number of σ bonds can be calculated in 4 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 6
    ♦ Number of $\sigma_{C-C}$ = 1
(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:
    ♦ $\sigma_{C-H}$ = 6
    ♦ $\sigma_{C-C}$ = 1
    ♦ $\sigma_{C=C}$ = 2
4. 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 two double bonds in this molecule is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the two double bonds = 2

Solved example 12.2
What is the type of hybridization of each carbon atom in the following compounds?
(a) CH₃Cl,   (b) (CH₃)₂CO,  (c) CH₃CN,  (d) HCONH₂,  (e) CH₃CH=CHCN
Solution:
• We are given the condensed formulae. Based on that, we will draw the structural formulae. In the structural formulae, all the single, double and triple bonds will become visible. They are shown in figs.12.2 and 12.3 below. Note that, we will learn the details about how to draw those structural formulae, in later sections of this chapter.

Fig.12.2

Part (a):
• Consider the C atom in fig.12.2(a) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The Cl and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the C atom is sp³

Part (b):
1. Consider the left side C atom in fig.12.2(b) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The middle C and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the left side C atom is sp³
2. Consider the middle side C atom in fig.12.2(b) above.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The left and right C atoms will enter into σ bonding with two of the three sp² hybrid orbitals.
• The O atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with O. This creates the double bond with O.
• So we can write:
The type of hybridization of the middle C atom is sp²
3. Consider the right side C atom.
• By symmetry, we can write the same steps as in (1).
• We can write:
The type of hybridization of the right side C atom is sp³

Part (c):
1. Consider the left side C atom in fig.12.2(c) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The right side C and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the left side C atom is sp³.
2. Consider the right side C atom in fig.12.2(c) above.
• There is one single bond and one triple bond around it. This is possible only if there are two orbitals around that C atom.
• Only sp hybridization can give two orbitals. The left C atom will enter into σ bonding with one of the two sp hybrid orbitals.
• The N atom will enter into σ bonding with the other sp hybrid orbital. Also, the p orbitals of the C atom will enter into two π bonding s with N. This creates the triple bond with N.
• So we can write:
The type of hybridization of right side C atom is sp.

Fig.12.3

Part (d):
• Consider the C atom in fig.12.3(d) above.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The left H atom and right NH₂ group will enter into σ bonding with two of the three sp² hybrid orbitals.
• The O atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with O. This creates the double bond with O.
• So we can write:
The type of hybridization of the C atom is sp²

Part (e):
1. Consider the first (from left) C atom in fig.12.3(e) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The second C atom and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the left side C atom is sp³
2. Consider the second C atom.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The first C atom and the H atom will enter into σ bonding with two of the three sp² hybrid orbitals.
• The third C atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with the third C. This creates the double bond with the third C.
• So we can write:
The type of hybridization of the second C atom is sp²
3. Consider the third C atom.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The fourth C atom and the H atom will enter into σ bonding with two of the three sp² hybrid orbitals.
• The second C atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with the second C. This creates the double bond with the second C.
• So we can write:
The type of hybridization of the third C atom is sp²
4. Consider the fourth C atom.
• There is one single bond and one triple bond around it. This is possible only if there are two orbitals around that C atom.
• Only sp hybridization can give two orbitals. The third C atom will enter into σ bonding with one of the two sp hybrid orbitals.
• The N atom will enter into σ bonding with the other sp hybrid orbital. Also, the p orbitals of the C atom will enter into two π bonding s with N. This creates the triple bond with N.
• So we can write:
The type of hybridization of the fourth C atom is sp.

Solved example 12.3
Write the state of hybridization of carbon in the following compounds and shape of each of the molecules.
(a) H₂C=O,   (b) CH₃F,  (c) HC☰N
Solution:
• We are given the condensed formulae. Based on that, we will draw the structural formulae. In the structural formulae, all the single, double and triple bonds will become visible. They are shown in figs.12.2 and 12.3 below. Note that, we will learn the details about how to draw those structural formulae, in later sections of this chapter.

Fig.12.4

Part (a):
1. Consider the C atom in fig.12.4(a) above.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The left and right H atoms will enter into σ bonding with two of the three sp² hybrid orbitals.
• The O atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with O. This creates the double bond with O.
• So we can write:
The type of hybridization of the C atom is sp²
2. In sp² hybridization, there are three orbitals. Those orbitals are directed towards the three corners of an equilateral triangle.
• So we can write:
H₂C=O will have a trigonal planar structure

Part (b):
1. Consider the C atom in fig.12.4(b) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The F and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the C atom is sp³
2. In sp³ hybridization, there are four orbitals. Those orbitals are directed towards the four corners of a tetrahedron.
• So we can write:
CH₃F will have a tetrahedral structure

Part (c):
1. Consider the C atom in fig.12.4(c) above.
• There is one single bond and one triple bond around it. This is possible only if there are two orbitals around that C atom.
• Only sp hybridization can give two orbitals. The left C atom will enter into σ bonding with one of the two sp hybrid orbitals.
• The N atom will enter into σ bonding with the other sp hybrid orbital. Also, the p orbitals of the C atom will enter into two π bonding s with N. This creates the triple bond with N.
• So we can write:
The type of hybridization of right side C atom is sp.
2. In sp hybridization, there are two orbitals. Those orbitals are directed along a straight line with an angle of 180^o between them.
• So we can write:
HC☰N will have a linear structure.

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In the next section, we will see Structural representations of Organic compounds.


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