Chapter 13.22 - Directive Influence of Functional Groups

In the previous section, we completed a discussion on electrophilic substitution reactions of aromatic compounds. In this section, we will see addition reactions. Later in this section, we will see the directive influence of functional groups.

• We have seen that, benzene has unusual stability. It prefers to undergo substitution reactions rather than addition reactions. This is to retain the ring structure.
• But benzene can be forced to undergo addition reaction by providing suitable conditions.
• We have to learn about two addition reactions.

I. Addition of H atoms to the benzene ring
This can be written in 3 steps:
1. For addition of H atoms, we must provide high temperature and/or pressure.
• Nickel should be present as a catalyst.
2. Three hydrogen molecules add to one benzene ring.
• The product is cyclohexane. The reaction is shown in the fig.13.127 below:

Fig.13.127

3. This process is known as hydrogenation of benzene.

II. Addition of Cl atoms to the benzene ring
This can be written in 2 steps:
1. For addition of Cl atoms, we must provide ultraviolet light.
2. Three chlorine molecules add to one benzene ring.
• The product is benzene hexachloride.
   ♦ Another name for benzene hexachloride is gammaxane.
• The reaction is shown in the fig.13.128 below:

Fig.13.128

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Combustion of benzene

This can be written in 2 steps:
1. When benzene is heated in air, it burns with a sooty flame.
• The equation is:
C₆H₆ + 15/2 O₂ ⟶ 6CO₂ + 3H₂O
2. The general equation for the combustion of any hydrocarbon can be written as:
C_xH_y + (x+y/4) O₂ ⟶ xCO₂ + y/2 H₂O
• Recall that, we have already seen this general equation when we saw the quantitative analysis of hydrocarbons [see section 12.22].

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Directive influence of a functional group

This can be explained in 6 steps:
1. Consider a monosubstituted benzene molecule.
• Monosubstituted indicates that, there is only one functional group in the benzene ring.
2. Now, a second functional group wants to get attached to the ring. In such a situation, there are three possibilities:
(i) The incoming group gets attached to the second C atom.
• This creates a 1,2-disubstituted product.
• We know that, a 1,2-disubstituted product is also known as ortho product. See fig.12.58 of section 12.8.
(ii) The incoming group gets attached to the third C atom.
• This creates a 1,3-disubstituted product.
• We know that, a 1,3-disubstituted product is also known as meta product.
(iii) The incoming group gets attached to the fourth C atom.
• This creates a 1,4-disubstituted product.
• We know that, a 1,4-disubstituted product is also known as para product.
3. The above three are the only possibilities.
• So we would expect one third of the product to be ortho, another one third to be meta and the remaining one third to be para.
4. But in reality, we never get such equal products.
(i) In some reactions, ortho and para products are the major products. The meta product will be formed only in small quantities.
(ii) In some reactions, meta product is the major product. The ortho and para products will be formed only in small quantities.
5. This is because, the incoming group is directed towards particular positions.
• In 4(i), the incoming group is directed towards ortho and para positions.
• In 4(ii), the incoming group is directed towards meta position.
6. The incoming group has no role in deciding the positions.
• The functional group already present in the benzene ring is responsible for directing the incoming group towards particular positions.
• This is known as directive influence of functional groups.

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◼ We have seen that:
   ♦ The group which is already present
   ♦ is able to direct the incoming group
   ♦ towards particular positions.
• Let us see how such an ability is acquired. It can be written in 6 steps:
1. Consider the molecule of phenol.
• Phenol is a resonance hybrid as shown in the fig.13.129 below:

Fig.13.129

• The reader must be able to realize the significance of each curved arrow in the above fig. The details related to fig.12.89(a) in section 12.14 can be used as a guide.
• The reason for the charges in various C atoms also must be verified. For example, the second C atom in II has five electrons around it. An independent C atom will have only four electrons. So this second C atom has a -ve charge.  
2. We see that:
   ♦ In II, the electron density is greater at the ortho position.
   ♦ In III, the electron density is greater at the para position.
   ♦ In IV, the electron density is greater at the ortho position.
3. So we can write:
Due to the presence of the -OH group, electron density is greater at the ortho and para positions. As a consequence, substitution takes place mainly at these positions.
4. The -OH group has a tendency to pull electrons towards itself. As a result, the successive C atoms gain small +ve charges δ⁺, δδ⁺, δδδ⁺ so on.
• This is known as -ve inductive effect (-I effect). See fig.12.88 in section 12.13. Due to this effect, the electron density is slightly reduced at ortho and para positions.
• But the resonance effect predominates the -I effect. Due to resonance, the electron density will be greater at ortho and para positions.
5. Recall that, an ordinary benzene ring has the same electron density at all the six C atoms.
• But now we see that, when the -OH group is present in the benzene ring, electron density increases at the ortho and para positions.
• We can say that, the -OH group activates the benzene ring in such a way that, the ring becomes vulnerable to attack by an electrophile.
• Like the -OH group, other groups like -NH₂, -NHR, -NHCOCH₃, -OCH₃, -CH₃, -C₂H₅ etc., also activates the benzene ring. They are all ortho/para directing groups.
6. If instead of the above groups, suppose that, a halogen atom is present.
• Halogen atoms are highly electronegative. So the -I effect will be higher.
• That means, the δ⁺, δδ⁺, δδδ⁺ charges acquired by successive C atoms will be higher. As a consequence, the electron density will decrease at the various C atoms.
• However, due to resonance, the density will be greater (when compared to meta position) at ortho and para positions. So halogens are also ortho/para directing groups.

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◼ Now we will see the meta directing groups. It can be written in 6 steps:
1. Consider the molecule of nitromethane.
• Nitromethane is a resonance hybrid as shown in the fig.13.130 below:

Fig.13.130

• The reader must be able to realize the significance of each curved arrow in the above fig. The details related to fig.12.89(a) in section 12.14 can be used as a guide.
• The reason for the charges in various C atoms also must be verified. For example, the second C atom in II has only three electrons around it. An independent C atom will have four electrons. So this second C atom has a +ve charge.  
2. We see that:
   ♦ In II, the electron density is lesser at the ortho position.
   ♦ In III, the electron density is lesser at the para position.
   ♦ In IV, the electron density is lesser at the ortho position.
3. So we can write:
• Due to the presence of the -NO₂ group, electron density is lesser at the ortho and para positions.
• Conversely, due to the presence of the -NO₂ group, electron density is greater at the meta position (3 and 5).
• As a consequence, substitution takes place mainly at meta position.
4. The -NO₂ group has a strong tendency to pull electrons towards itself. As a result, the successive C atoms gain small +ve charges δ⁺, δδ⁺, δδδ⁺ so on.
• This is known as -ve inductive effect (-I effect). See fig.12.88 in section 12.13. Due to this effect, the electron density is slightly reduced at meta position.
• But the resonance effect predominates the -I effect. Due to resonance, the electron density will be greater at meta position.
5. Recall that, an ordinary benzene ring has the same electron density at all the six C atoms.
• When the -NO₂ group is present in the benzene ring, electron density decreases at various C atoms in the ring. This is due to the strong -I effect.
• Due to this decrease in electron densities, the electrophiles will find it very difficult to get attached to the benzene ring.
• We can say that, the -NO₂ group deactivates the benzene ring in such a way that, the ring becomes stable against attack by an electrophile.
• Like the -NO₂ group, other groups like -CN, -CHO, -COR, -COOH, -COOR, -SO₃H etc., also deactivates the benzene ring.
6. Even when such deactivation takes place, the electron density at meta position will be greater (when compared to ortho/para positions).
So the electrophile attacks the meta position, resulting in meta substitution. So these groups are meta directing groups.

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Carcinogenicity and Toxicity

This can be explained in 4 steps:
1. We know that a benzene molecule contains only one ring.
• In some molecules, two or more benzene rings will be fused together. Such molecules are called polynuclear aromatic hydrocarbons (PAH).
2. Benzene and polynuclear aromatic hydrocarbons are carcinogens.
• A carcinogen is a substance, organism or agent that cause cancer.
3. Carcinogenic substances are produced during incomplete combustion of tobacco, coal, petroleum etc.,
• They enter into human body and undergo various bio chemical reactions. Such reactions will damage the DNA and cause cancer.
4. A few carcinogens are shown in fig.13.131 below:

Fig.13.131

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The link below gives more solved examples related to the topics that we saw in this chapter.

Exercises on chapter 13

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We have completed a discussion on hydrocarbons. In the next chapter, we will see environmental chemistry.

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Chapter 13.17 - Aromatic Hydrocarbons

In the previous section, we completed a discussion on alkynes. In this section, we will see aromatic hydrocarbons.

Three basic features of aromatic hydrocarbons are written below:
1. Aromatic hydrocarbons derived their name from the Greek word ‘aroma’ which means ‘pleasant smelling’.
• Another name for aromatic hydrocarbons is Arenes.
2. Recall that, we have seen some details about the benzene ring in the previous chapter [see fig.12.89 in section 12.14].
• Most of the aromatic compounds contain one or more benzene rings. We know that benzene contains double bonds. That means, benzene is an unsaturated hydrocarbon.
• We also know that, unsaturated hydrocarbons are easily attacked by other reagents. This is because, the loosely held π-electrons are readily available for the attacking reagents. So during the reaction, the double or triple bonds will be converted into single bonds.
• But in the case of aromatic hydrocarbons, the benzene ring is retained even after the reaction. That means, the π-electrons of the benzene ring are not easily available for the attacking reagents. We will see the reason in the next section.
3. Most aromatic hydrocarbons contain one or more benzene rings. But there are a few which do not contain any benzene rings.
• Aromatic compounds which contain benzene rings are known as benzenoids.  
• Aromatic compounds which do not contain benzene rings are known as non-benzenoids.

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Nomenclature and Isomerism

• We have seen the details about the nomenclature and isomerism of aromatic hydrocarbons in the previous chapter [see section 12.8].
• A few more details are written in 2 steps below:
1. All the six H atoms in benzene are equivalent. So if we want a monosubstituted benzene molecule, we will get only one type.
• ‘Monosubstituted’ means:
    ♦ one H atom is removed
    ♦ a group like -CH₃ or -OH takes it’s place.
• We can remove any one of the six H atoms. Then we can put a suitable group in it’s place. We will always get the same product. This is because, all six H atoms are equivalent.
• We have already seen the explanation in the previous chapter. There we wrote:
When there is only one branch, the position of the C atom is not important [see the animation in fig.12.55 in section 12.8].
• A similar example is shown in fig.13.96 below.

Fig.13.96

    ♦ Consider the structure II
    ♦ Imagine an axis such that:
        ✰ It is perpendicular to the plane of the paper.
        ✰ It passes through the center of II.
    ♦ If we rotate II about that axis, we will get I.
    ♦ Both the structures in the fig., represent toluene.
• So we can write:
    ♦ If the benzene ring is monosubstituted, six different arrangements are possible.
    ♦ But all those six arrangements will represent the same molecule.
• We explained the 'equality of the six structures' by using the concept of rotation. However, a more scientific explanation can be given by using the concept of resonance. Due to resonance in benzene, there will not be any difference between single bonds and double bonds. All six bonds are identical. So all the six H atoms are in identical positions. We will see more details about resonance in the next section. 
2. In the case of disubstituted benzene ring, five different arrangements are possible. But there will be only three different molecules. This can be explained in 4 steps:
(i) Fig.13.97(a) below shows two arrangements.

Fig.13.97

• In I, the substituents are in 1, 2 positions.
    ♦ The substituents are at the two ends of a double bond.
• In II, the substituents are in 1, 6 positions.
    ♦ The substituents are at the two ends of a single bond.
• But thanks to resonance, there is no difference between single bonds and double bonds in benzene. All bonds are equivalent. So, the substituents are in identical positions. Consequently, the two structures are the same.
• So if the substituents are in 1, 2 or 1, 6, we give a common name: ortho. It is abbreviated as: o-.
• So we can write:
    ♦ Both structures in fig.13.97(a) represent the same molecule.
    ♦ It’s common name is o-xylene.
• When we use the word ortho, what really matters is: The substituents are attached to adjacent C atoms.
    ♦ It does not matter whether it is 1,2 or 1,6.
• Based on the rules that we saw in the previous chapter, we can write the IUPAC name of this o-xylene. It is 1,2-Dimethylbenzene.
(ii) Fig.13.97(b) above shows two arrangements.
• In III, the substituents are in 1, 3 positions.
• In II, the substituents are in 1, 5 positions.
• Comparing I and II, we see that, the arrangement of single and double bonds between the two substituents is different.
• But thanks to resonance, there is no difference between single bonds and double bonds in benzene. All bonds are equivalent. So, the substituents are in identical positions. Consequently, the two structures are the same.
• So if the substituents are in 1, 3 or 1, 5, we give a common name: meta. It is abbreviated as: m-.
• So we can write:
    ♦ Both structures in fig.13.97(b) represent the same molecule.
    ♦ It’s common name is m-xylene.
• When we use the word meta, what really matters is: The substituents are attached to two C atoms with a C atom in between.
    ♦ It does not matter whether it is 1,3 or 1,5.
• Based on the rules that we saw in the previous chapter, we can write the IUPAC name of this o-xylene. It is 1,3-Dimethylbenzene.
(iii) Fig.13.97(c) above shows the last of the five arrangements.
• In V, the substituents are in 1, 4 positions.
• In this case, we give a common name: para. It is abbreviated as: p-.
• So we can write:
    ♦ The common name of the structure is p-xylene.
• When we use the word para, what really matters is: The substituents are attached to two C atoms with two C atoms in between.
• Based on the rules that we saw in the previous chapter, we can write the IUPAC name of this p-xylene. It is 1,4-Dimethylbenzene.
(iv) Consider the above three molecules again:
    ♦ 1,2-Dimethylbenzene
    ♦ 1,3-Dimethylbenzene
    ♦ 1,4-Dimethylbenzene
• It is clear that, the three are position isomers.

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Structure of Benzene

Some basics about the structure can be written in 5 steps:
1. Benzene was first isolated by Michael Faraday in 1825. At that time, it’s structural details were not known.
2. We have seen the methods that scientists use to determine molecular formula of unknown compounds [see section 12.22]. Through those methods, the molecular formula of benzene was found to be C₆H₆.
• Six H atoms will not be able to satisfy the valencies of six C atoms through single bonds. So scientists assumed that, double or triple bonds are present in a benzene molecule.
3. To confirm the presence of double or triple bonds, bromine solution was added.
• We have seen test for unsaturation in previous sections [see fig.13.89 in section 13.15].
• We would expect the double or triple bonds to break and add Br atoms. This is shown in fig.13.98(a) below.

Fig.13.98

• But surprisingly, benzene was unreactive to bromine. No such product was obtained.
• Scientists came to this conclusion:
Though double or triple bonds are present, benzene has a stable structure.
4. Later scientists forced benzene to react with bromine. They used a catalyst for this purpose.
• The actual result obtained was unexpected. The double bond did not break. One H atom was removed and a Br atom was substituted for that H atom. This is shown in fig.13.98(b) above.
• We see that, by preferring substitution reaction, the benzene molecule is able to retain the double bonds.
5. After years of research, the German scientist August Kekule proposed the familiar structure (with alternate single and double bonds) that we see today in our basic courses in organic chemistry.
• But this structure was not able to explain why the double bond does not add bromine atoms.
• In other words, the Kekule structure was not able to explain the unusual stability of benzene.

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Scientists were able to give a satisfactory explanation for the stability of benzene by using the concept of resonance. We will see it in the next section.

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Chapter 13.16 - Polymerisation of Alkynes

In the previous section, we saw the preparation of alkynes. We also saw some chemical properties of alkynes. In this section, we will see two more chemical properties.

Addition of water

Let us see the reaction between an alkyne and water. It can be written in 5 steps:
1. Alkynes react with water. Mercuric sulphate and concentrated sulphuric acid is also required for the reaction. The reaction mixture should be warmed to a temperature of 333 K. Carbonyl compounds will be obtained as products. An example is shown in fig.13.92 below:

 

Fig.13.92

 

• First the water molecule splits to give two parts:
    ♦ H⁺ ion, which is the +ve part.
    ♦ OH^- ion, which is the -ve part.
2. The H⁺ ion thus produced, will get attached to one of the two C atoms of the ethyne molecule. This is shown in fig.a.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the triple bond in the alkyne. That is why both electrons in the bond are shown in red color.
3. When the H⁺ ion leaves the water molecule, the remaining portion will have an extra electron. So it will have a -ve charge.
• This remaining portion is the -ve part of the addendum. This -ve part (OH^-) gets attached to the other C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
4. The alcohol thus formed will undergo isomerisation. As a result we get ethanal. This is shown in fig.b.
• Isomerisation is a reaction in which a molecule gets transformed into it's isomer.
• We saw a type of isomerisation in an earlier section [see fig.13.38 in section 13.6].
5. In this example, we see that, the two species H⁺ and OH^- get added to the original alkyne molecule.
• So we can write:
The reaction between alkynes and water is an addition reaction.

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Let us see another example. It is shown in fig.13.93 below. It can be written in 6 steps.

 

Fig.13.93

1. First the water molecule splits to give two parts:
    ♦ H⁺ ion, which is the +ve part.
    ♦ OH^- ion, which is the -ve part.
2. The H+' ion thus produced, will get attached to the first C atom of the propyne molecule. This is shown in fig.a.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the triple bond in the alkyne. That is why both electrons in the bond are shown in red color.
3. When the H⁺ ion leaves the water molecule, the remaining portion will have an extra electron. So it will have a -ve charge.
• This remaining portion is the -ve part of the addendum. This -ve part (OH^-) gets attached to the second C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
4. Here we see that, the -ve part gets attached to the C atom with the least number of H atoms.
• So we can write:
The reaction between alkenes and water obeys Markovnikov’s rule.
5. The alcohol thus formed will undergo isomerisation. As a result we get propanone. This is shown in fig.b.
6. In this example, we see that, the two species H⁺ and OH^- get added to the original alkyne molecule.
• So we can write:
The reaction between alkynes and water is an addition reaction.

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Polymerization of alkynes

• First we will see linear polymerization. It can be written in 6 steps:
1. Consider the molecule of ethyne ($\rm{CH ☰ CH}$).
• We know that in ethyne, the triple bond is necessary to satisfy the valencies of C and H atoms.
2. If we break the triple bond, the structure would look like this: $\rm{-CH = CH -}$
• The ‘${}-{}$’ on the sides indicate that, the structure is looking for electrons.
3. Consider the structure shown in the above step (2).
• If there are a large number of such structures, they can join together to satisfy the valencies. The joined structure would look like this:
$\rm{-CH = CH - CH = CH - CH = CH - CH = CH -}$
• The process of making the joined structure is known as polymerization. We saw this in the case of alkenes also.
4. In our present polymerization, we subject ethyne molecules to high pressure and high temperature. Presence of a suitable catalyst is also necessary.
• Due to the high pressure and temperature, ethyne molecules will change into $\rm{-CH = CH -}$
• These structures will join together to form a large molecule.
   ♦ The large molecule is called a polymer.
   ♦ The simple molecule from which polymer is obtained is called a monomer.
• In our present case,
   ♦ Ethyne is the monomer.
   ♦ The polymer obtained has a common name: polyacetylene.
        ✰ It's IUPAC name is polyethyne.
5. Under special conditions, polyethyne conducts electricity. Thin films of this polymer can be used as electrodes in batteries. These films are good conductors. They are lighter and cheaper than metal conductors. Some images can be seen here.
6. The joined structure in step (3) can be written in short form as: $\rm{-(CH = CH - CH = CH )_n -}$
• So the process of this polymerization to make polyethyne can be written as:
$\rm{n(CH ☰ CH)~ \color {green}{\xrightarrow[{\text{Catalyst}}]{{\text{High temp./pressure}}}} ~ -(CH = CH - CH = CH )_n -}$

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Now we will see cyclic polymerization. It can be written in 4 steps:

1. We have already seen the details about the structure of the benzene ring [see fig.12.53 of section12.8].
• We see that, there are six CH groups in benzene. Can three ethyne molecules join together to form a benzene ring?
2. Fig.13.94(a) below shows three ethyne molecules aligned together in favorable positions.

Fig.13.94

• If two electrons in the triple bonds can shift, new single bonds will be formed between the three ethyne molecules. The shifting of electrons are indicated by the three curved red arrows in fig.b.
3. This is a cyclic polymerization. For this polymerization to take place, we allow the ethyne gas to pass through red hot iron tube kept at 873 K.
4. So we have an easy method to prepare benzene. Once we obtain benzene, we can make a variety of useful products like benzene derivatives, dyes, drugs etc.,
• This method helps us to enter the world of aromatic compounds from the world of aliphatic compounds.
• We know that:
    ♦ Aliphatic compounds have an open chain structure.
    ♦ Aromatic compounds have a closed chain structure.
• So we enter the world of closed chain structures from the world of open chain structures.

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Let us see a solved example.

Solved example 13.14
How will you convert ethanoic acid into benzene ?
Solution:
1. We can start from benzene and think in a reverse direction.
• To obtain benzene, we must have ethyne (CH☰CH). Because, ethyne when passed through a red hot iron tube will give benzene. See fig.13.94 above.
• This process is marked as the last step (X) in fig.13.95 below.
• So our aim must be to convert the given ethanoic acid to ethyne.

Fig.13.95

2. Ethyne can be obtained from alkenyl halides. We saw this in the preparation of alkynes. See the topic at the beginning of the previous section.
• In our present case, the alkenyl halide CH₂=CHBr can be used to obtain ethyne.
• This process is marked as (IX) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkenyl halide CH₂=CHBr.
3. Alkenyl halides can be obtained from vicinal dihalides. We saw this in the preparation of alkynes. See the topic at the beginning of the previous section.
• In our present case, CH₂Br-CH₂Br can be used to obtain the alkenyl halide CH₂=CHBr.
• This process is marked as (VIII) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the vicinal dihalide CH₂Br-CH₂Br.
4. The above vicinal dihalide can be prepared by the addition reaction between ethene and Br₂. We saw this in the chemical properties of alkenes. See fig.13.68 of section 13.11.
• This process is marked as (VII) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to ethene.
5. Ethene can be obtained from the alkyl halide CH₃-CH₂Cl. We saw this in the preparation of alkenes from alkyl halides. See fig.13.65 of section 13.10.
• This process is marked as (VI) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkyl halide CH₃-CH₂Cl.
6. The above alkyl halide can be prepared from the alkane CH₃-CH₃. We saw this in the addition reaction of alkanes. See substitution reactions in section 13.4.
• This process is marked as (V) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkane CH₃-CH₃.
7. The above alkane can be obtained using CH₃Cl by Wurtz reaction, We saw this in the preparation of alkanes from alkyl halides. See section 13.3.
• This process is marked as (IV) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkyl halide CH₃Cl.
8. The above alkyl halide can be obtained from CH₄. We saw this in the chemical properties of alkanes. See substitution reactions in section 13.4.
• This process is marked as (III) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkane CH₄.
9. CH₄ can be obtained from the sodium salt of ethanoic acid. We saw this in the preparation of alkanes. See preparation of alkanes from carboxylic acids in section 13.3.
• This process is marked as (II) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the sodium salt of the ethanoic acid.
10. The preparation of the above sodium salt is a simple process.
• Ethanoic acid is an acid. It will react with the base NaOH to give the sodium salt.
• This process is marked as (I) in fig.13.95 above.
• So we have worked in the reverse order from benzene to ethanoic acid.

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In the next section we will see aromatic hydrocarbons.

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Chapter 13.15 - Preparation and Properties of Alkynes

In the previous section, we saw nomenclature and isomerism in alkynes. We also saw the structure of the triple bond. In this section, we will see preparation of alkynes.

• We will see two methods for preparing ethyne.
   ♦ From calcium carbide
   ♦ From vicinal dihalides

From calcium carbide

This can be written in 3 steps:
1. First, lime stone is heated to obtain quick lime (CaO).
• The equation is:
$\rm{CaCO_3~ \color {green}{\xrightarrow[{}]{Δ}} ~ CaO~+~CO_2}$
2. The quick lime is heated with coke to obtain calcium carbide (CaC₂)
• The equation is:
$\rm{CaO~+~3C~ \color {green}{\xrightarrow[{}]{}} ~ CaC_2~+~CO}$
3. The calcium carbide is treated with water to obtain ethyne.
• The equation is:
$\rm{CaC_2~+~2H_2O~ \color {green}{\xrightarrow[{}]{}} ~ Ca(OH)_2~+~C_2H_2}$

From vicinal dihalides

This can be written in 2 steps:
1. The vicinal dihalide is first treated with alcoholic potassium hydroxide.
• One atom of hydrogen and one atom of halogen will be removed from the vicinal dihalide. This removal is known as dehydrohalogenation.
• The product will contain a double bond between the two carbon atoms.
• So the product is not an alkyl halide, but an alkenyl halide.
• The equation is:
$\rm{CH_2 Br - CH_2 Br~+~KOH~ \color {green}{\xrightarrow[{-KBr~and~-H_2O}]{alcohol}} ~ CH_2 = CHBr}$
2. The alkenyl halide is treated with sodamide to obtain the alkyne.
• The equation is:
$\rm{CH_2 = CHBr~ \color {green}{\xrightarrow[{-NaBr~and~-NH_3}]{Na^{+} {NH_2}^{-}}} ~ CH ≡ CH}$

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Physical properties of alkynes.

This can be written in 5 steps:
1. We have already seen the physical properties of alkanes [see section 13.4] and those of alkenes [see section 13.11].
• The physical properties of alkynes follow a similar trend as alkanes and alkenes.
2. The first three members of the alkyne series are gases.
• The next eight members are liquids.
• The members coming after that are solids.
3. All alkynes are colorless.
• All alkynes except ethyne are odour less. Ethyne has a characteristic odour.
4. Alkynes are insoluble in water. But they are soluble in non-polar solvents like carbon tetrachloride, benzene and petroleum ether.
(Petroleum ether is obtained from petroleum. It is used as a laboratory solvent)
5. The members of the alkyne series show a regular increase in melting point, boiling point and density.

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Chemical properties of alkynes

• We have to learn about three chemical properties of alkynes. They are:

A. Acidic character

B. Addition reaction

C. Polymerization

A. Acidic character

• Before discussing about the acidic character of alkynes, we must consider the electronegativity of C atom.
• For that, we make a statement related to the electronegativity of C. The statement can be written in 3 steps:
(i) Consider a molecule containing C atom.
(ii) Suppose that, the C atom is sp hybridized.
Then that C atom will be highly electronegative.
(iii) If that C atom is sp² hybridized or sp³ hybridized, then it will not be so much electronegative.

 

• Now we will see the proof for the statement. It can be written in 5 steps:
1. Consider the sp³ hybridized orbitals. Each of those orbitals will be having 25% s-characteristics.
2. Consider the sp² hybridized orbitals. Each of those orbitals will be having 33% s-characteristics.
3. Consider the sp hybridized orbitals. Each of those orbitals will be having 50% s-characteristics.
4. We have already seen the above details in an earlier section [see fig.4.135 in section 4.24].
• So it is clear that, sp hybridized orbitals have greater s-characteristics.
5. s-orbitals are closer to the nucleus. So the electrons in the s-orbitals will be attracted more towards the nucleus of the atom.
• So in our present case, if the C atom is sp hybridized, then the shared electrons around that C atom will be attracted more towards the nucleus of the C atom.
• Consequently, such a C atom will be more electronegative.

Now we can discuss about the acidic character of alkynes. It can be written in 4 steps:

1. Consider a C☰H bond in ethyne. The C atom here is sp hybridized. So it will pull the electrons in the bonds. The H will become +ve charged.

2. The H can be released as H⁺. That means, ethyne can act as a proton donor. We know that acids are proton donors. So now we understand why ethyne is acidic.

3. In alkanes, the C atoms are sp³ hybridized. Those C atoms do not have high electronegativity. So alkanes are not acidic.

4. In alkenes, the C atoms are sp² hybridized. Those C atoms do not have high electronegativity. So alkenes are not acidic.

Note:
In an alkyne, there may be more than one triple bonds. There may be other double and single bonds also. Only those H atoms in the triple bonds are available for release as protons. We must not expect the other H atoms to contribute to the acidic character of that alkyne.

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Let us see two reactions where alkynes show their acidic character.

Reaction 1:

This can be written in three steps

1. We know that, acids react with sodium (Na) to release hydrogen gas.

• An example is:
$\rm{HCl~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ Na^{+} Cl^{-}~+~\frac{1}{2}H_2}$

2. In a similar way, ethyne reacts with Na to release hydrogen gas. The equation is:

$\rm{HC☰CH~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ HC☰C^{-}Na^{+}~+~\frac{1}{2}H_2}$

• HC☰C^-Na⁺ obtained above is monosodium ethynide.

3. This monosodium ethynide reacts with another atom of Na to give disodium ethynide. The equation is:

$\rm{HC☰C^{-}Na^{+}~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ Na^{+}C^{-} ☰ C^{-}Na^{+}~+~\frac{1}{2}H_2}$

 

Reaction 2:

• Here we consider the reaction between propyne and sodamide. The equation is:

$\rm{CH_3 - C☰CH~+~Na^{+} {NH_2}^{-}~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - C☰C^{-} Na^{+}~+~NH_3}$

• $\rm{CH_3 - C☰C^{-} Na^{+}}$ is sodium propynide 

• The above reactions are not shown by alkanes and alkenes. So we can test a sample of unknown hydrocarbons by adding Na or sodamide. If a reaction takes place, we can confirm that, the sample contains alkynes.

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Let us compare the acidic characters of but-1-yne and but-2-yne. The comparison can be written in 3 steps:
1. First we write the structures:
   ♦ The structure of but-1-yne is: HC☰C-CH₂-CH₃
   ♦ The structure of but-2-yne is: CH₃-C☰C-CH₃
2. Consider the structure of but-1-yne.
• There are two C atoms on either sides of the triple bond.
   ♦ Both of them will be sp hybridized.
• One of those C atoms have a H atom.
   ♦ This particular C atom can pull the electrons away from the H atom.
   ♦ As a result, there is one H atom available to be donated as a proton.
• Thus but-1-yne gets it’s acidic character.
3. Consider the structure of but-2-yne.
• There are two C atoms on either sides of the triple bond.
   ♦ Both of them will be sp hybridized.
• None of those C atoms have any H atoms.
   ♦ So the C atoms cannot pull electrons.
   ♦ As a consequence, there are no H atoms available to be donated as protons.
• Thus but-2-yne gets no acidic character.

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B. Addition reaction

Some basics can be written in 2 steps:
1. We know that, each triple bond in alkynes consist of two π-bonds. The electrons in the π-bonds are loosely held. So those electrons are easily available for ‘electron seeking species’ (electrophiles).
2. Due to this availability of electrons, the electrophiles will get attached to the alkynes, resulting in new compounds. We call such reactions as addition reactions. We saw this situation in the case of alkenes also. Now we will see different types of addition reactions in alkynes.

I. Addition of dihydrogen
This can be written in 6 steps:
1. Each triple bond in an alkyne can add two molecules of dihydrogen.
2. The dihydrogen molecule first splits into two H atoms. This is indicated by the blue dashed curves in fig.13.88(a) below:

Fig.13.88

• Each newly formed H atom will have only one electron (yellow dot). So each H atom will be looking for one more electron to complete octet.
3. The loosely held electrons at the π-bonds will supply the required electrons for the H atoms.
• That means, two of the six red dots in the triple bond, will leave the triple bond. Those two red dots will help the new H atoms to form single bonds with C atoms.
• Thus we get two new C-H bonds.
4. So we now know how the H atoms add up to the alkyne. The alkyne then will no longer require the triple bond. It will be converted to an alkene.
5. The newly formed alkene has a π-bond. So two more atoms of H can be added. The result will be an alkane. This is shown in fig.b above. We saw this process in the case of alkenes [see fig.13.67 of section 13.11].
6. In an earlier section, we saw this process as a method for preparing alkanes [see section 13.3].
• For this process, finely divided nickel, palladium or platinum is required as catalyst.

II. Addition of halogens
This can be written in 7 steps:
1. Each triple bond in an alkyne can add two molecules of halogen.
2. First the alkyne gets converted into an alkene. Two individual halogen atoms (F, Cl, Br or I) are required for this process. This is indicated by the two Br atoms in fig.13.89(a) below:

Fig.13.89

• Each individual X atom will have only seven electrons in the outer most shell.
• This is indicated by the seven dots around the Br atoms.
   ♦ Seven grey dots for the first Br atom
   ♦ Seven yellow dots for the second Br atom.
• So each X atom will be looking for one more electron to complete octet.
3. The loosely held electrons at the π-bond will supply the required electrons for the X atoms.
• That means, two of the six red dots in the triple bond, will leave the triple bond. Those two red dots will help the two new X atoms to form single bonds with C atoms.
• Thus we get two new C-Br bonds to form 1,2-Dibromopropene.
4. So we now know how the X atoms add up to the alkyne. The alkyne then will no longer require the triple bond. It will be converted to an alkene.
5. The newly formed alkene has a π-bond. So two more atoms of Br can be added.
• The result will be $\rm{CH_3 - CBr_2=CHBr_2}$. This is shown in fig.b above. We saw this process in the case of alkenes [see fig.13.69 of section 13.11].
6. The equations of the reactions in fig.13.89 above can be written as:
• $\rm{CH_3 - C☰CH~+~Br-Br~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CBr=CHBr}$
• $\rm{CH_3 - CBr=CHBr~+~Br-Br~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CBr_2 - CHBr_2}$
7. Let us see how this reaction can be used as a test for the presence of double or triple bonds (test for unsaturation). It can be written in 4 steps:
(i) Bromine solution has a reddish orange color. This color is due to the presence of Br^- ions.
(ii) We add this bromine solution to a solution which is to be tested. If the solution contains any triple bonds, those triple bonds will break. Each triple bond will take up four Br atoms.
(iii) Thus all the Br^- ions will be used up. The reddish orange color will disappear.
(iv) So, if the reddish orange color disappear, we will get an indication that, triple bonds are present.
• We saw these details in the case of alkenes also.

III. Addition of hydrogen halides
We will consider the reaction between but-2-yne and HBr. It can be written in 8 steps:
1. First the HBr molecule splits into two parts. We get H⁺ and Br^-. It is shown in fig.13.90 (a) below:

Fig.13.90

2. The H⁺ attacks the alkyne molecule. This is shown in fig.13.90 (b) above.
• At the product side, we see that, the H⁺ is attached to the third C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the triple bond. That is why we see two red dots in the new C-H bond.
3. But now, the triple bond is broken (the π-electrons in the triple bond were utilized for the new C-H bond).
• When the triple bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a vinylic cation is formed.
• A vinylic cation is a carbocation in which the +ve charge is possessed by a C atom in a double bond. 
4. The Br^- now attacks the newly formed vinylic cation. This is shown in fig.13.90 (c) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2-Bromobut-2-ene.
5. Now we have a molecule with a double bond. This double bond has to be converted into a single bond.
• For that, a new H⁺ attacks the 2-Bromo-but-2-ene. This is shown in fig.d above.
• At the product side, we see that, the H⁺ is attached to the third C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the double bond. That is why we see two red dots in the new C-H bond.
6. But now, the double bond is broken (the π-electrons in the double bond were utilized for the new C-H bond).
• When the double bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
7. The Br^- now attacks the newly formed carbocation. This is shown in fig.13.90 (e) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2,2-Dibromobutane.
8. We note an interesting point here. It can be written in 5 steps:
(i) In this reaction, there are two stages.
• In the first stage, the triple bond is converted to a double bond.
• In the second stage. the double bond is converted to a single bond.
(ii) A molecule of HBr is added in each stage.
• In the first stage, a molecule of HBr is added to the triple bond. 
• In the second stage, another molecule of HBr is added to the double bond.
(iii) H atoms are being added to the same C atom
• In the first stage, the H atom is added to the third C atom. This is shown in fig.13.90(b)  
• In the second stage, the H atom is added to the same third C atom. This is shown in fig.13.90(d)
(iv) Br atoms are being added to the same C atom
• In the first stage, the Br atom is added to the second C atom. This is shown in fig.13.90(c)  
• In the second stage, the Br atom is added to the same second C atom. This is shown in fig.13.90(e)
(v) So we get an end product in which two halogen atoms are attached to the same C atom. A dihalide in which two halogen atoms are attached to the same C atom is called gem dihalide.

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• In the above example, the alkyne was symmetric. Now we will consider an unsymmetrical alkyne. For that, we will see the reaction between propyne and HBr. It can be written in steps:
1. First the HBr molecule splits into two parts. We get H⁺ and Br^-. It is shown in fig.13.91 (a) below:

Fig.13.91

2. The H⁺ attacks the alkyne molecule. This is shown in fig.13.91 (b) above.
• At the product side, we see that, the H⁺ is attached to the first C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the triple bond. That is why we see two red dots in the new C-H bond.
3. But now, the triple bond is broken (the π-electrons in the triple bond were utilized for the new C-H bond).
• When the triple bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a vinylic cation is formed.
4. The Br^- now attacks the newly formed vinylic cation. This is shown in fig.13.91 (c) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2-Bromopropene.
5. Now we have a molecule with a double bond. This double bond has to be converted into a single bond.
• For that, a new H⁺ attacks the 2-Bromopropene. This is shown in fig.d above.
• At the product side, we see that, the H⁺ is attached to the first C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the double bond. That is why we see two red dots in the new C-H bond.
6. But now, the double bond is broken (the π-electrons in the double bond were utilized for the new C-H bond).
• When the double bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
7. The Br^- now attacks the newly formed carbocation. This is shown in fig.13.90 (e) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2,2-Dibromopropane.
8. We note an interesting point here. It can be written in 5 steps:
(i) In this reaction, there are two stages.
• In the first stage, the triple bond is converted to a double bond.
• In the second stage. the double bond is converted to a single bond.
(ii) A molecule of HBr is added in each stage.
• In the first stage, a molecule of HBr is added to the triple bond. 
• In the second stage, another molecule of HBr is added to the double bond.
(iii) H atoms are being added to the same C atom
• In the first stage, the H atom is added to the first C atom. This is shown in fig.13.91(b)  
• In the second stage, the H atom is added to the same first C atom. This is shown in fig.13.91(d)
(iv) Br atoms are being added to the same C atom
• In the first stage, the Br atom is added to the second C atom. This is shown in fig.13.91(c)  
• In the second stage, the Br atom is added to the same second C atom. This is shown in fig.13.91(e)
(v) So we get a gem dihalide.

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• Based on the two examples, we can write:
An alkyne may be symmetric or unsymmetric. When the addition of hydrogen halides occur, we always get a gem dihalide.

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We have seen that Markovnikov's rule is applicable when addition of hydrogen halides to unsymmetrical alkenes occur. Let us see whether the rule is applicable to unsymmetrical alkynes. It can be written in steps:
1. In example 2, HBr is being added to CH₃ - C☰CH
2. The product is: CH₃ - CBr₂ - CH₃
3. It is clear that, the -ve part (Br^-) is being added to the C atom with the least number of H atoms.
4. So Markovnikov's rule is applicable here.

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In the next section we will see addition of water and polymerisation.

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Chapter 13.12 - Anti-Markovnikov Rule

In the previous section, we completed a discussion on some properties of alkenes and Markovnikov's rule. In this section, we will see anti Markovnikov's rule.

Some basics can be written in 7 steps:
1. In the previous section, we have seen the addition reaction between HBr and an unsymmetrical alkene. This reaction follows the Markovnikov's rule.
2. But if peroxide is present in the reaction mixture, the reaction will take place contrary to the Markovnikov’s rule.
• This effect was discovered in 1933 by M.S. Kharash and F.R. Mayo at the university of Chicago.
   ♦ This effect is known as Peroxide effect.
   ♦ This effect is also known as Kharash effect.
   ♦ This effect is also known as addition reaction anti to Markovnikov’s rule.
3. Consider the reaction between HBr and prop-1-ene.
• Based on the discussions in the previous section, we know that, the product will be 2-Bromopropane.
• But if benzoyl peroxide is present, the product will be 1-Bromopropane.
4. The benzoyl peroxide molecule consists of two C₆H₅-C=O groups attached together by a -O-O- group.
• The peroxide undergoes homolysis as shown in fig.13.74 (a) below:

Fig.13.74

• We see that, as a result of the homolysis, two free radicals are obtained as products. Each is a free radical because, the O atom has an unpaired electron.
5. This free radical will again undergo homolysis.
• The homolysis takes place at the bond between the C₆H₅ group and the CO₂ group. This is shown in fig.13.74(b) above.
• We see that, the product is again a free radical ($\rm{\dot{C}_6 H_5}$) and a molecule of CO₂.
• We have seen the Lewis structure of CO₂ [See fig.4.11 in section 4.1]. It is clear that, the homolysis in fig.b enables the CO₂ group to acquire an extra electron  needed for the new double bond.
6. The $\rm{\dot{C}_6 H_5}$ radical attacks the HBr molecule.
• The HBr undergoes homolysis. This is shown in fig.13.74(c) above.
• The products are $\rm{\dot{H}}$ radical and $\rm{\dot{Br}}$ radical.
7. The $\rm{\dot{H}}$ radical combines with $\rm{\dot{C}_6 H_5}$ radical to form C₆H₆. This is shown in fig.13.74(d) above.
• We want the $\rm{\dot{Br}}$ radical. It is required for our further discussions.

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◼ The $\rm{\dot{Br}}$ radical attacks the molecule of prop-1-ene, resulting in the formation of 1-Bromopropane or 2-Bromopropane. We want to know which one is the major product.
• Let us first see the mechanism of the reaction which produces 1-Bromopropane. It can be written in 3 steps:
1. First the $\rm{\dot{Br}}$ radical attacks the prop-1-ene molecule.
• The radical gets attached to the first C atom. This is shown in fig.13.75(a) below:

Fig.13.75

• Note that, the $\rm{\dot{Br}}$ radical brought one electron required for the bond. The other electron on the bond is obtained by breaking the double bond of the prop-1-ene.
• When the double bond is broken in this way, the second C atom has now an unpaired electron.
• A new radical is formed in this way. The newly formed radical is a secondary free radical. This is because, the "C atom with the unpaired electron" is attached to two other C atoms.
2. The secondary free radical now attacks a HBr molecule. It results in the homolysis of HBr. This is shown in fig.13.75(b) above. Thus we get a $\rm{\dot{H}}$ radical and a new $\rm{\dot{Br}}$ free radical.
• We must note a point here. It can be written in 3 steps:
(i) The $\rm{\dot{Br}}$ free radical in fig.13.75(a) is obtained by the action of the peroxide.
(ii) But the $\rm{\dot{Br}}$ free radical in fig.13.75(b) is obtained by the action of the secondary free radical. This radical can attack another prop-1-ene molecule.
(iii) So a chain reaction will be set up.
3. The $\rm{\dot{H}}$ free radical attacks the secondary free radical. This is shown in fig.c above.
• The $\rm{\dot{H}}$ free radical gets attached to the second C atom.
• The $\rm{\dot{H}}$ free radical brought one electron required for the bond. The second electron is already present as an unpaired electron.
• Thus we get 1-Bromopropane.

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• Next, let us see the mechanism of the reaction which produces 2-Bromopropane. It can be written in 3 steps:
1. First the $\rm{\dot{Br}}$ radical attacks the prop-1-ene molecule.
• The radical gets attached to the second C atom. This is shown in fig.13.76(a) below:

Fig.13.76

• Note that, the $\rm{\dot{Br}}$ radical brought one electron required for the bond. The other electron on the bond is obtained by breaking the double bond of the prop-1-ene.
• When the double bond is broken in this way, the first C atom has now an unpaired electron.
• A new radical is formed in this way. The newly formed radical is a primary free radical. This is because, the "C atom with the unpaired electron" is attached to only one other C atom.
2. The primary free radical now attacks a HBr molecule. It results in the homolysis of HBr. This is shown in fig.13.76(b) above. Thus we get a H radical and a new $\rm{\dot{Br}}$ free radical.
• As before, we must note a point here. It can be written in 3 steps:
(i) The $\rm{\dot{Br}}$ free radical in fig.13.76(a) is obtained by the action of the peroxide.
(ii) But the $\rm{\dot{Br}}$ free radical in fig.13.76(b) is obtained by the action of the primary free radical. This radical can attack another prop-1-ene molecule.
(iii) So a chain reaction will be set up.
3. The $\rm{\dot{H}}$ free radical attacks the primary free radical. This is shown in fig.c above.
• The $\rm{\dot{H}}$ free radical gets attached to the first C atom.
• The $\rm{\dot{H}}$ free radical brought one electron required for the bond. The second electron is already present as an unpaired electron.
• Thus we get 2-Bromopropane.

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◼ We want to know this:
In an actual reaction, when peroxide is present, what will be the major product? 1-Bromopropane or 2-Bromopropane?
• We can now write the answer. It can be written in 3 steps:
1. We have seen the two possibilities. Let us compare them.
• In fig.13.75, we see that, the prop-1-ene changes into a secondary free radical.
• In fig.13.76, we see that, the prop-1-ene changes into a primary free radical.
2. Secondary free radicals are more stable than primary free radicals.
• So in the reaction mixture, there will be a greater quantity of secondary free radicals.
• Consequently, the $\rm{\dot{H}}$ free radical will be reacting more with secondary free radicals. This will result in a greater quantity of 1-Bromopropane.
3. So the answer is:
In an actual reaction, when peroxide is present, the principal product will be 1-Bromopropane.
• This result is contrary to Markovnikov’s rule. If we apply Markovnikov’s rule, we would be expecting 2-Bromopropane. It is the presence of peroxide which causes the contradicting result.

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• Based on the discussions so far, we can write a comparison between two cases: Case I and Case II. The comparison can be written in 5 steps:
1. Which are the reactants in each case?
• Case I is the reaction between prop-1-ene and HBr.
• Case II is also the reaction between the same prop-1-ene and HBr. But benzoylperoxide is also present.
2. Which is the attacking reagent?
• In case I, the H⁺ attacks the prop-1-ene molecule.
• In case II, the $\rm{\dot{Br}}$ free radical attacks the prop-1-ene molecule.
3. What is the result of the attack?
• As a result of the attack,
   ♦ In case I, prop-1-ene changes to primary/secondary carbocation.
   ♦ In case II, prop-1-ene changes to primary/secondary free radical.
4. What is the end product?
• In case I, the end product is 2-Bromopropene.
• In case II, the end product is 1-Bromopropene.
5. How is Markovnikov’s rule applicable to the two cases?
• Markovnikov’s rule states that:
The negative part of the addendum gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
• In case I, the Br^- indeed gets attached to the second C atom which has only one H atom.
• In case II, the Br^- gets attached to the first C atom which has two H atoms. This is contrary to the Markovnikov’s rule.

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• The anti-Markovnikov’s effect is not observed if we use HCl or HI in the place of HBr.
• First we will discuss about HCl. It can be written in 2 steps:
1. In fig.13.74(c) above, we saw that the free radicals formed from the peroxide, causes the homolysis of HBr.
• But such a homolysis is not possible in the case of HCl
2. The reason can be understood if we compare the bond strengths.
   ♦ H-Cl bond has a bond strength of 430.5 kJ mol^-1.      
   ♦ H-Br bond has a bond strength of 363.7 kJ mol^-1.
• Obviously, H-Cl bond is much stronger. It cannot be subjected to homolysis easily.

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• Next we will discuss about HI. It can be written in 2 steps:
1. The H-I bond strength (296.8 kJ mol^-1) is comparatively low.        
• So The H-I bond can be easily subjected to homolysis by the radicals formed from the peroxide.
2. But the $\rm{\dot{I}}$ free radicals so formed will combine together to form I₂ molecules. Those $\rm{\dot{I}}$ free radicals will not attack the alkene.

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Now we will see a solved example
Solved example 13.12
Write IUPAC names of products obtained by addition reaction of HBr to hex-1-ene.
(i) in the absence of peroxide.
(ii) in the presence of peroxide.
Solution:
1. The condensed formula of hex-1-ene is shown in fig.13.77(a) below:

Fig.13.77

2. The H atom will get attached to either the first C atom or the second C atom.
• Similarly, the Br atom will get attached to either the first C atom or the second C atom.
• This is because, those two C atoms are the ones which form the double bond.
3. When peroxide is absent, we can apply the Markovnikov’s rule.
• So Br^- will get attached to that C atom with the lesser number of H atoms. That is the second C atom.
• So the H atom gets attached to the first C atom.
• Thus we get the product: 2-Bromohexane. This is shown in fig.b.
4. When peroxide is present, we can apply the anti-Markovnikov’s rule.
• So Br^- will get attached to that C atom with the greater number of H atoms. That is the first C atom.
• So the H atom gets attached to the second C atom.
• Thus we get the product: 1-Bromohexane. This is shown in fig.c

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We have completed a discussion on anti-Markovnikov's rule. In the next section we will see a few more chemical properties of alkenes.

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